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No algebraic realization of these orthogonal projectors is presumed.","Cycle frame & $S=CC^*$. Positivity spans a target packet when every supplied algebraic vector lies in it.","Unconditional rank test & $\\rank_{\\Q}[C\\ H_{\\Q}]=\\rank_{\\Q}C$ tests target containment in the actual cycle span.","Polyphase assembly & $S_{M,r}=\\sum_{\\ell=0}^{L-1}S_{LM,r+\\ell M}$. These are frame sums, not sums of span projectors.","Support test & $j^*\\alpha=0$. Purity turns it into cycle algebraicity when the support has codimension at least $p$.","Divisor transfer & $T=\\sum_rB_rA_r$ with actual Chow correspondences, packet preservation and invertibility.","Positive transfer & $T=\\sum_rA_r^\\dagger A_r$ only with algebraic realizations of those adjoints.","Balanced Fermat tuple & $\\sum_i\\langle ua_i\\rangle_m=3m$ for every unit $u$.","Conductor & $c(a)=m/\\gcd(m,a_0,\\ldots,a_5)$.","Matrix capacity & $C(a)=\\sup\\{\\sum w_v:\\sum w_vv=\\mathbf1,\\ w_v\\ge0\\}$.","Global dual & $y\\cdot v\\ge1$ for every invariant column implies $C(a)\\le y\\cdot\\mathbf1$.","Five-phase surface & $e_1=e_2=P-\\kappa z^5=0$, with $\\kappa^q=-1/5$.","Weil projection & $P_W(c)=-(\\langle c,u\\rangle u+\\langle c,v\\rangle v)/32$.","\\chapter{Correspondence degrees}\nLet $X$ have dimension $n$ and $Y$ dimension $m$. A correspondence $\\Gamma\\in\\CH^r(X\\times Y)$ acts by\n\\[\n\\Gamma_*:H^i(X)\\longrightarrow H^{i+2(r-n)}(Y)\n\\]\nand shifts Hodge type by $(r-n,r-n)$. To lower $H^{2p}(X)$ to $H^2(Y)$, take $r=n-p+1$. To return degree two to degree $2p$ on $X$, a correspondence on $Y\\times X$ has codimension $m+p-1$.","For a finite generically finite morphism of degree $d$, the push-pull identity on rational cohomology is $f_*f^*=d\\id$. Division by $d$ transfers algebraicity downward. Pullback transfers it upward without division. Coordinate-power maps preserve the old conductor of an inflated character, even though the ambient Fermat degree grows.","\\chapter{The rational Hodge target}\n\\section{Statement and status}\nLet $X$ be a smooth projective complex variety. For each codimension $p$, define\n\\[\n\\Hdg^p(X)=H^{2p}(X,\\Q)\\cap H^{p,p}(X)\n\\]\nand\n\\[\n\\Alg^p(X)=\\im\\bigl(\\CH^p(X)_{\\Q}\\xrightarrow{\\cl}H^{2p}(X,\\Q)\\bigr).\n\\]\nEvery algebraic cycle class has type $(p,p)$, so $\\Alg^p(X)\\subseteq\\Hdg^p(X)$. The rational Hodge conjecture asserts equality for every smooth projective $X$ and every $p$ \\cite{ClayHodge,VoisinHodge}.","target}\n\\section{Statement and status}\nLet $X$ be a smooth projective complex variety. For each codimension $p$, define\n\\[\n\\Hdg^p(X)=H^{2p}(X,\\Q)\\cap H^{p,p}(X)\n\\]\nand\n\\[\n\\Alg^p(X)=\\im\\bigl(\\CH^p(X)_{\\Q}\\xrightarrow{\\cl}H^{2p}(X,\\Q)\\bigr).\n\\]\nEvery algebraic cycle class has type $(p,p)$, so $\\Alg^p(X)\\subseteq\\Hdg^p(X)$. The rational Hodge conjecture asserts equality for every smooth projective $X$ and every $p$ \\cite{ClayHodge,VoisinHodge}.","ion{Statement and status}\nLet $X$ be a smooth projective complex variety. For each codimension $p$, define\n\\[\n\\Hdg^p(X)=H^{2p}(X,\\Q)\\cap H^{p,p}(X)\n\\]\nand\n\\[\n\\Alg^p(X)=\\im\\bigl(\\CH^p(X)_{\\Q}\\xrightarrow{\\cl}H^{2p}(X,\\Q)\\bigr).\n\\]\nEvery algebraic cycle class has type $(p,p)$, so $\\Alg^p(X)\\subseteq\\Hdg^p(X)$. The rational Hodge conjecture asserts equality for every smooth projective $X$ and every $p$ \\cite{ClayHodge,VoisinHodge}.","Choose a positive inner product on $H^{2p}(X,\\R)$. The orthogonal projectors $P_{\\Hdg}^p$ and $P_{\\Alg}^p$ act on the real scalar extensions $\\Hdg^p(X)\\otimes_{\\Q}\\R$ and $\\Alg^p(X)\\otimes_{\\Q}\\R$. These real orthogonal projectors provide a reformulation. They are not asserted to be rational matrices or algebraic correspondences. Since the algebraic space is nested in the Hodge space,\n\\[\nP_{\\Hdg}^pP_{\\Alg}^p=P_{\\Alg}^p.\n\\]","\\begin{proposition}[Master obstruction projector]\nThe operator\n\\[\n\\mathcal O_X^p=P_{\\Hdg}^p-P_{\\Alg}^p\n\\]\nis the orthogonal projector onto $(\\Hdg^p(X)\\otimes_{\\Q}\\R)\\cap(\\Alg^p(X)\\otimes_{\\Q}\\R)^\\perp$. Therefore\n\\[\n\\mathcal O_X^p=0\n\\quad\\Longleftrightarrow\\quad\n\\Hdg^p(X)=\\Alg^p(X).\n\\]\n\\end{proposition}\n\\begin{proof}\nFor nested orthogonal projectors $Q\\le P$, the difference $P-Q$ is self-adjoint and\n\\[\n(P-Q)^2=P-PQ-QP+Q=P-Q.\n\\]\nIts range is $\\im P\\cap(\\im Q)^\\perp$.\n\\end{proof}","n{proposition}[Master obstruction projector]\nThe operator\n\\[\n\\mathcal O_X^p=P_{\\Hdg}^p-P_{\\Alg}^p\n\\]\nis the orthogonal projector onto $(\\Hdg^p(X)\\otimes_{\\Q}\\R)\\cap(\\Alg^p(X)\\otimes_{\\Q}\\R)^\\perp$. Therefore\n\\[\n\\mathcal O_X^p=0\n\\quad\\Longleftrightarrow\\quad\n\\Hdg^p(X)=\\Alg^p(X).\n\\]\n\\end{proposition}\n\\begin{proof}\nFor nested orthogonal projectors $Q\\le P$, the difference $P-Q$ is self-adjoint and\n\\[\n(P-Q)^2=P-PQ-QP+Q=P-Q.\n\\]\nIts range is $\\im P\\cap(\\im Q)^\\perp$.\n\\end{proof}","or]\nThe operator\n\\[\n\\mathcal O_X^p=P_{\\Hdg}^p-P_{\\Alg}^p\n\\]\nis the orthogonal projector onto $(\\Hdg^p(X)\\otimes_{\\Q}\\R)\\cap(\\Alg^p(X)\\otimes_{\\Q}\\R)^\\perp$. Therefore\n\\[\n\\mathcal O_X^p=0\n\\quad\\Longleftrightarrow\\quad\n\\Hdg^p(X)=\\Alg^p(X).\n\\]\n\\end{proposition}\n\\begin{proof}\nFor nested orthogonal projectors $Q\\le P$, the difference $P-Q$ is self-adjoint and\n\\[\n(P-Q)^2=P-PQ-QP+Q=P-Q.\n\\]\nIts range is $\\im P\\cap(\\im Q)^\\perp$.\n\\end{proof}","\\chapter{Primitive and coniveau reductions}\n\\section{Lefschetz packet decomposition}\nLet $h\\in H^2(X,\\Q)$ be the class of an ample divisor and let $L(\\alpha)=h\\smile\\alpha$. Hard Lefschetz and primitive decomposition yield\n\\[\nH^{2p}(X,\\Q)=\\bigoplus_{r\\ge0}L^rH_{\\mathrm{prim}}^{2p-2r}(X,\\Q).\n\\]\nThe decomposition respects Hodge type.","\\begin{theorem}[Primitive reduction]\nIf every primitive rational Hodge class is algebraic, then every rational Hodge class is algebraic.\n\\end{theorem}\n\\begin{proof}\nWrite $\\alpha=\\sum_rL^r\\alpha_r$ with each $\\alpha_r$ primitive and of Hodge type. The class $h$ is algebraic. If $\\alpha_r$ is algebraic, then $L^r\\alpha_r=h^r\\smile\\alpha_r$ is algebraic. Summing gives an algebraic representative of $\\alpha$.\n\\end{proof}","begin{theorem}[Primitive reduction]\nIf every primitive rational Hodge class is algebraic, then every rational Hodge class is algebraic.\n\\end{theorem}\n\\begin{proof}\nWrite $\\alpha=\\sum_rL^r\\alpha_r$ with each $\\alpha_r$ primitive and of Hodge type. The class $h$ is algebraic. If $\\alpha_r$ is algebraic, then $L^r\\alpha_r=h^r\\smile\\alpha_r$ is algebraic. Summing gives an algebraic representative of $\\alpha$.\n\\end{proof}","For a fourfold and $p=2$,\n\\[\n\\alpha=L\\beta+\\alpha_{\\mathrm{prim}},\n\\qquad\n\\beta\\in H^{1,1}(X)\\cap H^2(X,\\Q).\n\\]\nThe Lefschetz $(1,1)$ theorem makes $\\beta$ algebraic. Only $\\alpha_{\\mathrm{prim}}$ remains.","\\section{Support localization and coniveau}\nLet $N^pH^{2p}(X,\\Q)$ be the geometric coniveau-$p$ subspace. In degree $2p$, purity forces support contributions of codimension $p$ to be generated by fundamental classes of codimension-$p$ components.","\\begin{theorem}[Degree-$2p$ coniveau identification]\nFor smooth projective $X$,\n\\[\nN^pH^{2p}(X,\\Q)=\\Alg^p(X).\n\\]\n\\end{theorem}\n\\begin{proof}[Proof sketch]\nThe inclusion from right to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","rem}[Degree-$2p$ coniveau identification]\nFor smooth projective $X$,\n\\[\nN^pH^{2p}(X,\\Q)=\\Alg^p(X).\n\\]\n\\end{theorem}\n\\begin{proof}[Proof sketch]\nThe inclusion from right to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","$X$,\n\\[\nN^pH^{2p}(X,\\Q)=\\Alg^p(X).\n\\]\n\\end{theorem}\n\\begin{proof}[Proof sketch]\nThe inclusion from right to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","\\Q)=\\Alg^p(X).\n\\]\n\\end{theorem}\n\\begin{proof}[Proof sketch]\nThe inclusion from right to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","gin{proof}[Proof sketch]\nThe inclusion from right to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","t to left is immediate because a codimension-$p$ cycle is supported on its underlying algebraic set. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","et. Conversely, a class in coniveau $p$ is in the image of cohomology with supports in an algebraic subset $Z$ of codimension at least $p$. Contributions from components of codimension greater than $p$ vanish in degree $2p$. Purity identifies the degree-$2p$ supported contribution of each codimension-$p$ irreducible component with its fundamental class. Resolution and additivity reduce a general $Z$ to these components.\n\\end{proof}","Thus the conjecture is equivalently the support statement\n\\[\n\\Hdg^p(X)\\subseteq N^pH^{2p}(X,\\Q).\n\\]","\\chapter{Positive cycle-frame calculus}\n\\section{Analysis and synthesis}\nFix a finite family of rational algebraic cycle classes\n\\[\nc_1,\\ldots,c_N\\in H^{2p}(X,\\Q).\n\\]\nLet $V\\subseteq H^{2p}(X,\\R)$ be the real extension of the rational Hodge packet under study. For the positive frame certificate below, require every $c_j$ to belong to $V$. This can be achieved by a rational algebraic packet projector when such a projector has been constructed. Projection through an arbitrary orthogonal Hodge projector does not establish that the projected vectors are algebraic. Define\n\\[\nC:\\R^N\\to H^{2p}(X,\\R),\n\\qquad Ce_j=c_j,\n\\]\nand the positive frame operator\n\\[\nS=CC^*=\\sum_{j=1}^N|c_j\\rangle\\langle c_j|.\n\\]","bseteq H^{2p}(X,\\R)$ be the real extension of the rational Hodge packet under study. For the positive frame certificate below, require every $c_j$ to belong to $V$. This can be achieved by a rational algebraic packet projector when such a projector has been constructed. Projection through an arbitrary orthogonal Hodge projector does not establish that the projected vectors are algebraic. Define\n\\[\nC:\\R^N\\to H^{2p}(X,\\R),\n\\qquad Ce_j=c_j,\n\\]\nand the positive frame operator\n\\[\nS=CC^*=\\sum_{j=1}^N|c_j\\rangle\\langle c_j|.\n\\]","tudy. For the positive frame certificate below, require every $c_j$ to belong to $V$. This can be achieved by a rational algebraic packet projector when such a projector has been constructed. Projection through an arbitrary orthogonal Hodge projector does not establish that the projected vectors are algebraic. Define\n\\[\nC:\\R^N\\to H^{2p}(X,\\R),\n\\qquad Ce_j=c_j,\n\\]\nand the positive frame operator\n\\[\nS=CC^*=\\sum_{j=1}^N|c_j\\rangle\\langle c_j|.\n\\]","\\begin{theorem}[Finite positive cycle-frame certificate]\nLet $A=\\Span_{\\R}\\{c_1,\\ldots,c_N\\}$. Then\n\\[\n\\im S=A,\n\\qquad\n\\ker S=A^\\perp.\n\\]\nUnder the stated containment $c_j\\in V$, the selected cycles span $V$ if and only if $S|_V$ is positive definite. Without containment, positivity only detects the absence of vectors in $V$ orthogonal to the cycle span and does not prove that span contains $V$.\n\\end{theorem}\n\\begin{proof}\nThe equality $\\im(CC^*)=\\im C$ is standard. Moreover\n\\[\n\\langle v,Sv\\rangle=\\|C^*v\\|^2=\\sum_j|\\langle c_j,v\\rangle|^2.\n\\]\nThis vanishes exactly when $v$ is orthogonal to every cycle class.\n\\end{proof}","me certificate]\nLet $A=\\Span_{\\R}\\{c_1,\\ldots,c_N\\}$. Then\n\\[\n\\im S=A,\n\\qquad\n\\ker S=A^\\perp.\n\\]\nUnder the stated containment $c_j\\in V$, the selected cycles span $V$ if and only if $S|_V$ is positive definite. Without containment, positivity only detects the absence of vectors in $V$ orthogonal to the cycle span and does not prove that span contains $V$.\n\\end{theorem}\n\\begin{proof}\nThe equality $\\im(CC^*)=\\im C$ is standard. Moreover\n\\[\n\\langle v,Sv\\rangle=\\|C^*v\\|^2=\\sum_j|\\langle c_j,v\\rangle|^2.\n\\]\nThis vanishes exactly when $v$ is orthogonal to every cycle class.\n\\end{proof}","dots,c_N\\}$. Then\n\\[\n\\im S=A,\n\\qquad\n\\ker S=A^\\perp.\n\\]\nUnder the stated containment $c_j\\in V$, the selected cycles span $V$ if and only if $S|_V$ is positive definite. Without containment, positivity only detects the absence of vectors in $V$ orthogonal to the cycle span and does not prove that span contains $V$.\n\\end{theorem}\n\\begin{proof}\nThe equality $\\im(CC^*)=\\im C$ is standard. Moreover\n\\[\n\\langle v,Sv\\rangle=\\|C^*v\\|^2=\\sum_j|\\langle c_j,v\\rangle|^2.\n\\]\nThis vanishes exactly when $v$ is orthogonal to every cycle class.\n\\end{proof}",", the selected cycles span $V$ if and only if $S|_V$ is positive definite. Without containment, positivity only detects the absence of vectors in $V$ orthogonal to the cycle span and does not prove that span contains $V$.\n\\end{theorem}\n\\begin{proof}\nThe equality $\\im(CC^*)=\\im C$ is standard. Moreover\n\\[\n\\langle v,Sv\\rangle=\\|C^*v\\|^2=\\sum_j|\\langle c_j,v\\rangle|^2.\n\\]\nThis vanishes exactly when $v$ is orthogonal to every cycle class.\n\\end{proof}","If $H$ is a matrix whose columns form an orthonormal basis of $V$, the finite Gram certificate is\n\\[\nG=H^*SH.\n\\]\nUnder $A\\subseteq V$, we have $A=V$ if and only if $\\lambda_{\\min}(G)>0$. Without that containment, use a rational basis matrix $H_{\\Q}$ for the target and the exact certificate $\\operatorname{rank}_{\\Q}[C\\ H_{\\Q}]=\\operatorname{rank}_{\\Q}C$ to test $V\\subseteq A$.","The orthogonal synthesis of a class $\\alpha$ from the cycle span is\n\\[\nP_A\\alpha=S S^\\dagger\\alpha.\n\\]\nIf $\\alpha\\in A$, this is exact.","\\section{Packet assembly without false orthogonality}\nPartition the cycle indices into packets $I_\\lambda$. Define\n\\[\nS_\\lambda=\\sum_{j\\in I_\\lambda}|c_j\\rangle\\langle c_j|.\n\\]\nThen\n\\[\nS=\\sum_\\lambda S_\\lambda.\n\\]\nNo orthogonality of the child spans is assumed.","\\begin{theorem}[Positive child-kernel intersection]\nFor any subspace $V$,\n\\[\n\\ker(S|_V)=\\bigcap_\\lambda\\ker(S_\\lambda|_V).\n\\]\n\\end{theorem}\n\\begin{proof}\nIf $Sv=0$, positivity gives\n\\[\n0=\\langle v,Sv\\rangle=\\sum_\\lambda\\langle v,S_\\lambda v\\rangle.\n\\]\nEach nonnegative summand vanishes, which implies $S_\\lambda v=0$. The converse is immediate.\n\\end{proof}","\\chapter{Degree-residue polyphase refinement}\n\\section{Exact parent-child frame law}\nFix a projective embedding and assign each cycle its degree. For modulus $M$, define the residue packet\n\\[\nI_{M,r}=\\{j:\\deg Z_j\\equiv r\\pmod M\\},\n\\qquad\nS_{M,r}=\\sum_{j\\in I_{M,r}}|c_j\\rangle\\langle c_j|.\n\\]","\\begin{theorem}[Exact degree polyphase refinement]\nFor every positive integer $L$,\n\\[\nS_{M,r}=\\sum_{\\ell=0}^{L-1}S_{LM,r+\\ell M}.\n\\]\n\\end{theorem}\n\\begin{proof}\nThe congruence class $r$ modulo $M$ is the disjoint union of the classes $r+\\ell M$ modulo $LM$. The rank-one frame terms are partitioned accordingly.\n\\end{proof}","\\section{Multiaxial packet addresses}\nDegree residue is only one address. A refined cycle packet may be labeled by\n\\[\n\\lambda=(r_{\\mathrm{Lef}},r_{\\deg},\\chi_{\\mathrm{mon}},s_{\\mathrm{supp}},\\nu_{\\mathrm{Frob}}).\n\\]\nThe entries record Lefschetz depth, degree residue, monodromy or symmetry type, support stratum, and arithmetic phase. The total positive frame remains the sum of all leaves. This allows independent methods to solve different geometric sectors before exact synthesis.","\\chapter{Finite saturation and exact rational closure}\nFor each degree cutoff $D$, let $A_D^p(X)$ be the rational span of codimension-$p$ cycle classes of degree at most $D$. Then\n\\[\nA_1^p\\subseteq A_2^p\\subseteq\\cdots\\subseteq H^{2p}(X,\\Q).\n\\]","\\begin{theorem}[Finite stabilization]\nThere exists $D_0$ such that $A_D^p(X)=A_{D_0}^p(X)$ for every $D\\ge D_0$.\n\\end{theorem}\n\\begin{proof}\nThe dimensions form a nondecreasing sequence of integers bounded by $\\dim H^{2p}(X,\\Q)$. Only finitely many strict increases are possible.\n\\end{proof}","\\begin{corollary}[Approximation becomes exact]\nLet $P_D$ be the orthogonal projector onto $A_D^p(X)\\otimes\\R$. For a rational class $\\alpha$,\n\\[\n\\lim_{D\\to\\infty}\\|(I-P_D)\\alpha\\|=0\n\\]\nif and only if $\\alpha\\in A_D^p(X)$ for some finite $D$.\n\\end{corollary}\n\\begin{proof}\nAfter stabilization, the residual is constant. A constant tail converging to zero is zero.\n\\end{proof}","Because the matrices and target vector can be chosen over $\\Q$, real membership implies rational membership by row reduction over $\\Q$.","\\begin{resultbox}[Finite completion certificate]\nIf $H$ is a rational basis matrix of $\\Hdg^p(X)$ and $C_D$ is the cycle matrix, then the Hodge conjecture for $(X,p)$ is equivalent to the existence of $D$ with\n\\[\n\\rank_{\\Q} C_D=\\rank_{\\Q}[C_D\\ H].\n\\]\nA positive compressed frame is equivalent only when the supplied cycle vectors lie in the target packet, or when a separately verified invariant restriction supplies the required containment. The rank equality is valid without that extra hypothesis.\n\\end{resultbox}","\\chapter{Support verification and multiplicity recovery}\nLet $Z\\subseteq X$ be a closed algebraic subset and $j:X\\setminus Z\\hookrightarrow X$. The support exact sequence contains\n\\[\nH_Z^{2p}(X,\\Q)\\longrightarrow H^{2p}(X,\\Q)\\xrightarrow{j^*}H^{2p}(X\\setminus Z,\\Q).\n\\]","\\begin{theorem}[Exact support test]\nA class $\\alpha\\in H^{2p}(X,\\Q)$ is supported on $Z$ if and only if $j^*\\alpha=0$.\n\\end{theorem}\n\\begin{proof}\nThis is exactness at the middle term.\n\\end{proof}","Suppose $Z$ has codimension at least $p$ and its codimension-$p$ components are $Z_1,\\ldots,Z_r$. Purity gives\n\\[\n\\alpha=\\sum_{i=1}^r q_i[Z_i].\n\\]\nChoose complementary test classes $\\gamma_k$ and define\n\\[\nM_{ki}=\\int_X[Z_i]\\smile\\gamma_k,\n\\qquad\nb_k=\\int_X\\alpha\\smile\\gamma_k.\n\\]\nThen $Mq=b$. Removing dependent columns yields exact rational multiplicities.","\\chapter{Divisor-lowering correspondence channels}\n\\section{Degree bookkeeping}\nLet $X$ have complex dimension $n$, let $V\\subseteq H_{\\mathrm{prim}}^{2p}(X,\\Q)$ be a rational Hodge substructure, and let $Y_r$ be smooth projective. Choose correspondences\n\\[\n\\Gamma_r\\in\\CH^{n-p+1}(X\\times Y_r)_{\\Q},\n\\qquad\n\\Lambda_r\\in\\CH^{\\dim Y_r+p-1}(Y_r\\times X)_{\\Q}.\n\\]\nThey induce\n\\[\nA_r=(\\Gamma_r)_*:V\\to H^2(Y_r,\\Q),\n\\qquad\nB_r=(\\Lambda_r)_*:H^2(Y_r,\\Q)\\to H^{2p}(X,\\Q).\n\\]\nAlgebraic correspondences preserve rationality and shift Hodge type by the required bidegree.","\\begin{theorem}[Invertible divisor-lowering transfer]\nSuppose\n\\[\nT=\\sum_{r=1}^M B_rA_r\n\\]\npreserves $V$ and is invertible on $V$. Then every rational Hodge class in $V$ is algebraic.\n\\end{theorem}\n\\begin{proof}\nLet $\\alpha\\in V$ be of type $(p,p)$. Each $A_r\\alpha$ is rational of type $(1,1)$ on $Y_r$, hence is the class of a rational divisor $D_r$ by the Lefschetz $(1,1)$ theorem. Therefore\n\\[\nT\\alpha=\\sum_r(\\Lambda_r)_*[D_r]\n\\]\nis algebraic on $X$.","r]\nSuppose\n\\[\nT=\\sum_{r=1}^M B_rA_r\n\\]\npreserves $V$ and is invertible on $V$. Then every rational Hodge class in $V$ is algebraic.\n\\end{theorem}\n\\begin{proof}\nLet $\\alpha\\in V$ be of type $(p,p)$. Each $A_r\\alpha$ is rational of type $(1,1)$ on $Y_r$, hence is the class of a rational divisor $D_r$ by the Lefschetz $(1,1)$ theorem. Therefore\n\\[\nT\\alpha=\\sum_r(\\Lambda_r)_*[D_r]\n\\]\nis algebraic on $X$.","Let $\\chi_T(t)=t^d+c_{d-1}t^{d-1}+\\cdots+c_0$ be the characteristic polynomial on $V$. Invertibility means $c_0\\ne0$. Cayley-Hamilton gives\n\\[\nT^{-1}=-\\frac{1}{c_0}\\bigl(T^{d-1}+c_{d-1}T^{d-2}+\\cdots+c_1I\\bigr).\n\\]\nEach power of $T$ is induced by compositions and rational sums of algebraic correspondences. Thus $T^{-1}$ preserves algebraic cycle classes. Hence $\\alpha=T^{-1}(T\\alpha)$ is algebraic.\n\\end{proof}","\\section{Positive multi-channel form}\nAssume the polarized adjoint $A_r^\\dagger$ is realized by an algebraic return correspondence $B_r$. This is an additional geometric hypothesis, not an automatic consequence in arbitrary settings. Then\n\\[\nT=\\sum_rA_r^\\dagger A_r\n\\]\nis positive and\n\\[\n\\langle v,Tv\\rangle=\\sum_r\\|A_rv\\|^2.\n\\]\nTherefore\n\\[\n\\ker T=\\bigcap_r\\ker A_r.\n\\]\nJoint injectivity of the lowering channels is enough.","\\begin{corollary}[Finite Gram certificate]\nChoose a rational basis $e_1,\\ldots,e_d$ of $V$. If\n\\[\nG_{ij}=\\sum_r\\langle A_re_i,A_re_j\\rangle_{Y_r}\n\\]\nhas nonzero determinant, every rational Hodge class in $V$ is algebraic.