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This appendix follows a state vector, a density matrix, and a retained part of a dynamical system. They may describe the same experiment, but they do not answer the same question.\n\\vfig{three_objects}\n\\head{Amplitudes retain relative phase}\nFor the Bell vector $\\psi=(\\ket{00}+\\ket{11})/\\sqrt2$, the two nonzero entries are amplitudes. The density matrix $\\rho=\\psi\\psi^*$ also records coherence in its off-diagonal entries. Measurements that compare the two addresses can detect this coherence. Deleting those entries applies a physical channel and changes the state.\n\\head{How hidden dynamics returns}\nWhen a Hamiltonian couples a retained state to a hidden state, the retained amplitude can leave and later return. Its exact equation then remembers the hidden coupling. A smaller matrix obtained by deleting rows and columns does not generally reproduce that equation.\n\\head{A route through the paper}\nPages \\pagerefV{3}--\\pagerefV{10} develop seeds, symmetry…","to follow when its objects remain visible. This appendix follows a state vector, a density matrix, and a retained part of a dynamical system. They may describe the same experiment, but they do not answer the same question.\n\\vfig{three_objects}\n\\head{Amplitudes retain relative phase}\nFor the Bell vector $\\psi=(\\ket{00}+\\ket{11})/\\sqrt2$, the two nonzero entries are amplitudes. The density matrix $\\rho=\\psi\\psi^*$ also records coherence in its off-diagonal entries. Measurements that compare the two addresses can detect this coherence. Deleting those entries applies a physical channel and changes the state.\n\\head{How hidden dynamics returns}\nWhen a Hamiltonian couples a retained state to a hidden state, the retained amplitude can leave and later return. Its exact equation then remembers the hidden coupling. A smaller matrix obtained by deleting rows and columns does not generally reproduce that equation.\n\\head{A route through the paper}\nPages \\pagerefV{3}--\\pagerefV{10} develop seeds, symmetry, and coherent circuit paths. Pages \\pagerefV{11}--\\pagerefV{15} track leakage and moving subsp…","\\vpage{2}{The objects on the page}{AN ATLAS OF TYPES AND CONVENTIONS}{Traditional edition: Definition II.2.1, Definition II.7.1.}\nThe same letter can stand for different kinds of data in different calculations. Keep the following types fixed while reading a formula. In this appendix, vectors are columns, operators compose from right to left, and $A^*$ denotes the conjugate transpose.\n\\par\\vspace{4pt}\n\\begin{tabularx}{\\linewidth}{@{}p{.13\\linewidth}YY@{}}\n\\toprule\nSymbol & Object & What it records","midrule\n$\\HH,\\psi$ & Hilbert space and state vector & They specify the space of addresses and its coherent amplitudes.","$\\AA$ & Declared operator algebra & It specifies the available linear operations.","$P_a$ & Orthogonal packet projector & It selects an entire subspace.","$\\rho$ & Positive matrix of trace one & It records populations and coherences.","$E_k$ & Positive measurement effect & It gives the outcome probability $\\Tr(E_k\\rho)$.","$K_a$ & Kraus operator & It describes one channel branch and need not be a projector.","$H$ & Self-adjoint Hamiltonian & It generates evolution in real time through $e^{-\\ii tH}$.","$C,\\nabla$ & Packet transport generator and connection & They describe moving subspaces and differentiation within their ranges.","$P,Q$ & Retained and hidden projectors & They define a decomposition with $Q=I-P$.","$\\Sigma(z)$ & Self-energy & It records the spectral influence of the hidden sector.","$L,T(\\tau)$ & Nonnegative operator and Euclidean transfer & They determine decay factors and finite heat trace diagnostics.","bottomrule\n\\end{tabularx}\n\\head{Normalization and indexing}\nThe tensor factors are ordered from left to right, so $X_1=X\\otimes I\\otimes I$. A normalized pure state gives $\\rho=\\psi\\psi^*$. A coefficient matrix used to count a seed orbit need not be normalized. Scaling a nonzero seed does not change its generated linear span.","A Hamiltonian and a Hadamard gate are both traditionally denoted $H$. Where both appear, the gate is $H_{\\rm Had}$ and the Hamiltonian keeps the letter $H$. Numerical labels on a drawing are coordinates or exact example values, not observations of a physical device.\n\\takeaway{Check the type before interpreting the picture. An idempotent projector, a density matrix, and a channel matrix have different defining equations.}","\\vpage{3}{One seed, eight addresses}{CYCLICITY DEPENDS ON THE GRAMMAR}{Traditional edition: Proposition II.2.5.}\nStart with $\\psi_0=\\ket{000}$ and allow the three bit flip operators $X_1,X_2,X_3$. For a binary address $a=(a_1,a_2,a_3)$, put\n\\[\nW_a=X_1^{a_1}X_2^{a_2}X_3^{a_3}.\n\\qquad W_a\\psi_0=\\ket{a_1a_2a_3}.\n\\]\nThe drawing shows the actual eight states reached by these words. Dashed edges connect the two squares by changing the first bit. The other edges change one of the remaining bits.\n\\vfig{addresses}\n\\head{Two different meanings of ``reach''}\nThe eight vectors are orthonormal, so their \\emph{linear span} is all of $\\CC^8$. This proves cyclicity for the algebra generated by the flips. Preparing every normalized vector requires more than products of these flips alone.","2,a_3)$, put\n\\[\nW_a=X_1^{a_1}X_2^{a_2}X_3^{a_3}.\n\\qquad W_a\\psi_0=\\ket{a_1a_2a_3}.\n\\]\nThe drawing shows the actual eight states reached by these words. Dashed edges connect the two squares by changing the first bit. The other edges change one of the remaining bits.\n\\vfig{addresses}\n\\head{Two different meanings of ``reach''}\nThe eight vectors are orthonormal, so their \\emph{linear span} is all of $\\CC^8$. This proves cyclicity for the algebra generated by the flips. Preparing every normalized vector requires more than products of these flips alone.","For example, no such word sends $\\ket{000}$ to $(\\ket{000}+\\ket{001})/\\sqrt2$. A Hadamard gate, a suitable linear combination in the observable algebra, or a larger preparation grammar is required for that operation. Counting linearly independent addressed vectors and synthesizing a physical gate sequence are separate tasks.","The exact check constructs each $8\\times8$ word matrix, applies it to the seed, and verifies that the matrix of all eight outputs is the identity. It also tests the displayed superposition against every word output. No search over an unspecified gate set is involved.\n\\takeaway{A cyclic seed generates a vector space under an algebra. That conclusion does not automatically provide a unitary preparation procedure for every vector in it.}","\\vpage{4}{Entanglement changes the local span}{SCHMIDT RANK AS ACCESSIBLE MULTIPLICITY}{Traditional edition: Theorem II.2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It","\\vpage{4}{Entanglement changes the local span}{SCHMIDT RANK AS ACCESSIBLE MULTIPLICITY}{Traditional edition: Theorem II.2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve S","\\vpage{4}{Entanglement changes the local span}{SCHMIDT RANK AS ACCESSIBLE MULTIPLICITY}{Traditional edition: Theorem II.2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt c","\\vpage{4}{Entanglement changes the local span}{SCHMIDT RANK AS ACCESSIBLE MULTIPLICITY}{Traditional edition: Theorem II.2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vect","ment changes the local span}{SCHMIDT RANK AS ACCESSIBLE MULTIPLICITY}{Traditional edition: Theorem II.2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vector into an arbitrary product vector by a deterministic transformation betwee…",".2.2, Corollary II.2.4.}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vector into an arbitrary product vector by a deterministic transformation between pure states.",".}\nWrite a bipartite vector as a coefficient matrix $C$, with the first subsystem indexing rows. Acting only on that subsystem transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vector into an arbitrary product vector by a deterministic transformation between pure states.","m transforms $C$ to $AC$. Every resulting row remains in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vector into an arbitrary product vector by a deterministic transformation between pure states.","in the row space of $C$, and every independent choice of such rows can be obtained in this way.\n\\[\n\\dim\\Span\\{(A\\otimes I)\\psi:A\\in M_d(\\CC)\\}\n=d\\,\\rank C.\n\\]\n\\vfig{local_orbits}\nThe product vector $\\ket{00}$ has matrix $C=\\diag(1,0)$. Its local orbit span allows arbitrary vectors in the first subsystem but only the second subsystem direction $\\ket0$, so its dimension is two. The Bell vector has $C=I/\\sqrt2$, rank two, and a four-dimensional local orbit span.\n\\head{What the formula does not say}\nThe action here allows arbitrary matrices and then takes a linear span. It is not the orbit under local unitary gates alone. Local unitary operations preserve Schmidt coefficients. They therefore cannot turn the Bell vector into an arbitrary product vector by a deterministic transformation between pure states.","For the exact calculation, apply the four matrix units $E_{ab}$ to each seed matrix, flatten the outputs, and compute their column rank. The observed dimensions are two and four, respectively. The calculation directly exhibits the independent rows counted by the theorem.\n\\takeaway{Schmidt rank measures the remote multiplicity directions accessible to the local operator span. It is not permission for unrestricted deterministic local state preparation.}","\\vpage{5}{Why one seed is not always enough}{AN EXACT MULTIPLICITY LOWER BOUND}{Traditional edition: Theorem II.2.3.}\nConsider $M_2(\\CC)\\otimes I_3$ acting on $\\CC^2\\otimes\\CC^3$. A seed is a $2\\times3$ coefficient matrix. Its two rows span at most two directions in the three-dimensional multiplicity factor. Therefore no single seed is cyclic for this algebra.\n\\[\nC_1=\\mat{1&0&0","\\vpage{5}{Why one seed is not always enough}{AN EXACT MULTIPLICITY LOWER BOUND}{Traditional edition: Theorem II.2.3.}\nConsider $M_2(\\CC)\\otimes I_3$ acting on $\\CC^2\\otimes\\CC^3$. A seed is a $2\\times3$ coefficient matrix. Its two rows span at most two directions in the three-dimensional multiplicity factor. Therefore no single seed is cyclic for this algebra.\n\\[\nC_1=\\mat{1&0&0\\\\0&1&0},\\qquad\nC_2=\\mat{0&0&1\\\\0&0&0}.\n\\]\nThese two matrices provide a complete set of row directions. Their individual orbit dimensions are four and two, and together those spans fill the six-dimensional state space.\n\\vfig{seed_rank}\n\\head{From the example to the finite formula}\nFor a block $M_d(\\CC)\\otimes I_m$, $k$ seed matrices contribute at most $kd$ independent multiplicity directions. Hence $k\\ge\\lceil m/d\\rceil$. Partitioning a multiplicity basis into groups of size at most $d$ attains this bound. For a direct sum of such blocks, the worst block determines the required count:\n\\[\n\\operatorname{srank}_{\\AA}(\\HH)=\\max_i\\lceil m_i/d_i\\rceil.\n\\]\nThis is an invariant of the declared representation of the alg…","0&0&0}.\n\\]\nThese two matrices provide a complete set of row directions. Their individual orbit dimensions are four and two, and together those spans fill the six-dimensional state space.\n\\vfig{seed_rank}\n\\head{From the example to the finite formula}\nFor a block $M_d(\\CC)\\otimes I_m$, $k$ seed matrices contribute at most $kd$ independent multiplicity directions. Hence $k\\ge\\lceil m/d\\rceil$. Partitioning a multiplicity basis into groups of size at most $d$ attains this bound. For a direct sum of such blocks, the worst block determines the required count:\n\\[\n\\operatorname{srank}_{\\AA}(\\HH)=\\max_i\\lceil m_i/d_i\\rceil.\n\\]\nThis is an invariant of the declared representation of the algebra, and depends on more than the dimension of the Hilbert space alone. If the available algebra were enlarged to the full matrix algebra on $\\CC^6$, every nonzero seed would instead be cyclic.","ow directions. Their individual orbit dimensions are four and two, and together those spans fill the six-dimensional state space.\n\\vfig{seed_rank}\n\\head{From the example to the finite formula}\nFor a block $M_d(\\CC)\\otimes I_m$, $k$ seed matrices contribute at most $kd$ independent multiplicity directions. Hence $k\\ge\\lceil m/d\\rceil$. Partitioning a multiplicity basis into groups of size at most $d$ attains this bound. For a direct sum of such blocks, the worst block determines the required count:\n\\[\n\\operatorname{srank}_{\\AA}(\\HH)=\\max_i\\lceil m_i/d_i\\rceil.\n\\]\nThis is an invariant of the declared representation of the algebra, and depends on more than the dimension of the Hilbert space alone. If the available algebra were enlarged to the full matrix algebra on $\\CC^6$, every nonzero seed would instead be cyclic.","dimensional state space.\n\\vfig{seed_rank}\n\\head{From the example to the finite formula}\nFor a block $M_d(\\CC)\\otimes I_m$, $k$ seed matrices contribute at most $kd$ independent multiplicity directions. Hence $k\\ge\\lceil m/d\\rceil$. Partitioning a multiplicity basis into groups of size at most $d$ attains this bound. For a direct sum of such blocks, the worst block determines the required count:\n\\[\n\\operatorname{srank}_{\\AA}(\\HH)=\\max_i\\lceil m_i/d_i\\rceil.\n\\]\nThis is an invariant of the declared representation of the algebra, and depends on more than the dimension of the Hilbert space alone. If the available algebra were enlarged to the full matrix algebra on $\\CC^6$, every nonzero seed would instead be cyclic.","o groups of size at most $d$ attains this bound. For a direct sum of such blocks, the worst block determines the required count:\n\\[\n\\operatorname{srank}_{\\AA}(\\HH)=\\max_i\\lceil m_i/d_i\\rceil.\n\\]\nThis is an invariant of the declared representation of the algebra, and depends on more than the dimension of the Hilbert space alone. If the available algebra were enlarged to the full matrix algebra on $\\CC^6$, every nonzero seed would instead be cyclic.","\\vpage{6}{Four phases resolve a cyclic shift}{CHARACTER PROJECTORS YOU CAN MULTIPLY}{Traditional edition: Corollary II.3.3.}\nOn four sites, let $S\\ket k=\\ket{k+1\\bmod4}$ and let $\\omega=\\ii$. The character projectors are the exact matrices\n\\[\nP_j=\\frac14\\sum_{r=0}^3\\ii^{-jr}S^r,\n\\qquad v_j(k)=\\frac12\\ii^{-jk},\\qquad P_j=v_jv_j^*.\n\\]\nAll entries belong to the Gaussian rationals. Thus no approximation to a root of unity is needed in this model.\n\\vfig{cyclic_four}\n\\head{Reading the columns}\nEach column $v_j$ is a normalized wave around the four sites. Advancing the address multiplies it by $\\ii^j$. The entries show phase differences rather than four different probability distributions: every column has uniform site probability $1/4$.","Corollary II.3.3.}\nOn four sites, let $S\\ket k=\\ket{k+1\\bmod4}$ and let $\\omega=\\ii$. The character projectors are the exact matrices\n\\[\nP_j=\\frac14\\sum_{r=0}^3\\ii^{-jr}S^r,\n\\qquad v_j(k)=\\frac12\\ii^{-jk},\\qquad P_j=v_jv_j^*.\n\\]\nAll entries belong to the Gaussian rationals. Thus no approximation to a root of unity is needed in this model.\n\\vfig{cyclic_four}\n\\head{Reading the columns}\nEach column $v_j$ is a normalized wave around the four sites. Advancing the address multiplies it by $\\ii^j$. The entries show phase differences rather than four different probability distributions: every column has uniform site probability $1/4$.","character projectors are the exact matrices\n\\[\nP_j=\\frac14\\sum_{r=0}^3\\ii^{-jr}S^r,\n\\qquad v_j(k)=\\frac12\\ii^{-jk},\\qquad P_j=v_jv_j^*.\n\\]\nAll entries belong to the Gaussian rationals. Thus no approximation to a root of unity is needed in this model.\n\\vfig{cyclic_four}\n\\head{Reading the columns}\nEach column $v_j$ is a normalized wave around the four sites. Advancing the address multiplies it by $\\ii^j$. The entries show phase differences rather than four different probability distributions: every column has uniform site probability $1/4$.","All entries belong to the Gaussian rationals. Thus no approximation to a root of unity is needed in this model.\n\\vfig{cyclic_four}\n\\head{Reading the columns}\nEach column $v_j$ is a normalized wave around the four sites. Advancing the address multiplies it by $\\ii^j$. The entries show phase differences rather than four different probability distributions: every column has uniform site probability $1/4$.","Multiplying the actual projectors gives $P_jP_k=0$ for $j\\ne k$, $P_j^2=P_j$, and $\\sum_jP_j=I$. In particular, the complete packet tuple $(P_j\\psi)_j$ retains the original vector exactly. Recording only the four packet norms is a different operation and can discard phase information between packets.","The same shift on four sites returns on \\pagerefV{27}, where it diagonalizes the Hamiltonian $L=2I-S-S^*$, and its four character phases explain perfect transfer between opposite sites. The same fixed projectors serve as coordinates, symmetry sectors, and a complete interference ledger.\n\\takeaway{A character projector retains a complex subspace component. Its norm summarizes the size of that component while leaving its phase and direction unspecified.}","\\vpage{7}{A coarse packet can still contain coherence}{TENSOR REFINEMENT BY TOTAL PARITY}{Traditional edition: Theorem II.3.7, Corollary II.3.8.}\nFor three qubits, the eight fine projectors are $P_a=\\ket a\\bra a$. Grouping addresses according to the parity of their Hamming weight gives\n\\[\nP_{\\rm even}=P_{000}+P_{011}+P_{101}+P_{110}\n=\\frac{I+Z\\otimes Z\\otimes Z}{2},\n\\qquad P_{\\rm odd}=I-P_{\\rm even}.\n\\]\nThis is a sum of subspaces, not an averaging of their coordinates. The rank-four even packet retains all complex amplitudes on its four children.\n\\vfig{parity}\n\\head{Refinement is not measurement}\nThe fine packet description and the coarse packet description each reconstruct the full vector when every component is retained. Replacing a state by the mixture $\\sum_aP_a\\rho P_a$ is instead a dephasing channel. That operation removes coherence between the fine packets and can differ from dephasing only","\\vpage{7}{A coarse packet can still contain coherence}{TENSOR REFINEMENT BY TOTAL PARITY}{Traditional edition: Theorem II.3.7, Corollary II.3.8.}\nFor three qubits, the eight fine projectors are $P_a=\\ket a\\bra a$. Grouping addresses according to the parity of their Hamming weight gives\n\\[\nP_{\\rm even}=P_{000}+P_{011}+P_{101}+P_{110}\n=\\frac{I+Z\\otimes Z\\otimes Z}{2},\n\\qquad P_{\\rm odd}=I-P_{\\rm even}.\n\\]\nThis is a sum of subspaces, not an averaging of their coordinates. The rank-four even packet retains all complex amplitudes on its four children.\n\\vfig{parity}\n\\head{Refinement is not measurement}\nThe fine packet description and the coarse packet description each reconstruct the full vector when every component is retained. Replacing a state by the mixture $\\sum_aP_a\\rho P_a$ is instead a dephasing channel. That operation removes coherence between the fine packets and can differ from dephasing only between the even and odd sectors.",",\n\\qquad P_{\\rm odd}=I-P_{\\rm even}.\n\\]\nThis is a sum of subspaces, not an averaging of their coordinates. The rank-four even packet retains all complex amplitudes on its four children.\n\\vfig{parity}\n\\head{Refinement is not measurement}\nThe fine packet description and the coarse packet description each reconstruct the full vector when every component is retained. Replacing a state by the mixture $\\sum_aP_a\\rho P_a$ is instead a dephasing channel. That operation removes coherence between the fine packets and can differ from dephasing only between the even and odd sectors.","For instance, a superposition of $\\ket{000}$ and $\\ket{011}$ lies entirely in the even packet. Its internal relative phase survives a parity measurement but is lost by a complete measurement in the computational basis if outcomes are forgotten. The two measurements use projectors from the same refinement hierarchy but expose different information.","The exact check verifies all eight addresses, their parity grouping, the rank of both coarse projectors, and equality with the tensor expression above. The general residue version replaces parity by total label modulo $N$ and groups the tensor descendants by their summed residue.\n\\takeaway{A coarse packet is a whole subspace. Internal coherence survives a change of address description. It is lost only when an operation discards it.}","\\vpage{8}{Track the Bell state through the circuit}{INTERMEDIATE PACKETS ARE NOT FINAL PACKETS}{Traditional edition: Theorem II.4.1, Example II.4.6.}\nThe input is $\\ket{00}$. First apply a Hadamard to the first qubit, then a controlled flip:\n\\[\n\\ket{00}\\ \\xmapsto{H_{\\rm Had}\\otimes I}\\\n\\frac{\\ket{00}+\\ket{10}}{\\sqrt2}\n\\ \\xmapsto{\\rm CNOT}\\\n\\frac{\\ket{00}+\\ket{11}}{\\sqrt2}.\n\\]\nThe displayed columns keep amplitudes, including their normalization, at every step. The middle vector occupies both parity packets. The final vector occupies only the even packet.\n\\vfig{bell_circuit}\n\\head{How the exact path expansion works}\nInsert $P_{\\rm even}+P_{\\rm odd}=I$ between the two gates. Each summand records a particular intermediate packet, while the boundary projectors select its input and output packets. Summing all terms recovers the original $4\\times4$ circuit matrix:\n\\[","\\vpage{8}{Track the Bell state through the circuit}{INTERMEDIATE PACKETS ARE NOT FINAL PACKETS}{Traditional edition: Theorem II.4.1, Example II.4.6.}\nThe input is $\\ket{00}$. First apply a Hadamard to the first qubit, then a controlled flip:\n\\[\n\\ket{00}\\ \\xmapsto{H_{\\rm Had}\\otimes I}\\\n\\frac{\\ket{00}+\\ket{10}}{\\sqrt2}\n\\ \\xmapsto{\\rm CNOT}\\\n\\frac{\\ket{00}+\\ket{11}}{\\sqrt2}.\n\\]\nThe displayed columns keep amplitudes, including their normalization, at every step. The middle vector occupies both parity packets. The final vector occupies only the even packet.\n\\vfig{bell_circuit}\n\\head{How the exact path expansion works}\nInsert $P_{\\rm even}+P_{\\rm odd}=I$ between the two gates. Each summand records a particular intermediate packet, while the boundary projectors select its input and output packets. Summing all terms recovers the original $4\\times4$ circuit matrix:\n\\[\n\\sum_{a,b,c}P_b\\,\\mathrm{CNOT}\\,P_a\n(H_{\\rm Had}\\otimes I)P_c\n=\\mathrm{CNOT}(H_{\\rm Had}\\otimes I).\n\\]\nThis expansion is an identity even though the intermediate parity is not measured. Physically insert","olled flip:\n\\[\n\\ket{00}\\ \\xmapsto{H_{\\rm Had}\\otimes I}\\\n\\frac{\\ket{00}+\\ket{10}}{\\sqrt2}\n\\ \\xmapsto{\\rm CNOT}\\\n\\frac{\\ket{00}+\\ket{11}}{\\sqrt2}.\n\\]\nThe displayed columns keep amplitudes, including their normalization, at every step. The middle vector occupies both parity packets. The final vector occupies only the even packet.\n\\vfig{bell_circuit}\n\\head{How the exact path expansion works}\nInsert $P_{\\rm even}+P_{\\rm odd}=I$ between the two gates. Each summand records a particular intermediate packet, while the boundary projectors select its input and output packets. Summing all terms recovers the original $4\\times4$ circuit matrix:\n\\[\n\\sum_{a,b,c}P_b\\,\\mathrm{CNOT}\\,P_a\n(H_{\\rm Had}\\otimes I)P_c\n=\\mathrm{CNOT}(H_{\\rm Had}\\otimes I).\n\\]\nThis expansion is an identity even though the intermediate parity is not measured. Physically inserting a measurement would alter the problem by removing interference between its alternatives. The derivation resolves intermediate coordinates without physically measuring them.","ion, at every step. The middle vector occupies both parity packets. The final vector occupies only the even packet.\n\\vfig{bell_circuit}\n\\head{How the exact path expansion works}\nInsert $P_{\\rm even}+P_{\\rm odd}=I$ between the two gates. Each summand records a particular intermediate packet, while the boundary projectors select its input and output packets. Summing all terms recovers the original $4\\times4$ circuit matrix:\n\\[\n\\sum_{a,b,c}P_b\\,\\mathrm{CNOT}\\,P_a\n(H_{\\rm Had}\\otimes I)P_c\n=\\mathrm{CNOT}(H_{\\rm Had}\\otimes I).\n\\]\nThis expansion is an identity even though the intermediate parity is not measured. Physically inserting a measurement would alter the problem by removing interference between its alternatives. The derivation resolves intermediate coordinates without physically measuring them.","le vector occupies both parity packets. The final vector occupies only the even packet.\n\\vfig{bell_circuit}\n\\head{How the exact path expansion works}\nInsert $P_{\\rm even}+P_{\\rm odd}=I$ between the two gates. Each summand records a particular intermediate packet, while the boundary projectors select its input and output packets. Summing all terms recovers the original $4\\times4$ circuit matrix:\n\\[\n\\sum_{a,b,c}P_b\\,\\mathrm{CNOT}\\,P_a\n(H_{\\rm Had}\\otimes I)P_c\n=\\mathrm{CNOT}(H_{\\rm Had}\\otimes I).\n\\]\nThis expansion is an identity even though the intermediate parity is not measured. Physically inserting a measurement would alter the problem by removing interference between its alternatives. The derivation resolves intermediate coordinates without physically measuring them.","\\vpage{9}{Why path probabilities do not add}{HOW TWO PATH AMPLITUDES CANCEL}{Traditional edition: Theorem II.4.2, Theorem II.4.4.}\nTwo consecutive Hadamard gates multiply to the identity. Starting from $\\ket0$, there are two resolved paths to output $\\ket1$. Their signed products are\n\\[\n\\bra1H_{\\rm Had}\\ket0\\bra0H_{\\rm Had}\\ket0=\\frac12,\n\\qquad\n\\bra1H_{\\rm Had}\\ket1\\bra1H_{\\rm Had}\\ket0=-\\frac12.\n\\]\nTheir sum is zero. Squaring each separately gives a different answer.\n\\vfig{interference}\n\\head{Where the missing term goes}\nFor path amplitudes $a,b$, the outcome probability is $|a+b|^2=|a|^2+|b|^2+2\\Re(a\\bar b)$. Here the last term is $-1/2$, canceling the first two terms. The path picture retains this information only if its edges carry complex amplitudes rather than unsigned reachability or probabilities.","damard gates multiply to the identity. Starting from $\\ket0$, there are two resolved paths to output $\\ket1$. Their signed products are\n\\[\n\\bra1H_{\\rm Had}\\ket0\\bra0H_{\\rm Had}\\ket0=\\frac12,\n\\qquad\n\\bra1H_{\\rm Had}\\ket1\\bra1H_{\\rm Had}\\ket0=-\\frac12.\n\\]\nTheir sum is zero. Squaring each separately gives a different answer.\n\\vfig{interference}\n\\head{Where the missing term goes}\nFor path amplitudes $a,b$, the outcome probability is $|a+b|^2=|a|^2+|b|^2+2\\Re(a\\bar b)$. Here the last term is $-1/2$, canceling the first two terms. The path picture retains this information only if its edges carry complex amplitudes rather than unsigned reachability or probabilities.","ity. Starting from $\\ket0$, there are two resolved paths to output $\\ket1$. Their signed products are\n\\[\n\\bra1H_{\\rm Had}\\ket0\\bra0H_{\\rm Had}\\ket0=\\frac12,\n\\qquad\n\\bra1H_{\\rm Had}\\ket1\\bra1H_{\\rm Had}\\ket0=-\\frac12.\n\\]\nTheir sum is zero. Squaring each separately gives a different answer.\n\\vfig{interference}\n\\head{Where the missing term goes}\nFor path amplitudes $a,b$, the outcome probability is $|a+b|^2=|a|^2+|b|^2+2\\Re(a\\bar b)$. Here the last term is $-1/2$, canceling the first two terms. The path picture retains this information only if its edges carry complex amplitudes rather than unsigned reachability or probabilities.","output $\\ket1$. Their signed products are\n\\[\n\\bra1H_{\\rm Had}\\ket0\\bra0H_{\\rm Had}\\ket0=\\frac12,\n\\qquad\n\\bra1H_{\\rm Had}\\ket1\\bra1H_{\\rm Had}\\ket0=-\\frac12.\n\\]\nTheir sum is zero. Squaring each separately gives a different answer.\n\\vfig{interference}\n\\head{Where the missing term goes}\nFor path amplitudes $a,b$, the outcome probability is $|a+b|^2=|a|^2+|b|^2+2\\Re(a\\bar b)$. Here the last term is $-1/2$, canceling the first two terms. The path picture retains this information only if its edges carry complex amplitudes rather than unsigned reachability or probabilities.","The packetwise unitarity identity is likewise an operator identity, including the off-diagonal cancellations:\n\\[\n\\sum_b U_{ba}^*U_{bc}=\\delta_{ac}P_a,\n\\qquad U_{ba}=P_bUP_a.\n\\]\nIt cannot be replaced by the assertion that every path individually has unit norm. The exact checks verify all four choices of $a,c$ for the two rank-one packets, as well as the two path products leading to the opposite output.","A measurement at the intermediate stage would remove the cross term. In that modified experiment the two alternatives really would contribute total probability $1/2$ to output one. Adding the two path probabilities would therefore describe the experiment with an intermediate measurement.\n\\takeaway{Insert projectors algebraically to resolve paths, then add amplitudes. Add probabilities only when the physical setup has actually made the alternatives incoherent.}","\\vpage{10}{A symmetry reveals blocks, not approximations}{ONE HAMILTONIAN IN TWO ORDERS}{Traditional edition: Theorem II.5.1.}\nTake the exact rational Hamiltonian\n\\[\nH=X\\otimes X+\\frac38(Z\\otimes I+I\\otimes Z).\n\\]\nIt commutes with total parity $Z\\otimes Z$. Reordering the computational basis into even addresses $(00,11)$ and odd addresses $(01,10)$ displays the invariant subspaces explicitly.\n\\vfig{hamiltonian}\n\\[\nH_{\\rm even}=\\mat{3/4&1","etry reveals blocks, not approximations}{ONE HAMILTONIAN IN TWO ORDERS}{Traditional edition: Theorem II.5.1.}\nTake the exact rational Hamiltonian\n\\[\nH=X\\otimes X+\\frac38(Z\\otimes I+I\\otimes Z).\n\\]\nIt commutes with total parity $Z\\otimes Z$. Reordering the computational basis into even addresses $(00,11)$ and odd addresses $(01,10)$ displays the invariant subspaces explicitly.\n\\vfig{hamiltonian}\n\\[\nH_{\\rm even}=\\mat{3/4&1\\\\1&-3/4},\\qquad\nH_{\\rm odd}=\\mat{0&1\\\\1&0}.\n\\]\nThe even block squares to $(25/16)I$, so its eigenvalues are $\\pm5/4$. The odd block squares to $I$, so its eigenvalues are $\\pm1$. Exponentiating either block gives the corresponding part of the exact evolution. Transfer between the packets remains zero for every real time.\n\\head{What has changed?}\nOnly the order of basis vectors has changed. No matrix entries, amplitudes, or energy levels were discarded. This is why the decomposition is exact. Keeping just the even block would be a separate restriction of the state space, justified for initial data known to be even and for dynamics that preserve it.","1&0}.\n\\]\nThe even block squares to $(25/16)I$, so its eigenvalues are $\\pm5/4$. The odd block squares to $I$, so its eigenvalues are $\\pm1$. Exponentiating either block gives the corresponding part of the exact evolution. Transfer between the packets remains zero for every real time.\n\\head{What has changed?}\nOnly the order of basis vectors has changed. No matrix entries, amplitudes, or energy levels were discarded. This is why the decomposition is exact. Keeping just the even block would be a separate restriction of the state space, justified for initial data known to be even and for dynamics that preserve it.","\\vpage{11}{Quantify the amplitude that escapes}{COMMUTATORS AND PROBABILITY BOUNDS}{Traditional edition: Theorem II.5.3, Corollary II.5.4.}\nFor a two-level system, choose $H=\\varepsilon X$, $P=\\ket0\\bra0$, and $Q=I-P$. Direct exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape prob","\\vpage{11}{Quantify the amplitude that escapes}{COMMUTATORS AND PROBABILITY BOUNDS}{Traditional edition: Theorem II.5.3, Corollary II.5.4.}\nFor a two-level system, choose $H=\\varepsilon X$, $P=\\ket0\\bra0$, and $Q=I-P$. Direct exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are se","\\vpage{11}{Quantify the amplitude that escapes}{COMMUTATORS AND PROBABILITY BOUNDS}{Traditional edition: Theorem II.5.3, Corollary II.5.4.}\nFor a two-level system, choose $H=\\varepsilon X$, $P=\\ket0\\bra0$, and $Q=I-P$. Direct exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. M","\\vpage{11}{Quantify the amplitude that escapes}{COMMUTATORS AND PROBABILITY BOUNDS}{Traditional edition: Theorem II.5.3, Corollary II.5.4.}\nFor a two-level system, choose $H=\\varepsilon X$, $P=\\ket0\\bra0$, and $Q=I-P$. Direct exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","ORS AND PROBABILITY BOUNDS}{Traditional edition: Theorem II.5.3, Corollary II.5.4.}\nFor a two-level system, choose $H=\\varepsilon X$, $P=\\ket0\\bra0$, and $Q=I-P$. Direct exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","irect exponentiation gives\n\\[\ne^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","e^{-\\ii tH}=\\cos(\\varepsilon t)I-\\ii\\sin(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","in(\\varepsilon t)X,\n\\quad\n\\norm{Qe^{-\\ii tH}P}=|\\sin(\\varepsilon t)|,\n\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","\\quad\\norm{[H,P]}=|\\varepsilon|.\n\\]\nThe general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","general commutator estimate becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","becomes the elementary bound $|\\sin u|\\le |u|$. Unitarity also bounds the amplitude by one.\n\\vfig{leakage}\n\\head{From a norm to a probability}\nFor this initial state, the escape probability is $\\sin^2(\\varepsilon t)$, not $|\\sin(\\varepsilon t)|$. Consequently\n\\[\n\\Pr(\\text{hidden packet at }t)\\le\\min\\{1,\\varepsilon^2t^2\\}.\n\\]\nAt $\\varepsilon t=\\pi/6$, the amplitude is $1/2$ and the probability is $1/4$. Near zero, amplitudes are first order while escape probabilities are second order. Mixing those two quantities produces an incorrect interpretation of a leakage plot.","\\vpage{12}{A moving frame creates a connection term}{NOT ALL MOTION IS PHYSICAL EVOLUTION}{Traditional edition: Theorem II.6.3.}\nSet the laboratory Hamiltonian to zero and keep $\\psi(t)=\\ket0$ fixed. Describe it in rotating coordinates,\n\\[\nU(t)=\\mat{\\cos t&-\\sin t","\\vpage{12}{A moving frame creates a connection term}{NOT ALL MOTION IS PHYSICAL EVOLUTION}{Traditional edition: Theorem II.6.3.}\nSet the laboratory Hamiltonian to zero and keep $\\psi(t)=\\ket0$ fixed. Describe it in rotating coordinates,\n\\[\nU(t)=\\mat{\\cos t&-\\sin t\\\\\\sin t&\\cos t},\n\\qquad y(t)=U(t)^*\\psi(t)=\\mat{\\cos t\\\\-\\sin t}.\n\\]\nThe vector has not moved in the laboratory, but its coordinate entries have changed.\n\\vfig{moving_frame}\n\\head{The missing term can be computed exactly}\nLet $J=\\mat{0&-1","-\\sin t}.\n\\]\nThe vector has not moved in the laboratory, but its coordinate entries have changed.\n\\vfig{moving_frame}\n\\head{The missing term can be computed exactly}\nLet $J=\\mat{0&-1\\\\1&0}$. Since $U^*\\dot U=J$, the coordinate Hamiltonian is\n\\[\nH_{\\rm frame}=-\\ii J=\\mat{0&\\ii","1&0}$. Since $U^*\\dot U=J$, the coordinate Hamiltonian is\n\\[\nH_{\\rm frame}=-\\ii J=\\mat{0&\\ii","1&0}$. Since $U^*\\dot U=J$, the coordinate Hamiltonian is\n\\[\nH_{\\rm frame}=-\\ii J=\\mat{0&\\ii\\\\-\\ii&0},\n\\qquad \\ii\\dot y=H_{\\rm frame}y.\n\\]\nThis matrix is Hermitian. Omitting it would predict a constant $y$, contradicting the displayed coordinates. In this example the norm of the omitted time derivative is exactly one, so the failure is not a small numerical residual.","-\\ii&0},\n\\qquad \\ii\\dot y=H_{\\rm frame}y.\n\\]\nThis matrix is Hermitian. Omitting it would predict a constant $y$, contradicting the displayed coordinates. In this example the norm of the omitted time derivative is exactly one, so the failure is not a small numerical residual.","For a nonzero laboratory Hamiltonian, the same product rule adds the transformed physical term $U^*HU$. Thus $U^*HU-\\ii U^*\\dot U$ separates the physical generator from the coordinate correction. Both pieces belong to the same equation but have different origins.","\\vpage{13}{Differentiate the subspace itself}{THE CANONICAL PACKET CONNECTION}{Traditional edition: Theorem II.6.1, Theorem II.6.2.}\nContinue with $P(t)=U(t)\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[","\\vpage{13}{Differentiate the subspace itself}{THE CANONICAL PACKET CONNECTION}{Traditional edition: Theorem II.6.1, Theorem II.6.2.}\nContinue with $P(t)=U(t)\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]","\\vpage{13}{Differentiate the subspace itself}{THE CANONICAL PACKET CONNECTION}{Traditional edition: Theorem II.6.1, Theorem II.6.2.}\nContinue with $P(t)=U(t)\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection re","\\vpage{13}{Differentiate the subspace itself}{THE CANONICAL PACKET CONNECTION}{Traditional edition: Theorem II.6.1, Theorem II.6.2.}\nContinue with $P(t)=U(t)\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No ba","ional edition: Theorem II.6.1, Theorem II.6.2.}\nContinue with $P(t)=U(t)\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.",")\\diag(1,0)U(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","(t)^*$ and $Q(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","(t)=I-P(t)$. Differentiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","entiating $P^2=P$ gives\n\\[\n\\dot P P+P\\dot P=\\dot P,\n\\qquad P\\dot P P=0.\n\\]\nThe second identity says that the derivative has no component entirely inside the current packet. It moves the subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","subspace relative to its complement.\n\\vfig{connection}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","n}\nFor this complete pair of packets, define\n\\[\nC=\\dot P P+\\dot Q Q.\n\\]\nExact calculation yields the constant matrix $C=J$, with $C^*=-C$ and $[C,P]=\\dot P$. At $t=0$, the displayed small matrices allow these identities to be checked by direct multiplication.\n\\head{From two packets to a complete family}\nFor differentiable orthogonal projectors $P_a$ summing to the identity, the paper constructs\n\\[\nC=\\sum_a\\dot P_aP_a=\\frac12\\sum_a[\\dot P_a,P_a].\n\\]\nThis single connection reproduces the motion of every packet in the family. No basis within a packet is needed for this formula. A basis choice would add internal coordinate data without changing which subspaces are selected.","The completeness of the family matters in deriving skew-Hermiticity and the simultaneous transport identities. Using only one summand of an incomplete list is not automatically the same connection. The example retains both $P$ and $Q$ throughout, even when the picture emphasizes only the first line.\n\\takeaway{The projector derivative describes motion between subspaces. The canonical connection reconstructs that motion from the complete family without choosing internal packet bases.}","\\vpage{14}{A chart failure is not a singular fiber}{THE RANGE BUNDLE IS CONSTRUCTED LOCALLY}{Traditional edition: Theorem II.6.7. Details of the range atlas: saved v0.6.0 supplement, pp. 2--4.}\nThe fiber of a projector field is the actual fixed subspace $E_x=\\{v:P(x)v=v\\}$. To compare it with a nearby anchor fiber $E_a$, the completion proof constructs\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nIt equals the identity at the anchor and intertwines the two projectors. Where it is invertible, its restriction provides local fiber coordinates.\n\\vfig{range_atlas}\n\\head{An exact example of the domain restriction}\nFor the rotating line and anchor $P(0)$,\n\\[\nT_0(t)=\\cos t\\,U(t),\\qquad \\det T_0(t)=\\cos^2t.\n\\]\nThe chart fails at $t=\\pi/2$, but $P(\\pi/2)=\\diag(0,1)$ is a perfectly regular rank-one projector. Choosing a vertical anchor supplies a new chart there. What failed was the comparison with one reference fiber, not the fiber itself.","subspace $E_x=\\{v:P(x)v=v\\}$. To compare it with a nearby anchor fiber $E_a$, the completion proof constructs\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nIt equals the identity at the anchor and intertwines the two projectors. Where it is invertible, its restriction provides local fiber coordinates.\n\\vfig{range_atlas}\n\\head{An exact example of the domain restriction}\nFor the rotating line and anchor $P(0)$,\n\\[\nT_0(t)=\\cos t\\,U(t),\\qquad \\det T_0(t)=\\cos^2t.\n\\]\nThe chart fails at $t=\\pi/2$, but $P(\\pi/2)=\\diag(0,1)$ is a perfectly regular rank-one projector. Choosing a vertical anchor supplies a new chart there. What failed was the comparison with one reference fiber, not the fiber itself.","$E_x=\\{v:P(x)v=v\\}$. To compare it with a nearby anchor fiber $E_a$, the completion proof constructs\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nIt equals the identity at the anchor and intertwines the two projectors. Where it is invertible, its restriction provides local fiber coordinates.\n\\vfig{range_atlas}\n\\head{An exact example of the domain restriction}\nFor the rotating line and anchor $P(0)$,\n\\[\nT_0(t)=\\cos t\\,U(t),\\qquad \\det T_0(t)=\\cos^2t.\n\\]\nThe chart fails at $t=\\pi/2$, but $P(\\pi/2)=\\diag(0,1)$ is a perfectly regular rank-one projector. Choosing a vertical anchor supplies a new chart there. What failed was the comparison with one reference fiber, not the fiber itself.","completion proof constructs\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nIt equals the identity at the anchor and intertwines the two projectors. Where it is invertible, its restriction provides local fiber coordinates.\n\\vfig{range_atlas}\n\\head{An exact example of the domain restriction}\nFor the rotating line and anchor $P(0)$,\n\\[\nT_0(t)=\\cos t\\,U(t),\\qquad \\det T_0(t)=\\cos^2t.\n\\]\nThe chart fails at $t=\\pi/2$, but $P(\\pi/2)=\\diag(0,1)$ is a perfectly regular rank-one projector. Choosing a vertical anchor supplies a new chart there. What failed was the comparison with one reference fiber, not the fiber itself.","of constructs\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nIt equals the identity at the anchor and intertwines the two projectors. Where it is invertible, its restriction provides local fiber coordinates.\n\\vfig{range_atlas}\n\\head{An exact example of the domain restriction}\nFor the rotating line and anchor $P(0)$,\n\\[\nT_0(t)=\\cos t\\,U(t),\\qquad \\det T_0(t)=\\cos^2t.\n\\]\nThe chart fails at $t=\\pi/2$, but $P(\\pi/2)=\\diag(0,1)$ is a perfectly regular rank-one projector. Choosing a vertical anchor supplies a new chart there. What failed was the comparison with one reference fiber, not the fiber itself.","In general, the open sets where $T_a(x)$ is invertible cover the base. The inverse chart maps and smooth overlap maps are constructed from the actual matrices. Their cocycle identities express consistent coordinate changes, rather than assuming a global basis exists.","\\vpage{15}{Curvature measures a local area response}{AN EXACT BLOCH PROJECTOR CALCULATION}{Traditional edition: Theorem II.6.7, Corollary II.6.8.}\nOn the unit sphere, set $n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta)$ and\n\\[\nP(\\theta,\\phi)=\\frac12(I+n\\cdot\\sigma).\n\\]\nThis is a smooth rank-one orthogonal projector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta","\\vpage{15}{Curvature measures a local area response}{AN EXACT BLOCH PROJECTOR CALCULATION}{Traditional edition: Theorem II.6.7, Corollary II.6.8.}\nOn the unit sphere, set $n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta)$ and\n\\[\nP(\\theta,\\phi)=\\frac12(I+n\\cdot\\sigma).\n\\]\nThis is a smooth rank-one orthogonal projector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta","\\vpage{15}{Curvature measures a local area response}{AN EXACT BLOCH PROJECTOR CALCULATION}{Traditional edition: Theorem II.6.7, Corollary II.6.8.}\nOn the unit sphere, set $n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta)$ and\n\\[\nP(\\theta,\\phi)=\\frac12(I+n\\cdot\\sigma).\n\\]\nThis is a smooth rank-one orthogonal projector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula a…","LOCH PROJECTOR CALCULATION}{Traditional edition: Theorem II.6.7, Corollary II.6.8.}\nOn the unit sphere, set $n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta)$ and\n\\[\nP(\\theta,\\phi)=\\frac12(I+n\\cdot\\sigma).\n\\]\nThis is a smooth rank-one orthogonal projector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","n\\cdot\\sigma).\n\\]\nThis is a smooth rank-one orthogonal projector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","ojector. The intrinsic curvature is the endomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","ndomorphism-valued two-form $P(\\dd P\\wedge\\dd P)P$. With the paper's scalar convention, its component is\n\\[\nF_{\\theta\\phi}=-\\ii\\Tr\\!\\left(P[\\partial_\\theta P,\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","\\partial_\\phi P]\\right)\n=\\frac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","rac12\\sin\\theta.\n\\]\n\\vfig{bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","bloch_curvature}\n\\head{A local frame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","rame gives the same answer}\nAway from the south pole, use $u_N=(\\cos(\\theta/2),e^{\\ii\\phi}\\sin(\\theta/2))^T$. Then $P=u_Nu_N^*$ and the scalar one-form is\n\\[\nA_N=-\\ii u_N^*\\dd u_N=\\frac{1-\\cos\\theta}{2}\\,\\dd\\phi,\n\\qquad \\dd A_N=\\frac12\\sin\\theta\\,\\dd\\theta\\wedge\\dd\\phi.\n\\]\nThus the curvature flux through the cap $0\\le\\theta\\le\\theta_0$ is $\\pi(1-\\cos\\theta_0)$. At $\\theta_0=\\pi/3$, it is exactly $\\pi/2$. The latitude drawn in the plate illustrates this cap. The exact flux comes from the formula above.","The vanishing of the coordinate density at a pole does not mean the geometric curvature disappears there. The spherical area form also carries the factor $\\sin\\theta$. Assigning a phase to parallel transport also requires an orientation and a sign convention for the connection. These choices determine how flux relates to phase.\n\\takeaway{The projector formula and a local frame calculation agree. Curvature is intrinsic even when no single smooth frame covers the whole sphere.}","\\vpage{16}{Which coherence can a measurement see?}{THE ANSWER DEPENDS ON ITS EFFECTS}{Traditional edition: Theorem II.7.2, Theorem II.7.4, Corollary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is no","\\vpage{16}{Which coherence can a measurement see?}{THE ANSWER DEPENDS ON ITS EFFECTS}{Traditional edition: Theorem II.7.2, Theorem II.7.4, Corollary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","NSWER DEPENDS ON ITS EFFECTS}{Traditional edition: Theorem II.7.2, Theorem II.7.4, Corollary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","eorem II.7.4, Corollary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","7.4, Corollary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","llary II.7.5.}\nFor the Bell density matrix $\\rho=\\ket{\\Phi^+}\\bra{\\Phi^+}$, parity dephasing does nothing: both nonzero amplitudes already lie in the even subspace. Dephasing in the computational basis removes the entries between addresses $00$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","$ and $11$.\n\\vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","vfig{dephasing}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","g}\nEvery effect in the computational basis $E_a=\\ket a\\bra a$ has the same expectation on both matrices, namely $1/2$ at $00$ and $11$ and zero elsewhere. These measurements read only the diagonal. They cannot distinguish the pure Bell state from this incoherent mixture.\n\\head{A measurement that resolves the difference}\nUse the valid two-outcome measurement with effects $E=\\rho$ and $I-E$. Then\n\\[\n\\Tr(E\\rho)=1,\\qquad\n\\Tr\\!\\left(E D_{\\rm computational}(\\rho)\\right)=\\frac12.\n\\]\nThe effect itself is not diagonal in the fine computational packets. It compares their coherent combination, so it falls outside the paper's invariance result for block-diagonal effects.","\\vpage{17}{A channel acts on four matrix entries}{POPULATIONS AND COHERENCES TRANSFORM DIFFERENTLY}{Traditional edition: Theorem II.7.6, Example II.7.12.}\nChoose amplitude damping with $\\gamma=9/25$:\n\\[\nK_0=\\mat{1&0","\\vpage{17}{A channel acts on four matrix entries}{POPULATIONS AND COHERENCES TRANSFORM DIFFERENTLY}{Traditional edition: Theorem II.7.6, Example II.7.12.}\nChoose amplitude damping with $\\gamma=9/25$:\n\\[\nK_0=\\mat{1&0\\\\0&4/5},\\qquad\nK_1=\\mat{0&3/5\\\\0&0},\\qquad K_0^*K_0+K_1^*K_1=I.\n\\]\nThe channel $\\Phi(\\rho)=\\sum_aK_a\\rho K_a^*$ is trace preserving and completely positive. The special parameter keeps every displayed coefficient rational.\n\\vfig{damping}\nThe superoperator uses the entry order $(\\rho_{00},\\rho_{01},\\rho_{10},\\rho_{11})$. Population leaves the excited state with factor $9/25$, while coherences contract by $4/5$. These are not the same rate or the same coordinate.","0&0},\\qquad K_0^*K_0+K_1^*K_1=I.\n\\]\nThe channel $\\Phi(\\rho)=\\sum_aK_a\\rho K_a^*$ is trace preserving and completely positive. The special parameter keeps every displayed coefficient rational.\n\\vfig{damping}\nThe superoperator uses the entry order $(\\rho_{00},\\rho_{01},\\rho_{10},\\rho_{11})$. Population leaves the excited state with factor $9/25$, while coherences contract by $4/5$. These are not the same rate or the same coordinate.","For $\\rho_+=\\frac12\\mat{1&1\\\\1&1}$, the exact output is\n\\[\n\\Phi(\\rho_+)=\\mat{17/25&2/5","1&1}$, the exact output is\n\\[\n\\Phi(\\rho_+)=\\mat{17/25&2/5\\\\2/5&8/25}.\n\\]\nIts trace is one and determinant is $36/625>0$, so it is a mixed positive state. Arrows that show population transfer alone would leave its off-diagonal entries unspecified.\n\\head{How complete positivity is certified}\nThe Choi matrix is an explicit sum of two outer products of vectorized Kraus operators. Its eigenvalues are $41/25,9/25,0,0$, all nonnegative. This gives an independent finite certificate in addition to Kraus normalization. The Choi matrix has trace two under the convention used by the check, which leaves the maximally entangled vector unnormalized.\n\\takeaway{A packet channel acts on pairs of packet addresses, because density matrices have two indices. Population transfer al","2/5&8/25}.\n\\]\nIts trace is one and determinant is $36/625>0$, so it is a mixed positive state. Arrows that show population transfer alone would leave its off-diagonal entries unspecified.\n\\head{How complete positivity is certified}\nThe Choi matrix is an explicit sum of two outer products of vectorized Kraus operators. Its eigenvalues are $41/25,9/25,0,0$, all nonnegative. This gives an independent finite certificate in addition to Kraus normalization. The Choi matrix has trace two under the convention used by the check, which leaves the maximally entangled vector unnormalized.\n\\takeaway{A packet channel acts on pairs of packet addresses, because density matrices have two indices. Population transfer alone does not reconstruct a quantum channel.}","\\vpage{18}{An ancilla tests more than positivity}{WHY COMPLETE POSITIVITY IS THE RELEVANT CONDITION}{Traditional edition: Classical input II.7.7.}\nOrdinary transpose $\\mathsf T$ sends a positive matrix on one system to another positive matrix and preserves its trace. Those properties are insufficient for a quantum channel on a subsystem of an entangled state. The test must include $\\id\\otimes\\mathsf T$ on the larger system.\n\\vfig{complete_positivity}\nApply this extended map to the Bell density matrix. The result is\n\\[\n(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})\n=\\frac12\\mat{1&0&0&0","lassical input II.7.7.}\nOrdinary transpose $\\mathsf T$ sends a positive matrix on one system to another positive matrix and preserves its trace. Those properties are insufficient for a quantum channel on a subsystem of an entangled state. The test must include $\\id\\otimes\\mathsf T$ on the larger system.\n\\vfig{complete_positivity}\nApply this extended map to the Bell density matrix. The result is\n\\[\n(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})\n=\\frac12\\mat{1&0&0&0\\\\0&0&1&0\\\\0&1&0&0\\\\0&0&0&1}.\n\\]\nIts middle block has one negative eigenvalue. More concretely, the normalized vector $w=(\\ket{01}-\\ket{10})/\\sqrt2$ satisfies\n\\[\nw^*(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})w=-\\frac12.\n\\]\nThis exact negative value of the quadratic form certifies failure of positivity on the composite system. No numerical eigenvalue tolerance is needed.\n\\head{Why the Kraus form avoids this problem}\nFor every auxiliary dimension and positive $X$, each term $(I\\otimes K_a)X(I\\otimes K_a)^*$ is positive. Their sum is positive as well. This argument explains the word \\emph{completely}. The conclusion holds for every auxiliar…","0&0&0&1}.\n\\]\nIts middle block has one negative eigenvalue. More concretely, the normalized vector $w=(\\ket{01}-\\ket{10})/\\sqrt2$ satisfies\n\\[\nw^*(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})w=-\\frac12.\n\\]\nThis exact negative value of the quadratic form certifies failure of positivity on the composite system. No numerical eigenvalue tolerance is needed.\n\\head{Why the Kraus form avoids this problem}\nFor every auxiliary dimension and positive $X$, each term $(I\\otimes K_a)X(I\\otimes K_a)^*$ is positive. Their sum is positive as well. This argument explains the word \\emph{completely}. The conclusion holds for every auxiliary dimension, which a test of one input density matrix on the original system cannot establish.","negative eigenvalue. More concretely, the normalized vector $w=(\\ket{01}-\\ket{10})/\\sqrt2$ satisfies\n\\[\nw^*(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})w=-\\frac12.\n\\]\nThis exact negative value of the quadratic form certifies failure of positivity on the composite system. No numerical eigenvalue tolerance is needed.\n\\head{Why the Kraus form avoids this problem}\nFor every auxiliary dimension and positive $X$, each term $(I\\otimes K_a)X(I\\otimes K_a)^*$ is positive. Their sum is positive as well. This argument explains the word \\emph{completely}. The conclusion holds for every auxiliary dimension, which a test of one input density matrix on the original system cannot establish.","alue. More concretely, the normalized vector $w=(\\ket{01}-\\ket{10})/\\sqrt2$ satisfies\n\\[\nw^*(\\id\\otimes\\mathsf T)(\\rho_{\\rm Bell})w=-\\frac12.\n\\]\nThis exact negative value of the quadratic form certifies failure of positivity on the composite system. No numerical eigenvalue tolerance is needed.\n\\head{Why the Kraus form avoids this problem}\nFor every auxiliary dimension and positive $X$, each term $(I\\otimes K_a)X(I\\otimes K_a)^*$ is positive. Their sum is positive as well. This argument explains the word \\emph{completely}. The conclusion holds for every auxiliary dimension, which a test of one input density matrix on the original system cannot establish.","\\vpage{19}{Find the error without reading the logical state}{FOUR EXACT SYNDROME PACKETS}{Traditional edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical cohere","\\vpage{19}{Find the error without reading the logical state}{FOUR EXACT SYNDROME PACKETS}{Traditional edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered a","\\vpage{19}{Find the error without reading the logical state}{FOUR EXACT SYNDROME PACKETS}{Traditional edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as","\\vpage{19}{Find the error without reading the logical state}{FOUR EXACT SYNDROME PACKETS}{Traditional edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","ing the logical state}{FOUR EXACT SYNDROME PACKETS}{Traditional edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","onal edition: Theorem II.8.2, Theorem II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","II.8.4, Example II.8.8.}\nEncode a qubit by $\\ket{0_L}=\\ket{000}$ and $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","d $\\ket{1_L}=\\ket{111}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","}$. Let $P$ project onto this code and take errors $E_a\\in\\{I,X_1,X_2,X_3\\}$. Their images are four orthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","rthogonal rank-two packets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","ets $Q_a=E_aPE_a^*$ that sum to the full identity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","dentity.\n\\vfig{syndromes}\nUse the logical state $\\psi=(3\\ket{0_L}+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","+4\\ket{1_L})/5$ and channel probabilities $w=(1,2,3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.",",3,4)/10$. The noisy state is $\\sum_aw_aE_a\\psi\\psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","psi^*E_a^*$. A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.",". A syndrome measurement returns label $a$ with probability $w_a$, independent of the logical amplitudes.\n\\head{The recovery is an actual channel}\nThe operators $R_a=E_a^*Q_a$ satisfy $\\sum_aR_a^*R_a=I$. Applying the recovery $\\mathcal R(X)=\\sum_aR_aXR_a^*$ to the noisy state gives exactly $\\psi\\psi^*$. The off-diagonal logical coherence $12/25$ is recovered as well as the populations $9/25$ and $16/25$.","The Knill--Laflamme compressions are $PE_a^*E_bP=\\delta_{ab}P$. Their scalar form says that the error labels do not resolve the encoded vector. The exact checks evaluate all sixteen compressions, all sixteen syndrome products, the four probabilities, the recovery normalization, and the full recovered matrix.","\\vpage{20}{An unchanged syndrome can hide a logical error}{THE SCALAR COMPRESSION TEST MATTERS}{Traditional edition: Theorem II.8.2, Proposition II.8.7.}\nApply $Z_1=Z\\otimes I\\otimes I$ to the same repetition code. It sends $\\ket{000}$ to itself and $\\ket{111}$ to its negative. Both states remain inside the original code, so the previous syndrome partition reports no bit flip.\n\\[\n\\frac{3\\ket{000}+4\\ket{111}}5\n\\ \\longmapsto\\\n\\frac{3\\ket{000}-4\\ket{111}}5.\n\\]\nThe populations are unchanged, but the relative phase is not.\n\\vfig{logical_error}\nLet $V$ be the code isometry with columns $\\ket{000},\\ket{111}$. For the proposed error family $\\{I,Z_1\\}$, its off-diagonal error compression is\n\\[\nV^*I^*Z_1V=\\mat{1&0","7.}\nApply $Z_1=Z\\otimes I\\otimes I$ to the same repetition code. It sends $\\ket{000}$ to itself and $\\ket{111}$ to its negative. Both states remain inside the original code, so the previous syndrome partition reports no bit flip.\n\\[\n\\frac{3\\ket{000}+4\\ket{111}}5\n\\ \\longmapsto\\\n\\frac{3\\ket{000}-4\\ket{111}}5.\n\\]\nThe populations are unchanged, but the relative phase is not.\n\\vfig{logical_error}\nLet $V$ be the code isometry with columns $\\ket{000},\\ket{111}$. For the proposed error family $\\{I,Z_1\\}$, its off-diagonal error compression is\n\\[\nV^*I^*Z_1V=\\mat{1&0","the same repetition code. It sends $\\ket{000}$ to itself and $\\ket{111}$ to its negative. Both states remain inside the original code, so the previous syndrome partition reports no bit flip.\n\\[\n\\frac{3\\ket{000}+4\\ket{111}}5\n\\ \\longmapsto\\\n\\frac{3\\ket{000}-4\\ket{111}}5.\n\\]\nThe populations are unchanged, but the relative phase is not.