\n\\end{corollary}","\\chapter{Family propagation}\nLet $\\pi:\\mathcal X\\to S$ and $\\rho_r:\\mathcal Y_r\\to S$ be smooth projective families over a connected base. Relative algebraic correspondences induce morphisms of Gauss-Manin local systems. In a flat rational trivialization, their transfer matrix is constant.","\\begin{theorem}[Propagation of invertibility]\nIf a relative transfer $T_s$ is invertible on one fiber $s_0$, it is invertible on every fiber in the connected base on which the local systems and correspondences are defined.\n\\end{theorem}\n\\begin{proof}\nThe determinant of the constant rational matrix representing the local-system morphism is locally constant. It is nonzero at $s_0$, hence everywhere on the connected base.\n\\end{proof}","\\chapter{Geometric closure rules}\n\\section{Affine pavings}\n\\begin{theorem}[Affine-paving branch]\nIf $X$ admits a finite filtration whose successive differences are affine spaces, then the cycle class map\n\\[\n\\CH^p(X)_{\\Q}\\to H^{2p}(X,\\Q)\n\\]\nis surjective for every $p$.\n\\end{theorem}\n\\begin{proof}[Proof sketch]\nInduct through the filtration using localization for Chow groups and cohomology. Each affine cell contributes one even-degree generator represented by the closure of that cell. No odd cohomology is introduced.\n\\end{proof}","\\section{Projective bundles, blow-ups, and descent}\n\\begin{proposition}[Projective-bundle closure]\nIf the rational Hodge conjecture holds for $X$, it holds for $\\mathbb P(E)\\to X$.\n\\end{proposition}\n\\begin{proof}\nThe projective-bundle formula expresses cohomology as a direct sum of powers of the algebraic hyperplane class times pullbacks of cohomology from $X$.\n\\end{proof}","\\begin{proposition}[Blow-up closure]\nIf it holds for $X$ and a smooth center $Z$, it holds for $\\operatorname{Bl}_ZX$.\n\\end{proposition}\n\\begin{proof}\nUse the standard blow-up cohomology decomposition. The new summands are Tate twists of cohomology from $Z$ multiplied by powers of the algebraic exceptional divisor.\n\\end{proof}","\\begin{proposition}[Generically finite descent]\nLet $f:Y\\to X$ be proper and generically finite of degree $d$ between smooth projective varieties of equal dimension. If the conjecture holds for $Y$, it holds for $X$.\n\\end{proposition}\n\\begin{proof}\nFor a Hodge class $\\alpha$ on $X$, the pullback $f^*\\alpha$ is algebraic on $Y$. Push forward and use $f_*f^*\\alpha=d\\alpha$.\n\\end{proof}","\\chapter{Cubic fourfolds as a validation of the transfer theorem}\nLet $X\\subset\\mathbb P^5$ be a smooth cubic fourfold and let $F(X)$ be its Fano variety of lines. The universal incidence variety\n\\[\nP=\\{(\\ell,x):x\\in\\ell\\}\\subset F(X)\\times X\n\\]\ninduces the Abel-Jacobi map\n\\[\nA=p_*q^*:H^4_{\\mathrm{prim}}(X,\\Q)\\to H^2_{\\mathrm{prim}}(F(X),\\Q).\n\\]\nBeauville and Donagi proved that this is a polarized Hodge isomorphism up to the standard twist \\cite{BeauvilleDonagi}. The polarized adjoint is represented by an algebraic incidence return after inserting the appropriate power of the Pluecker class. Therefore the divisor-lowering transfer is invertible.","\\begin{resultbox}[Classical validation]\nA primitive rational $(2,2)$ class on a cubic fourfold is sent to a rational $(1,1)$ class on $F(X)$, represented by a divisor. The algebraic return correspondence and its polynomial inverse reconstruct an algebraic codimension-two class on $X$. This is a validation of the packet criterion, not a claim of a new proof of the classical cubic-fourfold result.\n\\end{resultbox}","\\begin{itemize}\n\\item For suitable GM fourfolds, the vanishing $H^4$ packet is lowered by universal lines to divisor degree.\n\\item For suitable GM sixfolds, the vanishing $H^6$ packet is lowered by universal planes to divisor degree.\n\\item Ambient and tautological classes are treated separately and are algebraic.\n\\end{itemize}","\\chapter{Hodge packets in topological K-theory}\nFor finite CW complexes, the rational topological Chern character gives\n\\[\n\\operatorname{ch}_{\\mathrm{top}}:K^0_{\\mathrm{top}}(X)\\otimes\\Q\\xrightarrow{\\sim}H^{\\mathrm{even}}(X,\\Q).\n\\]\nFor smooth projective $X$, the algebraic Chern character identifies algebraic $K$-theory rationally with the Chow ring after the usual filtration. Thus the rational Hodge conjecture for all degrees is equivalent to algebraization of topological $K$-classes whose Chern-character components are rational Hodge classes \\cite{AtiyahHirzebruch}.","ational topological Chern character gives\n\\[\n\\operatorname{ch}_{\\mathrm{top}}:K^0_{\\mathrm{top}}(X)\\otimes\\Q\\xrightarrow{\\sim}H^{\\mathrm{even}}(X,\\Q).\n\\]\nFor smooth projective $X$, the algebraic Chern character identifies algebraic $K$-theory rationally with the Chow ring after the usual filtration. Thus the rational Hodge conjecture for all degrees is equivalent to algebraization of topological $K$-classes whose Chern-character components are rational Hodge classes \\cite{AtiyahHirzebruch}.","\\section{Adams degree projectors}\nFor integer $m\\ge2$, Adams operations satisfy\n\\[\n\\operatorname{ch}_q(\\psi^m\\xi)=m^q\\operatorname{ch}_q(\\xi).\n\\]\nDefine\n\\[\ne_p^{(m)}=\\prod_{\\substack{0\\le q\\le\\dim X","\\section{Adams degree projectors}\nFor integer $m\\ge2$, Adams operations satisfy\n\\[\n\\operatorname{ch}_q(\\psi^m\\xi)=m^q\\operatorname{ch}_q(\\xi).\n\\]\nDefine\n\\[\ne_p^{(m)}=\\prod_{\\substack{0\\le q\\le\\dim X\\\\q\\ne p}}\n\\frac{\\psi^m-m^qI}{m^p-m^q}.\n\\]\nThen\n\\[\n\\operatorname{ch}_q(e_p^{(m)}\\xi)=\\delta_{pq}\\operatorname{ch}_p(\\xi).\n\\]\nThis exactly isolates codimension $p$ inside a topological $K$-packet.","q\\ne p}}\n\\frac{\\psi^m-m^qI}{m^p-m^q}.\n\\]\nThen\n\\[\n\\operatorname{ch}_q(e_p^{(m)}\\xi)=\\delta_{pq}\\operatorname{ch}_p(\\xi).\n\\]\nThis exactly isolates codimension $p$ inside a topological $K$-packet.","\\chapter{Nonlinear Dolbeault obstruction packets}\nLet $E$ be a smooth complex bundle and consider a candidate $(0,1)$-connection. After gauge fixing, write the integrability equation abstractly as\n\\[\nF(a)=f+La+B(a,a)=0,\n\\]\nwhere $L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","er{Nonlinear Dolbeault obstruction packets}\nLet $E$ be a smooth complex bundle and consider a candidate $(0,1)$-connection. After gauge fixing, write the integrability equation abstractly as\n\\[\nF(a)=f+La+B(a,a)=0,\n\\]\nwhere $L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","sider a candidate $(0,1)$-connection. After gauge fixing, write the integrability equation abstractly as\n\\[\nF(a)=f+La+B(a,a)=0,\n\\]\nwhere $L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","ction. After gauge fixing, write the integrability equation abstractly as\n\\[\nF(a)=f+La+B(a,a)=0,\n\\]\nwhere $L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","write the integrability equation abstractly as\n\\[\nF(a)=f+La+B(a,a)=0,\n\\]\nwhere $L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","$L$ is the linearized operator and $B$ is a bounded bilinear term. Assume a Fredholm splitting of the target\n\\[\nI=Q+\\Pi,\n\\]\nwhere $Q$ projects onto the closed range of $L$ and $\\Pi$ onto a finite-dimensional obstruction space. Choose a gauge-fixed complement to $\\ker L$ as the domain and let $R$ be a right inverse of $L$ taking values in that complement. Use norms on the projected data for which $\\|QB(a,b)\\|\\le C\\|a\\|\\|b\\|$.","\\begin{theorem}[Range-packet contraction]\nAssume\n\\[\n\\|R\\|\\le M,\n\\qquad\n\\|B(a,b)\\|\\le C\\|a\\|\\|b\\|,\n\\qquad\n4CM^2\\|Qf\\|<1.\n\\]\nThen the map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Ap","\\begin{theorem}[Range-packet contraction]\nAssume\n\\[\n\\|R\\|\\le M,\n\\qquad\n\\|B(a,b)\\|\\le C\\|a\\|\\|b\\|,\n\\qquad\n4CM^2\\|Qf\\|<1.\n\\]\nThen the map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancell","\\begin{theorem}[Range-packet contraction]\nAssume\n\\[\n\\|R\\|\\le M,\n\\qquad\n\\|B(a,b)\\|\\le C\\|a\\|\\|b\\|,\n\\qquad\n4CM^2\\|Qf\\|<1.\n\\]\nThen the map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","em}[Range-packet contraction]\nAssume\n\\[\n\\|R\\|\\le M,\n\\qquad\n\\|B(a,b)\\|\\le C\\|a\\|\\|b\\|,\n\\qquad\n4CM^2\\|Qf\\|<1.\n\\]\nThen the map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","|\\|b\\|,\n\\qquad\n4CM^2\\|Qf\\|<1.\n\\]\nThen the map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","map\n\\[\n\\mathcal T(a)=-RQ(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","(f+B(a,a))\n\\]\nhas a unique fixed point in the gauge-fixed ball $\\|a\\|\\le2M\\|Qf\\|$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","$. If $Qf=0$, the ball is the singleton $\\{0\\}$ and the range solution is immediate. At the fixed point, the range component of $F(a)$ vanishes. The remaining obstruction is\n\\[\n\\Phi=\\Pi(f+B(a,a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","a)).\n\\]\nThe full integrability equation holds if and only if $\\Phi=0$.\n\\end{theorem}\n\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","\\begin{proof}\nFor $\\|a\\|\\le r=2M\\|Qf\\|$,\n\\[\n\\|\\mathcal T(a)\\|\\le M\\|Qf\\|+MC r^2\\le r\n\\]\nunder the stated bound. Moreover\n\\[\n\\|\\mathcal T(a)-\\mathcal T(b)\\|\n\\le MC(\\|a\\|+\\|b\\|)\\|a-b\\|.\n\\]\nThe Lipschitz constant is strictly less than one on the stated ball.\nBanach's fixed-point theorem gives the solution. Applying $Q$ to the equation proves range cancellation. The remaining component is precisely $\\Phi$.\n\\end{proof}","\\section{A positive spectral microscope}\nLarge ample twists can make integrability defects visible in the asymptotic degree distribution of Dolbeault-Dirac kernels. Charbonneau and Stern obtain asymptotic Hodge filtrations in which positive coefficients depend on powers of $F_A^{0,2}$ \\cite{CharbonneauStern}. Schematically,\n\\[\n\\Tr(P_{2j}\\Pi_{A,k})\n=c_{n,j}k^{n-2j}\\|(F_A^{0,2})^{\\wedge j}\\|_{L^2}^2+O(k^{n-2j-1}),\n\\qquad c_{n,j}>0.\n\\]\nThis detects nonintegrability. It does not by itself remove the defect.","\\section{Trace blindness and the true missing bridge}\nChern characters control traces of powers of curvature. An endomorphism-valued obstruction can be nonzero while all such trace moments vanish. For example, $N=\\left(\\begin{smallmatrix}0&1&0\\\\0&0&1\\\\0&0&0\\end{smallmatrix}\\right)$ is nonzero, while $N^3=0$ and $\\operatorname{Tr}(N^r)=0$ for every positive integer $r$. This is the elementary negative control.","0&0&0\\end{smallmatrix}\\right)$ is nonzero, while $N^3=0$ and $\\operatorname{Tr}(N^r)=0$ for every positive integer $r$. This is the elementary negative control.","\\begin{openproblem}[Analytic bridge]\nThe unresolved analytic theorem is not the range equation. It is a positive saturation statement forcing the finite obstruction packet $\\Phi$ to vanish for every rational Hodge $K$-packet after allowable stabilization. Proving that statement in general would be essentially a K-theoretic form of the Hodge conjecture.\n\\end{openproblem}","\\section{Consequences of the obstruction split}\nThe obstruction split prevents a common logical error. Solving the projected range equation does not solve the original integrability problem unless the complementary packet also vanishes. The calculus therefore records three distinct outcomes. First, if $\\Pi=0$, the problem is unobstructed and the contraction theorem gives an actual integrable structure. Second, if $\\Pi\\ne0$ but $\\Phi=0$ for geometric reasons, the correction again closes. Third, if $\\Phi\\ne0$, the method has produced a finite certificate explaining exactly why that representative cannot be algebraized by the chosen deformation.","ces of the obstruction split}\nThe obstruction split prevents a common logical error. Solving the projected range equation does not solve the original integrability problem unless the complementary packet also vanishes. The calculus therefore records three distinct outcomes. First, if $\\Pi=0$, the problem is unobstructed and the contraction theorem gives an actual integrable structure. Second, if $\\Pi\\ne0$ but $\\Phi=0$ for geometric reasons, the correction again closes. Third, if $\\Phi\\ne0$, the method has produced a finite certificate explaining exactly why that representative cannot be algebraized by the chosen deformation.","truction split}\nThe obstruction split prevents a common logical error. Solving the projected range equation does not solve the original integrability problem unless the complementary packet also vanishes. The calculus therefore records three distinct outcomes. First, if $\\Pi=0$, the problem is unobstructed and the contraction theorem gives an actual integrable structure. Second, if $\\Pi\\ne0$ but $\\Phi=0$ for geometric reasons, the correction again closes. Third, if $\\Phi\\ne0$, the method has produced a finite certificate explaining exactly why that representative cannot be algebraized by the chosen deformation.","Solving the projected range equation does not solve the original integrability problem unless the complementary packet also vanishes. The calculus therefore records three distinct outcomes. First, if $\\Pi=0$, the problem is unobstructed and the contraction theorem gives an actual integrable structure. Second, if $\\Pi\\ne0$ but $\\Phi=0$ for geometric reasons, the correction again closes. Third, if $\\Phi\\ne0$, the method has produced a finite certificate explaining exactly why that representative cannot be algebraized by the chosen deformation.","This creates a practical research program. One can refine the finite obstruction space by symmetry, monodromy, support, or arithmetic phase, obtaining\n\\[\n\\Phi=\\sum_{\\lambda}\\Phi_{\\lambda}.\n\\]\nSince the decomposition is complete, integrability is equivalent to the simultaneous vanishing of all children. Different geometric arguments may eliminate different children. A symmetry packet may vanish by representation theory, a support packet by localization, and an arithmetic packet by compatibility across reductions. The analytic branch therefore remains useful even before the general bridge is proved: it supplies exact special-case theorems whenever all obstruction children can be eliminated.","The negative controls are equally important. Ample twisting can amplify $F_A^{0,2}$ in a spectral observable without reducing it. Direct-sum stabilization adds positive defect energies for block-diagonal connections. Trace identities may miss nilpotent components. These facts rule out several tempting but invalid shortcuts and keep the remaining theorem sharply stated.","Degree-$2p$ coniveau identification & Proved & Rational coefficients and purity","\\chapter{A focused multi-incidence target}\nLet $X$ be a special smooth projective fourfold. Construct algebraic lowering channels from families of subvarieties:\n\\[\nA_{\\mathrm{line}},\nA_{\\mathrm{conic}},\nA_{\\mathrm{plane}},\nA_{\\mathrm{exceptional}}.\n\\]\nThe exact target is\n\\[\n\\ker A_{\\mathrm{line}}\n\\cap\\ker A_{\\mathrm{conic}}\n\\cap\\ker A_{\\mathrm{plane}}\n\\cap\\ker A_{\\mathrm{exceptional}}=\\{0\\}\n\\]\non primitive degree-four cohomology. Under algebraic adjoint hypotheses, equivalently prove\n\\[\nT_{\\mathrm{tot}}\n=\\sum_rA_r^\\dagger A_r\n\\quad\\text{satisfies}\\quad\n\\lambda_{\\min}(T_{\\mathrm{tot}})>0.\n\\]","Construct algebraic lowering channels from families of subvarieties:\n\\[\nA_{\\mathrm{line}},\nA_{\\mathrm{conic}},\nA_{\\mathrm{plane}},\nA_{\\mathrm{exceptional}}.\n\\]\nThe exact target is\n\\[\n\\ker A_{\\mathrm{line}}\n\\cap\\ker A_{\\mathrm{conic}}\n\\cap\\ker A_{\\mathrm{plane}}\n\\cap\\ker A_{\\mathrm{exceptional}}=\\{0\\}\n\\]\non primitive degree-four cohomology. Under algebraic adjoint hypotheses, equivalently prove\n\\[\nT_{\\mathrm{tot}}\n=\\sum_rA_r^\\dagger A_r\n\\quad\\text{satisfies}\\quad\n\\lambda_{\\min}(T_{\\mathrm{tot}})>0.\n\\]","\\section{Why this target is realistic}\nThe problem is finite after the geometry is chosen:\n\\begin{enumerate}\n\\item compute the primitive cohomology representation and its symmetry decomposition\n\\item construct the incidence correspondences and their action\n\\item split the action into monodromy or symmetry packets\n\\item compute the joint kernel or positive Gram matrix over $\\Q$\n\\item prove exact nonvanishing through rational rank, representation theory, or a dual certificate\n\\item invoke the divisor-lowering theorem to reconstruct cycles\n\\end{enumerate}\nNo single incidence family must be complete. Polyphase analysis permits separate geometric mechanisms to occupy separate primitive sectors.","\\section{Three exact certificate forms}\nOnce the incidence channels have been constructed, the saturation statement admits three finite certificates. The geometric form is the kernel identity\n\\[\n\\bigcap_r\\ker A_r=\\{0\\}.\n\\]\nWhen algebraic correspondences realize the polarized adjoints, the equivalent positive form is\n\\[\n\\lambda_{\\min}\\left(\\sum_rA_r^\\dagger A_r\\right)>0.\n\\]\nIn a rational basis, the final exact proof object is full column rank of the vertically stacked channel matrix. The kernel form explains the geometry, the positive form is stable for numerical scouting, and the rational rank form is appropriate for symbolic certification.","\\chapter{Computational and formal proof workflow}\nNumerical experiments can scout candidate families but cannot replace an exact rank certificate. The intended pipeline is:\n\\[\n\\begin{aligned}\n\\text{exact incidence data}\n&\\longrightarrow \\text{rational cohomology basis}\n\\longrightarrow \\text{packet transfer matrices}\\\\\n&\\longrightarrow \\text{rank or positivity certificate}\n\\longrightarrow \\text{symbolic theorem proof}.\n\\end{aligned}\n\\]","\\section{Reproducible finite checks}\nThe accompanying Python generator produces all figures and runs eighteen independent finite checks covering:\n\\begin{itemize}\n\\item frame idempotency and range rank\n\\item positive Hodge Gram spectrum\n\\item exact modulo-$2$ and modulo-$4$ polyphase assembly\n\\item positive transfer rank and energy identities\n\\item Cayley-Hamilton inversion\n\\item contraction of the removable nonlinear packet\n\\item Adams projector idempotency and degree isolation\n\\item nilpotent trace blindness\n\\item support multiplicity recovery\n\\item finite rank stabilization\n\\end{itemize}\nThese tests verify the finite examples and implementation. They do not establish the open geometric saturation theorem.","\\chapter{Correct scope and claim boundary}\nThe projector-resolved Hodge calculus naturally specializes to Fermat fourfolds\n\\[\nX_m^4=\\{x_0^m+x_1^m+x_2^m+x_3^m+x_4^m+x_5^m=0\\}\\subset\\mathbf P^5,\n\\qquad m\\ge 3.\n\\]\nThe diagonal symmetry group makes the primitive middle cohomology unusually explicit: it splits into one-dimensional complex character lines, and rational Hodge packets are finite Galois orbits of those lines. This permits exact symbolic analysis of individual addresses, conductors, local prime phases, and candidate algebraic correspondences. A complete classification at every mixed conductor is a separate coverage problem.","\\begin{claimboundary}\nThe general rational Hodge conjecture remains open. For Fermat fourfolds, however, Kang is cited in the checked later literature as proving the relevant generalized Hodge statement in arbitrary degree \\cite{KangFermat}. This complete edition therefore distinguishes two questions:\n\\begin{enumerate}\n\\item \\emph{Existence}: does every rational Fermat $(2,2)$ class have an algebraic representative?\n\\item \\emph{Construction}: can one write an explicit packet-resolved algebraic cycle or correspondence for every character orbit?\n\\end{enumerate}\nThe present Fermat program is primarily constructive. Its new proof objects are conductor projectors, exact packet identities, explicit Delsarte maps, weighted dyadic refinements, the proposed all-degree trivalent-tree classification, CRT--Kummer phase factorizations, and explicit CM carrier data. The relation between that existence theorem and later packet-specific combinatorial work should be stated explicitly in any publication.\n\\end{claimboundary}","The constructive objective is\n\\begin{resultbox}[Packet-resolved construction target]\nFor every rational packet\n\\[\nW\\subset H^4_{\\mathrm{prim}}(X_m^4,\\Q)\\cap H^{2,2},\n\\]\nconstruct\n\\[\nZ_W\\in\\CH^2(X_m^4)_\\Q\n\\qquad\\text{with}\\qquad\n[Z_W]\\ne0\\text{ in }W.\n\\]\n\\end{resultbox}\nOnce one nonzero algebraic vector is produced in a rational irreducible diagonal-symmetry packet, the entire rational orbit is algebraic.","\\chapter{Universal balanced-packet equations}\nLet\n\\[\nU_m=(\\mathbf Z/m\\mathbf Z)^\\times\n\\]\nand let \\(\\langle r\\rangle_m\\in\\{1,\\ldots,m\\}\\) denote the least positive residue, with \\(\\langle r\\rangle_m=m\\) when \\(m\\mid r\\). A primitive character address is a six-tuple\n\\[\na=(a_0,\\ldots,a_5),\\qquad 1\\le a_i<m,\n\\qquad \\sum_i a_i\\equiv0\\pmod m.