\n\\vfig{logical_error}\nLet $V$ be the code isometry with columns $\\ket{000},\\ket{111}$. For the proposed error family $\\{I,Z_1\\}$, its off-diagonal error compression is\n\\[\nV^*I^*Z_1V=\\mat{1&0","self and $\\ket{111}$ to its negative. Both states remain inside the original code, so the previous syndrome partition reports no bit flip.\n\\[\n\\frac{3\\ket{000}+4\\ket{111}}5\n\\ \\longmapsto\\\n\\frac{3\\ket{000}-4\\ket{111}}5.\n\\]\nThe populations are unchanged, but the relative phase is not.\n\\vfig{logical_error}\nLet $V$ be the code isometry with columns $\\ket{000},\\ket{111}$. For the proposed error family $\\{I,Z_1\\}$, its off-diagonal error compression is\n\\[\nV^*I^*Z_1V=\\mat{1&0","main inside the original code, so the previous syndrome partition reports no bit flip.\n\\[\n\\frac{3\\ket{000}+4\\ket{111}}5\n\\ \\longmapsto\\\n\\frac{3\\ket{000}-4\\ket{111}}5.\n\\]\nThe populations are unchanged, but the relative phase is not.\n\\vfig{logical_error}\nLet $V$ be the code isometry with columns $\\ket{000},\\ket{111}$. For the proposed error family $\\{I,Z_1\\}$, its off-diagonal error compression is\n\\[\nV^*I^*Z_1V=\\mat{1&0\\\\0&-1}.\n\\]\nThis matrix is not $cI$ for any scalar $c$. Thus the Knill--Laflamme condition fails for that family. Its trace is zero even though its action on the code is nonzero. A trace test alone would miss the obstruction.\n\\head{Known correction versus family recovery}\nIf it is known with certainty that $Z_1$ occurred, applying $Z_1$ again reverses it. The failed criterion concerns a \\emph{single recovery valid for every unknown error in the declared family}, not invertibility of each error considered separately. This distinction prevents a misleading conclusion that no phase error can ever be undone.","0&-1}.\n\\]\nThis matrix is not $cI$ for any scalar $c$. Thus the Knill--Laflamme condition fails for that family. Its trace is zero even though its action on the code is nonzero. A trace test alone would miss the obstruction.\n\\head{Known correction versus family recovery}\nIf it is known with certainty that $Z_1$ occurred, applying $Z_1$ again reverses it. The failed criterion concerns a \\emph{single recovery valid for every unknown error in the declared family}, not invertibility of each error considered separately. This distinction prevents a misleading conclusion that no phase error can ever be undone.","\\vpage{21}{Hidden states return as spectral dependence}{TWO INVERTIBILITY CONDITIONS, NOT ONE}{Traditional edition: Theorem II.9.2, Theorem II.9.3.}\nTake $H=\\mat{0&1\\\\1&2}$ and retain its first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as wr","1&2}$ and retain its first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entr","1&2}$ and retain its first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The r","1&2}$ and retain its first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression h","1&2}$ and retain its first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression has a continuation to this value. By contrast, $1\\pm\\sqrt2$ are poles of the full resolvent.","ts first coordinate. The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression has a continuation to this value. By contrast, $1\\pm\\sqrt2$ are poles of the full resolvent.",". The eliminated block is $H_{QQ}=2$. Whenever $z-2$ and the full $zI-H$ are invertible,\n\\[\nH_{\\rm eff}(z)=\\frac1{z-2},\\qquad\n\\bigl[(zI-H)^{-1}\\bigr]_{00}\n=\\frac1{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression has a continuation to this value. By contrast, $1\\pm\\sqrt2$ are poles of the full resolvent.","{z-1/(z-2)}=\\frac{z-2}{z^2-2z-1}.\n\\]\nThis is the corrected hypothesis set recorded in the formal release.\n\\vfig{feshbach}\nAt $z=3$, the exact retained resolvent is $1/2$. Bare truncation uses only $H_{PP}=0$ and gives $1/3$, an error of $1/6$. The omitted coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression has a continuation to this value. By contrast, $1\\pm\\sqrt2$ are poles of the full resolvent.","coupling has changed the response even though the retained diagonal entry did not change.\n\\head{A pole can belong to the elimination chart}\nAt $z=2$, the eliminated inverse does not exist, so the displayed Schur formula cannot be applied there as written. Nevertheless, the full resolvent exists and its retained entry is zero. The rational expression has a continuation to this value. By contrast, $1\\pm\\sqrt2$ are poles of the full resolvent.","Invertibility of the full matrix is an independent requirement. At $z=0$, $H=\\diag(0,-1)$ has an invertible eliminated block in $zI-H$, but the full matrix is singular. The original prose omitted this condition. The preserved amendments restore it. This appendix uses the corrected form throughout.\n\\takeaway{An exact elimination formula has a spectral domain. Check both inverses before evaluating it, and distinguish poles of an intermediate inverse from poles of the full physical resolvent.}","\\vpage{22}{The smaller system remembers the discarded one}{AN EXACT QUANTUM MEMORY EQUATION}{Traditional edition: Theorem II.9.8, Theorem II.9.11.}\nChoose $H=X$, initial state $(1,0)^T$, and retain the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidde","\\vpage{22}{The smaller system remembers the discarded one}{AN EXACT QUANTUM MEMORY EQUATION}{Traditional edition: Theorem II.9.8, Theorem II.9.11.}\nChoose $H=X$, initial state $(1,0)^T$, and retain the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add","\\vpage{22}{The smaller system remembers the discarded one}{AN EXACT QUANTUM MEMORY EQUATION}{Traditional edition: Theorem II.9.8, Theorem II.9.11.}\nChoose $H=X$, initial state $(1,0)^T$, and retain the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","MORY EQUATION}{Traditional edition: Theorem II.9.8, Theorem II.9.11.}\nChoose $H=X$, initial state $(1,0)^T$, and retain the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","T$, and retain the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","the first coordinate. The coupled equations and their solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","r solution are\n\\[\n\\ii\\dot x=y,\\qquad \\ii\\dot y=x,\\qquad\nx(t)=\\cos t,\\quad y(t)=-\\ii\\sin t.\n\\]\nThe hidden probability returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","returns periodically. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","ly. Deleting the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","the hidden row and column would instead leave the zero Hamiltonian and predict $x(t)=1$.\n\\vfig{memory}\n\\head{Eliminate exactly, not approximately}\nIntegrating the second equation with $y(0)=0$ gives $y(t)=-\\ii\\int_0^tx(s)\\,ds$. Substitution produces\n\\[\n\\dot x(t)=-\\int_0^tx(s)\\,ds.\n\\]\nThe memory kernel is the constant function one. Differentiating once more recovers $\\ddot x=-x$ with $x(0)=1$ and $\\dot x(0)=0$, explaining the cosine solution. A nonzero hidden initial state would add a forcing term that must also be retained.","The full propagator is a unitary group, but its retained scalar entry is not a semigroup. At $t=s=\\pi/4$,\n\\[\nx(t+s)-x(t)x(s)=0-\\frac12=-\\frac12.\n\\]\nThere is no contradiction: compression of a product need not equal the product of the compressed factors when the hidden subspace carries intermediate amplitude.","\\vpage{23}{Positive transfer is not a stochastic matrix}{SPECTRAL POSITIVITY IS BASIS INDEPENDENT}{Traditional edition: Theorem II.10.1, Theorem II.10.3.}\nFor $H=I-X$, the two eigenvalues are zero and two. Its Euclidean transfer is\n\\[\nT(\\tau)=e^{-\\tau H}\n=\\frac12\\mat{1+q&1-q","\\vpage{23}{Positive transfer is not a stochastic matrix}{SPECTRAL POSITIVITY IS BASIS INDEPENDENT}{Traditional edition: Theorem II.10.1, Theorem II.10.3.}\nFor $H=I-X$, the two eigenvalues are zero and two. Its Euclidean transfer is\n\\[\nT(\\tau)=e^{-\\tau H}\n=\\frac12\\mat{1+q&1-q\\\\1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\rig","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe fact","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite opera","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator iden","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure,","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity","1-q&1+q},\\qquad q=e^{-2\\tau}.\n\\]\nFor $\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these","\\tau\\ge0$, the eigenvalues $1,q$ lie in $(0,1]$, and multiplying two such matrices replaces $q,r$ by $qr$. This is the semigroup law in spectral coordinates.\n\\vfig{euclidean}\nAt $q=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these assumptions requires its own justification beyond the two positive eigenvalues.","=1/2$, the displayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these assumptions requires its own justification beyond the two positive eigenvalues.","splayed $T$ happens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these assumptions requires its own justification beyond the two positive eigenvalues.","ens also to have nonnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these assumptions requires its own justification beyond the two positive eigenvalues.","nnegative entries and column sums one. Conjugate by the unitary $Z$. The spectrum and positivity of the quadratic form are preserved, but both off-diagonal entries become $-1/4$. Thus positivity as an operator is not entrywise positivity and does not by itself give a Markov interpretation.\n\\head{A useful finite positive kernel}\nFor this nonnegative $H$, vectors $u_a$ and times $t_a\\ge0$ satisfy\n\\[\n\\sum_{a,b}\\langle u_a,e^{-(t_a+t_b)H}u_b\\rangle\n=\\left\\|\\sum_a e^{-t_aH}u_a\\right\\|^2\\ge0.\n\\]\nThe factorization makes positivity transparent. It remains a finite operator identity. A continuum reconstruction theorem requires a compatible field structure, limits, positivity hypotheses, and the other assumptions of the chosen reconstruction framework. Each of these assumptions requires its own justification beyond the two positive eigenvalues.","The exact check verifies the matrix semigroup law as a polynomial identity in $q,r$ and verifies the unchanged spectrum after conjugation. It rejects entrywise nonnegativity for the conjugated matrix.\n\\takeaway{Euclidean positivity refers to quadratic forms and spectral values. A stochastic interpretation or a continuum physical reconstruction needs additional structure.}","\\vpage{24}{A dimension curve is a scale diagnostic}{THE EXACT HEAT TRACE ON FOUR SITES}{Traditional edition: Definition II.11.1, Proposition II.11.2.}\nFor the Laplacian on a cycle with four sites $L=2I-S-S^*$, the eigenvalues are $0,2,2,4$. The normalized heat trace and its diagnostic of spectral dimension are therefore\n\\[\nK(\\tau)=\\frac{1+2e^{-2\\tau}+e^{-4\\tau}}4\n=\\frac{(1+e^{-2\\tau})^2}4,\n\\qquad\n d_s(\\tau)=-2\\tau\\frac{K'(\\tau)}{K(\\tau)}\n=\\frac{8\\tau}{e^{2\\tau}+1}.\n\\]\nNo numerical differentiation is needed for this finite example.\n\\vfig{spectral_dimension}\n\\head{Both ends reflect the finite size of the system}\nAs $\\tau\\to0$, every eigenmode contributes approximately one and the dimension diagnostic tends to zero. As $\\tau\\to\\infty$, the zero mode dominates, so it again tends to zero. A finite spectrum cannot sustain a nonzero plateau at all scales.","ues are $0,2,2,4$. The normalized heat trace and its diagnostic of spectral dimension are therefore\n\\[\nK(\\tau)=\\frac{1+2e^{-2\\tau}+e^{-4\\tau}}4\n=\\frac{(1+e^{-2\\tau})^2}4,\n\\qquad\n d_s(\\tau)=-2\\tau\\frac{K'(\\tau)}{K(\\tau)}\n=\\frac{8\\tau}{e^{2\\tau}+1}.\n\\]\nNo numerical differentiation is needed for this finite example.\n\\vfig{spectral_dimension}\n\\head{Both ends reflect the finite size of the system}\nAs $\\tau\\to0$, every eigenmode contributes approximately one and the dimension diagnostic tends to zero. As $\\tau\\to\\infty$, the zero mode dominates, so it again tends to zero. A finite spectrum cannot sustain a nonzero plateau at all scales.","\\tau)=\\frac{1+2e^{-2\\tau}+e^{-4\\tau}}4\n=\\frac{(1+e^{-2\\tau})^2}4,\n\\qquad\n d_s(\\tau)=-2\\tau\\frac{K'(\\tau)}{K(\\tau)}\n=\\frac{8\\tau}{e^{2\\tau}+1}.\n\\]\nNo numerical differentiation is needed for this finite example.\n\\vfig{spectral_dimension}\n\\head{Both ends reflect the finite size of the system}\nAs $\\tau\\to0$, every eigenmode contributes approximately one and the dimension diagnostic tends to zero. As $\\tau\\to\\infty$, the zero mode dominates, so it again tends to zero. A finite spectrum cannot sustain a nonzero plateau at all scales.","For the Cartesian product of two identical cycles with four sites, the Laplacian eigenvalues add and the heat traces multiply: $K_2=K^2$. Consequently $d_{s,2}=2d_s$ exactly. The product identity determines the second curve exactly from the first.","At $\\tau=\\log2/2$, the first curve equals $4\\log2/3$. This noninteger value is a spectral quantity that depends on scale, not a claim that the graph has a fractional topological dimension. Four sites are too few to promise a broad geometric plateau. Experiments on larger grids remain finite diagnostics. Convergence requires the analytic hypotheses developed on \\pagerefV{37}.\n\\takeaway{A convincing continuum dimension needs a stable scale window that persists under controlled refinement. One appealing finite curve is not that theorem.}","\\vpage{25}{Refine the grid while holding the mode fixed}{AN EXACT ERROR ENCLOSURE, NOT A FITTED SLOPE}{Traditional edition: Theorem II.11.4.}\nFor the circle symbol, the continuum value is $k^2$ and the grid value is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\!\\left(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","XACT ERROR ENCLOSURE, NOT A FITTED SLOPE}{Traditional edition: Theorem II.11.4.}\nFor the circle symbol, the continuum value is $k^2$ and the grid value is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\!\\left(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","}\nFor the circle symbol, the continuum value is $k^2$ and the grid value is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\!\\left(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","circle symbol, the continuum value is $k^2$ and the grid value is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\!\\left(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","tinuum value is $k^2$ and the grid value is\n\\[\n\\lambda_{k,h}=\\frac4{h^2}\\sin^2\\!\\left(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","eft(\\frac{kh}{2}\\right),\\qquad h\\ne0.\n\\]\nFor a fixed low mode, the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","the paper bounds the error by $0\\le k^2-\\lambda_{k,h}\\le k^4h^2/12$. The spacing condition matters because the displayed expression divides by $h^2$.\n\\vfig{dispersion}\n\\head{What is certified in the displayed table?}\nTake $k=1$ and $h=1/4,1/8,1/16$. For $x=h/2\\in[0,1]$, alternating Taylor sums give rational lower and upper bounds for $\\sin x$. Squaring these positive endpoints and multiplying by $4/h^2$ yields exact rational enclosures for each error. The displayed decimals only locate those much narrower rational intervals.","Each interval lies strictly inside $(0,h^2/12)$. Dividing successive interval endpoints proves that both error ratios are between $3.99$ and $4$. The check verifies these inequalities by rational arithmetic, without relying on approximate sine values or a fitted logarithmic slope.\n\\head{Why the mode regime matters}\nIf instead $k=\\pi/h$ moves with the grid, the ratio $\\lambda_{k,h}/k^2=4/\\pi^2$ stays constant, not one. The bound's factor $k^4$ warns that it is not uniform in arbitrary moving high modes. The displayed rational test evaluates the analytic symbol. It does not assert that every chosen spacing defines an integer periodic grid.\n\\takeaway{Second-order convergence is a quantified statement with a specified mode regime. Keep that","Each interval lies strictly inside $(0,h^2/12)$. Dividing successive interval endpoints proves that both error ratios are between $3.99$ and $4$. The check verifies these inequalities by rational arithmetic, without relying on approximate sine values or a fitted logarithmic slope.\n\\head{Why the mode regime matters}\nIf instead $k=\\pi/h$ moves with the grid, the ratio $\\lambda_{k,h}/k^2=4/\\pi^2$ stays constant, not one. The bound's factor $k^4$ warns that it is not uniform in arbitrary moving high modes. The displayed rational test evaluates the analytic symbol. It does not assert that every chosen spacing defines an integer periodic grid.\n\\takeaway{Second-order convergence is a quantified statement with a specified mode regime. Keep that regime fixed when reading an error curve or comparing refinements.}","/12)$. Dividing successive interval endpoints proves that both error ratios are between $3.99$ and $4$. The check verifies these inequalities by rational arithmetic, without relying on approximate sine values or a fitted logarithmic slope.\n\\head{Why the mode regime matters}\nIf instead $k=\\pi/h$ moves with the grid, the ratio $\\lambda_{k,h}/k^2=4/\\pi^2$ stays constant, not one. The bound's factor $k^4$ warns that it is not uniform in arbitrary moving high modes. The displayed rational test evaluates the analytic symbol. It does not assert that every chosen spacing defines an integer periodic grid.\n\\takeaway{Second-order convergence is a quantified statement with a specified mode regime. Keep that regime fixed when reading an error curve or comparing refinements.}","\\vpage{26}{A zero compressed anomaly can be automatic}{TEST THE CONSTRAINTS BEFORE THE PROJECTION}{Traditional edition: Theorem II.12.2, Proposition II.12.3.}\nTake the self-adjoint matrices\n\\[\nC_1=\\diag(0,1,-1),\\qquad\nC_2=\\mat{0&0&0\\\\0&0&1\\\\0&1&0},\\qquad\nP=\\diag(1,0,0).\n\\]\nBoth constraints annihilate $P\\CC^3$, and their common kernel is exactly that one-dimensional space. Nevertheless, their full commutator is nonzero.\n\\vfig{constraints}\n\\head{Why the compressed test cannot decide closure}\nSince $C_1P=C_2P=0$, each product $C_1C_2P$ and $C_2C_1P$ vanishes. Therefore $P[C_1,C_2]P=0$ without any requirement that the unprojected commutator close in the chosen constraint span. In this example the skew off-diagonal commutator is not a linear combination of the two displayed symmetric constraints.","0&1&0},\\qquad\nP=\\diag(1,0,0).\n\\]\nBoth constraints annihilate $P\\CC^3$, and their common kernel is exactly that one-dimensional space. Nevertheless, their full commutator is nonzero.\n\\vfig{constraints}\n\\head{Why the compressed test cannot decide closure}\nSince $C_1P=C_2P=0$, each product $C_1C_2P$ and $C_2C_1P$ vanishes. Therefore $P[C_1,C_2]P=0$ without any requirement that the unprojected commutator close in the chosen constraint span. In this example the skew off-diagonal commutator is not a linear combination of the two displayed symmetric constraints.","\\vpage{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephas","\\vpage{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\","\\vpage{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+","\\vpage{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le","\\vpage{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probabi","age{27}{One calculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3…","lculation of transfer to the opposite site}{SEEDS, CHARACTERS, INTERFERENCE, AND DEPHASING}{Traditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Compl…","ditional edition: Corollary II.3.3, Theorem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Complete momentum dephasing reduces it to $1/4$. This calculation isolates how discarded coherence c…","rem II.4.4, Theorem II.7.4.}\nReturn to the shift on four sites $S$ and Laplacian $L=2I-S-S^*$. The same character projectors have energies $(0,2,4,2)$. At time $t=\\pi/2$, their evolution phases are $(1,-1,1,-1)$, exactly the character eigenvalues of $S^2$. Hence\n\\[\ne^{-\\ii\\pi L/2}=S^2,\\qquad\n\\ket0\\longmapsto\\ket2.\n\\]\n\\vfig{perfect_transfer}\nAt halfway time $\\pi/4$, the actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Complete momentum dephasing reduces it to $1/4$. This calculation isolates how discarded coherence changes the interference result.","actual amplitude vector is $(-\\ii,1,\\ii,1)^T/2$, giving uniform site probabilities. The subsequent phases recombine those four amplitudes into a single output at the opposite site. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Complete momentum dephasing reduces it to $1/4$. This calculation isolates how discarded coherence changes the interference result.","ite. Uniform intermediate probabilities do not imply an incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Complete momentum dephasing reduces it to $1/4$. This calculation isolates how discarded coherence changes the interference result.","incoherent intermediate state.\n\\head{A channel changes the conclusion predictably}\nBefore evolving, mix the seed density matrix with its complete momentum dephasing:\n\\[\n\\rho_\\eta=(1-\\eta)\\ket0\\bra0+\\eta\\sum_jP_j\\ket0\\bra0P_j\n=(1-\\eta)\\ket0\\bra0+\\eta I/4,\n\\quad0\\le\\eta\\le1.\n\\]\nAt transfer time, the probability at the opposite site is exactly $1-3\\eta/4$. Complete momentum dephasing reduces it to $1/4$. This calculation isolates how discarded coherence changes the interference result.","Finally, the shift grammar generates four independent vectors from $\\ket0$, while the Hamiltonian powers generate only three: its repeated energy links two momentum directions. The same seed can therefore be cyclic for the shift algebra and noncyclic for the Hamiltonian algebra, even though perfect transfer occurs at one time.\n\\takeaway{Perfect transfer is an exact finite interference statement. It neither implies cyclicity for every operator grammar nor survives an unaccounted dephasing operation.}","imitation of testing only the common kernel on \\pagerefV{26}.\n\\takeaway{Exact finite structure is valuable because its boundary is explicit. The appendix adds understanding and reproducibility, not unproved continuum claims or a new theorem count.}\n\\vpage{29}{Nonabelian packets carry multiplicity}{A SIX-DIMENSIONAL REGULAR REPRESENTATION}{Traditional edition: Theorem II.3.1, Theorem II.3.2.}\nLet $\\HH=\\CC[S_3]$ have the six orthonormal basis vectors $\\ket g$, with left action $U(h)\\ket g=\\ket{hg}$. The irreducible characters on the identity, transposition, and three-cycle classes are\n\\[\n\\begin{array}{c|rrr|r}\n & e & (12)&(123)&d_\\lambda","fV{26}.\n\\takeaway{Exact finite structure is valuable because its boundary is explicit. The appendix adds understanding and reproducibility, not unproved continuum claims or a new theorem count.}\n\\vpage{29}{Nonabelian packets carry multiplicity}{A SIX-DIMENSIONAL REGULAR REPRESENTATION}{Traditional edition: Theorem II.3.1, Theorem II.3.2.}\nLet $\\HH=\\CC[S_3]$ have the six orthonormal basis vectors $\\ket g$, with left action $U(h)\\ket g=\\ket{hg}$. The irreducible characters on the identity, transposition, and three-cycle classes are\n\\[\n\\begin{array}{c|rrr|r}\n & e & (12)&(123)&d_\\lambda","ite structure is valuable because its boundary is explicit. The appendix adds understanding and reproducibility, not unproved continuum claims or a new theorem count.}\n\\vpage{29}{Nonabelian packets carry multiplicity}{A SIX-DIMENSIONAL REGULAR REPRESENTATION}{Traditional edition: Theorem II.3.1, Theorem II.3.2.}\nLet $\\HH=\\CC[S_3]$ have the six orthonormal basis vectors $\\ket g$, with left action $U(h)\\ket g=\\ket{hg}$. The irreducible characters on the identity, transposition, and three-cycle classes are\n\\[\n\\begin{array}{c|rrr|r}\n & e & (12)&(123)&d_\\lambda",", not unproved continuum claims or a new theorem count.}\n\\vpage{29}{Nonabelian packets carry multiplicity}{A SIX-DIMENSIONAL REGULAR REPRESENTATION}{Traditional edition: Theorem II.3.1, Theorem II.3.2.}\nLet $\\HH=\\CC[S_3]$ have the six orthonormal basis vectors $\\ket g$, with left action $U(h)\\ket g=\\ket{hg}$. The irreducible characters on the identity, transposition, and three-cycle classes are\n\\[\n\\begin{array}{c|rrr|r}\n & e & (12)&(123)&d_\\lambda\\\\\\hline\n\\text{trivial}&1&1&1&1\\\\\n\\text{sign}&1&-1&1&1\\\\\n\\text{standard}&2&0&-1&2\n\\end{array}\n\\qquad\nP_\\lambda=\\frac{d_\\lambda}{6}\\sum_{g\\in S_3}\\chi_\\lambda(g^{-1})U(g).\n\\]\nAll matrices in this example are rational. Multiplying them gives self-adjoint idempotents of ranks $1,1,4$, with pairwise zero products and sum $I_6$.\n\\xfig{s3}\n\\head{The standard packet has dimension four}\nThe standard representation occurs twice in the regular representation. Its isotypic component therefore has dimension $2\\cdot2=4$: one factor carries the two-dimensional irreducible action and the other records multiplicity. A character projector selects the whole i…","\\text{standard}&2&0&-1&2\n\\end{array}\n\\qquad\nP_\\lambda=\\frac{d_\\lambda}{6}\\sum_{g\\in S_3}\\chi_\\lambda(g^{-1})U(g).\n\\]\nAll matrices in this example are rational. Multiplying them gives self-adjoint idempotents of ranks $1,1,4$, with pairwise zero products and sum $I_6$.\n\\xfig{s3}\n\\head{The standard packet has dimension four}\nThe standard representation occurs twice in the regular representation. Its isotypic component therefore has dimension $2\\cdot2=4$: one factor carries the two-dimensional irreducible action and the other records multiplicity. A character projector selects the whole isotypic component. It does not select a particular copy or a particular internal vector.","_\\lambda=\\frac{d_\\lambda}{6}\\sum_{g\\in S_3}\\chi_\\lambda(g^{-1})U(g).\n\\]\nAll matrices in this example are rational. Multiplying them gives self-adjoint idempotents of ranks $1,1,4$, with pairwise zero products and sum $I_6$.\n\\xfig{s3}\n\\head{The standard packet has dimension four}\nThe standard representation occurs twice in the regular representation. Its isotypic component therefore has dimension $2\\cdot2=4$: one factor carries the two-dimensional irreducible action and the other records multiplicity. A character projector selects the whole isotypic component. It does not select a particular copy or a particular internal vector.","cible factor, while leaving the multiplicity factor untouched. Their multiplication law is stronger information than a list of packet ranks. A basis chosen inside this packet supplies coordinates. It does not canonically distinguish the two copies of the standard representation.\n\\head{What the retained check establishes}\nThe retained exact check constructs all six permutation matrices, tests all $36$ multiplication relations, computes the character averages, and checks ranks independently by Gaussian elimination and traces. It rejects the claim that the standard isotypic projector has rank two. This is one specified representation of $S_3$, not a computational classification of all finite group representations.\n\\takeaway{For a nonabelian symmetry, distinguish irreducible dimension from multiplicity. A central character projector retains both.}","pies of the standard representation.\n\\head{What the retained check establishes}\nThe retained exact check constructs all six permutation matrices, tests all $36$ multiplication relations, computes the character averages, and checks ranks independently by Gaussian elimination and traces. It rejects the claim that the standard isotypic projector has rank two. This is one specified representation of $S_3$, not a computational classification of all finite group representations.\n\\takeaway{For a nonabelian symmetry, distinguish irreducible dimension from multiplicity. A central character projector retains both.}","\\vpage{30}{Rotate whole packets by a polar factor}{AN EXACT LOCAL CHART AND ITS FAILURE}{Traditional edition: Theorem II.6.5, Theorem II.6.6.}\nOn $\\HH=\\CC^2$, put $P_0=\\diag(1,0)$, $P_1=I-P_0$, and\n\\[\nR=\\mat{3/5&-4/5","\\vpage{30}{Rotate whole packets by a polar factor}{AN EXACT LOCAL CHART AND ITS FAILURE}{Traditional edition: Theorem II.6.5, Theorem II.6.6.}\nOn $\\HH=\\CC^2$, put $P_0=\\diag(1,0)$, $P_1=I-P_0$, and\n\\[\nR=\\mat{3/5&-4/5\\\\4/5&3/5},\\qquad Q_a=RP_aR^*,\n\\qquad S=\\sum_{a=0}^1Q_aP_a=\\frac35R.\n\\]\nThe exact Gram matrices give\n\\[\n(S-I)^*(S-I)=\\frac{16}{25}I,\\quad S^*S=\\frac9{25}I,\n\\quad\\norm{S-I}=\\frac45<1.\n\\]\nConsequently the unitary polar factor is $S(S^*S)^{-1/2}=R$, and $Q_aR=RP_a$ for both complete packets.\n\\xfig{polar}\n\\head{The norm condition selects a chart}\nIf the target rank-one packet is orthogonal to the initial one, the two labels are exchanged and $S=0$. Its polar inverse square root is undefined. This is a failure of this direct local construction, although a unitary carrying one packet system to the other still exists. The strict norm assumption is sufficient, not a global existence criterion for all packet rotations.","4/5&3/5},\\qquad Q_a=RP_aR^*,\n\\qquad S=\\sum_{a=0}^1Q_aP_a=\\frac35R.\n\\]\nThe exact Gram matrices give\n\\[\n(S-I)^*(S-I)=\\frac{16}{25}I,\\quad S^*S=\\frac9{25}I,\n\\quad\\norm{S-I}=\\frac45<1.\n\\]\nConsequently the unitary polar factor is $S(S^*S)^{-1/2}=R$, and $Q_aR=RP_a$ for both complete packets.\n\\xfig{polar}\n\\head{The norm condition selects a chart}\nIf the target rank-one packet is orthogonal to the initial one, the two labels are exchanged and $S=0$. Its polar inverse square root is undefined. This is a failure of this direct local construction, although a unitary carrying one packet system to the other still exists. The strict norm assumption is sufficient, not a global existence criterion for all packet rotations.","m_{a=0}^1Q_aP_a=\\frac35R.\n\\]\nThe exact Gram matrices give\n\\[\n(S-I)^*(S-I)=\\frac{16}{25}I,\\quad S^*S=\\frac9{25}I,\n\\quad\\norm{S-I}=\\frac45<1.\n\\]\nConsequently the unitary polar factor is $S(S^*S)^{-1/2}=R$, and $Q_aR=RP_a$ for both complete packets.\n\\xfig{polar}\n\\head{The norm condition selects a chart}\nIf the target rank-one packet is orthogonal to the initial one, the two labels are exchanged and $S=0$. Its polar inverse square root is undefined. This is a failure of this direct local construction, although a unitary carrying one packet system to the other still exists. The strict norm assumption is sufficient, not a global existence criterion for all packet rotations.","Short steps can be composed before reaching a distant endpoint. For example, the rational step $R_1=\\mat{4/5&-3/5\\\\3/5&4/5}$ obeys\n\\[\nR_1^2=\\mat{7/25&-24/25","3/5&4/5}$ obeys\n\\[\nR_1^2=\\mat{7/25&-24/25\\\\24/25&7/25}.\n\\]\nThe multiplication is exact and each step has a valid local overlap. The convergence theorem has a separate analytic contract: a twice continuously differentiable complete projector path on a compact time interval, sufficiently fine partitions, and a constant depending on the path in an $O(|\\pi|)$ estimate in operator norm.\n\\head{Evidence boundary}\nThe retained run checks the Gram identities, polar factor, intertwining, and product of two rational steps. It also checks that the overlap has rank zero when the endpoint packets are orthogonal. It does not certify the analytic convergence theorem by sampling a path.\n\\takeaway{A local polar chart can fail while the subspaces remain regular. Us","24/25&7/25}.\n\\]\nThe multiplication is exact and each step has a valid local overlap. The convergence theorem has a separate analytic contract: a twice continuously differentiable complete projector path on a compact time interval, sufficiently fine partitions, and a constant depending on the path in an $O(|\\pi|)$ estimate in operator norm.\n\\head{Evidence boundary}\nThe retained run checks the Gram identities, polar factor, intertwining, and product of two rational steps. It also checks that the overlap has rank zero when the endpoint packets are orthogonal. It does not certify the analytic convergence theorem by sampling a path.\n\\takeaway{A local polar chart can fail while the subspaces remain regular. Use the stated chart hypothesis and the separate convergence theorem when composing small rotations.}","\\vpage{31}{Coherence order survives a covariant channel}{CHANNEL, CHOI, AND GENERATOR CHECKS}{Traditional edition: Theorem II.7.8, Theorem II.7.10, Theorem II.7.14.}\nTake $S=Z=\\diag(1,-1)$ on $\\CC^2$. Under conjugation $X\\mapsto ZXZ$, diagonal matrices have coherence order zero and off-diagonal matrices have coherence order one modulo two. Its projectors on Hilbert--Schmidt operator space are $\\Lcal_0(X)=(X+ZXZ)/2$ and $\\Lcal_1(X)=(X-ZXZ)/2$.","er survives a covariant channel}{CHANNEL, CHOI, AND GENERATOR CHECKS}{Traditional edition: Theorem II.7.8, Theorem II.7.10, Theorem II.7.14.}\nTake $S=Z=\\diag(1,-1)$ on $\\CC^2$. Under conjugation $X\\mapsto ZXZ$, diagonal matrices have coherence order zero and off-diagonal matrices have coherence order one modulo two. Its projectors on Hilbert--Schmidt operator space are $\\Lcal_0(X)=(X+ZXZ)/2$ and $\\Lcal_1(X)=(X-ZXZ)/2$.","Choose amplitude damping with $K_0=\\diag(1,4/5)$ and $K_1=(3/5)\\ket0\\bra1$. The channel is trace preserving, population flows from state one to state zero, and each coherence contracts by $4/5$.\n\\xfig{choi}\nWith the unnormalized Choi convention $J=\\sum_{a,b}\\Phi(\\ket a\\bra b)\\otimes\\ket a\\bra b$ and basis $(00,01,10,11)$, the exact matrix is\n\\[\nJ=\\mat{1&0&0&4/5","Choose amplitude damping with $K_0=\\diag(1,4/5)$ and $K_1=(3/5)\\ket0\\bra1$. The channel is trace preserving, population flows from state one to state zero, and each coherence contracts by $4/5$.\n\\xfig{choi}\nWith the unnormalized Choi convention $J=\\sum_{a,b}\\Phi(\\ket a\\bra b)\\otimes\\ket a\\bra b$ and basis $(00,01,10,11)$, the exact matrix is\n\\[\nJ=\\mat{1&0&0&4/5\\\\0&9/25&0&0\\\\0&0&0&0\\\\4/5&0&0&16/25},\n\\qquad [J,Z\\otimes Z]=0.\n\\]\nThe $00,11$ block is positive and has rank one. The other nonzero eigenvalue is $9/25$. The retained check also verifies that every principal minor is nonnegative.\n\\head{Symmetry in continuous time has its own hypotheses}\nFor the Lindblad generator with Hamiltonian $H=\\diag(0,2)$ and jump $L=(3/5)\\ket0\\bra1$, $H$ has charge zero and $L$ has charge one. Each matrix unit stays within its coherence order under the generator. Replacing $H$ by $X$ breaks that conclusion: the generator then sends a diagonal matrix to off-diagonal entries.","4/5&0&0&16/25},\n\\qquad [J,Z\\otimes Z]=0.\n\\]\nThe $00,11$ block is positive and has rank one. The other nonzero eigenvalue is $9/25$. The retained check also verifies that every principal minor is nonnegative.\n\\head{Symmetry in continuous time has its own hypotheses}\nFor the Lindblad generator with Hamiltonian $H=\\diag(0,2)$ and jump $L=(3/5)\\ket0\\bra1$, $H$ has charge zero and $L$ has charge one. Each matrix unit stays within its coherence order under the generator. Replacing $H$ by $X$ breaks that conclusion: the generator then sends a diagonal matrix to off-diagonal entries.","rank one. The other nonzero eigenvalue is $9/25$. The retained check also verifies that every principal minor is nonnegative.\n\\head{Symmetry in continuous time has its own hypotheses}\nFor the Lindblad generator with Hamiltonian $H=\\diag(0,2)$ and jump $L=(3/5)\\ket0\\bra1$, $H$ has charge zero and $L$ has charge one. Each matrix unit stays within its coherence order under the generator. Replacing $H$ by $X$ breaks that conclusion: the generator then sends a diagonal matrix to off-diagonal entries.","ne. The other nonzero eigenvalue is $9/25$. The retained check also verifies that every principal minor is nonnegative.\n\\head{Symmetry in continuous time has its own hypotheses}\nFor the Lindblad generator with Hamiltonian $H=\\diag(0,2)$ and jump $L=(3/5)\\ket0\\bra1$, $H$ has charge zero and $L$ has charge one. Each matrix unit stays within its coherence order under the generator. Replacing $H$ by $X$ breaks that conclusion: the generator then sends a diagonal matrix to off-diagonal entries.","\\vpage{32}{Null Gram directions are removed before recovery}{A TOTAL CHANNEL FROM SINGULAR ERROR DATA}{Traditional edition: Theorem II.8.3, Theorem II.8.4.}\nLet $\\HH=\\CC^3$, with nonzero code projector $P=\\diag(1,1,0)$. On the code use the error family $E_1=(3/5)P$, $E_2=(4/5)P$, and $E_3=0$. Its Knill--Laflamme Gram matrix is\n\\[\nc=\\frac1{25}\\mat{9&12&0","\\vpage{32}{Null Gram directions are removed before recovery}{A TOTAL CHANNEL FROM SINGULAR ERROR DATA}{Traditional edition: Theorem II.8.3, Theorem II.8.4.}\nLet $\\HH=\\CC^3$, with nonzero code projector $P=\\diag(1,1,0)$. On the code use the error family $E_1=(3/5)P$, $E_2=(4/5)P$, and $E_3=0$. Its Knill--Laflamme Gram matrix is\n\\[\nc=\\frac1{25}\\mat{9&12&0\\\\12&16&0\\\\0&0&0},\n\\qquad PE_a^*E_bP=c_{ab}P.\n\\]\nThe errors are redundant rather than three distinct syndromes. A real unitary change of error coordinates gives\n\\[\nW=\\mat{3/5&4/5&0","0&0&0},\n\\qquad PE_a^*E_bP=c_{ab}P.\n\\]\nThe errors are redundant rather than three distinct syndromes. A real unitary change of error coordinates gives\n\\[\nW=\\mat{3/5&4/5&0\\\\-4/5&3/5&0\\\\0&0&1},\n\\quad WcW^*=\\diag(1,0,0),\n\\quad(F_1,F_2,F_3)=(P,0,0).\n\\]\n\\xfig{gram}\n\\head{Normalize only positive eigenvalues}\nThe positive branch has $V_1=P$ as an isometry \\emph{on the code}, and syndrome support $Q=P$. No factor $\\lambda^{-1/2}$ is formed on either zero branch: those branches vanish on the code. Treating them as extra recoverable packets would divide by zero.","0&0&1},\n\\quad WcW^*=\\diag(1,0,0),\n\\quad(F_1,F_2,F_3)=(P,0,0).\n\\]\n\\xfig{gram}\n\\head{Normalize only positive eigenvalues}\nThe positive branch has $V_1=P$ as an isometry \\emph{on the code}, and syndrome support $Q=P$. No factor $\\lambda^{-1/2}$ is formed on either zero branch: those branches vanish on the code. Treating them as extra recoverable packets would divide by zero.","The syndrome support is smaller than the ambient space. Define a total recovery with Kraus operators\n\\[\nR_1=P,\\qquad R_2=\\ket0\\bra2,\n\\qquad R_1^*R_1+R_2^*R_2=I_3.\n\\]\nIt acts as the identity on operators supported on the code and sends the unused complement population into a chosen code state. The declared error map preserves the trace of code states, so recovery returns every code matrix unit, including both off-diagonal units, exactly. It need not preserve states on the unused complement.\n\\head{Exceptional cases and scope}\nThis construction assumes a nonzero code so that a complement replacement state exists. The case of a zero code is vacuous and has its own formal statement in the saved ledger. The retained exact run verifies all nine Gram compressions, the diagonalization, both null branches, total channel normalization, and all four code matrix u","\\vpage{33}{An approximate defect has a precise norm meaning}{THE OPTIMAL SCALAR FOR TWO SPECTRAL POINTS}{Traditional edition: Definition II.8.6, Proposition II.8.7.}\nOn a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,","\\vpage{33}{An approximate defect has a precise norm meaning}{THE OPTIMAL SCALAR FOR TWO SPECTRAL POINTS}{Traditional edition: Definition II.8.6, Proposition II.8.7.}\nOn a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bo","\\vpage{33}{An approximate defect has a precise norm meaning}{THE OPTIMAL SCALAR FOR TWO SPECTRAL POINTS}{Traditional edition: Definition II.8.6, Proposition II.8.7.}\nOn a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attai","ct has a precise norm meaning}{THE OPTIMAL SCALAR FOR TWO SPECTRAL POINTS}{Traditional edition: Definition II.8.6, Proposition II.8.7.}\nOn a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also ente…","inition II.8.6, Proposition II.8.7.}\nOn a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full","On a two-dimensional code, consider the overlap compression\n\\[\nA=PE_a^*E_bP=\\diag(11/10,9/10).\n\\]\nFor instance, within the code take $E_a=I$ and $E_b=A$. This declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL…","declares an error family, not by itself a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","self a normalized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","malized noise channel. The optimal scalar compression has\n\\[\n\\inf_{c\\in\\CC}\\norm{A-cI}=\\frac1{10},\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.",",\\qquad c=1\\ \\text{attains it}.\n\\]\nIndeed, for every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","or every complex $c$, the triangle inequality gives\n\\[\n\\frac15=\\left|\\frac{11}{10}-\\frac9{10}\\right|\n\\leq\\left|\\frac{11}{10}-c\\right|+\\left|\\frac9{10}-c\\right|.\n\\]\nThus the larger of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","er of the two distances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","istances is at least $1/10$, while both distances equal $1/10$ at $c=1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","1$.\n\\xfig{defect}\n\\head{What one pair contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","contributes to the family defect}\nFor unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","unit code vectors $\\psi,\\phi$, this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","this pair therefore satisfies\n\\[\n\\left|\\langle E_a\\psi,E_b\\phi\\rangle-\\langle\\psi,\\phi\\rangle\\right|\n\\leq\\frac1{10}.\n\\]\nThe bound is attained on either coordinate vector. For the \\emph{whole} declared family $\\{I,A\\}$, the $A^*A$ compression also enters the maximum: its diagonal entries are $121/100,81/100$, giving pair defect $1/5$. Hence the full family value is $\\varepsilon_{\\rm KL}=1/5$, not $1/10$.","This distinction prevents a tempting shortcut from one favorable overlap to an approximate correction claim. The quantity measures how close all overlap operators are to scalar action on the code. A recovery fidelity bound requires a declared normalized channel, a metric, and a separate continuity result.\n\\head{What is checked exactly}\nFor $\\varepsilon\\in\\{0,1/10,1/4,1\\}$ the script checks $(A-I)^*(A-I)=\\varepsilon^2I$ for $A=\\diag(1+\\varepsilon,1-\\varepsilon)$ and several rational scalar controls. The triangle argument proves the lower bound for every complex scalar. Checking a finite list of candidates would not establish that conclusion. The family maximum above is also recorded separately.\n\\takeaway{An overlap norm has a useful operational interpretation. Evaluate every error pair, then keep that statement separate from an unproved recovery fidelity estimate.}","tempting shortcut from one favorable overlap to an approximate correction claim. The quantity measures how close all overlap operators are to scalar action on the code. A recovery fidelity bound requires a declared normalized channel, a metric, and a separate continuity result.\n\\head{What is checked exactly}\nFor $\\varepsilon\\in\\{0,1/10,1/4,1\\}$ the script checks $(A-I)^*(A-I)=\\varepsilon^2I$ for $A=\\diag(1+\\varepsilon,1-\\varepsilon)$ and several rational scalar controls. The triangle argument proves the lower bound for every complex scalar. Checking a finite list of candidates would not establish that conclusion. The family maximum above is also recorded separately.\n\\takeaway{An overlap norm has a useful operational interpretation. Evaluate every error pair, then keep that statement separate from an unproved recovery fidelity estimate.}","\\vpage{34}{Control the hidden spectral response}{SIGN, STABILITY, AND A CONVERGENT EXPANSION}{Traditional edition: Theorem II.9.4, Proposition II.9.5, Theorem II.9.12.}\nRetain one coordinate in $H=\\mat{0&1\\\\1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\del","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\wi","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference","1&2}$, so $A=0$, $B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires unifo","$B=1$, and $D=2$. In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four t…","In its elimination domain, $\\Sigma(z)=1/(z-2)$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact re…",")$. At $z=3+\\ii$,\n\\[\n\\Sigma(3+\\ii)=\\frac{1-\\ii}{2},\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric…","\\qquad\n\\Im\\Sigma(z)=-(\\Im z)\\left|(z-2)^{-1}\\right|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained ch…","ht|^2\\leq0.\n\\]\nThe sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign id…","The sign in the upper half plane comes from self-adjoint hidden data. It is not an entrywise sign condition on an arbitrary effective matrix.\n\\xfig{stability}\n\\head{A stability estimate with a common spectral gap}\nCompare $(B,D)=(1,2)$ with $(\\widetilde B,\\widetilde D)=(9/8,17/8)$ at $z=3$. Both hidden spectral distances are at least $\\delta=3/4$, and both couplings are bounded by $g=9/8$. Then\n\\[\n|\\Sigma-\\widetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign identity, and the…","idetilde\\Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign identity, and the identities for the projected resolvent and bare truncation at four specified real spectral parameters.","Sigma|=\\left|1-\\frac{81}{56}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign identity, and the identities for the projected resolvent and bare truncation at four specified real spectral parameters.","}\\right|=\\frac{25}{56},\n\\qquad\n\\frac{2g}{\\delta}|B-\\widetilde B|\n+\\frac{g^2}{\\delta^2}|D-\\widetilde D|=\\frac{21}{32}.\n\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign identity, and the identities for the projected resolvent and bare truncation at four specified real spectral parameters.","\\]\nThe theorem bounds the difference using a common spectral gap. Applying the bound uniformly to a refinement family requires uniform spectral separation and control of the couplings.\n\\head{A series near zero energy and its exact remainder}\nAt $z=1/2$, $|z|\\,|D^{-1}|=1/4<1$, so\n\\[\n\\Sigma(z)=-\\sum_{n=0}^{\\infty}\\frac{z^n}{2^{n+1}}.\n\\]\nAfter the first four terms, the exact remainder has magnitude $1/384$. The geometric series controls the infinite tail. The retained check verifies that remainder, the complex sign identity, and the identities for the projected resolvent and bare truncation at four specified real spectral parameters.","At $z=2$ the hidden inverse fails. At an eigenvalue of the full Hamiltonian, the retained Feshbach inverse fails. Neither point belongs to the resolvent identity's domain. Analytic continuation of a scalar expression does not erase its theorem hypotheses.\n\\takeaway{A uniform effective response requires a uniform gap and coupling bound. Exact small examples show how those hypotheses enter each estimate.}","\\vpage{35}{Memory positivity is a Gram identity}{TWO HIDDEN FREQUENCIES AND A TEST OF LOCAL CLOSURE}{Traditional edition: Theorem II.9.9, Theorem II.9.11.}\nUse retained space $\\CC$ and hidden space $\\CC^2$, with $A=0$, $B=(3/5,4/5)$, and $D=\\diag(1,2)$. The full Hamiltonian is $\\mat{0&3/5&4/5","\\vpage{35}{Memory positivity is a Gram identity}{TWO HIDDEN FREQUENCIES AND A TEST OF LOCAL CLOSURE}{Traditional edition: Theorem II.9.9, Theorem II.9.11.}\nUse retained space $\\CC$ and hidden space $\\CC^2$, with $A=0$, $B=(3/5,4/5)$, and $D=\\diag(1,2)$. The full Hamiltonian is $\\mat{0&3/5&4/5\\\\3/5&1&0\\\\4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWit","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$.","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for","4/5&0&2}$. Its memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initia","ts memory kernel is\n\\[\nK(t)=\\frac9{25}e^{-\\ii t}+\\frac{16}{25}e^{-2\\ii t},\n\\qquad K(0)=1.\n\\]\nEach oscillatory coefficient is positive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, e…","ive although $K(t)$ need not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","ed not be real or pointwise nonnegative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","egative. At $t=\\pi/2$, it equals $-16/25-9\\ii/25$.\n\\xfig{memory_positive}\n\\head{Positive definite means a matrix test at arbitrary times}\nFor retained scalars $v_j$ and real times $t_j$, the structural identity is\n\\[\n\\sum_{j,k}\\overline v_jK(t_j-t_k)v_k\n=\\left\\|\\sum_k e^{\\ii t_kD}B^*v_k\\right\\|^2\\geq0.\n\\]\nThe exact worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","ct worked check specializes to $t=(0,\\pi/2,\\pi)$ and $v=(1,2,-1)$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.",")$. Its phases lie in $\\QQ(\\ii)$, so both sides are computed without evaluating trigonometric functions. This Gram test illustrates the factorization. The written identity proves positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","positivity for all choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","choices of times and vectors.\n\\head{Why this does not close to a bare local equation}\nWith hidden initial state zero, the full system has $\\ddot x(0)=-(A^2+BB^*)x(0)=-x(0)$. The bare retained equation $\\ii\\dot x=Ax$ would instead give zero acceleration. Therefore exact local closure for every retained initial state fails. More generally, equality of these accelerations forces $BB^*=0$ and hence $B=0$.","A nonzero initial hidden vector also supplies the forcing term $Be^{-\\ii tD}y(0)$. It is additional data, not part of the convolution kernel. The retained checks cover the displayed kernel values, exact Gram factorization, and acceleration defect. The integral equation follows from the written elimination argument.\n\\takeaway{Memory is positive definite through a Gram factorization, not through pointwise signs. Exact bare local closure requires the coupling to vanish.}","\\vpage{36}{A positive transfer determines its Hamiltonian}{STRICT POSITIVITY AND THE GROUND GAP}{Traditional edition: Theorem II.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to","\\vpage{36}{A positive transfer determines its Hamiltonian}{STRICT POSITIVITY AND THE GROUND GAP}{Traditional edition: Theorem II.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which","\\vpage{36}{A positive transfer determines its Hamiltonian}{STRICT POSITIVITY AND THE GROUND GAP}{Traditional edition: Theorem II.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","ransfer determines its Hamiltonian}{STRICT POSITIVITY AND THE GROUND GAP}{Traditional edition: Theorem II.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","ermines its Hamiltonian}{STRICT POSITIVITY AND THE GROUND GAP}{Traditional edition: Theorem II.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","I.10.2, Proposition II.10.5.}\nOn $\\HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","HH=\\CC^3$, take the strictly positive contraction\n\\[\nT=\\diag(1,1/2,1/4),\\qquad H=-\\log T=\\diag(0,\\log2,2\\log2).\n\\]\nAt unit Euclidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","lidean time, $T=e^{-H}$. The self-adjoint logarithm is unique because the logarithm acts on positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","positive spectral values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","al values. At integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","integer time $n\\geq0$,\n\\[\nT^n=\\diag(1,2^{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","{-n},4^{-n}),\\qquad\nP_0=\\diag(1,0,0),\\qquad\\norm{T^n-P_0}=2^{-n}.\n\\]\nThe gap $\\Delta=\\log2$ determines the rate $e^{-n\\Delta}$ exactly in this example.\n\\xfig{transfer_ground}\n\\head{Strict positivity is an essential finite hypothesis}\nFor $T'=\\diag(1,0)$, the operator is positive semidefinite, but there is no finite self-adjoint $H'$ on all of $\\CC^2$ with $T'=e^{-H'}$: every finite eigenvalue of $H'$ exponentiates to a strictly positive value. One may restrict to the positive support, but that changes the space on which the logarithm is defined.","For a family with nontrivial excited spectra, a uniform exponential rate needs a common positive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statem","For a family with nontrivial excited spectra, a uniform exponential rate needs a common positive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Gro","For a family with nontrivial excited spectra, a uniform exponential rate needs a common positive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Ground convergence needs a declared normalizatio","trivial excited spectra, a uniform exponential rate needs a common positive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Ground convergence needs a declared normalization and gap, while continuum reconstruction still needs its own structure.}","xponential rate needs a common positive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Ground convergence needs a declared normalization and gap, while continuum reconstruction still needs its own structure.}","tive lower bound on the gap. Ground multiplicity does not cause a problem: $P_0$ selects the entire eigenspace at the minimum energy. For a Hamiltonian whose minimum is $E_0\\ne0$, the normalized transfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Ground convergence needs a declared normalization and gap, while continuum reconstruction still needs its own structure.}","nsfer is $e^{-t(H-E_0I)}$ before taking the limit.\n\\head{Positivity is retained at several Euclidean times}\nFor arbitrary vectors $u_j$ and nonnegative integer times $n_j$, the reflected kernel obeys\n\\[\n\\sum_{j,k}\\langle u_j,T^{n_j+n_k}u_k\\rangle\n=\\left\\|\\sum_jT^{n_j}u_j\\right\\|^2.\n\\]\nThis is a finite Gram factorization. The retained run checks the integer semigroup and diagonal ground error for $n=0,\\ldots,12$, and rejects a logarithm on the full space for the singular control. The written spectral argument justifies the logarithms and extends the functional calculus statement to all times.\n\\takeaway{A positive definite transfer can be logarithmically reconstructed. Ground convergence needs a declared normalization and gap, while continuum reconstruction still needs its own structure.}","\\vpage{37}{The dimension window has an order of limits}{FINITE HEAT DATA AND THE QUANTIFIERS}{Traditional edition: Theorem II.11.5, Corollary II.11.8.}\nFor the unit circle grid with $N$ sites, the scaled eigenvalues are $\\lambda_N(k)=4N^2\\sin^2(\\pi k/N)$. Put $Z_N(t)=\\sum_ke^{-t\\lambda_N(k)}$ and $d_N(t)=2t\\sum_k\\lambda_N(k)e^{-t\\lambda_N(k)}/Z_N(t)$.","At the single time $t=\\log2/16$, two small grids admit exact rational heat weights:\n\\[\n\\begin{array}{c|c|c|c|c}\nN&\\text{spectrum}&Z_N&\\sum\\lambda e^{-t\\lambda}&d_N/\\log2","At the single time $t=\\log2/16$, two small grids admit exact rational heat weights:\n\\[\n\\begin{array}{c|c|c|c|c}\nN&\\text{spectrum}&Z_N&\\sum\\lambda e^{-t\\lambda}&d_N/\\log2\\\\\\hline\n2&(0,16)&3/2&8&2/3\\\\\n4&(0,32,32,64)&25/16&20&8/5\n\\end{array}\n\\]\nThese are unnormalized traces. Normalizing by $N$ changes $Z_N$ but leaves the logarithmic derivative at that fixed grid unchanged.\n\\xfig{heat_limits}\n\\head{The theorem uses two separate choices}\nFor each $\\varepsilon>0$, first choose a fixed positive time $t_\\varepsilon$ so that the continuum torus dimension differs from one by less than $\\varepsilon/2$. Then choose $N_\\varepsilon$ so that every $N\\geq N_\\varepsilon$ differs from that continuum value by less than $\\varepsilon/2$ at the \\emph{same time}. This proves one controlled intermediate time for all sufficiently fine grids.","4&(0,32,32,64)&25/16&20&8/5\n\\end{array}\n\\]\nThese are unnormalized traces. Normalizing by $N$ changes $Z_N$ but leaves the logarithmic derivative at that fixed grid unchanged.\n\\xfig{heat_limits}\n\\head{The theorem uses two separate choices}\nFor each $\\varepsilon>0$, first choose a fixed positive time $t_\\varepsilon$ so that the continuum torus dimension differs from one by less than $\\varepsilon/2$. Then choose $N_\\varepsilon$ so that every $N\\geq N_\\varepsilon$ differs from that continuum value by less than $\\varepsilon/2$ at the \\emph{same time}. This proves one controlled intermediate time for all sufficiently fine grids.","&25/16&20&8/5\n\\end{array}\n\\]\nThese are unnormalized traces. Normalizing by $N$ changes $Z_N$ but leaves the logarithmic derivative at that fixed grid unchanged.\n\\xfig{heat_limits}\n\\head{The theorem uses two separate choices}\nFor each $\\varepsilon>0$, first choose a fixed positive time $t_\\varepsilon$ so that the continuum torus dimension differs from one by less than $\\varepsilon/2$. Then choose $N_\\varepsilon$ so that every $N\\geq N_\\varepsilon$ differs from that continuum value by less than $\\varepsilon/2$ at the \\emph{same time}. This proves one controlled intermediate time for all sufficiently fine grids.","hese are unnormalized traces. Normalizing by $N$ changes $Z_N$ but leaves the logarithmic derivative at that fixed grid unchanged.\n\\xfig{heat_limits}\n\\head{The theorem uses two separate choices}\nFor each $\\varepsilon>0$, first choose a fixed positive time $t_\\varepsilon$ so that the continuum torus dimension differs from one by less than $\\varepsilon/2$. Then choose $N_\\varepsilon$ so that every $N\\geq N_\\varepsilon$ differs from that continuum value by less than $\\varepsilon/2$ at the \\emph{same time}. This proves one controlled intermediate time for all sufficiently fine grids.","aves the logarithmic derivative at that fixed grid unchanged.\n\\xfig{heat_limits}\n\\head{The theorem uses two separate choices}\nFor each $\\varepsilon>0$, first choose a fixed positive time $t_\\varepsilon$ so that the continuum torus dimension differs from one by less than $\\varepsilon/2$. Then choose $N_\\varepsilon$ so that every $N\\geq N_\\varepsilon$ differs from that continuum value by less than $\\varepsilon/2$ at the \\emph{same time}. This proves one controlled intermediate time for all sufficiently fine grids.","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle wit","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window res","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves con","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fix","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time.","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time. The exact table is a worked calculation","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time. The exact table is a worked calculation, not its proof.","It does not assert a uniform interval width from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time. The exact table is a worked calculation, not its proof.}","from that argument, a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time. The exact table is a worked calculation, not its proof.}","a moving time $t_N$ with no extra estimates, or a dimension plateau at every scale. At fixed finite $N$, $d_N(t)\\to0$ as $t\\downarrow0$ and as $t\\to\\infty$. By contrast, first taking $N\\to\\infty$ at positive $t$ and then taking $t\\downarrow0$ yields the continuum dimension.\n\\head{Where the infinite tail is controlled}\nAt a fixed time, the proof bounds centered modes below by $16k^2$ and above by $4\\pi^2k^2$. The summable Gaussian dominators control both the heat trace and its first moment. A finite grid table supplies neither those tails nor the quantifiers. The retained check verifies the two displayed spectra, rational weights, traces, and dimension coefficients. It also distinguishes the unscaled cycle with four sites from the scaled unit circle grid.\n\\takeaway{The strengthened window result proves convergence after fixing a suitable time. The exact table is a worked calculation, not its proof.}","\\vpage{38}{Transport constraints without hiding closure}{THE IMAGE OF AN ISOMETRY IS NOT THE WHOLE LEVEL}{Traditional edition: Theorem II.12.2, Theorem II.12.5, Proposition II.12.6.}\nOn $\\CC^3=\\CC\\ket0\\oplus\\CC^2$, use $C_a=0\\oplus\\sigma_a$ for $a=x,y,z$. The common kernel is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structu","\\vpage{38}{Transport constraints without hiding closure}{THE IMAGE OF AN ISOMETRY IS NOT THE WHOLE LEVEL}{Traditional edition: Theorem II.12.2, Theorem II.12.5, Proposition II.12.6.}\nOn $\\CC^3=\\CC\\ket0\\oplus\\CC^2$, use $C_a=0\\oplus\\sigma_a$ for $a=x,y,z$. The common kernel is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on","\\vpage{38}{Transport constraints without hiding closure}{THE IMAGE OF AN ISOMETRY IS NOT THE WHOLE LEVEL}{Traditional edition: Theorem II.12.2, Theorem II.12.5, Proposition II.12.6.}\nOn $\\CC^3=\\CC\\ket0\\oplus\\CC^2$, use $C_a=0\\oplus\\sigma_a$ for $a=x,y,z$. The common kernel is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fin","\\vpage{38}{Transport constraints without hiding closure}{THE IMAGE OF AN ISOMETRY IS NOT THE WHOLE LEVEL}{Traditional edition: Theorem II.12.2, Theorem II.12.5, Proposition II.12.6.}\nOn $\\CC^3=\\CC\\ket0\\oplus\\CC^2$, use $C_a=0\\oplus\\sigma_a$ for $a=x,y,z$. The common kernel is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate","\\vpage{38}{Transport constraints without hiding closure}{THE IMAGE OF AN ISOMETRY IS NOT THE WHOLE LEVEL}{Traditional edition: Theorem II.12.2, Theorem II.12.5, Proposition II.12.6.}\nOn $\\CC^3=\\CC\\ket0\\oplus\\CC^2$, use $C_a=0\\oplus\\sigma_a$ for $a=x,y,z$. The common kernel is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $","l is $\\CC\\ket0$, and the full algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same …","ull algebra closes:\n\\[\n[C_x,C_y]=2\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result f…","\\ii C_z,\\qquad[C_y,C_z]=2\\ii C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls …","C_x,\\qquad[C_z,C_x]=2\\ii C_y.\n\\]\nThese are uncompressed identities. They supply an actual closure test in the quotient by the constraint span, unlike the automatic compressed zero on the physical kernel.\n\\xfig{constraint_image}\n\\head{Exact transport holds on the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutato…","the coarse image}\nEmbed by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","by the coordinate isometry $R:\\CC^3\\to\\CC^4$ and set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","d set $\\widetilde C_a=C_a\\oplus1$. Then $\\widetilde C_aR=RC_a$, so\n\\[\n[\\widetilde C_a,\\widetilde C_b]R=R[C_a,C_b].\n\\]\nThe coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","e coarse physical vector maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","ctor maps into the fine common kernel. However, the same prescribed Pauli structure does \\emph{not} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","t} close on the whole fine level: on the added coordinate the commutator is zero, whereas $2\\ii\\widetilde C_z$ is $2\\ii$. Exact transport constrains the image of $R$ and leaves its orthogonal complement as separate data.\n\\head{An approximate residual is also explicit}\nPerturb only $\\widetilde C_x$ by $(1/10)\\ket0\\bra0$. For the unit coarse physical vector, the fine constraint residual is exactly $1/10$. The intertwining defect has the same norm. The general result for approximate transport controls these residuals and commutator errors using the defect sizes and a common bound on the constraint norms. A nonzero anomaly remains an error that must be controlled.","\\vpage{39}{Two routes to a bounded Hamiltonian limit}{EXACT EMBEDDINGS AND A SUMMABLE TAIL}{Traditional edition: Theorem II.13.2, Theorem II.13.4, Corollary II.13.5.}\nFor the exact route, let $\\HH_n=\\CC^n$ with coordinate inclusions and\n\\[\nH_n=\\diag(0,1,1/2,\\ldots,1/(n-1)),\\qquad H_1=(0).\n\\]\nThe old energies are retained at every level, $H_{n+1}R_n=R_nH_n$, and $\\norm{H_n}\\leq1$. On the Hilbert space of the direct limit $\\ell^2(\\NN_0)$, the diagonal sequence defines a bounded self-adjoint $H_\\infty$. Projectors onto even addresses also intertwine and give a fixed complete system of two packets in the limit.\n\\xfig{tower_tail}\n\\head{Compatibility and boundedness are different assumptions}\nThe equally compatible tower $\\diag(0,1,2,\\ldots,n-1)$ has growing norms. It does not satisfy the bounded direct limit theorem. Its natural diagonal operator needs a separately specified domain. Dropping the","\\vpage{39}{Two routes to a bounded Hamiltonian limit}{EXACT EMBEDDINGS AND A SUMMABLE TAIL}{Traditional edition: Theorem II.13.2, Theorem II.13.4, Corollary II.13.5.}\nFor the exact route, let $\\HH_n=\\CC^n$ with coordinate inclusions and\n\\[\nH_n=\\diag(0,1,1/2,\\ldots,1/(n-1)),\\qquad H_1=(0).\n\\]\nThe old energies are retained at every level, $H_{n+1}R_n=R_nH_n$, and $\\norm{H_n}\\leq1$. On the Hilbert space of the direct limit $\\ell^2(\\NN_0)$, the diagonal sequence defines a bounded self-adjoint $H_\\infty$. Projectors onto even addresses also intertwine and give a fixed complete system of two packets in the limit.\n\\xfig{tower_tail}\n\\head{Compatibility and boundedness are different assumptions}\nThe equally compatible tower $\\diag(0,1,2,\\ldots,n-1)$ has growing norms. It does not satisfy the bounded direct limit theorem. Its natural diagonal operator needs a separately specified domain. Dropping the uniform norm hypothesis does not supply that domain argument.","a bounded Hamiltonian limit}{EXACT EMBEDDINGS AND A SUMMABLE TAIL}{Traditional edition: Theorem II.13.2, Theorem II.13.4, Corollary II.13.5.}\nFor the exact route, let $\\HH_n=\\CC^n$ with coordinate inclusions and\n\\[\nH_n=\\diag(0,1,1/2,\\ldots,1/(n-1)),\\qquad H_1=(0).\n\\]\nThe old energies are retained at every level, $H_{n+1}R_n=R_nH_n$, and $\\norm{H_n}\\leq1$. On the Hilbert space of the direct limit $\\ell^2(\\NN_0)$, the diagonal sequence defines a bounded self-adjoint $H_\\infty$. Projectors onto even addresses also intertwine and give a fixed complete system of two packets in the limit.\n\\xfig{tower_tail}\n\\head{Compatibility and boundedness are different assumptions}\nThe equally compatible tower $\\diag(0,1,2,\\ldots,n-1)$ has growing norms. It does not satisfy the bounded direct limit theorem. Its natural diagonal operator needs a separately specified domain. Dropping the uniform norm hypothesis does not supply that domain argument.","ditional edition: Theorem II.13.2, Theorem II.13.4, Corollary II.13.5.}\nFor the exact route, let $\\HH_n=\\CC^n$ with coordinate inclusions and\n\\[\nH_n=\\diag(0,1,1/2,\\ldots,1/(n-1)),\\qquad H_1=(0).\n\\]\nThe old energies are retained at every level, $H_{n+1}R_n=R_nH_n$, and $\\norm{H_n}\\leq1$. On the Hilbert space of the direct limit $\\ell^2(\\NN_0)$, the diagonal sequence defines a bounded self-adjoint $H_\\infty$. Projectors onto even addresses also intertwine and give a fixed complete system of two packets in the limit.\n\\xfig{tower_tail}\n\\head{Compatibility and boundedness are different assumptions}\nThe equally compatible tower $\\diag(0,1,2,\\ldots,n-1)$ has growing norms. It does not satisfy the bounded direct limit theorem. Its natural diagonal operator needs a separately specified domain. Dropping the uniform norm hypothesis does not supply that domain argument.","every level, $H_{n+1}R_n=R_nH_n$, and $\\norm{H_n}\\leq1$. On the Hilbert space of the direct limit $\\ell^2(\\NN_0)$, the diagonal sequence defines a bounded self-adjoint $H_\\infty$. Projectors onto even addresses also intertwine and give a fixed complete system of two packets in the limit.\n\\xfig{tower_tail}\n\\head{Compatibility and boundedness are different assumptions}\nThe equally compatible tower $\\diag(0,1,2,\\ldots,n-1)$ has growing norms. It does not satisfy the bounded direct limit theorem. Its natural diagonal operator needs a separately specified domain. Dropping the uniform norm hypothesis does not supply that domain argument.","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed g","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A f","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspeci","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tow","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tower.\n\\takeaway{Exact embedded compatibility uses a uniform","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tower.\n\\takeaway{Exact embedded compatibility uses a uniform norm bound. Appr","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tower.\n\\takeaway{Exact embedded compatibility uses a uniform norm bound. Approximate convergence uses a common space and summable defects. Keep the hypotheses of th","For the norm route, work instead on the \\emph{single fixed space} $\\CC^2$ and set\n\\[\nH_n=\\diag(0,1+2^{-n}),\\qquad H=\\diag(0,1).\n\\]\nHere $\\norm{H_{n+1}-H_n}=2^{-(n+1)}$ and $\\norm{H_n-H}=2^{-n}$. The summable tail supplies an error bound for every $n$. For $|t|\\leq T$, the paper's Duhamel estimate gives\n\\[\n\\norm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tower.\n\\takeaway{Exact embedded compatibility uses a uniform norm bound. Approximate convergence uses a common space and summable defects. Keep the hypotheses of these two routes explicit.}","rm{e^{-\\ii tH_n}-e^{-\\ii tH}}\\leq T2^{-n}.\n\\]\nThe Euclidean counterpart holds on $0\\leq t\\leq T$ because both Hamiltonians are nonnegative. If packet projectors also converge in norm, leakage operators converge uniformly on these time intervals.\n\\head{Execution scope}\nExact inclusions and projector compatibility are checked through level eight. The two-dimensional tail identities are checked for $n=0,\\ldots,12$. Their infinite conclusion uses the displayed geometric series and the theorem's analytic estimate. A finite prefix cannot certify an unspecified continuation of a tower.\n\\takeaway{Exact embedded compatibility uses a uniform norm bound. Approximate convergence uses a common space and summable defects. Keep the hypotheses of these two routes explicit.}","\\vpage{40}{Choose the topology that preserves the quantity}{REFLECTION GRAMS, TRACES, AND FINAL SCOPE}{Traditional edition: Theorem II.13.6, Theorem II.13.8, Section II.14.}\nLet $q_n=1/2+2^{-(n+2)}$ and define the scalar reflected kernel $C_n(j,k)=1+q_n^{j+k}$ on nonnegative integers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their ze","\\vpage{40}{Choose the topology that preserves the quantity}{REFLECTION GRAMS, TRACES, AND FINAL SCOPE}{Traditional edition: Theorem II.13.6, Theorem II.13.8, Section II.14.}\nLet $q_n=1/2+2^{-(n+2)}$ and define the scalar reflected kernel $C_n(j,k)=1+q_n^{j+k}$ on nonnegative integers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","topology that preserves the quantity}{REFLECTION GRAMS, TRACES, AND FINAL SCOPE}{Traditional edition: Theorem II.13.6, Theorem II.13.8, Section II.14.}\nLet $q_n=1/2+2^{-(n+2)}$ and define the scalar reflected kernel $C_n(j,k)=1+q_n^{j+k}$ on nonnegative integers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","e quantity}{REFLECTION GRAMS, TRACES, AND FINAL SCOPE}{Traditional edition: Theorem II.13.6, Theorem II.13.8, Section II.14.}\nLet $q_n=1/2+2^{-(n+2)}$ and define the scalar reflected kernel $C_n(j,k)=1+q_n^{j+k}$ on nonnegative integers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","EFLECTION GRAMS, TRACES, AND FINAL SCOPE}{Traditional edition: Theorem II.13.6, Theorem II.13.8, Section II.14.}\nLet $q_n=1/2+2^{-(n+2)}$ and define the scalar reflected kernel $C_n(j,k)=1+q_n^{j+k}$ on nonnegative integers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","egers. Each finite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","nite Gram matrix is a sum of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","um of two positive rank-one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","one matrices, and pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","pointwise convergence gives $C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","$C(j,k)=1+(1/2)^{j+k}$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","$. For times $(0,1,2)$ and coefficients $(1,-2,1)$,\n\\[\n\\sum_{j,k}v_jC_n(j,k)v_k=(1-q_n)^4\\geq0.\n\\]\nThis exact check illustrates how finite quadratic forms pass positivity to the limit.\n\\xfig{trace_topology}\n\\head{Operator norm does not preserve trace in growing rank}\nOn $\\ell^2$, let $P_n$ select the first $n$ coordinates and set $A_n=P_n/n$. Then $\\norm{A_n}=1/n\\to0$ while $\\Tr A_n=1$ for every $n$. These positive trace-class operators show why the choice of topology matters. Their zero limit is not the heat operator of a finite Hamiltonian.","For transfers on two states $T_n=\\diag(1,q_n)=e^{-H_n}$, with $H_n=\\diag(0,-\\log q_n)$ at unit time, the error in trace norm is $2^{-(n+2)}$ and the heat traces converge to $3/2$. The criterion for convergence of spectral dimension additionally assumes convergence of the first heat moments in trace norm and a nonzero limiting heat trace. Convergence of the heat trace alone does not determine that quotient.\n\\head{The retained evidence and the remaining open targets}\nThe extension includes retained exact checks over specified finite ranges in $\\QQ(\\ii)$ using \\texttt{Fraction} from Python's standard library. The original appendix's record of 180 checks, including symbolic derivatives, is also retained. Both records come from earlier executions. T","For transfers on two states $T_n=\\diag(1,q_n)=e^{-H_n}$, with $H_n=\\diag(0,-\\log q_n)$ at unit time, the error in trace norm is $2^{-(n+2)}$ and the heat traces converge to $3/2$. The criterion for convergence of spectral dimension additionally assumes convergence of the first heat moments in trace norm and a nonzero limiting heat trace. Convergence of the heat trace alone does not determine that quotient.\n\\head{The retained evidence and the remaining open targets}\nThe extension includes retained exact checks over specified finite ranges in $\\QQ(\\ii)$ using \\texttt{Fraction} from Python's standard library. The original appendix's record of 180 checks, including symbolic derivatives, is also retained. Both records come from earlier executions. This prose revision adds no symb","For transfers on two states $T_n=\\diag(1,q_n)=e^{-H_n}$, with $H_n=\\diag(0,-\\log q_n)$ at unit time, the error in trace norm is $2^{-(n+2)}$ and the heat traces converge to $3/2$. The criterion for convergence of spectral dimension additionally assumes convergence of the first heat moments in trace norm and a nonzero limiting heat trace. Convergence of the heat trace alone does not determine that quotient.\n\\head{The retained evidence and the remaining open targets}\nThe extension includes retained exact checks over specified finite ranges in $\\QQ(\\ii)$ using \\texttt{Fraction} from Python's standard library. The original appendix's record of 180 checks, including symbolic derivatives, is also retained. Both records come from earlier executions. This prose revision adds no symbolic or Lean kernel run. All model and vector sources","For transfers on two states $T_n=\\diag(1,q_n)=e^{-H_n}$, with $H_n=\\diag(0,-\\log q_n)$ at unit time, the error in trace norm is $2^{-(n+2)}$ and the heat traces converge to $3/2$. The criterion for convergence of spectral dimension additionally assumes convergence of the first heat moments in trace norm and a nonzero limiting heat trace. Convergence of the heat trace alone does not determine that quotient.\n\\head{The retained evidence and the remaining open targets}\nThe extension includes retained exact checks over specified finite ranges in $\\QQ(\\ii)$ using \\texttt{Fraction} from Python's standard library. The original appendix's record of 180 checks, including symbolic derivatives, is also retained. Both records come from earlier executions. This prose revision adds no symbolic or Lean kernel run. All model and vector sources remain supplied.","heat traces converge to $3/2$. The criterion for convergence of spectral dimension additionally assumes convergence of the first heat moments in trace norm and a nonzero limiting heat trace. Convergence of the heat trace alone does not determine that quotient.\n\\head{The retained evidence and the remaining open targets}\nThe extension includes retained exact checks over specified finite ranges in $\\QQ(\\ii)$ using \\texttt{Fraction} from Python's standard library. The original appendix's record of 180 checks, including symbolic derivatives, is also retained. Both records come from earlier executions. This prose revision adds no symbolic or Lean kernel run. All model and vector sources remain supplied.","\\section*{Three exact finite models}\\addcontentsline{toc}{section}{Three exact finite models}\n\\subsection*{A seed remembers multiplicity support}\nFor $M_2(\\mathbb C)\\otimes I_3$, a coefficient matrix with rank one generates an orbit of dimension two. A matrix of rank two generates an orbit of dimension four. Neither single seed spans the full space of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not b","\\section*{Three exact finite models}\\addcontentsline{toc}{section}{Three exact finite models}\n\\subsection*{A seed remembers multiplicity support}\nFor $M_2(\\mathbb C)\\otimes I_3$, a coefficient matrix with rank one generates an orbit of dimension two. A matrix of rank two generates an orbit of dimension four. Neither single seed spans the full space of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1","thbb C)\\otimes I_3$, a coefficient matrix with rank one generates an orbit of dimension two. A matrix of rank two generates an orbit of dimension four. Neither single seed spans the full space of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1","imes I_3$, a coefficient matrix with rank one generates an orbit of dimension two. A matrix of rank two generates an orbit of dimension four. Neither single seed spans the full space of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1","l space of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1","of dimension six, because a $2\\times3$ matrix cannot have rank three. Two seeds can nevertheless cover all three multiplicity coordinates together. The example illustrates the finite seed rank formula with $d=2$ and $m=3$.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1","gures/exact_seed_support.pdf}\n\\caption{The figure displays exact coefficient matrices for $d=2$, $m=3$. The combined row support of two seeds has rank three. The bar heights record algebraic orbit dimensions and should not be read as probabilities of unitary state preparation.}\\label{monograph:fig:seed_support}\n\\end{figure}\n\\clearpage\n\\subsection*{An eliminated coordinate still contributes}\nTake $H=\\left(\\begin{smallmatrix}2&1\\\\1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compress","1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compression.pdf}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the e","1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compression.pdf}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the eliminated block","1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compression.pdf}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the eliminated block and the full operator","1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compression.pdf}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the eliminated block and the full operator are invertible, and the plotted values","1&3\\end{smallmatrix}\\right)$ and retain its first coordinate. The eliminated block is invertible and the Schur complement of $H$ is $2-1/3=5/3$. Its inverse is $3/5$, exactly the retained entry of $H^{-1}$. Inverting the compressed entry alone gives $1/2$. The difference records the contribution of a path through the eliminated coordinate. Direct compression omits that contribution, while block inversion retains it.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_feshbach_compression.pdf}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the eliminated block and the full operator are invertible, and the plotted values are exact fractions. The example shows why direct","df}\n\\caption{This rational model uses two coordinates to illustrate block inversion. Both the eliminated block and the full operator are invertible, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograp","model uses two coordinates to illustrate block inversion. Both the eliminated block and the full operator are invertible, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_c","ordinates to illustrate block inversion. Both the eliminated block and the full operator are invertible, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The fig","d block and the full operator are invertible, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported cod","ull operator are invertible, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The norm","e, and the plotted values are exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one","exact fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the u","fractions. The example shows why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormal","why direct compression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormalized positive map has output trace fo","mpression and resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormalized positive map has output trace four. T","resolvent elimination give different results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormalized positive map has output trace four. The illustration records this fi","ferent results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormalized positive map has output trace four. The illustration records this finite scaling","results.}\\label{monograph:fig:feshbach_compression}\n\\end{figure}\n\\clearpage\n\\subsection*{Normalization changes the map}\nLet $P=|0\\rangle\\langle0|$ on $\\mathbb C^2$ and $Z=\\operatorname{diag}(1,-1)$. The normalized Kraus family $\\sqrt{3/4}\\,I$, $\\sqrt{1/4}\\,Z$ satisfies $\\sum_j K_j^*K_j=I$ and maps $P$ to $P$. The unnormalized family $\\sqrt3\\,I$, $Z$ satisfies $\\sum_j E_j^*E_j=4I$ and maps $P$ to $4P$. The two families have the same relative weights, but their total normalization differs. Both outputs are proportional to the same projector, but their traces differ.\n\\begin{figure}[H]\\centering\n\\includegraphics[width=.94\\textwidth]{monograph_figures/exact_channel_scale.pdf}\n\\caption{The figure shows exact weights for one supported code state. The normalized map preserves trace one, while the unnormalized positive map has output trace four. The illustration records this finite scaling distinct","\\paragraph{Conventions.}\nUnless the zero space is explicitly included as a boundary case, quantum state spaces are nonzero complex Hilbert spaces of finite dimension. A complete packet family is a finite collection satisfying\n\\[\nP_\\alpha^*=P_\\alpha,\\qquad\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","re nonzero complex Hilbert spaces of finite dimension. A complete packet family is a finite collection satisfying\n\\[\nP_\\alpha^*=P_\\alpha,\\qquad\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","of finite dimension. A complete packet family is a finite collection satisfying\n\\[\nP_\\alpha^*=P_\\alpha,\\qquad\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","ction satisfying\n\\[\nP_\\alpha^*=P_\\alpha,\\qquad\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","ha,\\qquad\nP_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","P_\\alpha P_\\beta=\\delta_{\\alpha\\beta}P_\\alpha,\\qquad\n\\sum_\\alpha P_\\alpha=I.\n\\]\nAll operator norms in the analytic estimates are induced by the norm on the Hilbert space. The estimates do not use the entrywise maximum norm or the Frobenius norm. We write $[A,B]=AB-BA$ and use units with $\\hbar=1$. Parameter derivatives are real derivatives. A matrix $X$ is supported on a code projector $P$ when $PXP=X$.","The common operator sequence of the series becomes\n\\[\n\\boxed{\n\\begin{gathered}\n\\text{quantum seed}\n\\longrightarrow\n\\text{gate or observable address}\n\\longrightarrow\n\\text{symmetry projection}\\\\\n\\longrightarrow\n\\text{packet transfer}\n\\longrightarrow\n\\text{constraint}\n\\longrightarrow\n\\text{refinement}\n\\longrightarrow\n\\text{reconstruction}\n\\end{gathered}\n}\n\\]","\\begin{definition}[Finite quantum seed system]\\label{core:definition:001}\nA finite quantum seed system is a tuple\n\\[\n\\mathfrak Q=\n\\left(\n\\cH,\n\\psi_0,\n\\cO,\n\\cR,\n\\{P_{\\ell,\\alpha}\\},\nH,\n\\tau\n\\right),\n\\]\nwhere $\\cH$ is a finite-dimensional Hilbert space, $\\psi_0\\in\\cH$ is a nonzero seed, $\\cO$ is the declared operator grammar, $\\cR$ is a collection of algebraic or physical constraints, $\\{P_{\\ell,\\alpha}\\}$ is an exact packet system at each refinement level, $H=H^*$ is a Hamiltonian when dynamics are present, and $\\tau$ specifies the topology used in a later completion.\n\\end{definition}","inition:001}\nA finite quantum seed system is a tuple\n\\[\n\\mathfrak Q=\n\\left(\n\\cH,\n\\psi_0,\n\\cO,\n\\cR,\n\\{P_{\\ell,\\alpha}\\},\nH,\n\\tau\n\\right),\n\\]\nwhere $\\cH$ is a finite-dimensional Hilbert space, $\\psi_0\\in\\cH$ is a nonzero seed, $\\cO$ is the declared operator grammar, $\\cR$ is a collection of algebraic or physical constraints, $\\{P_{\\ell,\\alpha}\\}$ is an exact packet system at each refinement level, $H=H^*$ is a Hamiltonian when dynamics are present, and $\\tau$ specifies the topology used in a later completion.\n\\end{definition}","\\mathfrak Q=\n\\left(\n\\cH,\n\\psi_0,\n\\cO,\n\\cR,\n\\{P_{\\ell,\\alpha}\\},\nH,\n\\tau\n\\right),\n\\]\nwhere $\\cH$ is a finite-dimensional Hilbert space, $\\psi_0\\in\\cH$ is a nonzero seed, $\\cO$ is the declared operator grammar, $\\cR$ is a collection of algebraic or physical constraints, $\\{P_{\\ell,\\alpha}\\}$ is an exact packet system at each refinement level, $H=H^*$ is a Hamiltonian when dynamics are present, and $\\tau$ specifies the topology used in a later completion.\n\\end{definition}","The generated state space is\n\\[\n\\cH(\\psi_0;\\cO)\n=\n\\operatorname{span}\n\\left\\{\nO_r\\cdots O_1\\psi_0:\nO_j\\in\\cO,\n\\ r\\ge0\n\\right\\}.\n\\]\nThe quantum seed is cyclic for the grammar when\n\\[\n\\cH(\\psi_0;\\cO)=\\cH.\n\\]\nThis is a statement about the pair $(\\psi_0,\\cO)$, not about the vector alone. A state may be cyclic for the full matrix algebra and noncyclic for a restricted physical gate set.","Let a finite-dimensional unital $*$-subalgebra of operators act in its standard multiplicity form\n\\[\n\\cA\n\\cong\n\\bigoplus_{i=1}^r\n\\left(M_{d_i}(\\bbC)\\otimes I_{m_i}\\right)\n\\]\non\n\\[\n\\cH\n\\cong\n\\bigoplus_{i=1}^r\n\\left(\\bbC^{d_i}\\otimes\\bbC^{m_i}\\right).\n\\]\nHere every $d_i$ is a positive integer and every multiplicity $m_i$ is a nonnegative integer. An empty block family represents the zero space and has seed rank zero. The maximum over an empty block family is also assigned the value zero. A vector $\\psi_i\\in\\bbC^{d_i}\\otimes\\bbC^{m_i}$ is identified with a $d_i\\times m_i$ coefficient matrix $C_i$.","\\cA\n\\cong\n\\bigoplus_{i=1}^r\n\\left(M_{d_i}(\\bbC)\\otimes I_{m_i}\\right)\n\\]\non\n\\[\n\\cH\n\\cong\n\\bigoplus_{i=1}^r\n\\left(\\bbC^{d_i}\\otimes\\bbC^{m_i}\\right).\n\\]\nHere every $d_i$ is a positive integer and every multiplicity $m_i$ is a nonnegative integer. An empty block family represents the zero space and has seed rank zero. The maximum over an empty block family is also assigned the value zero. A vector $\\psi_i\\in\\bbC^{d_i}\\otimes\\bbC^{m_i}$ is identified with a $d_i\\times m_i$ coefficient matrix $C_i$.","i}\\right)\n\\]\non\n\\[\n\\cH\n\\cong\n\\bigoplus_{i=1}^r\n\\left(\\bbC^{d_i}\\otimes\\bbC^{m_i}\\right).\n\\]\nHere every $d_i$ is a positive integer and every multiplicity $m_i$ is a nonnegative integer. An empty block family represents the zero space and has seed rank zero. The maximum over an empty block family is also assigned the value zero. A vector $\\psi_i\\in\\bbC^{d_i}\\otimes\\bbC^{m_i}$ is identified with a $d_i\\times m_i$ coefficient matrix $C_i$.","lus_{i=1}^r\n\\left(\\bbC^{d_i}\\otimes\\bbC^{m_i}\\right).\n\\]\nHere every $d_i$ is a positive integer and every multiplicity $m_i$ is a nonnegative integer. An empty block family represents the zero space and has seed rank zero. The maximum over an empty block family is also assigned the value zero. A vector $\\psi_i\\in\\bbC^{d_i}\\otimes\\bbC^{m_i}$ is identified with a $d_i\\times m_i$ coefficient matrix $C_i$.","\\begin{theorem}[Orbit dimension of one quantum seed]\\label{core:theorem:001}\\label{core:thm:seed-orbit}\nFor integers $d\\ge1$ and $m\\ge0$, consider the action of $M_d(\\bbC)\\otimes I_m$ on $\\bbC^d\\otimes\\bbC^m$. Let $C$ be the coefficient matrix of a vector $\\psi$. Then\n\\[\n\\boxed{\n\\dim\\operatorname{span}\\{(A\\otimes I_m)\\psi:A\\in M_d(\\bbC)\\}\n=d\\,\\rank C.\n}\n\\]\nThe seed is cyclic if and only if $\\rank C=m$, which is possible only when $m\\le d$.\n\\end{theorem}","one quantum seed]\\label{core:theorem:001}\\label{core:thm:seed-orbit}\nFor integers $d\\ge1$ and $m\\ge0$, consider the action of $M_d(\\bbC)\\otimes I_m$ on $\\bbC^d\\otimes\\bbC^m$. Let $C$ be the coefficient matrix of a vector $\\psi$. Then\n\\[\n\\boxed{\n\\dim\\operatorname{span}\\{(A\\otimes I_m)\\psi:A\\in M_d(\\bbC)\\}\n=d\\,\\rank C.\n}\n\\]\nThe seed is cyclic if and only if $\\rank C=m$, which is possible only when $m\\le d$.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:001}\nUnder the identification by coefficient matrices, $(A\\otimes I_m)\\psi$ corresponds to $AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","f}\\phantomsection\\label{core:proof:001}\nUnder the identification by coefficient matrices, $(A\\otimes I_m)\\psi$ corresponds to $AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","1}\nUnder the identification by coefficient matrices, $(A\\otimes I_m)\\psi$ corresponds to $AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","ficient matrices, $(A\\otimes I_m)\\psi$ corresponds to $AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","times I_m)\\psi$ corresponds to $AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","$AC$. Every row of $AC$ lies in the row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","row space of $C$. Conversely, any $d\\times m$ matrix whose rows lie in the row space of $C$ equals $AC$ for a suitable $A$: choose each row of $A$ to produce the desired row combination. The orbit therefore consists of all matrices with $d$ independently chosen rows in the $\\rank C$-dimensional row space of $C$, and has dimension $d\\rank C$. It equals the full $dm$-dimensional space exactly when $\\rank C=m$.\n\\phantomsection\\label{core:proof:001:end}\\end{proof}","\\begin{theorem}[Quantum seed-rank formula]\\label{core:theorem:002}\\label{core:thm:seed-rank}\nFor\n\\[\n\\cA=\\bigoplus_i(M_{d_i}(\\bbC)\\otimes I_{m_i})\n\\quad\\text{on}\\quad\n\\cH=\\bigoplus_i(\\bbC^{d_i}\\otimes\\bbC^{m_i}),\n\\]\nthe minimum number of seeds required to generate $\\cH$ is\n\\[\n\\boxed{\n\\operatorname{srank}_{\\cA}(\\cH)\n=\n\\max_i\\left\\lceil\\frac{m_i}{d_i}\\right\\rceil.\n}\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:002}\nEach seed contributes a coefficient matrix of rank at most $d_i$ in the $i$th block. Thus $k$ seeds can cover at most a $kd_i$-dimensional subspace of the multiplicity factor $\\bbC^{m_i}$, so $k\\ge\\lceil m_i/d_i\\rceil$ for every $i$.","For the converse, take\n\\[\nk=\\max_i\\left\\lceil\\frac{m_i}{d_i}\\right\\rceil.\n\\]\nIn each block, partition a basis of $\\bbC^{m_i}$ into at most $k$ groups of size at most $d_i$. Choose the coefficient matrix of seed $a$ so that its nonzero rows span the $a$th group. The combined row spaces of the $k$ seeds then span $\\bbC^{m_i}$ in every block. The preceding theorem shows that their $\\cA$-orbits generate the whole direct sum.\n\\phantomsection\\label{core:proof:002:end}\\end{proof}","\\begin{corollary}[Local orbit and Schmidt rank]\\label{core:corollary:001}\\label{core:thm:schmidt}\nLet $\\psi\\in\\cH_A\\otimes\\cH_B$, with $\\dim\\cH_A=d_A\\ge1$, and let $r$ be its Schmidt rank. The reduced positive operator on $B$ is $\\rho_B=\\Tr_A(\\ket\\psi\\bra\\psi)$. Normalization of $\\psi$ does not affect its support. Then\n\\[\n\\boxed{\n\\dim\\operatorname{span}\\{(A\\otimes I_B)\\psi:A\\in M_{d_A}(\\bbC)\\}\n=d_A r.\n}\n\\]\nIn particular, local operators on $A$ generate the complete bipartite Hilbert space exactly when the reduced state on $B$ has full support.\n\\end{corollary}","k]\\label{core:corollary:001}\\label{core:thm:schmidt}\nLet $\\psi\\in\\cH_A\\otimes\\cH_B$, with $\\dim\\cH_A=d_A\\ge1$, and let $r$ be its Schmidt rank. The reduced positive operator on $B$ is $\\rho_B=\\Tr_A(\\ket\\psi\\bra\\psi)$. Normalization of $\\psi$ does not affect its support. Then\n\\[\n\\boxed{\n\\dim\\operatorname{span}\\{(A\\otimes I_B)\\psi:A\\in M_{d_A}(\\bbC)\\}\n=d_A r.\n}\n\\]\nIn particular, local operators on $A$ generate the complete bipartite Hilbert space exactly when the reduced state on $B$ has full support.\n\\end{corollary}","with $\\dim\\cH_A=d_A\\ge1$, and let $r$ be its Schmidt rank. The reduced positive operator on $B$ is $\\rho_B=\\Tr_A(\\ket\\psi\\bra\\psi)$. Normalization of $\\psi$ does not affect its support. Then\n\\[\n\\boxed{\n\\dim\\operatorname{span}\\{(A\\otimes I_B)\\psi:A\\in M_{d_A}(\\bbC)\\}\n=d_A r.\n}\n\\]\nIn particular, local operators on $A$ generate the complete bipartite Hilbert space exactly when the reduced state on $B$ has full support.\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{core:proof:003}\nA Schmidt decomposition of length $r$ gives a coefficient matrix of rank $r$. Applying Theorem~\\ref{core:thm:seed-orbit} gives the orbit dimension $d_A r$. The range of $\\rho_B$ is the span of the $r$ Schmidt vectors on $B$, so full support is equivalent to $r=\\dim\\cH_B$. Because $d_A>0$, this is equivalent to the orbit having the full bipartite dimension.\n\\phantomsection\\label{core:proof:003:end}\\end{proof}","\\begin{theorem}[One seed for the computational basis]\\label{core:theorem:003}\\label{core:thm:computational}\nFor an integer $n\\ge0$, let\n\\[\n\\psi_0=\\ket{0}^{\\otimes n}\n\\]\nand let $X_j$ denote the Pauli $X$ operator on qubit $j$. Then\n\\[\n\\left\\{\nX_1^{a_1}\\cdots X_n^{a_n}\\psi_0:\n(a_1,\\ldots,a_n)\\in\\{0,1\\}^n\n\\right\\}\n\\]\nis the complete computational basis. Hence the seed is cyclic under the algebra generated by the $X_j$.\n\\end{theorem}","[One seed for the computational basis]\\label{core:theorem:003}\\label{core:thm:computational}\nFor an integer $n\\ge0$, let\n\\[\n\\psi_0=\\ket{0}^{\\otimes n}\n\\]\nand let $X_j$ denote the Pauli $X$ operator on qubit $j$. Then\n\\[\n\\left\\{\nX_1^{a_1}\\cdots X_n^{a_n}\\psi_0:\n(a_1,\\ldots,a_n)\\in\\{0,1\\}^n\n\\right\\}\n\\]\nis the complete computational basis. Hence the seed is cyclic under the algebra generated by the $X_j$.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:004}\nFor a binary address $a=(a_1,\\ldots,a_n)$,\n\\[\nX_1^{a_1}\\cdots X_n^{a_n}\\ket{0\\cdots0}\n=\n\\ket{a_1\\cdots a_n}.\n\\]\nDistinct binary addresses produce distinct orthonormal basis vectors. There are $2^n$ of them, equal to the dimension of the Hilbert space.\n\\phantomsection\\label{core:proof:004:end}\\end{proof}","\\begin{certificate}[Three-qubit seed orbit]\\label{core:certificate:001}\nThe original numerical implementation generated all eight computational basis states from $\\ket{000}$ using the three local $X$ address operators. It reported an orbit rank of $8/8$.\n\\end{certificate}","Let $N\\ge1$ be an integer, and let $S$ be a unitary symmetry whose order divides $N$:\n\\[\nS^N=I.\n\\]\nSet $\\omega=e^{2\\pi i/N}$.","\\begin{theorem}[Cyclic quantum projectors]\\label{core:theorem:004}\\label{core:thm:cyclic}\nThe operators\n\\[\n\\boxed{\nP_j\n=\n\\frac1N\\sum_{r=0}^{N-1}\\omega^{-jr}S^r,\n\\qquad\n0\\le j<N,\n}\n\\]\nare self-adjoint, pairwise orthogonal projectors satisfying\n\\[\nP_jP_k=\\delta_{jk}P_j,\n\\qquad\n\\sum_{j=0}^{N-1}P_j=I,\n\\qquad\nSP_j=\\omega^jP_j.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:005}\nUsing $S^{r+s}=S^{r+s\\bmod N}$ and the finite character orthogonality relation,\n\\[\n\\frac1N\\sum_{r=0}^{N-1}\\omega^{(k-j)r}=\\delta_{jk},\n\\]\nwe obtain $P_jP_k=\\delta_{jk}P_j$. Summing over $j$ cancels every term indexed by $r\\ne0$ and retains the $r=0$ term, which proves completeness. Unitarity and the change of residue $r\\mapsto-r$ give\n\\[\nP_j^*=\\frac1N\\sum_{r=0}^{N-1}\\omega^{jr}S^{-r}=P_j.\n\\]\nFinally, multiplication by $S$ shifts the summation index and contributes the factor $\\omega^j$.\n\\phantomsection\\label{core:proof:005:end}\\end{proof}","mod N}$ and the finite character orthogonality relation,\n\\[\n\\frac1N\\sum_{r=0}^{N-1}\\omega^{(k-j)r}=\\delta_{jk},\n\\]\nwe obtain $P_jP_k=\\delta_{jk}P_j$. Summing over $j$ cancels every term indexed by $r\\ne0$ and retains the $r=0$ term, which proves completeness. Unitarity and the change of residue $r\\mapsto-r$ give\n\\[\nP_j^*=\\frac1N\\sum_{r=0}^{N-1}\\omega^{jr}S^{-r}=P_j.\n\\]\nFinally, multiplication by $S$ shifts the summation index and contributes the factor $\\omega^j$.\n\\phantomsection\\label{core:proof:005:end}\\end{proof}","[\n\\frac1N\\sum_{r=0}^{N-1}\\omega^{(k-j)r}=\\delta_{jk},\n\\]\nwe obtain $P_jP_k=\\delta_{jk}P_j$. Summing over $j$ cancels every term indexed by $r\\ne0$ and retains the $r=0$ term, which proves completeness. Unitarity and the change of residue $r\\mapsto-r$ give\n\\[\nP_j^*=\\frac1N\\sum_{r=0}^{N-1}\\omega^{jr}S^{-r}=P_j.\n\\]\nFinally, multiplication by $S$ shifts the summation index and contributes the factor $\\omega^j$.\n\\phantomsection\\label{core:proof:005:end}\\end{proof}","For parity,\n\\[\nS=Z^{\\otimes n},\n\\qquad\nP_{\\mathrm{even}}=\\frac{I+S}{2},\n\\qquad\nP_{\\mathrm{odd}}=\\frac{I-S}{2}.\n\\]","Fix $N\\ge1$. Suppose every factor $a$ has a complete orthogonal family $P^{(a)}_0,\\ldots,P^{(a)}_{N-1}$. The fine tensor packets are\n\\[\nP_{\\mathbf j}^{\\mathrm{fine}}\n=\nP^{(1)}_{j_1}\\otimes\\cdots\\otimes P^{(m)}_{j_m}.\n\\]","\\begin{theorem}[Tensor/polyphase refinement]\\label{core:theorem:005}\\label{core:thm:tensor}\nFor each residue $r\\in\\bbZ/N\\bbZ$, define\n\\[\n\\boxed{\nP_r^{\\mathrm{coarse}}\n=\n\\sum_{j_1+\\cdots+j_m\\equiv r\\pmod N}\nP^{(1)}_{j_1}\\otimes\\cdots\\otimes P^{(m)}_{j_m}.\n}\n\\]\nThen the $P_r^{\\mathrm{coarse}}$ are complete orthogonal projectors, and every coarse packet is the exact sum of its fine descendants.\n\\end{theorem}","\\begin{certificate}[Exact parity refinement]\\label{core:certificate:002}\nFor three qubits, the local computational projectors were grouped by the parity of their Hamming weight. The resulting coarse projectors agreed with $(I\\pm Z^{\\otimes3})/2$ with residual $0$.\n\\end{certificate}","Let\n\\[\nU=U_mU_{m-1}\\cdots U_1\n\\]\nbe a circuit with $m\\ge1$, and let $\\{P_\\alpha\\}$ be a complete packet system. For the empty circuit $m=0$, set $U=I$. Its boundary block is $P_\\beta P_\\alpha=\\delta_{\\beta\\alpha}P_\\alpha$.","\\begin{theorem}[Exact circuit path expansion]\\label{core:theorem:006}\\label{core:thm:paths}\nFor source packet $\\alpha_0$ and target packet $\\alpha_m$,\n\\[\n\\boxed{\nP_{\\alpha_m}UP_{\\alpha_0}\n=\n\\sum_{\\alpha_1,\\ldots,\\alpha_{m-1}}\nP_{\\alpha_m}U_mP_{\\alpha_{m-1}}\n\\cdots\nP_{\\alpha_1}U_1P_{\\alpha_0}.\n}\n\\]\nSumming over the source and target packets reconstructs the complete circuit operator.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:007}\nInsert the identity resolution $I=\\sum_\\alpha P_\\alpha$ between every adjacent pair of gates. Distributing the finite sums gives the path expansion. Summing the left and right boundary projectors reconstructs $U$.\n\\phantomsection\\label{core:proof:007:end}\\end{proof}","\\begin{theorem}[Packetwise unitarity conservation]\\label{core:theorem:007}\\label{core:thm:unitarity}\nIf $U$ is unitary and $U_{\\beta\\alpha}=P_\\beta UP_\\alpha$, then\n\\[\n\\boxed{\n\\sum_\\beta U_{\\beta\\alpha}^*U_{\\beta\\gamma}\n=\n\\delta_{\\alpha\\gamma}P_\\alpha.\n}\n\\]\nSimilarly,\n\\[\n\\sum_\\alpha U_{\\beta\\alpha}U_{\\delta\\alpha}^*\n=\n\\delta_{\\beta\\delta}P_\\beta.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:008}\nUsing completeness and unitarity,\n\\[\n\\sum_\\beta U_{\\beta\\alpha}^*U_{\\beta\\gamma}\n=\nP_\\alpha U^*\\left(\\sum_\\beta P_\\beta\\right)UP_\\gamma\n=\nP_\\alpha P_\\gamma.\n\\]\nFor the second identity, sum over the source packets and use unitarity in the reverse order.\n\\phantomsection\\label{core:proof:008:end}\\end{proof}","\\begin{example}[Bell seed circuit]\\label{core:example:001}\nIn this example only, $H$ denotes the Hadamard gate, not a Hamiltonian. Starting from $\\ket{00}$,\n\\[\n(H\\otimes I)\\ket{00}\n=\n\\frac{\\ket{00}+\\ket{10}}{\\sqrt2},\n\\]\nand then\n\\[\n\\operatorname{CNOT}(H\\otimes I)\\ket{00}\n=\n\\frac{\\ket{00}+\\ket{11}}{\\sqrt2}.\n\\]\nThe parity packets show that the final state lies entirely in the even sector, while the intermediate circuit path passes through both even and odd addresses.\n\\end{example}","\\begin{figure}[htbp]\n\\centering\n\\begin{minipage}[t]{0.48\\textwidth}\n\\centering\n\\includegraphics[width=\\textwidth]{02_bell_seed_probabilities.pdf}\n\\captionof{figure}{The Bell state generated from the seed $\\ket{00}$.}\n\\label{core:fig:bell}\n\\end{minipage}\\hfill\n\\begin{minipage}[t]{0.48\\textwidth}\n\\centering\n\\includegraphics[width=\\textwidth]{04_circuit_packet_paths.pdf}\n\\captionof{figure}{Circuit reconstruction as a sum over packet paths.}\n\\label{core:fig:circuit-paths}\n\\end{minipage}\n\\end{figure}","\\begin{certificate}[Bell and path reconstruction]\\label{core:certificate:003}\nThe original numerical implementation reported residuals of $0$ for both the Bell state and its complete reconstruction from parity paths.\n\\end{certificate}","Let $H=H^*$ and let $P$ be a packet projector. If $[H,P]=0$, then $P\\cH$ is invariant under $e^{-itH}$.","\\begin{theorem}[Exact symmetry packets]\\label{core:theorem:008}\\label{core:thm:symmetry}\nIf $[H,P_\\alpha]=0$ for every packet, then\n\\[\nH=\\sum_\\alpha P_\\alpha HP_\\alpha\n\\]\nand\n\\[\ne^{-itH}\n=\n\\sum_\\alpha P_\\alpha e^{-itH}P_\\alpha.\n\\]\nNo amplitude transfers between distinct packets.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:009}\nThe commutation relation gives $P_\\beta HP_\\alpha=0$ for $\\beta\\ne\\alpha$. Every polynomial in $H$ has the same block form, and the finite-dimensional exponential is the norm limit of its polynomial series.\n\\phantomsection\\label{core:proof:009:end}\\end{proof}","For two qubits and real parameters $J,h$, consider\n\\[\nH\n=\nJX\\otimes X\n+h(Z\\otimes I+I\\otimes Z).\n\\]\nIn the ordered even basis $\\{\\ket{00},\\ket{11}\\}$ and odd basis $\\{\\ket{01},\\ket{10}\\}$,\n\\[\nH_{\\mathrm{even}}\n=\n\\begin{pmatrix}\n2h&J","For two qubits and real parameters $J,h$, consider\n\\[\nH\n=\nJX\\otimes X\n+h(Z\\otimes I+I\\otimes Z).\n\\]\nIn the ordered even basis $\\{\\ket{00},\\ket{11}\\}$ and odd basis $\\{\\ket{01},\\ket{10}\\}$,\n\\[\nH_{\\mathrm{even}}\n=\n\\begin{pmatrix}\n2h&J\\\\\nJ&-2h\n\\end{pmatrix},\n\\qquad\nH_{\\mathrm{odd}}\n=\n\\begin{pmatrix}\n0&J\\\\\nJ&0\n\\end{pmatrix}.\n\\]\nTherefore\n\\[\n\\Spec H_{\\mathrm{even}}\n=\\{\\pm\\sqrt{4h^2+J^2}\\},\n\\qquad\n\\Spec H_{\\mathrm{odd}}=\\{\\pm J\\}.\n\\]","J&0\n\\end{pmatrix}.\n\\]\nTherefore\n\\[\n\\Spec H_{\\mathrm{even}}\n=\\{\\pm\\sqrt{4h^2+J^2}\\},\n\\qquad\n\\Spec H_{\\mathrm{odd}}=\\{\\pm J\\}.\n\\]","Set $Q=I-P$.","\\begin{theorem}[Commutator leakage bound]\\label{core:theorem:009}\\label{core:thm:leakage}\nFor every self-adjoint $H$, orthogonal projector $P$, and real time $t$,\n\\[\n\\boxed{\n\\norm{Qe^{-itH}P}\n\\le\n|t|\\,\\norm{[H,P]}.\n}\n\\]\nIn particular, exact commutation gives zero leakage for all time.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:010}\nSet $F(s)=e^{isH}Pe^{-isH}$. Differentiating the exponential gives\n\\[\nF'(s)=i\\,e^{isH}[H,P]e^{-isH},\\qquad F(0)=P.\n\\]\nUnitary conjugation preserves the operator norm, so $\\|F'(s)\\|=\\|[H,P]\\|$. The mean-value estimate on the interval between $0$ and $t$ yields\n\\[\n\\|F(t)-P\\|\\le |t|\\,\\|[H,P]\\|.\n\\]\nFinally, $e^{itH}Qe^{-itH}P=(P-F(t))P$ and $\\|P\\|\\le1$. Left multiplication by $e^{itH}$ preserves the norm and gives the desired inequality. This argument includes positive, negative and zero time.\n\\phantomsection\\label{core:proof:010:end}\\end{proof}","n\\label{core:proof:010}\nSet $F(s)=e^{isH}Pe^{-isH}$. Differentiating the exponential gives\n\\[\nF'(s)=i\\,e^{isH}[H,P]e^{-isH},\\qquad F(0)=P.\n\\]\nUnitary conjugation preserves the operator norm, so $\\|F'(s)\\|=\\|[H,P]\\|$. The mean-value estimate on the interval between $0$ and $t$ yields\n\\[\n\\|F(t)-P\\|\\le |t|\\,\\|[H,P]\\|.\n\\]\nFinally, $e^{itH}Qe^{-itH}P=(P-F(t))P$ and $\\|P\\|\\le1$. Left multiplication by $e^{itH}$ preserves the norm and gives the desired inequality. This argument includes positive, negative and zero time.\n\\phantomsection\\label{core:proof:010:end}\\end{proof}","\\begin{certificate}[Symmetry and negative control]\\label{core:certificate:004}\nFor the Hamiltonian that preserves parity, the norm of the block between distinct packets and the maximum measured leakage were both $0$. Adding an $X\\otimes I$ perturbation broke parity and produced a maximum leakage of approximately $0.79196$. The maximum violation of the proved bound remained $0$.\n\\end{certificate}","\\begin{theorem}[Moving-frame Hamiltonian]\\label{core:theorem:010}\\label{core:thm:moving}\nLet $U(t)$ and $\\psi(t)$ be differentiable on an open real time interval, let $U(t)$ be unitary, and let $H(t)=H(t)^*$. Suppose\n\\[\ni\\dot\\psi(t)=H(t)\\psi(t).\n\\]\nWriting\n\\[\n\\psi(t)=U(t)y(t)\n\\]\ngives\n\\[\n\\boxed{\ni\\dot y(t)\n=\n\\left(U(t)^*H(t)U(t)-iU(t)^*\\dot U(t)\\right)y(t).\n}\n\\]\nBoth the connection term $-iU^*\\dot U$ and the effective Hamiltonian $U^*HU-iU^*\\dot U$ are Hermitian.\n\\end{theorem}","onian]\\label{core:theorem:010}\\label{core:thm:moving}\nLet $U(t)$ and $\\psi(t)$ be differentiable on an open real time interval, let $U(t)$ be unitary, and let $H(t)=H(t)^*$. Suppose\n\\[\ni\\dot\\psi(t)=H(t)\\psi(t).\n\\]\nWriting\n\\[\n\\psi(t)=U(t)y(t)\n\\]\ngives\n\\[\n\\boxed{\ni\\dot y(t)\n=\n\\left(U(t)^*H(t)U(t)-iU(t)^*\\dot U(t)\\right)y(t).\n}\n\\]\nBoth the connection term $-iU^*\\dot U$ and the effective Hamiltonian $U^*HU-iU^*\\dot U$ are Hermitian.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:011}\nDifferentiate $\\psi=Uy$:\n\\[\ni\\dot U y+iU\\dot y=HUy.\n\\]\nMultiplying by $U^*$ gives the formula. Differentiating $U^*U=I$ shows $U^*\\dot U$ is skew-Hermitian, so $-iU^*\\dot U$ is Hermitian.\n\\phantomsection\\label{core:proof:011:end}\\end{proof}","\\begin{certificate}[Moving-frame correction]\\label{core:certificate:005}\nThe reported residual for the complete equation in the moving frame was $1.0645\\times10^{-10}$. Omitting the connection increased the residual to $0.7000$.\n\\end{certificate}","Let $P_1(t),\\ldots,P_m(t)$ be differentiable on an open real time interval, with $P_a(t)^*=P_a(t)$ and\n\\[\nP_aP_b=\\delta_{ab}P_a,\n\\qquad\n\\sum_aP_a=I.\n\\]","\\begin{theorem}[Canonical packet connection]\\label{core:theorem:011}\\label{core:thm:canonical}\nDefine\n\\[\n\\boxed{\nC(t)\n=\n\\frac12\\sum_a[\\dot P_a(t),P_a(t)]\n=\n\\sum_a\\dot P_a(t)P_a(t).\n}\n\\]\nThen\n\\[\nC(t)^*=-C(t)\n\\]\nand\n\\[\n\\boxed{\n[C(t),P_b(t)]=\\dot P_b(t)\n}\n\\]\nfor every packet $b$.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:012}\nDifferentiating $P_a^2=P_a$ gives\n\\[\n\\dot P_aP_a+P_a\\dot P_a=\\dot P_a.\n\\]\nSumming over $a$ and using $\\sum_a\\dot P_a=0$ yields\n\\[\n\\sum_aP_a\\dot P_a=-\\sum_a\\dot P_aP_a,\n\\]\nwhich proves both displayed formulas for $C$ and its skew-Hermiticity.","For fixed $b$,\n\\[\nCP_b=\\dot P_bP_b.\n\\]\nWhen $a\\ne b$, differentiating $P_bP_a=0$ gives\n\\[\nP_b\\dot P_a=-\\dot P_bP_a.\n\\]\nHence\n\\[\nP_bC\n=\\sum_aP_b\\dot P_aP_a\n=-\\dot P_b(I-P_b),\n\\]\nand subtraction gives $[C,P_b]=\\dot P_b$.\n\\phantomsection\\label{core:proof:012:end}\\end{proof}","\\begin{classical}[Projected connection]\\label{core:classical:001}\\label{core:thm:projected}\nLet $M$ be a smooth finite-dimensional real manifold without boundary, let $V$ be a finite-dimensional real normed vector space, and let $P:M\\to\\operatorname{End}_{\\bbR}(V)$ be smooth with $P(x)^2=P(x)$. In the quantum application, $V$ is the underlying real space of a finite complex Hilbert space and $P$ is an orthogonal complex-linear projector.","The rank of $P$ is locally constant. On each open rank region, its ranges form a smooth vector bundle. For a smooth range section $s$, so that $Ps=s$, define\n\\[\n\\nabla_Xs=P\\,ds(X).\n\\]\nThis is a connection on smooth range sections. For smooth vector fields $X,Y$, its curvature satisfies\n\\[\n\\boxed{\n(\\nabla_X\\nabla_Y-\\nabla_Y\\nabla_X-\\nabla_{[X,Y]})s\n=P[dP(X),dP(Y)]P\\,s.\n}\n\\]\nEquivalently, its endomorphism-valued curvature is\n\\[\n\\boxed{F=P(dP\\wedge dP)P.}\n\\]\nNo global frame or globally constant rank is required, and the vector fields need not commute. The geometric formalization concerns real smooth structures. It does not cover complex analytic manifolds or extensions to a boundary.\n\\end{classical}","nstant. On each open rank region, its ranges form a smooth vector bundle. For a smooth range section $s$, so that $Ps=s$, define\n\\[\n\\nabla_Xs=P\\,ds(X).\n\\]\nThis is a connection on smooth range sections. For smooth vector fields $X,Y$, its curvature satisfies\n\\[\n\\boxed{\n(\\nabla_X\\nabla_Y-\\nabla_Y\\nabla_X-\\nabla_{[X,Y]})s\n=P[dP(X),dP(Y)]P\\,s.\n}\n\\]\nEquivalently, its endomorphism-valued curvature is\n\\[\n\\boxed{F=P(dP\\wedge dP)P.}\n\\]\nNo global frame or globally constant rank is required, and the vector fields need not commute. The geometric formalization concerns real smooth structures. It does not cover complex analytic manifolds or extensions to a boundary.\n\\end{classical}","\\begin{proof}\\phantomsection\\label{core:proof:013}\n\\textit{Range coordinates.} Fix $a\\in M$ and put\n\\[\nT_a(x)=P(x)P(a)+(I-P(x))(I-P(a)).\n\\]\nThen $T_a(a)=I$ and $P(x)T_a(x)=T_a(x)P(a)$. Invertibility is open, so on a neighborhood of $a$, $T_a(x)$ restricts to an isomorphism from $\\Ran P(a)$ to $\\Ran P(x)$. This proves local constancy of rank. The local coordinates and their transitions are\n\\[\n(x,v)\\longmapsto (x,T_a(x)^{-1}v),\\qquad\nu\\longmapsto T_b(x)^{-1}T_a(x)u.\n\\]\nTheir smoothness follows from smoothness of $P$ and inversion. The transition maps satisfy the cocycle identity by cancellation. These charts construct the smooth range bundles on the open rank regions, including when different connected components have different ranks.",")+(I-P(x))(I-P(a)).\n\\]\nThen $T_a(a)=I$ and $P(x)T_a(x)=T_a(x)P(a)$. Invertibility is open, so on a neighborhood of $a$, $T_a(x)$ restricts to an isomorphism from $\\Ran P(a)$ to $\\Ran P(x)$. This proves local constancy of rank. The local coordinates and their transitions are\n\\[\n(x,v)\\longmapsto (x,T_a(x)^{-1}v),\\qquad\nu\\longmapsto T_b(x)^{-1}T_a(x)u.\n\\]\nTheir smoothness follows from smoothness of $P$ and inversion. The transition maps satisfy the cocycle identity by cancellation. These charts construct the smooth range bundles on the open rank regions, including when different connected components have different ranks.","\\textit{Connection rules.} Smoothness of $P\\,ds(X)$ and $P^2=P$ show that $\\nabla_Xs$ is a smooth range section. Differentiation and projection give\n\\[\n\\begin{aligned}\n\\nabla_{X+Y}s&=\\nabla_Xs+\\nabla_Ys,&\n\\nabla_{fX}s&=f\\nabla_Xs,","\\textit{Connection rules.} Smoothness of $P\\,ds(X)$ and $P^2=P$ show that $\\nabla_Xs$ is a smooth range section. Differentiation and projection give\n\\[\n\\begin{aligned}\n\\nabla_{X+Y}s&=\\nabla_Xs+\\nabla_Ys,&\n\\nabla_{fX}s&=f\\nabla_Xs,\\\\\n\\nabla_X(s+t)&=\\nabla_Xs+\\nabla_Xt,&\n\\nabla_X(fs)&=f\\nabla_Xs+(Xf)s\n\\end{aligned}\n\\]\nfor smooth real $f$. The range coordinates above identify ambient smooth range sections with smooth sections of the constructed bundle.","\\nabla_X(s+t)&=\\nabla_Xs+\\nabla_Xt,&\n\\nabla_X(fs)&=f\\nabla_Xs+(Xf)s\n\\end{aligned}\n\\]\nfor smooth real $f$. The range coordinates above identify ambient smooth range sections with smooth sections of the constructed bundle.","\\textit{Curvature.} Differentiating $P^2=P$ gives $P(dP(X))P=0$. Write $D_Xs=ds(X)$. For ambient smooth sections,\n\\[\nD_XD_Ys-D_YD_Xs-D_{[X,Y]}s=0.\n\\]\nExpansion of the projected derivatives therefore leaves\n\\[\nR(X,Y)s=P(dP(X))D_Ys-P(dP(Y))D_Xs.\n\\]\nSince $Ps=s$, differentiation gives $D_Ys=(dP(Y))s+PD_Ys$. The terms with a middle $P$ vanish. Thus\n\\[\nR(X,Y)s=P\\bigl(dP(X)dP(Y)-dP(Y)dP(X)\\bigr)P\\,s.\n\\]\nThis expression is alternating and bilinear in the tangent vectors, and it preserves the range. The ordinary chain rule and the transformation law for the Lie bracket make the local calculations agree on chart overlaps, which gives the global formula.\n\\phantomsection\\label{core:proof:013:end}\\end{proof}","For an orthogonal rank-one complex projector, use the scalar connection convention $A=-i\\langle u,du\\rangle$ in a local unit frame. The associated scalar curvature density is\n\\[\n\\cF_{\\mu\\nu}\n=-i\\Tr\\left(P[\\partial_\\mu P,\\partial_\\nu P]\\right).\n\\]\nOn the Bloch sphere,\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\math","For an orthogonal rank-one complex projector, use the scalar connection convention $A=-i\\langle u,du\\rangle$ in a local unit frame. The associated scalar curvature density is\n\\[\n\\cF_{\\mu\\nu}\n=-i\\Tr\\left(P[\\partial_\\mu P,\\partial_\\nu P]\\right).\n\\]\nOn the Bloch sphere,\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a se","For an orthogonal rank-one complex projector, use the scalar connection convention $A=-i\\langle u,du\\rangle$ in a local unit frame. The associated scalar curvature density is\n\\[\n\\cF_{\\mu\\nu}\n=-i\\Tr\\left(P[\\partial_\\mu P,\\partial_\\nu P]\\right).\n\\]\nOn the Bloch sphere,\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a separate entry among the 19 items in the Lean register.","n convention $A=-i\\langle u,du\\rangle$ in a local unit frame. The associated scalar curvature density is\n\\[\n\\cF_{\\mu\\nu}\n=-i\\Tr\\left(P[\\partial_\\mu P,\\partial_\\nu P]\\right).\n\\]\nOn the Bloch sphere,\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a separate entry among the 19 items in the Lean register.","P,\\partial_\\nu P]\\right).\n\\]\nOn the Bloch sphere,\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a separate entry among the 19 items in the Lean register.","I+\\mathbf n(\\theta,\\phi)\\cdot\\boldsymbol\\sigma\\right),\n\\]\nwhere\n\\[\n\\mathbf n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nThe Pauli multiplication law gives\n\\[\n[\\partial_\\theta P,\\partial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a separate entry among the 19 items in the Lean register.","rtial_\\phi P]\n=\\frac{i}{2}(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\\cdot\\boldsymbol\\sigma.\n\\]\nUsing $\\Tr(P\\,\\mathbf a\\cdot\\boldsymbol\\sigma)=\\mathbf n\\cdot\\mathbf a$ therefore yields\n\\[\n\\boxed{\\cF_{\\theta\\phi}\n=\\frac12\\mathbf n\\cdot(\\partial_\\theta\\mathbf n\\mathbin{\\times}\\partial_\\phi\\mathbf n)\n=\\frac12\\sin\\theta.}\n\\]\nThe scalar sign changes if the opposite convention for $A$ is chosen. This explicit calculation is not a separate entry among the 19 items in the Lean register.","\\begin{figure}[htbp]\n\\centering\n\\includegraphics[width=0.76\\textwidth]{09_bloch_curvature.pdf}\n\\caption{The curvature of the Bloch projector, evaluated by finite differences, is compared with the exact density $\\frac12\\sin\\theta$.}\n\\label{core:fig:bloch}\n\\end{figure}","\\begin{certificate}[Geometric packet identities]\\label{core:certificate:006}\nThe canonical connection reconstructed every packet derivative with residual $0$. The maximum residual for the Bloch curvature was $2.0803\\times10^{-11}$.\n\\end{certificate}","Let $\\rho$ be a density matrix and let $\\{E_k\\}$ be a POVM:\n\\[\nE_k\\ge0,\n\\qquad\n\\sum_kE_k=I.\n\\]\nThe outcome probabilities are\n\\[\np_k=\\Tr(E_k\\rho).\n\\]","\\begin{theorem}[Block-diagonal measurement invariance]\\label{core:theorem:012}\\label{core:thm:measurement}\nLet $\\{P_\\alpha\\}$ be a complete packet family, and suppose every measurement effect is packet diagonal:\n\\[\nE_k=\\sum_\\alpha P_\\alpha E_kP_\\alpha.\n\\]\nDefine the packet-dephased state\n\\[\n\\cD_P(\\rho)=\\sum_\\alpha P_\\alpha\\rho P_\\alpha.\n\\]\nThen\n\\[\n\\boxed{\n\\Tr(E_k\\rho)\n=\n\\Tr(E_k\\cD_P(\\rho))\n}\n\\]\nfor every outcome $k$.\n\\end{theorem}","lock-diagonal measurement invariance]\\label{core:theorem:012}\\label{core:thm:measurement}\nLet $\\{P_\\alpha\\}$ be a complete packet family, and suppose every measurement effect is packet diagonal:\n\\[\nE_k=\\sum_\\alpha P_\\alpha E_kP_\\alpha.\n\\]\nDefine the packet-dephased state\n\\[\n\\cD_P(\\rho)=\\sum_\\alpha P_\\alpha\\rho P_\\alpha.\n\\]\nThen\n\\[\n\\boxed{\n\\Tr(E_k\\rho)\n=\n\\Tr(E_k\\cD_P(\\rho))\n}\n\\]\nfor every outcome $k$.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:014}\nInsert the packet decomposition of $E_k$ and use cyclicity of the trace:\n\\[\n\\Tr(E_k\\rho)\n=\n\\sum_\\alpha\\Tr(P_\\alpha E_kP_\\alpha\\rho)\n=\n\\sum_\\alpha\\Tr(E_kP_\\alpha\\rho P_\\alpha).\n\\]\n\\phantomsection\\label{core:proof:014:end}\\end{proof}","\\begin{classical}[Finite Kraus channel]\\label{core:classical:002}\\label{core:thm:kraus}\nLet $K_1,\\ldots,K_r$ satisfy\n\\[\n\\sum_aK_a^*K_a=I.\n\\]\nThen\n\\[\n\\boxed{\n\\Phi(\\rho)=\\sum_aK_a\\rho K_a^*\n}\n\\]\nis completely positive and trace preserving.\n\\end{classical}","\\begin{proof}\\phantomsection\\label{core:proof:015}\nTrace preservation follows from\n\\[\n\\Tr\\Phi(\\rho)\n=\\Tr\\left(\\rho\\sum_aK_a^*K_a\\right)\n=\\Tr\\rho.\n\\]\nFor any auxiliary dimension and positive matrix $X$, each term\n\\[\n(I\\otimes K_a)X(I\\otimes K_a)^*\n\\]\nis positive, so their sum is positive. This is complete positivity.\n\\phantomsection\\label{core:proof:015:end}\\end{proof}","Packet superoperator blocks are defined by\n\\[\n\\Phi_{\\alpha\\beta;\\gamma\\delta}(X)\n=\nP_\\alpha\n\\Phi(P_\\gamma XP_\\delta)\nP_\\beta.