\n\\]\nThe Hodge level of the conjugate indexed by \\(u\\in U_m\\) is\n\\[\n\\ell_a(u)=\\frac1m\\sum_{i=0}^5\\langle ua_i\\rangle_m.\n\\]\nThe character line has type \\((2,2)\\) exactly when \\(\\ell_a(1)=3\\). Its rational Galois orbit is a Hodge packet exactly when\n\\begin{equation}\\label{eq:legacy-7.1}\n\\boxed{\n\\ell_a(u)=3\\qquad\\forall u\\in U_m.\n}\\end{equation}","ter{Universal balanced-packet equations}\nLet\n\\[\nU_m=(\\mathbf Z/m\\mathbf Z)^\\times\n\\]\nand let \\(\\langle r\\rangle_m\\in\\{1,\\ldots,m\\}\\) denote the least positive residue, with \\(\\langle r\\rangle_m=m\\) when \\(m\\mid r\\). A primitive character address is a six-tuple\n\\[\na=(a_0,\\ldots,a_5),\\qquad 1\\le a_i<m,\n\\qquad \\sum_i a_i\\equiv0\\pmod m.\n\\]\nThe Hodge level of the conjugate indexed by \\(u\\in U_m\\) is\n\\[\n\\ell_a(u)=\\frac1m\\sum_{i=0}^5\\langle ua_i\\rangle_m.\n\\]\nThe character line has type \\((2,2)\\) exactly when \\(\\ell_a(1)=3\\). Its rational Galois orbit is a Hodge packet exactly when\n\\begin{equation}\\label{eq:legacy-7.1}\n\\boxed{\n\\ell_a(u)=3\\qquad\\forall u\\in U_m.\n}\\end{equation}","nd let \\(\\langle r\\rangle_m\\in\\{1,\\ldots,m\\}\\) denote the least positive residue, with \\(\\langle r\\rangle_m=m\\) when \\(m\\mid r\\). A primitive character address is a six-tuple\n\\[\na=(a_0,\\ldots,a_5),\\qquad 1\\le a_i<m,\n\\qquad \\sum_i a_i\\equiv0\\pmod m.\n\\]\nThe Hodge level of the conjugate indexed by \\(u\\in U_m\\) is\n\\[\n\\ell_a(u)=\\frac1m\\sum_{i=0}^5\\langle ua_i\\rangle_m.\n\\]\nThe character line has type \\((2,2)\\) exactly when \\(\\ell_a(1)=3\\). Its rational Galois orbit is a Hodge packet exactly when\n\\begin{equation}\\label{eq:legacy-7.1}\n\\boxed{\n\\ell_a(u)=3\\qquad\\forall u\\in U_m.\n}\\end{equation}","le_m\\in\\{1,\\ldots,m\\}\\) denote the least positive residue, with \\(\\langle r\\rangle_m=m\\) when \\(m\\mid r\\). A primitive character address is a six-tuple\n\\[\na=(a_0,\\ldots,a_5),\\qquad 1\\le a_i<m,\n\\qquad \\sum_i a_i\\equiv0\\pmod m.\n\\]\nThe Hodge level of the conjugate indexed by \\(u\\in U_m\\) is\n\\[\n\\ell_a(u)=\\frac1m\\sum_{i=0}^5\\langle ua_i\\rangle_m.\n\\]\nThe character line has type \\((2,2)\\) exactly when \\(\\ell_a(1)=3\\). Its rational Galois orbit is a Hodge packet exactly when\n\\begin{equation}\\label{eq:legacy-7.1}\n\\boxed{\n\\ell_a(u)=3\\qquad\\forall u\\in U_m.\n}\\end{equation}","when \\(m\\mid r\\). A primitive character address is a six-tuple\n\\[\na=(a_0,\\ldots,a_5),\\qquad 1\\le a_i<m,\n\\qquad \\sum_i a_i\\equiv0\\pmod m.\n\\]\nThe Hodge level of the conjugate indexed by \\(u\\in U_m\\) is\n\\[\n\\ell_a(u)=\\frac1m\\sum_{i=0}^5\\langle ua_i\\rangle_m.\n\\]\nThe character line has type \\((2,2)\\) exactly when \\(\\ell_a(1)=3\\). Its rational Galois orbit is a Hodge packet exactly when\n\\begin{equation}\\label{eq:legacy-7.1}\n\\boxed{\n\\ell_a(u)=3\\qquad\\forall u\\in U_m.\n}\\end{equation}","To remove coordinate permutations, introduce the count vector\n\\[\nn(a)=(n_1,\\ldots,n_{m-1})\\in\\mathbf N^{m-1},\n\\qquad n_r=\\#\\{i:a_i=r\\}.\n\\]\nLet\n\\[\nR_m(u,r)=\\langle ur\\rangle_m.\n\\]\nThen the complete balanced six-packet space is the finite integer set\n\\begin{equation}\\label{eq:legacy-7.2}\n\\boxed{\n\\mathcal B_m=\n\\left\\{\n n\\in\\mathbf N^{m-1}:\n \\mathbf1^Tn=6,\n \\quad R_mn=3m\\,\\mathbf1\n\\right\\}.\n}\\end{equation}\nThe unit group permutes the residue coordinates. After additionally identifying coordinate permutations by algebraic automorphisms, the count-vector descriptions correspond to \\(U_m\\)-orbits in \\(\\mathcal B_m\\). The dimension of a fixed rational character packet is computed from distinct ordered unit conjugates before this permutation quotient.","the complete balanced six-packet space is the finite integer set\n\\begin{equation}\\label{eq:legacy-7.2}\n\\boxed{\n\\mathcal B_m=\n\\left\\{\n n\\in\\mathbf N^{m-1}:\n \\mathbf1^Tn=6,\n \\quad R_mn=3m\\,\\mathbf1\n\\right\\}.\n}\\end{equation}\nThe unit group permutes the residue coordinates. After additionally identifying coordinate permutations by algebraic automorphisms, the count-vector descriptions correspond to \\(U_m\\)-orbits in \\(\\mathcal B_m\\). The dimension of a fixed rational character packet is computed from distinct ordered unit conjugates before this permutation quotient.","ced six-packet space is the finite integer set\n\\begin{equation}\\label{eq:legacy-7.2}\n\\boxed{\n\\mathcal B_m=\n\\left\\{\n n\\in\\mathbf N^{m-1}:\n \\mathbf1^Tn=6,\n \\quad R_mn=3m\\,\\mathbf1\n\\right\\}.\n}\\end{equation}\nThe unit group permutes the residue coordinates. After additionally identifying coordinate permutations by algebraic automorphisms, the count-vector descriptions correspond to \\(U_m\\)-orbits in \\(\\mathcal B_m\\). The dimension of a fixed rational character packet is computed from distinct ordered unit conjugates before this permutation quotient.","\\section{Centered Fourier form}\nDefine\n\\[\n\\rho_m(r)=\\langle r\\rangle_m-\\frac m2,\n\\qquad\nF_a(u)=\\sum_i\\rho_m(ua_i).\n\\]\nThen \\(a\\) is balanced if and only if \\(F_a(u)=0\\) for all units. Moreover,\n\\[\nF_a(-u)=-F_a(u),\n\\]\nso only odd characters occur. For a Dirichlet character \\(\\chi\\) on \\(U_m\\), put\n\\[\n\\mathcal B_{m,\\chi}(r)=\n\\sum_{u\\in U_m}\\rho_m(ur)\\overline{\\chi(u)}.\n\\]\nFinite Fourier inversion gives the exact criterion\n\\begin{equation}\\label{eq:legacy-7.3}\n\\boxed{\n n(a)\\in\\mathcal B_m\n \\Longleftrightarrow\n \\sum_i\\mathcal B_{m,\\chi}(a_i)=0\n \\quad\\text{for every odd }\\chi.\n}\\end{equation}\nHere $n(a)$ is the residue count vector. This formulation underlies the prime-power filtration and the proposed conductor-defect tree.","[\n\\rho_m(r)=\\langle r\\rangle_m-\\frac m2,\n\\qquad\nF_a(u)=\\sum_i\\rho_m(ua_i).\n\\]\nThen \\(a\\) is balanced if and only if \\(F_a(u)=0\\) for all units. Moreover,\n\\[\nF_a(-u)=-F_a(u),\n\\]\nso only odd characters occur. For a Dirichlet character \\(\\chi\\) on \\(U_m\\), put\n\\[\n\\mathcal B_{m,\\chi}(r)=\n\\sum_{u\\in U_m}\\rho_m(ur)\\overline{\\chi(u)}.\n\\]\nFinite Fourier inversion gives the exact criterion\n\\begin{equation}\\label{eq:legacy-7.3}\n\\boxed{\n n(a)\\in\\mathcal B_m\n \\Longleftrightarrow\n \\sum_i\\mathcal B_{m,\\chi}(a_i)=0\n \\quad\\text{for every odd }\\chi.\n}\\end{equation}\nHere $n(a)$ is the residue count vector. This formulation underlies the prime-power filtration and the proposed conductor-defect tree.","-u)=-F_a(u),\n\\]\nso only odd characters occur. For a Dirichlet character \\(\\chi\\) on \\(U_m\\), put\n\\[\n\\mathcal B_{m,\\chi}(r)=\n\\sum_{u\\in U_m}\\rho_m(ur)\\overline{\\chi(u)}.\n\\]\nFinite Fourier inversion gives the exact criterion\n\\begin{equation}\\label{eq:legacy-7.3}\n\\boxed{\n n(a)\\in\\mathcal B_m\n \\Longleftrightarrow\n \\sum_i\\mathcal B_{m,\\chi}(a_i)=0\n \\quad\\text{for every odd }\\chi.\n}\\end{equation}\nHere $n(a)$ is the residue count vector. This formulation underlies the prime-power filtration and the proposed conductor-defect tree.","\\chapter{Exact conductor polyphase}\nFor a packet \\(a\\), define\n\\[\ng(a)=\\gcd(m,a_0,\\ldots,a_5),\n\\qquad\nc(a)=\\frac{m}{g(a)}.\n\\]\nFor a count vector, $L_k$ denotes scaling each residue entry by $k$ while multiplying the modulus by $k$. Writing \\(a=g(a)b\\), the coordinate-power map\n\\[\n\\pi_{g(a)}:X_m^4\\longrightarrow X_{c(a)}^4,\n\\qquad [x_i]\\longmapsto[x_i^{g(a)}],\n\\]\nsends the degree-\\(c(a)\\) character packet \\(b\\) to the degree-\\(m\\) packet \\(a\\).","xact conductor polyphase}\nFor a packet \\(a\\), define\n\\[\ng(a)=\\gcd(m,a_0,\\ldots,a_5),\n\\qquad\nc(a)=\\frac{m}{g(a)}.\n\\]\nFor a count vector, $L_k$ denotes scaling each residue entry by $k$ while multiplying the modulus by $k$. Writing \\(a=g(a)b\\), the coordinate-power map\n\\[\n\\pi_{g(a)}:X_m^4\\longrightarrow X_{c(a)}^4,\n\\qquad [x_i]\\longmapsto[x_i^{g(a)}],\n\\]\nsends the degree-\\(c(a)\\) character packet \\(b\\) to the degree-\\(m\\) packet \\(a\\).","hase}\nFor a packet \\(a\\), define\n\\[\ng(a)=\\gcd(m,a_0,\\ldots,a_5),\n\\qquad\nc(a)=\\frac{m}{g(a)}.\n\\]\nFor a count vector, $L_k$ denotes scaling each residue entry by $k$ while multiplying the modulus by $k$. Writing \\(a=g(a)b\\), the coordinate-power map\n\\[\n\\pi_{g(a)}:X_m^4\\longrightarrow X_{c(a)}^4,\n\\qquad [x_i]\\longmapsto[x_i^{g(a)}],\n\\]\nsends the degree-\\(c(a)\\) character packet \\(b\\) to the degree-\\(m\\) packet \\(a\\).","acket \\(a\\), define\n\\[\ng(a)=\\gcd(m,a_0,\\ldots,a_5),\n\\qquad\nc(a)=\\frac{m}{g(a)}.\n\\]\nFor a count vector, $L_k$ denotes scaling each residue entry by $k$ while multiplying the modulus by $k$. Writing \\(a=g(a)b\\), the coordinate-power map\n\\[\n\\pi_{g(a)}:X_m^4\\longrightarrow X_{c(a)}^4,\n\\qquad [x_i]\\longmapsto[x_i^{g(a)}],\n\\]\nsends the degree-\\(c(a)\\) character packet \\(b\\) to the degree-\\(m\\) packet \\(a\\).","\\begin{theorem}[Conductor descent]\nEvery imprimitive balanced packet at degree \\(m\\) is the coordinate-power pullback of a unique full-conductor packet at its exact conductor \\(c\\mid m\\). Consequently\n\\[\n\\mathcal B_m=\\bigsqcup_{c\\mid m}L_{m/c}\\mathcal B_c^{\\mathrm{full}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nIf every occupied residue is divisible by \\(g\\), divide the six entries and the modulus by \\(g\\). Since\n\\[\n\\langle ugb_i\\rangle_{gc}=g\\langle \\bar u b_i\\rangle_c,\n\\]\nthe balance equation descends. The maximal common divisor is unique. The converse follows by scaling.\n\\end{proof}","orem}[Conductor descent]\nEvery imprimitive balanced packet at degree \\(m\\) is the coordinate-power pullback of a unique full-conductor packet at its exact conductor \\(c\\mid m\\). Consequently\n\\[\n\\mathcal B_m=\\bigsqcup_{c\\mid m}L_{m/c}\\mathcal B_c^{\\mathrm{full}}.\n\\]\n\\end{theorem}\n\\begin{proof}\nIf every occupied residue is divisible by \\(g\\), divide the six entries and the modulus by \\(g\\). Since\n\\[\n\\langle ugb_i\\rangle_{gc}=g\\langle \\bar u b_i\\rangle_c,\n\\]\nthe balance equation descends. The maximal common divisor is unique. The converse follows by scaling.\n\\end{proof}","For \\(d\\mid m\\), let\n\\[\nE_{m,d}=\\frac1{(m/d)^5}\\pi_{m/d}^*\\pi_{m/d*}.\n\\]\nOn a character of exact conductor \\(c\\),\n\\[\nE_{m,d}=\\begin{cases}1,&c\\mid d,","For \\(d\\mid m\\), let\n\\[\nE_{m,d}=\\frac1{(m/d)^5}\\pi_{m/d}^*\\pi_{m/d*}.\n\\]\nOn a character of exact conductor \\(c\\),\n\\[\nE_{m,d}=\\begin{cases}1,&c\\mid d,\\\\0,&c\\nmid d.\\end{cases}\n\\]\nTherefore\n\\begin{equation}\\label{eq:legacy-8.1}\n\\boxed{E_{m,d}E_{m,e}=E_{m,\\gcd(d,e)}.}\\end{equation}\nM\\\"obius inversion produces pairwise orthogonal exact-conductor projectors\n\\begin{equation}\\label{eq:legacy-8.2}\n\\boxed{\nQ_{m,c}=\\sum_{d\\mid c}\\mu(c/d)E_{m,d}.\n}\\end{equation}\nThey satisfy\n\\[\nQ_{m,c}Q_{m,e}=\\delta_{ce}Q_{m,c},\n\\qquad\n\\sum_{c\\mid m}Q_{m,c}=I.\n\\]\nHence\n\\begin{equation}\\label{eq:legacy-8.3}\n\\boxed{\nH^4_{\\mathrm{prim}}(X_m^4,\\Q)\n=\\bigoplus_{c\\mid m}\\operatorname{im}Q_{m,c}.\n}\\end{equation}\nEach \\(Q_{m,c}\\) is a rational linear combination of graph correspondences, so the conductor decomposition is itself algebraic.","0,&c\\nmid d.\\end{cases}\n\\]\nTherefore\n\\begin{equation}\\label{eq:legacy-8.1}\n\\boxed{E_{m,d}E_{m,e}=E_{m,\\gcd(d,e)}.}\\end{equation}\nM\\\"obius inversion produces pairwise orthogonal exact-conductor projectors\n\\begin{equation}\\label{eq:legacy-8.2}\n\\boxed{\nQ_{m,c}=\\sum_{d\\mid c}\\mu(c/d)E_{m,d}.\n}\\end{equation}\nThey satisfy\n\\[\nQ_{m,c}Q_{m,e}=\\delta_{ce}Q_{m,c},\n\\qquad\n\\sum_{c\\mid m}Q_{m,c}=I.\n\\]\nHence\n\\begin{equation}\\label{eq:legacy-8.3}\n\\boxed{\nH^4_{\\mathrm{prim}}(X_m^4,\\Q)\n=\\bigoplus_{c\\mid m}\\operatorname{im}Q_{m,c}.\n}\\end{equation}\nEach \\(Q_{m,c}\\) is a rational linear combination of graph correspondences, so the conductor decomposition is itself algebraic.","end{cases}\n\\]\nTherefore\n\\begin{equation}\\label{eq:legacy-8.1}\n\\boxed{E_{m,d}E_{m,e}=E_{m,\\gcd(d,e)}.}\\end{equation}\nM\\\"obius inversion produces pairwise orthogonal exact-conductor projectors\n\\begin{equation}\\label{eq:legacy-8.2}\n\\boxed{\nQ_{m,c}=\\sum_{d\\mid c}\\mu(c/d)E_{m,d}.\n}\\end{equation}\nThey satisfy\n\\[\nQ_{m,c}Q_{m,e}=\\delta_{ce}Q_{m,c},\n\\qquad\n\\sum_{c\\mid m}Q_{m,c}=I.\n\\]\nHence\n\\begin{equation}\\label{eq:legacy-8.3}\n\\boxed{\nH^4_{\\mathrm{prim}}(X_m^4,\\Q)\n=\\bigoplus_{c\\mid m}\\operatorname{im}Q_{m,c}.\n}\\end{equation}\nEach \\(Q_{m,c}\\) is a rational linear combination of graph correspondences, so the conductor decomposition is itself algebraic.","rsion produces pairwise orthogonal exact-conductor projectors\n\\begin{equation}\\label{eq:legacy-8.2}\n\\boxed{\nQ_{m,c}=\\sum_{d\\mid c}\\mu(c/d)E_{m,d}.\n}\\end{equation}\nThey satisfy\n\\[\nQ_{m,c}Q_{m,e}=\\delta_{ce}Q_{m,c},\n\\qquad\n\\sum_{c\\mid m}Q_{m,c}=I.\n\\]\nHence\n\\begin{equation}\\label{eq:legacy-8.3}\n\\boxed{\nH^4_{\\mathrm{prim}}(X_m^4,\\Q)\n=\\bigoplus_{c\\mid m}\\operatorname{im}Q_{m,c}.\n}\\end{equation}\nEach \\(Q_{m,c}\\) is a rational linear combination of graph correspondences, so the conductor decomposition is itself algebraic.","\\chapter{Algebraic certificate grammar}\nFix a full-conductor orbit in \\(\\mathcal B_m\\). The constructive program tests a hierarchy of increasingly general certificate channels.","\\section{Complementary pairs}\nFor \\(1\\le r<m\\), let\n\\[\np_r=e_r+e_{m-r}.\n\\]\nA packet is pair-decomposable when\n\\begin{equation}\\label{eq:legacy-9.1}\nn=p_{r_1}+p_{r_2}+p_{r_3}.\\end{equation}\nThese are the standard linear-cycle or Tate-pair packets.","\\section{Balanced four-packets and saturation}\nLet\n\\[\n\\mathcal F_m=\n\\left\\{\n\\gamma\\in\\mathbf N^{m-1}:\n\\mathbf1^T\\gamma=4,\n\\quad R_m\\gamma=2m\\,\\mathbf1\n\\right\\}.\n\\]\nA saturated certificate is an exact nonnegative identity\n\\begin{equation}\\label{eq:legacy-9.2}\n\\boxed{\n n+\\sum_{j=1}^{2N-3}p_{r_j}\n =\\sum_{i=1}^N\\gamma_i,\n \\qquad \\gamma_i\\in\\mathcal F_m.\n}\\end{equation}\nThe case \\(N=2\\) is one-step quasi-decomposition. Higher \\(N\\) allows several surface packets to reconstruct one six-packet after Tate stabilization.","ets and saturation}\nLet\n\\[\n\\mathcal F_m=\n\\left\\{\n\\gamma\\in\\mathbf N^{m-1}:\n\\mathbf1^T\\gamma=4,\n\\quad R_m\\gamma=2m\\,\\mathbf1\n\\right\\}.\n\\]\nA saturated certificate is an exact nonnegative identity\n\\begin{equation}\\label{eq:legacy-9.2}\n\\boxed{\n n+\\sum_{j=1}^{2N-3}p_{r_j}\n =\\sum_{i=1}^N\\gamma_i,\n \\qquad \\gamma_i\\in\\mathcal F_m.\n}\\end{equation}\nThe case \\(N=2\\) is one-step quasi-decomposition. Higher \\(N\\) allows several surface packets to reconstruct one six-packet after Tate stabilization.","The algebraic implication must use the actual classical correspondence operations. A balanced four-entry packet is algebraic by Lefschetz $(1,1)$ on a Fermat surface. Concatenate the $N$ surface characters using the Shioda--Ran construction, then remove precisely the $2N-3$ complementary pairs by the classical pair-removal theorem. The ambient dimension rises to $4N-2$ before cancellation. This proves algebraicity of a packet admitting such an identity once those cited operations and their character conventions have been checked. It does not authorize cancellation of arbitrary tensor factors. The August 15 status note correctly demanded a complete correspondence-level account of the proposed general grammar. The concrete degree-210 application below spells out the necessary stabilization and uses only the checked pair-removal theore","The algebraic implication must use the actual classical correspondence operations. A balanced four-entry packet is algebraic by Lefschetz $(1,1)$ on a Fermat surface. Concatenate the $N$ surface characters using the Shioda--Ran construction, then remove precisely the $2N-3$ complementary pairs by the classical pair-removal theorem. The ambient dimension rises to $4N-2$ before cancellation. This proves algebraicity of a packet admitting such an identity once those cited operations and their character conventions have been checked. It does not authorize cancellation of arbitrary tensor factors. The August 15 status note correctly demanded a complete correspondence-level account of the proposed general grammar. The concrete degree-210 application below spells out the necessary stabilization and uses only the checked pair-removal theorem.","\\section{Dual charges}\nAn antisymmetric charge is a rational function \\(w\\) on residues satisfying\n\\[\nw(m-r)=-w(r)\n\\]\nand\n\\[\n\\sum_r\\gamma_r w(r)=0\n\\qquad\\forall \\gamma\\in\\mathcal F_m.\n\\]\nSuch a charge vanishes on every complementary pair and on every saturated identity. Writing $w(n)=\\sum_r n_r w(r)$, if \\(w(n)\\ne0\\), the entire lower-packet saturation semigroup is excluded. This is a universal obstruction to a grammar, not a nonalgebraicity theorem.","\\section{Curve, Delsarte, and incidence channels}\nA curve certificate splits the six entries into triples\n\\[\na=A\\sqcup B,\n\\qquad \\sum A=m,\n\\quad \\sum B=2m,\n\\quad B=uA\\pmod m.\n\\]\nThe resulting \\((1,1)\\) class on a product of Fermat curves is algebraic by Lefschetz \\((1,1)\\), and the join correspondence returns a surface class.","A Delsarte certificate consists of an exponent matrix and a resolved monomial map to a lower-degree hypersurface carrying an already algebraic packet. An incidence certificate uses algebraic lowering maps\n\\[\nA_r:V\\to H^2(Y_r)\n\\]\nwhose combined positive transfer\n\\[\nT=\\sum_rA_r^\\dagger A_r\n\\]\nis invertible on the packet. Divisor lowering and Cayley--Hamilton reconstruction then produce the cycle.","\\section{The exceptional packet}\nA representative is\n\\[\na=(7,10,13,19,22,28),\n\\]\nor after permutation,\n\\[\n\\widetilde a=(7,19,28,10,13,22).\n\\]\nEvery unit conjugate has residue sum \\(99=3\\cdot33\\), so the orbit is rational of type \\((2,2)\\). It is not totally pair-decomposable. The previously proposed symmetric surface\n\\[\nW=\\{p_1(x^{11})=p_2(x^{11})=p_3(x^{11})=0\\}\n\\]\ndoes not carry the primitive packet: Newton's identity makes the third equation redundant on \\(X_{33}^4\\), and\n\\[\n[W]=242h^2.\n\\]\nThus \\(W\\) is purely ambient.","13,19,22,28),\n\\]\nor after permutation,\n\\[\n\\widetilde a=(7,19,28,10,13,22).\n\\]\nEvery unit conjugate has residue sum \\(99=3\\cdot33\\), so the orbit is rational of type \\((2,2)\\). It is not totally pair-decomposable. The previously proposed symmetric surface\n\\[\nW=\\{p_1(x^{11})=p_2(x^{11})=p_3(x^{11})=0\\}\n\\]\ndoes not carry the primitive packet: Newton's identity makes the third equation redundant on \\(X_{33}^4\\), and\n\\[\n[W]=242h^2.\n\\]\nThus \\(W\\) is purely ambient.","r after permutation,\n\\[\n\\widetilde a=(7,19,28,10,13,22).\n\\]\nEvery unit conjugate has residue sum \\(99=3\\cdot33\\), so the orbit is rational of type \\((2,2)\\). It is not totally pair-decomposable. The previously proposed symmetric surface\n\\[\nW=\\{p_1(x^{11})=p_2(x^{11})=p_3(x^{11})=0\\}\n\\]\ndoes not carry the primitive packet: Newton's identity makes the third equation redundant on \\(X_{33}^4\\), and\n\\[\n[W]=242h^2.\n\\]\nThus \\(W\\) is purely ambient.","\\widetilde a=(7,19,28,10,13,22).\n\\]\nEvery unit conjugate has residue sum \\(99=3\\cdot33\\), so the orbit is rational of type \\((2,2)\\). It is not totally pair-decomposable. The previously proposed symmetric surface\n\\[\nW=\\{p_1(x^{11})=p_2(x^{11})=p_3(x^{11})=0\\}\n\\]\ndoes not carry the primitive packet: Newton's identity makes the third equation redundant on \\(X_{33}^4\\), and\n\\[\n[W]=242h^2.\n\\]\nThus \\(W\\) is purely ambient.","\\section{Klein--Delsarte cubic}\nConsider the smooth cubic fourfold\n\\[\nY:\\quad y_0^2y_4+y_1^2y_0+y_2^2y_1+y_3^2y_2+y_4^2y_3+y_5^3=0.\n\\]\nAn explicit monomial exponent matrix gives a dominant rational map\n\\[\n\\phi:X_{33}^4\\dashrightarrow Y\n\\]\nwhose projective-torus lattice map has Smith diagonal\n\\[\n(1,33,33,33,33),\n\\]\nso the generic torus degree is \\(33^4\\).","Take the cubic Jacobian-ring class\n\\[\nP(y)=y_1y_2y_5.\n\\]\nIts exponent vector is $b=(0,1,1,0,0,1)$. Write $C=33A^{-1}$ for the cubic exponent matrix $A$. The class is nonzero in the degree-three Jacobian ring. The residue pullback has the exact exponent identity\n\\begin{equation}\\label{eq:legacy-10.1}\nC^Tb+10\\mathbf1=(6,18,27,9,12,21)=\\widetilde a-\\mathbf1.