\n\\]\nSumming all four packet indices reconstructs $\\Phi(X)$ exactly.","For amplitude damping with a real parameter $0\\le\\gamma\\le1$,\n\\[\nK_0=\n\\begin{pmatrix}\n1&0","For amplitude damping with a real parameter $0\\le\\gamma\\le1$,\n\\[\nK_0=\n\\begin{pmatrix}\n1&0\\\\0&\\sqrt{1-\\gamma}\n\\end{pmatrix},\n\\qquad\nK_1=\n\\begin{pmatrix}\n0&\\sqrt\\gamma\\\\0&0\n\\end{pmatrix}.\n\\]\nThe excited state becomes\n\\[\n\\Phi(\\ket1\\bra1)\n=\n\\gamma\\ket0\\bra0+(1-\\gamma)\\ket1\\bra1.\n\\]","0&0\n\\end{pmatrix}.\n\\]\nThe excited state becomes\n\\[\n\\Phi(\\ket1\\bra1)\n=\n\\gamma\\ket0\\bra0+(1-\\gamma)\\ket1\\bra1.\n\\]","\\begin{certificate}[Channel reconstruction]\\label{core:certificate:007}\nFor $\\gamma=0.23$, the residuals for Kraus completeness and the output trace were $0$. The computed output from $\\ket1\\bra1$ was $\\diag(0.23,0.77)$. In the Bell state test, packet dephasing left every measurement probability in the computational basis unchanged.\n\\end{certificate}","Let $P=P^*=P^2$ project onto a code space $\\cC\\subset\\cH$, and let $E_1,\\ldots,E_r$ be a finite family of complex error matrices. Neither a nonzero code nor channel normalization is assumed unless stated. The operators\n\\[\nP E_a^*E_bP\n\\]\nare the obstruction packets on the code. If they retain information about the logical state beyond one scalar multiple of $P$, the environment can distinguish logical states and exact correction is impossible.","We call a recovery complete when it is a Kraus map\n\\[\n\\cR(X)=\\sum_jK_jXK_j^*,\\qquad \\sum_jK_j^*K_j=I.\n\\]\nThe family is \\emph{exactly correctable} when one such recovery works for every coherent error combination $A_w=\\sum_aw_aE_a$: for every $w$ there is a real $t_w\\ge0$, independent of the logical input, such that\n\\[\nPXP=X\\quad\\Longrightarrow\\quad\n\\cR(A_wXA_w^*)=t_wX.\n\\]\nThe coefficient is an unnormalized weight. It is not necessarily a probability for an arbitrary matrix $A_w$.","quad \\sum_jK_j^*K_j=I.\n\\]\nThe family is \\emph{exactly correctable} when one such recovery works for every coherent error combination $A_w=\\sum_aw_aE_a$: for every $w$ there is a real $t_w\\ge0$, independent of the logical input, such that\n\\[\nPXP=X\\quad\\Longrightarrow\\quad\n\\cR(A_wXA_w^*)=t_wX.\n\\]\nThe coefficient is an unnormalized weight. It is not necessarily a probability for an arbitrary matrix $A_w$.","\\begin{classical}[Finite Knill--Laflamme criterion]\\label{core:classical:003}\\label{core:thm:kl}\nThe finite error family is exactly correctable in the sense just defined if and only if there is a positive semidefinite complex matrix $C=(c_{ab})$ such that\n\\[\n\\boxed{PE_a^*E_bP=c_{ab}P\\qquad\\text{for every }a,b.}\n\\]\nIf $\\sum_aE_a^*E_a=I$, the equivalent normalized-channel formulation is the existence of a complete recovery satisfying\n\\[\n\\cR\\!\\left(\\sum_aE_aXE_a^*\\right)=X\n\\quad\\text{for every }X\\text{ with }PXP=X.\n\\]\nThe zero code and an empty error family are included. An empty family cannot define a channel that preserves trace on a nonzero ambient space.\n\\end{classical}","me criterion]\\label{core:classical:003}\\label{core:thm:kl}\nThe finite error family is exactly correctable in the sense just defined if and only if there is a positive semidefinite complex matrix $C=(c_{ab})$ such that\n\\[\n\\boxed{PE_a^*E_bP=c_{ab}P\\qquad\\text{for every }a,b.}\n\\]\nIf $\\sum_aE_a^*E_a=I$, the equivalent normalized-channel formulation is the existence of a complete recovery satisfying\n\\[\n\\cR\\!\\left(\\sum_aE_aXE_a^*\\right)=X\n\\quad\\text{for every }X\\text{ with }PXP=X.\n\\]\nThe zero code and an empty error family are included. An empty family cannot define a channel that preserves trace on a nonzero ambient space.\n\\end{classical}","\\begin{proof}\\phantomsection\\label{core:proof:016}\n\\textit{Necessity.} Fix an error $E_a$ and let $L_j=K_jE_aP$. Perfect recovery on rank-one code operators gives\n\\[\n\\sum_j\\ket{L_j\\psi}\\bra{L_j\\psi}\n=t_a\\ket\\psi\\bra\\psi\\qquad(\\psi\\in\\cC).\n\\]\nFor every vector orthogonal to $\\psi$, the quadratic form on the left is a sum of nonnegative squared absolute values and equals zero. Each $L_j\\psi$ therefore lies in the span of $\\psi$. A linear operator that preserves every line in $\\cC$ is scalar there: apply it to two independent vectors and their sum. The one-dimensional case is immediate, and the claim is vacuous for the zero code. Consequently\n\\[\nK_jE_aP=d_{ja}P.\n\\]\nRecovery completeness now gives\n\\[\n\\begin{aligned}\nPE_a^*E_bP\n&=\\sum_j (K_jE_aP)^*(K_jE_bP)","ction\\label{core:proof:016}\n\\textit{Necessity.} Fix an error $E_a$ and let $L_j=K_jE_aP$. Perfect recovery on rank-one code operators gives\n\\[\n\\sum_j\\ket{L_j\\psi}\\bra{L_j\\psi}\n=t_a\\ket\\psi\\bra\\psi\\qquad(\\psi\\in\\cC).\n\\]\nFor every vector orthogonal to $\\psi$, the quadratic form on the left is a sum of nonnegative squared absolute values and equals zero. Each $L_j\\psi$ therefore lies in the span of $\\psi$. A linear operator that preserves every line in $\\cC$ is scalar there: apply it to two independent vectors and their sum. The one-dimensional case is immediate, and the claim is vacuous for the zero code. Consequently\n\\[\nK_jE_aP=d_{ja}P.\n\\]\nRecovery completeness now gives\n\\[\n\\begin{aligned}\nPE_a^*E_bP\n&=\\sum_j (K_jE_aP)^*(K_jE_bP)","error $E_a$ and let $L_j=K_jE_aP$. Perfect recovery on rank-one code operators gives\n\\[\n\\sum_j\\ket{L_j\\psi}\\bra{L_j\\psi}\n=t_a\\ket\\psi\\bra\\psi\\qquad(\\psi\\in\\cC).\n\\]\nFor every vector orthogonal to $\\psi$, the quadratic form on the left is a sum of nonnegative squared absolute values and equals zero. Each $L_j\\psi$ therefore lies in the span of $\\psi$. A linear operator that preserves every line in $\\cC$ is scalar there: apply it to two independent vectors and their sum. The one-dimensional case is immediate, and the claim is vacuous for the zero code. Consequently\n\\[\nK_jE_aP=d_{ja}P.\n\\]\nRecovery completeness now gives\n\\[\n\\begin{aligned}\nPE_a^*E_bP\n&=\\sum_j (K_jE_aP)^*(K_jE_bP)","ry vector orthogonal to $\\psi$, the quadratic form on the left is a sum of nonnegative squared absolute values and equals zero. Each $L_j\\psi$ therefore lies in the span of $\\psi$. A linear operator that preserves every line in $\\cC$ is scalar there: apply it to two independent vectors and their sum. The one-dimensional case is immediate, and the claim is vacuous for the zero code. Consequently\n\\[\nK_jE_aP=d_{ja}P.\n\\]\nRecovery completeness now gives\n\\[\n\\begin{aligned}\nPE_a^*E_bP\n&=\\sum_j (K_jE_aP)^*(K_jE_bP)","he left is a sum of nonnegative squared absolute values and equals zero. Each $L_j\\psi$ therefore lies in the span of $\\psi$. A linear operator that preserves every line in $\\cC$ is scalar there: apply it to two independent vectors and their sum. The one-dimensional case is immediate, and the claim is vacuous for the zero code. Consequently\n\\[\nK_jE_aP=d_{ja}P.\n\\]\nRecovery completeness now gives\n\\[\n\\begin{aligned}\nPE_a^*E_bP\n&=\\sum_j (K_jE_aP)^*(K_jE_bP)\\\\\n&=\\left(\\sum_j\\overline{d_{ja}}d_{jb}\\right)P.\n\\end{aligned}\n\\]\nThe coefficients form a Gram matrix, hence are positive semidefinite. For a normalized channel, apply the same rank-one argument to all composite branches $K_jE_aP$ at once to obtain the necessary condition.","&=\\left(\\sum_j\\overline{d_{ja}}d_{jb}\\right)P.\n\\end{aligned}\n\\]\nThe coefficients form a Gram matrix, hence are positive semidefinite. For a normalized channel, apply the same rank-one argument to all composite branches $K_jE_aP$ at once to obtain the necessary condition.","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. E","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every reco","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent co","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for t","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space v","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices,","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting reco","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement witho","\\textit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\","tit{Sufficiency.} Diagonalize $C=U\\operatorname{diag}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, su…","}(\\lambda_r)U^*$ with $U$ unitary and $\\lambda_r\\ge0$. Define compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so …","efine compressed transformed errors\n\\[\nF_r=\\sum_aU_{ar}E_aP.\n\\]\nThen $F_rP=F_r$ and $F_r^*F_s=\\lambda_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every …","_r\\delta_{rs}P$. If $\\lambda_r=0$, $F_r^*F_r=0$ forces $F_r=0$. For positive eigenvalues only, set $V_r=F_r/\\sqrt{\\lambda_r}$. These satisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completene…","tisfy\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phant…","r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nLet $S=\\sum_{\\lambda_r>0}V_rV_r^*$. Orthogonality makes $S$ an orthogonal projector. Use the recovery branches\n\\[\nK_r=V_r^*,\\qquad K_{\\mathrm{unused}}=I-S.\n\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{…","\\]\nThey are complete on the whole ambient space because\n\\[\n\\sum_rK_r^*K_r+K_{\\mathrm{unused}}^*K_{\\mathrm{unused}}\n=S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","S+(I-S)^2=I.\n\\]\nThe inverse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","erse unitary change of coordinates reconstructs each $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","ach $E_aP$ from the $F_r$. Every recovery branch of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","h of a coherent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","erent combination is therefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","erefore scalar on the code, while the branch for the unused space vanishes. For all supported matrices, the resulting recovery is\n\\[\n\\cR(A_wXA_w^*)=(w^*Cw)X.\n\\]\nThis establishes the arbitrary-family statement without assuming normalization. Under $\\sum_aE_a^*E_a=I$ and $P\\ne0$, summing the diagonal KL equations gives $\\Tr C=1$, so the normalized-channel output is $X$. When $P=0$, every supported $X$ is zero and the conclusion is automatic. The recovery is trace preserving by completeness and completely positive at every finite ancillary dimension by the Kraus calculation in Classical Input~\\ref{core:thm:kraus}.\n\\phantomsection\\label{core:proof:016:end}\\end{proof}","For the three-qubit bit-flip code,\n\\[\n\\ket{0_L}=\\ket{000},\n\\qquad\n\\ket{1_L}=\\ket{111},\n\\]\nand the error family is\n\\[\nI,\n\\quad X\\otimes I\\otimes I,\n\\quad I\\otimes X\\otimes I,\n\\quad I\\otimes I\\otimes X.\n\\]\nThe four error images of the code are pairwise orthogonal two-dimensional syndrome packets whose projectors sum to the identity on the eight-dimensional Hilbert space.","\\begin{certificate}[Exact bit-flip correction]\\label{core:certificate:008}\nThe residuals for the Knill--Laflamme condition and completeness of the syndrome packets were both $0$. A mixed channel of single bit flip errors, applied to a nontrivial logical superposition, was reversed with a residual of $1.0385\\times10^{-16}$ in the density matrix.\n\\end{certificate}","Let $P$ be a retained packet projector and $Q=I-P$. In packet coordinates,\n\\[\nH=\n\\begin{pmatrix}\nH_{PP}&H_{PQ}","Let $P$ be a retained packet projector and $Q=I-P$. In packet coordinates,\n\\[\nH=\n\\begin{pmatrix}\nH_{PP}&H_{PQ}\\\\\nH_{QP}&H_{QQ}\n\\end{pmatrix}.\n\\]","\\begin{theorem}[Exact Feshbach resolvent]\\label{core:theorem:013}\\label{core:thm:feshbach}\nAssume that both $zI_Q-H_{QQ}$ and $zI-H$ are invertible. Define\n\\[\n\\boxed{\nH_{\\mathrm{eff}}(z)\n=\nH_{PP}\n+H_{PQ}(z-H_{QQ})^{-1}H_{QP}.\n}\n\\]\nThen\n\\[\n\\boxed{\nP(z-H)^{-1}P\n=\n\\left(z-H_{\\mathrm{eff}}(z)\\right)^{-1}\n}\n\\]\nwhere the left side is restricted to $P\\cH$, the right side is an inverse on that retained space, and both are written in the same orthonormal basis. Here $z-H_{QQ}$ and $z-H_{\\mathrm{eff}}$ abbreviate the appropriately sized scalar matrices minus the indicated blocks.\n\\end{theorem}","3}\\label{core:thm:feshbach}\nAssume that both $zI_Q-H_{QQ}$ and $zI-H$ are invertible. Define\n\\[\n\\boxed{\nH_{\\mathrm{eff}}(z)\n=\nH_{PP}\n+H_{PQ}(z-H_{QQ})^{-1}H_{QP}.\n}\n\\]\nThen\n\\[\n\\boxed{\nP(z-H)^{-1}P\n=\n\\left(z-H_{\\mathrm{eff}}(z)\\right)^{-1}\n}\n\\]\nwhere the left side is restricted to $P\\cH$, the right side is an inverse on that retained space, and both are written in the same orthonormal basis. Here $z-H_{QQ}$ and $z-H_{\\mathrm{eff}}$ abbreviate the appropriately sized scalar matrices minus the indicated blocks.\n\\end{theorem}","hm:feshbach}\nAssume that both $zI_Q-H_{QQ}$ and $zI-H$ are invertible. Define\n\\[\n\\boxed{\nH_{\\mathrm{eff}}(z)\n=\nH_{PP}\n+H_{PQ}(z-H_{QQ})^{-1}H_{QP}.\n}\n\\]\nThen\n\\[\n\\boxed{\nP(z-H)^{-1}P\n=\n\\left(z-H_{\\mathrm{eff}}(z)\\right)^{-1}\n}\n\\]\nwhere the left side is restricted to $P\\cH$, the right side is an inverse on that retained space, and both are written in the same orthonormal basis. Here $z-H_{QQ}$ and $z-H_{\\mathrm{eff}}$ abbreviate the appropriately sized scalar matrices minus the indicated blocks.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:017}\nThe block in the lower right corner of $z-H$ is $z-H_{QQ}$. Gaussian block elimination produces the Schur complement\n\\[\nz-H_{PP}-H_{PQ}(z-H_{QQ})^{-1}H_{QP}.\n\\]\nThe assumed invertibility of this block and of the full matrix implies that the Schur complement is invertible. The block in the upper left corner of the full inverse is the inverse of that Schur complement, which is exactly the displayed operator.\n\\phantomsection\\label{core:proof:017:end}\\end{proof}","The second premise cannot be dropped: at $z=0$ and $H=\\operatorname{diag}(0,-1)$, the eliminated block is $1$ but $zI-H$ is singular. This counterexample is also included in the Lean module.","The self-energy\n\\[\n\\Sigma_P(z)\n=H_{PQ}(z-H_{QQ})^{-1}H_{QP}\n\\]\ncontains the complete influence of the eliminated packet. Bare truncation drops $\\Sigma_P(z)$ and is therefore exact only when the coupling between packets vanishes or the self-energy cancels for another stated reason.","\\begin{certificate}[Effective resolvent]\\label{core:certificate:009}\nIn the reported test with a random Hermitian matrix of dimension six, the exact Feshbach residual was $3.2291\\times10^{-17}$. The corresponding residual for bare truncation was $1.3091\\times10^{-2}$.\n\\end{certificate}","For a time-independent finite Hamiltonian $H$ and fixed projectors $P,Q$, write $\\psi=x+y$, with $x=P\\psi$ and $y=Q\\psi$. Then\n\\[\n\\begin{aligned}\ni\\dot x&=H_{PP}x+H_{PQ}y,","For a time-independent finite Hamiltonian $H$ and fixed projectors $P,Q$, write $\\psi=x+y$, with $x=P\\psi$ and $y=Q\\psi$. Then\n\\[\n\\begin{aligned}\ni\\dot x&=H_{PP}x+H_{PQ}y,\\\\\ni\\dot y&=H_{QP}x+H_{QQ}y.\n\\end{aligned}\n\\]\nSolving the second equation gives\n\\[\ny(t)\n=e^{-itH_{QQ}}y(0)\n-i\\int_0^t e^{-i(t-s)H_{QQ}}H_{QP}x(s)\\,ds.\n\\]\nSubstitution yields the exact Volterra equation\n\\[\n\\boxed{\n\\begin{aligned}\ni\\dot x(t)\n={}&H_{PP}x(t)\n+H_{PQ}e^{-itH_{QQ}}y(0)","i\\dot y&=H_{QP}x+H_{QQ}y.\n\\end{aligned}\n\\]\nSolving the second equation gives\n\\[\ny(t)\n=e^{-itH_{QQ}}y(0)\n-i\\int_0^t e^{-i(t-s)H_{QQ}}H_{QP}x(s)\\,ds.\n\\]\nSubstitution yields the exact Volterra equation\n\\[\n\\boxed{\n\\begin{aligned}\ni\\dot x(t)\n={}&H_{PP}x(t)\n+H_{PQ}e^{-itH_{QQ}}y(0)","i\\dot y&=H_{QP}x+H_{QQ}y.\n\\end{aligned}\n\\]\nSolving the second equation gives\n\\[\ny(t)\n=e^{-itH_{QQ}}y(0)\n-i\\int_0^t e^{-i(t-s)H_{QQ}}H_{QP}x(s)\\,ds.\n\\]\nSubstitution yields the exact Volterra equation\n\\[\n\\boxed{\n\\begin{aligned}\ni\\dot x(t)\n={}&H_{PP}x(t)\n+H_{PQ}e^{-itH_{QQ}}y(0)\\\\\n&-i\\int_0^t\nK(t-s)x(s)\\,ds,\n\\end{aligned}\n}\n\\]\nwith memory kernel\n\\[\n\\boxed{\nK(t)=H_{PQ}e^{-itH_{QQ}}H_{QP}.\n}\n\\]\nThe influence of the eliminated degrees of freedom is therefore retained in a self-energy that depends on frequency, or equivalently in an exact memory term. For forward evolution the integral is causal. When the Hamiltonian depends on time, the corresponding propagator between two times replaces the stationary exponential used here. This Volterra derivation is not a separate entry in the finite Lean register.","&-i\\int_0^t\nK(t-s)x(s)\\,ds,\n\\end{aligned}\n}\n\\]\nwith memory kernel\n\\[\n\\boxed{\nK(t)=H_{PQ}e^{-itH_{QQ}}H_{QP}.\n}\n\\]\nThe influence of the eliminated degrees of freedom is therefore retained in a self-energy that depends on frequency, or equivalently in an exact memory term. For forward evolution the integral is causal. When the Hamiltonian depends on time, the corresponding propagator between two times replaces the stationary exponential used here. This Volterra derivation is not a separate entry in the finite Lean register.","At resolution $\\ell$, define\n\\[\n\\boxed{\n\\mathfrak Q_\\ell\n=\n\\left(\n\\cA_\\ell,\n\\cH_\\ell,\nD_\\ell,\n\\{P_{\\ell,\\alpha}\\},\nE_\\ell,\nC_\\ell,\n\\Pi_\\ell^{\\mathrm{phys}},\nT_\\ell\n\\right).\n}\n\\]\nIn order, the entries specify the observable algebra, Hilbert space, geometric or Dirac-type operator, packet projectors, embedding into the fine space, reconstruction onto the coarse space, physical constraint projector, and transfer operator. The maps have types $E_\\ell:\\cH_\\ell\\to\\cH_{\\ell+1}$ and $C_\\ell:\\cH_{\\ell+1}\\to\\cH_\\ell$. The left inverse relation $C_\\ell E_\\ell=I$ alone does not imply $C_\\ell=E_\\ell^*$ or make the embedding isometric.","hfrak Q_\\ell\n=\n\\left(\n\\cA_\\ell,\n\\cH_\\ell,\nD_\\ell,\n\\{P_{\\ell,\\alpha}\\},\nE_\\ell,\nC_\\ell,\n\\Pi_\\ell^{\\mathrm{phys}},\nT_\\ell\n\\right).\n}\n\\]\nIn order, the entries specify the observable algebra, Hilbert space, geometric or Dirac-type operator, packet projectors, embedding into the fine space, reconstruction onto the coarse space, physical constraint projector, and transfer operator. The maps have types $E_\\ell:\\cH_\\ell\\to\\cH_{\\ell+1}$ and $C_\\ell:\\cH_{\\ell+1}\\to\\cH_\\ell$. The left inverse relation $C_\\ell E_\\ell=I$ alone does not imply $C_\\ell=E_\\ell^*$ or make the embedding isometric.","ell,\nD_\\ell,\n\\{P_{\\ell,\\alpha}\\},\nE_\\ell,\nC_\\ell,\n\\Pi_\\ell^{\\mathrm{phys}},\nT_\\ell\n\\right).\n}\n\\]\nIn order, the entries specify the observable algebra, Hilbert space, geometric or Dirac-type operator, packet projectors, embedding into the fine space, reconstruction onto the coarse space, physical constraint projector, and transfer operator. The maps have types $E_\\ell:\\cH_\\ell\\to\\cH_{\\ell+1}$ and $C_\\ell:\\cH_{\\ell+1}\\to\\cH_\\ell$. The left inverse relation $C_\\ell E_\\ell=I$ alone does not imply $C_\\ell=E_\\ell^*$ or make the embedding isometric.","\\mathrm{phys}},\nT_\\ell\n\\right).\n}\n\\]\nIn order, the entries specify the observable algebra, Hilbert space, geometric or Dirac-type operator, packet projectors, embedding into the fine space, reconstruction onto the coarse space, physical constraint projector, and transfer operator. The maps have types $E_\\ell:\\cH_\\ell\\to\\cH_{\\ell+1}$ and $C_\\ell:\\cH_{\\ell+1}\\to\\cH_\\ell$. The left inverse relation $C_\\ell E_\\ell=I$ alone does not imply $C_\\ell=E_\\ell^*$ or make the embedding isometric.","order, the entries specify the observable algebra, Hilbert space, geometric or Dirac-type operator, packet projectors, embedding into the fine space, reconstruction onto the coarse space, physical constraint projector, and transfer operator. The maps have types $E_\\ell:\\cH_\\ell\\to\\cH_{\\ell+1}$ and $C_\\ell:\\cH_{\\ell+1}\\to\\cH_\\ell$. The left inverse relation $C_\\ell E_\\ell=I$ alone does not imply $C_\\ell=E_\\ell^*$ or make the embedding isometric.","A coherent refinement should satisfy\n\\[\nC_\\ell E_\\ell=I\n\\]\nand\n\\[\n\\boxed{\nP_{\\ell,\\alpha}\n=\nC_\\ell\n\\left(\n\\sum_{\\beta\\succ\\alpha}P_{\\ell+1,\\beta}\n\\right)\nE_\\ell.\n}\n\\]\nFor an operator $D$, define the refinement defect\n\\[\n\\Delta_\\ell(D)\n=C_\\ell D_{\\ell+1}E_\\ell-D_\\ell.\n\\]\nExact resolution independence is $\\Delta_\\ell(D)=0$. When exact equality is too strong, the correct continuum problem is to prove convergence of the resolvents and uniform control of the effective self-energies.","\\begin{theorem}[Finite Euclidean transfer]\\label{core:theorem:014}\\label{core:thm:euclidean}\nLet $H$ be Hermitian on a nonzero finite-dimensional Hilbert space. For all real $\\tau,\\sigma\\ge0$,\n\\[\nT(\\tau)=e^{-\\tau(H-E_0I)},\n\\qquad\nE_0=\\min\\Spec H,\n\\]\nis positive definite and satisfies\n\\[\n\\boxed{\nT(\\tau+\\sigma)=T(\\tau)T(\\sigma).\n}\n\\]\nIts spectrum is contained in $(0,1]$.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:018}\nDiagonalize $H=\\sum_jE_jP_j$. Then\n\\[\nT(\\tau)=\\sum_j e^{-\\tau(E_j-E_0)}P_j.\n\\]\nEvery eigenvalue is strictly positive and at most one. Multiplication adds the exponents, proving the semigroup law and $T(0)=I$. Positivity here is operator positivity, not an assertion that every matrix entry is nonnegative.\n\\phantomsection\\label{core:proof:018:end}\\end{proof}","The paper tests a controlled volume-state model\n\\[\nH_{nn}=2\\kappa+V(n),\n\\qquad\nH_{n,n+1}=H_{n+1,n}=-\\kappa,\n\\]\nwith $\\kappa>0$ and a real confining potential. This is a finite transfer-matrix test only. Operator positivity alone is not reflection positivity of a Euclidean field measure. It is not derived from a gravitational path integral and is not evidence for a unique microscopic spacetime dynamics.","\\begin{certificate}[Finite transfer model]\\label{core:certificate:010}\nThe minimum transfer eigenvalue was $0.0767054$, the semigroup residual was $0$, and reconstruction from four volume-residue packets had residual $0$.\n\\end{certificate}","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior a","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptot","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conc","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate wi","Let $L\\ge0$ be a finite Laplacian on $N\\ge1$ states, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate window and a controlled refinement limit.","tes, with eigenvalues $\\lambda_j$ and at least one zero eigenvalue. For diffusion time $\\tau>0$, define the normalized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate window and a controlled refinement limit.","lized heat trace\n\\[\nK(\\tau)=\\frac1N\\sum_j e^{-\\tau\\lambda_j}\n\\]\nand the scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate window and a controlled refinement limit.","e scale-dependent spectral dimension\n\\[\n\\boxed{\nd_s(\\tau)\n=-2\\frac{d\\log K(\\tau)}{d\\log\\tau}.\n}\n\\]\nFor a finite system, $d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate window and a controlled refinement limit.","$d_s(\\tau)$ is a diagnostic, not a topological invariant. Indeed, finite differentiation gives\n\\[\nd_s(\\tau)=2\\tau\\,\n\\frac{\\sum_j\\lambda_j e^{-\\tau\\lambda_j}}{\\sum_j e^{-\\tau\\lambda_j}}.\n\\]\nIt tends to zero as $\\tau\\downarrow0$ and, under the stated assumption of a zero mode, as $\\tau\\to\\infty$. If a general positive operator has no zero mode, its behavior at large times is instead asymptotic to $2\\tau\\lambda_{\\min}$. The conclusion for operators with a zero mode therefore does not apply to that case. Inferring a geometric dimension requires a stable intermediate window and a controlled refinement limit.","For a periodic grid with spacing $h>0$ and physical wave number $k\\in\\bbR$,\n\\[\n\\lambda_{k,h}\n=\n\\frac4{h^2}\\sin^2\\left(\\frac{kh}{2}\\right).\n\\]","\\begin{theorem}[Low-mode second-order bound]\\label{core:theorem:015}\\label{core:thm:lowmode}\nFor every real $k$ and every nonzero real $h$, define $\\lambda_{k,h}=4h^{-2}\\sin^2(kh/2)$. Then\n\\[\n\\boxed{\n0\\le k^2-\\lambda_{k,h}\n\\le\n\\frac{k^4h^2}{12}.\n}\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{core:proof:019}\nPut $x=kh/2$. The inequalities $|\\sin x|\\le|x|$ and $|x-\\sin x|\\le |x|^3/6$, which hold for every real argument, give\n\\[\n0\\le x^2-\\sin^2x\n=|x-\\sin x|\\,|x+\\sin x|\n\\le\\frac{|x|^3}{6}\\,2|x|\n=\\frac{x^4}{3}.\n\\]\nMultiplying by $4/h^2$ proves both bounds. The estimate holds for every frequency. Its interpretation as convergence of second order keeps $k$ fixed while $h\\to0$, so it does not assert a uniformly small relative error across all grid modes.\n\\phantomsection\\label{core:proof:019:end}\\end{proof}","\\phantomsection\\label{core:proof:019}\nPut $x=kh/2$. The inequalities $|\\sin x|\\le|x|$ and $|x-\\sin x|\\le |x|^3/6$, which hold for every real argument, give\n\\[\n0\\le x^2-\\sin^2x\n=|x-\\sin x|\\,|x+\\sin x|\n\\le\\frac{|x|^3}{6}\\,2|x|\n=\\frac{x^4}{3}.\n\\]\nMultiplying by $4/h^2$ proves both bounds. The estimate holds for every frequency. Its interpretation as convergence of second order keeps $k$ fixed while $h\\to0$, so it does not assert a uniformly small relative error across all grid modes.\n\\phantomsection\\label{core:proof:019:end}\\end{proof}","\\begin{certificate}[Continuum dispersion]\\label{core:certificate:011}\nThe reported convergence order was $1.99936$. For every tested mode, the error was below the bound, so none of those tests violated the theorem.\n\\end{certificate}","\\begin{certificate}[Spectral dimension]\\label{core:certificate:012}\nThe median intermediate plateaus were\n\\[\nd_s^{(1)}=1.00062,\n\\qquad\nd_s^{(2)}=2.03781.\n\\]\nThese tests recover the expected dimensions of the chosen finite geometries. They do not select a physical theory of quantum spacetime.\n\\end{certificate}","At one finite level, let $\\widehat C_A$ be a finite family of matrix constraints, with fixed scalar structure coefficients, and let $\\Pi^{\\mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This","At one finite level, let $\\widehat C_A$ be a finite family of matrix constraints, with fixed scalar structure coefficients, and let $\\Pi^{\\mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression","At one finite level, let $\\widehat C_A$ be a finite family of matrix constraints, with fixed scalar structure coefficients, and let $\\Pi^{\\mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0","At one finite level, let $\\widehat C_A$ be a finite family of matrix constraints, with fixed scalar structure coefficients, and let $\\Pi^{\\mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is n…","ite family of matrix constraints, with fixed scalar structure coefficients, and let $\\Pi^{\\mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficient…","mathrm{phys}}$ be the orthogonal projector onto their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","o their exact common kernel. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","l. If the classical constraint algebra has structure coefficients $f_{AB}^{\\ \\ C}$, define the anomaly operator\n\\[\n\\boxed{\n\\mathfrak A_{AB}\n=\n[\\widehat C_A,\\widehat C_B]\n-i\\sum_C f_{AB}^{\\ \\ C}\\widehat C_C.\n}\n\\]\nIts physical packet is\n\\[\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}.\n\\]\nIn this exact finite setting, $\\widehat C_C\\Pi^{\\mathrm{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","{phys}}=0$ for every $C$. Associativity gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","ty gives\n\\[\n\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0,\n\\qquad\n\\Pi^{\\mathrm{phys}}\\mathfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","hfrak A_{AB}\\Pi^{\\mathrm{phys}}=0\n\\]\nautomatically. This compression therefore cannot certify operator closure away from the physical range. A nonzero measured compression contradicts the exact-kernel assumptions or indicates an approximation or implementation error. For example, $C_1=\\operatorname{diag}(1,0,0)$ and $C_2=E_{12}+E_{21}$ annihilate the same third coordinate, while their commutator $E_{12}-E_{21}$ is nonzero. With zero proposed structure coefficients, the anomaly is nonzero but its physical compression vanishes.","\\begin{conjecture}[Anomaly-free refinement]\\label{core:conjecture:001}\nFor a specified nontrivial quantum-geometric constraint model, there exists a coherent packet tower with nonzero physical spaces, compatible physical projectors and exact operator closure\n\\[\n\\mathfrak A^{(\\ell)}_{AB}=0\n\\]\nat every finite level. The compatible physical compression then vanishes as a consequence, not as a substitute for closure.\n\\end{conjecture}\nThe substantive research target is the operator condition. The earlier condition stated only in terms of compression was too weak. A choice of model, dynamics, and refinement maps is still needed before this program becomes a fully specified physical existence problem.","\\begin{conjecture}[Einstein infrared limit]\\label{core:conjecture:004}\nThere exist semiclassical packet states $\\Psi_{g,\\ell}$ such that the effective action converges to\n\\[\n\\frac1{16\\pi G}\n\\int\\sqrt{-g}\\,(R-2\\Lambda)\\,d^4x\n\\]\nwith controlled higher-order corrections.\n\\end{conjecture}","\\midrule\nBell-state residual & $0$","Three-qubit seed-orbit rank & $8/8$","Parity off-block norm & $0$","Circuit path reconstruction residual & $0$","Exact-symmetry maximum leakage & $0$","Maximum leakage-bound violation & $0$","Feshbach projected-resolvent residual & $3.2291\\times10^{-17}$","Bare-truncation projected-resolvent residual & $1.3091\\times10^{-2}$","Moving-frame residual with connection & $1.0645\\times10^{-10}$","Moving-frame residual without connection & $7.0000\\times10^{-1}$","Canonical packet-connection residual & $0$","Bloch-curvature residual & $2.0803\\times10^{-11}$","Kraus completeness residual & $0$","Output trace residual & $0$","Knill--Laflamme residual & $0$","Bit-flip recovery residual & $1.0385\\times10^{-16}$","Tensor parity-refinement residual & $0$","Circle low-mode convergence order & $1.99936$","One-dimensional spectral plateau & $1.00062$","Two-dimensional spectral plateau & $2.03781$","Minimum Euclidean-transfer eigenvalue & $0.0767054$","Transfer semigroup residual & $0$","Volume-packet reconstruction residual & $0$","The original reported count is\n\\[\n\\boxed{22/22\\text{ historical numerical tests, not rerun here}.}\n\\]\nThe 19 original figures are retained from the recovered numerical companion. The separate visual guide supplies a reading route through these constructions and the later strengthened results.","The finite theory therefore provides a coherent language for\n\\[\n\\boxed{\n\\begin{gathered}\n\\text{seed reachability}\n+\n\\text{symmetry resolution}\n+\n\\text{circuit paths}\\\\\n+\n\\text{effective dynamics}\n+\n\\text{geometric transport}\n+\n\\text{constraint diagnostics}\n\\end{gathered}\n}\n\\]\nA physical continuum theorem remains open. The next required results are anomaly-free refinement, uniform control of effective operators, Lorentzian reconstruction, the Einstein infrared limit, and a positive massless spin-two sector. Each requires its own hypotheses and proof. Agreement in a finite numerical model cannot replace those arguments.","The elementary coordinate object is a complete family of orthogonal projectors\n\\[\nP_aP_b=\\delta_{ab}P_a,\n\\qquad\n\\sum_{a\\in I}P_a=I.\n\\]\nEvery vector has packet coordinates \\(P_a\\psi\\). Every operator has transfer blocks\n\\[\nA_{ba}=P_bAP_a.\n\\]\nThese identities are simple. Their value appears when the same coordinates are used across representation theory, circuits, channels, error correction, effective dynamics, and geometric limits.","ative to a declared observable algebra. For an algebra in multiplicity form, the local orbit dimension is controlled by the rank of a coefficient matrix. This recovers the relation between Schmidt rank and the dimension reached by local operations. The finite group section then constructs central isotypic projectors. Cyclic symmetry gives a sharper charge selection rule. An operator with charge \\(q\\) can move a state only from packet \\(j\\) to packet \\(j+q\\).","For an algebra in multiplicity form, the local orbit dimension is controlled by the rank of a coefficient matrix. This recovers the relation between Schmidt rank and the dimension reached by local operations. The finite group section then constructs central isotypic projectors. Cyclic symmetry gives a sharper charge selection rule. An operator with charge \\(q\\) can move a state only from packet \\(j\\) to packet \\(j+q\\).","n multiplicity form, the local orbit dimension is controlled by the rank of a coefficient matrix. This recovers the relation between Schmidt rank and the dimension reached by local operations. The finite group section then constructs central isotypic projectors. Cyclic symmetry gives a sharper charge selection rule. An operator with charge \\(q\\) can move a state only from packet \\(j\\) to packet \\(j+q\\).","The moving packet section treats time dependent projectors. The canonical generator\n\\[\nK(t)=\\sum_a\\dot P_a(t)P_a(t)\n\\]\nis skew adjoint and transports every packet. In the moving frame, the physical Hamiltonian acquires the geometric term \\(-\\ii W(t)^*\\dot W(t)\\). The leakage estimate must account for the motion of the packet itself. The physical coupling \\(QHP\\) is therefore replaced by\n\\[\nQ(t)\\bigl(H(t)-\\ii K(t)\\bigr)P(t).\n\\]\nThis is the exact finite distinction between following a moving subspace and remaining in a fixed one.","Density matrices require separate packet labels for their output and input coordinates. The superoperators\n\\[\n\\Pcal_{ab}(X)=P_aXP_b\n\\]\nform a complete orthogonal packet system on Hilbert Schmidt operator space. Dephasing is the diagonal projection. Every channel has input and output superblocks, with the same composition law as ordinary matrix blocks. A covariant channel preserves the isotypic packets of the conjugation action. In the cyclic case these are coherence orders. The Choi matrix carries the equivalent symmetry under \\(U\\otimes\\overline U\\).","form a complete orthogonal packet system on Hilbert Schmidt operator space. Dephasing is the diagonal projection. Every channel has input and output superblocks, with the same composition law as ordinary matrix blocks. A covariant channel preserves the isotypic packets of the conjugation action. In the cyclic case these are coherence orders. The Choi matrix carries the equivalent symmetry under \\(U\\otimes\\overline U\\).","Error correction is expressed through compression to a code. Let \\(P\\) be the code projector. The class of \\(PAP\\) modulo scalar multiples of \\(P\\) measures the obstruction to making \\(A\\) undetectable within the code. The Knill Laflamme criterion says that this obstruction vanishes for every product \\(E_a^*E_b\\) from the error family. Diagonalizing the resulting positive coefficient matrix produces orthogonal syndrome packets and an explicit recovery on the error span.","Constraint closure needs a different obstruction. If \\(C_a\\) are self adjoint constraints and \\(P_{\\phys}\\) projects onto their common kernel, then\n\\[\nP_{\\phys}[C_a,C_b]P_{\\phys}=0\n\\]\nholds automatically. This compression cannot test closure. The correct finite obstruction is the class of \\([C_a,C_b]\\) in the quotient of \\(\\End(\\Hcal)\\) by the span of the constraints. Vanishing means that the constraint span is closed under commutators.","\\midrule\nState & \\(P_a\\psi\\) & completeness of the projector family","Operator & \\(P_bAP_a\\) & block multiplication","Density matrix & \\(P_a\\rho P_b\\) & two sided operator packets","\\begin{definition}[Finite quantum seed system]\\label{strong:definition:001}\\label{strong:def:quantum_seed}\nA finite quantum seed system is a triple\n\\[\n(\\Hcal,\\Acal,S),\n\\]\nwhere \\(\\Hcal\\) is a finite dimensional Hilbert space, \\(\\Acal\\subseteq\\End(\\Hcal)\\) is a unital star subalgebra, and \\(S\\subseteq\\Hcal\\) is a finite set. Its generated hull is\n\\[\n\\Acal S\n=\n\\operatorname{span}\\{A\\psi:A\\in\\Acal,\\ \\psi\\in S\\}.\n\\]\nThe seed set is cyclic when \\(\\Acal S=\\Hcal\\). The quantum seed rank is the least cardinality of a cyclic seed set.\n\\end{definition}","\\label{strong:definition:001}\\label{strong:def:quantum_seed}\nA finite quantum seed system is a triple\n\\[\n(\\Hcal,\\Acal,S),\n\\]\nwhere \\(\\Hcal\\) is a finite dimensional Hilbert space, \\(\\Acal\\subseteq\\End(\\Hcal)\\) is a unital star subalgebra, and \\(S\\subseteq\\Hcal\\) is a finite set. Its generated hull is\n\\[\n\\Acal S\n=\n\\operatorname{span}\\{A\\psi:A\\in\\Acal,\\ \\psi\\in S\\}.\n\\]\nThe seed set is cyclic when \\(\\Acal S=\\Hcal\\). The quantum seed rank is the least cardinality of a cyclic seed set.\n\\end{definition}","The algebra is part of the data. A vector may be cyclic for \\(\\End(\\Hcal)\\) and fail to be cyclic for a restricted observable or gate algebra.","Let\n\\[\n\\Acal\n\\cong\n\\bigoplus_{r=1}^{s}\n\\bigl(M_{d_r}(\\CC)\\otimes I_{m_r}\\bigr)\n\\]\nact on\n\\[\n\\Hcal\n\\cong\n\\bigoplus_{r=1}^{s}\n\\bigl(\\CC^{d_r}\\otimes\\CC^{m_r}\\bigr).\n\\]\nAssume \\(d_r,m_r\\ge1\\) for each displayed block. For an empty block family, the seed rank and the displayed finite supremum are defined as zero. A vector in one block is identified with a \\(d_r\\) by \\(m_r\\) coefficient matrix.","\\begin{theorem}[Orbit dimension of one seed]\\label{strong:theorem:001}\\label{strong:thm:orbit_dimension}\nFor integers \\(d\\ge1\\), \\(m\\ge0\\), and the action of \\(M_d(\\CC)\\otimes I_m\\) on \\(\\CC^d\\otimes\\CC^m\\), let \\(C_\\psi\\) be the coefficient matrix of \\(\\psi\\). Then\n\\[\n\\dim\\operatorname{span}\\{(A\\otimes I_m)\\psi:A\\in M_d(\\CC)\\}\n=\n d\\rank C_\\psi.\n\\]\nThe vector is cyclic exactly when \\(\\rank C_\\psi=m\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:001}\nUnder the coefficient matrix identification, \\((A\\otimes I_m)\\psi\\) corresponds to \\(AC_\\psi\\). Every row of \\(AC_\\psi\\) lies in the row space of \\(C_\\psi\\). Conversely, a matrix whose rows lie in that row space equals \\(AC_\\psi\\) for a suitable \\(A\\). There are \\(d\\) independent row choices in a space of dimension \\(\\rank C_\\psi\\). The orbit dimension is therefore \\(d\\rank C_\\psi\\). It equals \\(dm\\) exactly when \\(C_\\psi\\) has full column rank.\n\\phantomsection\\label{strong:proof:001:end}\\end{proof}","ction\\label{strong:proof:001}\nUnder the coefficient matrix identification, \\((A\\otimes I_m)\\psi\\) corresponds to \\(AC_\\psi\\). Every row of \\(AC_\\psi\\) lies in the row space of \\(C_\\psi\\). Conversely, a matrix whose rows lie in that row space equals \\(AC_\\psi\\) for a suitable \\(A\\). There are \\(d\\) independent row choices in a space of dimension \\(\\rank C_\\psi\\). The orbit dimension is therefore \\(d\\rank C_\\psi\\). It equals \\(dm\\) exactly when \\(C_\\psi\\) has full column rank.\n\\phantomsection\\label{strong:proof:001:end}\\end{proof}","Under the coefficient matrix identification, \\((A\\otimes I_m)\\psi\\) corresponds to \\(AC_\\psi\\). Every row of \\(AC_\\psi\\) lies in the row space of \\(C_\\psi\\). Conversely, a matrix whose rows lie in that row space equals \\(AC_\\psi\\) for a suitable \\(A\\). There are \\(d\\) independent row choices in a space of dimension \\(\\rank C_\\psi\\). The orbit dimension is therefore \\(d\\rank C_\\psi\\). It equals \\(dm\\) exactly when \\(C_\\psi\\) has full column rank.\n\\phantomsection\\label{strong:proof:001:end}\\end{proof}","nt matrix identification, \\((A\\otimes I_m)\\psi\\) corresponds to \\(AC_\\psi\\). Every row of \\(AC_\\psi\\) lies in the row space of \\(C_\\psi\\). Conversely, a matrix whose rows lie in that row space equals \\(AC_\\psi\\) for a suitable \\(A\\). There are \\(d\\) independent row choices in a space of dimension \\(\\rank C_\\psi\\). The orbit dimension is therefore \\(d\\rank C_\\psi\\). It equals \\(dm\\) exactly when \\(C_\\psi\\) has full column rank.\n\\phantomsection\\label{strong:proof:001:end}\\end{proof}","\\begin{theorem}[Quantum seed rank]\\label{strong:theorem:002}\\label{strong:thm:quantum_seed_rank}\nFor the displayed multiplicity decomposition,\n\\[\n\\operatorname{srank}_{\\Acal}(\\Hcal)\n=\n\\max_{1\\le r\\le s}\n\\left\\lceil\\frac{m_r}{d_r}\\right\\rceil.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:002}\nA single seed contributes a row space of dimension at most \\(d_r\\) to the \\(r\\)th multiplicity factor. Hence \\(k\\) seeds cover at most \\(kd_r\\) multiplicity directions, which gives the lower bound.","Set\n\\[\nk=\\max_r\\left\\lceil\\frac{m_r}{d_r}\\right\\rceil.\n\\]\nFor each \\(r\\), partition a basis of \\(\\CC^{m_r}\\) into at most \\(k\\) groups of size at most \\(d_r\\). Choose the coefficient matrix of the \\(j\\)th seed so that its row space spans the \\(j\\)th group. The combined row spaces span every multiplicity factor. The preceding theorem then shows that the \\(k\\) seeds generate every summand.\n\\phantomsection\\label{strong:proof:002:end}\\end{proof}","\\begin{corollary}[Local orbit and Schmidt rank]\\label{strong:corollary:001}\\label{strong:cor:schmidt_orbit}\nLet \\(\\psi\\in\\Hcal_A\\otimes\\Hcal_B\\), let \\(d_A=\\dim\\Hcal_A\\ge1\\), and let \\(r\\) be the Schmidt rank of \\(\\psi\\). Then\n\\[\n\\dim\\operatorname{span}\\{(A\\otimes I_B)\\psi:A\\in\\End(\\Hcal_A)\\}\n=\n d_A r.\n\\]\nLocal operators on \\(A\\) generate the whole bipartite Hilbert space exactly when the reduced state on \\(B\\) has full support.\n\\end{corollary}","llary}[Local orbit and Schmidt rank]\\label{strong:corollary:001}\\label{strong:cor:schmidt_orbit}\nLet \\(\\psi\\in\\Hcal_A\\otimes\\Hcal_B\\), let \\(d_A=\\dim\\Hcal_A\\ge1\\), and let \\(r\\) be the Schmidt rank of \\(\\psi\\). Then\n\\[\n\\dim\\operatorname{span}\\{(A\\otimes I_B)\\psi:A\\in\\End(\\Hcal_A)\\}\n=\n d_A r.\n\\]\nLocal operators on \\(A\\) generate the whole bipartite Hilbert space exactly when the reduced state on \\(B\\) has full support.\n\\end{corollary}","\\begin{proposition}[Computational address orbit]\\label{strong:proposition:001}\\label{strong:prop:computational_orbit}\nFor \\(n\\) qubits, let\n\\[\n\\psi_0=\\ket{0}^{\\otimes n}\n\\]\nand let \\(X_j\\) be the Pauli \\(X\\) operator on the \\(j\\)th qubit. Then\n\\[\n\\left\\{\nX_1^{a_1}\\cdots X_n^{a_n}\\psi_0:\n(a_1,\\ldots,a_n)\\in\\{0,1\\}^n\n\\right\\}\n\\]\nis the computational basis.\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:004}\nThe addressed vector is \\(\\ket{a_1\\cdots a_n}\\). Distinct binary addresses produce the \\(2^n\\) distinct basis vectors.\n\\phantomsection\\label{strong:proof:004:end}\\end{proof}","Let \\(G\\) be a finite group and let\n\\[\nU:G\\longrightarrow\\Ucal(\\Hcal)\n\\]\nbe a unitary representation. Write \\(\\widehat G\\) for the set of irreducible unitary representations. For \\(\\lambda\\in\\widehat G\\), let \\(d_\\lambda\\) and \\(\\chi_\\lambda\\) denote its dimension and character.","\\begin{theorem}[Finite group isotypic projectors]\\label{strong:theorem:003}\\label{strong:thm:isotypic_projectors}\nThe operators\n\\[\nP_\\lambda\n=\n\\frac{d_\\lambda}{\\abs{G}}\n\\sum_{g\\in G}\n\\overline{\\chi_\\lambda(g)}U(g)\n\\]\nare orthogonal projectors. They satisfy\n\\[\nP_\\lambda P_\\mu=\\delta_{\\lambda\\mu}P_\\lambda,\n\\qquad\n\\sum_{\\lambda\\in\\widehat G}P_\\lambda=I_{\\Hcal}.\n\\]\nThe range of \\(P_\\lambda\\) is the \\(\\lambda\\) isotypic component of \\(\\Hcal\\).\n\\end{theorem}","pic projectors]\\label{strong:theorem:003}\\label{strong:thm:isotypic_projectors}\nThe operators\n\\[\nP_\\lambda\n=\n\\frac{d_\\lambda}{\\abs{G}}\n\\sum_{g\\in G}\n\\overline{\\chi_\\lambda(g)}U(g)\n\\]\nare orthogonal projectors. They satisfy\n\\[\nP_\\lambda P_\\mu=\\delta_{\\lambda\\mu}P_\\lambda,\n\\qquad\n\\sum_{\\lambda\\in\\widehat G}P_\\lambda=I_{\\Hcal}.\n\\]\nThe range of \\(P_\\lambda\\) is the \\(\\lambda\\) isotypic component of \\(\\Hcal\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:005}\nComplete reducibility gives a unitary decomposition\n\\[\n\\Hcal\n\\cong\n\\bigoplus_{\\mu\\in\\widehat G}\nV_\\mu\\otimes M_\\mu,\n\\qquad\nU(g)\n\\cong\n\\bigoplus_\\mu\n\\pi_\\mu(g)\\otimes I_{M_\\mu}.\n\\]\nOn the \\(\\mu\\) summand, the displayed average is central in the irreducible representation \\(\\pi_\\mu\\), so it is a scalar by Schur's lemma. Taking its trace and applying character orthogonality shows that the scalar is one when \\(\\mu=\\lambda\\) and zero otherwise. The asserted identities follow on the direct sum.\n\\phantomsection\\label{strong:proof:005:end}\\end{proof}","educibility gives a unitary decomposition\n\\[\n\\Hcal\n\\cong\n\\bigoplus_{\\mu\\in\\widehat G}\nV_\\mu\\otimes M_\\mu,\n\\qquad\nU(g)\n\\cong\n\\bigoplus_\\mu\n\\pi_\\mu(g)\\otimes I_{M_\\mu}.\n\\]\nOn the \\(\\mu\\) summand, the displayed average is central in the irreducible representation \\(\\pi_\\mu\\), so it is a scalar by Schur's lemma. Taking its trace and applying character orthogonality shows that the scalar is one when \\(\\mu=\\lambda\\) and zero otherwise. The asserted identities follow on the direct sum.\n\\phantomsection\\label{strong:proof:005:end}\\end{proof}","\\begin{theorem}[Representation matrix units]\\label{strong:theorem:004}\\label{strong:thm:representation_matrix_units}\nFix an irreducible unitary representation \\(\\pi_\\lambda:G\\to U(d_\\lambda)\\) and an orthonormal basis of its representation space. Define\n\\[\nE_{ij}^{\\lambda}\n=\n\\frac{d_\\lambda}{\\abs{G}}\n\\sum_{g\\in G}\n\\bigl[\\pi_\\lambda(g^{-1})\\bigr]_{ji}U(g).\n\\]\nThen\n\\[\nE_{ij}^{\\lambda}E_{k\\ell}^{\\mu}\n=\n\\delta_{\\lambda\\mu}\\delta_{jk}E_{i\\ell}^{\\lambda},\n\\qquad\n\\bigl(E_{ij}^{\\lambda}\\bigr)^*=E_{ji}^{\\lambda},\n\\]\nand\n\\[\nP_\\lambda=\\sum_{i=1}^{d_\\lambda}E_{ii}^{\\lambda}.\n\\]\n\\end{theorem}","x an irreducible unitary representation \\(\\pi_\\lambda:G\\to U(d_\\lambda)\\) and an orthonormal basis of its representation space. Define\n\\[\nE_{ij}^{\\lambda}\n=\n\\frac{d_\\lambda}{\\abs{G}}\n\\sum_{g\\in G}\n\\bigl[\\pi_\\lambda(g^{-1})\\bigr]_{ji}U(g).\n\\]\nThen\n\\[\nE_{ij}^{\\lambda}E_{k\\ell}^{\\mu}\n=\n\\delta_{\\lambda\\mu}\\delta_{jk}E_{i\\ell}^{\\lambda},\n\\qquad\n\\bigl(E_{ij}^{\\lambda}\\bigr)^*=E_{ji}^{\\lambda},\n\\]\nand\n\\[\nP_\\lambda=\\sum_{i=1}^{d_\\lambda}E_{ii}^{\\lambda}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:006}\nApply the matrix coefficient form of Schur orthogonality to the two group averages. It gives the matrix unit multiplication law. Unitarity gives the adjoint relation after replacing \\(g\\) by \\(g^{-1}\\). Summing the diagonal coefficients gives the character average in \\cref{strong:thm:isotypic_projectors}.\n\\phantomsection\\label{strong:proof:006:end}\\end{proof}","Let \\(N\\ge1\\) be an integer, \\(S\\) unitary with \\(S^N=I\\), and \\(\\omega=\\e^{2\\pi\\ii/N}\\).","\\begin{corollary}[Cyclic charge projectors]\\label{strong:corollary:002}\\label{strong:cor:cyclic_projectors}\nThe operators\n\\[\nP_j\n=\n\\frac1N\n\\sum_{r=0}^{N-1}\n\\omega^{-jr}S^r,\n\\qquad\nj\\in\\ZZ/N\\ZZ,\n\\]\nform a complete orthogonal packet system and satisfy\n\\[\nSP_j=\\omega^jP_j.\n\\]\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:007}\nApply \\cref{strong:thm:isotypic_projectors} to the cyclic group. Its irreducible representations are the one dimensional characters \\(r\\mapsto\\omega^{jr}\\).\n\\phantomsection\\label{strong:proof:007:end}\\end{proof}","\\begin{definition}[Operator charge]\\label{strong:definition:002}\\label{strong:def:operator_charge}\nAn operator \\(A\\in\\End(\\Hcal)\\) has cyclic charge \\(q\\in\\ZZ/N\\ZZ\\) when\n\\[\nSAS^*=\\omega^qA.\n\\]\n\\end{definition}","\\begin{theorem}[Charge selection rule]\\label{strong:theorem:005}\\label{strong:thm:charge_selection}\nIf \\(A\\) has charge \\(q\\), then\n\\[\nP_kAP_j=0\n\\]\nunless \\(k=j+q\\) in \\(\\ZZ/N\\ZZ\\). In fact,\n\\[\nAP_j=P_{j+q}AP_j.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:008}\nFor \\(x\\in P_j\\Hcal\\),\n\\[\nS(Ax)\n=(SAS^*)Sx\n=\\omega^{q+j}Ax.\n\\]\nThus \\(Ax\\in P_{j+q}\\Hcal\\). The block statement follows.\n\\phantomsection\\label{strong:proof:008:end}\\end{proof}","\\begin{corollary}[Charge addition]\\label{strong:corollary:003}\\label{strong:cor:charge_addition}\nIf \\(A\\) and \\(B\\) have charges \\(q\\) and \\(r\\), then \\(AB\\) has charge \\(q+r\\). The adjoint \\(A^*\\) has charge \\(-q\\).\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:009}\nConjugation by \\(S\\) is a star automorphism. Applying it to \\(AB\\) adds the two charges, while applying it to \\(A^*\\) changes the sign of the charge.\n\\phantomsection\\label{strong:proof:009:end}\\end{proof}","Suppose the \\(a\\)th tensor factor carries a cyclic packet family\n\\[\n\\bigl(P_j^{(a)}\\bigr)_{j\\in\\ZZ/N\\ZZ}.\n\\]\nThe fine tensor packets are\n\\[\nP_{\\bm j}^{\\rm fine}\n=\nP_{j_1}^{(1)}\\otimes\\cdots\\otimes P_{j_m}^{(m)}.\n\\]","\\begin{theorem}[Tensor charge refinement]\\label{strong:theorem:006}\\label{strong:thm:tensor_refinement}\nFor \\(r\\in\\ZZ/N\\ZZ\\), define\n\\[\nP_r^{\\rm total}\n=\n\\sum_{j_1+\\cdots+j_m=r}\nP_{j_1}^{(1)}\\otimes\\cdots\\otimes P_{j_m}^{(m)}.\n\\]\nThen the \\(P_r^{\\rm total}\\) form a complete orthogonal packet system. Every total charge packet is the exact sum of its fine tensor descendants.\n\\end{theorem}","\\begin{corollary}[Tensor selection rule]\\label{strong:corollary:004}\\label{strong:cor:tensor_selection}\nIf \\(A_a\\) has local charge \\(q_a\\), then\n\\[\nA_1\\otimes\\cdots\\otimes A_m\n\\]\nhas total charge \\(q_1+\\cdots+q_m\\).\n\\end{corollary}","Let\n\\[\nU=U_mU_{m-1}\\cdots U_1\n\\]\nbe a finite circuit with \\(m\\ge1\\), and let \\((P_a)_{a\\in I}\\) be a complete orthogonal packet family. For the empty circuit, \\(U=I\\) and its transfer block is \\(P_bP_a=\\delta_{ba}P_a\\), whose endpoint sum reconstructs \\(I\\).","\\begin{theorem}[Exact circuit path expansion]\\label{strong:theorem:007}\\label{strong:thm:circuit_path}\nFor source packet \\(a_0\\) and target packet \\(a_m\\),\n\\[\nP_{a_m}UP_{a_0}\n=\n\\sum_{a_1,\\ldots,a_{m-1}}\nP_{a_m}U_mP_{a_{m-1}}\n\\cdots\nP_{a_1}U_1P_{a_0}.\n\\]\nSumming over \\(a_0\\) and \\(a_m\\) reconstructs \\(U\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:012}\nInsert \\(I=\\sum_aP_a\\) between every adjacent pair of gates and distribute the finite sums. Completeness at the two endpoints gives the full operator.\n\\phantomsection\\label{strong:proof:012:end}\\end{proof}","For a packet path\n\\[\n\\gamma=(a_0,a_1,\\ldots,a_m),\n\\]\nwrite\n\\[\nU_\\gamma\n=\nP_{a_m}U_mP_{a_{m-1}}\n\\cdots\nP_{a_1}U_1P_{a_0}.\n\\]","\\begin{theorem}[Packet unitarity identities]\\label{strong:theorem:008}\\label{strong:thm:packet_unitarity}\nIf \\(U\\) is unitary and \\(U_{ba}=P_bUP_a\\), then\n\\[\n\\sum_bU_{ba}^*U_{bc}=\\delta_{ac}P_a\n\\]\nand\n\\[\n\\sum_aU_{ba}U_{da}^*=\\delta_{bd}P_b.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:013}\nUse completeness and unitarity:\n\\[\n\\sum_bU_{ba}^*U_{bc}\n=\nP_aU^*\\left(\\sum_bP_b\\right)UP_c\n=P_aP_c.\n\\]\nSumming instead over the source packets and using unitarity in the reverse order gives the second identity.\n\\phantomsection\\label{strong:proof:013:end}\\end{proof}","\\begin{corollary}[Probability conservation from one packet]\\label{strong:corollary:005}\\label{strong:cor:packet_probability}\nIf \\(\\psi\\in P_a\\Hcal\\), then\n\\[\n\\sum_b\\norm{P_bU\\psi}^2=\\norm{\\psi}^2.\n\\]\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:014}\nTake the inner product of the first identity in \\cref{strong:thm:packet_unitarity} with \\(\\psi\\) and set \\(c=a\\).\n\\phantomsection\\label{strong:proof:014:end}\\end{proof}","\\begin{theorem}[Path interference formula]\\label{strong:theorem:009}\\label{strong:thm:path_interference}\nFix a source vector \\(\\psi\\in P_{a_0}\\Hcal\\) and a target packet \\(b\\). Let \\(\\Gamma_b\\) be the set of packet paths from \\(a_0\\) to \\(b\\), and set\n\\[\n\\eta_\\gamma=U_\\gamma\\psi.\n\\]\nChoose any total order on this finite path set. Then\n\\[\n\\norm{P_bU\\psi}^2\n=\n\\sum_{\\gamma\\in\\Gamma_b}\\norm{\\eta_\\gamma}^2\n+\n2\\operatorname{Re}\n\\sum_{\\substack{\\gamma,\\gamma'\\in\\Gamma_b","\\begin{theorem}[Path interference formula]\\label{strong:theorem:009}\\label{strong:thm:path_interference}\nFix a source vector \\(\\psi\\in P_{a_0}\\Hcal\\) and a target packet \\(b\\). Let \\(\\Gamma_b\\) be the set of packet paths from \\(a_0\\) to \\(b\\), and set\n\\[\n\\eta_\\gamma=U_\\gamma\\psi.\n\\]\nChoose any total order on this finite path set. Then\n\\[\n\\norm{P_bU\\psi}^2\n=\n\\sum_{\\gamma\\in\\Gamma_b}\\norm{\\eta_\\gamma}^2\n+\n2\\operatorname{Re}\n\\sum_{\\substack{\\gamma,\\gamma'\\in\\Gamma_b\\\\ \\gamma<\\gamma'}}\n\\ip{\\eta_\\gamma}{\\eta_{\\gamma'}}.\n\\]\nConsequently,\n\\[\n\\norm{P_bU\\psi}\n\\leq\n\\sum_{\\gamma\\in\\Gamma_b}\\norm{\\eta_\\gamma}.\n\\]\nIf the path outputs are pairwise orthogonal, the target probability is the sum of the path probabilities.\n\\end{theorem}","\\gamma<\\gamma'}}\n\\ip{\\eta_\\gamma}{\\eta_{\\gamma'}}.\n\\]\nConsequently,\n\\[\n\\norm{P_bU\\psi}\n\\leq\n\\sum_{\\gamma\\in\\Gamma_b}\\norm{\\eta_\\gamma}.\n\\]\nIf the path outputs are pairwise orthogonal, the target probability is the sum of the path probabilities.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:015}\nThe path expansion gives\n\\[\nP_bU\\psi=\\sum_{\\gamma\\in\\Gamma_b}\\eta_\\gamma.\n\\]\nExpanding the squared norm and grouping conjugate cross terms gives the interference formula. The triangle inequality gives the norm bound. When the path outputs are orthogonal, every cross term vanishes and only the individual path probabilities remain.\n\\phantomsection\\label{strong:proof:015:end}\\end{proof}","\\begin{proposition}[Charge pruning of circuit paths]\\label{strong:proposition:002}\\label{strong:prop:charge_pruning}\nSuppose every gate has a finite charge decomposition\n\\[\nU_r=\\sum_qU_r^{(q)},\n\\qquad\nSU_r^{(q)}S^*=\\omega^qU_r^{(q)}.\n\\]\nA term with gate charges \\(q_1,\\ldots,q_m\\) can contribute from source charge \\(a_0\\) only to target charge\n\\[\na_m=a_0+q_1+\\cdots+q_m.\n\\]\nEvery packet path violating one of the intermediate charge updates vanishes.\n\\end{proposition}","\\begin{example}[Bell circuit]\\label{strong:example:001}\\label{strong:ex:bell_circuit}\nStarting from \\(\\ket{00}\\), apply a Hadamard gate to the first qubit and then a controlled NOT gate. The final state is\n\\[\n\\frac{\\ket{00}+\\ket{11}}{\\sqrt2}.\n\\]\nIt lies in the even parity packet. The intermediate state has one even and one odd computational contribution, so the final even amplitude is reconstructed by more than one packet path.\n\\end{example}","Let \\(H=H^*\\) and let \\((P_a)_{a\\in I}\\) be a complete orthogonal packet system.","\\begin{theorem}[Exact invariant packets]\\label{strong:theorem:010}\\label{strong:thm:invariant_packets}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item The Hamiltonian commutes with each projector: \\([H,P_a]=0\\) for every \\(a\\).\n\\item Transfer blocks between distinct packets vanish: \\(P_bHP_a=0\\) whenever \\(a\\neq b\\).\n\\item Every packet \\(P_a\\Hcal\\) is invariant under \\(\\e^{-\\ii tH}\\) for every real \\(t\\).\n\\end{enumerate}\nWhen they hold,\n\\[\nH=\\sum_aP_aHP_a\n\\]\nand\n\\[\n\\e^{-\\ii tH}=\\sum_aP_a\\e^{-\\ii tH}P_a.\n\\]\n\\end{theorem}","invariant packets]\\label{strong:theorem:010}\\label{strong:thm:invariant_packets}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item The Hamiltonian commutes with each projector: \\([H,P_a]=0\\) for every \\(a\\).\n\\item Transfer blocks between distinct packets vanish: \\(P_bHP_a=0\\) whenever \\(a\\neq b\\).\n\\item Every packet \\(P_a\\Hcal\\) is invariant under \\(\\e^{-\\ii tH}\\) for every real \\(t\\).\n\\end{enumerate}\nWhen they hold,\n\\[\nH=\\sum_aP_aHP_a\n\\]\nand\n\\[\n\\e^{-\\ii tH}=\\sum_aP_a\\e^{-\\ii tH}P_a.\n\\]\n\\end{theorem}","m:010}\\label{strong:thm:invariant_packets}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item The Hamiltonian commutes with each projector: \\([H,P_a]=0\\) for every \\(a\\).\n\\item Transfer blocks between distinct packets vanish: \\(P_bHP_a=0\\) whenever \\(a\\neq b\\).\n\\item Every packet \\(P_a\\Hcal\\) is invariant under \\(\\e^{-\\ii tH}\\) for every real \\(t\\).\n\\end{enumerate}\nWhen they hold,\n\\[\nH=\\sum_aP_aHP_a\n\\]\nand\n\\[\n\\e^{-\\ii tH}=\\sum_aP_a\\e^{-\\ii tH}P_a.\n\\]\n\\end{theorem}","variant_packets}\nThe following conditions are equivalent:\n\\begin{enumerate}\n\\item The Hamiltonian commutes with each projector: \\([H,P_a]=0\\) for every \\(a\\).\n\\item Transfer blocks between distinct packets vanish: \\(P_bHP_a=0\\) whenever \\(a\\neq b\\).\n\\item Every packet \\(P_a\\Hcal\\) is invariant under \\(\\e^{-\\ii tH}\\) for every real \\(t\\).\n\\end{enumerate}\nWhen they hold,\n\\[\nH=\\sum_aP_aHP_a\n\\]\nand\n\\[\n\\e^{-\\ii tH}=\\sum_aP_a\\e^{-\\ii tH}P_a.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:017}\nCommutation is equivalent to vanishing of every off diagonal block. The same block decomposition passes to polynomials in \\(H\\), then to the exponential by norm convergence. Invariance of every packet implies invariance under both \\(\\e^{-\\ii tH}\\) and its adjoint. Differentiation at \\(t=0\\) gives commutation with \\(H\\).\n\\phantomsection\\label{strong:proof:017:end}\\end{proof}","Fix one packet projector \\(P\\) and set \\(Q=I-P\\).","\\begin{lemma}[Off diagonal norm]\\label{strong:lemma:001}\\label{strong:lem:off_diagonal_norm}\nFor self adjoint \\(H\\),\n\\[\n\\norm{[H,P]}=\\norm{QHP}=\\norm{PHQ}.\n\\]\n\\end{lemma}","\\begin{proof}\\phantomsection\\label{strong:proof:018}\nRelative to \\(P\\Hcal\\oplus Q\\Hcal\\),\n\\[\n[H,P]\n=\n\\begin{pmatrix}\n0&-PHQ","\\begin{proof}\\phantomsection\\label{strong:proof:018}\nRelative to \\(P\\Hcal\\oplus Q\\Hcal\\),\n\\[\n[H,P]\n=\n\\begin{pmatrix}\n0&-PHQ\\\\\nQHP&0\n\\end{pmatrix}.\n\\]\nIts square has diagonal blocks \\(-PHQHP\\) and \\(-QHPHQ\\). The norm is therefore the largest singular value of \\(QHP\\).\n\\phantomsection\\label{strong:proof:018:end}\\end{proof}","QHP&0\n\\end{pmatrix}.\n\\]\nIts square has diagonal blocks \\(-PHQHP\\) and \\(-QHPHQ\\). The norm is therefore the largest singular value of \\(QHP\\).\n\\phantomsection\\label{strong:proof:018:end}\\end{proof}","\\begin{theorem}[Fixed packet leakage]\\label{strong:theorem:011}\\label{strong:thm:fixed_leakage}\nFor every real \\(t\\),\n\\[\n\\norm{Q\\e^{-\\ii tH}P}\n\\leq\n\\abs{t}\\norm{QHP}\n=\n\\abs{t}\\norm{[H,P]}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:019}\nLet\n\\[\nH_{\\rm d}=PHP+QHQ,\n\\qquad\nH_{\\rm o}=PHQ+QHP.\n\\]\nThe propagator generated by \\(H_{\\rm d}\\) preserves \\(P\\Hcal\\) and \\(Q\\Hcal\\). Duhamel's formula relative to \\(H_{\\rm d}\\) gives\n\\[\nQ\\e^{-\\ii tH}P\n=\n-\\ii\n\\int_0^t\nQ\\e^{-\\ii(t-s)H}H_{\\rm o}\\e^{-\\ii sH_{\\rm d}}P\n\\dd s.\n\\]\nSince \\(H_{\\rm o}P=QHP\\) and both propagators are unitary, the norm is at most \\(\\abs{t}\\norm{QHP}\\). Use \\cref{strong:lem:off_diagonal_norm} for the last equality.\n\\phantomsection\\label{strong:proof:019:end}\\end{proof}","oof}\\phantomsection\\label{strong:proof:019}\nLet\n\\[\nH_{\\rm d}=PHP+QHQ,\n\\qquad\nH_{\\rm o}=PHQ+QHP.\n\\]\nThe propagator generated by \\(H_{\\rm d}\\) preserves \\(P\\Hcal\\) and \\(Q\\Hcal\\). Duhamel's formula relative to \\(H_{\\rm d}\\) gives\n\\[\nQ\\e^{-\\ii tH}P\n=\n-\\ii\n\\int_0^t\nQ\\e^{-\\ii(t-s)H}H_{\\rm o}\\e^{-\\ii sH_{\\rm d}}P\n\\dd s.\n\\]\nSince \\(H_{\\rm o}P=QHP\\) and both propagators are unitary, the norm is at most \\(\\abs{t}\\norm{QHP}\\). Use \\cref{strong:lem:off_diagonal_norm} for the last equality.\n\\phantomsection\\label{strong:proof:019:end}\\end{proof}","\\begin{corollary}[Leakage probability]\\label{strong:corollary:006}\\label{strong:cor:leakage_probability}\nIf \\(\\psi\\in P\\Hcal\\) is a unit vector, then\n\\[\n\\norm{Q\\e^{-\\ii tH}\\psi}^2\n\\leq\n t^2\\norm{QHP}^2.\n\\]\n\\end{corollary}","Let \\((P_a(t))_{a\\in I}\\) be a twice continuously differentiable family of complete orthogonal packet systems on a finite dimensional Hilbert space.","\\begin{theorem}[Canonical packet connection]\\label{strong:theorem:012}\\label{strong:thm:kato_connection}\nDefine\n\\[\nK(t)\n=\n\\sum_a\\dot P_a(t)P_a(t).\n\\]\nThen\n\\[\nK(t)^*=-K(t)\n\\]\nand\n\\[\n[K(t),P_a(t)]=\\dot P_a(t)\n\\]\nfor every \\(a\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:021}\nDifferentiating \\(P_a^2=P_a\\) gives\n\\[\n\\dot P_aP_a+P_a\\dot P_a=\\dot P_a.\n\\]\nSumming over \\(a\\) and using \\(\\sum_a\\dot P_a=0\\) yields\n\\[\n\\sum_aP_a\\dot P_a=-\\sum_a\\dot P_aP_a.\n\\]\nThis proves skew adjointness. For fixed \\(b\\),\n\\[\nKP_b=\\dot P_bP_b.\n\\]\nDifferentiating \\(P_bP_a=0\\) for \\(a\\neq b\\) gives\n\\[\nP_b\\dot P_a=-\\dot P_bP_a.\n\\]\nHence\n\\[\nP_bK=-\\dot P_b(I-P_b).\n\\]\nSubtracting gives \\([K,P_b]=\\dot P_b\\).\n\\phantomsection\\label{strong:proof:021:end}\\end{proof}","ng:proof:021}\nDifferentiating \\(P_a^2=P_a\\) gives\n\\[\n\\dot P_aP_a+P_a\\dot P_a=\\dot P_a.\n\\]\nSumming over \\(a\\) and using \\(\\sum_a\\dot P_a=0\\) yields\n\\[\n\\sum_aP_a\\dot P_a=-\\sum_a\\dot P_aP_a.\n\\]\nThis proves skew adjointness. For fixed \\(b\\),\n\\[\nKP_b=\\dot P_bP_b.\n\\]\nDifferentiating \\(P_bP_a=0\\) for \\(a\\neq b\\) gives\n\\[\nP_b\\dot P_a=-\\dot P_bP_a.\n\\]\nHence\n\\[\nP_bK=-\\dot P_b(I-P_b).\n\\]\nSubtracting gives \\([K,P_b]=\\dot P_b\\).\n\\phantomsection\\label{strong:proof:021:end}\\end{proof}","\\begin{theorem}[Exact Kato transport]\\label{strong:theorem:013}\\label{strong:thm:kato_transport}\nLet \\(W(t)\\) solve\n\\[\n\\dot W(t)=K(t)W(t),\n\\qquad\nW(0)=I.\n\\]\nThen \\(W(t)\\) is unitary and\n\\[\nP_a(t)W(t)=W(t)P_a(0)\n\\]\nfor every packet \\(a\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:022}\nSkew adjointness gives\n\\[\n\\frac{\\dd}{\\dd t}W(t)^*W(t)=0.\n\\]\nThus \\(W(t)\\) is unitary. Next,\n\\[\n\\frac{\\dd}{\\dd t}\n\\left(W(t)^*P_a(t)W(t)\\right)\n=\nW(t)^*\n\\left(\n\\dot P_a-[K,P_a]\n\\right)\nW(t)=0.\n\\]\nThe expression equals \\(P_a(0)\\) at \\(t=0\\), which proves the intertwining identity.\n\\phantomsection\\label{strong:proof:022:end}\\end{proof}","\\begin{theorem}[Moving frame Hamiltonian]\\label{strong:theorem:014}\\label{strong:thm:moving_frame}\nLet \\(U(t)\\) solve\n\\[\n\\ii\\dot U(t)=H(t)U(t),\n\\qquad\nU(0)=I,\n\\]\nwhere \\(H(t)=H(t)^*\\) is continuous on the compact time interval under consideration. Set\n\\[\nV(t)=W(t)^*U(t).\n\\]\nThen\n\\[\n\\ii\\dot V(t)=H_{\\rm mov}(t)V(t),\n\\]\nwhere\n\\[\nH_{\\rm mov}(t)\n=\nW(t)^*H(t)W(t)-\\ii W(t)^*\\dot W(t)\n=\nW(t)^*\\bigl(H(t)-\\ii K(t)\\bigr)W(t).\n\\]\nThe moving frame Hamiltonian is self adjoint.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:023}\nDifferentiate \\(V=W^*U\\). Since \\(\\dot W=KW\\) and \\(K^*=-K\\),\n\\[\n\\dot W^*=-W^*K.\n\\]\nSubstitution gives the equation. The first term in \\(H_{\\rm mov}\\) is self adjoint. The operator \\(-\\ii W^*\\dot W\\) is self adjoint because \\(W^*\\dot W\\) is skew adjoint.\n\\phantomsection\\label{strong:proof:023:end}\\end{proof}","\\begin{theorem}[Moving packet leakage]\\label{strong:theorem:015}\\label{strong:thm:moving_leakage}\nFix a packet \\(P(t)=P_a(t)\\) and put \\(Q(t)=I-P(t)\\). For \\(0\\leq t\\leq T\\),\n\\[\n\\norm{Q(t)U(t)P(0)}\n\\leq\n\\int_0^t\n\\norm{Q(s)\\bigl(H(s)-\\ii K(s)\\bigr)P(s)}\n\\dd s.\n\\]\nIn particular, the physical evolution follows the moving packet exactly on \\([0,T]\\) when the displayed off packet block vanishes for every time in that interval.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:024}\nIn the Kato frame, \\(P(0)\\) is fixed and\n\\[\nQ(t)U(t)P(0)\n=\nW(t)Q(0)V(t)P(0).\n\\]\nApply the time dependent version of the Duhamel argument in \\cref{strong:thm:fixed_leakage} to \\(H_{\\rm mov}(t)\\). The off diagonal block is\n\\[\nQ(0)H_{\\rm mov}(s)P(0)\n=\nW(s)^*Q(s)\\bigl(H(s)-\\ii K(s)\\bigr)P(s)W(s).\n\\]\nUnitary conjugation preserves the norm.\n\\phantomsection\\label{strong:proof:024:end}\\end{proof}","Let \\((P_a)\\) and \\((Q_a)\\) be two complete packet systems with the same index set. Define\n\\[\nS(\\Qcal,\\Pcal)=\\sum_aQ_aP_a.\n\\]","\\begin{theorem}[Direct packet rotation]\\label{strong:theorem:016}\\label{strong:thm:direct_rotation}\nAssume\n\\[\n\\norm{S(\\Qcal,\\Pcal)-I}<1.\n\\]\nThen \\(S(\\Qcal,\\Pcal)\\) is invertible. Its unitary polar factor\n\\[\n\\Ucal(\\Qcal,\\Pcal)\n=\nS(S^*S)^{-1/2}\n\\]\nsatisfies\n\\[\nQ_a\\Ucal(\\Qcal,\\Pcal)\n=\n\\Ucal(\\Qcal,\\Pcal)P_a\n\\]\nfor every \\(a\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:025}\nThe norm condition gives invertibility by the Neumann series. The identity\n\\[\nQ_aS=SP_a=Q_aP_a\n\\]\nshows that \\(S^*S\\) commutes with every \\(P_a\\). Its inverse square root also commutes with every \\(P_a\\). Hence\n\\[\nQ_a\\Ucal\n=Q_aS(S^*S)^{-1/2}\n=SP_a(S^*S)^{-1/2}\n=\\Ucal P_a.\n\\]\nThe polar factor is unitary because \\(S\\) is invertible.\n\\phantomsection\\label{strong:proof:025:end}\\end{proof}","\\begin{theorem}[Convergence of short packet rotations]\\label{strong:theorem:017}\\label{strong:thm:short_rotation_convergence}\nLet \\((P_a(t))\\) be twice continuously differentiable on \\([0,T]\\). For a partition\n\\[\n0=t_0<t_1<\\cdots<t_n=T,\n\\]\nlet\n\\[\nW_\\pi\n=\n\\Ucal\\bigl(\\Pcal(t_n),\\Pcal(t_{n-1})\\bigr)\n\\cdots\n\\Ucal\\bigl(\\Pcal(t_1),\\Pcal(t_0)\\bigr).\n\\]\nFor sufficiently fine partitions the factors are defined, and\n\\[\n\\norm{W_\\pi-W(T)}\n\\leq\nC_T\\abs{\\pi}\n\\]\nfor a constant depending on the packet path but not on the partition. Here the size of the partition is measured by its mesh, which is the length of its largest subinterval.\n\\end{theorem}","em}[Convergence of short packet rotations]\\label{strong:theorem:017}\\label{strong:thm:short_rotation_convergence}\nLet \\((P_a(t))\\) be twice continuously differentiable on \\([0,T]\\). For a partition\n\\[\n0=t_0<t_1<\\cdots<t_n=T,\n\\]\nlet\n\\[\nW_\\pi\n=\n\\Ucal\\bigl(\\Pcal(t_n),\\Pcal(t_{n-1})\\bigr)\n\\cdots\n\\Ucal\\bigl(\\Pcal(t_1),\\Pcal(t_0)\\bigr).\n\\]\nFor sufficiently fine partitions the factors are defined, and\n\\[\n\\norm{W_\\pi-W(T)}\n\\leq\nC_T\\abs{\\pi}\n\\]\nfor a constant depending on the packet path but not on the partition. Here the size of the partition is measured by its mesh, which is the length of its largest subinterval.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:026}\nTaylor expansion gives\n\\[\nS\\bigl(\\Pcal(t+h),\\Pcal(t)\\bigr)\n=\nI+hK(t)+O(h^2)\n\\]\nuniformly on \\([0,T]\\). Since \\(K(t)^*=-K(t)\\),\n\\[\nS^*S=I+O(h^2),\n\\]\nso the polar factor satisfies\n\\[\n\\Ucal\\bigl(\\Pcal(t+h),\\Pcal(t)\\bigr)\n=\nI+hK(t)+O(h^2).\n\\]\nThis gives a first order approximation to \\(\\dot W=KW\\) over a single time step. A telescoping product estimate over the partition gives a total error bounded by a constant times \\(\\sum_j(t_{j+1}-t_j)^2\\), which is at most \\(T\\abs{\\pi}\\).\n\\phantomsection\\label{strong:proof:026:end}\\end{proof}","es\n\\[\nS\\bigl(\\Pcal(t+h),\\Pcal(t)\\bigr)\n=\nI+hK(t)+O(h^2)\n\\]\nuniformly on \\([0,T]\\). Since \\(K(t)^*=-K(t)\\),\n\\[\nS^*S=I+O(h^2),\n\\]\nso the polar factor satisfies\n\\[\n\\Ucal\\bigl(\\Pcal(t+h),\\Pcal(t)\\bigr)\n=\nI+hK(t)+O(h^2).\n\\]\nThis gives a first order approximation to \\(\\dot W=KW\\) over a single time step. A telescoping product estimate over the partition gives a total error bounded by a constant times \\(\\sum_j(t_{j+1}-t_j)^2\\), which is at most \\(T\\abs{\\pi}\\).\n\\phantomsection\\label{strong:proof:026:end}\\end{proof}","K(t)+O(h^2)\n\\]\nuniformly on \\([0,T]\\). Since \\(K(t)^*=-K(t)\\),\n\\[\nS^*S=I+O(h^2),\n\\]\nso the polar factor satisfies\n\\[\n\\Ucal\\bigl(\\Pcal(t+h),\\Pcal(t)\\bigr)\n=\nI+hK(t)+O(h^2).\n\\]\nThis gives a first order approximation to \\(\\dot W=KW\\) over a single time step. A telescoping product estimate over the partition gives a total error bounded by a constant times \\(\\sum_j(t_{j+1}-t_j)^2\\), which is at most \\(T\\abs{\\pi}\\).\n\\phantomsection\\label{strong:proof:026:end}\\end{proof}","Let \\(P(x)\\) be a smooth rank \\(r\\) orthogonal projector over a parameter manifold. Choose a local isometry\n\\[\nV(x):\\CC^r\\longrightarrow\\Hcal,\n\\qquad\nV(x)V(x)^*=P(x).\n\\]","\\begin{theorem}[Nonabelian packet connection]\\label{strong:theorem:018}\\label{strong:thm:nonabelian_connection}\nThe one form with matrix values\n\\[\n\\Acal=V^*\\dd V\n\\]\nis skew adjoint. Under a frame change \\(V'=VG\\), where \\(G(x)\\in U(r)\\),\n\\[\n\\Acal'\n=\nG^*\\Acal G+G^*\\dd G.\n\\]\nIts curvature\n\\[\n\\Fcal=\\dd\\Acal+\\Acal\\wedge\\Acal\n\\]\nsatisfies\n\\[\n\\Fcal=V^*(\\dd P\\wedge\\dd P)V.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:027}\nDifferentiate \\(V^*V=I\\) to obtain \\(\\Acal^*=-\\Acal\\). The gauge transformation law follows by differentiating \\(VG\\). Since \\(P=VV^*\\) and \\((I-P)\\dd V=\\dd P\\,V\\), direct expansion gives\n\\[\n\\dd\\Acal+\\Acal\\wedge\\Acal\n=\n\\dd V^*(I-P)\\wedge\\dd V\n=\nV^*(\\dd P\\wedge\\dd P)V.\n\\]\n\\phantomsection\\label{strong:proof:027:end}\\end{proof}","\\begin{corollary}[Bloch sphere curvature]\\label{strong:corollary:007}\\label{strong:cor:bloch_curvature}\nLet\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\bm n(\\theta,\\phi)\\cdot\\bm\\sigma\\right),\n\\]\nwhere\n\\[\n\\bm n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nWith the convention\n\\[\n\\Fcal_{\\theta\\phi}\n=-\\ii\\Tr\\bigl(P[\\partial_\\theta P,\\partial_\\phi P]\\bigr),\n\\]\none has\n\\[\n\\Fcal_{\\theta\\phi}=\\frac12\\sin\\theta\n\\]\nand\n\\[\n\\frac1{2\\pi}\n\\int_0^{2\\pi}\\int_0^\\pi\n\\Fcal_{\\theta\\phi}\\dd\\theta\\dd\\phi\n=1.\n\\]\n\\end{corollary}","[Bloch sphere curvature]\\label{strong:corollary:007}\\label{strong:cor:bloch_curvature}\nLet\n\\[\nP(\\theta,\\phi)\n=\n\\frac12\\left(I+\\bm n(\\theta,\\phi)\\cdot\\bm\\sigma\\right),\n\\]\nwhere\n\\[\n\\bm n=(\\sin\\theta\\cos\\phi,\\sin\\theta\\sin\\phi,\\cos\\theta).\n\\]\nWith the convention\n\\[\n\\Fcal_{\\theta\\phi}\n=-\\ii\\Tr\\bigl(P[\\partial_\\theta P,\\partial_\\phi P]\\bigr),\n\\]\none has\n\\[\n\\Fcal_{\\theta\\phi}=\\frac12\\sin\\theta\n\\]\nand\n\\[\n\\frac1{2\\pi}\n\\int_0^{2\\pi}\\int_0^\\pi\n\\Fcal_{\\theta\\phi}\\dd\\theta\\dd\\phi\n=1.\n\\]\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:028}\nThe Pauli identity\n\\[\n(\\bm a\\cdot\\bm\\sigma)(\\bm b\\cdot\\bm\\sigma)\n=(\\bm a\\cdot\\bm b)I\n+\\ii(\\bm a\\times\\bm b)\\cdot\\bm\\sigma\n\\]\ngives\n\\[\n-\\ii\\Tr\\bigl(P[\\partial_\\theta P,\\partial_\\phi P]\\bigr)\n=\\frac12\n\\bm n\\cdot\n(\\partial_\\theta\\bm n\\times\\partial_\\phi\\bm n)\n=\\frac12\\sin\\theta.\n\\]\nIntegrating the displayed curvature over the sphere and applying the stated normalization gives one.\n\\phantomsection\\label{strong:proof:028:end}\\end{proof}","Density matrices and observables carry both an input packet and an output packet. Equip \\(\\End(\\Hcal)\\) with the Hilbert Schmidt inner product\n\\[\n\\ip{X}{Y}_{\\HS}=\\Tr(X^*Y).\n\\]","\\begin{definition}[Two sided operator packet]\\label{strong:definition:003}\\label{strong:def:operator_packet}\nFor a complete orthogonal family \\((P_a)\\), define\n\\[\n\\Pcal_{ab}(X)=P_aXP_b.\n\\]\nThe range\n\\[\n\\Bcal_{ab}=P_a\\End(\\Hcal)P_b\n\\]\nis the operator packet with output address \\(a\\) and input address \\(b\\).\n\\end{definition}","\\begin{theorem}[Complete operator packet system]\\label{strong:theorem:019}\\label{strong:thm:operator_packets}\nThe superoperators \\((\\Pcal_{ab})_{a,b\\in I}\\) are mutually orthogonal projections on Hilbert Schmidt operator space. They satisfy\n\\[\nX=\\sum_{a,b}P_aXP_b\n\\]\nand\n\\[\n\\norm{X}_{\\HS}^2\n=\n\\sum_{a,b}\\norm{P_aXP_b}_{\\HS}^2.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:029}\nIdempotence and completeness follow from the projector identities. For Hilbert Schmidt orthogonality,\n\\[\n\\begin{aligned}\n\\ip{P_aXP_b}{P_cYP_d}_{\\HS}\n&=\n\\Tr(P_bX^*P_aP_cYP_d)\\\\\n&=\n\\delta_{ac}\\delta_{bd}\n\\Tr(P_bX^*P_aYP_b).\n\\end{aligned}\n\\]\nSince the packets are orthogonal and sum to the original operator, their squared norms add to its squared norm.\n\\phantomsection\\label{strong:proof:029:end}\\end{proof}","\\begin{definition}[Packet dephasing]\\label{strong:definition:004}\\label{strong:def:dephasing}\nThe packet dephasing map is\n\\[\n\\Dcal_{\\Pcal}(X)=\\sum_aP_aXP_a.\n\\]\n\\end{definition}","\\begin{theorem}[Dephasing as conditional expectation]\\label{strong:theorem:020}\\label{strong:thm:dephasing_expectation}\nThe map \\(\\Dcal_{\\Pcal}\\) is completely positive, trace preserving, unital, and idempotent. It is the Hilbert Schmidt orthogonal projection onto the block diagonal algebra\n\\[\n\\Acal_{\\Pcal}\n=\n\\{X:[X,P_a]=0\\text{ for every }a\\}.\n\\]\nIn particular,\n\\[\n\\norm{\\Dcal_{\\Pcal}(X)}_{\\HS}\n\\leq\n\\norm{X}_{\\HS}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:030}\nThe projectors \\(P_a\\) form a Kraus family because \\(\\sum_aP_a^*P_a=I\\). Thus the map is completely positive and trace preserving. It is unital for the same reason. Orthogonality and idempotence follow from \\cref{strong:thm:operator_packets}. Its range is the sum of the diagonal operator packets. This is exactly the commutant of the projector family.\n\\phantomsection\\label{strong:proof:030:end}\\end{proof}","\\begin{corollary}[Block diagonal measurement invariance]\\label{strong:corollary:008}\\label{strong:cor:measurement_invariance}\nLet \\((E_k)\\) be a positive operator valued measurement, and suppose\n\\[\nE_k=\\Dcal_{\\Pcal}(E_k)\n\\]\nfor every \\(k\\). Then every density matrix \\(\\rho\\) satisfies\n\\[\n\\Tr(E_k\\rho)\n=\n\\Tr\\bigl(E_k\\Dcal_{\\Pcal}(\\rho)\\bigr).\n\\]\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:031}\nSince \\(\\Dcal_{\\Pcal}\\) is self adjoint for the Hilbert Schmidt inner product,\n\\[\n\\Tr(E_k\\rho)\n=\n\\Tr\\bigl(\\Dcal_{\\Pcal}(E_k)\\rho\\bigr)\n=\n\\Tr\\bigl(E_k\\Dcal_{\\Pcal}(\\rho)\\bigr).\n\\]\n\\phantomsection\\label{strong:proof:031:end}\\end{proof}","Let \\(\\Phi:\\End(\\Hcal_X)\\to\\End(\\Hcal_Y)\\) be linear. Choose packet systems \\((P_a)\\) on \\(\\Hcal_X\\) and \\((Q_c)\\) on \\(\\Hcal_Y\\). Define\n\\[\n\\Phi_{cd,ab}\n=\n\\Qcal_{cd}\\Phi\\Pcal_{ab}.\n\\]","\\begin{theorem}[Exact channel block calculus]\\label{strong:theorem:021}\\label{strong:thm:channel_blocks}\nEvery linear map has the exact reconstruction\n\\[\n\\Phi=\\sum_{a,b,c,d}\\Phi_{cd,ab}.\n\\]\nIf \\(\\Psi:\\End(\\Hcal_Y)\\to\\End(\\Hcal_Z)\\), then\n\\[\n(\\Psi\\Phi)_{ef,ab}\n=\n\\sum_{c,d}\n\\Psi_{ef,cd}\\Phi_{cd,ab}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:032}\nInsert the complete operator packet resolutions before and after \\(\\Phi\\). For composition, insert the middle resolution between \\(\\Psi\\) and \\(\\Phi\\) and collect the output and input packets.\n\\phantomsection\\label{strong:proof:032:end}\\end{proof}","\\begin{standardfact}[Finite Kraus and Choi criteria]\\label{strong:standardfact:001}\\label{strong:fact:kraus_choi}\nA linear map \\(\\Phi:M_n(\\CC)\\to M_m(\\CC)\\) is completely positive exactly when it has a Kraus representation\n\\[\n\\Phi(X)=\\sum_rK_rXK_r^*.\n\\]\nIt is trace preserving exactly when\n\\[\n\\sum_rK_r^*K_r=I.\n\\]\nEquivalently, its Choi matrix\n\\[\nJ(\\Phi)\n=\n\\sum_{i,j}\\Phi(\\ket{i}\\bra{j})\\otimes\\ket{i}\\bra{j}\n\\]\nis positive, and trace preservation is\n\\[\n\\Tr_{\\rm out}J(\\Phi)=I_{\\rm in}.\n\\]\n\\end{standardfact}","label{strong:standardfact:001}\\label{strong:fact:kraus_choi}\nA linear map \\(\\Phi:M_n(\\CC)\\to M_m(\\CC)\\) is completely positive exactly when it has a Kraus representation\n\\[\n\\Phi(X)=\\sum_rK_rXK_r^*.\n\\]\nIt is trace preserving exactly when\n\\[\n\\sum_rK_r^*K_r=I.\n\\]\nEquivalently, its Choi matrix\n\\[\nJ(\\Phi)\n=\n\\sum_{i,j}\\Phi(\\ket{i}\\bra{j})\\otimes\\ket{i}\\bra{j}\n\\]\nis positive, and trace preservation is\n\\[\n\\Tr_{\\rm out}J(\\Phi)=I_{\\rm in}.\n\\]\n\\end{standardfact}","Let \\(S\\) be unitary and \\(N\\ge1\\) an integer with \\(S^N=I\\), and define the conjugation action\n\\[\n\\Ad_S(X)=SXS^*.\n\\]","\\begin{theorem}[Coherence order projectors]\\label{strong:theorem:022}\\label{strong:thm:coherence_projectors}\nFor \\(q\\in\\ZZ/N\\ZZ\\), define\n\\[\n\\Lcal_q(X)\n=\n\\frac1N\n\\sum_{r=0}^{N-1}\n\\omega^{-qr}S^rXS^{-r}.\n\\]\nThen the \\(\\Lcal_q\\) form a complete orthogonal packet system on Hilbert Schmidt operator space. Their ranges are the operators satisfying\n\\[\nSXS^*=\\omega^qX.\n\\]\nIf \\(P_j\\) are the state charge projectors, then\n\\[\n\\Ran\\Lcal_q\n=\n\\bigoplus_jP_{j+q}\\End(\\Hcal)P_j.\n\\]\n\\end{theorem}","erence order projectors]\\label{strong:theorem:022}\\label{strong:thm:coherence_projectors}\nFor \\(q\\in\\ZZ/N\\ZZ\\), define\n\\[\n\\Lcal_q(X)\n=\n\\frac1N\n\\sum_{r=0}^{N-1}\n\\omega^{-qr}S^rXS^{-r}.\n\\]\nThen the \\(\\Lcal_q\\) form a complete orthogonal packet system on Hilbert Schmidt operator space. Their ranges are the operators satisfying\n\\[\nSXS^*=\\omega^qX.\n\\]\nIf \\(P_j\\) are the state charge projectors, then\n\\[\n\\Ran\\Lcal_q\n=\n\\bigoplus_jP_{j+q}\\End(\\Hcal)P_j.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:033}\nApply the cyclic character projector formula to the unitary representation \\(\\Ad_S\\) on operator space. The last identity follows from \\cref{strong:thm:charge_selection} applied to each operator packet.\n\\phantomsection\\label{strong:proof:033:end}\\end{proof}","\\begin{theorem}[Covariant channels preserve coherence order]\\label{strong:theorem:023}\\label{strong:thm:covariant_channel}\nSuppose\n\\[\n\\Phi(SXS^*)=S\\Phi(X)S^*\n\\]\nfor every \\(X\\). Then\n\\[\n\\Phi\\Lcal_q=\\Lcal_q\\Phi\n\\]\nfor every \\(q\\). Thus \\(\\Phi\\) cannot mix distinct coherence orders.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:034}\nThe covariance identity says that \\(\\Phi\\) commutes with every power of \\(\\Ad_S\\). It therefore commutes with each finite character average \\(\\Lcal_q\\).\n\\phantomsection\\label{strong:proof:034:end}\\end{proof}","\\begin{theorem}[Choi symmetry criterion]\\label{strong:theorem:024}\\label{strong:thm:choi_covariance}\nLet \\(U:G\\to\\Ucal(\\Hcal)\\) be a finite group representation. A channel \\(\\Phi\\) is covariant,\n\\[\n\\Phi\\bigl(U(g)XU(g)^*\\bigr)\n=\nU(g)\\Phi(X)U(g)^*,\n\\]\nif and only if\n\\[\n\\bigl(U(g)\\otimes\\overline{U(g)}\\bigr)\nJ(\\Phi)\n\\bigl(U(g)^*\\otimes U(g)^{\\mathsf T}\\bigr)\n=\nJ(\\Phi)\n\\]\nfor every \\(g\\in G\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:035}\nLet\n\\[\n\\Omega=\\sum_i\\ket{i}\\otimes\\ket{i}.\n\\]\nThen\n\\[\nJ(\\Phi)=(\\Phi\\otimes\\id)(\\ket{\\Omega}\\bra{\\Omega}).\n\\]\nThe vector \\(\\Omega\\) is invariant under \\(U\\otimes\\overline U\\). Covariance therefore implies the displayed Choi invariance. Conversely, the displayed transformation is the Choi matrix of\n\\[\n\\Ad_{U(g)}\\circ\\Phi\\circ\\Ad_{U(g)^*}.\n\\]\nThe Choi correspondence is injective, so invariance implies that this map equals \\(\\Phi\\), which is covariance.\n\\phantomsection\\label{strong:proof:035:end}\\end{proof}","i}\\otimes\\ket{i}.\n\\]\nThen\n\\[\nJ(\\Phi)=(\\Phi\\otimes\\id)(\\ket{\\Omega}\\bra{\\Omega}).\n\\]\nThe vector \\(\\Omega\\) is invariant under \\(U\\otimes\\overline U\\). Covariance therefore implies the displayed Choi invariance. Conversely, the displayed transformation is the Choi matrix of\n\\[\n\\Ad_{U(g)}\\circ\\Phi\\circ\\Ad_{U(g)^*}.\n\\]\nThe Choi correspondence is injective, so invariance implies that this map equals \\(\\Phi\\), which is covariance.\n\\phantomsection\\label{strong:proof:035:end}\\end{proof}","\\begin{corollary}[Choi packet decomposition]\\label{strong:corollary:009}\\label{strong:cor:choi_packets}\nThe Choi matrix of a covariant channel is block diagonal with respect to the isotypic decomposition of \\(U\\otimes\\overline U\\).\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:036}\nApply \\cref{strong:thm:isotypic_projectors} to the representation \\(U\\otimes\\overline U\\) and use \\cref{strong:thm:choi_covariance}.\n\\phantomsection\\label{strong:proof:036:end}\\end{proof}","\\begin{example}[Amplitude damping]\\label{strong:example:002}\\label{strong:ex:amplitude_damping}\nFor \\(0\\leq\\gamma\\leq1\\), let\n\\[\nK_0=\n\\begin{pmatrix}\n1&0","\\begin{example}[Amplitude damping]\\label{strong:example:002}\\label{strong:ex:amplitude_damping}\nFor \\(0\\leq\\gamma\\leq1\\), let\n\\[\nK_0=\n\\begin{pmatrix}\n1&0\\\\\n0&\\sqrt{1-\\gamma}\n\\end{pmatrix},\n\\qquad\nK_1=\n\\begin{pmatrix}\n0&\\sqrt\\gamma\\\\\n0&0\n\\end{pmatrix}.\n\\]\nThe channel\n\\[\n\\Phi_\\gamma(\\rho)=K_0\\rho K_0^*+K_1\\rho K_1^*\n\\]\nis trace preserving. It sends the off diagonal packet \\(\\ket0\\bra1\\) to \\(\\sqrt{1-\\gamma}\\ket0\\bra1\\), while the excited population transfers partly into the ground population.\n\\end{example}","0&0\n\\end{pmatrix}.\n\\]\nThe channel\n\\[\n\\Phi_\\gamma(\\rho)=K_0\\rho K_0^*+K_1\\rho K_1^*\n\\]\nis trace preserving. It sends the off diagonal packet \\(\\ket0\\bra1\\) to \\(\\sqrt{1-\\gamma}\\ket0\\bra1\\), while the excited population transfers partly into the ground population.\n\\end{example}","\\begin{standardfact}[Finite Lindblad generator]\\label{strong:standardfact:002}\\label{strong:fact:lindblad}\nA norm continuous semigroup of completely positive trace preserving maps on a finite matrix algebra, in the Schr\\\"odinger picture, has a generator of the form\n\\[\n\\Gcal(X)\n=\n-\\ii[H,X]\n+\n\\sum_r\n\\left(\nL_rXL_r^*\n-\n\\frac12L_r^*L_rX\n-\n\\frac12XL_r^*L_r\n\\right),\n\\]\nwhere \\(H=H^*\\). Conversely, every operator of this form generates a completely positive trace preserving semigroup \\cite{Lindblad1976}.\n\\end{standardfact}","\\begin{theorem}[Charge preservation by a Lindblad generator]\\label{strong:theorem:025}\\label{strong:thm:lindblad_charge}\nLet \\(S\\) be unitary and \\(N\\ge1\\) an integer with \\(S^N=I\\). Assume\n\\[\nSHS^*=H\n\\]\nand, for every jump operator, there is a charge \\(q_r\\) such that\n\\[\nSL_rS^*=\\omega^{q_r}L_r.\n\\]\nThen the Lindblad generator \\(\\Gcal\\) preserves every coherence order:\n\\[\n\\Gcal\\Lcal_p=\\Lcal_p\\Gcal.\n\\]\nThe quantum dynamical semigroup \\(\\e^{t\\Gcal}\\) has the same property.\n\\end{theorem}","a Lindblad generator]\\label{strong:theorem:025}\\label{strong:thm:lindblad_charge}\nLet \\(S\\) be unitary and \\(N\\ge1\\) an integer with \\(S^N=I\\). Assume\n\\[\nSHS^*=H\n\\]\nand, for every jump operator, there is a charge \\(q_r\\) such that\n\\[\nSL_rS^*=\\omega^{q_r}L_r.\n\\]\nThen the Lindblad generator \\(\\Gcal\\) preserves every coherence order:\n\\[\n\\Gcal\\Lcal_p=\\Lcal_p\\Gcal.\n\\]\nThe quantum dynamical semigroup \\(\\e^{t\\Gcal}\\) has the same property.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:037}\nLet \\(X\\) have coherence order \\(p\\). The Hamiltonian commutator has order \\(p\\) because \\(H\\) has charge zero. The jump term \\(L_rXL_r^*\\) has charge\n\\[\nq_r+p-q_r=p.\n\\]\nAlso \\(L_r^*L_r\\) has charge zero, so both anticommutator terms have charge \\(p\\). Thus \\(\\Gcal\\) preserves \\(\\Ran\\Lcal_p\\). The exponential series then shows that \\(\\e^{t\\Gcal}\\) preserves the same packets.\n\\phantomsection\\label{strong:proof:037:end}\\end{proof}","Let \\(P\\) be the orthogonal projector onto a code space\n\\[\n\\Ccal=P\\Hcal,\n\\qquad\nk=\\dim\\Ccal\\ge1.\n\\]\nThe compression of \\(A\\in\\End(\\Hcal)\\) to the code is \\(PAP\\in\\End(\\Ccal)\\). The zero code has only the zero supported operator and no normalized code state. Part III treats that branch in the mathematical completion.","\\begin{definition}[Code compression obstruction]\\label{strong:definition:005}\\label{strong:def:code_obstruction}\nLet\n\\[\nq_P:\\End(\\Ccal)\\longrightarrow\\End(\\Ccal)/(\\CC I_{\\Ccal})\n\\]\nbe the quotient map. Define\n\\[\n\\Ob_P(A)=q_P(PAP).\n\\]\nEquivalently, one may use the traceless representative\n\\[\n\\ob_P(A)\n=\nPAP-\n\\frac{\\Tr(PAP)}{k}P.\n\\]\n\\end{definition}","The class vanishes exactly when \\(A\\) acts as a scalar after compression to the code. Such an operator cannot distinguish normalized code vectors by expectation value.","Exact correction of an error span means that one completely positive trace preserving recovery $\\Rcal$ satisfies $\\Rcal(F\\rho F^*)=q_F\\rho$ for every $F$ in that span and every code-supported matrix $\\rho$, with a nonnegative scalar $q_F$ depending on $F$ but not on the logical input. For every normalized channel with Kraus operators in the same span, the same recovery restores the code input with total scalar one. Individual unnormalized errors are not required to have unit recovery weight.","\\begin{theorem}[Knill Laflamme as obstruction vanishing]\\label{strong:theorem:026}\\label{strong:thm:kl_obstruction}\nLet \\(E_1,\\ldots,E_r\\) be an error family. The code \\(\\Ccal\\) corrects their linear span exactly if and only if\n\\[\n\\Ob_P(E_a^*E_b)=0\n\\]\nfor all \\(a,b\\). Equivalently, there is a matrix \\(c=(c_{ab})\\) such that\n\\[\nPE_a^*E_bP=c_{ab}P\n\\]\nfor all \\(a,b\\).\n\\end{theorem}","\\begin{theorem}[Orthogonal syndrome construction]\\label{strong:theorem:027}\\label{strong:thm:syndrome_construction}\nAssume the Knill Laflamme identities. Then the coefficient matrix \\(c\\) is positive semidefinite. Choose a unitary matrix \\(W\\) such that\n\\[\nWcW^*=\\diag(\\lambda_1,\\ldots,\\lambda_r),\n\\qquad\n\\lambda_j\\geq0,\n\\]\nand define\n\\[\nF_j=\\sum_a\\overline{W_{ja}}E_a.\n\\]\nFor every \\(j\\) with \\(\\lambda_j>0\\), the operator\n\\[\nV_j=\\lambda_j^{-1/2}F_jP\n\\]\nis an isometry from the code into \\(\\Hcal\\). Its range projector\n\\[\nQ_j=V_jV_j^*\n\\]\nis a syndrome packet, and the nonzero \\(Q_j\\) are pairwise orthogonal.\n\\end{theorem}","syndrome construction]\\label{strong:theorem:027}\\label{strong:thm:syndrome_construction}\nAssume the Knill Laflamme identities. Then the coefficient matrix \\(c\\) is positive semidefinite. Choose a unitary matrix \\(W\\) such that\n\\[\nWcW^*=\\diag(\\lambda_1,\\ldots,\\lambda_r),\n\\qquad\n\\lambda_j\\geq0,\n\\]\nand define\n\\[\nF_j=\\sum_a\\overline{W_{ja}}E_a.\n\\]\nFor every \\(j\\) with \\(\\lambda_j>0\\), the operator\n\\[\nV_j=\\lambda_j^{-1/2}F_jP\n\\]\nis an isometry from the code into \\(\\Hcal\\). Its range projector\n\\[\nQ_j=V_jV_j^*\n\\]\nis a syndrome packet, and the nonzero \\(Q_j\\) are pairwise orthogonal.\n\\end{theorem}","m:syndrome_construction}\nAssume the Knill Laflamme identities. Then the coefficient matrix \\(c\\) is positive semidefinite. Choose a unitary matrix \\(W\\) such that\n\\[\nWcW^*=\\diag(\\lambda_1,\\ldots,\\lambda_r),\n\\qquad\n\\lambda_j\\geq0,\n\\]\nand define\n\\[\nF_j=\\sum_a\\overline{W_{ja}}E_a.\n\\]\nFor every \\(j\\) with \\(\\lambda_j>0\\), the operator\n\\[\nV_j=\\lambda_j^{-1/2}F_jP\n\\]\nis an isometry from the code into \\(\\Hcal\\). Its range projector\n\\[\nQ_j=V_jV_j^*\n\\]\nis a syndrome packet, and the nonzero \\(Q_j\\) are pairwise orthogonal.\n\\end{theorem}","the Knill Laflamme identities. Then the coefficient matrix \\(c\\) is positive semidefinite. Choose a unitary matrix \\(W\\) such that\n\\[\nWcW^*=\\diag(\\lambda_1,\\ldots,\\lambda_r),\n\\qquad\n\\lambda_j\\geq0,\n\\]\nand define\n\\[\nF_j=\\sum_a\\overline{W_{ja}}E_a.\n\\]\nFor every \\(j\\) with \\(\\lambda_j>0\\), the operator\n\\[\nV_j=\\lambda_j^{-1/2}F_jP\n\\]\nis an isometry from the code into \\(\\Hcal\\). Its range projector\n\\[\nQ_j=V_jV_j^*\n\\]\nis a syndrome packet, and the nonzero \\(Q_j\\) are pairwise orthogonal.\n\\end{theorem}","ix \\(c\\) is positive semidefinite. Choose a unitary matrix \\(W\\) such that\n\\[\nWcW^*=\\diag(\\lambda_1,\\ldots,\\lambda_r),\n\\qquad\n\\lambda_j\\geq0,\n\\]\nand define\n\\[\nF_j=\\sum_a\\overline{W_{ja}}E_a.\n\\]\nFor every \\(j\\) with \\(\\lambda_j>0\\), the operator\n\\[\nV_j=\\lambda_j^{-1/2}F_jP\n\\]\nis an isometry from the code into \\(\\Hcal\\). Its range projector\n\\[\nQ_j=V_jV_j^*\n\\]\nis a syndrome packet, and the nonzero \\(Q_j\\) are pairwise orthogonal.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:040}\nFor \\(z\\in\\CC^r\\), let \\(F_z=\\sum_az_aE_a\\). Then\n\\[\nPF_z^*F_zP=(z^*cz)P\\geq0,\n\\]\nso \\(c\\geq0\\). The unitary change of error coordinates gives\n\\[\nPF_j^*F_\\ell P\n=\n\\lambda_j\\delta_{j\\ell}P.\n\\]\nThus\n\\[\nV_j^*V_\\ell=\\delta_{j\\ell}P.\n\\]\nEach \\(V_j\\) is an isometry on \\(P\\Hcal\\), and distinct ranges are orthogonal. Their range projectors are therefore pairwise orthogonal.\n\\phantomsection\\label{strong:proof:040:end}\\end{proof}","\\begin{theorem}[Exact recovery on the error span]\\label{strong:theorem:028}\\label{strong:thm:exact_recovery}\nUnder the hypotheses of \\cref{strong:thm:syndrome_construction}, define on the syndrome support\n\\[\nQ=\\sum_{\\lambda_j>0}Q_j\n\\]\nthe completely positive map\n\\[\n\\Rcal_0(X)\n=\n\\sum_{\\lambda_j>0}\nV_j^*Q_jXQ_jV_j.\n\\]\nIt extends to a quantum channel \\(\\Rcal\\) on \\(\\End(\\Hcal)\\). If\n\\[\n\\Phi(X)=\\sum_aE_aXE_a^*\n\\]\nis trace preserving on code states, then\n\\[\n\\Rcal\\Phi(\\rho)=\\rho\n\\]\nfor every density matrix \\(\\rho=P\\rho P\\).\n\\end{theorem}","strong:theorem:028}\\label{strong:thm:exact_recovery}\nUnder the hypotheses of \\cref{strong:thm:syndrome_construction}, define on the syndrome support\n\\[\nQ=\\sum_{\\lambda_j>0}Q_j\n\\]\nthe completely positive map\n\\[\n\\Rcal_0(X)\n=\n\\sum_{\\lambda_j>0}\nV_j^*Q_jXQ_jV_j.\n\\]\nIt extends to a quantum channel \\(\\Rcal\\) on \\(\\End(\\Hcal)\\). If\n\\[\n\\Phi(X)=\\sum_aE_aXE_a^*\n\\]\nis trace preserving on code states, then\n\\[\n\\Rcal\\Phi(\\rho)=\\rho\n\\]\nfor every density matrix \\(\\rho=P\\rho P\\).\n\\end{theorem}","Under the hypotheses of \\cref{strong:thm:syndrome_construction}, define on the syndrome support\n\\[\nQ=\\sum_{\\lambda_j>0}Q_j\n\\]\nthe completely positive map\n\\[\n\\Rcal_0(X)\n=\n\\sum_{\\lambda_j>0}\nV_j^*Q_jXQ_jV_j.\n\\]\nIt extends to a quantum channel \\(\\Rcal\\) on \\(\\End(\\Hcal)\\). If\n\\[\n\\Phi(X)=\\sum_aE_aXE_a^*\n\\]\nis trace preserving on code states, then\n\\[\n\\Rcal\\Phi(\\rho)=\\rho\n\\]\nfor every density matrix \\(\\rho=P\\rho P\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:041}\nThe map \\(\\Rcal_0\\) is completely positive. It is trace preserving on operators supported in \\(Q\\Hcal\\) because\n\\[\n\\Tr\\bigl(V_j^*Q_jXQ_jV_j\\bigr)\n=\n\\Tr(Q_jXQ_j).\n\\]\nFix a code state \\(\\rho_0\\) and write \\(Q^\\perp=I-Q\\). Then\n\\[\n\\Rcal(X)\n=\n\\Rcal_0(QXQ)\n+\n\\Tr(Q^\\perp X)\\rho_0\n\\]\nis completely positive and trace preserving. It extends \\(\\Rcal_0\\) on operators supported in \\(Q\\Hcal\\).","n\\label{strong:proof:041}\nThe map \\(\\Rcal_0\\) is completely positive. It is trace preserving on operators supported in \\(Q\\Hcal\\) because\n\\[\n\\Tr\\bigl(V_j^*Q_jXQ_jV_j\\bigr)\n=\n\\Tr(Q_jXQ_j).\n\\]\nFix a code state \\(\\rho_0\\) and write \\(Q^\\perp=I-Q\\). Then\n\\[\n\\Rcal(X)\n=\n\\Rcal_0(QXQ)\n+\n\\Tr(Q^\\perp X)\\rho_0\n\\]\nis completely positive and trace preserving. It extends \\(\\Rcal_0\\) on operators supported in \\(Q\\Hcal\\).","Unitary mixing of a Kraus family does not change the channel, so\n\\[\n\\Phi(\\rho)\n=\n\\sum_jF_j\\rho F_j^*\n=\n\\sum_{\\lambda_j>0}\n\\lambda_jV_j\\rho V_j^*.\n\\]\nRecovery gives\n\\[\n\\Rcal\\Phi(\\rho)\n=\n\\left(\\sum_j\\lambda_j\\right)\\rho.\n\\]\nTrace preservation on a code state implies \\(\\sum_j\\lambda_j=1\\).\n\\phantomsection\\label{strong:proof:041:end}\\end{proof}","\\begin{corollary}[Syndrome measurement preserves logical information]\\label{strong:corollary:011}\\label{strong:cor:syndrome_preserves_logic}\nFor a channel trace preserving on code states as in the preceding theorem, the completed projective measurement consisting of \\((Q_j)_{\\lambda_j>0}\\) and \\(I-\\sum_{\\lambda_j>0}Q_j\\) identifies the orthogonal error syndrome without resolving the encoded state inside the code. The complement outcome has probability zero on the corrupted code state.\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:042}\nOn syndrome \\(j\\), the encoded state is carried by the isometry \\(V_j\\). The outcome probability is \\(\\lambda_j\\), independent of the normalized code state, while the postmeasurement state is \\(V_j\\rho V_j^*\\).\n\\phantomsection\\label{strong:proof:042:end}\\end{proof}","\\begin{definition}[Approximate Knill Laflamme defect]\\label{strong:definition:006}\\label{strong:def:approximate_kl}\nFor a finite error family, define the following defect. When the family is empty, assign the maximum the value zero.\n\\[\n\\varepsilon_{\\rm KL}\n=\n\\max_{a,b}\n\\inf_{c\\in\\CC}\n\\norm{PE_a^*E_bP-cP}.\n\\]\n\\end{definition}","\\begin{proposition}[Meaning of the approximate defect]\\label{strong:proposition:003}\\label{strong:prop:approximate_kl_meaning}\nFor each pair \\((a,b)\\), choose a minimizing scalar \\(c_{ab}\\). For unit code vectors \\(\\psi,\\phi\\),\n\\[\n\\abs{\n\\ip{E_a\\psi}{E_b\\phi}\n-c_{ab}\\ip{\\psi}{\\phi}\n}\n\\leq\n\\varepsilon_{\\rm KL}.\n\\]\nThus \\(\\varepsilon_{\\rm KL}\\) controls the deviation of error overlaps from their scalar code form.\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:043}\nThe left side is\n\\[\n\\abs{\n\\ip{\\psi}{(PE_a^*E_bP-c_{ab}P)\\phi}\n},\n\\]\nwhich is bounded by the operator norm.\n\\phantomsection\\label{strong:proof:043:end}\\end{proof}","\\begin{example}[Three qubit repetition code]\\label{strong:example:003}\\label{strong:ex:bit_flip_code}\nLet\n\\[\n\\ket{0_L}=\\ket{000},\n\\qquad\n\\ket{1_L}=\\ket{111},\n\\]\nand take the errors\n\\[\nI,\n\\quad\nX\\otimes I\\otimes I,\n\\quad\nI\\otimes X\\otimes I,\n\\quad\nI\\otimes I\\otimes X.\n\\]\nTheir images of the code are four pairwise orthogonal two dimensional syndrome packets whose sum is the full eight dimensional Hilbert space.\n\\end{example}","\\begin{standardfact}[Correctable observable algebras]\\label{strong:standardfact:003}\\label{strong:fact:operator_qec}\nLet \\(\\Phi\\) be a finite completely positive trace preserving channel with Kraus operators \\((E_a)\\). A finite dimensional unital star subalgebra \\(\\Mcal\\subseteq P\\End(\\Hcal)P\\) is correctable when a completely positive trace preserving recovery \\(\\Rcal\\) satisfies \\(P(\\Phi^*\\circ\\Rcal^*)(X)P=X\\) for every \\(X\\in\\Mcal\\), where stars on maps denote Heisenberg adjoints. This holds exactly when every compressed product \\(PE_a^*E_bP\\) belongs to the commutant \\(\\Mcal'\\) on the code. The scalar code criterion is the special case \\(\\Mcal=P\\End(\\Hcal)P\\) \\cite{BenyKempfKribs2007}.\n\\end{standardfact}","able observable algebras]\\label{strong:standardfact:003}\\label{strong:fact:operator_qec}\nLet \\(\\Phi\\) be a finite completely positive trace preserving channel with Kraus operators \\((E_a)\\). A finite dimensional unital star subalgebra \\(\\Mcal\\subseteq P\\End(\\Hcal)P\\) is correctable when a completely positive trace preserving recovery \\(\\Rcal\\) satisfies \\(P(\\Phi^*\\circ\\Rcal^*)(X)P=X\\) for every \\(X\\in\\Mcal\\), where stars on maps denote Heisenberg adjoints. This holds exactly when every compressed product \\(PE_a^*E_bP\\) belongs to the commutant \\(\\Mcal'\\) on the code. The scalar code criterion is the special case \\(\\Mcal=P\\End(\\Hcal)P\\) \\cite{BenyKempfKribs2007}.\n\\end{standardfact}","e completely positive trace preserving channel with Kraus operators \\((E_a)\\). A finite dimensional unital star subalgebra \\(\\Mcal\\subseteq P\\End(\\Hcal)P\\) is correctable when a completely positive trace preserving recovery \\(\\Rcal\\) satisfies \\(P(\\Phi^*\\circ\\Rcal^*)(X)P=X\\) for every \\(X\\in\\Mcal\\), where stars on maps denote Heisenberg adjoints. This holds exactly when every compressed product \\(PE_a^*E_bP\\) belongs to the commutant \\(\\Mcal'\\) on the code. The scalar code criterion is the special case \\(\\Mcal=P\\End(\\Hcal)P\\) \\cite{BenyKempfKribs2007}.\n\\end{standardfact}","hannel with Kraus operators \\((E_a)\\). A finite dimensional unital star subalgebra \\(\\Mcal\\subseteq P\\End(\\Hcal)P\\) is correctable when a completely positive trace preserving recovery \\(\\Rcal\\) satisfies \\(P(\\Phi^*\\circ\\Rcal^*)(X)P=X\\) for every \\(X\\in\\Mcal\\), where stars on maps denote Heisenberg adjoints. This holds exactly when every compressed product \\(PE_a^*E_bP\\) belongs to the commutant \\(\\Mcal'\\) on the code. The scalar code criterion is the special case \\(\\Mcal=P\\End(\\Hcal)P\\) \\cite{BenyKempfKribs2007}.\n\\end{standardfact}","r subalgebra \\(\\Mcal\\subseteq P\\End(\\Hcal)P\\) is correctable when a completely positive trace preserving recovery \\(\\Rcal\\) satisfies \\(P(\\Phi^*\\circ\\Rcal^*)(X)P=X\\) for every \\(X\\in\\Mcal\\), where stars on maps denote Heisenberg adjoints. This holds exactly when every compressed product \\(PE_a^*E_bP\\) belongs to the commutant \\(\\Mcal'\\) on the code. The scalar code criterion is the special case \\(\\Mcal=P\\End(\\Hcal)P\\) \\cite{BenyKempfKribs2007}.\n\\end{standardfact}","Let \\(P\\) be an orthogonal projector, put \\(Q=I-P\\), and write a self adjoint Hamiltonian as\n\\[\nH=\n\\begin{pmatrix}\nA&B","Let \\(P\\) be an orthogonal projector, put \\(Q=I-P\\), and write a self adjoint Hamiltonian as\n\\[\nH=\n\\begin{pmatrix}\nA&B\\\\\nB^*&D\n\\end{pmatrix}\n\\]\nrelative to \\(P\\Hcal\\oplus Q\\Hcal\\).","B^*&D\n\\end{pmatrix}\n\\]\nrelative to \\(P\\Hcal\\oplus Q\\Hcal\\).","\\begin{definition}[Feshbach map and self energy]\\label{strong:definition:007}\\label{strong:def:feshbach_map}\nFor \\(z\\notin\\Spec D\\), define\n\\[\n\\Sigma(z)=B(z-D)^{-1}B^*\n\\]\nand\n\\[\nF_P(z)=z-A-\\Sigma(z).\n\\]\n\\end{definition}","\\begin{theorem}[Exact projected resolvent]\\label{strong:theorem:029}\\label{strong:thm:feshbach_resolvent}\nIf \\(z\\notin\\Spec D\\) and \\(F_P(z)\\) is invertible, then\n\\[\nP(z-H)^{-1}P=F_P(z)^{-1}\n\\]\nas an operator on \\(P\\Hcal\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:044}\nThe lower right block of \\(z-H\\) is \\(z-D\\). Block Gaussian elimination gives\n\\[\n\\begin{pmatrix}\nI&B(z-D)^{-1}","\\begin{proof}\\phantomsection\\label{strong:proof:044}\nThe lower right block of \\(z-H\\) is \\(z-D\\). Block Gaussian elimination gives\n\\[\n\\begin{pmatrix}\nI&B(z-D)^{-1}\\\\\n0&I\n\\end{pmatrix}\n(z-H)\n\\begin{pmatrix}\nI&0\\\\\n(z-D)^{-1}B^*&I\n\\end{pmatrix}\n=\n\\begin{pmatrix}\nF_P(z)&0\\\\\n0&z-D\n\\end{pmatrix}.\n\\]\nInvert the factors. The retained resolvent is the block in the upper left corner of the resulting inverse.\n\\phantomsection\\label{strong:proof:044:end}\\end{proof}","\\begin{theorem}[Spectral equivalence]\\label{strong:theorem:030}\\label{strong:thm:feshbach_spectrum}\nLet \\(z\\notin\\Spec D\\). Then\n\\[\nz\\in\\Spec H\n\\quad\\Longleftrightarrow\\quad\n0\\in\\Spec F_P(z).\n\\]\nThe map\n\\[\nx\\longmapsto\n\\begin{pmatrix}\nx","\\begin{theorem}[Spectral equivalence]\\label{strong:theorem:030}\\label{strong:thm:feshbach_spectrum}\nLet \\(z\\notin\\Spec D\\). Then\n\\[\nz\\in\\Spec H\n\\quad\\Longleftrightarrow\\quad\n0\\in\\Spec F_P(z).\n\\]\nThe map\n\\[\nx\\longmapsto\n\\begin{pmatrix}\nx\\\\\n(z-D)^{-1}B^*x\n\\end{pmatrix}\n\\]\nis an isomorphism from \\(\\Ker F_P(z)\\) onto \\(\\Ker(z-H)\\).\n\\end{theorem}","(z-D)^{-1}B^*x\n\\end{pmatrix}\n\\]\nis an isomorphism from \\(\\Ker F_P(z)\\) onto \\(\\Ker(z-H)\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:045}\nThe equation \\((z-H)(x,y)^{\\mathsf T}=0\\) is\n\\[\n(z-A)x-By=0,\n\\qquad\n-B^*x+(z-D)y=0.\n\\]\nThe second equation gives \\(y=(z-D)^{-1}B^*x\\). Substitution into the first gives \\(F_P(z)x=0\\). The construction is reversible.\n\\phantomsection\\label{strong:proof:045:end}\\end{proof}","\\begin{theorem}[Upper half plane sign]\\label{strong:theorem:031}\\label{strong:thm:self_energy_sign}\nIf \\(\\operatorname{Im}z>0\\), then\n\\[\n\\operatorname{Im}\\Sigma(z)\n=\n-(\\operatorname{Im}z)\nB(z-D)^{-1}(\\overline z-D)^{-1}B^*\n\\leq0.\n\\]\nThus \\(-\\Sigma\\) is an operator valued Herglotz function.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:046}\nFor \\(R(z)=(z-D)^{-1}\\), the resolvent identity gives\n\\[\nR(z)-R(z)^*\n=(\\overline z-z)R(z)R(z)^*.\n\\]\nDivide by \\(2\\ii\\), then multiply the result on the left by \\(B\\) and on the right by \\(B^*\\).\n\\phantomsection\\label{strong:proof:046:end}\\end{proof}","\\begin{proposition}[Uniform self energy bounds]\\label{strong:proposition:004}\\label{strong:prop:self_energy_bounds}\nLet\n\\[\n\\delta=\\dist(z,\\Spec D)>0,\n\\qquad\ng=\\norm B.\n\\]\nThen\n\\[\n\\norm{\\Sigma(z)}\\leq\\frac{g^2}{\\delta}\n\\]\nand\n\\[\n\\norm{\\Sigma'(z)}\\leq\\frac{g^2}{\\delta^2}.\n\\]\nFor real \\(E,E'\\) in an interval at distance at least \\(\\delta\\) from \\(\\Spec D\\),\n\\[\n\\norm{\\Sigma(E)-\\Sigma(E')}\n\\leq\n\\frac{g^2}{\\delta^2}\\abs{E-E'}.\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:047}\nThe self adjoint resolvent satisfies\n\\[\n\\norm{(z-D)^{-1}}=\\frac1{\\dist(z,\\Spec D)}.\n\\]\nDifferentiate the resolvent to obtain\n\\[\n\\Sigma'(z)=-B(z-D)^{-2}B^*.\n\\]\nTaking operator norms gives the first two estimates. Applying the derivative bound along the real interval gives the final mean value estimate.\n\\phantomsection\\label{strong:proof:047:end}\\end{proof}","\\begin{proposition}[Low energy expansion]\\label{strong:proposition:005}\\label{strong:prop:low_energy_expansion}\nAssume \\(D\\) is invertible and \\(\\abs z\\norm{D^{-1}}<1\\). Then\n\\[\n\\Sigma(z)\n=-\n\\sum_{n=0}^{\\infty}\nz^nBD^{-n-1}B^*\n\\]\nwith norm convergence. In particular,\n\\[\nF_P(z)\n=\n z-A+BD^{-1}B^*+O\\bigl(\\abs z\\norm B^2\\norm{D^{-1}}^2\\bigr).\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:048}\nWrite\n\\[\n(z-D)^{-1}\n=-D^{-1}(I-zD^{-1})^{-1}\n=-\\sum_{n=0}^{\\infty}z^nD^{-n-1}\n\\]\nand multiply on the left by \\(B\\) and on the right by \\(B^*\\).\n\\phantomsection\\label{strong:proof:048:end}\\end{proof}","\\begin{proposition}[Bare truncation error identity]\\label{strong:proposition:006}\\label{strong:prop:bare_truncation}\nIf both \\(F_P(z)\\) and \\(z-A\\) are invertible, then\n\\[\nP(z-H)^{-1}P-(z-A)^{-1}\n=\nF_P(z)^{-1}\\Sigma(z)(z-A)^{-1}.\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:049}\nUse the inverse difference identity\n\\[\nX^{-1}-Y^{-1}=X^{-1}(Y-X)Y^{-1}\n\\]\nwith \\(X=F_P(z)\\) and \\(Y=z-A\\).\n\\phantomsection\\label{strong:proof:049:end}\\end{proof}","Write a Schrödinger state as \\(\\psi=x+y\\), where \\(x=P\\psi\\) and \\(y=Q\\psi\\).","\\begin{theorem}[Exact retained memory equation]\\label{strong:theorem:032}\\label{strong:thm:memory_equation}\nThe retained coordinate satisfies\n\\[\n\\begin{aligned}\n\\ii\\dot x(t)\n={}&Ax(t)+B\\e^{-\\ii tD}y(0)\\\\\n&-\\ii\\int_0^tK(t-s)x(s)\\dd s,\n\\end{aligned}\n\\]\nwhere\n\\[\nK(t)=B\\e^{-\\ii tD}B^*.\n\\]\nFor \\(\\operatorname{Re}s>0\\),\n\\[\n\\int_0^\\infty\\e^{-st}K(t)\\dd t\n=\nB(s+\\ii D)^{-1}B^*.\n\\]\n\\end{theorem}","&-\\ii\\int_0^tK(t-s)x(s)\\dd s,\n\\end{aligned}\n\\]\nwhere\n\\[\nK(t)=B\\e^{-\\ii tD}B^*.\n\\]\nFor \\(\\operatorname{Re}s>0\\),\n\\[\n\\int_0^\\infty\\e^{-st}K(t)\\dd t\n=\nB(s+\\ii D)^{-1}B^*.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:050}\nThe hidden equation is\n\\[\n\\ii\\dot y=B^*x+Dy.\n\\]\nVariation of constants gives\n\\[\ny(t)=\\e^{-\\ii tD}y(0)\n-\\ii\\int_0^t\\e^{-\\ii(t-s)D}B^*x(s)\\dd s.\n\\]\nSubstitute into \\(\\ii\\dot x=Ax+By\\). The Laplace formula follows by diagonalizing \\(D\\) or integrating the norm convergent finite dimensional exponential.\n\\phantomsection\\label{strong:proof:050:end}\\end{proof}","\\begin{theorem}[Positive definite quantum memory]\\label{strong:theorem:033}\\label{strong:thm:positive_memory}\nThe kernel \\(K(t-s)\\) is positive definite. For times \\(t_1,\\ldots,t_m\\) and retained vectors \\(v_1,\\ldots,v_m\\),\n\\[\n\\sum_{j,k=1}^{m}\n\\ip{v_j}{K(t_j-t_k)v_k}\n\\geq0.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:051}\nSet\n\\[\nw=\\sum_{k=1}^{m}\\e^{\\ii t_kD}B^*v_k.\n\\]\nThen\n\\[\n\\norm w^2\n=\n\\sum_{j,k}\n\\ip{B^*v_j}{\\e^{-\\ii(t_j-t_k)D}B^*v_k}\n=\n\\sum_{j,k}\n\\ip{v_j}{K(t_j-t_k)v_k}.\n\\]\n\\phantomsection\\label{strong:proof:051:end}\\end{proof}","\\begin{corollary}[Finite oscillatory memory expansion]\\label{strong:corollary:012}\\label{strong:cor:oscillatory_memory}\nIf\n\\[\nD=\\sum_{r}E_rQ_r\n\\]\nis the spectral decomposition of the hidden Hamiltonian, then\n\\[\nK(t)=\\sum_r\\e^{-\\ii tE_r}BQ_rB^*.\n\\]\nEvery coefficient \\(BQ_rB^*\\) is positive semidefinite.\n\\end{corollary}","\\begin{theorem}[Exact local closure criterion]\\label{strong:theorem:034}\\label{strong:thm:local_closure}\nAssume \\(y(0)=0\\). The retained equation is local and equal to\n\\[\n\\ii\\dot x=Ax\n\\]\nfor every retained initial state exactly when \\(B=0\\). Equivalently, exact local closure holds exactly when the retained and hidden packets reduce \\(H\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:053}\nIf \\(B=0\\), the retained and hidden equations decouple. Conversely, suppose that every solution with \\(y(0)=0\\) satisfies \\(x(t)=\\e^{-\\ii tA}x(0)\\). The full block equation gives\n\\[\n\\ddot x(0)\n=\n-\\bigl(A^2+BB^*\\bigr)x(0),\n\\]\nwhile the declared local equation gives\n\\[\n\\ddot x(0)=-A^2x(0).\n\\]\nSince this holds for every retained initial vector, \\(BB^*=0\\). Hence \\(B=0\\).\n\\phantomsection\\label{strong:proof:053:end}\\end{proof}","ntomsection\\label{strong:proof:053}\nIf \\(B=0\\), the retained and hidden equations decouple. Conversely, suppose that every solution with \\(y(0)=0\\) satisfies \\(x(t)=\\e^{-\\ii tA}x(0)\\). The full block equation gives\n\\[\n\\ddot x(0)\n=\n-\\bigl(A^2+BB^*\\bigr)x(0),\n\\]\nwhile the declared local equation gives\n\\[\n\\ddot x(0)=-A^2x(0).\n\\]\nSince this holds for every retained initial vector, \\(BB^*=0\\). Hence \\(B=0\\).\n\\phantomsection\\label{strong:proof:053:end}\\end{proof}","\\begin{theorem}[Stability of the self energy]\\label{strong:theorem:035}\\label{strong:thm:self_energy_stability}\nLet \\(D=D^*\\) and \\(\\widetilde D=\\widetilde D^*\\) act on one fixed hidden Hilbert space, and let \\(B,\\widetilde B\\) map that hidden space to one fixed retained Hilbert space. Define\n\\[\n\\Sigma(z)=B(z-D)^{-1}B^*,\n\\qquad\n\\widetilde\\Sigma(z)=\\widetilde B(z-\\widetilde D)^{-1}\\widetilde B^*.\n\\]\nAssume\n\\[\n\\norm B,\\norm{\\widetilde B}\\leq g\n\\]\nand\n\\[\n\\dist(z,\\Spec D),\n\\dist(z,\\Spec\\widetilde D)\n\\geq\\delta>0.\n\\]\nThen\n\\[\n\\norm{\\Sigma(z)-\\widetilde\\Sigma(z)}\n\\leq\n\\frac{2g}{\\delta}\\norm{B-\\widetilde B}\n+\n\\frac{g^2}{\\delta^2}\\norm{D-\\widetilde D}.\n\\]\n\\end{theorem}","rem}[Stability of the self energy]\\label{strong:theorem:035}\\label{strong:thm:self_energy_stability}\nLet \\(D=D^*\\) and \\(\\widetilde D=\\widetilde D^*\\) act on one fixed hidden Hilbert space, and let \\(B,\\widetilde B\\) map that hidden space to one fixed retained Hilbert space. Define\n\\[\n\\Sigma(z)=B(z-D)^{-1}B^*,\n\\qquad\n\\widetilde\\Sigma(z)=\\widetilde B(z-\\widetilde D)^{-1}\\widetilde B^*.\n\\]\nAssume\n\\[\n\\norm B,\\norm{\\widetilde B}\\leq g\n\\]\nand\n\\[\n\\dist(z,\\Spec D),\n\\dist(z,\\Spec\\widetilde D)\n\\geq\\delta>0.\n\\]\nThen\n\\[\n\\norm{\\Sigma(z)-\\widetilde\\Sigma(z)}\n\\leq\n\\frac{2g}{\\delta}\\norm{B-\\widetilde B}\n+\n\\frac{g^2}{\\delta^2}\\norm{D-\\widetilde D}.\n\\]\n\\end{theorem}","ong:theorem:035}\\label{strong:thm:self_energy_stability}\nLet \\(D=D^*\\) and \\(\\widetilde D=\\widetilde D^*\\) act on one fixed hidden Hilbert space, and let \\(B,\\widetilde B\\) map that hidden space to one fixed retained Hilbert space. Define\n\\[\n\\Sigma(z)=B(z-D)^{-1}B^*,\n\\qquad\n\\widetilde\\Sigma(z)=\\widetilde B(z-\\widetilde D)^{-1}\\widetilde B^*.\n\\]\nAssume\n\\[\n\\norm B,\\norm{\\widetilde B}\\leq g\n\\]\nand\n\\[\n\\dist(z,\\Spec D),\n\\dist(z,\\Spec\\widetilde D)\n\\geq\\delta>0.\n\\]\nThen\n\\[\n\\norm{\\Sigma(z)-\\widetilde\\Sigma(z)}\n\\leq\n\\frac{2g}{\\delta}\\norm{B-\\widetilde B}\n+\n\\frac{g^2}{\\delta^2}\\norm{D-\\widetilde D}.\n\\]\n\\end{theorem}","and \\(\\widetilde D=\\widetilde D^*\\) act on one fixed hidden Hilbert space, and let \\(B,\\widetilde B\\) map that hidden space to one fixed retained Hilbert space. Define\n\\[\n\\Sigma(z)=B(z-D)^{-1}B^*,\n\\qquad\n\\widetilde\\Sigma(z)=\\widetilde B(z-\\widetilde D)^{-1}\\widetilde B^*.\n\\]\nAssume\n\\[\n\\norm B,\\norm{\\widetilde B}\\leq g\n\\]\nand\n\\[\n\\dist(z,\\Spec D),\n\\dist(z,\\Spec\\widetilde D)\n\\geq\\delta>0.\n\\]\nThen\n\\[\n\\norm{\\Sigma(z)-\\widetilde\\Sigma(z)}\n\\leq\n\\frac{2g}{\\delta}\\norm{B-\\widetilde B}\n+\n\\frac{g^2}{\\delta^2}\\norm{D-\\widetilde D}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:054}\nAdd and subtract \\(\\widetilde B(z-D)^{-1}B^*\\) and \\(\\widetilde B(z-D)^{-1}\\widetilde B^*\\). The first two differences give the coupling term. For the last, use\n\\[\n(z-D)^{-1}-(z-\\widetilde D)^{-1}\n=(z-D)^{-1}(D-\\widetilde D)(z-\\widetilde D)^{-1}.\n\\]\nApply the resolvent bounds.\n\\phantomsection\\label{strong:proof:054:end}\\end{proof}","\\begin{corollary}[Uniform Feshbach control]\\label{strong:corollary:013}\\label{strong:cor:uniform_feshbach}\nFor a family represented on fixed retained and hidden Hilbert spaces, with self adjoint hidden Hamiltonians, a common hidden spectral separation \\(\\delta\\), and a common coupling bound \\(g\\), the self energies are uniformly bounded and uniformly Lipschitz on the declared spectral region. If the couplings and hidden Hamiltonians converge in norm, then the self energies converge uniformly there.\n\\end{corollary}","Let \\(H=H^*\\) on a nonzero finite dimensional Hilbert space and let\n\\[\nE_0=\\min\\Spec H.\n\\]\nFor a lattice time step \\(a>0\\), define\n\\[\nT=\\e^{-a(H-E_0I)}.\n\\]","\\begin{theorem}[Finite Euclidean transfer]\\label{strong:theorem:036}\\label{strong:thm:euclidean_transfer}\nThe operator \\(T\\) is positive definite, self adjoint, and contractive. Its spectrum lies in \\((0,1]\\), and\n\\[\nT^n=\\e^{-na(H-E_0I)}\n\\]\nfor every \\(n\\in\\NN\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:056}\nDiagonalize \\(H\\). Each eigenvalue of \\(T\\) is\n\\[\n\\e^{-a(E_j-E_0)}\n\\]\nand therefore lies in \\((0,1]\\). The power law follows from the functional calculus.\n\\phantomsection\\label{strong:proof:056:end}\\end{proof}","\\begin{theorem}[Hamiltonian reconstruction from transfer]\\label{strong:theorem:037}\\label{strong:thm:transfer_logarithm}\nLet \\(T\\) be positive definite, self adjoint, and contractive. Then\n\\[\nH_T=-a^{-1}\\log T\n\\]\nis a nonnegative self adjoint operator and\n\\[\nT^n=\\e^{-naH_T}.\n\\]\nIt is the unique nonnegative self adjoint operator with \\(T=\\e^{-aH_T}\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:057}\nApply the real logarithm to the positive spectrum of \\(T\\). The spectral calculus gives existence and the power identity. Injectivity of the scalar exponential on \\([0,\\infty)\\) gives uniqueness.\n\\phantomsection\\label{strong:proof:057:end}\\end{proof}","\\begin{theorem}[Finite reflection positivity]\\label{strong:theorem:038}\\label{strong:thm:finite_reflection_positivity}\nFor vectors \\(v_1,\\ldots,v_m\\in\\Hcal\\) and nonnegative integers \\(n_1,\\ldots,n_m\\), the matrix\n\\[\nG_{jk}=\\ip{v_j}{T^{n_j+n_k}v_k}\n\\]\nis positive semidefinite. Equivalently,\n\\[\n\\sum_{j,k}\\overline{c_j}c_k\n\\ip{v_j}{T^{n_j+n_k}v_k}\n\\geq0\n\\]\nfor every \\(c\\in\\CC^m\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:058}\nSince \\(T\\) is positive and self adjoint,\n\\[\n\\ip{v_j}{T^{n_j+n_k}v_k}\n=\n\\ip{T^{n_j}v_j}{T^{n_k}v_k}.\n\\]\nThe quadratic form is the squared norm of\n\\[\n\\sum_kc_kT^{n_k}v_k.\n\\]\n\\phantomsection\\label{strong:proof:058:end}\\end{proof}","\\begin{corollary}[Reflection positive correlators]\\label{strong:corollary:014}\\label{strong:cor:reflection_correlators}\nFix a vector \\(\\Omega\\) and operators \\(A_1,\\ldots,A_m\\). The matrix\n\\[\nC_{jk}\n=\n\\ip{A_j\\Omega}{T^{n_j+n_k}A_k\\Omega}\n\\]\nis positive semidefinite.\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:059}\nApply \\cref{strong:thm:finite_reflection_positivity} with \\(v_j=A_j\\Omega\\).\n\\phantomsection\\label{strong:proof:059:end}\\end{proof}","\\begin{proposition}[Ground packet convergence]\\label{strong:proposition:007}\\label{strong:prop:ground_packet}\nLet \\(P_0\\) be the spectral projector of \\(H\\) at \\(E_0\\), and let\n\\[\n\\Delta=\\min\\{E_j-E_0:E_j>E_0\\}.\n\\]\nIf an excited eigenvalue exists, then\n\\[\n\\norm{T^n-P_0}=\\e^{-na\\Delta}.\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:060}\nIn the spectral decomposition, \\(T^n-P_0\\) has eigenvalues zero on the ground space and \\(\\e^{-na(E_j-E_0)}\\) on the excited eigenspaces. The largest is \\(\\e^{-na\\Delta}\\).\n\\phantomsection\\label{strong:proof:060:end}\\end{proof}","\\begin{proposition}[Symmetry resolved transfer]\\label{strong:proposition:008}\\label{strong:prop:symmetry_transfer}\nIf \\([H,P_a]=0\\) for a complete packet family, then\n\\[\nT=\\sum_aP_aTP_a\n\\]\nand each reflection positive Gram matrix decomposes as the sum of its packet Gram matrices.\n\\end{proposition}","A finite spectral datum consists of a finite dimensional Hilbert space, a nonnegative self adjoint operator \\(L\\), and any declared packet family. One may take \\(L=D^2\\) for a finite Dirac type operator or a graph Laplacian. Related geometric viewpoints appear in \\cite{Connes1994,Vassilevich2003}.","\\begin{definition}[Heat trace and spectral dimension]\\label{strong:definition:008}\\label{strong:def:spectral_dimension}\nLet the eigenvalues of \\(L\\) be \\(\\lambda_1,\\ldots,\\lambda_N\\), with at least one zero eigenvalue. Define\n\\[\nZ_L(t)=\\Tr(\\e^{-tL})\n\\]\nand\n\\[\nd_L(t)\n=-2\\frac{\\dd\\log Z_L(t)}{\\dd\\log t}\n=\n2t\n\\frac{\\Tr(L\\e^{-tL})}{\\Tr(\\e^{-tL})}.\n\\]\n\\end{definition}","\\begin{proposition}[Finite scale limits]\\label{strong:proposition:009}\\label{strong:prop:finite_spectral_limits}\nFor every finite nonnegative \\(L\\) with nontrivial kernel,\n\\[\n\\lim_{t\\downarrow0}d_L(t)=0\n\\]\nand\n\\[\n\\lim_{t\\to\\infty}d_L(t)=0.\n\\]\nFor every \\(t>0\\),\n\\[\n0\\leq d_L(t)\\leq2t\\norm L.\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:062}\nThe formula in \\cref{strong:def:spectral_dimension} is twice \\(t\\) times a weighted average of the eigenvalues. This gives the bound and the limit at zero. At large time, the denominator converges to \\(\\dim\\Ker L\\), while every positive eigenvalue contribution in the numerator decays exponentially.\n\\phantomsection\\label{strong:proof:062:end}\\end{proof}","\\begin{proposition}[Heat packet decomposition]\\label{strong:proposition:010}\\label{strong:prop:heat_packets}\nIf \\([L,P_a]=0\\) for every packet, then\n\\[\nZ_L(t)=\\sum_aZ_{L,a}(t),\n\\qquad\nZ_{L,a}(t)=\\Tr(P_a\\e^{-tL}).\n\\]\nThe global spectral dimension is the average of the packet dimensions weighted by their heat traces:\n\\[\nd_L(t)\n=\n\\sum_{a:Z_{L,a}(t)>0}\n\\frac{Z_{L,a}(t)}{Z_L(t)}\n d_{L,a}(t),\n\\]\nwhere\n\\[\nd_{L,a}(t)\n=2t\n\\frac{\\Tr(P_aL\\e^{-tL})}{Z_{L,a}(t)}\n\\]\nfor nonempty packet heat traces.\n\\end{proposition}","begin{proposition}[Heat packet decomposition]\\label{strong:proposition:010}\\label{strong:prop:heat_packets}\nIf \\([L,P_a]=0\\) for every packet, then\n\\[\nZ_L(t)=\\sum_aZ_{L,a}(t),\n\\qquad\nZ_{L,a}(t)=\\Tr(P_a\\e^{-tL}).\n\\]\nThe global spectral dimension is the average of the packet dimensions weighted by their heat traces:\n\\[\nd_L(t)\n=\n\\sum_{a:Z_{L,a}(t)>0}\n\\frac{Z_{L,a}(t)}{Z_L(t)}\n d_{L,a}(t),\n\\]\nwhere\n\\[\nd_{L,a}(t)\n=2t\n\\frac{\\Tr(P_aL\\e^{-tL})}{Z_{L,a}(t)}\n\\]\nfor nonempty packet heat traces.\n\\end{proposition}","Let \\(N\\geq2\\) and let \\(\\Lambda_N=(\\ZZ/N\\ZZ)^d\\) be the unit flat torus grid with spacing \\(1/N\\). The standard periodic difference Laplacian has eigenvalues\n\\[\n\\lambda_N(k)\n=\n4N^2\n\\sum_{j=1}^{d}\n\\sin^2\\left(\\frac{\\pi k_j}{N}\\right),\n\\]\nwhere \\(k\\) runs through centered residue representatives. This is the standard periodic finite difference Laplacian \\cite{Strikwerda2004}.","\\begin{theorem}[Low mode dispersion in dimension \\(d\\)]\\label{strong:theorem:039}\\label{strong:thm:dispersion_bound}\nFor every centered mode \\(k\\),\n\\[\n0\n\\leq\n4\\pi^2\\abs{k}^2-\\lambda_N(k)\n\\leq\n\\frac{4\\pi^4}{3N^2}\n\\sum_{j=1}^{d}k_j^4.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:064}\nFor \\(x\\in[0,\\pi/2]\\),\n\\[\n\\sin x\\leq x\n\\]\nand\n\\[\n\\sin x\\geq x-\\frac{x^3}{6}.\n\\]\nThus\n\\[\n0\\leq x^2-\\sin^2x\\leq\\frac{x^4}{3}.\n\\]\nApply this with \\(x=\\pi\\abs{k_j}/N\\), multiply by \\(4N^2\\), and sum over \\(j\\).\n\\phantomsection\\label{strong:proof:064:end}\\end{proof}","\\begin{theorem}[Fixed time heat trace convergence]\\label{strong:theorem:040}\\label{strong:thm:heat_trace_convergence}\nFor every \\(t>0\\),\n\\[\n\\sum_{k\\in\\Lambda_N^*}\n\\e^{-t\\lambda_N(k)}\n\\longrightarrow\n\\sum_{k\\in\\ZZ^d}\n\\e^{-4\\pi^2t\\abs{k}^2}\n\\]\nas \\(N\\to\\infty\\). The first spectral moments also converge:\n\\[\n\\sum_{k\\in\\Lambda_N^*}\n\\lambda_N(k)\\e^{-t\\lambda_N(k)}\n\\longrightarrow\n\\sum_{k\\in\\ZZ^d}\n4\\pi^2\\abs{k}^2\\e^{-4\\pi^2t\\abs{k}^2}.\n\\]\nConsequently, the discrete spectral dimensions converge to the continuum torus spectral dimension at every fixed positive time.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:065}\nFor a fixed integer mode, \\cref{strong:thm:dispersion_bound} gives\n\\[\n\\lambda_N(k)\\longrightarrow4\\pi^2\\abs{k}^2.\n\\]\nFor centered modes, the inequality\n\\[\n\\sin x\\geq\\frac{2x}{\\pi},\n\\qquad\n0\\leq x\\leq\\frac\\pi2,\n\\]\ngives\n\\[\n\\lambda_N(k)\\geq16\\abs{k}^2.\n\\]\nAlso \\(\\lambda_N(k)\\leq4\\pi^2\\abs{k}^2\\). Extend the discrete summands by zero outside the centered residue box. The heat trace terms are dominated by\n\\[\n\\e^{-16t\\abs{k}^2},\n\\]\nand the moment terms by\n\\[\n4\\pi^2\\abs{k}^2\\e^{-16t\\abs{k}^2}.