\\end{equation}\nTherefore the pullback of the corresponding primitive cubic \\((2,2)\\)-class lies in the one-dimensional Fermat line \\(V(\\widetilde a)\\).","\\[\nP(y)=y_1y_2y_5.\n\\]\nIts exponent vector is $b=(0,1,1,0,0,1)$. Write $C=33A^{-1}$ for the cubic exponent matrix $A$. The class is nonzero in the degree-three Jacobian ring. The residue pullback has the exact exponent identity\n\\begin{equation}\\label{eq:legacy-10.1}\nC^Tb+10\\mathbf1=(6,18,27,9,12,21)=\\widetilde a-\\mathbf1.\\end{equation}\nTherefore the pullback of the corresponding primitive cubic \\((2,2)\\)-class lies in the one-dimensional Fermat line \\(V(\\widetilde a)\\).","0,0,1)$. Write $C=33A^{-1}$ for the cubic exponent matrix $A$. The class is nonzero in the degree-three Jacobian ring. The residue pullback has the exact exponent identity\n\\begin{equation}\\label{eq:legacy-10.1}\nC^Tb+10\\mathbf1=(6,18,27,9,12,21)=\\widetilde a-\\mathbf1.\\end{equation}\nTherefore the pullback of the corresponding primitive cubic \\((2,2)\\)-class lies in the one-dimensional Fermat line \\(V(\\widetilde a)\\).","The Laurent exponent match is a useful exact certificate, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply t","The Laurent exponent match is a useful exact certificate, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned cla","e Laurent exponent match is a useful exact certificate, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","h is a useful exact certificate, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","eful exact certificate, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","e, but by itself it does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","does not prove that a rational class lifts through a compact graph. A complete manuscript argument instead uses the torus-carrier criterion developed in Chapter~\\ref{ch:carrier70}. For the reordered character, the exponent matrix $A$ of the cubic satisfies\n\\[\nA^T\\widetilde a/33=\\beta=(1,2,2,1,1,2),\\qquad C=33A^{-1}.\n\\]\nWith $r_0$ the first row of $A$, put $u=\\beta-3r_0$. Then $\\mathbf1^Tu=0$ and $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","nd $C^Tu=\\widetilde a-99e_0$. These identities establish triviality of the selected character on the torus deck group. Every selected character has full support, so its restriction to the coordinate torus is an isomorphism on the packet. Finite étale descent gives a pure rational weight-four Hodge substructure on the cubic torus. The lowest-weight image theorem and semisimplicity give a rational $(2,2)$ lift to the smooth cubic. Classical cubic algebraicity then supplies a surface class. Pull back through a resolved graph, apply the rational orbit projector, and use boundary injectivity to identify the returned class.","\\begin{manuscriptclaim}[Degree-33 packet construction]\nSubject to the resolved graph and the cited mixed-Hodge inputs in the preceding argument, the rational packet generated by $(7,10,13,19,22,28)$ is algebraic on $X_{33}^4$. The exponent identities are finite exact certificates. The compact geometric implication is a manuscript argument and is not certified by those identities alone.\n\\end{manuscriptclaim}","\\begin{theorem}[Infinite packet tower]\nLet \\(W\\) be an algebraic rational character packet on \\(X_d^{2p}\\). For every \\(k\\ge1\\), its coordinate-power pullback is an algebraic packet on \\(X_{kd}^{2p}\\). Applied to the preceding degree-33 construction under its geometric hypotheses, this gives a 20-dimensional algebraic packet on every $X_{33k}^4$.\n\\end{theorem}","The tower packet remains an old conductor-33 sector. A stronger construction appends a complementary pair\n\\[\n(1,m-1)\n\\]\nafter scaling a solved seed to modulus \\(m=kd\\). If\n\\[\nb=(b_0,\\ldots,b_{2p+1})\n\\]\nis algebraic at degree \\(d\\), then\n\\[\n\\Theta_{m,b}=(kb_0,\\ldots,kb_{2p+1},1,m-1)\n\\]\nhas full raw conductor \\(m\\) and middle Hodge type two dimensions higher. Its rational orbit has dimension \\(\\varphi(m)\\). This produces full-raw-conductor algebraic packets of unbounded rank.","\\section{Valuation strata and orbit mass}\nLet \\(m=p^r\\) with \\(p\\) odd. Decompose the entries by exact \\(p\\)-adic valuation. At one stratum, write \\(a=p^jb\\), with \\(b\\in U_{p^{r-j}}\\), and define the odd count\n\\[\n\\delta_j(b)=n_j(b)-n_j(-b).\n\\]\nSince the packet has six entries,\n\\begin{equation}\\label{eq:legacy-12.1}\n\\|\\delta_j\\|_1\\le 12.\\end{equation}\nFor $s\\ge2$, primitive odd characters force \\(\\delta_j\\) to be constant on the kernel cosets of\n\\[\nU_{p^s}\\to U_{p^{s-1}}.\n\\]\nA nonzero odd kernel-orbit defect has mass at least \\(2p\\).","r\\) with \\(p\\) odd. Decompose the entries by exact \\(p\\)-adic valuation. At one stratum, write \\(a=p^jb\\), with \\(b\\in U_{p^{r-j}}\\), and define the odd count\n\\[\n\\delta_j(b)=n_j(b)-n_j(-b).\n\\]\nSince the packet has six entries,\n\\begin{equation}\\label{eq:legacy-12.1}\n\\|\\delta_j\\|_1\\le 12.\\end{equation}\nFor $s\\ge2$, primitive odd characters force \\(\\delta_j\\) to be constant on the kernel cosets of\n\\[\nU_{p^s}\\to U_{p^{s-1}}.\n\\]\nA nonzero odd kernel-orbit defect has mass at least \\(2p\\).","ratum, write \\(a=p^jb\\), with \\(b\\in U_{p^{r-j}}\\), and define the odd count\n\\[\n\\delta_j(b)=n_j(b)-n_j(-b).\n\\]\nSince the packet has six entries,\n\\begin{equation}\\label{eq:legacy-12.1}\n\\|\\delta_j\\|_1\\le 12.\\end{equation}\nFor $s\\ge2$, primitive odd characters force \\(\\delta_j\\) to be constant on the kernel cosets of\n\\[\nU_{p^s}\\to U_{p^{s-1}}.\n\\]\nA nonzero odd kernel-orbit defect has mass at least \\(2p\\).","At $s=1$, all odd characters modulo $p$ are primitive. Their sawtooth Fourier coefficients are nonzero, so the odd count vector has every Fourier coefficient zero and is zero. This gives complementary pairing at the base prime level. The higher-level mass argument applies only for $s\\ge2$. The needed odd-character nonvanishing is a classical analytic leaf, checked through Sharifi's cyclotomic-field notes \\cite{SharifiDirichlet}.","\\section{Large odd primes}\nIf \\(p\\ge7\\), then \\(2p>12\\), so every odd defect vanishes. Each residue occurs with the same multiplicity as its complement. Therefore every packet is three complementary pairs.","\\section{The three-primary branch}\nFor $p=3$ and $r\\ge2$, a higher-level kernel orbit has three elements. Define\n\\[\nO_b=\\{b,b+q,b+2q\\},\n\\qquad q=3^{r-1},\n\\]\nand\n\\[\n\\Gamma_b=O_b\\sqcup\\{3(q-b)\\}.\n\\]\nTake $1\\le b<q$ with $3\\nmid b$, ensuring all entries are nonzero. Then \\(\\Gamma_b\\) is a balanced four-packet. A case analysis on the number of oriented kernel orbits gives exact identities expressing every full-conductor nonpair packet as a pair/surface packet or a one-step saturated sum of two balanced four-packets.","\\section{The five-primary branch}\nFor $p=5$ and $r\\ge2$, a nonzero odd defect consumes exactly five unit entries. The base case $r=1$ was already handled by the base-prime Fourier argument. The sixth entry is forced by balance. Up to the unit action there is one full-conductor exceptional orbit\n\\begin{equation}\\label{eq:legacy-12.2}\n\\boxed{\n\\alpha_{5^r}=\n(1,1+q,1+2q,1+3q,1+4q,5q-5),\n\\qquad q=5^{r-1}.\n}\\end{equation}\nAll other packets lie in the pair/surface grammar. The family \\eqref{eq:legacy-12.2} is the pure five-primary Hasse--Davenport branch.","\\chapter{Weighted dyadic reconstruction}\nPrime \\(2\\) cannot be handled by inserting \\(p=2\\) into the odd-prime orbit-mass proof. The unit group has the separate form\n\\[\n\\mathbf Z_2^\\times\\cong\\{\\pm1\\}\\times(1+4\\mathbf Z_2).\n\\]\nThe first address is a sign packet, only afterward does the principal-unit refinement begin.","Let $s\\ge3$, \\(m=2^s\\), \\(h=2^{s-1}\\), and write $\\rho_s=\\rho_{2^s}$. For an odd residue \\(b\\pmod h\\), its two lifts are \\(b\\) and \\(b+h\\). Define\n\\[\nL_sn(b)=n(b)+n(b+h),\n\\qquad\nD_sn(b)=n(b)-n(b+h).\n\\]\nFor every odd unit \\(u\\), one has the exact weighted identity\n\\begin{equation}\\label{eq:legacy-13.1}\n\\boxed{\n\\begin{aligned}\n&n(b)\\rho_s(ub)+n(b+h)\\rho_s(u(b+h))","Let $s\\ge3$, \\(m=2^s\\), \\(h=2^{s-1}\\), and write $\\rho_s=\\rho_{2^s}$. For an odd residue \\(b\\pmod h\\), its two lifts are \\(b\\) and \\(b+h\\). Define\n\\[\nL_sn(b)=n(b)+n(b+h),\n\\qquad\nD_sn(b)=n(b)-n(b+h).\n\\]\nFor every odd unit \\(u\\), one has the exact weighted identity\n\\begin{equation}\\label{eq:legacy-13.1}\n\\boxed{\n\\begin{aligned}\n&n(b)\\rho_s(ub)+n(b+h)\\rho_s(u(b+h))\\\\\n&\\qquad=L_sn(b)\\rho_{s-1}(\\bar ub)\n+2^{s-2}D_sn(b)\\varepsilon_s(u,b),\n\\end{aligned}\n}\\end{equation}\nwhere \\(\\varepsilon_s(u,b)\\in\\{\\pm1\\}\\). Repetition yields a Haar-like expansion whose detail packets carry visible powers of two. The Hodge balance equation becomes a binary carry relation between weighted details.","&\\qquad=L_sn(b)\\rho_{s-1}(\\bar ub)\n+2^{s-2}D_sn(b)\\varepsilon_s(u,b),\n\\end{aligned}\n}\\end{equation}\nwhere \\(\\varepsilon_s(u,b)\\in\\{\\pm1\\}\\). Repetition yields a Haar-like expansion whose detail packets carry visible powers of two. The Hodge balance equation becomes a binary carry relation between weighted details.","Exact enumeration gives:\n\\begin{center}\n\\small\n\\input{tables/dyadic_classification.tex}\n\\end{center}\nThe table suggests one residual orbit beginning at \\(m=32\\). The remaining derivation proposes an all-degree explanation. Its uniform status was downgraded by the later August 15 note, so this edition preserves the argument as a manuscript claim requiring a complete audit.","\\section{Complement defects and dyadic triangles}\nPut\n\\[\nq=\\frac m4,\n\\qquad h=2q,\n\\qquad m=4q.\n\\]\nFor a packet count vector \\(n\\), define its oriented complement defect\n\\[\n\\partial n=\\sum_{r=1}^{h-1}d_n(r)[r],\n\\qquad\nd_n(r)=n_r-n_{m-r}.\n\\]\nThe self-complementary residue \\(h\\) contributes zero. Since\n\\(\\rho_m(m-r)=-\\rho_m(r)\\), the balanced-packet equation is equivalent to\n\\[\n\\sum_{r=1}^{h-1}d_n(r)\\rho_m(ur)=0\n\\qquad\\text{for every odd unit }u.\n\\]\nMoreover,\n\\[\n\\|\\partial n\\|_1=\\sum_{r=1}^{h-1}|d_n(r)|\\le 6.\n\\]\nThus the dyadic problem is a short-vector problem in an integral defect lattice.","c m4,\n\\qquad h=2q,\n\\qquad m=4q.\n\\]\nFor a packet count vector \\(n\\), define its oriented complement defect\n\\[\n\\partial n=\\sum_{r=1}^{h-1}d_n(r)[r],\n\\qquad\nd_n(r)=n_r-n_{m-r}.\n\\]\nThe self-complementary residue \\(h\\) contributes zero. Since\n\\(\\rho_m(m-r)=-\\rho_m(r)\\), the balanced-packet equation is equivalent to\n\\[\n\\sum_{r=1}^{h-1}d_n(r)\\rho_m(ur)=0\n\\qquad\\text{for every odd unit }u.\n\\]\nMoreover,\n\\[\n\\|\\partial n\\|_1=\\sum_{r=1}^{h-1}|d_n(r)|\\le 6.\n\\]\nThus the dyadic problem is a short-vector problem in an integral defect lattice.","For \\(1\\le b<q\\), define the triangle defect\n\\[\n\\kappa_b=[2b]-[b]+[h-b].\n\\]\nIt is realized by the four-packet\n\\[\n\\Gamma_b=(2b,h-b,h,m-b).\n\\]","\\begin{manuscriptclaim}[Dyadic triangle relation]\nFor every odd unit \\(u\\) and every \\(1\\le b<q\\),\n\\[\n\\rho_m(2ub)-\\rho_m(ub)+\\rho_m(u(h-b))=0.\n\\]\nConsequently \\(\\kappa_b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","riptclaim}[Dyadic triangle relation]\nFor every odd unit \\(u\\) and every \\(1\\le b<q\\),\n\\[\n\\rho_m(2ub)-\\rho_m(ub)+\\rho_m(u(h-b))=0.\n\\]\nConsequently \\(\\kappa_b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","triangle relation]\nFor every odd unit \\(u\\) and every \\(1\\le b<q\\),\n\\[\n\\rho_m(2ub)-\\rho_m(ub)+\\rho_m(u(h-b))=0.\n\\]\nConsequently \\(\\kappa_b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","ion]\nFor every odd unit \\(u\\) and every \\(1\\le b<q\\),\n\\[\n\\rho_m(2ub)-\\rho_m(ub)+\\rho_m(u(h-b))=0.\n\\]\nConsequently \\(\\kappa_b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","dd unit \\(u\\) and every \\(1\\le b<q\\),\n\\[\n\\rho_m(2ub)-\\rho_m(ub)+\\rho_m(u(h-b))=0.\n\\]\nConsequently \\(\\kappa_b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","b\\) is balanced and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","d and \\(\\Gamma_b\\in\\mathcal F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","l F_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nLet \\(x=\\langle ub\\rangle_m\\). If \\(0<x<h\\), then\n\\[\n\\rho_m(2x)=2x-h,\n\\qquad\n\\rho_m(x)=x-h,\n\\qquad\n\\rho_m(h-x)=-x.\n\\]\nTheir indicated combination is zero. If \\(h<x<m\\), write \\(x=m-y\\) with \\(0<y<h\\), then\n\\[\n\\rho_m(2x)=h-2y,\n\\qquad\n\\rho_m(x)=h-y,\n\\qquad\n\\rho_m(h-x)=y,\n\\]\nand the same cancellation occurs. Since \\(u\\) is odd, \\(uh\\equiv h\\pmod m\\), so \\(u(h-b)\\equiv h-ub\\pmod m\\). This proves the identity.\n\\end{proof}","\\section{The integral triangle basis}\nLet\n\\[\n\\mathcal K_m=\n\\left\\{\n d\\in\\mathbf Z^{h-1}:\n \\sum_{r=1}^{h-1}d(r)\\rho_m(ur)=0\n \\text{ for every odd }u\n\\right\\}.\n\\]\nThe triangle relations do not merely generate examples, they form the whole balanced defect lattice.","\\begin{manuscriptclaim}[Integral dyadic triangle basis]\nFor \\(m=2^s\\), \\(s\\ge3\\), the vectors\n\\[\n\\{\\kappa_b:1\\le b<q\\}\n\\]\nform a \\(\\mathbf Z\\)-basis of \\(\\mathcal K_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nSet \\(r_0=q/2\\). On the vertex set\n\\[\nV_m=\\{1,2,\\ldots,q-1\\},\n\\]\ndefine, for \\(b\\ne r_0\\),\n\\[\np(b)=\n\\begin{cases}\n2b,&b<r_0,","\\begin{manuscriptclaim}[Integral dyadic triangle basis]\nFor \\(m=2^s\\), \\(s\\ge3\\), the vectors\n\\[\n\\{\\kappa_b:1\\le b<q\\}\n\\]\nform a \\(\\mathbf Z\\)-basis of \\(\\mathcal K_m\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nSet \\(r_0=q/2\\). On the vertex set\n\\[\nV_m=\\{1,2,\\ldots,q-1\\},\n\\]\ndefine, for \\(b\\ne r_0\\),\n\\[\np(b)=\n\\begin{cases}\n2b,&b<r_0,\\\\\n2(q-b),&b>r_0,\n\\end{cases}\n\\qquad\n\\epsilon_b=\n\\begin{cases}\n+1,&b<r_0,\\\\\n-1,&b>r_0.\n\\end{cases}\n\\]\nThe map \\(p\\) raises the two-adic valuation by one, so repeated application reaches \\(r_0\\). It therefore defines a rooted tree.","-1,&b>r_0.\n\\end{cases}\n\\]\nThe map \\(p\\) raises the two-adic valuation by one, so repeated application reaches \\(r_0\\). It therefore defines a rooted tree.","Suppose \\(d=\\sum_{b=1}^{q-1}z_b\\kappa_b\\). Its even coordinates are\n\\[\nd(q)=z_{r_0},\n\\qquad\nd(2b)=z_b-\\epsilon_bz_{p(b)}\n\\quad(b\\ne r_0).\n\\]\nStarting at the root and moving toward the leaves, these equations recover every \\(z_b\\) integrally. Hence the triangle vectors are linearly independent and their even-coordinate map is unimodular.","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, th","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\)","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,Sha","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathc","Now take \\(d\\in\\mathcal K_m\\), recover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).","ecover the unique integers \\(z_b\\) from its even coordinates, and subtract \\(\\sum_bz_b\\kappa_b\\). The residual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","sidual balanced vector is supported only on the odd residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","residues \\(1\\le r<h\\). The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","The corresponding square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","ing square sawtooth matrix is nonsingular. Indeed, extend an odd-support vector to an odd function on \\(U_m\\). Convolution with \\(\\rho_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","_m\\) is diagonalized by the odd Dirichlet characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","characters, with eigenvalues\n\\[\n\\lambda_\\chi=\n\\sum_{a\\in U_m}\\rho_m(a)\\overline{\\chi(a)}\n=\n\\sum_{a\\in U_m}a\\overline{\\chi(a)}.\n\\]\nIf \\(\\chi\\) is induced from a primitive odd character \\(\\chi^*\\) of conductor \\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","\\(2^t\\), then\n\\[\n\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","\\lambda_{\\chi,2^s}=2^{s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","s-t}\\lambda_{\\chi^*,2^t}.\n\\]\nThe primitive value is, up to a nonzero scalar, the generalized Bernoulli value \\(B_{1,\\overline{\\chi^*}}\\), equivalently \\(-L(0,\\overline{\\chi^*})\\), and is nonzero for every odd primitive character \\cite{WashingtonCyclotomic,SharifiDirichlet}. Thus every odd Fourier eigenvalue is nonzero. The odd residual must vanish, so the \\(\\kappa_b\\) span \\(\\mathcal K_m\\) over \\(\\mathbf Z\\).\n\\end{proof}","\\section{The trivalent defect tree}\nThe integral basis turns defect mass into exact tree variation. Define a sign gauge by\n\\[\n\\tau_{r_0}=1,\n\\qquad\n\\tau_b=\\epsilon_b\\tau_{p(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","tree variation. Define a sign gauge by\n\\[\n\\tau_{r_0}=1,\n\\qquad\n\\tau_b=\\epsilon_b\\tau_{p(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","ne a sign gauge by\n\\[\n\\tau_{r_0}=1,\n\\qquad\n\\tau_b=\\epsilon_b\\tau_{p(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","gauge by\n\\[\n\\tau_{r_0}=1,\n\\qquad\n\\tau_b=\\epsilon_b\\tau_{p(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","0}=1,\n\\qquad\n\\tau_b=\\epsilon_b\\tau_{p(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","(b)},\n\\qquad\ny_b=\\tau_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","au_bz_b.\n\\]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","]\nAttach one zero-valued boundary leaf to the root and two zero-valued boundary leaves to every odd vertex. Every vertex of the resulting augmented tree \\(\\widetilde T_m\\) has degree three. The boundary edges correspond bijectively to defect coordinates: the root edge is labelled \\(q\\), the edge from \\(b\\) to \\(p(b)\\) is labelled \\(2b\\), and the two leaves at an odd \\(b\\) are labelled \\(b\\) and \\(h-b\\).","\\begin{manuscriptclaim}[Exact trivalent-tree norm]\nIf \\(d=\\sum_bz_b\\kappa_b\\in\\mathcal K_m\\), then\n\\begin{equation}\\label{eq:legacy-13.2}\n\\boxed{\n\\|d\\|_1\n=|y_{r_0}|\n+\\sum_{b\\ne r_0}|y_b-y_{p(b)}|\n+2\\sum_{\\substack{1\\le b<q","\\begin{manuscriptclaim}[Exact trivalent-tree norm]\nIf \\(d=\\sum_bz_b\\kappa_b\\in\\mathcal K_m\\), then\n\\begin{equation}\\label{eq:legacy-13.2}\n\\boxed{\n\\|d\\|_1\n=|y_{r_0}|\n+\\sum_{b\\ne r_0}|y_b-y_{p(b)}|\n+2\\sum_{\\substack{1\\le b<q\\\\ b\\text{ odd}}}|y_b|.\n}\\end{equation}\nThe right side is the total variation of the integer function \\(y\\) on \\(\\widetilde T_m\\) with zero boundary values.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe root coordinate is \\(d(q)=z_{r_0}\\). For \\(b\\ne r_0\\), multiplying\n\\(d(2b)=z_b-\\epsilon_bz_{p(b)}\\) by \\(\\tau_b\\) gives\n\\[\n\\tau_bd(2b)=y_b-y_{p(b)}.\n\\]\nFor odd \\(b\\), the two odd coordinates are\n\\[\nd(b)=-z_b,\n\\qquad\nd(h-b)=z_b.\n\\]\nTaking absolute values and summing all defect coordinates gives \\eqref{eq:legacy-13.2}.\n\\end{proof}","b\\text{ odd}}}|y_b|.\n}\\end{equation}\nThe right side is the total variation of the integer function \\(y\\) on \\(\\widetilde T_m\\) with zero boundary values.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe root coordinate is \\(d(q)=z_{r_0}\\). For \\(b\\ne r_0\\), multiplying\n\\(d(2b)=z_b-\\epsilon_bz_{p(b)}\\) by \\(\\tau_b\\) gives\n\\[\n\\tau_bd(2b)=y_b-y_{p(b)}.\n\\]\nFor odd \\(b\\), the two odd coordinates are\n\\[\nd(b)=-z_b,\n\\qquad\nd(h-b)=z_b.\n\\]\nTaking absolute values and summing all defect coordinates gives \\eqref{eq:legacy-13.2}.\n\\end{proof}","For an integer-valued \\(y\\), discrete coarea gives\n\\begin{equation}\\label{eq:legacy-13.3}\n\\operatorname{TV}(y)\n=\n\\sum_{k\\ge1}|\\partial\\{y\\ge k\\}|\n+\n\\sum_{k\\ge1}|\\partial\\{y\\le-k\\}|.\\end{equation}\nIf \\(S\\) is a finite forest of internal vertices of a trivalent tree and \\(c(S)\\) is its number of connected components, then\n\\begin{equation}\\label{eq:legacy-13.4}\n|\\partial S|=|S|+2c(S),\\end{equation}\nbecause \\(3|S|=2(|S|-c(S))+|\\partial S|\\).","teger-valued \\(y\\), discrete coarea gives\n\\begin{equation}\\label{eq:legacy-13.3}\n\\operatorname{TV}(y)\n=\n\\sum_{k\\ge1}|\\partial\\{y\\ge k\\}|\n+\n\\sum_{k\\ge1}|\\partial\\{y\\le-k\\}|.