\n\\]\nBoth dominating sequences are summable on \\(\\ZZ^d\\). Dominated convergence proves the two limits. Their quotient gives spectral dimension convergence.\n\\phantomsection\\label{strong:proof:065:end}\\end{proof}","ixed integer mode, \\cref{strong:thm:dispersion_bound} gives\n\\[\n\\lambda_N(k)\\longrightarrow4\\pi^2\\abs{k}^2.\n\\]\nFor centered modes, the inequality\n\\[\n\\sin x\\geq\\frac{2x}{\\pi},\n\\qquad\n0\\leq x\\leq\\frac\\pi2,\n\\]\ngives\n\\[\n\\lambda_N(k)\\geq16\\abs{k}^2.\n\\]\nAlso \\(\\lambda_N(k)\\leq4\\pi^2\\abs{k}^2\\). Extend the discrete summands by zero outside the centered residue box. The heat trace terms are dominated by\n\\[\n\\e^{-16t\\abs{k}^2},\n\\]\nand the moment terms by\n\\[\n4\\pi^2\\abs{k}^2\\e^{-16t\\abs{k}^2}.\n\\]\nBoth dominating sequences are summable on \\(\\ZZ^d\\). Dominated convergence proves the two limits. Their quotient gives spectral dimension convergence.\n\\phantomsection\\label{strong:proof:065:end}\\end{proof}","nd} gives\n\\[\n\\lambda_N(k)\\longrightarrow4\\pi^2\\abs{k}^2.\n\\]\nFor centered modes, the inequality\n\\[\n\\sin x\\geq\\frac{2x}{\\pi},\n\\qquad\n0\\leq x\\leq\\frac\\pi2,\n\\]\ngives\n\\[\n\\lambda_N(k)\\geq16\\abs{k}^2.\n\\]\nAlso \\(\\lambda_N(k)\\leq4\\pi^2\\abs{k}^2\\). Extend the discrete summands by zero outside the centered residue box. The heat trace terms are dominated by\n\\[\n\\e^{-16t\\abs{k}^2},\n\\]\nand the moment terms by\n\\[\n4\\pi^2\\abs{k}^2\\e^{-16t\\abs{k}^2}.\n\\]\nBoth dominating sequences are summable on \\(\\ZZ^d\\). Dominated convergence proves the two limits. Their quotient gives spectral dimension convergence.\n\\phantomsection\\label{strong:proof:065:end}\\end{proof}","inequality\n\\[\n\\sin x\\geq\\frac{2x}{\\pi},\n\\qquad\n0\\leq x\\leq\\frac\\pi2,\n\\]\ngives\n\\[\n\\lambda_N(k)\\geq16\\abs{k}^2.\n\\]\nAlso \\(\\lambda_N(k)\\leq4\\pi^2\\abs{k}^2\\). Extend the discrete summands by zero outside the centered residue box. The heat trace terms are dominated by\n\\[\n\\e^{-16t\\abs{k}^2},\n\\]\nand the moment terms by\n\\[\n4\\pi^2\\abs{k}^2\\e^{-16t\\abs{k}^2}.\n\\]\nBoth dominating sequences are summable on \\(\\ZZ^d\\). Dominated convergence proves the two limits. Their quotient gives spectral dimension convergence.\n\\phantomsection\\label{strong:proof:065:end}\\end{proof}","\\begin{standardfact}[Theta transformation]\\label{strong:standardfact:004}\\label{strong:fact:theta_transform}\nThe continuum flat torus heat trace satisfies\n\\[\nZ_d(t)\n=\n\\sum_{k\\in\\ZZ^d}\\e^{-4\\pi^2t\\abs{k}^2}\n=\n(4\\pi t)^{-d/2}\n\\sum_{n\\in\\ZZ^d}\\e^{-\\abs{n}^2/(4t)}.\n\\]\nThis is the Poisson summation formula for a Gaussian.\n\\end{standardfact}","\\begin{theorem}[Continuum torus small time dimension]\\label{strong:theorem:041}\\label{strong:thm:continuum_dimension}\nLet\n\\[\nd_{\\rm torus}(t)\n=-2\\frac{\\dd\\log Z_d(t)}{\\dd\\log t}.\n\\]\nThen\n\\[\nd_{\\rm torus}(t)\n=\n d+O\\left(t^{-1}\\e^{-1/(4t)}\\right)\n\\]\nas \\(t\\downarrow0\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:066}\nBy \\cref{strong:fact:theta_transform},\n\\[\nZ_d(t)\n=(4\\pi t)^{-d/2}(1+R(t)),\n\\]\nwhere\n\\[\nR(t)=\\sum_{n\\neq0}\\e^{-\\abs{n}^2/(4t)}\n=O\\bigl(\\e^{-1/(4t)}\\bigr).\n\\]\nDifferentiation gives\n\\[\ntR'(t)=O\\left(t^{-1}\\e^{-1/(4t)}\\right).\n\\]\nSubstituting this estimate into the logarithmic derivative gives the stated small time expansion.\n\\phantomsection\\label{strong:proof:066:end}\\end{proof}","\\begin{corollary}[Rigorous intermediate dimension window]\\label{strong:corollary:015}\\label{strong:cor:dimension_window}\nFor every \\(\\varepsilon>0\\), there is a diffusion time \\(t_\\varepsilon>0\\) and an integer \\(N_\\varepsilon\\) such that\n\\[\n\\abs{d_{L_N}(t_\\varepsilon)-d}<\\varepsilon\n\\]\nfor every \\(N\\geq N_\\varepsilon\\).\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:067}\nChoose \\(t_\\varepsilon\\) so that \\cref{strong:thm:continuum_dimension} is within \\(\\varepsilon/2\\) of \\(d\\). Then apply \\cref{strong:thm:heat_trace_convergence} to choose \\(N_\\varepsilon\\) so that the discrete spectral dimension is within \\(\\varepsilon/2\\) of the continuum value.\n\\phantomsection\\label{strong:proof:067:end}\\end{proof}","The order of limits matters. Taking \\(t\\downarrow0\\) at fixed \\(N\\) gives zero. Taking \\(N\\to\\infty\\) at fixed positive \\(t\\), then taking \\(t\\downarrow0\\), gives the continuum dimension.","Let \\(C_1,\\ldots,C_m\\) be self adjoint constraint operators on a finite dimensional Hilbert space. The distinction between constraint action and constraint closure is standard in constrained Hamiltonian theory \\cite{Dirac1964,HenneauxTeitelboim1992}. Write\n\\[\n\\mathfrak c=\\operatorname{span}_{\\CC}\\{C_1,\\ldots,C_m\\}\n\\subseteq\\End(\\Hcal).\n\\]","\\begin{definition}[Universal closure obstruction]\\label{strong:definition:009}\\label{strong:def:closure_obstruction}\nLet\n\\[\nq_{\\mathfrak c}:\\End(\\Hcal)\\longrightarrow\\End(\\Hcal)/\\mathfrak c\n\\]\nbe the quotient map. Define\n\\[\n\\Ob_{ab}^{\\rm cl}\n=\nq_{\\mathfrak c}([C_a,C_b]).\n\\]\n\\end{definition}","\\begin{theorem}[Finite constraint closure criterion]\\label{strong:theorem:042}\\label{strong:thm:constraint_closure}\nThe constraint span \\(\\mathfrak c\\) is closed under commutators exactly when\n\\[\n\\Ob_{ab}^{\\rm cl}=0\n\\]\nfor all \\(a,b\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:068}\nThe quotient class vanishes exactly when \\([C_a,C_b]\\in\\mathfrak c\\). This holds for every pair exactly when \\(\\mathfrak c\\) is a Lie algebra under the commutator.\n\\phantomsection\\label{strong:proof:068:end}\\end{proof}","If structure coefficients \\(f_{ab}^{\\phantom{ab}c}\\) are prescribed, define the fixed structure anomaly\n\\[\n\\mathfrak A_{ab}\n=\n[C_a,C_b]\n-\\ii\\sum_cf_{ab}^{\\phantom{ab}c}C_c.\n\\]\nExact realization of the prescribed algebra is the stronger condition \\(\\mathfrak A_{ab}=0\\).","\\begin{proposition}[Physical compression is not a closure test]\\label{strong:proposition:011}\\label{strong:prop:physical_compression_trivial}\nLet\n\\[\n\\Hcal_{\\phys}=\\bigcap_a\\Ker C_a\n\\]\nand let \\(P_{\\phys}\\) be its orthogonal projector. Then\n\\[\nP_{\\phys}[C_a,C_b]P_{\\phys}=0\n\\]\nfor every pair \\(a,b\\), whether or not the constraint span closes.\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:069}\nSelf adjointness and \\(C_aP_{\\phys}=0\\) give \\(P_{\\phys}C_a=0\\). Both terms in the compressed commutator therefore vanish.\n\\phantomsection\\label{strong:proof:069:end}\\end{proof}","\\begin{theorem}[Invariance of the physical space]\\label{strong:theorem:043}\\label{strong:thm:physical_invariance}\nLet \\(H=H^*\\). Suppose\n\\[\n[H,C_a]\n=\n\\sum_cb_a^{\\phantom a c}C_c\n\\]\nfor every \\(a\\). Then \\(H\\) preserves \\(\\Hcal_{\\phys}\\), and so does \\(\\e^{-\\ii tH}\\) for every real \\(t\\).\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:070}\nFor \\(\\psi\\in\\Hcal_{\\phys}\\),\n\\[\nC_aH\\psi\n=HC_a\\psi-[H,C_a]\\psi\n=0.\n\\]\nThus \\(H\\psi\\in\\Hcal_{\\phys}\\). The exponential preserves every invariant finite dimensional subspace.\n\\phantomsection\\label{strong:proof:070:end}\\end{proof}","Let \\((\\Hcal_\\ell,C_{\\ell,a})\\) and \\((\\Hcal_{\\ell+1},C_{\\ell+1,a})\\) be two levels, and let\n\\[\nR_\\ell:\\Hcal_\\ell\\longrightarrow\\Hcal_{\\ell+1}\n\\]\nbe an isometry.","\\begin{theorem}[Exact constraint transport]\\label{strong:theorem:044}\\label{strong:thm:constraint_transport}\nAssume\n\\[\nC_{\\ell+1,a}R_\\ell=R_\\ell C_{\\ell,a}\n\\]\nfor every \\(a\\). Then\n\\[\n[C_{\\ell+1,a},C_{\\ell+1,b}]R_\\ell\n=\nR_\\ell[C_{\\ell,a},C_{\\ell,b}].\n\\]\nIf the same structure coefficients are used at both levels, then\n\\[\n\\mathfrak A_{\\ell+1,ab}R_\\ell\n=\nR_\\ell\\mathfrak A_{\\ell,ab}.\n\\]\nThe isometry maps coarse physical states into fine physical states.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:071}\nApply the intertwining identity twice to each product in the commutator. The anomaly identity follows by subtracting the same linear combination of constraints. If \\(C_{\\ell,a}\\psi=0\\), then\n\\[\nC_{\\ell+1,a}R_\\ell\\psi=R_\\ell C_{\\ell,a}\\psi=0.\n\\]\n\\phantomsection\\label{strong:proof:071:end}\\end{proof}","\\begin{proposition}[Approximate constraint transport]\\label{strong:proposition:012}\\label{strong:prop:approx_constraint_transport}\nAssume\n\\[\n\\norm{C_{\\ell+1,a}R_\\ell-R_\\ell C_{\\ell,a}}\n\\leq\\varepsilon_a\n\\]\nand\n\\[\n\\norm{C_{\\ell,a}},\\norm{C_{\\ell+1,a}}\\leq M.\n\\]\nThen a coarse physical unit vector \\(\\psi\\) satisfies\n\\[\n\\norm{C_{\\ell+1,a}R_\\ell\\psi}\\leq\\varepsilon_a.\n\\]\nAlso,\n\\[\n\\begin{aligned}\n&\\norm{\n[C_{\\ell+1,a},C_{\\ell+1,b}]R_\\ell\n-R_\\ell[C_{\\ell,a},C_{\\ell,b}]\n}","\\begin{proposition}[Approximate constraint transport]\\label{strong:proposition:012}\\label{strong:prop:approx_constraint_transport}\nAssume\n\\[\n\\norm{C_{\\ell+1,a}R_\\ell-R_\\ell C_{\\ell,a}}\n\\leq\\varepsilon_a\n\\]\nand\n\\[\n\\norm{C_{\\ell,a}},\\norm{C_{\\ell+1,a}}\\leq M.\n\\]\nThen a coarse physical unit vector \\(\\psi\\) satisfies\n\\[\n\\norm{C_{\\ell+1,a}R_\\ell\\psi}\\leq\\varepsilon_a.\n\\]\nAlso,\n\\[\n\\begin{aligned}\n&\\norm{\n[C_{\\ell+1,a},C_{\\ell+1,b}]R_\\ell\n-R_\\ell[C_{\\ell,a},C_{\\ell,b}]\n}\\\\\n&\\hspace{4em}\\leq\n2M(\\varepsilon_a+\\varepsilon_b).\n\\end{aligned}\n\\]\n\\end{proposition}","\\begin{proof}\\phantomsection\\label{strong:proof:072}\nThe first claim follows by applying the intertwining defect to \\(\\psi\\). Write\n\\[\nE_a=C_{\\ell+1,a}R_\\ell-R_\\ell C_{\\ell,a}.\n\\]\nExpanding the two commutators gives four error terms:\n\\[\nE_aC_{\\ell,b},\n\\quad\nC_{\\ell+1,a}E_b,\n\\quad\n-E_bC_{\\ell,a},\n\\quad\n-C_{\\ell+1,b}E_a.\n\\]\nThe triangle inequality and the stated bounds on the constraints and defects give the claimed estimate.\n\\phantomsection\\label{strong:proof:072:end}\\end{proof}","\\begin{definition}[Finite quantum refinement tower]\\label{strong:definition:010}\\label{strong:def:quantum_tower}\nA finite quantum refinement tower consists of Hilbert spaces \\(\\Hcal_\\ell\\), isometries\n\\[\nR_\\ell:\\Hcal_\\ell\\longrightarrow\\Hcal_{\\ell+1},\n\\]\nself adjoint Hamiltonians \\(H_\\ell\\), packet projectors \\(P_{\\ell,a}\\), constraints \\(C_{\\ell,r}\\), and positive transfer operators \\(T_\\ell\\). Exact compatibility means that every declared object intertwines with \\(R_\\ell\\). Approximate compatibility is recorded by the corresponding operator norm defects.\n\\end{definition}","tion:010}\\label{strong:def:quantum_tower}\nA finite quantum refinement tower consists of Hilbert spaces \\(\\Hcal_\\ell\\), isometries\n\\[\nR_\\ell:\\Hcal_\\ell\\longrightarrow\\Hcal_{\\ell+1},\n\\]\nself adjoint Hamiltonians \\(H_\\ell\\), packet projectors \\(P_{\\ell,a}\\), constraints \\(C_{\\ell,r}\\), and positive transfer operators \\(T_\\ell\\). Exact compatibility means that every declared object intertwines with \\(R_\\ell\\). Approximate compatibility is recorded by the corresponding operator norm defects.\n\\end{definition}","Let \\(\\Hcal_\\infty\\) be the Hilbert direct limit of the isometric system, and let\n\\[\n\\iota_\\ell:\\Hcal_\\ell\\longrightarrow\\Hcal_\\infty\n\\]\nbe the canonical isometries.","\\begin{theorem}[Bounded Hamiltonian direct limit]\\label{strong:theorem:045}\\label{strong:thm:bounded_direct_limit}\nAssume\n\\[\nH_{\\ell+1}R_\\ell=R_\\ell H_\\ell\n\\]\nand\n\\[\n\\sup_\\ell\\norm{H_\\ell}<\\infty.\n\\]\nThen there is a unique bounded self adjoint operator \\(H_\\infty\\) on \\(\\Hcal_\\infty\\) such that\n\\[\nH_\\infty\\iota_\\ell=\\iota_\\ell H_\\ell\n\\]\nfor every \\(\\ell\\). It satisfies\n\\[\n\\e^{-\\ii tH_\\infty}\\iota_\\ell\n=\n\\iota_\\ell\\e^{-\\ii tH_\\ell}\n\\]\nand\n\\[\n\\e^{-tH_\\infty}\\iota_\\ell\n=\n\\iota_\\ell\\e^{-tH_\\ell}\n\\]\nwhen the Hamiltonians are nonnegative.\n\\end{theorem}","imit]\\label{strong:theorem:045}\\label{strong:thm:bounded_direct_limit}\nAssume\n\\[\nH_{\\ell+1}R_\\ell=R_\\ell H_\\ell\n\\]\nand\n\\[\n\\sup_\\ell\\norm{H_\\ell}<\\infty.\n\\]\nThen there is a unique bounded self adjoint operator \\(H_\\infty\\) on \\(\\Hcal_\\infty\\) such that\n\\[\nH_\\infty\\iota_\\ell=\\iota_\\ell H_\\ell\n\\]\nfor every \\(\\ell\\). It satisfies\n\\[\n\\e^{-\\ii tH_\\infty}\\iota_\\ell\n=\n\\iota_\\ell\\e^{-\\ii tH_\\ell}\n\\]\nand\n\\[\n\\e^{-tH_\\infty}\\iota_\\ell\n=\n\\iota_\\ell\\e^{-tH_\\ell}\n\\]\nwhen the Hamiltonians are nonnegative.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:073}\nDefine \\(H_\\infty\\) on the algebraic union of the embedded levels by\n\\[\nH_\\infty\\iota_\\ell x=\\iota_\\ell H_\\ell x.\n\\]\nCompatibility makes the definition independent of the chosen level. The uniform norm bound makes the operator bounded on the dense union, so it extends uniquely to the completion. It is symmetric because every \\(H_\\ell\\) is self adjoint. A bounded symmetric operator defined on the whole Hilbert space is self adjoint. The functional calculus intertwining follows first for polynomials and then for exponentials by uniform convergence.\n\\phantomsection\\label{strong:proof:073:end}\\end{proof}","\\begin{theorem}[Exact packet direct limit]\\label{strong:theorem:046}\\label{strong:thm:packet_direct_limit}\nAssume the packet families have one fixed finite address set at every level and each packet address \\(a\\) is compatible in the sense\n\\[\nP_{\\ell+1,a}R_\\ell=R_\\ell P_{\\ell,a}.\n\\]\nThen there is an orthogonal projector \\(P_{\\infty,a}\\) on \\(\\Hcal_\\infty\\) satisfying\n\\[\nP_{\\infty,a}\\iota_\\ell=\\iota_\\ell P_{\\ell,a}.\n\\]\nFinite completeness and orthogonality pass to the direct limit.\n\\end{theorem}","\\begin{theorem}[Norm summable Hamiltonian tower]\\label{strong:theorem:047}\\label{strong:thm:norm_summable_tower}\nLet \\((H_n)\\) be bounded self adjoint operators on one Hilbert space and assume\n\\[\n\\sum_{n=0}^{\\infty}\\norm{H_{n+1}-H_n}<\\infty.\n\\]\nThen \\(H_n\\) converges in norm to a bounded self adjoint operator \\(H\\). For every \\(T>0\\),\n\\[\n\\sup_{\\abs t\\leq T}\n\\norm{\\e^{-\\ii tH_n}-\\e^{-\\ii tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\nIf every \\(H_n\\geq0\\), then\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{\\e^{-tH_n}-\\e^{-tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\n\\end{theorem}","Hamiltonian tower]\\label{strong:theorem:047}\\label{strong:thm:norm_summable_tower}\nLet \\((H_n)\\) be bounded self adjoint operators on one Hilbert space and assume\n\\[\n\\sum_{n=0}^{\\infty}\\norm{H_{n+1}-H_n}<\\infty.\n\\]\nThen \\(H_n\\) converges in norm to a bounded self adjoint operator \\(H\\). For every \\(T>0\\),\n\\[\n\\sup_{\\abs t\\leq T}\n\\norm{\\e^{-\\ii tH_n}-\\e^{-\\ii tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\nIf every \\(H_n\\geq0\\), then\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{\\e^{-tH_n}-\\e^{-tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\n\\end{theorem}","\\label{strong:theorem:047}\\label{strong:thm:norm_summable_tower}\nLet \\((H_n)\\) be bounded self adjoint operators on one Hilbert space and assume\n\\[\n\\sum_{n=0}^{\\infty}\\norm{H_{n+1}-H_n}<\\infty.\n\\]\nThen \\(H_n\\) converges in norm to a bounded self adjoint operator \\(H\\). For every \\(T>0\\),\n\\[\n\\sup_{\\abs t\\leq T}\n\\norm{\\e^{-\\ii tH_n}-\\e^{-\\ii tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\nIf every \\(H_n\\geq0\\), then\n\\[\n\\sup_{0\\leq t\\leq T}\n\\norm{\\e^{-tH_n}-\\e^{-tH}}\n\\leq\nT\\norm{H_n-H}.\n\\]\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:075}\nThe summability condition makes \\((H_n)\\) Cauchy in operator norm. The limit is self adjoint. Duhamel's identity gives\n\\[\n\\e^{-\\ii tH_n}-\\e^{-\\ii tH}\n=-\\ii\\int_0^t\n\\e^{-\\ii(t-s)H_n}(H_n-H)\\e^{-\\ii sH}\\dd s.\n\\]\nBoth unitary factors have norm one, which gives the first estimate. The same argument gives the Euclidean estimate because the semigroups generated by nonnegative Hamiltonians are contractions.\n\\phantomsection\\label{strong:proof:075:end}\\end{proof}","\\begin{corollary}[Limit of packet leakage]\\label{strong:corollary:016}\\label{strong:cor:limit_leakage}\nLet \\(P_n\\) be orthogonal projectors and \\(H_n\\) bounded self adjoint operators on one Hilbert space, with \\(P_n\\to P\\) and \\(H_n\\to H\\) in norm. Then for every bounded time interval,\n\\[\n(I-P_n)\\e^{-\\ii tH_n}P_n\n\\longrightarrow\n(I-P)\\e^{-\\ii tH}P\n\\]\nuniformly in time.\n\\end{corollary}","\\begin{proof}\\phantomsection\\label{strong:proof:076}\nAdd and subtract terms so that one factor changes at a time. The projector terms converge uniformly because \\(P_n\\to P\\) in norm. Duhamel's identity gives\n\\[\n\\sup_{\\abs t\\leq T}\n\\norm{\\e^{-\\ii tH_n}-\\e^{-\\ii tH}}\n\\leq\nT\\norm{H_n-H},\n\\]\nwhich proves uniform convergence of the middle factor.\n\\phantomsection\\label{strong:proof:076:end}\\end{proof}","\\begin{theorem}[Reflection positivity is closed under limits]\\label{strong:theorem:048}\\label{strong:thm:reflection_limit}\nSuppose \\(C_n(j,k)\\) are reflection positive kernels on \\(\\NN\\), meaning every finite matrix\n\\[\n\\bigl(C_n(j_r,j_s)\\bigr)_{r,s}\n\\]\nis positive semidefinite. If\n\\[\nC_n(j,k)\\longrightarrow C(j,k)\n\\]\nfor every pair, then \\(C\\) is reflection positive.\n\\end{theorem}","\\begin{proof}\\phantomsection\\label{strong:proof:077}\nFor a fixed finite set of times and coefficients, the corresponding quadratic forms are nonnegative and converge to the quadratic form of \\(C\\). The limit is nonnegative.\n\\phantomsection\\label{strong:proof:077:end}\\end{proof}","\\begin{theorem}[Constraint closure under norm convergence]\\label{strong:theorem:049}\\label{strong:thm:constraint_limit}\nLet \\(C_{n,a}\\to C_a\\) in norm for every \\(a\\), and fix structure coefficients \\(f_{ab}^{\\phantom{ab}c}\\). If\n\\[\n\\norm{\n[C_{n,a},C_{n,b}]\n-\\ii\\sum_cf_{ab}^{\\phantom{ab}c}C_{n,c}\n}\n\\longrightarrow0,\n\\]\nthen\n\\[\n[C_a,C_b]\n=\\ii\\sum_cf_{ab}^{\\phantom{ab}c}C_c.\n\\]\n\\end{theorem}","\\begin{theorem}[Heat trace convergence from trace norm]\\label{strong:theorem:050}\\label{strong:thm:trace_norm_heat}\nLet \\(L_n,L\\) be nonnegative self adjoint operators on one Hilbert space. For a fixed \\(t>0\\), assume their heat operators are trace class and that\n\\[\n\\e^{-tL_n}\\longrightarrow\\e^{-tL}\n\\]\nin trace norm. Then\n\\[\n\\Tr(\\e^{-tL_n})\\longrightarrow\\Tr(\\e^{-tL}).\n\\]\nIf also the heat moment operators are trace class and\n\\[\nL_n\\e^{-tL_n}\\longrightarrow L\\e^{-tL}\n\\]\nin trace norm and the limiting heat trace is nonzero, then the spectral dimensions defined by the heat moment ratio converge at that time.\n\\end{theorem}","ergence from trace norm]\\label{strong:theorem:050}\\label{strong:thm:trace_norm_heat}\nLet \\(L_n,L\\) be nonnegative self adjoint operators on one Hilbert space. For a fixed \\(t>0\\), assume their heat operators are trace class and that\n\\[\n\\e^{-tL_n}\\longrightarrow\\e^{-tL}\n\\]\nin trace norm. Then\n\\[\n\\Tr(\\e^{-tL_n})\\longrightarrow\\Tr(\\e^{-tL}).\n\\]\nIf also the heat moment operators are trace class and\n\\[\nL_n\\e^{-tL_n}\\longrightarrow L\\e^{-tL}\n\\]\nin trace norm and the limiting heat trace is nonzero, then the spectral dimensions defined by the heat moment ratio converge at that time.\n\\end{theorem}","\\begin{definition}[Anomaly free continuum constraint system]\\label{strong:definition:011}\\label{strong:def:anomaly_free_continuum}\nA continuum constraint system consists of a Hilbert space, a common invariant dense domain \\(\\Dcal\\), and operators \\(C_a\\) on \\(\\Dcal\\). It is anomaly free for prescribed structure coefficients when\n\\[\n[C_a,C_b]\\psi\n=\n\\ii\\sum_cf_{ab}^{\\phantom{ab}c}C_c\\psi\n\\]\nfor every \\(\\psi\\in\\Dcal\\), and when the associated gauge evolution is defined on the physical quotient or physical subspace.\n\\end{definition}","in{definition}[Anomaly free continuum constraint system]\\label{strong:definition:011}\\label{strong:def:anomaly_free_continuum}\nA continuum constraint system consists of a Hilbert space, a common invariant dense domain \\(\\Dcal\\), and operators \\(C_a\\) on \\(\\Dcal\\). It is anomaly free for prescribed structure coefficients when\n\\[\n[C_a,C_b]\\psi\n=\n\\ii\\sum_cf_{ab}^{\\phantom{ab}c}C_c\\psi\n\\]\nfor every \\(\\psi\\in\\Dcal\\), and when the associated gauge evolution is defined on the physical quotient or physical subspace.\n\\end{definition}","\\begin{definition}[Einstein infrared criterion]\\label{strong:definition:013}\\label{strong:def:einstein_criterion}\nA quantum geometric tower meets an Einstein infrared criterion only if an effective action is controlled at long wavelengths and has an expansion whose leading geometric terms take the form\n\\[\n\\Gamma_{\\rm eff}[g]\n=\n\\int\n\\sqrt{\\abs g}\n\\left(\n\\Lambda\n+\n\\frac{1}{16\\pi G}R\n+\n\\text{controlled higher order terms}\n\\right)\n\\dd^dx.\n\\]\nHere the volume coefficient includes the sign and normalization written explicitly in the cosmological term in Part I. The expansion must respect the limiting gauge identities and must include an estimate for the omitted terms on the declared infrared regime.\n\\end{definition}","\\section{Exact packets and arbitrary circuit length}\\label{supp1:section:001}\nLet $R$ be a unital ring, not necessarily commutative, and let $I$ be a finite address set. An exact packet family consists of elements $P_a\\in R$ satisfying\n\\[\nP_aP_b=\\begin{cases}P_a,&a=b,","\\section{Exact packets and arbitrary circuit length}\\label{supp1:section:001}\nLet $R$ be a unital ring, not necessarily commutative, and let $I$ be a finite address set. An exact packet family consists of elements $P_a\\in R$ satisfying\n\\[\nP_aP_b=\\begin{cases}P_a,&a=b,\\\\0,&a\\ne b,\\end{cases}\n\\qquad \\sum_{a\\in I}P_a=1.\n\\]\nAdjoint conditions are added when needed. For an operator $U$, define the transfer corner\n\\[\nU_{ba}=P_bUP_a.\n\\]\nThe target index comes first. Completeness and distributivity give\n\\[\n\\sum_{b,a}U_{ba}=U,\\qquad\n(UV)_{ba}=\\sum_c U_{bc}V_{ca}.\n\\]\nThe second identity uses $P_c^2=P_c$ when adjacent transfer corners are multiplied.","0,&a\\ne b,\\end{cases}\n\\qquad \\sum_{a\\in I}P_a=1.\n\\]\nAdjoint conditions are added when needed. For an operator $U$, define the transfer corner\n\\[\nU_{ba}=P_bUP_a.\n\\]\nThe target index comes first. Completeness and distributivity give\n\\[\n\\sum_{b,a}U_{ba}=U,\\qquad\n(UV)_{ba}=\\sum_c U_{bc}V_{ca}.\n\\]\nThe second identity uses $P_c^2=P_c$ when adjacent transfer corners are multiplied.","\\subsection{Circuit paths: source Theorem 4.1}\nFor $U=U_m\\cdots U_1$, the checked result is\n\\[\nP_{a_m}UP_{a_0}\n=\\sum_{a_1,\\ldots,a_{m-1}}\nP_{a_m}U_mP_{a_{m-1}}\\cdots P_{a_1}U_1P_{a_0}.\n\\]\nThe Lean implementation defines a recursive path sum and proves the identity by induction over an arbitrary finite gate list. It is not restricted to a fixed circuit depth. The empty list has circuit operator $1$ and path value $P_bP_a$. Summing the source and target indices reconstructs the full circuit.","it paths: source Theorem 4.1}\nFor $U=U_m\\cdots U_1$, the checked result is\n\\[\nP_{a_m}UP_{a_0}\n=\\sum_{a_1,\\ldots,a_{m-1}}\nP_{a_m}U_mP_{a_{m-1}}\\cdots P_{a_1}U_1P_{a_0}.\n\\]\nThe Lean implementation defines a recursive path sum and proves the identity by induction over an arbitrary finite gate list. It is not restricted to a fixed circuit depth. The empty list has circuit operator $1$ and path value $P_bP_a$. Summing the source and target indices reconstructs the full circuit.","\\subsection{Adjoint conservation: source Theorem 4.2}\nAssume a star operation with $P_a^*=P_a$. Then\n\\[\n(U_{ba})^*=P_aU^*P_b.\n\\]\nThe two checked conservation laws are\n\\begin{align*}\nU^*U=1&\\quad\\Longrightarrow\\quad\n\\sum_b(U_{ba})^*U_{bc}=\\delta_{ac}P_a,","\\subsection{Adjoint conservation: source Theorem 4.2}\nAssume a star operation with $P_a^*=P_a$. Then\n\\[\n(U_{ba})^*=P_aU^*P_b.\n\\]\nThe two checked conservation laws are\n\\begin{align*}\nU^*U=1&\\quad\\Longrightarrow\\quad\n\\sum_b(U_{ba})^*U_{bc}=\\delta_{ac}P_a,\\\\\nUU^*=1&\\quad\\Longrightarrow\\quad\n\\sum_aU_{ba}(U_{da})^*=\\delta_{bd}P_b.\n\\end{align*}\nFor the first law, expand each corner, collapse $P_b^2$ and sum:\n\\[\n\\sum_bP_aU^*P_bUP_c\n=P_aU^*\\left(\\sum_bP_b\\right)UP_c=P_aP_c.\n\\]\nThe second law follows by the analogous calculation and has its own formal proof. For complex matrices, the adjoint is the conjugate transpose. The order of the operator products is preserved throughout.","UU^*=1&\\quad\\Longrightarrow\\quad\n\\sum_aU_{ba}(U_{da})^*=\\delta_{bd}P_b.\n\\end{align*}\nFor the first law, expand each corner, collapse $P_b^2$ and sum:\n\\[\n\\sum_bP_aU^*P_bUP_c\n=P_aU^*\\left(\\sum_bP_b\\right)UP_c=P_aP_c.\n\\]\nThe second law follows by the analogous calculation and has its own formal proof. For complex matrices, the adjoint is the conjugate transpose. The order of the operator products is preserved throughout.","\\section{Symmetry and measurement}\\label{supp1:section:002}\n\\subsection{Exact symmetry: source Theorem 5.1}\nIf $HP_a=P_aH$ for every packet, then\n\\[\nP_bHP_a=0\\quad(b\\ne a),\\qquad H=\\sum_aP_aHP_a.\n\\]\nIn a complex Banach algebra, commutation also propagates to the actual exponential:\n\\[\ne^{zH}P_a=P_ae^{zH}\\qquad(z\\in\\C).\n\\]\nConsequently,\n\\[\ne^{zH}=\\sum_aP_ae^{zH}P_a,\n\\qquad P_be^{zH}P_a=0\\quad(b\\ne a).\n\\]\nTaking $z=-it$ gives evolution in real time. The Lean proof applies the exponential commutation theorem directly, so invariance of the packets is a proved consequence of commutation. Self-adjointness of $H$ is not needed for the algebraic reducing property, so this result applies more broadly than the Hermitian specialization in the paper.","ry and measurement}\\label{supp1:section:002}\n\\subsection{Exact symmetry: source Theorem 5.1}\nIf $HP_a=P_aH$ for every packet, then\n\\[\nP_bHP_a=0\\quad(b\\ne a),\\qquad H=\\sum_aP_aHP_a.\n\\]\nIn a complex Banach algebra, commutation also propagates to the actual exponential:\n\\[\ne^{zH}P_a=P_ae^{zH}\\qquad(z\\in\\C).\n\\]\nConsequently,\n\\[\ne^{zH}=\\sum_aP_ae^{zH}P_a,\n\\qquad P_be^{zH}P_a=0\\quad(b\\ne a).\n\\]\nTaking $z=-it$ gives evolution in real time. The Lean proof applies the exponential commutation theorem directly, so invariance of the packets is a proved consequence of commutation. Self-adjointness of $H$ is not needed for the algebraic reducing property, so this result applies more broadly than the Hermitian specialization in the paper.","Banach algebra, commutation also propagates to the actual exponential:\n\\[\ne^{zH}P_a=P_ae^{zH}\\qquad(z\\in\\C).\n\\]\nConsequently,\n\\[\ne^{zH}=\\sum_aP_ae^{zH}P_a,\n\\qquad P_be^{zH}P_a=0\\quad(b\\ne a).\n\\]\nTaking $z=-it$ gives evolution in real time. The Lean proof applies the exponential commutation theorem directly, so invariance of the packets is a proved consequence of commutation. Self-adjointness of $H$ is not needed for the algebraic reducing property, so this result applies more broadly than the Hermitian specialization in the paper.","\\subsection{Measurements: source Theorem 7.1}\nFor finite complex matrices, define packet dephasing by\n\\[\n\\deph(X)=\\sum_aP_aXP_a.\n\\]\nThe code proves idempotence and trace duality:\n\\[\n\\deph(\\deph(X))=\\deph(X),\\qquad\n\\tr\\bigl(\\deph(E)\\rho\\bigr)=\\tr\\bigl(E\\deph(\\rho)\\bigr).\n\\]\nTrace duality follows term by term from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","nts: source Theorem 7.1}\nFor finite complex matrices, define packet dephasing by\n\\[\n\\deph(X)=\\sum_aP_aXP_a.\n\\]\nThe code proves idempotence and trace duality:\n\\[\n\\deph(\\deph(X))=\\deph(X),\\qquad\n\\tr\\bigl(\\deph(E)\\rho\\bigr)=\\tr\\bigl(E\\deph(\\rho)\\bigr).\n\\]\nTrace duality follows term by term from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","For finite complex matrices, define packet dephasing by\n\\[\n\\deph(X)=\\sum_aP_aXP_a.\n\\]\nThe code proves idempotence and trace duality:\n\\[\n\\deph(\\deph(X))=\\deph(X),\\qquad\n\\tr\\bigl(\\deph(E)\\rho\\bigr)=\\tr\\bigl(E\\deph(\\rho)\\bigr).\n\\]\nTrace duality follows term by term from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","otence and trace duality:\n\\[\n\\deph(\\deph(X))=\\deph(X),\\qquad\n\\tr\\bigl(\\deph(E)\\rho\\bigr)=\\tr\\bigl(E\\deph(\\rho)\\bigr).\n\\]\nTrace duality follows term by term from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","nd trace duality:\n\\[\n\\deph(\\deph(X))=\\deph(X),\\qquad\n\\tr\\bigl(\\deph(E)\\rho\\bigr)=\\tr\\bigl(E\\deph(\\rho)\\bigr).\n\\]\nTrace duality follows term by term from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","from cyclicity:\n\\[\n\\tr(P_aEP_a\\rho)=\\tr(EP_a\\rho P_a).\n\\]\nTherefore an effect that is diagonal in the packets, $\\deph(E)=E$, satisfies\n\\[\n\\boxed{\\tr(E\\rho)=\\tr\\bigl(E\\deph(\\rho)\\bigr).}\n\\]\nThe proof holds for arbitrary finite complex matrices $E$ and $\\rho$. Density matrices and positive measurement effects are valid specializations rather than assumptions needed for this trace identity. Taking $E=1$ also gives trace preservation of packet dephasing.","\\section{Corrected Feshbach resolvent}\\label{supp1:section:003}\nWrite a finite complex matrix in retained and eliminated coordinates:\n\\[\nH=\\begin{pmatrix}A&B\\\\C&D\\end{pmatrix},\\qquad\nzI-H=\\begin{pmatrix}zI_P-A&-B\\\\-C&zI_Q-D\\end{pmatrix}.\n\\]\nSet\n\\[\nD_z=zI_Q-D,\\qquad S_z=zI_P-A-BD_z^{-1}C.\n\\]\nThe effective Hamiltonian is $H_{\\mathrm{eff}}(z)=A+BD_z^{-1}C$, so\n$S_z=zI_P-H_{\\mathrm{eff}}(z)$.","-C&zI_Q-D\\end{pmatrix}.\n\\]\nSet\n\\[\nD_z=zI_Q-D,\\qquad S_z=zI_P-A-BD_z^{-1}C.\n\\]\nThe effective Hamiltonian is $H_{\\mathrm{eff}}(z)=A+BD_z^{-1}C$, so\n$S_z=zI_P-H_{\\mathrm{eff}}(z)$.","\\subsection{Correct statement: source Theorem 9.1}\nAssume both $D_z$ and $zI-H$ are invertible. Then $S_z$ is invertible and\n\\[\n\\boxed{\\bigl((zI-H)^{-1}\\bigr)_{PP}=S_z^{-1}.}\n\\]\nThe left side is the submatrix of the full inverse on the retained coordinates. This matches the paper's compressed operator when written in an orthonormal basis of the retained subspace. The proof is valid for arbitrary complex blocks of compatible sizes and does not require Hermiticity.","The checked invertibility criterion is\n\\[\nD_z\\text{ invertible}\\quad\\Longrightarrow\\quad\n\\bigl(zI-H\\text{ invertible}\\ \\Longleftrightarrow\\ S_z\\text{ invertible}\\bigr).\n\\]\nThus one may equivalently assume invertibility of $D_z$ and $S_z$.","\\textbf{Why the original assumption is insufficient.}\nTake $z=0$ and $H=\\diag(0,-1)$, retaining the first coordinate. Then\n\\[\nD_z=1,\\qquad zI-H=\\begin{pmatrix}0&0","\\textbf{Why the original assumption is insufficient.}\nTake $z=0$ and $H=\\diag(0,-1)$, retaining the first coordinate. Then\n\\[\nD_z=1,\\qquad zI-H=\\begin{pmatrix}0&0\\\\0&1\\end{pmatrix}.\n\\]\nThe eliminated block is invertible, but the full matrix is singular. Hence the original hypothesis does not ensure the existence of its stated resolvent. The Lean counterexample proves invertibility of the scalar $1$ and noninvertibility of this full $2\\times2$ matrix.","0&1\\end{pmatrix}.\n\\]\nThe eliminated block is invertible, but the full matrix is singular. Hence the original hypothesis does not ensure the existence of its stated resolvent. The Lean counterexample proves invertibility of the scalar $1$ and noninvertibility of this full $2\\times2$ matrix.","\\section{Dispersion with an explicit spacing condition}\\label{supp1:section:004}\nThe grid eigenvalue in source Theorem 11.1 is\n\\[\n\\lambda_{k,h}=\\frac{4}{h^2}\\sin^2\\left(\\frac{kh}{2}\\right).\n\\]\nThe original low-mode condition is $0\\le kh/2\\le1$. A nonzero spacing is also required for the quotient to have its intended meaning. For a physical grid, one normally takes $h>0$.","\\subsection{Checked stronger statement}\nFor every $k,h\\in\\R$ with $h\\ne0$,\n\\[\n\\boxed{0\\le k^2-\\lambda_{k,h}\\le\\frac{k^4h^2}{12}.}\n\\]\nThe code proves this without the low-mode restriction. The bound may be loose away from the low-mode regime, but it remains valid.","The proof begins with an estimate valid for every real argument:\n\\[\n0\\le x^2-\\sin^2x\\le\\frac{x^4}{3}.\n\\]\nThe lower bound follows from $|\\sin x|\\le|x|$. For the upper bound, use\n\\[\n|x-\\sin x|\\le\\frac{|x|^3}{6},\\qquad\n|x+\\sin x|\\le2|x|.\n\\]\nFactoring the difference of squares gives\n\\[\n0\\le x^2-\\sin^2x\n=|x-\\sin x|\\,|x+\\sin x|\n\\le\\frac{x^4}{3}.\n\\]\nWith $x=kh/2$ and $h\\ne0$,\n\\begin{align*}\nk^2-\\lambda_{k,h}\n&=\\frac4{h^2}\\left(x^2-\\sin^2x\\right),","The proof begins with an estimate valid for every real argument:\n\\[\n0\\le x^2-\\sin^2x\\le\\frac{x^4}{3}.\n\\]\nThe lower bound follows from $|\\sin x|\\le|x|$. For the upper bound, use\n\\[\n|x-\\sin x|\\le\\frac{|x|^3}{6},\\qquad\n|x+\\sin x|\\le2|x|.\n\\]\nFactoring the difference of squares gives\n\\[\n0\\le x^2-\\sin^2x\n=|x-\\sin x|\\,|x+\\sin x|\n\\le\\frac{x^4}{3}.\n\\]\nWith $x=kh/2$ and $h\\ne0$,\n\\begin{align*}\nk^2-\\lambda_{k,h}\n&=\\frac4{h^2}\\left(x^2-\\sin^2x\\right),\\\\\n\\frac4{h^2}\\frac{x^4}{3}\n&=\\frac{k^4h^2}{12}.\n\\end{align*}\nEvery algebraic transformation and inequality in the active source was accepted by Lean.","\\section{A constraint diagnostic that vanishes automatically}\\label{supp1:section:005}\nSection 12 defines the physical projector using the exact common kernel of the constraints. In finite dimensions this gives $C_aP=0$ for every $a$. Absorb the factor $i$ and any structure coefficient into $f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","s automatically}\\label{supp1:section:005}\nSection 12 defines the physical projector using the exact common kernel of the constraints. In finite dimensions this gives $C_aP=0$ for every $a$. Absorb the factor $i$ and any structure coefficient into $f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","bel{supp1:section:005}\nSection 12 defines the physical projector using the exact common kernel of the constraints. In finite dimensions this gives $C_aP=0$ for every $a$. Absorb the factor $i$ and any structure coefficient into $f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","ection:005}\nSection 12 defines the physical projector using the exact common kernel of the constraints. In finite dimensions this gives $C_aP=0$ for every $a$. Absorb the factor $i$ and any structure coefficient into $f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","constraints. In finite dimensions this gives $C_aP=0$ for every $a$. Absorb the factor $i$ and any structure coefficient into $f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","$f_c$, and write\n\\[\n\\mathcal A_{ab}=C_aC_b-C_bC_a-\\sum_cf_cC_c.\n\\]\n\\begin{proposition}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","n}[Exact common-kernel compression]\\label{supp1:proposition:001}\nIf $C_cP=0$ for every $c$, then\n\\[\n\\mathcal A_{ab}P=0,\\qquad P\\mathcal A_{ab}P=0.\n\\]\nNo idempotence or adjoint property of $P$ is needed for this implication.\n\\end{proposition}\n\\begin{proof}\\phantomsection\\label{supp1:proof:001}\nAssociativity gives\n\\[\n\\mathcal A_{ab}P=C_a(C_bP)-C_b(C_aP)-\\sum_cf_c(C_cP)=0.\n\\]\nLeft multiplication by $P$ gives the second identity.\n\\phantomsection\\label{supp1:proof:001:end}\\end{proof}","\\subsection{Compressed vanishing without operator closure}\nOver the rationals, take\n\\[\nP=\\begin{pmatrix}1&0&0\\\\0&0&0\\\\0&0&0\\end{pmatrix},\\quad\nX=\\begin{pmatrix}0&0&0\\\\0&0&1\\\\0&1&0\\end{pmatrix},\\quad\nZ=\\begin{pmatrix}0&0&0\\\\0&1&0\\\\0&0&-1\\end{pmatrix}.\n\\]\nThese displayed matrices are symmetric. The code checks that $P^2=P$, $P\\ne0$ and $XP=ZP=0$, but\n\\[\n[X,Z]=\\begin{pmatrix}0&0&0","0&0&-1\\end{pmatrix}.\n\\]\nThese displayed matrices are symmetric. The code checks that $P^2=P$, $P\\ne0$ and $XP=ZP=0$, but\n\\[\n[X,Z]=\\begin{pmatrix}0&0&0","0&0&-1\\end{pmatrix}.\n\\]\nThese displayed matrices are symmetric. The code checks that $P^2=P$, $P\\ne0$ and $XP=ZP=0$, but\n\\[\n[X,Z]=\\begin{pmatrix}0&0&0\\\\0&0&-2\\\\0&2&0\\end{pmatrix}\\ne0,\n\\qquad P[X,Z]P=0.\n\\]\nThe commutator therefore violates the closure relation with zero structure coefficients outside the physical range, even though its physical compression vanishes exactly.","For $n\\geq 0$, let\n\\[\n A_n=(\\mathbb Z/2\\mathbb Z)^{\\{0,\\ldots,n-1\\}},\\qquad\n V_n=\\{v:A_n\\to\\C\\}.\n\\]\nFor an address $a\\in A_n$, define the standard coordinate vector\n\\[\n \\ket a(x)=\\begin{cases}1&x=a,","For $n\\geq 0$, let\n\\[\n A_n=(\\mathbb Z/2\\mathbb Z)^{\\{0,\\ldots,n-1\\}},\\qquad\n V_n=\\{v:A_n\\to\\C\\}.\n\\]\nFor an address $a\\in A_n$, define the standard coordinate vector\n\\[\n \\ket a(x)=\\begin{cases}1&x=a,\\\\0&x\\ne a.\\end{cases}\n\\]\nThe seed is $\\psi_0=\\ket0$. Define a translation operator by\n\\[\n (T_a v)(x)=v(x-a).\n\\]\nAt position $j$, the local Pauli $X$ is $X_j=T_{e_j}$, where $e_j$ has a single nonzero binary coordinate. The formal circuit uses the identity when $a_j=0$ and $X_j$ when $a_j=1$.","0&x\\ne a.\\end{cases}\n\\]\nThe seed is $\\psi_0=\\ket0$. Define a translation operator by\n\\[\n (T_a v)(x)=v(x-a).\n\\]\nAt position $j$, the local Pauli $X$ is $X_j=T_{e_j}$, where $e_j$ has a single nonzero binary coordinate. The formal circuit uses the identity when $a_j=0$ and $X_j$ when $a_j=1$.","\\begin{statement}\\label{supp2:statement:001}\nFor every binary address $a$,\n\\[\n W_a=\\prod_{j=0}^{n-1}X_j^{a_j}=T_a,\n \\qquad W_a\\psi_0=\\ket a.\n\\]\nThe family $(W_a\\psi_0)_{a\\in A_n}$ is linearly independent, spans $V_n$, and has exactly $2^n$ elements. It is the standard computational basis.\n\\end{statement}","\\begin{proof}\\phantomsection\\label{supp2:proof:001}\nComposition gives $T_aT_b=T_{a+b}$, and direct evaluation gives $T_a\\ket b=\\ket{a+b}$. Every binary address satisfies $a+a=0$, so each translation is an involution. In particular, the local operator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","2:proof:001}\nComposition gives $T_aT_b=T_{a+b}$, and direct evaluation gives $T_a\\ket b=\\ket{a+b}$. Every binary address satisfies $a+a=0$, so each translation is an involution. In particular, the local operator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","mposition gives $T_aT_b=T_{a+b}$, and direct evaluation gives $T_a\\ket b=\\ket{a+b}$. Every binary address satisfies $a+a=0$, so each translation is an involution. In particular, the local operator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","particular, the local operator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","r, the local operator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","perator changes precisely the addressed bit:\n\\[\n X_j\\ket a=\\ket{a\\text{ with coordinate }j\\text{ replaced by }a_j+1}.\n\\]\nThe selected factor at $j$ equals translation by the address supported at $j$ with value $a_j$. Induction on a list proves that the product of translations is translation by their sum. Each of these addresses contributes one coordinate, and their sum is $a$. Thus $W_a=T_a$ and $W_a\\psi_0=\\ket a$.","Lean identifies the resulting family with \\lean{Pi.basisFun} over $\\C$. Its standard basis theorems supply linear independence and full span. The finite function-space cardinality gives $|A_n|=2^n$.\n\\phantomsection\\label{supp2:proof:001:end}\\end{proof}","\\textbf{Formal scope.} The proof constructs the sequence of gates that produces each addressed state and then identifies the resulting family with the standard coordinate basis. It includes the empty circuit for $n=0$. Orthogonality belongs to the usual computational basis interpretation. The formal endpoint records the basis identification, together with separate explicit theorems for span and linear independence.","Let $F$ be an arbitrary finite set of factors and let $D_i$ be a finite coordinate index for factor $i$. For matrix families $A_i$, define the coordinate tensor product on the product index space by\n\\[\n \\mathsf T(A)(x,y)=\\prod_{i\\in F} A_i(x_i,y_i).\n\\]\nThe proof first establishes the laws for multiplication, identity, adjoints, and finite sums. These laws apply to any finite number of factors with the stated finite coordinate sets.","Suppose $N>0$ and each factor has a complete orthogonal self-adjoint projector family $(P^{(i)}_j)_{j\\in\\mathbb Z/N\\mathbb Z}$. For a fine address $a:F\\to\\mathbb Z/N\\mathbb Z$, set\n\\[\n Q_a=\\mathsf T\\bigl(i\\mapsto P^{(i)}_{a_i}\\bigr),\n \\qquad\n C_r=\\sum_{\\sum_i a_i=r}Q_a.\n\\]","\\begin{statement}\\label{supp2:statement:002}\nThe fine and coarse families are complete orthogonal self-adjoint projector systems. In particular,\n\\[\n C_rC_s=\\delta_{rs}C_r,\\qquad\n \\sum_r C_r=I,\\qquad C_r^*=C_r.\n\\]\nEach coarse projector is exactly the displayed sum of its fine descendants.\n\\end{statement}","\\begin{proof}\\phantomsection\\label{supp2:proof:002}\nExpanding the finite product of sums gives\n\\[\n \\mathsf T(A)\\mathsf T(B)=\\mathsf T(i\\mapsto A_iB_i).\n\\]\nConsequently, if two fine addresses differ in any coordinate, the corresponding product contains a zero factor. If the addresses agree, the product is the same fine projector. The adjoint passes through the finite coordinate product, so $Q_a^*=Q_a$. Summing over all address functions and using the same expansion yields\n\\[\n \\sum_a Q_a=\\mathsf T\\left(i\\mapsto\\sum_jP^{(i)}_j\\right)=I.\n\\]\nA separate algebraic regrouping theorem applies to any finite orthogonal family and any address map into a finite set. In the double sum for $C_rC_s$, orthogonality removes all terms with different fine addresses. A surviving address can lie in both classes only when $r=s$. Completeness follows by summing ea","\\begin{proof}\\phantomsection\\label{supp2:proof:002}\nExpanding the finite product of sums gives\n\\[\n \\mathsf T(A)\\mathsf T(B)=\\mathsf T(i\\mapsto A_iB_i).\n\\]\nConsequently, if two fine addresses differ in any coordinate, the corresponding product contains a zero factor. If the addresses agree, the product is the same fine projector. The adjoint passes through the finite coordinate product, so $Q_a^*=Q_a$. Summing over all address functions and using the same expansion yields\n\\[\n \\sum_a Q_a=\\mathsf T\\left(i\\mapsto\\sum_jP^{(i)}_j\\right)=I.\n\\]\nA separate algebraic regrouping theorem applies to any finite orthogonal family and any address map into a finite set. In the double sum for $C_rC_s$, orthogonality removes all terms with different fine addresses. A surviving address can lie in both classes only when $r=s$. Completeness follows by summing each fine projector exactly once. Adjoint preservation follows term by term. Specializing the address map to total residue proves the claim.\n\\phantomsection\\label{supp2:proof:002:end}\\end{proof}","ct of sums gives\n\\[\n \\mathsf T(A)\\mathsf T(B)=\\mathsf T(i\\mapsto A_iB_i).\n\\]\nConsequently, if two fine addresses differ in any coordinate, the corresponding product contains a zero factor. If the addresses agree, the product is the same fine projector. The adjoint passes through the finite coordinate product, so $Q_a^*=Q_a$. Summing over all address functions and using the same expansion yields\n\\[\n \\sum_a Q_a=\\mathsf T\\left(i\\mapsto\\sum_jP^{(i)}_j\\right)=I.\n\\]\nA separate algebraic regrouping theorem applies to any finite orthogonal family and any address map into a finite set. In the double sum for $C_rC_s$, orthogonality removes all terms with different fine addresses. A surviving address can lie in both classes only when $r=s$. Completeness follows by summing each fine projector exactly once. Adjoint preservation follows term by term. Specializing the address map to total residue proves the claim.\n\\phantomsection\\label{supp2:proof:002:end}\\end{proof}","gree, the product is the same fine projector. The adjoint passes through the finite coordinate product, so $Q_a^*=Q_a$. Summing over all address functions and using the same expansion yields\n\\[\n \\sum_a Q_a=\\mathsf T\\left(i\\mapsto\\sum_jP^{(i)}_j\\right)=I.\n\\]\nA separate algebraic regrouping theorem applies to any finite orthogonal family and any address map into a finite set. In the double sum for $C_rC_s$, orthogonality removes all terms with different fine addresses. A surviving address can lie in both classes only when $r=s$. Completeness follows by summing each fine projector exactly once. Adjoint preservation follows term by term. Specializing the address map to total residue proves the claim.\n\\phantomsection\\label{supp2:proof:002:end}\\end{proof}","over all address functions and using the same expansion yields\n\\[\n \\sum_a Q_a=\\mathsf T\\left(i\\mapsto\\sum_jP^{(i)}_j\\right)=I.\n\\]\nA separate algebraic regrouping theorem applies to any finite orthogonal family and any address map into a finite set. In the double sum for $C_rC_s$, orthogonality removes all terms with different fine addresses. A surviving address can lie in both classes only when $r=s$. Completeness follows by summing each fine projector exactly once. Adjoint preservation follows term by term. Specializing the address map to total residue proves the claim.\n\\phantomsection\\label{supp2:proof:002:end}\\end{proof}","Fix a real time $t$ and a finite complex coordinate space. Let $U(s)$ be a unitary matrix curve, differentiable entrywise at $t$, and let $\\psi(s)$ be differentiable componentwise at $t$. Write $\\dot U$ and $\\dot\\psi$ for those derivatives. Suppose\n\\[\n i\\dot\\psi=H\\psi(t).\n\\]\nOnly the value of the Hamiltonian at the chosen time is needed. No derivative of $H$ is assumed.","\\begin{statement}\\label{supp2:statement:003}\nThe explicitly defined coordinate curve\n\\[\n y(s)=U(s)^*\\psi(s)\n\\]\nis differentiable at $t$, with\n\\[\n \\dot y=\\dot U^*\\psi(t)+U(t)^*\\dot\\psi.\n\\]\nIt satisfies\n\\[\n \\boxed{\\quad i\\dot y=H_{\\mathrm{eff}}y(t),\\qquad\n H_{\\mathrm{eff}}=U(t)^*HU(t)-iU(t)^*\\dot U.\\quad}\n\\]\nThe connection term $-iU(t)^*\\dot U$ is Hermitian. If $H^*=H$, then $H_{\\mathrm{eff}}^*=H_{\\mathrm{eff}}$.\n\\end{statement}","\\begin{proof}\\phantomsection\\label{supp2:proof:003}\nApplying the product rule entry by entry proves the differentiation formulas for matrix products and for matrices acting on vectors. Since the parameter is real, conjugation commutes with differentiation, giving the stated derivative of $U^*$ and hence of $y$.","Differentiate $U^*U=I$ using those rules and uniqueness of derivatives:\n\\[\n \\dot U^*U+U^*\\dot U=0.\n\\]\nThus $(U^*\\dot U)^*=-U^*\\dot U$, so multiplication by $-i$ produces a Hermitian matrix.","The identity $UU^*=I$ gives $\\psi=Uy$ for the constructed coordinate curve. Differentiating and inserting the original equation yields\n\\[\n i\\dot U y+iU\\dot y=HUy.\n\\]\nMultiplying by $U^*$ and using $U^*U=I$ gives the transformed equation, with the negative sign on the connection term. Finally, conjugation by $U$ preserves Hermiticity of $H$, and the sum of the two Hermitian terms is Hermitian.\n\\phantomsection\\label{supp2:proof:003:end}\\end{proof}","\\textbf{Formal scope.} The main certificate constructs $y$ and its derivative rather than assuming a separate derivative witness for the desired answer. Both unitary identities are explicit. The proofs are pointwise and apply at any time where the stated hypotheses hold. They do not assert existence of a global solution to an evolution equation.","Let $P_a(s)$ be a finite family in a normed real algebra with a continuous compatible adjoint. Assume actual derivatives $D_a=\\dot P_a(t)$ exist and, for all parameter values under consideration,\n\\[\n P_aP_b=\\delta_{ab}P_a,\\qquad \\sum_aP_a=I,\\qquad P_a^*=P_a.\n\\]\nIn the supplied Lean theorem, the curves are defined on $\\R$ and the projector laws hold at every real parameter. The algebraic formulation applies to matrices and bounded operators whenever they satisfy the stated typeclass hypotheses.","\\begin{statement}\\label{supp2:statement:004}\nDefine $C=\\sum_aD_aP_a(t)$. Then\n\\[\n \\boxed{\\quad C=\\frac12\\sum_a\\comm{D_a}{P_a(t)},\\qquad C^*=-C,\\quad}\n\\]\nand, for every $b$,\n\\[\n \\boxed{\\comm{C}{P_b(t)}=D_b.}\n\\]\n\\end{statement}","\\begin{proof}\\phantomsection\\label{supp2:proof:004}\nThe derivative product rule and uniqueness of derivatives give\n\\[\n D_aP_b+P_aD_b=\\delta_{ab}D_a.\n\\]\nDifferentiating completeness and adjoint invariance gives\n\\[\n \\sum_aD_a=0,\\qquad D_a^*=D_a.\n\\]\nThese are proved derivative consequences, not additional axioms supplied in place of differentiation.","On the diagonal, $D_aP_a+P_aD_a=D_a$. Summing shows\n\\[\n \\sum_aP_aD_a=-\\sum_aD_aP_a=-C.\n\\]\nThis proves the formula involving half the sum of commutators, together with $C^*=-C$.","Orthogonality directly yields $CP_b=D_bP_b$. For the left product, multiply the differentiated identity for $(b,a)$ on the right by $P_a$ and sum over $a$. Completeness reduces the result to\n\\[\n D_b+P_bC=D_bP_b.\n\\]\nSubtracting this equation from the identity for multiplication on the right gives $CP_b-P_bC=D_b$.\n\\phantomsection\\label{supp2:proof:004:end}\\end{proof}","For a finite family of complex matrices $K_a$, define\n\\[\n \\Phi(\\rho)=\\sum_a K_a\\rho K_a^*.\n\\]\nThe formalization constructs this as a complex-linear map and proves the following properties directly in the finite matrix representation.","\\begin{statement}\\label{supp2:statement:005}\nThe map $\\Phi$ is completely positive. If\n\\[\n \\sum_aK_a^*K_a=I,\n\\]\nthen $\\tr\\Phi(\\rho)=\\tr\\rho$ for every matrix $\\rho$. In particular, positive semidefinite matrices of trace one are mapped to positive semidefinite matrices of trace one.\n\\end{statement}","\\begin{proof}\\phantomsection\\label{supp2:proof:005}\nMatrix multiplication distributes across addition and scalar multiplication, which gives complex linearity. If $\\rho\\succeq0$, each term $K_a\\rho K_a^*$ is positive semidefinite, hence so is their finite sum.","To prove complete positivity rather than positivity alone, let $J$ be an arbitrary finite auxiliary index set. For a matrix $X$ indexed by $J\\times n$, define its blockwise amplification by\n\\[\n [\\Phi^{(J)}(X)]_{(i,u),(j,v)}\n =\\Phi\\bigl(X_{(i,\\cdot),(j,\\cdot)}\\bigr)_{u,v}.\n\\]\nThe explicit index calculation proves the identity\n\\[\n \\boxed{\\quad\n \\Phi^{(J)}(X)=\\sum_a (I_J\\otimes K_a)X(I_J\\otimes K_a)^*.\n \\quad}\n\\]\nThe displayed identity identifies the defined amplification with tensoring by the identity, which is the operation needed for complete positivity. Every summand on the right is positive when $X$ is positive, and this holds for every finite $J$.","X)]_{(i,u),(j,v)}\n =\\Phi\\bigl(X_{(i,\\cdot),(j,\\cdot)}\\bigr)_{u,v}.\n\\]\nThe explicit index calculation proves the identity\n\\[\n \\boxed{\\quad\n \\Phi^{(J)}(X)=\\sum_a (I_J\\otimes K_a)X(I_J\\otimes K_a)^*.\n \\quad}\n\\]\nThe displayed identity identifies the defined amplification with tensoring by the identity, which is the operation needed for complete positivity. Every summand on the right is positive when $X$ is positive, and this holds for every finite $J$.","ot)}\\bigr)_{u,v}.\n\\]\nThe explicit index calculation proves the identity\n\\[\n \\boxed{\\quad\n \\Phi^{(J)}(X)=\\sum_a (I_J\\otimes K_a)X(I_J\\otimes K_a)^*.\n \\quad}\n\\]\nThe displayed identity identifies the defined amplification with tensoring by the identity, which is the operation needed for complete positivity. Every summand on the right is positive when $X$ is positive, and this holds for every finite $J$.","Finally, cyclicity and linearity of the trace give\n\\begin{align*}\n \\tr\\Phi(\\rho)\n &=\\sum_a\\tr(K_a\\rho K_a^*)\\\\\n &=\\tr\\left(\\left(\\sum_aK_a^*K_a\\right)\\rho\\right)\n =\\tr\\rho.\n\\end{align*}\nPositivity together with trace preservation gives the stated conclusion for density matrices.\n\\phantomsection\\label{supp2:proof:005:end}\\end{proof}","\\section{Actual orbits and sharp seed counts}\\label{supp3:section:001}\nLet $C\\in M_{d\\times m}(\\C)$ be the coefficient matrix of one seed. Define its orbit by actual matrix multiplication:\n\\[\n\\mathcal O(C)=\\{AC:A\\in M_d(\\C)\\}.\n\\]\nThis set is already a linear subspace. The source proves that it equals the linear span of the orbit before computing its dimension.","A matrix $X$ belongs to $\\mathcal O(C)$ exactly when each of its rows belongs to $\\operatorname{Row}(C)$. One direction follows by expanding $AC$. For the converse, choose the coefficients producing each row separately and place them in the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","ngs to $\\mathcal O(C)$ exactly when each of its rows belongs to $\\operatorname{Row}(C)$. One direction follows by expanding $AC$. For the converse, choose the coefficients producing each row separately and place them in the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","when each of its rows belongs to $\\operatorname{Row}(C)$. One direction follows by expanding $AC$. For the converse, choose the coefficients producing each row separately and place them in the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","e{Row}(C)$. One direction follows by expanding $AC$. For the converse, choose the coefficients producing each row separately and place them in the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","C$. For the converse, choose the coefficients producing each row separately and place them in the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","the corresponding row of $A$. An explicit linear equivalence therefore gives\n\\[\n\\mathcal O(C)\\simeq \\operatorname{Row}(C)^d,\n\\qquad\n\\boxed{\\dim_\\C\\mathcal O(C)=d\\,\\rank C.}\n\\]\nIf $d>0$, equality with the full $d\\times m$ coefficient space is equivalent to $\\rank C=m$. Consequently a cyclic seed requires $m\\le d$. The assumption that there is at least one row is essential for canceling the factor $d$.","\\subsection{Several seeds: both necessity and construction}\nContinue under the assumption above that the number of rows is positive. For seeds $C_1,\\ldots,C_k$, generation means the concrete statement\n\\[\n\\forall X\\in M_{d\\times m}(\\C),\\quad\n\\exists A_1,\\ldots,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","truction}\nContinue under the assumption above that the number of rows is positive. For seeds $C_1,\\ldots,C_k$, generation means the concrete statement\n\\[\n\\forall X\\in M_{d\\times m}(\\C),\\quad\n\\exists A_1,\\ldots,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","sitive. For seeds $C_1,\\ldots,C_k$, generation means the concrete statement\n\\[\n\\forall X\\in M_{d\\times m}(\\C),\\quad\n\\exists A_1,\\ldots,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","crete statement\n\\[\n\\forall X\\in M_{d\\times m}(\\C),\\quad\n\\exists A_1,\\ldots,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","\\in M_{d\\times m}(\\C),\\quad\n\\exists A_1,\\ldots,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","s,A_k\\in M_d(\\C):\\quad\nX=\\sum_{a=1}^k A_aC_a.\n\\]\nStack the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","the $C_a$ vertically into a $kd\\times m$ matrix $D$. The source identifies the action on the stacked matrix and proves the equivalence\n\\[\n\\text{the seeds generate}\\quad\\Longleftrightarrow\\quad\\rank D=m.\n\\]\nThus $m\\le kd$ is necessary. It is sufficient as well: embed the $m$ coordinate columns into $kd$ coordinate positions to obtain a matrix $D$ of full column rank, then split its rows back into $k$ seeds. This construction proves that generating matrices with the required dimensions exist.","\\subsection{Simultaneous generation in a direct sum}\nFor the stipulated block representation\n\\[\n\\mathcal A=\\bigoplus_i\\bigl(M_{d_i}(\\C)\\otimes I_{m_i}\\bigr),\\qquad d_i>0,\n\\]\nthe proof establishes that $k$ seeds generate all blocks simultaneously exactly when $m_i\\le kd_i$ for every $i$. It follows that\n\\[\n\\boxed{\\srank_{\\mathcal A}(\\mathcal H)=\\max_i\\left\\lceil\\frac{m_i}{d_i}\\right\\rceil.}\n\\]\nThe final Lean proposition is an \\code{IsLeast} statement over the set of feasible seed counts. It proves both attainment and minimality. For an empty block family the finite supremum is defined as $0$, matching the zero module.","that $k$ seeds generate all blocks simultaneously exactly when $m_i\\le kd_i$ for every $i$. It follows that\n\\[\n\\boxed{\\srank_{\\mathcal A}(\\mathcal H)=\\max_i\\left\\lceil\\frac{m_i}{d_i}\\right\\rceil.}\n\\]\nThe final Lean proposition is an \\code{IsLeast} statement over the set of feasible seed counts. It proves both attainment and minimality. For an empty block family the finite supremum is defined as $0$, matching the zero module.","\\section{Schmidt rank and the reduced state}\\label{supp3:section:002}\nUse coordinates $\\psi_{ij}$ in the standard product basis and let $C_{ij}=\\psi_{ij}$. An explicit linear equivalence identifies the state vector on the product index set with its coefficient matrix. The implementation checks the actual Kronecker action:\n\\[\n\\operatorname{coeff}\\bigl((A\\otimes I_B)\\psi\\bigr)=AC.\n\\]\nIt also proves that the transported matrix orbit equals the span of the state vectors obtained by this Kronecker action. The dimension formula therefore applies to the physical local action on the state space.","\\subsection{An independent definition of Schmidt rank}\nThe source defines Schmidt rank as the least number of summands that are product vectors. A decomposition of length $s$ is the concrete equality\n\\[\nC_{ij}=\\sum_{a=1}^{s}u_a(i)v_a(j).\n\\]\nIt factors $C=UV$ through an $s$-dimensional intermediate space, so $\\rank C\\le s$. Conversely, choose a basis of the row space of $C$. Expanding each row in this basis constructs a decomposition with exactly $\\rank C$ product terms. Thus\n\\[\n\\boxed{r(\\psi)=\\rank C.}\n\\]\nThe definition uses a least natural number, and the equality with matrix rank is a theorem. This argument establishes the minimum number of product terms. It does not construct an orthonormal Schmidt decomposition or prove a new theorem on singular value decomposition.","nk}\nThe source defines Schmidt rank as the least number of summands that are product vectors. A decomposition of length $s$ is the concrete equality\n\\[\nC_{ij}=\\sum_{a=1}^{s}u_a(i)v_a(j).\n\\]\nIt factors $C=UV$ through an $s$-dimensional intermediate space, so $\\rank C\\le s$. Conversely, choose a basis of the row space of $C$. Expanding each row in this basis constructs a decomposition with exactly $\\rank C$ product terms. Thus\n\\[\n\\boxed{r(\\psi)=\\rank C.}\n\\]\nThe definition uses a least natural number, and the equality with matrix rank is a theorem. This argument establishes the minimum number of product terms. It does not construct an orthonormal Schmidt decomposition or prove a new theorem on singular value decomposition.","rank as the least number of summands that are product vectors. A decomposition of length $s$ is the concrete equality\n\\[\nC_{ij}=\\sum_{a=1}^{s}u_a(i)v_a(j).