\\end{equation}\nIf \\(S\\) is a finite forest of internal vertices of a trivalent tree and \\(c(S)\\) is its number of connected components, then\n\\begin{equation}\\label{eq:legacy-13.4}\n|\\partial S|=|S|+2c(S),\\end{equation}\nbecause \\(3|S|=2(|S|-c(S))+|\\partial S|\\).","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces bot","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the sam","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each,","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has on","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k","\\begin{manuscriptclaim}[All defect shapes of mass at most six]\nLet \\(d\\in\\mathcal K_m\\) satisfy \\(\\|d\\|_1\\le6\\), and let \\(y\\) be its gauged coefficient function. Exactly one of the following occurs:\n\\begin{enumerate}\n\\item \\(y=0\\)\n\\item \\(y=\\pm2\\) at one vertex and is zero elsewhere\n\\item \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\","em \\(y\\) has two singleton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","leton components, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","ents, each with value \\(\\pm1\\)\n\\item \\(y=\\pm1\\) on one connected subtree containing at most four vertices and is zero elsewhere.\n\\end{enumerate}\nFor a balanced four-packet, only the zero shape, a single vertex, or one tree edge can occur.\n\\end{manuscriptclaim}\n\\begin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","egin{proof}\nApply \\eqref{eq:legacy-13.3}--\\eqref{eq:legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","legacy-13.4}. A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}",". A nonempty level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","level-set component with \\(k\\) vertices costs \\(k+2\\). If \\(|y|\\ge2\\), two nonempty nested level sets already cost at least six, equality forces both to be the same singleton. If positive and negative values both occur, their first level sets cost at least three each, equality forces two singleton components. Otherwise \\(y\\) has one sign, and a connected support of size \\(k\\) costs \\(k+2\\), so \\(k\\le4\\). Two same-sign components can have total cost at most six only when both are singletons. Replacing six by four leaves only a vertex or an edge.\n\\end{proof}","\\section{SAT(2) closure and the four-star obstruction}\nFor a defect \\(d\\), let \\(\\operatorname{red}(d)\\) be the unmatched residue multiset: a positive coefficient at \\(r\\) contributes copies of \\(r\\), while a negative coefficient contributes copies of \\(m-r\\). Its length is \\(\\|d\\|_1\\). Every packet with defect \\(d\\) is \\(\\operatorname{red}(d)\\) together with complementary pairs and copies of \\(h\\).","A one-vertex tree atom has defect \\(\\pm\\kappa_b\\) after the gauge sign and is represented by \\(\\Gamma_b\\) or its conjugate, its reduced length is three and its fourth entry is \\(h\\). A one-edge atom has defect mass four, so its reduced multiset itself is a balanced four-packet. A zero atom is any union of two complementary pairs.","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings l","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings lift to exact","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings lift to exact nonnegative multiset identities. For a vertex, the neutral remainder of the six-packet","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings lift to exact nonnegative multiset identities. For a vertex, the neutral remainder of the six-packet is \\(h\\) plus one complementary p","\\begin{manuscriptclaim}[Exact SAT(2) criterion]\nLet \\(n\\) be a balanced six-packet at modulus \\(m=2^s\\). Then \\(n\\) is pair-decomposable or has an SAT(2) certificate unless its gauged defect is \\(\\pm1\\) on a four-vertex star \\(K_{1,3}\\) in the internal tree \\(T_m\\). A four-star has no SAT(2) certificate.\n\\end{manuscriptclaim}\n\\begin{proof}\nEvery nonstar shape in the preceding proposition splits into at most two four-packet atoms:\n\\begin{itemize}\n\\item a vertex or edge is paired with a zero atom\n\\item a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings lift to exact nonnegative multiset identities. For a vertex, the neutral remainder of the six-packet is \\(h\\) plus one complementary pair, for an edge it is one co","a three-vertex path is an edge plus its remaining terminal vertex\n\\item a four-vertex path is the union of two disjoint edges\n\\item a doubled singleton is two copies of one vertex atom\n\\item two singleton components are the corresponding two vertex atoms.\n\\end{itemize}\nThese defect splittings lift to exact nonnegative multiset identities. For a vertex, the neutral remainder of the six-packet is \\(h\\) plus one complementary pair, for an edge it is one complementary pair. A zero four-packet absorbs that pair and the stabilizing pair. Cutting a three- or four-vertex path produces exactly one oppositely oriented boundary edge, hence exactly one complementary pair in the union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.","omplementary pair, for an edge it is one complementary pair. A zero four-packet absorbs that pair and the stabilizing pair. Cutting a three- or four-vertex path produces exactly one oppositely oriented boundary edge, hence exactly one complementary pair in the union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.\n\\]",". A zero four-packet absorbs that pair and the stabilizing pair. Cutting a three- or four-vertex path produces exactly one oppositely oriented boundary edge, hence exactly one complementary pair in the union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.\n\\]","oppositely oriented boundary edge, hence exactly one complementary pair in the union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.\n\\]","ntary pair in the union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.\n\\]","union of the two atom packets, this is the SAT(2) stabilizer. In the three-vertex case, the vertex atom also supplies the unique neutral \\(h\\). For two vertex atoms, their two neutral entries are \\((h,h)=p_h\\). The zero-defect case is already a union of three complementary pairs. Thus in every nonstar case there are \\(\\gamma_1,\\gamma_2\\in\\mathcal F_m\\) and a complementary pair \\(p_r\\) such that\n\\[\nn+p_r=\\gamma_1+\\gamma_2.\n\\]","Conversely, suppose a four-star defect had an SAT(2) certificate. Each balanced four-packet has tree support equal to zero, one vertex, or one edge. The sum of two such supports must produce four nonzero star vertices. Therefore both supports must be disjoint edges covering all four vertices. This would be a perfect matching of \\(K_{1,3}\\), which does not exist. Hence the star is the unique tree obstruction to SAT(2).\n\\end{proof}","\\begin{manuscriptclaim}[Star conductor]\nLet a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a","\\begin{manuscriptclaim}[Star conductor]\nLet a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\)","\\begin{manuscriptclaim}[Star conductor]\nLet a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entr","\\begin{manuscriptclaim}[Star conductor]\nLet a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuati","\\begin{manuscriptclaim}[Star conductor]\nLet a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","Let a four-star have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","tar have center \\(c\\), and write \\(j=v_2(c)\\). The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","The gcd of \\(m\\) with its six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","ts six packet entries is\n\\[\n2^{j-1}.\n\\]\nHence a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","a star has full conductor exactly when \\(v_2(c)=1\\).\n\\end{manuscriptclaim}\n\\begin{proof}\nUse the boundary-edge labels of \\(\\widetilde T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","e T_m\\). The two children of \\(c\\) have valuation \\(j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","j-1\\). The four boundary edges leaving those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","those children have labels of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","of exact valuation \\(j-1\\): for \\(j=1\\) they are the odd leaf labels, and for \\(j>1\\) they are labels \\(2b\\) coming from grandchildren of valuation \\(j-2\\). The two boundary edges leaving the parent side have larger valuation. Thus every nonzero defect coordinate is divisible by \\(2^{j-1}\\), and at least four have exact valuation \\(j-1\\). Replacing a defect coordinate \\(r\\) by the packet entry \\(m-r\\) does not change its two-adic valuation. The stated gcd follows.\n\\end{proof}","Assume now \\(s\\ge5\\) and let a full-conductor star have center \\(c=2a\\) with \\(a\\) odd. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus","Assume now \\(s\\ge5\\) and let a full-conductor star have center \\(c=2a\\) with \\(a\\) odd. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-o","Assume now \\(s\\ge5\\) and let a full-conductor star have center \\(c=2a\\) with \\(a\\) odd. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","ull-conductor star have center \\(c=2a\\) with \\(a\\) odd. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","have center \\(c=2a\\) with \\(a\\) odd. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","d. Up to global conjugation, expanding its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","its four triangle atoms gives the unordered packet\n\\begin{equation}\\label{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","abel{eq:legacy-13.5}\nP_a=\n(8a,q-a,2q-a,2q-4a,3q-a,4q-a).\\end{equation}\nThe cases \\(c<q/2\\) and \\(c>q/2\\) give the same unordered tuple after the internal-edge cancellations. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","ons. Choose an odd \\(u\\) satisfying\n\\[\nua\\equiv1\\pmod{2q}.\n\\]\nSince \\(uq\\equiv q\\) or \\(3q\\pmod{4q}\\), multiplication by \\(u\\) sends \\eqref{eq:legacy-13.5} to\n\\[\nP_1=(8,q-1,2q-4,2q-1,3q-1,4q-1).\n\\]\nIts conjugate is\n\\begin{equation}\\label{eq:legacy-13.6}\n\\boxed{\nA_m=\n\\left(\n1,\\frac m4+1,\\frac m2+1,\\frac m2+4,\n\\frac{3m}4+1,m-8\n\\right).\n}\\end{equation}\nThus every full-conductor star belongs to one \\(U_m\\)-orbit.","The packet \\(A_m\\) obeys the exact uniform SAT(3) identity\n\\begin{equation}\\label{eq:legacy-13.7}\n\\begin{aligned}\n&A_m+(2,m-2)+(4,m-4)+(m/2,m/2)","The packet \\(A_m\\) obeys the exact uniform SAT(3) identity\n\\begin{equation}\\label{eq:legacy-13.7}\n\\begin{aligned}\n&A_m+(2,m-2)+(4,m-4)+(m/2,m/2)\\\\\n&=(1,m/2,m/2+1,m-2)\\\\\n&\\quad +(2,m/4+1,3m/4+1,m-4)\\\\\n&\\quad +(4,m/2,m/2+4,m-8).\n\\end{aligned}\\end{equation}\nThe first and third packets on the right have defects \\(-\\kappa_1\\) and \\(-\\kappa_4\\). The middle packet has defect \\(-\\kappa_2+\\kappa_{q-1}\\), an edge atom. Hence all three are balanced four-packets, and \\eqref{eq:legacy-13.7} is an identity of multisets.","&\\quad +(4,m/2,m/2+4,m-8).\n\\end{aligned}\\end{equation}\nThe first and third packets on the right have defects \\(-\\kappa_1\\) and \\(-\\kappa_4\\). The middle packet has defect \\(-\\kappa_2+\\kappa_{q-1}\\), an edge atom. Hence all three are balanced four-packets, and \\eqref{eq:legacy-13.7} is an identity of multisets.","\\begin{manuscriptclaim}[All-degree dyadic uniqueness]\nLet \\(m=2^s\\), \\(s\\ge3\\). Every full-conductor balanced six-packet is pair-decomposable or SAT(2), except that for \\(s\\ge5\\) there is exactly one residual unit orbit, represented by \\(A_m\\). This orbit is SAT(3) by \\eqref{eq:legacy-13.7}. For \\(m=8\\) and \\(m=16\\), no four-star exists and there is no residual orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\","\\begin{manuscriptclaim}[All-degree dyadic uniqueness]\nLet \\(m=2^s\\), \\(s\\ge3\\). Every full-conductor balanced six-packet is pair-decomposable or SAT(2), except that for \\(s\\ge5\\) there is exactly one residual unit orbit, represented by \\(A_m\\). This orbit is SAT(3) by \\eqref{eq:legacy-13.7}. For \\(m=8\\) and \\(m=16\\), no four-star exists and there is no residual orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The tree","\\begin{manuscriptclaim}[All-degree dyadic uniqueness]\nLet \\(m=2^s\\), \\(s\\ge3\\). Every full-conductor balanced six-packet is pair-decomposable or SAT(2), except that for \\(s\\ge5\\) there is exactly one residual unit orbit, represented by \\(A_m\\). This orbit is SAT(3) by \\eqref{eq:legacy-13.7}. For \\(m=8\\) and \\(m=16\\), no four-star exists and there is no residual orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq","\\begin{manuscriptclaim}[All-degree dyadic uniqueness]\nLet \\(m=2^s\\), \\(s\\ge3\\). Every full-conductor balanced six-packet is pair-decomposable or SAT(2), except that for \\(s\\ge5\\) there is exactly one residual unit orbit, represented by \\(A_m\\). This orbit is SAT(3) by \\eqref{eq:legacy-13.7}. For \\(m=8\\) and \\(m=16\\), no four-star exists and there is no residual orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq:legacy-13.7} closes the unique residual orbit at saturation dept","\\begin{manuscriptclaim}[All-degree dyadic uniqueness]\nLet \\(m=2^s\\), \\(s\\ge3\\). Every full-conductor balanced six-packet is pair-decomposable or SAT(2), except that for \\(s\\ge5\\) there is exactly one residual unit orbit, represented by \\(A_m\\). This orbit is SAT(3) by \\eqref{eq:legacy-13.7}. For \\(m=8\\) and \\(m=16\\), no four-star exists and there is no residual orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq:legacy-13.7} closes the unique residual orbit at saturation depth three.\n\\end{proof}","al orbit.\n\\end{manuscriptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq:legacy-13.7} closes the unique residual orbit at saturation depth three.\n\\end{proof}","ptclaim}\n\\begin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq:legacy-13.7} closes the unique residual orbit at saturation depth three.\n\\end{proof}","gin{proof}\nThe trivalent-tree norm and coarea classification list every defect of mass at most six. The SAT(2) criterion closes every listed shape except a four-star. The star-conductor lemma forces a full-conductor star center to have valuation one, and the calculation \\eqref{eq:legacy-13.5}--\\eqref{eq:legacy-13.6} places all such stars in the single orbit of \\(A_m\\). The trees for \\(m=8\\) and \\(m=16\\) have no internal vertex with three internal neighbors. Finally, \\eqref{eq:legacy-13.7} closes the unique residual orbit at saturation depth three.\n\\end{proof}","\\begin{manuscriptclaim}[Complete dyadic grammar]\nAfter exact conductor descent, every balanced packet at every dyadic modulus is generated by complementary pairs, SAT(2) surface packets, and coordinate-power pullbacks of the single SAT(3) family \\(A_{2^t}\\). No second residual dyadic family occurs at any depth.\n\\end{manuscriptclaim}","The archived companion dyadic verification report records complete orbit counts through $m=128$ and structural identities through $2^{14}$. The October replay freshly reconstructs the complete census through 128 and the triangle/tree identities through $2^9$. The report file is preserved evidence, rather than an executable verifier or a current replay of the larger historical range. These computations support the finite identities. They do not supply a current independent proof of the all-degree claim.","\\chapter{Finite global audit through degree 100}\nThe finite audit exactly enumerates balanced packets, reduces them by unit orbits, checks pair and SAT(2) certificates, and searches higher saturated identities with \\(N\\le8\\). Every returned identity is rechecked by exact integer arithmetic.","The archived named verification groups report the following outcomes:\n\\begin{center}\n\\small\n\\input{tables/fermat_named_checks.tex}\n\\end{center}\nThe archived dyadic report records \\(89{,}478{,}480\\) coefficient checks through \\(2^{14}\\). The October replay independently checked 87,376 coefficients for exponents $3\\le s\\le9$. The larger historical count was not rerun in full. The archived packet audit reports completeness through degree \\(100\\) for its implemented grammar, which is not a completeness theorem for all correspondence channels.","port the following outcomes:\n\\begin{center}\n\\small\n\\input{tables/fermat_named_checks.tex}\n\\end{center}\nThe archived dyadic report records \\(89{,}478{,}480\\) coefficient checks through \\(2^{14}\\). The October replay independently checked 87,376 coefficients for exponents $3\\le s\\le9$. The larger historical count was not rerun in full. The archived packet audit reports completeness through degree \\(100\\) for its implemented grammar, which is not a completeness theorem for all correspondence channels.","\\chapter{Local mod-5 phase grammar}\nAssume \\(m=5q\\) and first suppose \\((5,q)=1\\). For \\(b\\pmod q\\) and a phase \\(r\\in\\{0,1,2,3,4\\}\\), let \\(\\lambda_r(b)\\) be the CRT lift satisfying\n\\[\n\\lambda_r(b)\\equiv r\\pmod5,\n\\qquad\n\\lambda_r(b)\\equiv b\\pmod q.\n\\]\nThe finite audit reveals four canonical multiplicity profiles, up to multiplication by a unit modulo five:\n\\[\n(1,0,0,0,5),\n\\quad\n(2,1,1,1,1),\n\\quad\n(1,0,2,1,2),\n\\quad\n(1,0,1,3,1).\n\\]\nThe first is the complete-five-phase K5 packet, the second is a four-phase K4 lift, and the final two are one K23 pair--triple mechanism.","\\section{K4 four-phase reduction}\nChoose nonzero residues \\(b,c,d\\pmod q\\) and put\n\\[\ne\\equiv-5^{-1}b\\pmod q.\n\\]\nDefine\n\\[\nA_{b,c,d}=\n(\\lambda_1(b),\\lambda_2(b),\\lambda_3(b),\\lambda_4(b),5c,5d)\n\\]\nand\n\\[\n\\Gamma_{b,c,d}=(b,c,d,e).\n\\]","\\begin{theorem}[Four-phase CRT reduction]\nThe six-packet \\(A_{b,c,d}\\) is balanced modulo \\(5q\\) if and only if the four-packet \\(\\Gamma_{b,c,d}\\) is balanced modulo \\(q\\).\n\\end{theorem}\n\\begin{proof}\nFor a unit \\(u\\), let \\(s=\\langle ub\\rangle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","ase CRT reduction]\nThe six-packet \\(A_{b,c,d}\\) is balanced modulo \\(5q\\) if and only if the four-packet \\(\\Gamma_{b,c,d}\\) is balanced modulo \\(q\\).\n\\end{theorem}\n\\begin{proof}\nFor a unit \\(u\\), let \\(s=\\langle ub\\rangle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","if and only if the four-packet \\(\\Gamma_{b,c,d}\\) is balanced modulo \\(q\\).\n\\end{theorem}\n\\begin{proof}\nFor a unit \\(u\\), let \\(s=\\langle ub\\rangle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","f the four-packet \\(\\Gamma_{b,c,d}\\) is balanced modulo \\(q\\).\n\\end{theorem}\n\\begin{proof}\nFor a unit \\(u\\), let \\(s=\\langle ub\\rangle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","\\).\n\\end{theorem}\n\\begin{proof}\nFor a unit \\(u\\), let \\(s=\\langle ub\\rangle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","angle_q\\) and \\(z=\\langle u5^{-1}b\\rangle_q\\). The complete five-element fiber over \\(s\\) has sum \\(5s+10q\\), its phase-zero lift is \\(5z\\). The four nonzero phases therefore sum to \\(5s+10q-5z\\). Adding the last two entries and imposing the middle value \\(15q\\) gives\n\\[\ns+\\langle uc\\rangle_q+\\langle ud\\rangle_q-z=q.\n\\]\nSince \\(\\langle ue\\rangle_q=q-z\\), this is exactly the balance equation for \\((b,c,d,e)\\).\n\\end{proof}","\\chapter{K23 pair--triple factorization}\nLet a K23 packet have one phase-zero entry and five nonzero phases:\n\\[\nA=P\\sqcup T\\sqcup\\{5e\\},\n\\qquad |P|=2,\n\\quad |T|=3,\n\\]\nwith\n\\[\n\\sum P\\equiv0\\pmod5,\n\\qquad\n\\sum T\\equiv0\\pmod5.\n\\]\nWrite\n\\[\n\\sum P=5C,\n\\qquad\n\\sum T=5D,\n\\]\nand reduce \\(C,D\\) modulo \\(q\\) to representatives \\(c,d\\). Require $c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","2,\n\\quad |T|=3,\n\\]\nwith\n\\[\n\\sum P\\equiv0\\pmod5,\n\\qquad\n\\sum T\\equiv0\\pmod5.\n\\]\nWrite\n\\[\n\\sum P=5C,\n\\qquad\n\\sum T=5D,\n\\]\nand reduce \\(C,D\\) modulo \\(q\\) to representatives \\(c,d\\). Require $c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","[\n\\sum P=5C,\n\\qquad\n\\sum T=5D,\n\\]\nand reduce \\(C,D\\) modulo \\(q\\) to representatives \\(c,d\\). Require $c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","ad\n\\sum T=5D,\n\\]\nand reduce \\(C,D\\) modulo \\(q\\) to representatives \\(c,d\\). Require $c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","entatives \\(c,d\\). Require $c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","$c,d,e$ to be nonzero modulo $q$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","$ so that all three resulting character blocks have full support. Degenerate cases must leave this construction and be handled through pair removal or another channel. For example $(1,14,2,9,14,5)$ at modulus 15 is balanced and admits the indicated phase split, but $P=(1,14)$ gives $c=0$. The phase condition alone is insufficient. Define\n\\[\nC_P=P\\sqcup\\{m-5c\\},\n\\]\n\\[\nS_T=T\\sqcup\\{m-5d\\},\n\\]\nand\n\\[\nL=(5c,5d,5e).