\n\\]\nIt factors $C=UV$ through an $s$-dimensional intermediate space, so $\\rank C\\le s$. Conversely, choose a basis of the row space of $C$. Expanding each row in this basis constructs a decomposition with exactly $\\rank C$ product terms. Thus\n\\[\n\\boxed{r(\\psi)=\\rank C.}\n\\]\nThe definition uses a least natural number, and the equality with matrix rank is a theorem. This argument establishes the minimum number of product terms. It does not construct an orthonormal Schmidt decomposition or prove a new theorem on singular value decomposition.","The matrix-orbit dimension and the coefficient equivalence now yield\n\\[\n\\boxed{\\dim_\\C\\spanof\\{(A\\otimes I_B)\\psi:A\\in M_{d_A}(\\C)\\}\n=d_A r(\\psi).}\n\\]","\\subsection{The actual partial trace and its support}\nThe reduced matrix is defined by tracing the addressed index of the density matrix of the pure state:\n\\[\n(\\rho_B)_{j\\ell}=\\sum_i\\psi_{ij}\\overline{\\psi_{i\\ell}}.\n\\]\nFor the coefficient convention used here, the exact Gram identity is\n\\[\n\\rho_B=C^{\\mathsf T}(C^{\\mathsf T})^*,\n\\qquad \\rank\\rho_B=\\rank C=r(\\psi).\n\\]\nThe transpose is important. With this convention, writing $C^*C$ instead would conjugate the reduced matrix. The Lean source checks the entrywise identity and positivity of the resulting matrix.","al trace and its support}\nThe reduced matrix is defined by tracing the addressed index of the density matrix of the pure state:\n\\[\n(\\rho_B)_{j\\ell}=\\sum_i\\psi_{ij}\\overline{\\psi_{i\\ell}}.\n\\]\nFor the coefficient convention used here, the exact Gram identity is\n\\[\n\\rho_B=C^{\\mathsf T}(C^{\\mathsf T})^*,\n\\qquad \\rank\\rho_B=\\rank C=r(\\psi).\n\\]\nThe transpose is important. With this convention, writing $C^*C$ instead would conjugate the reduced matrix. The Lean source checks the entrywise identity and positivity of the resulting matrix.","Full support is expressed as surjectivity of the reduced matrix's linear map. Assuming $d_A>0$, the checked equivalence is\n\\[\n\\boxed{\\mathcal O_A(\\psi)=\\mathcal H_A\\otimes\\mathcal H_B\n\\quad\\Longleftrightarrow\\quad \\Ran\\rho_B=\\mathcal H_B.}\n\\]\nNormalization of $\\psi$ is unnecessary for these dimension and support statements. The zero state has rank $0$ because its decomposition needs no product terms.","\\section{Constructed cyclic character projectors}\\label{supp3:section:003}\nLet $N>0$ and let $S$ be a unitary complex matrix with $S^N=I$. The implementation uses the standard additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","structed cyclic character projectors}\\label{supp3:section:003}\nLet $N>0$ and let $S$ be a unitary complex matrix with $S^N=I$. The implementation uses the standard additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","003}\nLet $N>0$ and let $S$ be a unitary complex matrix with $S^N=I$. The implementation uses the standard additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","he implementation uses the standard additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","ses the standard additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","additive character of $\\mathbb Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","Z/N\\mathbb Z$ and proves its expression through a root of unity:\n\\[\n\\omega=\\exp(2\\pi i/N),\\qquad\n\\chi(r)=\\omega^{\\operatorname{val}(r)}.\n\\]\nHere $\\operatorname{val}(r)$ is the canonical representative in $\\{0,\\ldots,N-1\\}$. The representation is constructed as\n\\[\n\\rho(r)=S^{\\operatorname{val}(r)}.\n\\]\nReduction of exponents modulo $N$ is proved from $S^N=I$, giving $\\rho(r+s)=\\rho(r)\\rho(s)$ and $\\rho(0)=I$. The final matrix theorem derives the representation law and character orthogonality from the stated inputs.","\\subsection{Exact coefficient formula}\nThe packet definition is a finite sum over residue classes:\n\\[\nP_j=\\frac1N\\sum_{r\\in\\mathbb Z/N\\mathbb Z}\\chi(-rj)\\rho(r).\n\\]\nThe source proves\n\\[\n\\chi(-rj)=\\bigl(\\omega^{\\operatorname{val}(r)\\operatorname{val}(j)}\\bigr)^{-1},\n\\]\nso this is precisely the manuscript's $N$-term Fourier projector formula, with canonical representatives.","\\subsection{Algebraic identities}\nPrimitivity of the standard character yields the checked finite sum\n\\[\n\\sum_r\\chi(tr)=\\begin{cases}N,&t=0,\\\\0,&t\\ne0.\\end{cases}\n\\]\nReindexing finite sums then gives\n\\[\n\\boxed{P_jP_k=\\delta_{jk}P_j,\\qquad \\sum_jP_j=I,\\qquad SP_j=\\omega^jP_j.}\n\\]\nMultiplicativity and completeness are first proved for a general representation in a complex algebra, then instantiated with the actual powers of $S$. This organization makes the algebra of finite sums reusable while preserving the final hypotheses.","0,&t\\ne0.\\end{cases}\n\\]\nReindexing finite sums then gives\n\\[\n\\boxed{P_jP_k=\\delta_{jk}P_j,\\qquad \\sum_jP_j=I,\\qquad SP_j=\\omega^jP_j.}\n\\]\nMultiplicativity and completeness are first proved for a general representation in a complex algebra, then instantiated with the actual powers of $S$. This organization makes the algebra of finite sums reusable while preserving the final hypotheses.","\\subsection{Orthogonality of the projectors}\nUnitarity supplies $\\rho(r)^*=\\rho(-r)$, while the character obeys $\\overline{\\chi(r)}=\\chi(-r)$. Reindexing $r\\mapsto-r$ proves $P_j^*=P_j$. The final concrete matrix theorem includes Hermiticity together with multiplication, completeness, and eigenaction. Together these identities establish orthogonal spectral packets.","The proof includes the degenerate modulus $N=1$: finite order then forces $S=I$ and the sole projector is $I$. It does not need an artificial assumption $N>1$.","\\section{The actual Euclidean transfer exponential}\\label{supp3:section:004}\nLet $H$ be a Hermitian matrix on a nonzero complex state space of finite dimension. The source defines\n\\[\nT_e(t)=\\exp\\bigl(-t(H-eI)\\bigr)\n\\]\nusing Lean's exponential defined by a power series. The proof then relates this defined operator to its spectral expansion.","\\subsection{Connecting power series and spectral calculus}\nThe shifted generator is proved Hermitian. The implementation then identifies the actual matrix exponential with continuous functional calculus:\n\\[\n\\boxed{T_e(t)=f_t(H),\\qquad f_t(x)=e^{-t(x-e)}.}\n\\]\nThe formal proof uses composition, scalar multiplication, subtraction, and the established agreement between the exponential in functional calculus and the exponential defined by a power series. Spectral mapping is consequently available for the operator that was actually defined:\n\\[\n\\Spec_\\R T_e(t)=\\{e^{-t(x-e)}:x\\in\\Spec_\\R H\\}.\n\\]\nHermiticity also identifies the full complex spectrum with the image of these real eigenvalues. A separate checked theorem states the interval containing the complex spectrum explicitly.","tual matrix exponential with continuous functional calculus:\n\\[\n\\boxed{T_e(t)=f_t(H),\\qquad f_t(x)=e^{-t(x-e)}.}\n\\]\nThe formal proof uses composition, scalar multiplication, subtraction, and the established agreement between the exponential in functional calculus and the exponential defined by a power series. Spectral mapping is consequently available for the operator that was actually defined:\n\\[\n\\Spec_\\R T_e(t)=\\{e^{-t(x-e)}:x\\in\\Spec_\\R H\\}.\n\\]\nHermiticity also identifies the full complex spectrum with the image of these real eigenvalues. A separate checked theorem states the interval containing the complex spectrum explicitly.","\\subsection{The minimum eigenvalue}\nThe ground energy is the minimum of the finite, nonempty eigenvalue set. The proof establishes both\n\\[\nE_0\\in\\Spec_\\R H,\\qquad E_0\\le x\\quad(x\\in\\Spec_\\R H).\n\\]\nThe assumption that the state space is nonzero is recorded in the theorem because the minimum of an empty spectrum is not defined by the manuscript's formula.","\\subsection{Positivity, semigroup composition, and the interval}\nEvery eigenvalue of $T_e(t)$ is strictly positive, for arbitrary real $e,t$, since the real exponential is positive. For $e=E_0$ and $t\\ge0$ it is also at most one. Therefore\n\\[\n\\boxed{T_{E_0}(t)>0,\\qquad \\Spec T_{E_0}(t)\\subseteq(0,1].}\n\\]\nThe two scalar multiples of $H-eI$ commute, so the exponential defined by the power series obeys\n\\[\n\\boxed{T_e(t+s)=T_e(t)T_e(s),\\qquad T_e(0)=I.}\n\\]\nThe semigroup identity is proved for all real $t,s$. Nonnegative time is needed for the upper spectral bound, not for the composition identity or strict positivity.","ection{Positivity, semigroup composition, and the interval}\nEvery eigenvalue of $T_e(t)$ is strictly positive, for arbitrary real $e,t$, since the real exponential is positive. For $e=E_0$ and $t\\ge0$ it is also at most one. Therefore\n\\[\n\\boxed{T_{E_0}(t)>0,\\qquad \\Spec T_{E_0}(t)\\subseteq(0,1].}\n\\]\nThe two scalar multiples of $H-eI$ commute, so the exponential defined by the power series obeys\n\\[\n\\boxed{T_e(t+s)=T_e(t)T_e(s),\\qquad T_e(0)=I.}\n\\]\nThe semigroup identity is proved for all real $t,s$. Nonnegative time is needed for the upper spectral bound, not for the composition identity or strict positivity.","ly positive, for arbitrary real $e,t$, since the real exponential is positive. For $e=E_0$ and $t\\ge0$ it is also at most one. Therefore\n\\[\n\\boxed{T_{E_0}(t)>0,\\qquad \\Spec T_{E_0}(t)\\subseteq(0,1].}\n\\]\nThe two scalar multiples of $H-eI$ commute, so the exponential defined by the power series obeys\n\\[\n\\boxed{T_e(t+s)=T_e(t)T_e(s),\\qquad T_e(0)=I.}\n\\]\nThe semigroup identity is proved for all real $t,s$. Nonnegative time is needed for the upper spectral bound, not for the composition identity or strict positivity.","\\section{Quantitative leakage from the actual exponential}\\label{supp4:section:001}\nLet $A$ be a unital complex C-star algebra, let $H=H^*$, and let $P=P^*=P^2$. Put $Q=1-P$. The checked statement is\n\\[\n\\boxed{\\|Q\\exp(-itH)P\\|\\le |t|\\,\\|HP-PH\\|\\qquad(t\\in\\R).}\n\\]\nThe matrix specialization uses the operator norm on the Hilbert space, implemented by \\path{Matrix.Norms.L2Operator}. It is not an entrywise norm or the row-sum operator norm. The theorem includes the zero projector and empty matrix index type. The theorem also applies at time zero.","\\subsection{Deriving the estimate by differentiation}\nFor a skew-adjoint generator $G$, define the actual conjugation path\n\\[\nT_P(t)=\\exp(tG)P\\exp(-tG).\n\\]\nThe formalization derives its derivative from the exponential power series and the product rule:\n\\[\nT_P'(t)=\\exp(tG)(GP-PG)\\exp(-tG).\n\\]\nBoth exponential factors are proved unitary. Left and right multiplication by a unitary preserve the C-star norm. Therefore\n\\[\n\\|T_P'(t)\\|=\\|GP-PG\\|,\n\\qquad\n\\|T_P(t)-T_P(s)\\|\\le |t-s|\\,\\|GP-PG\\|.\n\\]\nThe second inequality follows from the mean value inequality in normed spaces and covers either ordering of $s$ and $t$. It is not a finite-difference estimate or a numerical time discretization.","g the estimate by differentiation}\nFor a skew-adjoint generator $G$, define the actual conjugation path\n\\[\nT_P(t)=\\exp(tG)P\\exp(-tG).\n\\]\nThe formalization derives its derivative from the exponential power series and the product rule:\n\\[\nT_P'(t)=\\exp(tG)(GP-PG)\\exp(-tG).\n\\]\nBoth exponential factors are proved unitary. Left and right multiplication by a unitary preserve the C-star norm. Therefore\n\\[\n\\|T_P'(t)\\|=\\|GP-PG\\|,\n\\qquad\n\\|T_P(t)-T_P(s)\\|\\le |t-s|\\,\\|GP-PG\\|.\n\\]\nThe second inequality follows from the mean value inequality in normed spaces and covers either ordering of $s$ and $t$. It is not a finite-difference estimate or a numerical time discretization.","the exponential power series and the product rule:\n\\[\nT_P'(t)=\\exp(tG)(GP-PG)\\exp(-tG).\n\\]\nBoth exponential factors are proved unitary. Left and right multiplication by a unitary preserve the C-star norm. Therefore\n\\[\n\\|T_P'(t)\\|=\\|GP-PG\\|,\n\\qquad\n\\|T_P(t)-T_P(s)\\|\\le |t-s|\\,\\|GP-PG\\|.\n\\]\nThe second inequality follows from the mean value inequality in normed spaces and covers either ordering of $s$ and $t$. It is not a finite-difference estimate or a numerical time discretization.","onential power series and the product rule:\n\\[\nT_P'(t)=\\exp(tG)(GP-PG)\\exp(-tG).\n\\]\nBoth exponential factors are proved unitary. Left and right multiplication by a unitary preserve the C-star norm. Therefore\n\\[\n\\|T_P'(t)\\|=\\|GP-PG\\|,\n\\qquad\n\\|T_P(t)-T_P(s)\\|\\le |t-s|\\,\\|GP-PG\\|.\n\\]\nThe second inequality follows from the mean value inequality in normed spaces and covers either ordering of $s$ and $t$. It is not a finite-difference estimate or a numerical time discretization.","Multiplication by $\\exp(tG)$ gives\n\\[\n(T_P(t)-P)\\exp(tG)=\\exp(tG)P-P\\exp(tG).\n\\]\nFor every orthogonal projector, $\\|P\\|\\le1$ and $\\|1-P\\|\\le1$. The exact corner identity\n\\[\n(1-P)UP=(1-P)(UP-PU)P\n\\]\nthen bounds leakage by the exponential commutator. Taking $G=-iH$ gives the stated theorem because $\\|[G,P]\\|=\\|[H,P]\\|$.","\\subsection{Checked consequences}\nUnitarity also gives the capped estimate\n\\[\n\\|Q\\exp(-itH)P\\|\\le\\min\\{1,\\ |t|\\,\\|[H,P]\\|\\}.\n\\]\nIf $H=H_0+V$, both summands are self-adjoint, and $[H_0,P]=0$, then\n\\[\n\\|Q\\exp(-it(H_0+V))P\\|\\le2|t|\\,\\|V\\|.\n\\]\nThe latter uses $\\|[V,P]\\|\\le2\\|V\\|$. Its constant is not claimed to be optimal. Exact commutation gives zero leakage. The work does not include unbounded generators or time-dependent Hamiltonians.","\\section{Actual projected differentiation and curvature}\\label{supp4:section:002}\nLet $E,F$ be real normed spaces. Let $P:E\\to\\mathcal L(F,F)$ be differentiable with $P(x)^2=P(x)$ for every $x$. Let $s:E\\to F$ be $C^2$ with $P(x)s(x)=s(x)$, and let $X,Y:E\\to E$ be differentiable vector fields. The checked connection is the actual expression\n\\[\n\\nabla_Xs(x)=P(x)\\,Ds(x)[X(x)].\n\\]\nIt is proved to preserve the range condition. It is linear in the vector field direction and obeys the scalar Leibniz rule\n\\[\n\\nabla_X(cs)=Dc[X]\\,s+c\\nabla_Xs\n\\]\nfor differentiable real scalar functions $c$. No differentiated projector identities are inserted as hypotheses in the main curvature theorem.","spaces. Let $P:E\\to\\mathcal L(F,F)$ be differentiable with $P(x)^2=P(x)$ for every $x$. Let $s:E\\to F$ be $C^2$ with $P(x)s(x)=s(x)$, and let $X,Y:E\\to E$ be differentiable vector fields. The checked connection is the actual expression\n\\[\n\\nabla_Xs(x)=P(x)\\,Ds(x)[X(x)].\n\\]\nIt is proved to preserve the range condition. It is linear in the vector field direction and obeys the scalar Leibniz rule\n\\[\n\\nabla_X(cs)=Dc[X]\\,s+c\\nabla_Xs\n\\]\nfor differentiable real scalar functions $c$. No differentiated projector identities are inserted as hypotheses in the main curvature theorem.","very $x$. Let $s:E\\to F$ be $C^2$ with $P(x)s(x)=s(x)$, and let $X,Y:E\\to E$ be differentiable vector fields. The checked connection is the actual expression\n\\[\n\\nabla_Xs(x)=P(x)\\,Ds(x)[X(x)].\n\\]\nIt is proved to preserve the range condition. It is linear in the vector field direction and obeys the scalar Leibniz rule\n\\[\n\\nabla_X(cs)=Dc[X]\\,s+c\\nabla_Xs\n\\]\nfor differentiable real scalar functions $c$. No differentiated projector identities are inserted as hypotheses in the main curvature theorem.","\\subsection{Differentiated projector and section identities}\nDifferentiating $P^2=P$ and $Ps=s$ produces\n\\begin{align*}\nDP[v]P+PDP[v]&=DP[v],","\\subsection{Differentiated projector and section identities}\nDifferentiating $P^2=P$ and $Ps=s$ produces\n\\begin{align*}\nDP[v]P+PDP[v]&=DP[v],\\\\\nPDP[v]P&=0,\\\\\nDs[v]&=DP[v]s+P Ds[v].\n\\end{align*}\nThe zero sandwich follows algebraically from the first identity. All products preserve their noncommutative order.","\\subsection{Retaining the bracket of vector fields}\nDefine the actual curvature action by\n\\[\nR_P(X,Y)s=\\nabla_X\\nabla_Ys-\\nabla_Y\\nabla_Xs-\\nabla_{[X,Y]}s.\n\\]\nThe source uses the Lie bracket defined through Fr\\'echet derivatives. Symmetry of second derivatives is supplied by the $C^2$ hypothesis, not postulated separately. Expanding the connection gives\n\\[\nR_P(X,Y)s=P\\bigl(DP[X]Ds[Y]-DP[Y]Ds[X]\\bigr).\n\\]\nSubstituting the section identity and removing the zero sandwich terms proves\n\\[\n\\boxed{R_P(X,Y)s=P\\bigl(DP[X]DP[Y]-DP[Y]DP[X]\\bigr)P s.}\n\\]\nThus the coefficient is precisely $P(\\dd P\\wedge\\dd P)P$ with the convention\n$(\\dd P\\wedge\\dd P)(v,w)=DP[v]DP[w]-DP[w]DP[v]$.","y\n\\[\nR_P(X,Y)s=\\nabla_X\\nabla_Ys-\\nabla_Y\\nabla_Xs-\\nabla_{[X,Y]}s.\n\\]\nThe source uses the Lie bracket defined through Fr\\'echet derivatives. Symmetry of second derivatives is supplied by the $C^2$ hypothesis, not postulated separately. Expanding the connection gives\n\\[\nR_P(X,Y)s=P\\bigl(DP[X]Ds[Y]-DP[Y]Ds[X]\\bigr).\n\\]\nSubstituting the section identity and removing the zero sandwich terms proves\n\\[\n\\boxed{R_P(X,Y)s=P\\bigl(DP[X]DP[Y]-DP[Y]DP[X]\\bigr)P s.}\n\\]\nThus the coefficient is precisely $P(\\dd P\\wedge\\dd P)P$ with the convention\n$(\\dd P\\wedge\\dd P)(v,w)=DP[v]DP[w]-DP[w]DP[v]$.","racket defined through Fr\\'echet derivatives. Symmetry of second derivatives is supplied by the $C^2$ hypothesis, not postulated separately. Expanding the connection gives\n\\[\nR_P(X,Y)s=P\\bigl(DP[X]Ds[Y]-DP[Y]Ds[X]\\bigr).\n\\]\nSubstituting the section identity and removing the zero sandwich terms proves\n\\[\n\\boxed{R_P(X,Y)s=P\\bigl(DP[X]DP[Y]-DP[Y]DP[X]\\bigr)P s.}\n\\]\nThus the coefficient is precisely $P(\\dd P\\wedge\\dd P)P$ with the convention\n$(\\dd P\\wedge\\dd P)(v,w)=DP[v]DP[w]-DP[w]DP[v]$.","s. Symmetry of second derivatives is supplied by the $C^2$ hypothesis, not postulated separately. Expanding the connection gives\n\\[\nR_P(X,Y)s=P\\bigl(DP[X]Ds[Y]-DP[Y]Ds[X]\\bigr).\n\\]\nSubstituting the section identity and removing the zero sandwich terms proves\n\\[\n\\boxed{R_P(X,Y)s=P\\bigl(DP[X]DP[Y]-DP[Y]DP[X]\\bigr)P s.}\n\\]\nThus the coefficient is precisely $P(\\dd P\\wedge\\dd P)P$ with the convention\n$(\\dd P\\wedge\\dd P)(v,w)=DP[v]DP[w]-DP[w]DP[v]$.","\\subsection{Tensor properties and the remaining global boundary}\nThe coefficient is proved bilinear, alternating, and supported on the range on both sides. The chain rule proves the pullback identity\n\\[\nF_{P\\circ f}(x)(v,w)=F_P(f(x))(Df(x)v,Df(x)w).\n\\]\nThe main theorem has global regularity hypotheses on normed parameter spaces. The pullback identity alone does not supply localization, smooth local trivializations, or gluing. The final supplement constructs the global range bundle and proves that the local connections glue.","\\section{A constructive and total syndrome recovery}\\label{supp4:section:003}\nLet $P$ be an orthogonal code projector. The input certificate consists of a finite family $V_a$ satisfying\n\\[\nV_aP=V_a,\n\\qquad\nV_a^*V_b=\\delta_{ab}P.\n\\]\nThese hypotheses state that the operators are partial isometries with orthogonal syndromes. The next supplement derives such a family from arbitrary Knill--Laflamme Gram data.","\\subsection{Constructing the recovery on the whole ambient space}\nDefine the syndrome projectors and their complement by\n\\[\nS_a=V_aV_a^*,\\qquad S=\\sum_aS_a,\\qquad Q=1-S.\n\\]\nThe source proves self-adjointness, $S_aS_b=\\delta_{ab}S_a$, $S^2=S$, and $Q^2=Q=Q^*$. It also proves $QV_a=0$ and $V_a^*Q=0$.","The recovery is not left undefined outside the syndrome subspaces. Its explicit Kraus operators are $R_a=V_a^*$ together with $R_{\\mathrm{extra}}=Q$, giving\n\\[\n\\mathcal R(\\rho)=\\sum_aV_a^*\\rho V_a+Q\\rho Q.\n\\]\nThey satisfy exact completeness:\n\\[\n\\sum_aR_a^*R_a+R_{\\mathrm{extra}}^*R_{\\mathrm{extra}}\n=\\sum_aV_aV_a^*+Q=S+Q=I.\n\\]\nFor finite complex matrices, trace preservation and complete positivity at every finite auxiliary dimension are proved for this actual map. The amplified map is identified with the Kraus family $I_J\\otimes R_a$, including the complement term. Positivity is not merely checked on a single matrix size.","ith $R_{\\mathrm{extra}}=Q$, giving\n\\[\n\\mathcal R(\\rho)=\\sum_aV_a^*\\rho V_a+Q\\rho Q.\n\\]\nThey satisfy exact completeness:\n\\[\n\\sum_aR_a^*R_a+R_{\\mathrm{extra}}^*R_{\\mathrm{extra}}\n=\\sum_aV_aV_a^*+Q=S+Q=I.\n\\]\nFor finite complex matrices, trace preservation and complete positivity at every finite auxiliary dimension are proved for this actual map. The amplified map is identified with the Kraus family $I_J\\otimes R_a$, including the complement term. Positivity is not merely checked on a single matrix size.","\\subsection{Exact correction, including coherent errors}\nFor every operator supported on the code, $P\\rho P=\\rho$, the central identity is\n\\[\n\\boxed{\\mathcal R(V_a\\rho V_b^*)=\\delta_{ab}\\rho.}\n\\]\nConsequently normalized mixtures satisfy\n\\[\n\\mathcal R\\!\\left(\\sum_ap_aV_a\\rho V_a^*\\right)=\\rho\n\\quad\\text{when}\\quad\\sum_ap_a=1.\n\\]\nThe identity is algebraic and is proved for complex weights. Physical mixtures use the nonnegative real case. For a coherent error $E_z=\\sum_a z_aV_a$, the calculation of the cross terms yields\n\\[\n\\mathcal R(E_z\\rho E_z^*)=\\left(\\sum_a z_a\\overline{z_a}\\right)\\rho.\n\\]\nThus the same map also corrects normalized coherent superpositions. This construction supplies the recovery stage used in the full Knill--Laflamme proof that follows.","very operator supported on the code, $P\\rho P=\\rho$, the central identity is\n\\[\n\\boxed{\\mathcal R(V_a\\rho V_b^*)=\\delta_{ab}\\rho.}\n\\]\nConsequently normalized mixtures satisfy\n\\[\n\\mathcal R\\!\\left(\\sum_ap_aV_a\\rho V_a^*\\right)=\\rho\n\\quad\\text{when}\\quad\\sum_ap_a=1.\n\\]\nThe identity is algebraic and is proved for complex weights. Physical mixtures use the nonnegative real case. For a coherent error $E_z=\\sum_a z_aV_a$, the calculation of the cross terms yields\n\\[\n\\mathcal R(E_z\\rho E_z^*)=\\left(\\sum_a z_a\\overline{z_a}\\right)\\rho.\n\\]\nThus the same map also corrects normalized coherent superpositions. This construction supplies the recovery stage used in the full Knill--Laflamme proof that follows.","identity is\n\\[\n\\boxed{\\mathcal R(V_a\\rho V_b^*)=\\delta_{ab}\\rho.}\n\\]\nConsequently normalized mixtures satisfy\n\\[\n\\mathcal R\\!\\left(\\sum_ap_aV_a\\rho V_a^*\\right)=\\rho\n\\quad\\text{when}\\quad\\sum_ap_a=1.\n\\]\nThe identity is algebraic and is proved for complex weights. Physical mixtures use the nonnegative real case. For a coherent error $E_z=\\sum_a z_aV_a$, the calculation of the cross terms yields\n\\[\n\\mathcal R(E_z\\rho E_z^*)=\\left(\\sum_a z_a\\overline{z_a}\\right)\\rho.\n\\]\nThus the same map also corrects normalized coherent superpositions. This construction supplies the recovery stage used in the full Knill--Laflamme proof that follows.","\\section{The precise finite channel statement}\\label{supp5:section:001}\nLet $\\mathcal H=\\C^n$ be a finite coordinate space and let $P=P^*=P^2$. A matrix is code supported when $P\\rho P=\\rho$. For a finite family of square matrices $E_a$, put\n\\[\n\\mathcal N_E(\\rho)=\\sum_a E_a\\rho E_a^*.\n\\]\nThe input channel normalization is the actual matrix identity\n\\[\n\\sum_a E_a^*E_a=I.\n\\]\nA finite Kraus recovery has the form $\\mathcal R(\\sigma)=\\sum_r R_r\\sigma R_r^*$, with $\\sum_rR_r^*R_r=I$. This representation, together with the checked amplification results, provides complete positivity and trace preservation on the entire ambient matrix space.","recise finite channel statement}\\label{supp5:section:001}\nLet $\\mathcal H=\\C^n$ be a finite coordinate space and let $P=P^*=P^2$. A matrix is code supported when $P\\rho P=\\rho$. For a finite family of square matrices $E_a$, put\n\\[\n\\mathcal N_E(\\rho)=\\sum_a E_a\\rho E_a^*.\n\\]\nThe input channel normalization is the actual matrix identity\n\\[\n\\sum_a E_a^*E_a=I.\n\\]\nA finite Kraus recovery has the form $\\mathcal R(\\sigma)=\\sum_r R_r\\sigma R_r^*$, with $\\sum_rR_r^*R_r=I$. This representation, together with the checked amplification results, provides complete positivity and trace preservation on the entire ambient matrix space.","on:001}\nLet $\\mathcal H=\\C^n$ be a finite coordinate space and let $P=P^*=P^2$. A matrix is code supported when $P\\rho P=\\rho$. For a finite family of square matrices $E_a$, put\n\\[\n\\mathcal N_E(\\rho)=\\sum_a E_a\\rho E_a^*.\n\\]\nThe input channel normalization is the actual matrix identity\n\\[\n\\sum_a E_a^*E_a=I.\n\\]\nA finite Kraus recovery has the form $\\mathcal R(\\sigma)=\\sum_r R_r\\sigma R_r^*$, with $\\sum_rR_r^*R_r=I$. This representation, together with the checked amplification results, provides complete positivity and trace preservation on the entire ambient matrix space.","\\subsection{Necessary and sufficient conditions}\nUnder the stated normalization, the following equivalence is proved:\n\\[\n\\boxed{\\begin{gathered}\n\\exists\\,\\mathcal R\\text{ of the above finite Kraus form,}\\\\[-2pt]\n\\mathcal R\\mathcal N_E(\\rho)=\\rho\n\\quad\\text{for every }\\rho\\succeq0,\\ \\Tr\\rho=1,\\ P\\rho P=\\rho\n\\end{gathered}}\n\\quad\\Longleftrightarrow\\quad\n\\boxed{\\begin{gathered}\n\\exists C\\succeq0,\\\\\nPE_a^*E_bP=C_{ab}P\\quad\\text{for all }a,b.\n\\end{gathered}}\n\\]\nThe statement applies throughout the code space. The formalization separately proves equivalence with correction of \\emph{every} complex matrix supported on the code. It also proves preservation with arbitrary finite reference systems, as detailed in the discussion of density matrices and reference systems below.","\\subsection{Including degenerate cases}\nThe matrix $C$ need not be positive definite. The proof includes zero eigenvalues, redundant error operators, an empty set of positive eigenvalues, and the zero projector. If $P=0$, the only supported matrix is zero and no trace-one code density matrix exists. The proofs handle that case separately rather than dividing by the code rank or by $\\Tr P$.","For an unnormalized error family, the correct statement is generally\n\\[\n\\mathcal R\\mathcal N_E(\\rho)=\\Tr(C)\\rho,\n\\]\nnot unconditional equality to $\\rho$. Normalization forces $\\Tr C=1$ on a nonzero code. The discussion of coherent errors below treats unrestricted error families and shows when the same decoder works throughout their span.","\\section{From Gram data to an actual total recovery}\\label{supp5:section:002}\nAssume $C\\succeq0$ and $PE_a^*E_bP=C_{ab}P$. First restrict the error operators to the code, $F_a=E_aP$. The finite Hermitian spectral theorem supplies a unitary $U$ such that\n\\[\nU^*CU=\\operatorname{diag}(\\lambda_r),\\qquad \\lambda_r\\ge0.\n\\]","Define the transformed errors by an actual finite sum,\n\\[\nB_r=\\sum_a U_{ar}F_a.\n\\]\nThe formal module proves the identities\n\\[\nB_rP=B_r,\\qquad B_r^*B_s=\\delta_{rs}\\lambda_rP,\n\\qquad \\sum_rB_r\\rho B_r^*=\\sum_aF_a\\rho F_a^*.\n\\]\nFor $\\rho$ supported on the code, the restricted and unrestricted error sums are equal. Thus the unitary change of error coordinates preserves the actual channel on the code.","\\subsection{Null directions are removed by proof}\nIf $\\lambda_r=0$, then $B_r^*B_r=0$. Positivity of the ordinary complex matrix inner product gives $B_r=0$. This is the checked declaration \\path{DiagonalErrors.zero_weight_error}. The normalized syndrome operators are defined only on\n\\[\nA_+=\\{r:\\lambda_r>0\\},\\qquad V_r=\\lambda_r^{-1/2}B_r.\n\\]\nConsequently no inverse square root at a null eigenvalue is needed. Lean proves\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nThe previous block's recovery is now instantiated from these derived syndromes, rather than from an assumed orthogonal error family.","are removed by proof}\nIf $\\lambda_r=0$, then $B_r^*B_r=0$. Positivity of the ordinary complex matrix inner product gives $B_r=0$. This is the checked declaration \\path{DiagonalErrors.zero_weight_error}. The normalized syndrome operators are defined only on\n\\[\nA_+=\\{r:\\lambda_r>0\\},\\qquad V_r=\\lambda_r^{-1/2}B_r.\n\\]\nConsequently no inverse square root at a null eigenvalue is needed. Lean proves\n\\[\nV_rP=V_r,\\qquad V_r^*V_s=\\delta_{rs}P.\n\\]\nThe previous block's recovery is now instantiated from these derived syndromes, rather than from an assumed orthogonal error family.","\\subsection{A channel on the whole ambient space}\nPut $Q=I-\\sum_{r\\in A_+}V_rV_r^*$. The recovery is\n\\[\n\\boxed{\\mathcal R(\\sigma)=\\sum_{r\\in A_+}V_r^*\\sigma V_r+Q\\sigma Q.}\n\\]\nThe complement is an orthogonal projector and\n\\[\n\\sum_{r\\in A_+}V_rV_r^*+Q^*Q=I.\n\\]\nHence the recovery is trace preserving. Its finite Kraus form also proves positivity after amplification by the identity on every finite auxiliary space. The complement term matters: the map is defined as a channel even outside the syndrome ranges.","Finally the checked calculation gives $\\mathcal R\\mathcal N_E(\\rho)=\\sum_r\\lambda_r\\rho=\\Tr(C)\\rho$. Under normalization of the input channel and $P\\ne0$, summing the diagonal Gram identities gives $\\Tr(C)P=P$, hence $\\Tr C=1$. The proof for the zero code does not use this cancellation.","\\section{Deriving the Gram matrix from recovery}\\label{supp5:section:003}\nSuppose a finite recovery that preserves trace corrects every operator supported on the code. To derive the Knill--Laflamme condition without assuming a syndrome decomposition, the proof expands the composite channel into the finite Kraus family\n\\[\nA_{ra}=R_rE_aP.\n\\]\nFor every ambient matrix $\\rho$, correction on the code implies\n\\[\n\\sum_{r,a} A_{ra}\\rho A_{ra}^*=P\\rho P.\n\\]\nThis equality on all matrices follows by applying correction to the sandwich supported on the code, $P\\rho P$, using the projector identities.","{Deriving the Gram matrix from recovery}\\label{supp5:section:003}\nSuppose a finite recovery that preserves trace corrects every operator supported on the code. To derive the Knill--Laflamme condition without assuming a syndrome decomposition, the proof expands the composite channel into the finite Kraus family\n\\[\nA_{ra}=R_rE_aP.\n\\]\nFor every ambient matrix $\\rho$, correction on the code implies\n\\[\n\\sum_{r,a} A_{ra}\\rho A_{ra}^*=P\\rho P.\n\\]\nThis equality on all matrices follows by applying correction to the sandwich supported on the code, $P\\rho P$, using the projector identities.","d on the code. To derive the Knill--Laflamme condition without assuming a syndrome decomposition, the proof expands the composite channel into the finite Kraus family\n\\[\nA_{ra}=R_rE_aP.\n\\]\nFor every ambient matrix $\\rho$, correction on the code implies\n\\[\n\\sum_{r,a} A_{ra}\\rho A_{ra}^*=P\\rho P.\n\\]\nThis equality on all matrices follows by applying correction to the sandwich supported on the code, $P\\rho P$, using the projector identities.","\\subsection{Rank-one rigidity of the corrected channel}\nThe module tests the map equality on actual matrix units and flattens each $A_{ra}$ into a row. If $X$ is the matrix of those rows and $p$ is the vector of entries of $P$, the resulting Gram identity is\n\\[\nX^*X=\\overline p\\,p^{\\mathsf T}.\n\\]\nThe checked rank-one Gram lemma proves that every row of $X$ is proportional to $p$. For nonzero $p$, a nonzero coordinate is selected and a test vector supported on two coordinates is placed in the null space of the Gram matrix. The implication $X^*Xv=0\\Rightarrow Xv=0$ then gives the required proportionality. When $p=0$, the Gram identity forces $X=0$ directly.","a row. If $X$ is the matrix of those rows and $p$ is the vector of entries of $P$, the resulting Gram identity is\n\\[\nX^*X=\\overline p\\,p^{\\mathsf T}.\n\\]\nThe checked rank-one Gram lemma proves that every row of $X$ is proportional to $p$. For nonzero $p$, a nonzero coordinate is selected and a test vector supported on two coordinates is placed in the null space of the Gram matrix. The implication $X^*Xv=0\\Rightarrow Xv=0$ then gives the required proportionality. When $p=0$, the Gram identity forces $X=0$ directly.","es of $P$, the resulting Gram identity is\n\\[\nX^*X=\\overline p\\,p^{\\mathsf T}.\n\\]\nThe checked rank-one Gram lemma proves that every row of $X$ is proportional to $p$. For nonzero $p$, a nonzero coordinate is selected and a test vector supported on two coordinates is placed in the null space of the Gram matrix. The implication $X^*Xv=0\\Rightarrow Xv=0$ then gives the required proportionality. When $p=0$, the Gram identity forces $X=0$ directly.","entity is\n\\[\nX^*X=\\overline p\\,p^{\\mathsf T}.\n\\]\nThe checked rank-one Gram lemma proves that every row of $X$ is proportional to $p$. For nonzero $p$, a nonzero coordinate is selected and a test vector supported on two coordinates is placed in the null space of the Gram matrix. The implication $X^*Xv=0\\Rightarrow Xv=0$ then gives the required proportionality. When $p=0$, the Gram identity forces $X=0$ directly.","Applied to the corrected Kraus family, this yields scalars $z_{ra}$ with\n\\[\nR_rE_aP=z_{ra}P.\n\\]\nThe essential declarations are \\path{rank_one_gram_rigidity}, \\path{kraus_rank_one_rigidity} and \\path{corrected_kraus_family}. No rank-one Choi characterization is inserted as an axiom or as the desired conclusion in hypothesis form.","\\subsection{Trace-preserving recovery supplies the original Gram data}\nLet $Z_{ra}=z_{ra}$. The recovery normalization gives\n\\begin{align*}\nPE_a^*E_bP\n&=PE_a^*\\Bigl(\\sum_rR_r^*R_r\\Bigr)E_bP","\\subsection{Trace-preserving recovery supplies the original Gram data}\nLet $Z_{ra}=z_{ra}$. The recovery normalization gives\n\\begin{align*}\nPE_a^*E_bP\n&=PE_a^*\\Bigl(\\sum_rR_r^*R_r\\Bigr)E_bP\\\\\n&=\\sum_r(R_rE_aP)^*(R_rE_bP)\\\\\n&=\\Bigl(\\sum_r\\overline{z_{ra}}z_{rb}\\Bigr)P\n=(Z^*Z)_{ab}P.\n\\end{align*}\nThus $C=Z^*Z$ is positive semidefinite and satisfies the required compressed identities. The theorem \\path{gram_of_recovery} implements this calculation with the original error and recovery matrices.","&=\\Bigl(\\sum_r\\overline{z_{ra}}z_{rb}\\Bigr)P\n=(Z^*Z)_{ab}P.\n\\end{align*}\nThus $C=Z^*Z$ is positive semidefinite and satisfies the required compressed identities. The theorem \\path{gram_of_recovery} implements this calculation with the original error and recovery matrices.","A second rigidity argument handles a nonnegative scalar multiple of the identity code map. It uses $\\sqrt t\\,P$ as the rank-one map generator and includes $t=0$. That argument proves necessity for unnormalized error families, which need not define channels.","\\subsection{Normalization is not an extra assumption}\nA nonzero positive semidefinite matrix has strictly positive real trace. If a complex-linear map fixes every normalized code density matrix, apply that hypothesis to\n\\[\n\\frac{\\rho}{\\Tr\\rho}\n\\]\nfor each nonzero positive code-supported $\\rho$. Linearity then gives equality on $\\rho$ itself. The zero matrix is handled separately. The source proves that the trace is real and nonzero, then justifies cancellation of the scalar. These steps establish that normalization is valid.","\\subsection{Complex polarization reaches the whole operator space}\nWrite $q(x)=xx^*$. The module proves the exact identity\n\\[\n4xy^*=q(x+y)-q(x-y)\n+i\\bigl(q(x+iy)-q(x-iy)\\bigr).\n\\]\nAgreement on all positive pure outer products therefore implies agreement on every outer product. Actual matrix units then span the full matrix space. Applying this argument to the sandwich by $P$ proves equality on every code-supported operator, including non-Hermitian ones.","\\subsection{An arbitrary finite reference system}\nFor any finite reference index set $A$, the amplified channel is defined using the actual Kronecker Kraus operators,\n\\[\n\\widetilde{\\mathcal N}_E(\\rho)\n=\\sum_a(I_A\\otimes E_a)\\rho(I_A\\otimes E_a)^*.\n\\]\nIts reference blocks are proved equal to the original channel applied blockwise. If\n\\[\n(I_A\\otimes P)\\rho(I_A\\otimes P)=\\rho,\n\\]\nthen every reference block is code supported. Applying the checked correction theorem for operators to those blocks gives\n\\[\n\\boxed{\\widetilde{\\mathcal R}\\,\\widetilde{\\mathcal N}_E(\\rho)=\\rho.}\n\\]\nThis includes entangled density matrices supported on the combined reference and code space. In fact, the identity holds for every jointly supported matrix and every finite reference dimension.","channel is defined using the actual Kronecker Kraus operators,\n\\[\n\\widetilde{\\mathcal N}_E(\\rho)\n=\\sum_a(I_A\\otimes E_a)\\rho(I_A\\otimes E_a)^*.\n\\]\nIts reference blocks are proved equal to the original channel applied blockwise. If\n\\[\n(I_A\\otimes P)\\rho(I_A\\otimes P)=\\rho,\n\\]\nthen every reference block is code supported. Applying the checked correction theorem for operators to those blocks gives\n\\[\n\\boxed{\\widetilde{\\mathcal R}\\,\\widetilde{\\mathcal N}_E(\\rho)=\\rho.}\n\\]\nThis includes entangled density matrices supported on the combined reference and code space. In fact, the identity holds for every jointly supported matrix and every finite reference dimension.","\\subsection{Cross terms for the original errors}\nFor arbitrary positive Gram data, the same recovery satisfies\n\\[\n\\boxed{\\mathcal R(E_a\\rho E_b^*)=C_{ba}\\rho\\qquad(P\\rho P=\\rho).}\n\\]\nThe index order $ba$ is intentional: the defining Gram relation is $PE_a^*E_bP=C_{ab}P$. The proof first derives scalar corrected Kraus composites for the constructed recovery and then uses its trace-preserving completeness to identify the original Gram entries.","For $E(c)=\\sum_a c_aE_a$, finite-sum expansion gives\n\\[\n\\mathcal R\\bigl(E(c)\\rho E(c)^*\\bigr)\n=\\left(\\sum_{a,b}c_a\\overline{c_b}C_{ba}\\right)\\rho\n=(c^*Cc)\\rho.\n\\]\nThe formula with a double sum is the literal checked conclusion. The final notation as a quadratic form follows by renaming finite indices. It is not restricted to orthogonal errors or real coefficients.","\\subsection{Every normalized channel in the same span}\nSuppose $F_j=\\sum_a c_{ja}E_a$ and $\\sum_jF_j^*F_j=I$. The decoder is \\emph{not} reconstructed for the $F_j$. Summing the coherent identities gives a scalar multiple $q\\rho$. Trace preservation of both the $F$ channel and the recovery forces $q=1$ on a nonzero code. The proof uses positivity of $P$ to justify cancellation of its trace and handles $P=0$ separately. Therefore\n\\[\n\\boxed{\\mathcal R\\Bigl(\\sum_jF_j\\rho F_j^*\\Bigr)=\\rho.}\n\\]\nThis is \\path{Data.recovery_span_channel}, a quantified statement over arbitrary finite $F$ families and their complex coefficients.","ection{Every normalized channel in the same span}\nSuppose $F_j=\\sum_a c_{ja}E_a$ and $\\sum_jF_j^*F_j=I$. The decoder is \\emph{not} reconstructed for the $F_j$. Summing the coherent identities gives a scalar multiple $q\\rho$. Trace preservation of both the $F$ channel and the recovery forces $q=1$ on a nonzero code. The proof uses positivity of $P$ to justify cancellation of its trace and handles $P=0$ separately. Therefore\n\\[\n\\boxed{\\mathcal R\\Bigl(\\sum_jF_j\\rho F_j^*\\Bigr)=\\rho.}\n\\]\nThis is \\path{Data.recovery_span_channel}, a quantified statement over arbitrary finite $F$ families and their complex coefficients.","el in the same span}\nSuppose $F_j=\\sum_a c_{ja}E_a$ and $\\sum_jF_j^*F_j=I$. The decoder is \\emph{not} reconstructed for the $F_j$. Summing the coherent identities gives a scalar multiple $q\\rho$. Trace preservation of both the $F$ channel and the recovery forces $q=1$ on a nonzero code. The proof uses positivity of $P$ to justify cancellation of its trace and handles $P=0$ separately. Therefore\n\\[\n\\boxed{\\mathcal R\\Bigl(\\sum_jF_j\\rho F_j^*\\Bigr)=\\rho.}\n\\]\nThis is \\path{Data.recovery_span_channel}, a quantified statement over arbitrary finite $F$ families and their complex coefficients.","nt identities gives a scalar multiple $q\\rho$. Trace preservation of both the $F$ channel and the recovery forces $q=1$ on a nonzero code. The proof uses positivity of $P$ to justify cancellation of its trace and handles $P=0$ separately. Therefore\n\\[\n\\boxed{\\mathcal R\\Bigl(\\sum_jF_j\\rho F_j^*\\Bigr)=\\rho.}\n\\]\nThis is \\path{Data.recovery_span_channel}, a quantified statement over arbitrary finite $F$ families and their complex coefficients.","\\subsection{The equivalence for unnormalized error families}\nFor arbitrary finite $E$, define exact family recovery to mean that a finite trace-preserving Kraus recovery and a single $t\\ge0$ exist such that\n\\[\n\\mathcal R\\mathcal N_E(\\rho)=t\\rho\n\\quad\\text{for every code-supported }\\rho.\n\\]\nThen \\path{error_family_correctable_iff_gram} proves that this condition is equivalent to the existence of $C\\succeq0$ satisfying the Gram identities. Sufficiency chooses $t=\\Tr C$. Necessity constructs $C$ from the scaled rank-one rigidity argument.","r arbitrary finite $E$, define exact family recovery to mean that a finite trace-preserving Kraus recovery and a single $t\\ge0$ exist such that\n\\[\n\\mathcal R\\mathcal N_E(\\rho)=t\\rho\n\\quad\\text{for every code-supported }\\rho.\n\\]\nThen \\path{error_family_correctable_iff_gram} proves that this condition is equivalent to the existence of $C\\succeq0$ satisfying the Gram identities. Sufficiency chooses $t=\\Tr C$. Necessity constructs $C$ from the scaled rank-one rigidity argument.","xact family recovery to mean that a finite trace-preserving Kraus recovery and a single $t\\ge0$ exist such that\n\\[\n\\mathcal R\\mathcal N_E(\\rho)=t\\rho\n\\quad\\text{for every code-supported }\\rho.\n\\]\nThen \\path{error_family_correctable_iff_gram} proves that this condition is equivalent to the existence of $C\\succeq0$ satisfying the Gram identities. Sufficiency chooses $t=\\Tr C$. Necessity constructs $C$ from the scaled rank-one rigidity argument.","The scale is essential. Zero errors can give $t=0$ and cannot be inverted to a nonzero normalized state. An input with a larger total trace is not silently declared a channel. For normalized input channels, the separate channel theorem removes the scale exactly as described in the sufficiency argument above.","\\section{Exact examples, trust audit and remaining work}\\label{supp5:section:006}\n\\subsection{Singular, nonorthogonal errors are accepted}\nOn a qubit with full code $P=I$, take\n\\[\nE_0=\\frac35I,\\qquad E_1=\\frac45I,\\qquad\nC=\\begin{pmatrix}9/25&12/25","\\section{Exact examples, trust audit and remaining work}\\label{supp5:section:006}\n\\subsection{Singular, nonorthogonal errors are accepted}\nOn a qubit with full code $P=I$, take\n\\[\nE_0=\\frac35I,\\qquad E_1=\\frac45I,\\qquad\nC=\\begin{pmatrix}9/25&12/25\\\\12/25&16/25\\end{pmatrix}.\n\\]\nLean checks the channel normalization, the Gram identities, positivity of $C$, $\\det C=0$, and the nonzero off-diagonal entry. It then applies the criterion for density matrices to prove correctability. The calculation is exact and uses rational complex entries.","12/25&16/25\\end{pmatrix}.\n\\]\nLean checks the channel normalization, the Gram identities, positivity of $C$, $\\det C=0$, and the nonzero off-diagonal entry. It then applies the criterion for density matrices to prove correctability. The calculation is exact and uses rational complex entries.","\\subsection{A genuine channel can still be uncorrectable}\nFor full qubit dephasing, $E_0=\\operatorname{diag}(1,0)$ and $E_1=\\operatorname{diag}(0,1)$, the Kraus normalization is checked. But $E_0^*E_0$ cannot be a scalar multiple of $I$, as its two diagonal entries differ. Necessity therefore proves that no finite recovery channel corrects the full code space. The case of the zero code is also checked independently.","\\section{The global projected connection}\\label{supp6:section:001}\nLet $M$ be a smooth real manifold without boundary, $F$ a complete normed real vector space and $P:M\\to\\End(F)$ a smooth family of bounded idempotents. For smooth range sections $s$, define\n\\[\n\\nabla_Xs=P\\,\\dd s[X].\n\\]\nThe proof begins by constructing the range bundle from $P$. Its local connection formulas glue, and the actual covariant commutator satisfies\n\\[\n\\boxed{(\\nabla_X\\nabla_Y-\\nabla_Y\\nabla_X-\\nabla_{[X,Y]})s\n=P\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P\\,s.}\n\\]\nThe derivative and Lie bracket are Mathlib's actual manifold operators. No global coordinate chart is assumed. The result requires idempotence, not an additional orthogonality hypothesis.","connection}\\label{supp6:section:001}\nLet $M$ be a smooth real manifold without boundary, $F$ a complete normed real vector space and $P:M\\to\\End(F)$ a smooth family of bounded idempotents. For smooth range sections $s$, define\n\\[\n\\nabla_Xs=P\\,\\dd s[X].\n\\]\nThe proof begins by constructing the range bundle from $P$. Its local connection formulas glue, and the actual covariant commutator satisfies\n\\[\n\\boxed{(\\nabla_X\\nabla_Y-\\nabla_Y\\nabla_X-\\nabla_{[X,Y]})s\n=P\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P\\,s.}\n\\]\nThe derivative and Lie bracket are Mathlib's actual manifold operators. No global coordinate chart is assumed. The result requires idempotence, not an additional orthogonality hypothesis.","\\section{Local products constructed from the projector}\\label{supp6:section:002}\nThe fiber over $x$ is the actual fixed subspace\n\\[\n\\mathcal E_x=\\{v\\in F:P(x)v=v\\}=\\Ran P(x).\n\\]\nThe total space is the subset $\\mathcal E=\\{(x,v):P(x)v=v\\}\\subseteq M\\times F$, with the induced topology. The construction provides local homeomorphisms of this total space as well as the transition functions between them.","\\subsection{An invertible intertwiner near each anchor}\nFor idempotents $p,q$, set\n\\[\nT(p,q)=pq+(I-p)(I-q).\n\\]\nLean proves $T(p,p)=I$ and $pT(p,q)=T(p,q)q$. For an anchor $a\\in M$, define\n\\[\nU_a=\\{x:T(P(x),P(a))\\text{ is invertible}\\}.\n\\]\nThe invertible elements form an open subset of the algebra of bounded operators. Consequently $U_a$ is open and contains $a$. The family $(U_a)$ covers the whole base. Invertibility is \\emph{derived locally}, not assumed at every pair of points.","\\subsection{The mutually inverse local maps}\nWrite $T_a(x)=T(P(x),P(a))$. For $x\\in U_a$,\n\\[\nB_a(x):\\mathcal E_a\\longrightarrow\\mathcal E_x,\n\\quad v\\longmapsto T_a(x)v,\n\\]\n\\[\nA_a(x):\\mathcal E_x\\longrightarrow\\mathcal E_a,\n\\quad w\\longmapsto P(a)T_a(x)^{-1}w.\n\\]\nThe intertwining identity and its inverse counterpart show that the displayed targets are correct. Lean proves both inverse laws, continuity, linearity in the fiber and smooth dependence on the base. Thus\n\\[\n\\mathcal E|_{U_a}\\ \\cong\\ U_a\\times\\mathcal E_a\n\\]\nis a local product homeomorphism that preserves the base point.",")$. For $x\\in U_a$,\n\\[\nB_a(x):\\mathcal E_a\\longrightarrow\\mathcal E_x,\n\\quad v\\longmapsto T_a(x)v,\n\\]\n\\[\nA_a(x):\\mathcal E_x\\longrightarrow\\mathcal E_a,\n\\quad w\\longmapsto P(a)T_a(x)^{-1}w.\n\\]\nThe intertwining identity and its inverse counterpart show that the displayed targets are correct. Lean proves both inverse laws, continuity, linearity in the fiber and smooth dependence on the base. Thus\n\\[\n\\mathcal E|_{U_a}\\ \\cong\\ U_a\\times\\mathcal E_a\n\\]\nis a local product homeomorphism that preserves the base point.","\\subsection{Smooth overlap maps and rank}\nThe coordinate transition is $g_{ab}(x)=A_b(x)B_a(x)$. It is smooth on $U_a\\cap U_b$, agrees with the actual change between local homeomorphisms, and satisfies\n\\[\ng_{aa}=I,\\qquad g_{bc}g_{ab}=g_{ac}.\n\\]\nA continuous linear equivalence between $\\mathcal E_a$ and $\\mathcal E_x$ also gives local constancy of fiber dimension. No single constant rank across disconnected components is assumed.","\\section{A connection on the constructed bundle}\\label{supp6:section:003}\nThe global formula is defined using the actual manifold derivative:\n\\[\n\\nabla_Xs(x)=P(x)\\,D_Ms(x)[X(x)].\n\\]\nHere $D_M$ is represented by \\path{mvfderiv}. A second definition, \\path{GlobalConnection.covariant}, has values in the actual fiber:\n\\[\n\\nabla s(x):T_xM\\longrightarrow\\mathcal E_x.\n\\]\nIt is not only an ambient expression asserted to land in the range.","\\subsection{Connection axioms and smoothness}\nThe formal proofs establish range preservation, additivity in the section, linearity over functions in the vector field argument, and the scalar Leibniz identity\n\\[\n\\nabla_X(fs)=\\dd f[X]\\,s+f\\nabla_Xs.\n\\]\nSmooth $P$, smooth sections and smooth vector fields give smooth covariant derivatives. Equality of section germs gives equality of covariant derivatives at the point, so the operation is local.","\\subsection{Smoothness of sections in bundle coordinates}\nLet $x\\in U_a$. If $P(y)s(y)=s(y)$ in a neighborhood of $x$, then\n\\[\ns\\text{ is smooth at }x\n\\quad\\Longleftrightarrow\\quad\n\\bigl(y\\mapsto A_a(y)s(y)\\bigr)\\text{ is smooth at }x.\n\\]\nThe range condition is only local. The proof uses the actual inverse local maps and equality of germs. It follows that smoothness in bundle coordinates does not depend on the chosen chart.","\\subsection{The local connection and its change of coordinates}\nFor a coordinate function $u$ valued in $\\mathcal E_a$, define\n\\[\n\\nabla^{(a)}_Xu=A_a\\bigl(P\\,D_M(B_a u)[X]\\bigr).\n\\]\nFor every overlap point, Lean proves the operator identity\n\\[\n\\boxed{\\nabla^{(b)}_X(g_{ab}u)=g_{ab}\\nabla^{(a)}_Xu.}\n\\]\nThis is proved for arbitrary local coordinate functions. It does not assume that $u$ was obtained from a globally smooth section or that a smooth extension has been supplied. Together with the local smoothness criterion, this closes the gluing step left open in the fifth block.","\\subsection{The flat ambient identity is proved in charts}\nFor a smooth vector-valued function $s$, the proof identifies directional derivatives with the pullback into each extended chart. It then proves\n\\[\nD_Ms([X,Y])=D_M(D_Ms[Y])[X]-D_M(D_Ms[X])[Y].\n\\]\nThis uses the chart pullback of the actual vector fields and the ordinary differential identity, then transfers the result back to the manifold. The auxiliary module works locally at a point, so the argument applies within arbitrary open chart domains.","\\subsection{Projector and section identities are derived}\nDifferentiating $P^2=P$ in a tangent direction $v$ yields\n\\[\n(\\dd P[v])P+P\\dd P[v]=\\dd P[v],\n\\qquad P\\dd P[v]P=0.\n\\]\nDifferentiating $Ps=s$ gives\n\\[\n\\dd s[v]=\\dd P[v]s+P\\dd s[v].\n\\]\nBoth statements are consequences of actual derivative rules. They are not additional assumptions.","\\subsection{Cancellation of the second derivatives}\nThe product rule and the identity for the ambient Lie bracket reduce curvature to\n\\[\nR^\\nabla(X,Y)s=P\\dd P[X]\\dd s[Y]-P\\dd P[Y]\\dd s[X].\n\\]\nSubstitute the identity obtained by differentiating the range condition for the section. The terms containing $P\\dd P[v]P$ vanish, leaving\n\\[\nR^\\nabla(X,Y)s=\nP\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P s.\n\\]\nLean also proves that this depends only on $X(x)$, $Y(x)$ and $s(x)$, not on their extensions. The operator-valued form is alternating, additive and homogeneous, and lands in the actual fiber. Constant projector bundles have zero curvature form.","es}\nThe product rule and the identity for the ambient Lie bracket reduce curvature to\n\\[\nR^\\nabla(X,Y)s=P\\dd P[X]\\dd s[Y]-P\\dd P[Y]\\dd s[X].\n\\]\nSubstitute the identity obtained by differentiating the range condition for the section. The terms containing $P\\dd P[v]P$ vanish, leaving\n\\[\nR^\\nabla(X,Y)s=\nP\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P s.\n\\]\nLean also proves that this depends only on $X(x)$, $Y(x)$ and $s(x)$, not on their extensions. The operator-valued form is alternating, additive and homogeneous, and lands in the actual fiber. Constant projector bundles have zero curvature form.","product rule and the identity for the ambient Lie bracket reduce curvature to\n\\[\nR^\\nabla(X,Y)s=P\\dd P[X]\\dd s[Y]-P\\dd P[Y]\\dd s[X].\n\\]\nSubstitute the identity obtained by differentiating the range condition for the section. The terms containing $P\\dd P[v]P$ vanish, leaving\n\\[\nR^\\nabla(X,Y)s=\nP\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P s.\n\\]\nLean also proves that this depends only on $X(x)$, $Y(x)$ and $s(x)$, not on their extensions. The operator-valued form is alternating, additive and homogeneous, and lands in the actual fiber. Constant projector bundles have zero curvature form.","e and the identity for the ambient Lie bracket reduce curvature to\n\\[\nR^\\nabla(X,Y)s=P\\dd P[X]\\dd s[Y]-P\\dd P[Y]\\dd s[X].\n\\]\nSubstitute the identity obtained by differentiating the range condition for the section. The terms containing $P\\dd P[v]P$ vanish, leaving\n\\[\nR^\\nabla(X,Y)s=\nP\\bigl(\\dd P[X]\\dd P[Y]-\\dd P[Y]\\dd P[X]\\bigr)P s.\n\\]\nLean also proves that this depends only on $X(x)$, $Y(x)$ and $s(x)$, not on their extensions. The operator-valued form is alternating, additive and homogeneous, and lands in the actual fiber. Constant projector bundles have zero curvature form.","s retain valid mathematical arguments from the original manuscript. Their hypotheses and notation agree with the corrected finite core. The earlier abbreviated argument for error correction is replaced by the complete normalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{","manuscript. Their hypotheses and notation agree with the corrected finite core. The earlier abbreviated argument for error correction is replaced by the complete normalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of proje","Their hypotheses and notation agree with the corrected finite core. The earlier abbreviated argument for error correction is replaced by the complete normalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curv","ses and notation agree with the corrected finite core. The earlier abbreviated argument for error correction is replaced by the complete normalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior f","omplete normalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential fo","ormalized and scaled recovery proofs above, and its substantive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bund","ive conclusions are included there.\n\\subsection{Duhamel proof of leakage from a fixed packet}\\label{legacy:duhamel}\nLet $H=H^*$ on a finite-dimensional complex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square,","lex Hilbert space, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof","pace, $P=P^*=P^2$, $Q=I-P$, and $t\\in\\mathbb R$.\n\\begin{proof}[Alternative proof of the commutator leakage bound]\\phantomsection\\label{legacy:proof:001}\nBecause $QP=0$,\n\\[\nQe^{-itH}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formu","}P=Q[e^{-itH},P]P.\n\\]\nThe Duhamel identity gives\n\\[\n[e^{-itH},P]=-i\\int_0^t e^{-i(t-s)H}[H,P]e^{-isH}\\,ds.\n\\]\nBoth unitary factors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and","tors have norm one, so taking norms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\su","rms and using the integration interval's length $|t|$ yields\n$\\|Qe^{-itH}P\\|\\le |t|\\,\\|[H,P]\\|$.\n\\phantomsection\\label{legacy:proof:001:end}\\end{proof}\n\\subsection{Proof of projected curvature with exterior forms}\\label{legacy:exterior}\nLet $P$ be a smooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restrict","ooth idempotent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1","potent field and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin","eld and extend the projected connection to differential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\s","ential forms with values in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6","in the range bundle by $\\nabla\\omega=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x","a=P\\,d\\omega$. Thus $\\nabla^2$ denotes the exterior covariant square, including the Lie bracket term in the exterior derivative when evaluated on vector fields.\n\\begin{proof}[Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4","Alternative proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","ve proof of the curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","e curvature formula]\\phantomsection\\label{legacy:proof:002}\nFor a section $s$ with $Ps=s$,\n\\[\n\\nabla^2s=P\\,d(P\\,ds)=P(dP)\\wedge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","dge ds.\n\\]\nSince $ds=d(Ps)=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","P=0$, the surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","surviving term is\n\\[\nP(dP\\wedge dP)Ps.\n\\]\n\\phantomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","tomsection\\label{legacy:proof:002:end}\\end{proof}\n\\par\\Needspace{24\\baselineskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","neskip}\n\\subsection{Restricted low-mode proof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","oof}\\label{legacy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","cy:lowmode}\nLet $h\\ne0$, $x=kh/2$, and $0\\le x\\le1$.\n\\begin{proof}[Alternative proof in the original low-mode regime]\\phantomsection\\label{legacy:proof:003}\nSince $0\\le\\sin x\\le x$, the error is nonnegative. On $[0,1]$,\n\\[\n\\sin x\\ge x-\\frac{x^3}{6}\\ge0.\n\\]\nTherefore\n\\[\n\\sin^2x\\ge\\left(x-\\frac{x^3}{6}\\right)^2\\ge x^2-\\frac{x^4}{3}.\n\\]\nHence $x^2-\\sin^2x\\le x^4/3$. Multiplication by $4/h^2$ gives\n\\[\n0\\le k^2-\\frac4{h^2}\\sin^2(kh/2)\\le\\frac{k^4h^2}{12}.