\n\\]","For any nonzero-entry block $B$ at modulus $m$, put $h_B(u)=m^{-1}\\sum_{r\\in B}\\langle ur\\rangle_m$, with multiplicities retained.","\\begin{theorem}[Exact K23 packet identity]\nOne has the exact multiset equality\n\\begin{equation}\\label{eq:legacy-15.1}\n\\boxed{\nA+(5c,m-5c)+(5d,m-5d)=C_P+S_T+L.\n}\\end{equation}\nFor every unit \\(u\\pmod m\\),\n\\[\nh_{C_P}(u)+h_{S_T}(u)+h_L(u)=5,\n\\]\nand \\(c+d+e\\equiv0\\pmod q\\).\n\\end{theorem}","After fixing a common cyclotomic coefficient field and a diagonal character projector, write the selected factor schematically as\n\\[\nN=e_{\\mathrm{diag}}\\bigl(M(C_P)\\otimes M(S_T)\\bigr).\n\\]\nAn ordinary tensor product of rational packets contains off-diagonal embedding pairs and is generally not rank one. Rank one here refers only to a selected eigenspace after scalar extension. The coefficient field and projector are part of the unresolved correspondence specification.\nThe Hodge-level identity implies that \\(N(1)\\) has weight one and has Hodge type opposite to the lower curve packet \\(M(L)\\) at every embedding.","schematically as\n\\[\nN=e_{\\mathrm{diag}}\\bigl(M(C_P)\\otimes M(S_T)\\bigr).\n\\]\nAn ordinary tensor product of rational packets contains off-diagonal embedding pairs and is generally not rank one. Rank one here refers only to a selected eigenspace after scalar extension. The coefficient field and projector are part of the unresolved correspondence specification.\nThe Hodge-level identity implies that \\(N(1)\\) has weight one and has Hodge type opposite to the lower curve packet \\(M(L)\\) at every embedding.","ordinary tensor product of rational packets contains off-diagonal embedding pairs and is generally not rank one. Rank one here refers only to a selected eigenspace after scalar extension. The coefficient field and projector are part of the unresolved correspondence specification.\nThe Hodge-level identity implies that \\(N(1)\\) has weight one and has Hodge type opposite to the lower curve packet \\(M(L)\\) at every embedding.","\\chapter{Imaginary CM selector}\nWrite the lower curve packet as\n\\[\n\\ell=(c,d,e)\\pmod q,\n\\qquad c+d+e\\equiv0\\pmod q.\n\\]\nLet \\(K=\\Q(\\zeta_q)\\) and define\n\\begin{equation}\\label{eq:legacy-16.1}\n\\boxed{\n\\Delta_\\ell=(1-\\zeta_q^c)(1-\\zeta_q^d)(1-\\zeta_q^e).\n}\\end{equation}","\\begin{theorem}[Pure imaginary selector]\n\\[\n\\overline{\\Delta_\\ell}=-\\Delta_\\ell.\n\\]\n\\end{theorem}\n\\begin{proof}\nSince\n\\[\n1-\\zeta_q^{-a}=-\\zeta_q^{-a}(1-\\zeta_q^a),\n\\]\ncomplex conjugation multiplies the product by\n\\[\n(-1)^3\\zeta_q^{-(c+d+e)}=-1.\n\\]\n\\end{proof}","For an embedding \\(\\sigma_u:\\zeta_q\\mapsto\\zeta_q^u\\), let\n\\[\nh_\\ell(u)=\\frac{\\langle uc\\rangle_q+\\langle ud\\rangle_q+\\langle ue\\rangle_q}{q}\\in\\{1,2\\}.\n\\]\nUsing\n\\[\n1-e^{2\\pi ir/q}=-2i\\,e^{\\pi ir/q}\\sin\\frac{\\pi r}{q},\n\\]\none obtains\n\\begin{equation}\\label{eq:legacy-16.2}\n\\sigma_u(\\Delta_\\ell)\n=8i(-1)^{h_\\ell(u)}\n\\prod_{j=1}^3\n\\sin\\frac{\\pi\\langle u\\ell_j\\rangle_q}{q}.\\end{equation}\nHere $(\\ell_1,\\ell_2,\\ell_3)=(c,d,e)$ is an indexed triple, so repeated residues contribute repeated factors. Every sine factor is positive, so the sign of the imaginary part is exactly the Hodge sign. Consequently the CM type of \\(N(1)\\) is the conjugate of the CM type of the lower curve packet:\n\\begin{equation}\\label{eq:legacy-16.3}\n\\boxed{\n\\Sigma_N=\\overline{\\","For an embedding \\(\\sigma_u:\\zeta_q\\mapsto\\zeta_q^u\\), let\n\\[\nh_\\ell(u)=\\frac{\\langle uc\\rangle_q+\\langle ud\\rangle_q+\\langle ue\\rangle_q}{q}\\in\\{1,2\\}.\n\\]\nUsing\n\\[\n1-e^{2\\pi ir/q}=-2i\\,e^{\\pi ir/q}\\sin\\frac{\\pi r}{q},\n\\]\none obtains\n\\begin{equation}\\label{eq:legacy-16.2}\n\\sigma_u(\\Delta_\\ell)\n=8i(-1)^{h_\\ell(u)}\n\\prod_{j=1}^3\n\\sin\\frac{\\pi\\langle u\\ell_j\\rangle_q}{q}.\\end{equation}\nHere $(\\ell_1,\\ell_2,\\ell_3)=(c,d,e)$ is an indexed triple, so repeated residues contribute repeated factors. Every sine factor is positive, so the sign of the imaginary part is exactly the Hodge sign. Consequently the CM type of \\(N(1)\\) is the conjugate of the CM type of the lower curve packet:\n\\begin{equation}\\label{eq:legacy-16.3}\n\\boxed{\n\\Sigma_N=\\overline{\\Sigma_\\ell}.\n}\\end{equation}\nThis is the rigorous form of the statement that the half-twist occurs on","For an embedding \\(\\sigma_u:\\zeta_q\\mapsto\\zeta_q^u\\), let\n\\[\nh_\\ell(u)=\\frac{\\langle uc\\rangle_q+\\langle ud\\rangle_q+\\langle ue\\rangle_q}{q}\\in\\{1,2\\}.\n\\]\nUsing\n\\[\n1-e^{2\\pi ir/q}=-2i\\,e^{\\pi ir/q}\\sin\\frac{\\pi r}{q},\n\\]\none obtains\n\\begin{equation}\\label{eq:legacy-16.2}\n\\sigma_u(\\Delta_\\ell)\n=8i(-1)^{h_\\ell(u)}\n\\prod_{j=1}^3\n\\sin\\frac{\\pi\\langle u\\ell_j\\rangle_q}{q}.\\end{equation}\nHere $(\\ell_1,\\ell_2,\\ell_3)=(c,d,e)$ is an indexed triple, so repeated residues contribute repeated factors. Every sine factor is positive, so the sign of the imaginary part is exactly the Hodge sign. Consequently the CM type of \\(N(1)\\) is the conjugate of the CM type of the lower curve packet:\n\\begin{equation}\\label{eq:legacy-16.3}\n\\boxed{\n\\Sigma_N=\\overline{\\Sigma_\\ell}.\n}\\end{equation}\nThis is the rigorous form of the statement that the half-twist occurs on the imaginary plane.","\\]\none obtains\n\\begin{equation}\\label{eq:legacy-16.2}\n\\sigma_u(\\Delta_\\ell)\n=8i(-1)^{h_\\ell(u)}\n\\prod_{j=1}^3\n\\sin\\frac{\\pi\\langle u\\ell_j\\rangle_q}{q}.\\end{equation}\nHere $(\\ell_1,\\ell_2,\\ell_3)=(c,d,e)$ is an indexed triple, so repeated residues contribute repeated factors. Every sine factor is positive, so the sign of the imaginary part is exactly the Hodge sign. Consequently the CM type of \\(N(1)\\) is the conjugate of the CM type of the lower curve packet:\n\\begin{equation}\\label{eq:legacy-16.3}\n\\boxed{\n\\Sigma_N=\\overline{\\Sigma_\\ell}.\n}\\end{equation}\nThis is the rigorous form of the statement that the half-twist occurs on the imaginary plane.","1)^{h_\\ell(u)}\n\\prod_{j=1}^3\n\\sin\\frac{\\pi\\langle u\\ell_j\\rangle_q}{q}.\\end{equation}\nHere $(\\ell_1,\\ell_2,\\ell_3)=(c,d,e)$ is an indexed triple, so repeated residues contribute repeated factors. Every sine factor is positive, so the sign of the imaginary part is exactly the Hodge sign. Consequently the CM type of \\(N(1)\\) is the conjugate of the CM type of the lower curve packet:\n\\begin{equation}\\label{eq:legacy-16.3}\n\\boxed{\n\\Sigma_N=\\overline{\\Sigma_\\ell}.\n}\\end{equation}\nThis is the rigorous form of the statement that the half-twist occurs on the imaginary plane.","In the audited cases, the minimal CM conductors are \\(3,5,7,\\) or \\(9\\), even when the original fourfold modulus is \\(50,70,\\) or \\(90\\). Thus the weight-one carrier is not arbitrary: it is a character factor of the Jacobian of a lower-degree Fermat curve.","\\chapter{Explicit carrier and remaining Chow map}\nFor an audited CM type, the lower Fermat curve provides a candidate rational character-orbit factor of its Jacobian. Write $e_{\\mathrm{orb}}J(F_d)$ for that factor, up to isogeny. The idempotent selects the complete rational Galois orbit. A projector onto just one holomorphic CM half-plane would not be rational and is not an algebraic rational packet projector.","The desired correspondence is schematically\n\\[\n\\Gamma\\in\\CH^{\\ast}(F_d\\times C_P\\times S_T)_{\\Q},\\qquad\n\\Gamma_*:e_{\\mathrm{orb}}H^1(F_d)\\longrightarrow\ne_{\\mathrm{diag}}\\bigl(M(C_P)\\otimes M(S_T)\\bigr)(1).\n\\]\nThis formula is a target specification, not a constructed map. Its coefficient field, diagonal projector, rational multiplicities, and dimensions must be fixed before an isomorphism can be asserted. The smaller CM conductor can have a smaller rational degree than the original packet. Matching sign functions alone does not resolve that difference.","phism over the splitting field. Proving that the map is induced by an algebraic correspondence remains essential. One needs a Kummer or half-twist correspondence with the correct covariance and a nonzero period or intersection coefficient. Tensoring a verified map with the conjugate lower curve packet can then place a class in divisor degree on a product of curves or abelian varieties. Lefschetz $(1,1)$ and actual Fermat join maps would transport it back.","\\chapter{Correct literature and novelty map}\nKang is cited in the checked later literature for the all-degree Fermat-fourfold Hodge result, while its original exact theorem and proof were not extracted here \\cite{KangFermat}. Separately, da Silva studies which packets are generated by Shioda-style induced-cycle criteria and records the degree-$33$ failure of that criterion \\cite{DaSilvaFermat}. The October source contracts authenticate the Kang attribution through the checked later literature while recording that its original proof was not extracted. No historical priority claim is made for this constructive program.","d in the current evidence chapter:\n\\begin{itemize}\n\\item 14 exact degree-33 Klein checks\n\\item 4 infinite tower checks\n\\item 4 exact conductor-polyphase checks\n\\item 13 odd prime-power checks\n\\item 89,478,480 dyadic weighted coefficient identities\n\\item exact packet classifications at dyadic degrees through 128\n\\item an independent trivalent-tree/SAT(2) audit and structural dyadic checks through $2^{14}$\n\\item exact balanced-packet audit through degree 100\n\\item 364 K23 CM-type certificates\n\\end{itemize}","K23 CM sign descent & Complete & Imaginary selector $\\Delta_\\ell$","\\bibitem{MarkmanWeil}\nE. Markman.\n\\newblock Cycles on abelian $2n$-folds of Weil type from secant sheaves on abelian $n$-folds.\n\\newblock arXiv:2502.03415v2, 8 June 2025.\n\\newblock Theorem 1.5.1 and Corollary 1.6.1.","\\bibitem{DLMFDirichlet}\nNIST Digital Library of Mathematical Functions.\n\\newblock \\S25.15, Dirichlet $L$-functions, equations 25.15.5 and 25.15.9.\n\\newblock \\url{https://dlmf.nist.gov/25.15}, accessed 9 October 2026.","\\[\nX_m=\\{F_m=x_0^m+\\cdots+x_5^m=0\\}\\subset\\mathbf P^5_{\\mathbf C},\n\\qquad G_m=(\\mu_m)^6/\\mu_m.\n\\]","Write \\(\\langle r\\rangle_m\\) for the least-positive residue of a nonzero\nresidue class. A full-support character is","\\[\na=(a_0,\\ldots,a_5),\\qquad 0<a_i<m,\\qquad \\sum a_i\\equiv0\\pmod m.\n\\]","The Griffiths residue basis of the Fermat hypersurface diagonalizes the\naction of \\(G_m\\). The monomial \\(x_0^{a_0-1}\\cdots x_5^{a_5-1}\\) at\npole order three gives the character of middle Hodge type when\n\\(\\sum a_i=3m\\). The entire rational orbit has type \\((2,2)\\) when","\\begin{equation}\\label{eq:capacity_cycles-1.1}\n\\sum_i\\langle ta_i\\rangle_m=3m\n\\quad\\text{for every }t\\in(\\mathbf Z/m\\mathbf Z)^\\times.\\end{equation}","\\begin{equation}\\label{eq:capacity_cycles-1.2}\nJ_{F_m}=(x_0^{m-1},\\ldots,x_5^{m-1}),\\qquad\nS_m=\\prod_{i=0}^5x_i^{m-2}.\\end{equation}","\\[\n\\Omega=\\sum_{i=0}^5(-1)^ix_i\\,dx_0\\wedge\\cdots\\widehat{dx_i}\\cdots\\wedge dx_5\n\\]","Fix a character \\(a\\) modulo \\(m\\). Its invariant exponent columns form","\\[\n\\mathcal L_a=\\{v\\in\\mathbf Z_{\\ge0}^6\\setminus\\{0\\}:a\\cdot v\\equiv0\\pmod m\\}.\n\\]","\\begin{equation}\\label{eq:capacity_cycles-2.1}\nC(a)=\\sup\\left\\{\\sum_vw_v:\n\\sum_vw_vv=\\mathbf1,\\quad w_v\\ge0,\n\\quad\\text{only finitely many }w_v\\ne0\\right\\}.\\end{equation}","Every nonsingular nonnegative integral matrix \\(A\\) satisfying\n\\(A^Ta\\in m\\mathbf Z^6\\), with \\(w=A^{-1}\\mathbf1>0\\), gives one\nfeasible point. Hence","\\begin{equation}\\label{eq:capacity_cycles-2.2}\n\\mathbf1^TA^{-1}\\mathbf1\\le C(a).\\end{equation}","This is a necessary numerical bound, not a sufficient geometric\ncriterion. A capacity maximizer can use fewer than six columns. A\nsix-column maximizer can fail normality or well-formedness. No claim\nabout a normal carrier follows merely from \\(C(a)>1\\).","Suppose \\(y\\in\\mathbf Q_{>0}^6\\) and \\(y\\cdot v\\ge1\\) for every\n\\(v\\in\\mathcal L_a\\). Then","\\[\nC(a)\\le y\\cdot\\mathbf1.\n\\]","\\[\n\\sum_vw_v\\le\\sum_vw_v(y\\cdot v)\n=y\\cdot\\sum_vw_vv=y\\cdot\\mathbf1.\n\\]","The infinitely many inequalities can be certified by a finite residue\ngraph. Write \\(y_i=c_i/L\\), where \\(c_i,L\\) are positive integers. On\n\\(\\mathbf Z/m\\mathbf Z\\), an edge \\(r\\to r+a_i\\) has cost \\(c_i\\). An\ninteger function \\(d\\) satisfying","\\begin{equation}\\label{eq:capacity_cycles-2.3}\nd(0)=0,\\qquad d(r+a_i)\\le d(r)+c_i,\n\\qquad d(-a_i)+c_i\\ge L\\end{equation}","for every \\(r,i\\) certifies the dual inequality for \\textbf{every}\ninvariant column, regardless of its total degree.","To prove this, order the multiset described by a nonzero invariant\ncolumn into a nonempty zero-sum word. Repeated application of the first\ninequalities bounds the cost of its prefix below by \\(d(-a_i)\\), where\n\\(i\\) is its last letter. Adding the last cost gives at least \\(L\\).\nDivide by \\(L\\).","\\[\na=(1,20,24,42,61,62)\\pmod{70},\n\\]","\\begin{equation}\\label{eq:capacity_cycles-2.4}\ny=\\frac1{70}(11,10,19,14,6,17).\\end{equation}","\\[\nC(a)\\le\\sum_i y_i=\\frac{11}{10}.\n\\]","\\[\nA=\\begin{pmatrix}\n0&0&2&0&0&3\\\\\n0&0&1&2&7&0\\\\\n1&0&2&2&0&1\\\\\n0&5&0&0&0&0\\\\\n0&0&0&2&0&3\\\\\n3&0&0&0&0&0\n\\end{pmatrix},\n\\quad\nA^{-1}\\mathbf1=(1/3,1/5,1/10,1/10,1/10,4/15).\n\\]","It is feasible and its weights sum to \\(11/10\\). Therefore","\\begin{equation}\\label{eq:capacity_cycles-2.5}\n\\boxed{C(1,20,24,42,61,62\\bmod70)=11/10.}\\end{equation}","For \\(q\\ge2\\) and \\(1\\le b<q\\), set","\\begin{equation}\\label{eq:capacity_cycles-3.1}\nm=5q,\\qquad\n a_{q,b}=(b,b+q,b+2q,b+3q,b+4q,5(q-b)).\\end{equation}","Every character in \\eqref{eq:capacity_cycles-3.1} is full support and its rational orbit is of\ntype \\((2,2)\\).","Let \\(t\\) be a unit modulo \\(5q\\), and let \\(b'=tb\\bmod q\\), with\n\\(1\\le b'<q\\). Multiplication by \\(t\\) permutes the five residues in the\nfiber over \\(b'\\), because \\(t\\) is a unit modulo 5. The first five\nleast-positive residues are therefore, in some order,","\\[\nb',b'+q,b'+2q,b'+3q,b'+4q.\n\\]","The last residue is \\(5(q-b')\\). Their sum is \\(15q=3m\\). This is a\nproof for all parameters, separate from the finite checks in the bundle.","Assume \\(q>5\\) and \\(\\gcd(b,q)=1\\). For every nonsingular nonnegative\nintegral matrix \\(A\\) such that","\\[\nA^Ta_{q,b}\\in5q\\mathbf Z^6,\n\\qquad w=A^{-1}\\mathbf1>0,\n\\]","\\begin{equation}\\label{eq:capacity_cycles-3.2}\n\\boxed{\\sum_j w_j\\le1.}\\end{equation}","In fact \\(C(a_{q,b})=1\\).","\\[\ny_q=\\frac1q(1,1,1,1,1,q-5),\\qquad y_q\\cdot\\mathbf1=1.\n\\]","If \\(v\\in\\mathcal L_{a_{q,b}}\\), reduction modulo \\(q\\), followed by\ncancellation of the unit \\(b\\), gives","\\[\nv_0+v_1+v_2+v_3+v_4-5v_5\\equiv0\\pmod q.\n\\]","\\[\nN=v_0+v_1+v_2+v_3+v_4+(q-5)v_5\n\\]","is divisible by \\(q\\). It is a positive integer because \\(v\\ne0\\) and\n\\(q>5\\). Therefore \\(N\\ge q\\), or \\(y_q\\cdot v\\ge1\\). Lemma 2.1 proves\n\\(C(a_{q,b})\\le1\\).","Conversely \\(v=\\mathbf1\\) is invariant, because \\(\\sum a_i=3m\\), and\nputting weight 1 on it is a feasible point of \\eqref{eq:capacity_cycles-2.1}. Thus\n\\(C(a_{q,b})=1\\), proving \\eqref{eq:capacity_cycles-3.2}. \\(\\square\\)","The invariant column set is unchanged by multiplying a character by a\nunit modulo \\(m\\), or by replacing \\((m,a)\\) by \\((km,ka)\\). A\ncoordinate permutation merely permutes the columns' entries and the dual\ncosts. Consequently none of these changes evades the obstruction.","For example, at degree 35 the packet is \\((1,8,15,22,29,30)\\), and the\ndual is \\((1,1,1,1,1,2)/7\\). No increase in exponent-search bounds can\nproduce a matrix of the earlier type with normalized weight sum greater\nthan 1.","Fix \\(q\\ge2\\), put \\(m=5q\\), and choose \\(\\kappa\\in\\mathbf C\\) with","\\[\n\\kappa^q=-1/5.\n\\]","\\[\nz=x_5,\\quad t_i=x_i^q\\ (0\\le i\\le4),\\quad P=x_0x_1x_2x_3x_4,\n\\]","and let \\(e_j\\) be the elementary symmetric polynomial of degree \\(j\\)\nin \\(t_0,\\ldots,t_4\\). Define","\\begin{equation}\\label{eq:capacity_cycles-4.1}\n\\boxed{\nZ_\\kappa=\\{e_1=0,\\ e_2=0,\\ P-\\kappa z^5=0\\}\\subset\\mathbf P^5.\n}\\end{equation}","\\begin{equation}\\label{eq:capacity_cycles-4.2}\np_5=e_1(e_1^4-5e_1^2e_2+5e_1e_3+5e_2^2-5e_4)-5e_2e_3+5e_5.\\end{equation}","Since \\(e_5=P^q\\), on \\(Z_\\kappa\\) one has","\\[\nF_m=p_5+z^{5q}=5P^q+z^{5q}\n=(5\\kappa^q+1)z^{5q}=0.\n\\]","Thus \\(Z_\\kappa\\subset X_{5q}\\). Its defining equations have degrees\n\\(q,2q,5\\).","\\[\nS_q=\\{e_1(x_0^q,\\ldots,x_4^q)=e_2(x_0^q,\\ldots,x_4^q)=0\\}\\subset\\mathbf P^4.\n\\]","The coordinate-power map to the \\(t\\)-space is finite, so the dimension\nis two. In characteristic zero, on \\(e_1=0\\) the equations are\nequivalent to","\\[\n\\sum x_i^q=0,\\qquad\\sum x_i^{2q}=0.\n\\]","At a hypothetical failure of Jacobian rank, there would be constants\n\\(\\lambda,\\mu\\), not both zero, such that for each nonzero coordinate","\\[\n\\lambda+2\\mu x_i^q=0.\n\\]","The case \\(\\mu=0\\) is impossible. Otherwise all nonzero \\(x_i^q\\) equal\nthe same nonzero constant. Their sum cannot vanish in characteristic\nzero. Thus \\(S_q\\) is smooth. It is a connected complete intersection of\npositive-dimensional ample divisors, hence irreducible.","The same argument with four coordinates shows that \\(S_q\\cap\\{x_i=0\\}\\)\nis a smooth reduced complete-intersection curve. Distinct such curves\nshare no irreducible component, since imposing two coordinate zeros\ngives a zero-dimensional intersection. In particular the divisor of\n\\(P\\) has valuation one at the generic point of each coordinate curve.","Forgetting \\(z\\) defines a finite flat cover \\(Z_\\kappa\\to S_q\\) of\ndegree five. On a chart trivializing \\(\\mathcal O_{S_q}(1)\\), its\nequation is \\(z^5=P/\\kappa\\), with the corresponding trivializations\nunderstood. At a generic coordinate curve its right side has valuation\none. The degree-five polynomial is Eisenstein over that discrete\nvaluation ring and is therefore irreducible over the function field of\n\\(S_q\\). Finite flatness makes the cover generically integral without\nvertical or embedded components, locally its monic algebra injects into\nthe generic field extension. Hence the cover is integral.","(z\\) defines a finite flat cover \\(Z_\\kappa\\to S_q\\) of\ndegree five. On a chart trivializing \\(\\mathcal O_{S_q}(1)\\), its\nequation is \\(z^5=P/\\kappa\\), with the corresponding trivializations\nunderstood. At a generic coordinate curve its right side has valuation\none. The degree-five polynomial is Eisenstein over that discrete\nvaluation ring and is therefore irreducible over the function field of\n\\(S_q\\). Finite flatness makes the cover generically integral without\nvertical or embedded components, locally its monic algebra injects into\nthe generic field extension. Hence the cover is integral.","The two equations of \\(S_q\\), followed by the monic equation in \\(z\\),\nform a homogeneous regular sequence. The resulting projective ideal is\nsaturated and defines the integral complete intersection \\eqref{eq:capacity_cycles-4.1}.\nTherefore","\\[\n\\dim Z_\\kappa=2,\\qquad\\deg Z_\\kappa=q(2q)5=10q^2.\n\\]","The surface need not be smooth. Villaflor Loyola's complete-intersection\ntheorem does not impose smoothness of \\(Z\\), its proof explicitly\nreduces to a smooth perturbation. We use this version, not an unverified\nsmooth-cycle variant.","\\[\n\\begin{aligned}\nf_1&=e_1,&g_1&=e_1^4-5e_1^2e_2+5e_1e_3+5e_2^2-5e_4,\\\\\nf_2&=e_2,&g_2&=-5e_3,\\\\\nf_3&=P-\\kappa z^5,&g_3&=5\\sum_{j=0}^{q-1}P^{q-1-j}\\kappa^jz^{5j}.\n\\end{aligned}\n\\]","\\begin{equation}\\label{eq:capacity_cycles-5.1}\nF_m=f_1g_1+f_2g_2+f_3g_3.\\end{equation}","\\[\nH=(f_1,g_1,f_2,g_2,f_3,g_3),\\qquad\nD_\\kappa=\\det\\operatorname{Jac}_{(x_0,\\ldots,x_4,z)}H,\n\\]","\\[\n\\Delta(t)=\\prod_{0\\le i<j\\le4}(t_i-t_j).\n\\]","\\begin{equation}\\label{eq:capacity_cycles-5.2}\n\\boxed{\nD_\\kappa=625q^5\\Delta(x_0^q,\\ldots,x_4^q)\n\\sum_{b=1}^{q-1}\\kappa^{q-b}P^{b-1}z^{5(q-b)-1}.\n}\\end{equation}","\\begin{equation}\\label{eq:capacity_cycles-5.3}\ndf_1\\wedge dg_1\\wedge df_2\\wedge dg_2\n=25\\,de_1\\wedge de_2\\wedge de_3\\wedge de_4.\\end{equation}","Only the term \\(-5de_4\\) in \\(dg_1\\) survives after wedging with\n\\(de_1,de_2,de_3\\), the row permutation has positive sign.","\\begin{equation}\\label{eq:capacity_cycles-5.4}\ndf_3\\wedge dg_3=\n25q\\sum_{j=1}^{q-1}\\kappa^jP^{q-1-j}z^{5j-1}\\,dP\\wedge dz.\\end{equation}","For each interior index \\(j\\), the two contributing coefficients add to\n\\(j+(q-j)=q\\). The endpoint terms have zero derivative coefficient. Thus\n\\eqref{eq:capacity_cycles-5.4} is an identity for every positive integer \\(q\\), not an\ninterpolation from checked values.","\\[\nde_1\\wedge\\cdots\\wedge de_5\n=\\Delta(t)\\,dt_0\\wedge\\cdots\\wedge dt_4.\n\\]","Use \\(e_5=P^q\\) and\n\\(dt_0\\wedge\\cdots\\wedge dt_4=q^5P^{q-1}\\,dx_0\\wedge\\cdots\\wedge dx_4\\).\nCancellation in the polynomial ring gives","\\begin{equation}\\label{eq:capacity_cycles-5.5}\nde_1\\wedge\\cdots\\wedge de_4\\wedge dP\n=q^4\\Delta(x^q)\\,dx_0\\wedge\\cdots\\wedge dx_4.\\end{equation}","Combining \\eqref{eq:capacity_cycles-5.3}, \\eqref{eq:capacity_cycles-5.4}, and \\eqref{eq:capacity_cycles-5.5}, and putting \\(b=q-j\\), proves \\eqref{eq:capacity_cycles-5.2}.\nThe identity extends across \\(P=0\\) because it is a polynomial identity.\n\\(\\square\\)","The degree of every monomial in \\eqref{eq:capacity_cycles-5.2} is \\(15q-6=3m-6\\), as required for\na middle residue numerator.","Fix \\(1\\le b<q\\). In the order used in \\eqref{eq:capacity_cycles-3.1}, the coefficient of","\\begin{equation}\\label{eq:capacity_cycles-6.1}\nx_0^{b-1}x_1^{b+q-1}x_2^{b+2q-1}x_3^{b+3q-1}x_4^{b+4q-1}\n z^{5(q-b)-1}\\end{equation}","in \\(D_\\kappa\\) is","\\begin{equation}\\label{eq:capacity_cycles-6.2}\n625q^5\\kappa^{q-b}\\ne0.\\end{equation}","All permutations of the first five exponents have the same nonzero\nmagnitude, with the Vandermonde sign. Different values of \\(b\\) and\ndifferent permutations give different monomials. Every exponent lies\nbetween zero and \\(m-2\\), so no term is killed by the Fermat Jacobian\nideal.","\\begin{equation}\\label{eq:capacity_cycles-6.3}\nQ_b=\\prod_{i=0}^5x_i^{m-a_{q,b,i}-1}.\\end{equation}","It has degree \\(3m-6\\). Because the product \\(Q_bD_\\kappa\\) has socle\ndegree, a surviving monomial modulo \\(J_{F_m}\\) must have all six\nexponents equal to \\(m-2\\). Formula \\eqref{eq:capacity_cycles-6.1} is the unique term with this\ncomplement. Therefore","\\begin{equation}\\label{eq:capacity_cycles-6.4}\nQ_bD_\\kappa\\equiv625q^5\\kappa^{q-b}S_m\\pmod{J_{F_m}}.\\end{equation}","\\[\n\\det\\operatorname{Hess}(F_m)=[m(m-1)]^6S_m.\n\\]","\\[\nc=\\frac{625q^5\\kappa^{q-b}}{[m(m-1)]^6}.\n\\]","\\begin{equation}\\label{eq:capacity_cycles-6.5}\n\\boxed{\n\\int_{Z_\\kappa}\\operatorname{res}\\left(\\frac{Q_b\\Omega}{F_m^3}\\right)\n=\\frac{(2\\pi i)^2}{50q}\\,\\kappa^{q-b}\\ne0.\n}\\end{equation}","The normalization follows from \\(625q^5/[2(5q)^6]=1/(50q)\\). In\nconventions using a differently normalized residue map, the displayed\nconstant changes with that convention, nonvanishing does not.","Let \\(h\\) denote the hyperplane class. Since \\(\\deg Z_\\kappa/m=2q\\), the\nprimitive part","\\[\n\\xi_\\kappa=[Z_\\kappa]-2q h^2\n\\]","is rational and algebraic. Villaflor Loyola's Theorem 1.1 identifies its\nprimitive residue numerator with \\(-2D_\\kappa/m\\). Thus its complex\neigenspace support consists of exactly the \\(120(q-1)\\) characters\nobtained from \\eqref{eq:capacity_cycles-3.1} by permuting the first five entries and varying\n\\(b\\).","That set is closed under multiplication by units modulo \\(m\\), by Lemma\n3.1. Each of its eigenspaces has dimension one. The\n\\(\\mathbf Q[G_m]\\)-submodule generated by \\(\\xi_\\kappa\\) therefore has\ncomplexification equal to the entire sum of those eigenlines: applying\nthe complex character idempotents isolates each nonzero component. Since\nthe generator and the group action are rational, this is the\ncomplexification of the claimed rational submodule.","Every vector in that rational submodule is a rational linear combination\nof classes of diagonal translates of \\(Z_\\kappa\\) and of \\(h^2\\). Hence\nevery rational packet in the family is algebraic. Its rational algebraic\nsubspace has dimension","\\[\n120(q-1)\n\\]","for this fixed choice of the last coordinate. This is not the dimension\nof all Hodge classes on \\(X_{5q}\\).","Take \\(q=7\\), \\(b=1\\), \\(\\kappa^7=-1/5\\). Then","\\[\na=(1,8,15,22,29,30),\\quad\nZ_\\kappa=\\left\\{\n\\begin{array}{l}\n\\sum_{i=0}^4x_i^7=0,\\\\\n\\sum_{0\\le i<j\\le4}x_i^7x_j^7=0,\\\\\nx_0x_1x_2x_3x_4=\\kappa x_5^5\n\\end{array}\\right\\}\\subset X_{35}.\n\\]","The cycle has degree 490. The rational packet has dimension\n\\(\\varphi(35)=24\\), since its first character entry is 1. The\ndeterminant coefficient is","\\[\n10,504,375\\,\\kappa^6.\n\\]","\\[\nQ_1=x_0^{33}x_1^{26}x_2^{19}x_3^{12}x_4^5x_5^4,\n\\]","\\[\n\\boxed{\n\\int_{Z_\\kappa}\\operatorname{res}\\left(\\frac{Q_1\\Omega}{F_{35}^3}\\right)\n=\\frac{(2\\pi i)^2\\kappa^6}{350}\\ne0.\n}\n\\]","At degree 70 the standard packet \\((1,15,29,43,57,65)\\) obtained by\nsetting \\(q=14\\) is \\textbf{different} from the earlier packet\n\\((1,20,24,42,61,62)\\). Their capacities are respectively 1 and\n\\(11/10\\). They must not be conflated.","\\begin{itemize}\n\\tightlist\n\\item\n  balance and exact primal/global-dual certificates for seven specified\n  packets,\n\\item\n  the predecessor archive checksum,\n\\item\n  Newton's identity and the elementary-symmetric Jacobian identity,\n\\item\n  exact six-by-six intermediate determinants for \\(q=2,3,4,5,7,11,17\\),\n\\item\n  independent socle-pairing checks for every \\(b\\) at \\(q=2,7,11\\),\n\\item\n  all \\(b\\), all units, and all first-five permutations for\n  \\(2\\le q\\le64\\), totaling 241,920 distinct support monomials and\n  218,836 balance/conjugacy instances.\n\\end{itemize}","rimal/global-dual certificates for seven specified\n  packets,\n\\item\n  the predecessor archive checksum,\n\\item\n  Newton's identity and the elementary-symmetric Jacobian identity,\n\\item\n  exact six-by-six intermediate determinants for \\(q=2,3,4,5,7,11,17\\),\n\\item\n  independent socle-pairing checks for every \\(b\\) at \\(q=2,7,11\\),\n\\item\n  all \\(b\\), all units, and all first-five permutations for\n  \\(2\\le q\\le64\\), totaling 241,920 distinct support monomials and\n  218,836 balance/conjugacy instances.\n\\end{itemize}","\\[\nX_m=\\{x_0^m+\\cdots+x_5^m=0\\}\\subset\\mathbf P^5_\\mathbf C.\n\\]","The diagonal group is \\(G_m=(\\mu_m)^6/\\mu_m\\). For a character\n\\(a=(a_0,\\ldots,a_5)\\) with all entries nonzero modulo \\(m\\) and sum\nzero modulo \\(m\\), let \\(W_{[a]}\\) denote the rational character-orbit\nsummand of primitive \\(H^4(X_m,\\mathbf Q)\\).","\\[\nm=70,\\qquad a=(1,20,24,42,61,62),\n\\]","all classes in the 24-dimensional rational space \\(W_{[a]}\\) are\nrational linear combinations of classes of algebraic surfaces on\n\\(X_{70}\\). For every positive integer \\(k\\), the corresponding\npulled-back packet \\(W_{[ka]}\\) on \\(X_{70k}\\) is also algebraic.","The primitive Fermat eigenspaces for full-support characters are\none-dimensional over the splitting field. The Griffiths residue\ndescription assigns type \\((2,2)\\) when the least-positive residues have\nsum \\(3m\\). A rational orbit is of type \\((2,2)\\) precisely when this\nholds for all its unit conjugates.","For every \\(t\\in(\\mathbf Z/70\\mathbf Z)^\\times\\), exact enumeration\ngives","\\[\n\\sum_{i=0}^{5}\\langle ta_i\\rangle_{70}=210.\n\\]","There are 24 conjugates, all distinct because \\(a_0=1\\). Thus the entire\nrational orbit, not just a single complex residue vector, has type\n\\((2,2)\\). The verifier stores every conjugate. The distinction between\na complex residue vector and a rational Hodge class is essential here.","\\[\nA=\\begin{pmatrix}\n0&0&2&0&0&3\\\\\n0&0&1&2&7&0\\\\\n1&0&2&2&0&1\\\\\n0&5&0&0&0&0\\\\\n0&0&0&2&0&3\\\\\n3&0&0&0&0&0\n\\end{pmatrix}.\n\\]","Rows are monomial exponents, columns correspond to \\(y_0,\\ldots,y_5\\).\nDirect rational arithmetic gives","\\[\n\\det A=-2100,\\qquad\nA^{-1}\\mathbf1=(1/3,1/5,1/10,1/10,1/10,4/15),\n\\]","\\[\n\\mathbf1^TA^{-1}\\mathbf1=11/10.\n\\]","Set \\(W=(10,6,3,3,3,8)\\) and \\(D=30\\), so \\(AW=D\\mathbf1\\). Define","\\[\nY=\\{F=0\\}\\subset\\mathbf P(10,6,3,3,3,8),\n\\]","\\[\nF=y_0^3+y_1^5+y_2^2y_5^3+y_2y_3^2y_4^7\n  +y_0y_2^2y_3^2y_5+y_3^2y_5^3.\n\\]","\\[\nD-\\sum W_i=30-33=-3.\n\\]","Let \\(M=210\\), \\(\\alpha=3a\\), and","\\[\nB=210A^{-1}=\\begin{pmatrix}\n0&0&0&0&0&70\\\\\n0&0&0&42&0&0\\\\\n42&0&63&0&-63&-21\\\\\n-63&0&63&0&42&-21\\\\\n12&30&-27&0&-3&9\\\\\n42&0&-42&0&42&14\n\\end{pmatrix}.\n\\]","\\[\nAB=210I,\\qquad B\\mathbf1=7W.\n\\]","On the projective tori define \\(y_i=\\prod_jx_j^{B_{ij}}\\). Negative\nexponents are allowed there. The second identity makes this well-defined\ninto the weighted projective torus, the first gives","\\[\nF(y(x))=x_0^{210}+\\cdots+x_5^{210}.\n\\]","\\[\nq:U_{210}\\longrightarrow U_Y,\n\\]","where \\(U_{210}=X_{210}\\cap\\{\\prod x_i\\ne0\\}\\) and\n\\(U_Y=Y\\cap\\{\\prod y_i\\ne0\\}\\).","The target torus hypersurface is smooth because its finite étale\npullback is smooth. It is irreducible because its Laurent polynomial\npulls back under an injective full-rank monomial substitution to an\nirreducible Fermat Laurent polynomial. Non-monomial factors remain\nnon-monomial under this substitution. The polynomial \\(F\\) has no\ncoordinate factor, so its projective closure \\(Y\\) is also irreducible.","Put \\((u,v,s,t,c,z)=(y_0,y_1,y_2,y_3,y_4,y_5)\\), so","\\[\nF=u^3+v^5+z^3(s^2+t^2)+st^2c^7+us^2t^2z.\n\\]","\\[\n\\begin{aligned}\nF_u&=3u^2+s^2t^2z, &F_v&=5v^4,\\\\\nF_s&=2sz^3+t^2c^7+2ust^2z,\n&F_t&=2t(z^3+sc^7+us^2z),\\\\\nF_c&=7st^2c^6,\n&F_z&=3z^2(s^2+t^2)+us^2t^2.\n\\end{aligned}\n\\]","\\[\n\\{y_0=y_1=y_2=y_3=0\\}\n\\ \\cup\\ \\{y_0=y_1=y_3=y_5=0\\}\n\\ \\cup\\ \\{y_0=y_1=y_4=y_5=0\\}.\n\\]","Here is a direct verification of exhaustion. At a singular point,\n\\(v=0\\) and \\(st^2c^6=0\\). The cases \\(s=0\\) or \\(t=0\\) give the\ndisplayed planes immediately from the other derivatives. In the\nremaining case \\(s,t\\ne0\\), one has \\(c=0\\). If \\(z=0\\), then \\(u=0\\),\ngiving the third plane. If \\(z\\ne0\\), the equations \\(F_s=F_t=0\\) give","\\[\nz^2+ut^2=z^2+us^2=0.\n\\]","Hence \\(s^2=t^2=R\\ne0\\) and \\(u=-z^2/R\\). But then \\(F_z=5Rz^2\\ne0\\), a\ncontradiction.","There is also an independent exact ideal certificate. For the gradient\nideal \\(J\\), the verifier proves","\\[\n\\sqrt J=(y_0,y_1,y_2y_5,y_3y_4,y_3y_5).\n\\]","\\[\ny_0^3,\\quad y_1^4,\\quad (y_2y_5)^3,\\quad\n(y_3y_4)^{13},\\quad (y_3y_5)^3\n\\]","belong to \\(J\\) by Gröbner reduction over \\(\\mathbf Q\\). The ideal on\nthe right is the intersection of the three coordinate prime ideals, so\nthese two inclusions prove equality of radicals.","The ambient weighted projective space is well formed. Its largest\nsingular stratum is the \\(\\mu_3\\)-stratum \\(y_0=y_5=0\\), of dimension\nthree. The restriction of \\(F\\) there is \\(y_1^5+y_2y_3^2y_4^7\\), not\nidentically zero. All other ambient singular strata have dimension at\nmost two. Thus \\(Y\\) meets the ambient singular locus in codimension at\nleast two. The verifier checks all coordinate strata, not just the\nlargest one.","On the smooth ambient locus, the weighted Euler sequence gives canonical\nclass \\(-33H\\). Ordinary hypersurface adjunction gives \\(-3H\\) on the\nsmooth part of \\(Y\\) there. Both complements have codimension at least\ntwo, so normality extends the equality as a canonical Weil divisor\nclass:","\\[\nK_Y\\sim_{\\mathbf Q}-3H.\n\\]","Here \\(H\\) is the ample rational hyperplane class, not an assertion that\n\\(\\mathcal O_Y(1)\\) is everywhere a line bundle. This argument\nestablishes the adjunction formula in the present non-quasismooth\nexample directly.","Take a projective resolution \\(r:\\widetilde Y\\to Y\\) which is an\nisomorphism over \\(U_Y\\). A general complete-intersection curve \\(C\\)\ncut out by three sufficiently ample Cartier divisors avoids the singular\nlocus of \\(Y\\), which has dimension at most two. Its strict transform\navoids the exceptional divisors. Therefore","\\[\nK_{\\widetilde Y}\\cdot\\widetilde C=K_Y\\cdot C=-3H\\cdot C<0.\n\\]","These curves move in a covering family. Thus \\(K_{\\widetilde Y}\\) is not\npseudoeffective. The theorem of Boucksom--Demailly--Păun--Peternell\nimplies that \\(\\widetilde Y\\) is uniruled.","The rational Hodge conjecture for smooth projective uniruled fourfolds\nis a proved theorem of Conte--Murre, also stated as Corollary 4.2 of\nArapura, arXiv:0706.2506. Therefore every rational \\((2,2)\\)-class on\n\\(\\widetilde Y\\) is algebraic.","For context, this theorem does not use the unknown general fourfold\ncase: a uniruled fourfold is dominated generically finitely by a\nresolution of \\(T\\times\\mathbf P^1\\), with \\(\\dim T=3\\). The relevant\nHodge classes on the product are algebraic by the threefold case and\nKünneth, blowup centers have dimension at most two, push-pull transfers\nalgebraicity.","\\[\nA^Ta=70\\beta,\\qquad \\beta=(3,3,1,3,2,3),\n\\]","and hence \\(B^T\\beta=\\alpha\\). To obtain a genuine character of the\nprojective torus, let \\(r_0=(0,0,2,0,0,3)\\) be the first row of \\(A\\)\nand set","\\[\nu=\\beta-3r_0=(3,3,-5,3,2,-6).\n\\]","\\[\nW\\cdot u=0,\\qquad B^Tu=\\alpha-630e_0.\n\\]","Thus \\(u\\) is a character of the weighted projective torus. Its pullback\nrestricts on \\(G_{210}\\) to the Fermat character \\(\\alpha\\), since\n\\(630=3\\cdot210\\).","Let \\(K\\) be the kernel of the torus isogeny. The relation \\(AB=210I\\)\nshows that \\(K\\) embeds in \\(G_{210}\\): an element in the kernel has\nrepresentatives whose 210th powers are all the same, and a common scalar\nmakes them roots of unity. The displayed character identity proves\n\\(\\chi_\\alpha|_K=1\\). The same holds for every rational conjugate.\nTherefore the whole rational packet is \\(K\\)-invariant.","The coordinate boundary of \\(X_{210}\\) is a simple normal crossing\ndivisor. Every nonempty coordinate stratum is fixed pointwise by at\nleast one coordinate scaling subgroup \\(\\mu_{210}\\). Our full-support\ncharacter is nontrivial on that subgroup. The same is true for every\ncharacter in its rational orbit.","\\[\ne_{[\\alpha]}H^4(X_{210},\\mathbf Q)\n\\xrightarrow{\\sim}\ne_{[\\alpha]}H^4(U_{210},\\mathbf Q).\n\\]","Because \\(q\\) is finite étale with deck group \\(K\\),","\\[\nq^*:H^4(U_Y,\\mathbf Q)\n\\xrightarrow{\\sim}H^4(U_{210},\\mathbf Q)^K.\n\\]","This is an isomorphism of mixed Hodge structures onto its image.\nNormalized transfer supplies the inverse. The rational pure\n\\((2,2)\\)-packet therefore descends to a pure weight-four Hodge\nsubstructure of \\(H^4(U_Y,\\mathbf Q)\\).","\\[\nW_4H^4(U_Y,\\mathbf Q)\n=\\operatorname{im}\\bigl(H^4(\\widetilde Y,\\mathbf Q)\n\\to H^4(U_Y,\\mathbf Q)\\bigr).\n\\]","Here \\(W_4\\) denotes the weight filtration, not the Fermat packet\nnotation. The descended classes have weight four, so belong to this\nimage. The category of polarizable pure rational Hodge structures is\nsemisimple. The surjection from \\(H^4(\\widetilde Y)\\) onto its image\nthus has a Hodge-theoretic section. Every descended rational\n\\((2,2)\\)-class has a rational \\((2,2)\\)-lift on \\(\\widetilde Y\\).","By the uniruled-fourfold theorem, each lifted class on \\(\\widetilde Y\\)\nis a rational combination of algebraic surface classes. The resolved\ngraph of the monomial rational map","\\[\nX_{210}\\dashrightarrow\\widetilde Y\n\\]","defines an algebraic correspondence. Its pullback on cohomology agrees\nwith \\(q^*\\) on the torus. Apply the rational orbit projector\n\\(e_{[\\alpha]}\\), which is a rational linear combination of graphs of\ndiagonal automorphisms. Therefore it preserves the span of algebraic\ncycle classes.","The resulting class and the prescribed class agree on \\(U_{210}\\), their\ndifference lies in the packet and restricts to zero. The preceding\nboundary isomorphism proves that difference is zero. This establishes\nalgebraicity of the entire \\(W_{[\\alpha]}\\) packet on \\(X_{210}\\).","\\[\n\\pi:X_{210}\\longrightarrow X_{70},\\qquad\n[x_i]\\longmapsto[x_i^3].\n\\]","Its pullback identifies \\(W_{[a]}\\) with \\(W_{[3a]}\\). The identity","\\[\n\\pi_*\\pi^*=(\\deg\\pi)\\operatorname{id}\n\\]","then transports the algebraic classes down to \\(X_{70}\\). Division by\nthe positive degree is legitimate because all cycle coefficients are\nrational.","For arbitrary \\(k\\ge1\\), the coordinate-power map \\(X_{70k}\\to X_{70}\\)\npulls the same packet back algebraically. This proves the stated\ninfinite family of packets. It does not prove the full Hodge conjecture\nfor those varieties.","\\textbf{Torus-carrier criterion.} Let \\(W\\) be a full-support rational\nmiddle Hodge packet on a Fermat fourfold. Suppose a finite torus isogeny\nidentifies its open Fermat variety with the inverse image of a smooth\ntorus hypersurface, the deck group acts trivially on \\(W\\), and that\ntorus hypersurface has a smooth projective compactification for which\nthe rational degree-four Hodge conjecture is known. Then \\(W\\) is\nalgebraic.","s-carrier criterion.} Let \\(W\\) be a full-support rational\nmiddle Hodge packet on a Fermat fourfold. Suppose a finite torus isogeny\nidentifies its open Fermat variety with the inverse image of a smooth\ntorus hypersurface, the deck group acts trivially on \\(W\\), and that\ntorus hypersurface has a smooth projective compactification for which\nthe rational degree-four Hodge conjecture is known. Then \\(W\\) is\nalgebraic.","For searching, choose nonnegative integral columns \\(v_j\\) satisfying\n\\(a\\cdot v_j\\equiv0\\pmod m\\). Seek a nonsingular six-column matrix \\(A\\)\nwith","\\[\nA^{-1}\\mathbf1>0,\\qquad \\mathbf1^TA^{-1}\\mathbf1>1.\n\\]","\\textbf{Matrix-carrier corollary.} Let \\(a\\) be a full-support balanced\nrational Hodge character modulo \\(m\\). Suppose \\(A\\) is a nonsingular\nnonnegative integral \\(6\\times6\\) matrix such that","\\[\nA^Ta\\in m\\mathbf Z^6,\\qquad A^{-1}\\mathbf1>0,\n\\qquad\\mathbf1^TA^{-1}\\mathbf1>1.\n\\]","Let \\(W/D=A^{-1}\\mathbf1\\), with primitive positive integral weights\n\\(W\\), and assume the row-monomial hypersurface is integral, normal, and\nwell formed in \\(\\mathbf P(W)\\). Then the rational Fermat packet\n\\(W_{[a]}\\) is algebraic.","Indeed, choose \\(M\\) divisible by \\(m\\) and all denominators of\n\\(A^{-1}\\). Put \\(B=MA^{-1}\\), \\(\\alpha=(M/m)a\\), and \\(\\beta=A^Ta/m\\).\nFor any row \\(r_i\\) of \\(A\\),","\\[\nu=\\beta-3r_i,\\qquad W\\cdot u=0,\\qquad\nB^Tu=\\alpha-3Me_i.\n\\]","Here \\(W\\cdot\\beta=3D\\) follows from \\(\\sum a_j=3m\\). Thus deck\ninvariance follows uniformly from the matrix congruence, it is not an\nextra conjecture. The positive weight sum gives negative canonical\ncoefficient, normality and well-formedness justify the canonical\ncomputation, and the uniruledness, descent, lifting and return arguments\nabove finish the proof. This corollary replaces an assumed cycle seed\nwith a finite exponent matrix and checkable geometric hypotheses.","The narrower atomic chain/loop search considered 6,600 labeled graphs\nand 24 unit conjugates, for 158,400 pairs. Its maximum normalized weight\nsum was \\(296/315<1\\), it found no Fano candidate in that implemented\nfamily. A broader functional-graph search also failed to find one. These\nfailures are not evidence against the Hodge conjecture.","Five-phase supports & 241,920 monomials & All $b$, all first-five permutations, and $2\\le q\\le64$. Distinct from all Fermat packets.","Dyadic census & Moduli $8,16,32,64,128$ & The complete finite packet census and implemented grammar at those moduli.","Dyadic weighted identity & 87,376 coefficients & Fresh range $3\\le s\\le9$. The historical larger range was not fully replayed.","The resulting counts by retained packet were\n\\[\n40,\\quad60,\\quad48,\\quad48,\\quad48,\\quad48,\\quad72,\n\\]\nwhich total 364. This stronger bounded replay guards against a certificate file whose generator omitted or misclassified a valid case. It still covers only the seven declared inputs. It neither proves the universal mixed-conductor grammar nor constructs a Chow correspondence.","\\section{Why the degree-210 dual is global}\nThe dual uses numerator vector $(31,87,57,101,84,60)$ and denominator 420. For invariant nonnegative columns of degrees 1 through 13, the exact enumerator checks all 27,131 compositions. Exactly 143 satisfy the congruence, and each has dual numerator at least 420.","For degree at least 14, every coordinate cost is at least 31, so the numerator is at least $14\\cdot31=434>420$. Thus the finite premise and the uniform tail cover every nonzero invariant column. The global impossibility statement is proved by this argument, rather than extrapolated from the degree-18 search.","\\section{Uniform formula versus finite benchmark}\nThe five-phase determinant formula is proved by wedge products and polynomial identities for every positive integer parameter in its stated range. Finite determinants at $q=2,3,4,5,7,11,17$ independently test the implementation and row sign. Socle pairing is separately checked for every $b$ at $q=2,7,11$.","Take $V=\\Span_{\\R}\\{e_1\\}$ in $\\R^2$ and let the only cycle vector be $c=e_1+e_2$. Then\n\\[\nS=cc^t=\\begin{pmatrix}1&1","Take $V=\\Span_{\\R}\\{e_1\\}$ in $\\R^2$ and let the only cycle vector be $c=e_1+e_2$. Then\n\\[\nS=cc^t=\\begin{pmatrix}1&1\\\\1&1\\end{pmatrix}.