\n\\]\n\\phantomsection\\label{legacy:proof:003:end}\\end{proof}","5,4.05);\n\\draw[dashed,black!55,line width=.6pt] (4.35,3.1) -- (6,4.05);\n\\draw[black!55,line width=.6pt] (2.65,1.6) -- (6,1.6);\n\\draw[black!55,line width=.6pt] (2.65,1.6) -- (2.65,4.05);\n\\draw[black!55,line width=.6pt] (6,1.6) -- (6,4.05);\n\\draw[black!55,line width=.6pt] (2.65,4.05) -- (6,4.05);\n\\path[draw,fill=white] (0.53,0.35) rectangle (1.47,0.95);\n\\node[font=\\small,align=center] at (1,0.65) {$\\ket{000}$};\n\\path[draw,fill=white] (3.88,0.35) rectangle (4.82,0.95);\n\\node[font=\\small,align=center] at (4.35,0.65) {$\\ket{001}$};\n\\path[draw,fill=white] (0.53,2.8) rectangle (1.47,3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53",",1.6);\n\\draw[black!55,line width=.6pt] (2.65,1.6) -- (2.65,4.05);\n\\draw[black!55,line width=.6pt] (6,1.6) -- (6,4.05);\n\\draw[black!55,line width=.6pt] (2.65,4.05) -- (6,4.05);\n\\path[draw,fill=white] (0.53,0.35) rectangle (1.47,0.95);\n\\node[font=\\small,align=center] at (1,0.65) {$\\ket{000}$};\n\\path[draw,fill=white] (3.88,0.35) rectangle (4.82,0.95);\n\\node[font=\\small,align=center] at (4.35,0.65) {$\\ket{001}$};\n\\path[draw,fill=white] (0.53,2.8) rectangle (1.47,3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75)","05);\n\\draw[black!55,line width=.6pt] (2.65,4.05) -- (6,4.05);\n\\path[draw,fill=white] (0.53,0.35) rectangle (1.47,0.95);\n\\node[font=\\small,align=center] at (1,0.65) {$\\ket{000}$};\n\\path[draw,fill=white] (3.88,0.35) rectangle (4.82,0.95);\n\\node[font=\\small,align=center] at (4.35,0.65) {$\\ket{001}$};\n\\path[draw,fill=white] (0.53,2.8) rectangle (1.47,3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle (3.12,4.35);\n\\node[font=\\small,align=center] at (2.65,4.05) {$\\ket{110}$};\n\\path[draw,fill=white] (5.53",");\n\\node[font=\\small,align=center] at (1,0.65) {$\\ket{000}$};\n\\path[draw,fill=white] (3.88,0.35) rectangle (4.82,0.95);\n\\node[font=\\small,align=center] at (4.35,0.65) {$\\ket{001}$};\n\\path[draw,fill=white] (0.53,2.8) rectangle (1.47,3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle (3.12,4.35);\n\\node[font=\\small,align=center] at (2.65,4.05) {$\\ket{110}$};\n\\path[draw,fill=white] (5.53,3.75) rectangle (6.47,4.35);\n\\node[font=\\small,align=center] at (6,4.05) {$\\ket{111}$};\n\\node[font=\\small,align=cent",");\n\\node[font=\\small,align=center] at (4.35,0.65) {$\\ket{001}$};\n\\path[draw,fill=white] (0.53,2.8) rectangle (1.47,3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle (3.12,4.35);\n\\node[font=\\small,align=center] at (2.65,4.05) {$\\ket{110}$};\n\\path[draw,fill=white] (5.53,3.75) rectangle (6.47,4.35);\n\\node[font=\\small,align=center] at (6,4.05) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,4.1) {Three actual words};\n\\node[font=\\small,align=center,font=\\small] at (10.15,3.1) {$W_{",",3.4);\n\\node[font=\\small,align=center] at (1,3.1) {$\\ket{010}$};\n\\path[draw,fill=white] (3.88,2.8) rectangle (4.82,3.4);\n\\node[font=\\small,align=center] at (4.35,3.1) {$\\ket{011}$};\n\\path[draw,fill=white] (2.18,1.3) rectangle (3.12,1.9);\n\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle (3.12,4.35);\n\\node[font=\\small,align=center] at (2.65,4.05) {$\\ket{110}$};\n\\path[draw,fill=white] (5.53,3.75) rectangle (6.47,4.35);\n\\node[font=\\small,align=center] at (6,4.05) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,4.1) {Three actual words};\n\\node[font=\\small,align=center,font=\\small] at (10.15,3.1) {$W_{101}=X_1X_3$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,2.2) 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{$\\","\\node[font=\\small,align=center] at (2.65,1.6) {$\\ket{100}$};\n\\path[draw,fill=white] (5.53,1.3) rectangle (6.47,1.9);\n\\node[font=\\small,align=center] at (6,1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle (3.12,4.35);\n\\node[font=\\small,align=center] at (2.65,4.05) {$\\ket{110}$};\n\\path[draw,fill=white] (5.53,3.75) rectangle (6.47,4.35);\n\\node[font=\\small,align=center] at (6,4.05) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,4.1) {Three actual words};\n\\node[font=\\small,align=center,font=\\small] at (10.15,3.1) {$W_{101}=X_1X_3$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,2.2) {$W_{011}=X_2X_3$};\n\\node[font=\\small,align=center,font=\\small] at (10.15,1.3) {$W_{110}=X_1X_2$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.1) {$\\dim\\Span\\{W_a\\ket{000}\\}=8$};\n\\end{tikzpicture}",",1.6) {$\\ket{101}$};\n\\path[draw,fill=white] (2.18,3.75) rectangle 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join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\path[draw,line width=.8pt] (2.8,2.45) circle (1.62);\n\\draw[black!35] (2.8,2.45) ellipse (1.62 and .45);\n\\draw[black!35] (2.8,2.45) ellipse (.6 and 1.62);\n\\draw[dashed,black!65] (2.8,3.25) ellipse (1.4 and .31);\n\\draw[->] (2.8,2.45) -- (2.8,4.08);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.8,0.2) {Cap flux at $\\theta=\\pi/3$ is $\\pi/2$.};\n\\draw[black!13,line width=.25pt] (6.4,0.85) -- (6.4,3.65);\n\\draw[] (6.4,0.79) -- (6.4,0.91);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.4,0.59) {0};\n\\draw[black!13,line width=.25pt] (6.4,0.85) -- (13.1,0.85);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,0.85) {0};\n\\draw[black!13,line width=.25pt] (6.4,2.016667) -- (13.1,2.016667);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,2.016667) {0.25};\n\\draw[black!13,line width=","d]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\path[draw,line width=.8pt] (2.8,2.45) circle (1.62);\n\\draw[black!35] (2.8,2.45) ellipse (1.62 and .45);\n\\draw[black!35] (2.8,2.45) ellipse (.6 and 1.62);\n\\draw[dashed,black!65] (2.8,3.25) ellipse (1.4 and .31);\n\\draw[->] (2.8,2.45) -- (2.8,4.08);\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.8,0.2) {Cap flux at $\\theta=\\pi/3$ is $\\pi/2$.};\n\\draw[black!13,line width=.25pt] (6.4,0.85) -- (6.4,3.65);\n\\draw[] (6.4,0.79) -- (6.4,0.91);\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.4,0.59) {0};\n\\draw[black!13,line width=.25pt] (6.4,0.85) -- (13.1,0.85);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,0.85) {0};\n\\draw[black!13,line width=.25pt] (6.4,2.016667) -- (13.1,2.016667);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,2.016667) {0.25};\n\\draw[black!13,line width=.25pt] (6.4","nter,anchor=east,font=\\scriptsize] at (6.24,2.016667) {0.25};\n\\draw[black!13,line width=.25pt] (6.4,3.183333) -- (13.1,3.183333);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,3.183333) {0.5};\n\\draw[->,line width=.6pt] (6.4,0.85) -- (13.22,0.85);\n\\draw[->,line width=.6pt] (6.4,0.85) -- (6.4,3.77);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.75,0.3) {polar angle $\\theta$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (6.4,4.05) {$F_{\\theta\\phi}=\\frac12\\sin\\theta$};\n\\draw[line width=1.1pt] plot coordinates {(6.4,0.85) (6.467,0.923292) (6.534,0.996511) (6.601,1.069586) (6.668,1.142444) (6.735,1.215014) (6.802,1.287223) (6.869,1.359001) (6.936,1.430276) (7.003,1.500979) (7.07,1.57104) (7.137,1.640388) (7.204,1.708957) (7.271,1.776678) (7.338,1.843485) (7.405,1.909311) (7.472,1.974092) (7.539,2.037763) (7.606,2.100263) (7.673,2.161528) (7.74,2.2","=.25pt] (6.4,3.183333) -- (13.1,3.183333);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (6.24,3.183333) {0.5};\n\\draw[->,line width=.6pt] 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(1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] a","ll] at (3,3.6) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,3.2) rectangle (4.2,4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,3.6) {$1$};\n\\path[draw=black!30,fill=black!8] (1,2.4) rectangle (1.8,3.2);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2.8) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,2.4) rectangle (2.6,3.2);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.8) {$- i$};\n\\path[draw=black!30,fill=black!8] (2.6,2.4) rectangle (3.4,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3,2.8) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,2.4) rectangle (4.2,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2","at (3.8,3.6) {$1$};\n\\path[draw=black!30,fill=black!8] (1,2.4) rectangle (1.8,3.2);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2.8) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,2.4) rectangle (2.6,3.2);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.8) {$- i$};\n\\path[draw=black!30,fill=black!8] (2.6,2.4) rectangle (3.4,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3,2.8) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,2.4) rectangle (4.2,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3","(1.4,2.8) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,2.4) rectangle (2.6,3.2);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.8) {$- i$};\n\\path[draw=black!30,fill=black!8] (2.6,2.4) rectangle (3.4,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3,2.8) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,2.4) rectangle (4.2,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2","at (2.2,2.8) {$- i$};\n\\path[draw=black!30,fill=black!8] (2.6,2.4) rectangle (3.4,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3,2.8) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,2.4) rectangle (4.2,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,","all] at (3,2.8) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,2.4) rectangle (4.2,3.2);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2","small] at (3.8,2.8) {$i$};\n\\path[draw=black!30,fill=black!8] (1,1.6) rectangle (1.8,2.4);\n\\node[font=\\small,align=center,font=\\small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at","small] at (1.4,2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,1.6) rectangle (2.6,2.4);\n\\node[font=\\small,align=center,font=\\small] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3,1.2) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,0.8) rectangle (4.2,1.6);\n\\node[font=\\small,align=center,font=\\small] at","mall] at (2.2,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (2.6,1.6) rectangle (3.4,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3,1.2) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,0.8) rectangle (4.2,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3.8,1.2) {$- i$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,3.6) {site 0};\n\\node[font=\\small,align","all] at (3,2) {$1$};\n\\path[draw=black!30,fill=black!8] (3.4,1.6) rectangle (4.2,2.4);\n\\node[font=\\small,align=center,font=\\small] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3,1.2) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,0.8) rectangle (4.2,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3.8,1.2) {$- i$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,3.6) {site 0};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.","l] at (3.8,2) {$-1$};\n\\path[draw=black!30,fill=black!8] (1,0.8) rectangle (1.8,1.6);\n\\node[font=\\small,align=center,font=\\small] at (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3,1.2) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,0.8) rectangle (4.2,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3.8,1.2) {$- i$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,3.6) {site 0};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=cen","t (1.4,1.2) {$1$};\n\\path[draw=black!30,fill=black!8] (1.8,0.8) rectangle (2.6,1.6);\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.2) {$i$};\n\\path[draw=black!30,fill=black!8] (2.6,0.8) rectangle (3.4,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3,1.2) {$-1$};\n\\path[draw=black!30,fill=black!8] (3.4,0.8) rectangle (4.2,1.6);\n\\node[font=\\small,align=center,font=\\small] at (3.8,1.2) {$- i$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,3.6) {site 0};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\sm","de[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,3.6) {site 0};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.","{site 0};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.","at (0.85,2.8) {site 1};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] a","=\\scriptsize] at (0.85,2) {site 2};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\sma","anchor=east,font=\\scriptsize] at (0.85,1.2) {site 3};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,1.25) {$S v_3=- iv_3$};\n\\node[font=\\small,align=center,font=\\","gn=center,font=\\scriptsize] at (1.4,4.23) {$j=0$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,1.25) {$S v_3=- iv_3$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,1.25) {$P_3=v_3v_3^*$};\n\\node[font=\\small,align=center,font=","nter,font=\\scriptsize] at (2.2,4.23) {$j=1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,1.25) {$S v_3=- iv_3$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,1.25) {$P_3=v_3v_3^*$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.8,0.12) {Each column has four entries of magnitude $1/2$",",font=\\scriptsize] at (3,4.23) {$j=2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,1.25) {$S v_3=- iv_3$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,1.25) {$P_3=v_3v_3^*$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.8,0.12) {Each column has four entries of magnitude $1/2$.};\n\\end{tikzpicture}","\\scriptsize] at (3.8,4.23) {$j=3$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S v_2=-1v_2$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.1) {$P_2=v_2v_2^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,1.25) {$S v_3=- iv_3$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,1.25) {$P_3=v_3v_3^*$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.8,0.12) {Each column has four entries of magnitude $1/2$.};\n\\end{tikzpicture}","ll] at (8.05,3.8) {$S v_0=1v_0$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,3.8) {$P_0=v_0v_0^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.95) {$S v_1=iv_1$};\n\\node[font=\\small,align=center,font=\\small] at (11.85,2.95) {$P_1=v_1v_1^*$};\n\\node[font=\\small,align=center,font=\\small] at (8.05,2.1) {$S 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(9.825,2.075) {$0$};\n\\path[draw=black!30,fill=white] (10.2,1.7) rectangle (10.95,2.45);\n\\node[font=\\small,align=center,font=\\small] at (10.575,2.075) {$0$};\n\\path[draw=black!30,fill=white] (10.95,1.7) rectangle (11.7,2.45);\n\\node[font=\\small,align=center,font=\\small] at (11.325,2.075) {$0$};\n\\path[draw=black!30,fill=white] (8.7,0.95) rectangle (9.45,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.075,1.325) {$0$};\n\\path[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,…","path[draw=black!30,fill=white] (9.45,1.7) rectangle (10.2,2.45);\n\\node[font=\\small,align=center,font=\\small] at (9.825,2.075) {$0$};\n\\path[draw=black!30,fill=white] (10.2,1.7) rectangle (10.95,2.45);\n\\node[font=\\small,align=center,font=\\small] at (10.575,2.075) {$0$};\n\\path[draw=black!30,fill=white] (10.95,1.7) rectangle (11.7,2.45);\n\\node[font=\\small,align=center,font=\\small] at (11.325,2.075) {$0$};\n\\path[draw=black!30,fill=white] (8.7,0.95) rectangle (9.45,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.075,1.325) {$0$};\n\\path[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,3.575) {00};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2…","\\path[draw=black!30,fill=white] (10.2,1.7) rectangle (10.95,2.45);\n\\node[font=\\small,align=center,font=\\small] at (10.575,2.075) {$0$};\n\\path[draw=black!30,fill=white] (10.95,1.7) rectangle (11.7,2.45);\n\\node[font=\\small,align=center,font=\\small] at (11.325,2.075) {$0$};\n\\path[draw=black!30,fill=white] (8.7,0.95) rectangle (9.45,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.075,1.325) {$0$};\n\\path[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,3.575) {00};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.825) {01};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.075) {10};\n\\node[font=\\small,align=center,anchor=…",";\n\\path[draw=black!30,fill=white] (10.95,1.7) rectangle (11.7,2.45);\n\\node[font=\\small,align=center,font=\\small] at (11.325,2.075) {$0$};\n\\path[draw=black!30,fill=white] (8.7,0.95) rectangle (9.45,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.075,1.325) {$0$};\n\\path[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,3.575) {00};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.825) {01};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.075) {10};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,1.325) {11};\n\\node[font=\\small,align=center,font=\\small] at (6.1,2.3) {same diagonal",";\n\\path[draw=black!30,fill=white] (8.7,0.95) rectangle (9.45,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.075,1.325) {$0$};\n\\path[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,3.575) {00};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.825) {01};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.075) {10};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,1.325) {11};\n\\node[font=\\small,align=center,font=\\small] at (6.1,2.3) {same diagonal","th[draw=black!30,fill=white] (9.45,0.95) rectangle (10.2,1.7);\n\\node[font=\\small,align=center,font=\\small] at (9.825,1.325) {$0$};\n\\path[draw=black!30,fill=white] (10.2,0.95) rectangle (10.95,1.7);\n\\node[font=\\small,align=center,font=\\small] at (10.575,1.325) {$0$};\n\\path[draw=black!30,fill=black!8] (10.95,0.95) rectangle (11.7,1.7);\n\\node[font=\\small,align=center,font=\\small] at (11.325,1.325) {$1$};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,3.575) {00};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.825) {01};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,2.075) {10};\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (8.55,1.325) {11};\n\\node[font=\\small,align=center,font=\\small] at (6.1,2.3) {same diagonal","different coherence};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {Bell-projector outcome: $1$ on the left, $1/2$ on the right.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (2.2,4.3) {$h=\\frac{1}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (2.2,4.3) {$h=\\frac{1}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[fo","e}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (2.2,4.3) {$h=\\frac{1}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,","]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center] at (2.2,4.3) {$h=\\frac{1}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\no","[font=\\small,align=center] at (2.2,4.3) {$h=\\frac{1}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\sim","all,align=center,font=\\small] at (2.2,2.65) {$\\simeq 0.005197495$};\n\\node[font=\\small,align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072","align=center,font=\\small] at (2.2,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive e","2$};\n\\node[font=\\small,align=center] at (2.2,1) {$\\frac{1}{192}$};\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikz",";\n\\node[font=\\small,align=center] at (7,4.3) {$h=\\frac{1}{8}$};\n\\node[font=\\small,align=center,font=\\small] at (7,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","align=center,font=\\small] at (7,2.65) {$\\simeq 0.001301405$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","ter,font=\\small] at (7,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","t=\\small,align=center] at (7,1) {$\\frac{1}{768}$};\n\\node[font=\\small,align=center] at (11.8,4.3) {$h=\\frac{1}{16}$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","e[font=\\small,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","ll,align=center,font=\\small] at (11.8,3.3) {error (approx.)};\n\\node[font=\\small,align=center,font=\\small] at (11.8,2.65) {$\\simeq 0.000325478$};\n\\node[font=\\small,align=center,font=\\small] at (11.8,1.65) {bound $h^2/12$};\n\\node[font=\\small,align=center] at (11.8,1) {$\\frac{1}{3072}$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.05) {Successive error ratios lie strictly between $3.99$ and $4$.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (2.25,4.2) {$T$ at $e^{-2\\tau}=1/2$};\n\\path[draw=black!30,fill=black!8] (1.1,2.35) rectangle (2.25,3.5);\n\\node[font=\\small,align=center,font=\\small] at (1.675,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (2.25,4.2) {$T$ at $e^{-2\\tau}=1/2$};\n\\path[draw=black!30,fill=black!8] (1.1,2.35) rectangle (2.25,3.5);\n\\node[font=\\small,align=center,font=\\small] at (1.675,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (2.25,4.2) {$T$ at $e^{-2\\tau}=1/2$};\n\\path[draw=black!30,fill=black!8] (1.1,2.35) rectangle (2.25,3.5);\n\\node[font=\\small,align=center,font=\\small] at (1.675,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black","unding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (2.25,4.2) {$T$ at $e^{-2\\tau}=1/2$};\n\\path[draw=black!30,fill=black!8] (1.1,2.35) rectangle (2.25,3.5);\n\\node[font=\\small,align=center,font=\\small] at (1.675,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=bla…","fill=black!8] (1.1,2.35) rectangle (2.25,3.5);\n\\node[font=\\small,align=center,font=\\small] at (1.675,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle (10.35,2.35);\n\\node[font=\\small,align=center,font=\\small] at (9.775,1.775) {$- \\frac{1}{4}$};\n\\path[draw=…","fill=black!8] (2.25,2.35) rectangle (3.4,3.5);\n\\node[font=\\small,align=center,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle (10.35,2.35);\n\\node[font=\\small,align=center,font=\\small] at (9.775,1.775) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,1.2) rectangle (11.5,2.35);\n\\node[font=\\small,align=center,font=\\small] at (10.925,1.775) {$\\frac{3}{4}$};\n\\node[fo…","r,font=\\small] at (2.825,2.925) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (1.1,1.2) rectangle (2.25,2.35);\n\\node[font=\\small,align=center,font=\\small] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle (10.35,2.35);\n\\node[font=\\small,align=center,font=\\small] at (9.775,1.775) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,1.2) rectangle (11.5,2.35);\n\\node[font=\\small,align=center,font=\\small] at (10.925,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (6.5,2.4) {Both have","all] at (1.675,1.775) {$\\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (2.25,1.2) rectangle (3.4,2.35);\n\\node[font=\\small,align=center,font=\\small] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle (10.35,2.35);\n\\node[font=\\small,align=center,font=\\small] at (9.775,1.775) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,1.2) rectangle (11.5,2.35);\n\\node[font=\\small,align=center,font=\\small] at (10.925,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (6.5,2.4) {Both have","] at (2.825,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (10.35,4.2) {$ZTZ$ in another basis};\n\\path[draw=black!30,fill=black!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle (10.35,2.35);\n\\node[font=\\small,align=center,font=\\small] at (9.775,1.775) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,1.2) rectangle (11.5,2.35);\n\\node[font=\\small,align=center,font=\\small] at (10.925,1.775) {$\\frac{3}{4}$};\n\\node[font=\\small,align=center,font=\\small] at (6.5,2.4) {Both have","ack!8] (9.2,2.35) rectangle (10.35,3.5);\n\\node[font=\\small,align=center,font=\\small] at (9.775,2.925) {$\\frac{3}{4}$};\n\\path[draw=black!30,fill=black!8] (10.35,2.35) rectangle (11.5,3.5);\n\\node[font=\\small,align=center,font=\\small] at (10.925,2.925) {$- \\frac{1}{4}$};\n\\path[draw=black!30,fill=black!8] (9.2,1.2) rectangle 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(3.975,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (1.7,2.2) -- (6.25,1.1);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,1.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,3.5) -- (12,2.4);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,1.1) -- (12,2.2);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,1.3) {$-1/\\sqrt2$};\n\\node[font=\\small,align=center,font=\\scriptsize] a","6,1) {$1$};\n\\path[draw,fill=white] (12.3,2.3) circle (0.27);\n\\node[font=\\small,align=center] at (12.3,2.3) {$1$};\n\\draw[->,line width=.8pt] (1.7,2.4) -- (6.25,3.5);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (1.7,2.2) -- (6.25,1.1);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,1.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,3.5) -- (12,2.4);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,1.1) -- (12,2.2);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,1.3) {$-1/\\sqrt2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.7,4.6) {Two actual Hadamard paths, not classical alternatives};\n\\node[font=\\small,align=center,font=\\small] at (6.8,-0.18) {$\\text{amplitude}=1/",") -- (6.25,3.5);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (1.7,2.2) -- (6.25,1.1);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,1.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,3.5) -- (12,2.4);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,1.1) -- (12,2.2);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,1.3) {$-1/\\sqrt2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.7,4.6) {Two actual Hadamard paths, not classical alternatives};\n\\node[font=\\small,align=center,font=\\small] at (6.8,-0.18) {$\\text{amplitude}=1/2-1/2=0\\quad\\ne\\quad\\text{sum of path probabilities}=1/2$};\n\\end{tikzpicture}","2) -- (6.25,1.1);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (3.975,1.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,3.5) -- (12,2.4);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,1.1) -- (12,2.2);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,1.3) {$-1/\\sqrt2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.7,4.6) {Two actual Hadamard paths, not classical alternatives};\n\\node[font=\\small,align=center,font=\\small] at (6.8,-0.18) {$\\text{amplitude}=1/2-1/2=0\\quad\\ne\\quad\\text{sum of path probabilities}=1/2$};\n\\end{tikzpicture}","ter,font=\\small,fill=white,inner sep=1pt] at (9.475,3.3) {$1/\\sqrt2$};\n\\draw[->,line width=.8pt] (6.95,1.1) -- (12,2.2);\n\\node[font=\\small,align=center,font=\\small,fill=white,inner sep=1pt] at (9.475,1.3) {$-1/\\sqrt2$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.7,4.6) {Two actual Hadamard paths, not classical alternatives};\n\\node[font=\\small,align=center,font=\\small] at (6.8,-0.18) {$\\text{amplitude}=1/2-1/2=0\\quad\\ne\\quad\\text{sum of path probabilities}=1/2$};\n\\end{tikzpicture}","ont=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.113043) {0.5};\n\\draw[black!13,line width=.25pt] (0.8,3.526087) -- (9.3,3.526087);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.526087) {1};\n\\draw[->,line width=.6pt] (0.8,0.7) -- (9.42,0.7);\n\\draw[->,line width=.6pt] (0.8,0.7) -- (0.8,4.07);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.15) {$u=|\\varepsilon t|$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.35) {leakage amplitude};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.7) (0.885,0.770645) (0.97,0.841245) (1.055,0.911758) (1.14,0.982138) (1.225,1.052342) (1.31,1.122325) (1.395,1.192045) (1.48,1.261457) (1.565,1.330518) (1.65,1.399185) (1.735,1.467415) (1.82,1.535166) (1.905,1.602394) (1.99,1.669059) (2.075,1.735118) (2.16,1.80053) (2.245,1.865254) (2.33,1.92925) (2.415,1.992478) (2.5,2.054898) (2.585,2.116472) (2.67,2.17716) (2.755,2.236924) (2.84,2.295729) (2.925,2.353536) (3.01,2.410309) (3.095,2.466014) (3.18,2.520615) (3.265,2.574078) (3.35,2.62637) (3.435,2.677458) (3.52,2.727311) (3.605,2…","087) (7.6,3.526087) (7.685,3.526087) (7.77,3.526087) (7.855,3.526087) (7.94,3.526087) (8.025,3.526087) (8.11,3.526087) (8.195,3.526087) (8.28,3.526087) (8.365,3.526087) (8.45,3.526087) (8.535,3.526087) (8.62,3.526087) (8.705,3.526087) (8.79,3.526087) (8.875,3.526087) (8.96,3.526087) (9.045,3.526087) (9.13,3.526087) (9.215,3.526087) (9.3,3.526087)};\n\\node[font=\\small,align=center] at (11.7,3.95) {$H=\\varepsilon X$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,2.95) {solid: $|\\sin u|$","8.025,3.526087) (8.11,3.526087) (8.195,3.526087) (8.28,3.526087) (8.365,3.526087) (8.45,3.526087) (8.535,3.526087) (8.62,3.526087) (8.705,3.526087) (8.79,3.526087) (8.875,3.526087) (8.96,3.526087) (9.045,3.526087) (9.13,3.526087) (9.215,3.526087) (9.3,3.526087)};\n\\node[font=\\small,align=center] at (11.7,3.95) {$H=\\varepsilon X$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,2.95) {solid: $|\\sin u|$","dashed: $\\min(1,u)$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.5) {Probability is $\\sin^2u$,","not $|\\sin u|$.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line 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{$0$};\n\\path[draw=black!30,fill=white] (7.6,1.45) rectangle (8.65,2.5);\n\\node[font=\\small,align=center,font=\\small] at (8.125,1.975) {$0$};\n\\path[draw=black!30,fill=black!8] (8.65,1.45) rectangle (9.7,2.5);\n\\node[font=\\small,align=center,font=\\small] at (9.175,1.975) {$1$};\n\\node[font=\\small,align=center,font=\\small] at (4.75,2.5) {row rank $1$","$};\n\\path[draw=black!30,fill=black!8] (7.6,2.5) rectangle (8.65,3.55);\n\\node[font=\\small,align=center,font=\\small] at (8.125,3.025) {$1$};\n\\path[draw=black!30,fill=white] (8.65,2.5) rectangle (9.7,3.55);\n\\node[font=\\small,align=center,font=\\small] at (9.175,3.025) {$0$};\n\\path[draw=black!30,fill=white] (7.6,1.45) rectangle (8.65,2.5);\n\\node[font=\\small,align=center,font=\\small] at (8.125,1.975) {$0$};\n\\path[draw=black!30,fill=black!8] (8.65,1.45) rectangle (9.7,2.5);\n\\node[font=\\small,align=center,font=\\small] at (9.175,1.975) {$1$};\n\\node[font=\\small,align=center,font=\\small] at (4.75,2.5) {row rank $1$","$};\n\\path[draw=black!30,fill=white] (8.65,2.5) rectangle (9.7,3.55);\n\\node[font=\\small,align=center,font=\\small] at (9.175,3.025) {$0$};\n\\path[draw=black!30,fill=white] (7.6,1.45) rectangle (8.65,2.5);\n\\node[font=\\small,align=center,font=\\small] at (8.125,1.975) {$0$};\n\\path[draw=black!30,fill=black!8] (8.65,1.45) rectangle (9.7,2.5);\n\\node[font=\\small,align=center,font=\\small] at (9.175,1.975) {$1$};\n\\node[font=\\small,align=center,font=\\small] at (4.75,2.5) {row rank $1$","[font=\\small,align=center,font=\\small] at (9.175,3.025) {$0$};\n\\path[draw=black!30,fill=white] (7.6,1.45) rectangle (8.65,2.5);\n\\node[font=\\small,align=center,font=\\small] at (8.125,1.975) {$0$};\n\\path[draw=black!30,fill=black!8] (8.65,1.45) rectangle (9.7,2.5);\n\\node[font=\\small,align=center,font=\\small] at (9.175,1.975) {$1$};\n\\node[font=\\small,align=center,font=\\small] at (4.75,2.5) {row rank $1$","orbit dimension $2$};\n\\node[font=\\small,align=center,font=\\small] at (11.4,2.5) {row rank $2$","orbit dimension $4$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.6) {$A$ acts on the left. It cannot invent a new direction in the row space.};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (1.76,4.15) {$V^*Z_1V$};\n\\path[draw=black!30,fill=black!8] (0.7,2.39) rectangle (1.76,3.45);\n\\node[font=\\small,align=center,font=\\small] at (1.23,2.92) {$1$};\n\\path[draw=black!30,fill=white] (1.76,2.39) rectangle (2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (1.76,4.15) {$V^*Z_1V$};\n\\path[draw=black!30,fill=black!8] (0.7,2.39) rectangle (1.76,3.45);\n\\node[font=\\small,align=center,font=\\small] at (1.23,2.92) {$1$};\n\\path[draw=black!30,fill=white] (1.76,2.39) rectangle (2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support","oin=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\node[font=\\small,align=center,font=\\small] at (1.76,4.15) {$V^*Z_1V$};\n\\path[draw=black!30,fill=black!8] (0.7,2.39) rectangle (1.76,3.45);\n\\node[font=\\small,align=center,font=\\small] at (1.23,2.92) {$1$};\n\\path[draw=black!30,fill=white] (1.76,2.39) rectangle (2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support","1V$};\n\\path[draw=black!30,fill=black!8] (0.7,2.39) rectangle (1.76,3.45);\n\\node[font=\\small,align=center,font=\\small] at (1.23,2.92) {$1$};\n\\path[draw=black!30,fill=white] (1.76,2.39) rectangle (2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support","1$};\n\\path[draw=black!30,fill=white] (1.76,2.39) rectangle (2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support","2.82,3.45);\n\\node[font=\\small,align=center,font=\\small] at (2.29,2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support",",2.92) {$0$};\n\\path[draw=black!30,fill=white] (0.7,1.33) rectangle (1.76,2.39);\n\\node[font=\\small,align=center,font=\\small] at (1.23,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (1.76,1.33) rectangle (2.82,2.39);\n\\node[font=\\small,align=center,font=\\small] at (2.29,1.86) {$-1$};\n\\node[font=\\small,align=center] at (7,3.65) {before: $(3,4)/5$};\n\\node[font=\\small,align=center] at (7,2.45) {after: $(3,-4)/5$};\n\\node[font=\\small,align=center,font=\\small] at (7,1.2) {same code support","different relative phase};\n\\node[font=\\small,align=center,font=\\small] at (11.86,4.15) {Required scalar form $cI$};\n\\path[draw=black!30,fill=black!8] (10.8,2.39) rectangle (11.86,3.45);\n\\node[font=\\small,align=center,font=\\small] at (11.33,2.92) {$c$};\n\\path[draw=black!30,fill=white] (11.86,2.39) rectangle (12.92,3.45);\n\\node[font=\\small,align=center,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small","different relative phase};\n\\node[font=\\small,align=center,font=\\small] at (11.86,4.15) {Required scalar form $cI$};\n\\path[draw=black!30,fill=black!8] (10.8,2.39) rectangle (11.86,3.45);\n\\node[font=\\small,align=center,font=\\small] at (11.33,2.92) {$c$};\n\\path[draw=black!30,fill=white] (11.86,2.39) rectangle (12.92,3.45);\n\\node[font=\\small,align=center,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small,align=center,font=\\small] at (12.39,1.86) {$c$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {A trace of zero does not make","different relative phase};\n\\node[font=\\small,align=center,font=\\small] at (11.86,4.15) {Required scalar form $cI$};\n\\path[draw=black!30,fill=black!8] (10.8,2.39) rectangle (11.86,3.45);\n\\node[font=\\small,align=center,font=\\small] at (11.33,2.92) {$c$};\n\\path[draw=black!30,fill=white] (11.86,2.39) rectangle (12.92,3.45);\n\\node[font=\\small,align=center,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small,align=center,font=\\small] at (12.39,1.86) {$c$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {A trace of zero does not make $\\diag(1,-1)$ the zero operator.};\n\\end{tikzpicture}","th[draw=black!30,fill=black!8] (10.8,2.39) rectangle (11.86,3.45);\n\\node[font=\\small,align=center,font=\\small] at (11.33,2.92) {$c$};\n\\path[draw=black!30,fill=white] (11.86,2.39) rectangle (12.92,3.45);\n\\node[font=\\small,align=center,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small,align=center,font=\\small] at (12.39,1.86) {$c$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {A trace of zero does not make $\\diag(1,-1)$ the zero operator.};\n\\end{tikzpicture}","h[draw=black!30,fill=white] (11.86,2.39) rectangle (12.92,3.45);\n\\node[font=\\small,align=center,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small,align=center,font=\\small] at (12.39,1.86) {$c$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {A trace of zero does not make $\\diag(1,-1)$ the zero operator.};\n\\end{tikzpicture}","er,font=\\small] at (12.39,2.92) {$0$};\n\\path[draw=black!30,fill=white] (10.8,1.33) rectangle (11.86,2.39);\n\\node[font=\\small,align=center,font=\\small] at (11.33,1.86) {$0$};\n\\path[draw=black!30,fill=black!8] (11.86,1.33) rectangle (12.92,2.39);\n\\node[font=\\small,align=center,font=\\small] at (12.39,1.86) {$c$};\n\\node[font=\\small,align=center,font=\\small] at (7,0.15) {A trace of zero does not make $\\diag(1,-1)$ the zero operator.};\n\\end{tikzpicture}","all,align=center,anchor=east,font=\\scriptsize] at (0.64,2.159091) {0.5};\n\\draw[black!13,line width=.25pt] (0.8,3.568182) -- (9.3,3.568182);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.568182) {1};\n\\draw[->,line width=.6pt] (0.8,0.75) -- (9.42,0.75);\n\\draw[->,line width=.6pt] (0.8,0.75) -- (0.8,3.97);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.05,0.2) {time $t$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.25) {probability};\n\\draw[line width=1.1pt] plot coordinates {(0.8,3.568182) (0.885,3.565401) (0.97,3.557071) (1.055,3.543223) (1.14,3.523913) (1.225,3.499216) (1.31,3.469231) (1.395,3.434074) (1.48,3.393887) (1.565,3.348826) (1.65,3.299069) (1.735,3.244814) (1.82,3.186274) (1.905,3.12368) (1.99,3.057279) (2.075,2.987334) (2.16,2.91412) (2.245,2.837926) (2.33,2.759053) (2.415,2.677812) (2.5,2.594524) (2.585,2.509518) (2.67,2.423128) (2.755,2.335697) (2.84,2.247568) (2.925,2.159091) (3.01,2.070613) (3.095,1.982485) (3.18,1.895054) (3.265,1.808664) (3.35,1.723658) (3.435,1.64037) (3.52,1.559129) (3.605,1.480256) (3.69,1…","dashed: hidden};\n\\node[font=\\small,align=center] at (11.7,2.4) {$K(t)=1$};\n\\node[font=\\small,align=center,font=\\small] at (11.7,1.2) {The two probabilities","sum to one.};\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.25,0.47) {$\\pi$};\n\\end{tikzpicture}","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[black!25] (0.75,2.3) -- (3.55,2.3);\n\\draw[black!25] (2,0.95) -- (2,3.7);\n\\draw[->,line width=1.1pt] (2,2.3) -- (3.45,2.3);\n\\draw[->,dashed] (2,2.3) -- (3.25,2.3);\n\\draw[->,dashed] (2,2.3) -- (2,3.35);\n\\node[font=\\small,align=center] at (2,4.2) {$\\theta=0$};\n\\node[font=\\small,align=center,font=\\small] at (2,0.25) {$(1,0)$};\n\\draw[black!25] (5.4,2.3) -- (8.2,2.3);\n\\draw[black!25] (6.65,0.95) -- (6.65,3.7);\n\\draw[->,line width=1.1pt] (6.65,2.3) -- (8.1,2.3);\n\\draw[->,dashed] (6.65,2.3) -- (7.732532,2.925);\n\\draw[->,dashed] (6.65,2.3) -- (6.125,3.209327);\n\\node[font=\\small,align=center] at (6.65,4.2) {$\\theta=\\pi/6$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[bl","round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\draw[black!25] (0.75,2.3) -- (3.55,2.3);\n\\draw[black!25] (2,0.95) -- (2,3.7);\n\\draw[->,line width=1.1pt] (2,2.3) -- (3.45,2.3);\n\\draw[->,dashed] (2,2.3) -- (3.25,2.3);\n\\draw[->,dashed] (2,2.3) -- (2,3.35);\n\\node[font=\\small,align=center] at (2,4.2) {$\\theta=0$};\n\\node[font=\\small,align=center,font=\\small] at (2,0.25) {$(1,0)$};\n\\draw[black!25] (5.4,2.3) -- (8.2,2.3);\n\\draw[black!25] (6.65,0.95) -- (6.65,3.7);\n\\draw[->,line width=1.1pt] (6.65,2.3) -- (8.1,2.3);\n\\draw[->,dashed] (6.65,2.3) -- (7.732532,2.925);\n\\draw[->,dashed] (6.65,2.3) -- (6.125,3.209327);\n\\node[font=\\small,align=center] at (6.65,4.2) {$\\theta=\\pi/6$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[black!25] (11.3,0.95) -- (11.3,3.7);\n\\draw[->,line width=1.1pt] (11.3","[font=\\small,align=center] at (2,4.2) {$\\theta=0$};\n\\node[font=\\small,align=center,font=\\small] at (2,0.25) {$(1,0)$};\n\\draw[black!25] (5.4,2.3) -- (8.2,2.3);\n\\draw[black!25] (6.65,0.95) -- (6.65,3.7);\n\\draw[->,line width=1.1pt] (6.65,2.3) -- (8.1,2.3);\n\\draw[->,dashed] (6.65,2.3) -- (7.732532,2.925);\n\\draw[->,dashed] (6.65,2.3) -- (6.125,3.209327);\n\\node[font=\\small,align=center] at (6.65,4.2) {$\\theta=\\pi/6$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[black!25] (11.3,0.95) -- (11.3,3.7);\n\\draw[->,line width=1.1pt] (11.3,2.3) -- (12.75,2.3);\n\\draw[->,dashed] (11.3,2.3) -- (11.925,3.382532);\n\\draw[->,dashed] (11.3,2.3) -- (10.390673,2.825);\n\\node[font=\\small,align=center] at (11.3,4.2) {$\\theta=\\pi/3$};\n\\node[font=\\small,align=center,font=\\small] at (11.3,0.25) {$(1/2,-\\sqrt3/2)$};\n\\node[font=\\small,align=center,","enter,font=\\small] at (2,0.25) {$(1,0)$};\n\\draw[black!25] (5.4,2.3) -- (8.2,2.3);\n\\draw[black!25] (6.65,0.95) -- (6.65,3.7);\n\\draw[->,line width=1.1pt] (6.65,2.3) -- (8.1,2.3);\n\\draw[->,dashed] (6.65,2.3) -- (7.732532,2.925);\n\\draw[->,dashed] (6.65,2.3) -- (6.125,3.209327);\n\\node[font=\\small,align=center] at (6.65,4.2) {$\\theta=\\pi/6$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[black!25] (11.3,0.95) -- (11.3,3.7);\n\\draw[->,line width=1.1pt] (11.3,2.3) -- (12.75,2.3);\n\\draw[->,dashed] (11.3,2.3) -- (11.925,3.382532);\n\\draw[->,dashed] (11.3,2.3) -- (10.390673,2.825);\n\\node[font=\\small,align=center] at (11.3,4.2) {$\\theta=\\pi/3$};\n\\node[font=\\small,align=center,font=\\small] at (11.3,0.25) {$(1/2,-\\sqrt3/2)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.75) {Solid vector: fixed in the laboratory. Dashed axe","t (6.65,4.2) {$\\theta=\\pi/6$};\n\\node[font=\\small,align=center,font=\\small] at (6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[black!25] (11.3,0.95) -- (11.3,3.7);\n\\draw[->,line width=1.1pt] (11.3,2.3) -- (12.75,2.3);\n\\draw[->,dashed] (11.3,2.3) -- (11.925,3.382532);\n\\draw[->,dashed] (11.3,2.3) -- (10.390673,2.825);\n\\node[font=\\small,align=center] at (11.3,4.2) {$\\theta=\\pi/3$};\n\\node[font=\\small,align=center,font=\\small] at (11.3,0.25) {$(1/2,-\\sqrt3/2)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.75) {Solid vector: fixed in the laboratory. Dashed axes: rotating frame.};\n\\end{tikzpicture}","(6.65,0.25) {$(\\sqrt3/2,-1/2)$};\n\\draw[black!25] (10.05,2.3) -- (12.85,2.3);\n\\draw[black!25] (11.3,0.95) -- (11.3,3.7);\n\\draw[->,line width=1.1pt] (11.3,2.3) -- (12.75,2.3);\n\\draw[->,dashed] (11.3,2.3) -- (11.925,3.382532);\n\\draw[->,dashed] (11.3,2.3) -- (10.390673,2.825);\n\\node[font=\\small,align=center] at (11.3,4.2) {$\\theta=\\pi/3$};\n\\node[font=\\small,align=center,font=\\small] at (11.3,0.25) {$(1/2,-\\sqrt3/2)$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.75) {Solid vector: fixed in the laboratory. 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{$0$};\n\\node[font=\\small,align=center,font=\\small] at (6.925,4.4) {$C_2$};\n\\path[draw=black!30,fill=white] (5.5,2.75) rectangle (6.45,3.7);\n\\node[font=\\small,align=center,font=\\small] at (5.975,3.225) {$0$};\n\\path[draw=black!30,fill=white] (6.45,2.75) rectangle (7.4,3.7);\n\\node[font=\\small,align=center,font=\\small] at (6.925,3.225) {$0$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center",");\n\\node[font=\\small,align=center,font=\\small] at (2.025,2.275) {$1$};\n\\path[draw=black!30,fill=white] (2.5,1.8) rectangle (3.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (2.975,2.275) {$0$};\n\\node[font=\\small,align=center,font=\\small] at (6.925,4.4) {$C_2$};\n\\path[draw=black!30,fill=white] (5.5,2.75) rectangle (6.45,3.7);\n\\node[font=\\small,align=center,font=\\small] at (5.975,3.225) {$0$};\n\\path[draw=black!30,fill=white] (6.45,2.75) rectangle (7.4,3.7);\n\\node[font=\\small,align=center,font=\\small] at (6.925,3.225) {$0$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center",");\n\\node[font=\\small,align=center,font=\\small] at (2.975,2.275) {$0$};\n\\node[font=\\small,align=center,font=\\small] at (6.925,4.4) {$C_2$};\n\\path[draw=black!30,fill=white] (5.5,2.75) rectangle (6.45,3.7);\n\\node[font=\\small,align=center,font=\\small] at (5.975,3.225) {$0$};\n\\path[draw=black!30,fill=white] (6.45,2.75) rectangle (7.4,3.7);\n\\node[font=\\small,align=center,font=\\small] at (6.925,3.225) {$0$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center","$};\n\\path[draw=black!30,fill=white] (5.5,2.75) rectangle (6.45,3.7);\n\\node[font=\\small,align=center,font=\\small] at (5.975,3.225) {$0$};\n\\path[draw=black!30,fill=white] (6.45,2.75) rectangle (7.4,3.7);\n\\node[font=\\small,align=center,font=\\small] at (6.925,3.225) {$0$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (","$};\n\\path[draw=black!30,fill=white] (6.45,2.75) rectangle (7.4,3.7);\n\\node[font=\\small,align=center,font=\\small] at (6.925,3.225) {$0$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small","$};\n\\path[draw=black!30,fill=black!8] (7.4,2.75) rectangle (8.35,3.7);\n\\node[font=\\small,align=center,font=\\small] at (7.875,3.225) {$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\s","$1$};\n\\path[draw=black!30,fill=white] (5.5,1.8) rectangle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,fo","ngle (6.45,2.75);\n\\node[font=\\small,align=center,font=\\small] at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\sm","at (5.975,2.275) {$0$};\n\\path[draw=black!30,fill=white] (6.45,1.8) rectangle (7.4,2.75);\n\\node[font=\\small,align=center,font=\\small] at (6.925,2.275) {$0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\","0$};\n\\path[draw=black!30,fill=white] (7.4,1.8) rectangle (8.35,2.75);\n\\node[font=\\small,align=center,font=\\small] at (7.875,2.275) {$0$};\n\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\CC^3$. Their local orbits span $\\CC^2\\otimes\\CC^3$.};\n\\end{tikzpicture}","\\node[font=\\small,align=center] at (11.5,3.4) {$d=2,\\ m=3$};\n\\node[font=\\small,align=center] at (11.5,2.35) {$\\lceil m/d\\rceil=2$};\n\\path[draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\CC^3$. Their local orbits span $\\CC^2\\otimes\\CC^3$.};\n\\end{tikzpicture}","draw,fill=black!10] (1,0.45) rectangle (4.5,1.2);\n\\node[font=\\small,align=center,font=\\small] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\CC^3$. Their local orbits span $\\CC^2\\otimes\\CC^3$.};\n\\end{tikzpicture}","all] at (2.75,0.83) {multiplicity $1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\CC^3$. Their local orbits span $\\CC^2\\otimes\\CC^3$.};\n\\end{tikzpicture}","$1$};\n\\path[draw,fill=black!10] (5.3,0.45) rectangle (8.8,1.2);\n\\node[font=\\small,align=center,font=\\small] at (7.05,0.83) {multiplicity $2$};\n\\path[draw,fill=black!28] (9.6,0.45) rectangle (13.1,1.2);\n\\node[font=\\small,align=center,font=\\small] at (11.35,0.83) {multiplicity $3$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,1.5) {The stacked rows span $\\CC^3$. Their local orbits span $\\CC^2\\otimes\\CC^3$.};\n\\end{tikzpicture}","125);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,2.1125) {1};\n\\draw[black!13,line width=.25pt] (0.8,3.425) -- (9.5,3.425);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.425) {2};\n\\draw[->,line width=.6pt] (0.8,0.8) -- (9.62,0.8);\n\\draw[->,line width=.6pt] (0.8,0.8) -- (0.8,4.07);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.15,0.25) {$\\log_{10}\\tau$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.35) {$d_s(\\tau)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.805245) (0.843719,0.805499) (0.887437,0.805766) (0.931156,0.806046) (0.974874,0.806339) (1.018593,0.806647) (1.062312,0.806969) (1.10603,0.807308) (1.149749,0.807662) (1.193467,0.808034) (1.237186,0.808423) (1.280905,0.808832) (1.324623,0.80926) (1.368342,0.809709) (1.41206,0.81018) (1.455779,0.810674) (1.499497,0.811191) (1.543216,0.811734) (1.586935,0.812302) (1.630653,0.812899) (1.674372,0.813524) (1.71809,0.814179) (1.761809,0.814866) (1.805528,0.815586) (1.849246,0.816341) (1.892965,0.817132) (1.936683,0.817962) (1.980402,0.818831) (…","[black!13,line width=.25pt] (0.8,3.425) -- (9.5,3.425);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (0.64,3.425) {2};\n\\draw[->,line width=.6pt] (0.8,0.8) -- (9.62,0.8);\n\\draw[->,line width=.6pt] (0.8,0.8) -- (0.8,4.07);\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.15,0.25) {$\\log_{10}\\tau$};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (0.8,4.35) {$d_s(\\tau)$};\n\\draw[line width=1.1pt] plot coordinates {(0.8,0.805245) (0.843719,0.805499) (0.887437,0.805766) (0.931156,0.806046) (0.974874,0.806339) (1.018593,0.806647) (1.062312,0.806969) (1.10603,0.807308) (1.149749,0.807662) (1.193467,0.808034) (1.237186,0.808423) (1.280905,0.808832) (1.324623,0.80926) (1.368342,0.809709) (1.41206,0.81018) (1.455779,0.810674) (1.499497,0.811191) (1.543216,0.811734) (1.586935,0.812302) (1.630653,0.812899) (1.674372,0.813524) (1.71809,0.814179) (1.761809,0.814866) (1.805528,0.815586) (1.849246,0.816341) (1.892965,0.817132) (1.936683,0.817962) (1.980402,0.818831) (2.024121,0.819742) (2.067839,0.820698) (2.111558,0.821699) (2.155276,0.822749) (2.198995,0.823…","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\path[draw,fill=black!4] (0.6,0.65) rectangle (3.35,3.9);\n\\node[font=\\small,align=center] at (1.975,3.45) {no error};\n\\node[font=\\small,align=center] at (1.975,2.6) {$\\ket{000}$};\n\\node[font=\\small,align=center] at (1.975,1.8) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65)","\\begin{tikzpicture}[x=1cm,y=1cm,>=Stealth,line cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\path[draw,fill=black!4] (0.6,0.65) rectangle (3.35,3.9);\n\\node[font=\\small,align=center] at (1.975,3.45) {no error};\n\\node[font=\\small,align=center] at (1.975,2.6) {$\\ket{000}$};\n\\node[font=\\small,align=center] at (1.975,1.8) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.","ne cap=round,line join=round]\n\\path[use as bounding box] (0,-.6) rectangle (14,4.8);\n\\path[draw,fill=black!4] (0.6,0.65) rectangle (3.35,3.9);\n\\node[font=\\small,align=center] at (1.975,3.45) {no error};\n\\node[font=\\small,align=center] at (1.975,2.6) {$\\ket{000}$};\n\\node[font=\\small,align=center] at (1.975,1.8) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) 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{$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at","er] at (1.975,3.45) {no error};\n\\node[font=\\small,align=center] at (1.975,2.6) {$\\ket{000}$};\n\\node[font=\\small,align=center] at (1.975,1.8) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) {$X_2$};\n\\node[font=\\small,align=center] at (8.875,2.6) {$\\ket{010}$};\n\\node[font=\\small,align=center] at (8.875,1.8) {$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.","] at (1.975,2.6) {$\\ket{000}$};\n\\node[font=\\small,align=center] at (1.975,1.8) {$\\ket{111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) {$X_2$};\n\\node[font=\\small,align=center] at (8.875,2.6) {$\\ket{010}$};\n\\node[font=\\small,align=center] at (8.875,1.8) {$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (","111}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (1.975,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (4.05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) {$X_2$};\n\\node[font=\\small,align=center] at (8.875,2.6) {$\\ket{010}$};\n\\node[font=\\small,align=center] at (8.875,1.8) {$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (12.325,3.45) {$X_3$};\n\\node[font=\\small,align=center] at (12.325,2.6) {$\\ket{001","05,0.65) rectangle (6.8,3.9);\n\\node[font=\\small,align=center] at (5.425,3.45) {$X_1$};\n\\node[font=\\small,align=center] at (5.425,2.6) {$\\ket{100}$};\n\\node[font=\\small,align=center] at (5.425,1.8) {$\\ket{011}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) {$X_2$};\n\\node[font=\\small,align=center] at (8.875,2.6) {$\\ket{010}$};\n\\node[font=\\small,align=center] at (8.875,1.8) {$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (12.325,3.45) {$X_3$};\n\\node[font=\\small,align=center] at (12.325,2.6) {$\\ket{001}$};\n\\node[font=\\small,align=center] at (12.325,1.8) {$\\ket{110}$};\n\\node[font=\\small,align=center,font=\\scriptsize]","nter] at (5.425,3.45) 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$1/10,","11}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (5.425,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (7.5,0.65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at (8.875,3.45) {$X_2$};\n\\node[font=\\small,align=center] at (8.875,2.6) {$\\ket{010}$};\n\\node[font=\\small,align=center] at (8.875,1.8) {$\\ket{101}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (12.325,3.45) {$X_3$};\n\\node[font=\\small,align=center] at (12.325,2.6) {$\\ket{001}$};\n\\node[font=\\small,align=center] at (12.325,1.8) {$\\ket{110}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.325,1.03) {ratio $3:4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.55) {Channel probabilities: $1/10,\\ 2/10,\\ 3/10,\\ 4/10$.};\n\\end{tikzpicture}",".65) rectangle (10.25,3.9);\n\\node[font=\\small,align=center] at 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$3:4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.55) {Channel probabilities: $1/10,\\ 2/10,\\ 3/10,\\ 4/10$.};\n\\end{tikzpicture}",";\n\\node[font=\\small,align=center,font=\\scriptsize] at (8.875,1.03) {ratio $3:4$};\n\\path[draw,fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (12.325,3.45) {$X_3$};\n\\node[font=\\small,align=center] at (12.325,2.6) {$\\ket{001}$};\n\\node[font=\\small,align=center] at (12.325,1.8) {$\\ket{110}$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (12.325,1.03) {ratio $3:4$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (7,4.55) {Channel probabilities: $1/10,\\ 2/10,\\ 3/10,\\ 4/10$.};\n\\end{tikzpicture}","fill=black!4] (10.95,0.65) rectangle (13.7,3.9);\n\\node[font=\\small,align=center] at (12.325,3.45) {$X_3$};\n\\node[font=\\small,align=center] at (12.325,2.6) {$\\ket{001}$};\n\\node[font=\\small,align=center] at (12.325,1.8) 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phase.};\n\\draw[black!30] (4.05,0.2) -- (4.05,4.35);\n\\draw[black!30] (9,0.2) -- (9,4.35);\n\\draw[black!13,line width=.25pt] (9.6,0.75) -- (9.6,3.4);\n\\draw[] (9.6,0.69) -- (9.6,0.81);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.6,0.49) {0};\n\\draw[black!13,line","$};\n\\path[draw=black!30,fill=white] (5.8,0.7) rectangle (6.55,1.45);\n\\node[font=\\small,align=center,font=\\small] at (6.175,1.075) {$0$};\n\\path[draw=black!30,fill=white] (6.55,0.7) rectangle (7.3,1.45);\n\\node[font=\\small,align=center,font=\\small] at (6.925,1.075) {$0$};\n\\path[draw=black!30,fill=black!8] (7.3,0.7) rectangle (8.05,1.45);\n\\node[font=\\small,align=center,font=\\small] at (7.675,1.075) {$1$};\n\\node[font=\\small,align=center,font=\\scriptsize] at (6.55,0.2) {Off-diagonal entries retain phase.};\n\\draw[black!30] (4.05,0.2) -- (4.05,4.35);\n\\draw[black!30] (9,0.2) -- (9,4.35);\n\\draw[black!13,line width=.25pt] (9.6,0.75) -- (9.6,3.4);\n\\draw[] (9.6,0.69) -- (9.6,0.81);\n\\node[font=\\small,align=center,font=\\scriptsize] at (9.6,0.49) {0};\n\\draw[black!13,line width=.25pt] (9.6,0.75) -- (13.4,0.75);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (9.44,0.75) {0};\n\\draw[black!13","75) {0};\n\\draw[black!13,line width=.25pt] (9.6,3.4) -- (13.4,3.4);\n\\node[font=\\small,align=center,anchor=east,font=\\scriptsize] at (9.44,3.4) {1};\n\\draw[->,line width=.6pt] (9.6,0.75) -- (13.52,0.75);\n\\draw[->,line width=.6pt] (9.6,0.75) -- (9.6,3.52);\n\\node[font=\\small,align=center,font=\\scriptsize] at (11.5,0.2) {time};\n\\node[font=\\small,align=center,anchor=west,font=\\scriptsize] at (9.6,3.8) {$|x(t)|^2$};\n\\draw[line width=1.1pt] plot coordinates {(9.6,3.4) (9.638,3.397385) (9.676,3.389552) (9.714,3.376531) (9.752,3.358373) (9.79,3.33515) (9.828,3.306954) (9.866,3.273896) (9.904,3.236106) (9.942,3.193735) (9.98,3.146948) (10.018,3.09593) (10.056,3.040883) (10.094,2.982025) (10.132,2.919587) (10.17,2.853815) (10.208,2.784971) (10.246,2.713324) (10.284,2.639158) (10.322,2.562765) (10.36,2.484448) (10.398,2.404514) (10.436,2.32328) (10.474,2.241067) (10.512,2.158197) (10.55,2.075) (10.588,1.99180","\\node[draw,fill=black!8,align=center,minimum width=4cm,minimum height=1.5cm] at (3.2,1.8) {even Choi addresses $00,11$","$\\begin{pmatrix}1&4/5\\\\4/5&16/25\\end{pmatrix}$};\n\\node[draw,fill=black!20,align=center,minimum width=4cm,minimum height=1.5cm] at (8.5,1.8) {odd Choi addresses $01,10$","4/5&16/25\\end{pmatrix}$};\n\\node[draw,fill=black!20,align=center,minimum width=4cm,minimum height=1.5cm] at (8.5,1.8) {odd Choi addresses $01,10$","$\\operatorname{diag}(9/25,0)$};\n\\node[font=\\small] at (6,0.5) {$\\Phi(\\ket0\\bra1)=(4/5)\\ket0\\bra1$};\n\\end{tikzpicture}","\\draw[fill=black!8] (1,0.6) rectangle (5,2.8);\\node[align=center] at (3,1.7) {coarse $\\CC^3$","$0\\oplus\\sigma_a$};\n\\draw[fill=black!8] (7,0.6) rectangle (10.8,2.8);\\draw[fill=black!25] (10.8,0.6) rectangle (12,2.8);\n\\draw[->] (5.2,1.7)--(6.8,1.7);\\node[above] at (6,1.7) {$R$};\n\\node[align=center] at (8.9,1.7) {embedded image","same constraints};\\node[rotate=90,font=\\footnotesize] at (11.4,1.7) {new coordinate};\n\\node[font=\\small] at (6,0.15) {closure on $\\operatorname{im}R$ transfers, and the new complement needs its own test};\n\\end{tikzpicture}","\\draw[->] (1,1.6)--(11,1.6);\\foreach \\x/\\t in {3/{9/10},6/{1},9/{11/10}} {\\fill (\\x,1.6) circle (0.08);\\node[below] at (\\x,1.5) {$\\t$};}\n\\draw[<->] (3,2.5)--(6,2.5);\\draw[<->] (6,2.5)--(9,2.5);\n\\node[above] at (4.5,2.5) {$1/10$};\\node[above] at (7.5,2.5) {$1/10$};\n\\node[font=\\small] at (6,0.45) {one pair: $1/10$, full family $\\{I,A\\}$: $1/5$};\n\\end{tikzpicture}","$(1,0,0)$};\n\\draw[->] (3.1,2)--(4.7,2);\\node[draw,fill=black!20,align=center] at (6,2) {one syndrome","$Q=P$, rank $2$};\n\\node[draw,align=center] at (10,2) {unused complement","$I-P$, rank $1$};\n\\draw[->] (10,1.5)--(6,0.6);\\node[draw,fill=black!8] at (6,0.35) {$\\ket0\\bra2$: replacement branch};\n\\end{tikzpicture}","\\node[draw,fill=black!8,align=center] (a) at (2.2,2.4) {fixed finite $N$","$t\\downarrow0$};\n\\node[draw,fill=black!8] (b) at (9.6,2.4) {$d_N(t)\\longrightarrow0$};\\draw[->] (a)--(b);\n\\node[draw,fill=black!20,align=center] (c) at (2.2,0.7) {fixed $t>0$","$N\\longrightarrow\\infty$};\n\\node[draw,fill=black!20,align=center] (d) at (9.6,0.7) {then $t\\downarrow0$","$d_{\\rm torus}(t)\\longrightarrow1$};\\draw[->] (c)--(d);\n\\end{tikzpicture}","\\node[draw,fill=black!7,align=center] at (3,2) {hidden energy $1$","weight $9/25$};\n\\node[draw,fill=black!20,align=center] at (9,2) {hidden energy $2$","weight $16/25$};\n\\draw[->] (3,1.4)--(5,0.6);\\draw[->] (9,1.4)--(7,0.6);\n\\node[draw,align=center] at (6,0.4) {$K(t)=(9/25)e^{-it}+(16/25)e^{-2it}$};\n\\end{tikzpicture}","\\draw[->] (1,0.5)--(4,0.5);\\draw[->] (1,0.5)--(1,3);\n\\draw[thick,->] (1,0.5)--(2.8,2.9);\\node[font=\\small] at (3.8,2.6) {$(3/5,4/5)$};\n\\node[draw,fill=black!6,align=center] at (8.4,2.3) {$S=(3/5)R$","$S^*S=(9/25)I$};\n\\draw[->] (8.4,1.7)--(8.4,1.1);\\node[draw,fill=black!18] at (8.4,0.7) {$S(S^*S)^{-1/2}=R$};\n\\end{tikzpicture}","\\draw[fill=black!6] (0.5,0.6) rectangle (2.1,2.2);\n\\draw[fill=black!17] (2.4,0.6) rectangle (4,2.2);\n\\draw[fill=black!30] (4.3,0.6) rectangle (11.5,2.2);\n\\node at (1.3,1.4) {$1$};\\node at (3.2,1.4) {$1$};\\node at (7.9,1.4) {$2\\cdot2=4$};\n\\node[font=\\small] at (1.3,2.55) {trivial};\\node[font=\\small] at (3.2,2.55) {sign};\\node[font=\\small] at (7.9,2.55) {standard isotypic};\n\\node[font=\\small] at (6,0.1) {$\\CC[S_3]\\cong\\CC\\oplus\\CC\\oplus(\\CC^2\\otimes\\CC^2)$};\n\\end{tikzpicture}","ill=black!6] (0.5,0.6) rectangle (2.1,2.2);\n\\draw[fill=black!17] (2.4,0.6) rectangle (4,2.2);\n\\draw[fill=black!30] (4.3,0.6) rectangle (11.5,2.2);\n\\node at (1.3,1.4) {$1$};\\node at (3.2,1.4) {$1$};\\node at (7.9,1.4) {$2\\cdot2=4$};\n\\node[font=\\small] at (1.3,2.55) {trivial};\\node[font=\\small] at (3.2,2.55) {sign};\\node[font=\\small] at (7.9,2.55) {standard isotypic};\n\\node[font=\\small] at (6,0.1) {$\\CC[S_3]\\cong\\CC\\oplus\\CC\\oplus(\\CC^2\\otimes\\CC^2)$};\n\\end{tikzpicture}","\\node[draw,fill=black!8,align=center] at (2.6,2) {$B=1$, $D=2$","$\\Sigma(3)=1$};\n\\node[draw,fill=black!20,align=center] at (8.9,2) {$\\widetilde B=9/8$, $\\widetilde D=17/8$","$\\widetilde\\Sigma(3)=81/56$};\n\\draw[<->] (4.2,2)--(6.9,2);\\node[below] at (5.6,1.6) {$25/56\\leq21/32$};\n\\node[font=\\small] at (6,0.5) {common hypotheses: $g=9/8$, $\\delta=3/4$};\n\\end{tikzpicture}","\\node[draw,fill=black!7,align=center] (a) at (2,2.2) {$\\CC^n$","$H_n=\\operatorname{diag}(0,1,1/2,\\ldots)$};\n\\node[draw,fill=black!7,align=center] (b) at (9.5,2.2) {$\\CC^{n+1}$","old energies retained};\\draw[->] (a)--node[above] {$R_n$}(b);\n\\node[draw,fill=black!20,align=center] at (6,0.6) {fixed $\\CC^2$: $H_n=\\operatorname{diag}(0,1+2^{-n})$","tail $\\|H_n-H\\|=2^{-n}$};\n\\end{tikzpicture}","\\node[draw,fill=black!8,align=center] at (2.8,1.9) {$A_n=P_n/n$","$n$ equal eigenvalues $1/n$};\n\\node[draw,fill=black!20,align=center] at (8.8,1.9) {$\\|A_n\\|=1/n\\to0$","$\\|A_n\\|_1=\\operatorname{Tr}A_n=1$};\n\\draw[->] (4.7,1.9)--(6.7,1.9);\\node[font=\\small] at (6,0.45) {positive trace-class control: norm convergence alone does not preserve a trace};\n\\end{tikzpicture}","\\foreach \\x/\\h/\\t in {2/2.4/{1},6/1.2/{1/2},10/0.6/{1/4}} {\\draw[fill=black!15] (\\x-0.8,0.5) rectangle (\\x+0.8,\\h+0.5);\\node[above] at (\\x,\\h+0.5) {$\\t$};}\n\\node[font=\\small] at (6,0.1) {$T^n:\\quad1,\\ 2^{-n},\\ 4^{-n}\\qquad\\|T^n-P_0\\|=2^{-n}$};\n\\end{tikzpicture}"]}