\n\\]\nThe compressed Gram matrix on $V$ is the positive $1\\times1$ matrix $(1)$, while $e_1$ does not belong to $\\Span(c)$. Positivity proves that the target direction is detected, not that synthesis returns it inside the original cycle span.","1&1\\end{pmatrix}.\n\\]\nThe compressed Gram matrix on $V$ is the positive $1\\times1$ matrix $(1)$, while $e_1$ does not belong to $\\Span(c)$. Positivity proves that the target direction is detected, not that synthesis returns it inside the original cycle span.","The corrected theorem requires $c_j\\in V$, obtainable through an explicitly algebraic rational packet projector when available. Without that condition, the exact containment test is\n\\[\n\\rank_{\\Q}[C\\ H_{\\Q}]=\\rank_{\\Q}C.\n\\]\nA rational target basis $H_{\\Q}$ is distinct from an orthonormal real basis, since orthonormalization can introduce irrational entries. The old source is preserved unchanged and this correction governs the current reader.","\\section{A parameterized weight sum did not identify a new packet family}\nAn exploratory modification replaced the exponent 7 in the old carrier matrix by a parameter. The formal normalized weights remained positive and summed to $1+7/(10q)$. That calculation alone did not verify a desired full-support character on the projective quotient. It established no uniform new primitive-packet family.","\\section{An invalid exploratory input}\nThe exploratory modulus-45 tuple\n\\[\n(1,10,19,28,36,37)\n\\]\nhas sum 131, whereas a middle-balanced six-tuple at that modulus must sum 135. It is not even an admissible starting Hodge character. Its reported candidate capacity $34/45$ is therefore not a Hodge result.","The invalid record remains in the historical discovery evidence and is excluded from the seven authoritative capacity inputs. The exact verifier checks every unit-conjugate sum before accepting a capacity claim. The general lower bound $C(a)\\ge1$ for a balanced character also flags the erroneous subunit value.","\\section{The five-phase cycles were already known}\nThe explicit $Z_\\kappa$ equations recover the Aoki standard family. The October source audit obtained the original article and checked that its elementary-symmetric construction becomes these equations through Newton identities. A direct determinant and period recomputation is valuable reproducible work, and is not a newly solved open Hodge family.","\\section{Arbitrary smooth projective varieties}\nThe real obstruction projector vanishes exactly when the algebraic and rational Hodge spaces agree. It provides no algorithm for computing the complete algebraic space. A finite cycle family, actual cohomology coordinates and an exact spanning certificate would close a specified $(X,p)$ case.","The analytic branch requires vanishing of the complete finite obstruction $\\Phi$, after the proposed stabilization. The range contraction solves only the range equation. Hodge-type Chern characters or trace moments do not force an endomorphism-valued defect to vanish.","\\section*{The fresh evidence}\nThe October replay passed the 24 grouped carrier checks, 46 grouped transfer checks and 137 grouped capacity/cycle assertions. It independently reconstructed 364 valid K23 sign certificates and all dyadic census records for moduli 8 through 128. It checked 87,376 weighted dyadic coefficient identities for $3\\le s\\le9$. The archival 89,478,480-count dyadic report has a larger range and was not replayed in full. The archived projector toy models use floating-point checks with explicit tolerances. The current figures use exact retained data or explicitly schematic structure, and are not themselves algebraic cycle constructions on an actual variety.","\\begin{tabular}{rrrrrr}\\toprule $m$ & Full orbits & Pair & SAT(2) & SAT(3) & Unresolved","\\begin{tabular}{rrrr}\\toprule $m$ & Full orbits & Pair/SAT(2) & Residual","\\[\nX_m^n=\\{x_0^m+\\cdots+x_{n+1}^m=0\\}\\subset\\mathbf P^{n+1}_{\\mathbf C}.\n\\]","\\begin{enumerate}\n\\def\\labelenumi{\\arabic{enumi}.}\n\\tightlist\n\\item\n  \\textbf{Algebraicity transfer.} The rational primitive Hodge packet\n  associated to \\[\n  b=(2,9,129,142,168,180)\\pmod {210}\n  \\] is algebraic, using the v0.1.0 degree-70 carrier proposition and\n  classical Fermat cycle correspondences. Its rational dimension is 48.\n  Its full conductor is 210, so it is not a coordinate-power inflation\n  of a lower-conductor packet. A 14-term signed identity supplies the\n  transfer.\n\\item\n  \\textbf{Global matrix obstruction.} No nonnegative integral matrix\n  capturing this packet can have positive normalized weights whose sum\n  exceeds one. This is proved for every matrix size-six candidate of the\n  previous criterion, not just for a bounded search. An exact dual\n  certificate has two independent finite verifications, with a uniform\n  argument covering all larger columns.\n\\item\n  \\t","A full-support character is an ordered tuple \\(a=(a_0,\\ldots,a_{n+1})\\)\nof nonzero residues modulo \\(m\\), with sum zero modulo \\(m\\). Its\neigenspace in primitive cohomology is a complex line. For even \\(n\\),\nits rational Galois packet is of middle Hodge type precisely when","\\begin{equation}\\label{eq:transfer210-2.1}\n\\sum_i\\langle ta_i\\rangle_m=\\frac{n+2}{2}m\n\\quad\\text{for every }t\\in(\\mathbf Z/m\\mathbf Z)^\\times.\\end{equation}","Write \\(\\nu(a)\\) for its integer residue-multiplicity vector. Write\n\\(a\\uplus b\\) for concatenation regarded as a multiset. Permuting\ncoordinates is an algebraic automorphism, so multiset identities suffice\nfor algebraicity transfer. Write \\(-a\\) for coordinatewise negation\nmodulo \\(m\\).","\\textbf{Complementary-pair removal.} If a character becomes algebraic\nafter concatenating any finite collection of pairs \\((r,m-r)\\), then it\nwas algebraic already.","\\textbf{Standard blocks.} For an odd prime \\(p\\mid m\\), put \\(d=m/p\\).\nAoki \\cite{Aoki}, Theorem 2-1 constructs an algebraic cycle representing","\\[\n\\sigma_{p,i}=(i,i+d,\\ldots,i+(p-1)d,m-pi)\n\\]","under the relevant nonzero-entry and coprimality conditions. We only use\n\\((m,p,i)=(35,5,5)\\) and \\((35,7,2)\\). Here \\(\\gcd(5,7)=\\gcd(2,5)=1\\),\nand all entries are nonzero.","\\textbf{Power transport.} The finite coordinate-power morphism\n\\(X_{km}^n\\to X_m^n\\) transports algebraic cycle classes by pullback.\nThe character is multiplied by \\(k\\).","\\begin{equation}\\label{eq:transfer210-4.1}\na=(1,20,24,42,61,62)\\pmod {70}.\\end{equation}","The proof uses a concrete monomial correspondence to a normal weighted\ndegree-30 hypersurface with weights \\((10,6,3,3,3,8)\\). Its resolution\nis shown to be uniruled, so the Conte--Murre Hodge theorem for uniruled\nfourfolds applies. See \\cite{ArapuraUniruled}, Corollary 4.2. Boundary\nvanishing, finite étale descent and lowest-weight lifting then transfer\ncycles back.","Pullback under \\(X_{210}^4\\to X_{70}^4\\) gives the algebraic source\npacket","\\begin{equation}\\label{eq:transfer210-4.2}\ns=3a=(3,60,72,126,183,186)\\pmod {210}.\\end{equation}","\\begin{equation}\\label{eq:transfer210-4.3}\nb=(2,9,129,142,168,180)\\pmod {210}\\end{equation}","\\[\nP_k=(k,210-k)\n\\]","\\[\n\\begin{aligned}\nQ_3&=(3,105,108,204),\\\\\nQ_9&=(9,105,114,192),\\\\\nQ_{24}&=(24,105,129,162),\\\\\nQ_{27}&=(27,105,132,156),\\\\\nR_2&=(2,72,142,204).\n\\end{aligned}\n\\]","\\[\nS_5=(30,60,72,114,156,198),\n\\qquad\nS_7=(12,42,72,102,126,132,162,192).\n\\]","\\[\n\\begin{aligned}\nS_5/6&=(5,10,12,19,26,33)=\\sigma_{5,5}\\quad\\text{at degree }35,\\\\\nS_7/6&=(2,7,12,17,21,22,27,32)=\\sigma_{7,2}\\quad\\text{at degree }35,\n\\end{aligned}\n\\]","\\begin{equation}\\label{eq:transfer210-5.1}\n\\boxed{\n\\begin{aligned}\n b\\uplus P_{24}\\uplus P_{27}\\uplus P_{105}\\uplus Q_3\\uplus S_5\\uplus S_7\n ={}&s\\uplus P_{12}\\uplus P_{30}\\uplus P_{42}\\uplus P_{102}\\\\\n &\\uplus Q_9\\uplus Q_{24}\\uplus Q_{27}\\uplus R_2.\n\\end{aligned}}\\end{equation}","Both sides contain exactly 30 entries, with equal multiplicity of every\nresidue. Equivalently, \\(\\nu(b)-\\nu(s)\\) is the signed sum of the\nfourteen displayed auxiliary generators, each with coefficient \\(+1\\) or\n\\(-1\\). The certificate is stored in \\texttt{results/transfer210.json}.","Assuming the inherited degree-70 carrier proposition, every rational\nclass in \\(W_{[b]}\\) is algebraic.","\\textbf{Proof.} Let \\(L\\) be the auxiliary multiset to the left of \\eqref{eq:transfer210-5.1}\nand \\(R\\) the auxiliary multiset to its right. Thus","\\[\nb\\uplus L=s\\uplus R,\n\\qquad |L|=|R|=24.\n\\]","All component packets of \\(L\\) and \\(R\\) are algebraic. Negation\npreserves their algebraicity, either by the rational Galois packet\nstatement or by conjugating the corresponding cycle class. Append \\(-L\\)\nto both sides:","\\begin{equation}\\label{eq:transfer210-5.2}\nb\\uplus L\\uplus(-L)=s\\uplus R\\uplus(-L).\\end{equation}","The right-hand side is algebraic by concatenation, because the source\nand every auxiliary component are algebraic. The left-hand auxiliary\npart \\(L\\uplus(-L)\\) is a multiset of exactly 24 complementary pairs.\nThe full character has 54 entries and is a character on\n\\(X_{210}^{52}\\). Complementary-pair removal therefore gives\nalgebraicity of \\(b\\) on \\(X_{210}^4\\).","The character projector and its rational Galois orbit transfer this\nstatement to every class in \\(W_{[b]}\\). Its dimension is 48. This\nproves the proposition. \\hfill\\ensuremath{\\square}","For every \\(k\\ge1\\), coordinate-power pullback gives the associated\npacket on \\(X_{210k}^4\\). It does not establish the Hodge conjecture for\nall packets on any of these fourfolds.","The previous search criterion required a nonsingular nonnegative\nintegral matrix \\(A\\), with columns \\(v_j\\), satisfying","\\begin{equation}\\label{eq:transfer210-6.1}\nb\\cdot v_j\\equiv0\\pmod {210},\\qquad\nw=A^{-1}\\mathbf1>0,\\qquad\n\\sum_jw_j>1.\\end{equation}","\\begin{equation}\\label{eq:transfer210-6.2}\n\\eta=\\frac1{420}(31,87,57,101,84,60).\\end{equation}","Its coordinates are positive and sum to one. For every nonzero\n\\(v\\in\\mathbf Z_{\\ge0}^6\\) satisfying \\(b\\cdot v\\equiv0\\pmod {210}\\),","\\begin{equation}\\label{eq:transfer210-6.3}\n\\boxed{\\eta\\cdot v\\ge1.}\\end{equation}","If \\(|v|\\ge14\\), the numerator of \\(\\eta\\cdot v\\) is at least\n\\(14\\cdot31=434>420\\). It therefore suffices to check \\(1\\le |v|\\le13\\).","As an independent implementation, place residue states in\n\\(\\mathbf Z/210\\) and give the six transitions \\(r\\mapsto r+b_i\\)\ninteger costs \\((31,87,57,101,84,60)\\). A nonempty zero-sum column is a\nreturn walk. Exact shortest-path computation finds minimum nonempty\nreturn cost 420. All costs are positive. This verifies the same\ninequality by a different method.","\\[\n1=\\eta^T\\mathbf1=\\eta^TAw\n =\\sum_jw_j(\\eta\\cdot v_j)\\ge\\sum_jw_j.\n\\]","This contradicts \\(\\sum_jw_j>1\\). \\hfill\\ensuremath{\\square}","The single invariant column \\(v=\\mathbf1\\), with coefficient one,\nsupplies a matching value-one witness for the relaxed column linear\nprogram. Thus its global optimum is exactly one. This is not a claim\nthat a nonsingular six-column positive solution attains that optimum.","The invariant-column condition is unchanged by unit-conjugating \\(b\\),\npermuting its coordinates with the corresponding dual permutation, or\ninflating \\(b\\) and the modulus together. The obstruction persists under\nthose operations.","For any integer \\(q>5\\), let","\\begin{equation}\\label{eq:transfer210-7.1}\na_q=(1,1+q,1+2q,1+3q,1+4q,5q-5)\\pmod {5q}.\\end{equation}","Every \\(a_q\\) is a full-support middle-balanced character. No matrix\nsatisfying the previous direct positive-weight criterion exists for\n\\(a_q\\).","\\textbf{Proof of balance.} Let \\(t\\) be a unit modulo \\(5q\\), and let\n\\(s\\) be its least positive residue modulo \\(q\\). The first five\nmultiplied residues permute","\\[\ns,s+q,s+2q,s+3q,s+4q.\n\\]","Their sum is \\(5s+10q\\). The last multiplied residue is \\(5(q-s)\\), so\nthe total is \\(15q=3(5q)\\).","\\[\n\\eta_q=\\frac1q(1,1,1,1,1,q-5).\n\\]","This vector is positive and has sum one. For a nonzero invariant column\n\\(v\\ge0\\), put \\(S=v_0+\\cdots+v_4\\). Reducing\n\\(a_q\\cdot v\\equiv0\\pmod {5q}\\) modulo \\(q\\) gives\n\\(S-5v_5\\equiv0\\pmod q\\). Therefore","\\[\n\\eta_q\\cdot v=\\frac{S+(q-5)v_5}{q}\n=\\frac{S-5v_5}{q}+v_5\n\\]","This is a proof for every \\(q>5\\), not an extrapolation from finite\nruns. The verifier's checks for \\(6\\le q\\le80\\) are benchmarks only.\nThese packets are themselves the standard family \\(\\sigma_{5,1}\\), so\nthe obstruction is to the method, not to algebraicity.","Let \\(a\\) be a full-support Fermat fourfold character. Suppose some\ninteger multiple","\\[\nc=ka\\pmod m\n\\]","has all entries nonzero and least-positive residue sum \\(m\\). A direct\nquotient by a subgroup of the diagonal automorphism group that captures\n\\(a\\) cannot have a smooth projective uniruled compactification.","\\textbf{Proof.} If \\(K\\) is a diagonal deck subgroup and\n\\(\\chi_a|_K=1\\), then","\\[\n\\chi_c|_K=\\chi_a^k|_K=1.\n\\]","The residue description places the \\(c\\)-line in \\(H^{4,0}\\) under the\ndirect-action convention (or its conjugate under the inverse-character\nconvention). Its rational Galois span contains a nonzero \\((4,0)\\) part\nand is also \\(K\\)-invariant.","For any smooth projective compactification, the lowest-weight image\ntheorem and the semisimplicity of polarizable pure Hodge structures lift\nthis space from open cohomology to compact cohomology. In particular the\ncompactification has a nonzero \\(H^{4,0}\\). These are the same classical\nmixed-Hodge inputs as in the predecessor, see \\cite{ArapuraUniruled},\n\\cite{DeligneHodgeII}.","A smooth projective uniruled fourfold has no nonzero holomorphic top\nform: a generically finite dominant parametrization from a resolution of\na threefold times \\(\\mathbf P^1\\) would pull such a form back\ninjectively, whereas the latter has none. This is a contradiction. \\hfill\\ensuremath{\\square}","The general Hodge conjecture is not established. This block does not\ncover all packets on \\(X_{210}^4\\), establish universal closure of the\nFermat packet calculus, or reduce arbitrary projective varieties to\ndiagonal hypersurfaces. It does not complete the earlier central-defect\nrealization lemma or a general CM half-twist Chow correspondence.","\\section{The lattice, polarization and seed}\nLet $K=\\Q(i)$ and take homology lattice $\\Z[i]^6$. Use the conjugate-linear-first Hermitian form\n\\[\nh(v,w)=\\overline v^{\\,t}\\operatorname{diag}(1,1,1,-1,-1,-3)w,\n\\qquad E=\\Im h.\n\\]\nThe seed complex structure is\n\\[\nJ_0=\\operatorname{diag}(i,i,i,-i,-i,-i).\n\\]\nIt preserves the alternating form $E$, and $E(v,J_0v)>0$ for nonzero $v$. These are the Riemann positivity conditions for the seed abelian variety. The sign depends on the declared Hermitian convention. Omitting that convention would leave the polarization claim ambiguous.","Write $(x_1,y_1,\\ldots,x_6,y_6)$ for the integral dual basis. Define\n\\[\n\\Omega_6=\\bigwedge_{j=1}^{6}(x_j+iy_j),\\qquad\nu=\\Re\\Omega_6,\\qquad v=\\Im\\Omega_6.\n\\]\nBoth $u$ and $v$ are explicit integral six-forms, with 32 nonzero terms each. An unspecified complex determinant-line generator would not by itself define these rational classes. The chosen expansion removes that ambiguity.","For complex structures in the Weil domain with three positive and three negative $K$-embedding directions, $\\Omega_6$ has type $(3,3)$. Indeed, the $K$-determinant contains three one-forms from each Hodge side. Thus the rational span\n\\[\nW=\\Span_{\\Q}\\{u,v\\}\n\\]\nis a Hodge substructure of degree six across the domain. The first-order computation confirms that persistence in all 18 real tangent directions at the seed. The all-domain assertion comes from the determinant Hodge-type argument, rather than extrapolation from those tangent tests.","\\section{The intersection form and primitive directions}\nUse the orientation\n\\[\n\\int x_1y_1\\cdots x_6y_6=-1\n\\]\nfor this seed. If $\\theta$ is the polarization class specified by $E$, the exact Gram matrix on $(\\theta^3,u,v)$ is\n\\[\n\\begin{pmatrix}\n2160&0&0","\\section{The intersection form and primitive directions}\nUse the orientation\n\\[\n\\int x_1y_1\\cdots x_6y_6=-1\n\\]\nfor this seed. If $\\theta$ is the polarization class specified by $E$, the exact Gram matrix on $(\\theta^3,u,v)$ is\n\\[\n\\begin{pmatrix}\n2160&0&0\\\\\n0&-32&0\\\\\n0&0&-32\n\\end{pmatrix}.\n\\]\nThe exterior calculation also gives $\\theta\\wedge u=\\theta\\wedge v=0$, so the two Weil directions are primitive. The negative signs are compatible with the middle-degree Hodge--Riemann convention and the stated orientation.","0&0&-32\n\\end{pmatrix}.\n\\]\nThe exterior calculation also gives $\\theta\\wedge u=\\theta\\wedge v=0$, so the two Weil directions are primitive. The negative signs are compatible with the middle-degree Hodge--Riemann convention and the stated orientation.","For any class $c$ in degree six, the orthogonal projection onto $W$ under this intersection pairing is\n\\begin{equation}\\label{eq:weilprojection}\nP_W(c)=-\\frac{\\langle c,u\\rangle u+\\langle c,v\\rangle v}{32}.\n\\end{equation}\nThe verifier checked this identity on every one of the $\\binom{12}{6}=924$ rational exterior basis vectors. Its output has image of dimension two, with image basis $u,v$. This is an exact finite-dimensional statement.","Let\n\\[\nS=[2+i]^*+[2-i]^*\n\\]\nact on $H^6$. The exact eigenvalues and dimensions are\n\\[\n\\begin{array}{c|rrrr}\n\\text{eigenvalue}&-234&-70&150&250","Let\n\\[\nS=[2+i]^*+[2-i]^*\n\\]\nact on $H^6$. The exact eigenvalues and dimensions are\n\\[\n\\begin{array}{c|rrrr}\n\\text{eigenvalue}&-234&-70&150&250\\\\\n\\text{dimension}&2&72&450&400 .\n\\end{array}\n\\]\nTheir dimensions sum to 924. The eigenvalue $-234$ selects $W$, because\n\\[\n(2+i)^6=-117+44i,\\qquad\n(2-i)^6=-117-44i.\n\\]\nThe rational polynomial\n\\begin{equation}\\label{eq:weilisogeny}\nP_W=\\frac{(S+70I)(S-150I)(S-250I)}{-30480384}\n\\end{equation}\nequals \\eqref{eq:weilprojection}. The denominator is the product of the three eigenvalue differences at $-234$. It is nonzero, and the polynomial takes value one on $W$ and zero on the other three eigenspaces.","\\text{dimension}&2&72&450&400 .\n\\end{array}\n\\]\nTheir dimensions sum to 924. The eigenvalue $-234$ selects $W$, because\n\\[\n(2+i)^6=-117+44i,\\qquad\n(2-i)^6=-117-44i.\n\\]\nThe rational polynomial\n\\begin{equation}\\label{eq:weilisogeny}\nP_W=\\frac{(S+70I)(S-150I)(S-250I)}{-30480384}\n\\end{equation}\nequals \\eqref{eq:weilprojection}. The denominator is the product of the three eigenvalue differences at $-234$. It is nonzero, and the polynomial takes value one on $W$ and zero on the other three eigenspaces.","nsions sum to 924. The eigenvalue $-234$ selects $W$, because\n\\[\n(2+i)^6=-117+44i,\\qquad\n(2-i)^6=-117-44i.\n\\]\nThe rational polynomial\n\\begin{equation}\\label{eq:weilisogeny}\nP_W=\\frac{(S+70I)(S-150I)(S-250I)}{-30480384}\n\\end{equation}\nequals \\eqref{eq:weilprojection}. The denominator is the product of the three eigenvalue differences at $-234$. It is nonzero, and the polynomial takes value one on $W$ and zero on the other three eigenspaces.","The maps $[2\\pm i]$ are algebraic endomorphisms of the abelian variety. Their graphs induce the pullback operators. Hence \\eqref{eq:weilisogeny} preserves rational algebraic cycle classes. A nonzero $P_W$ component of any algebraic threefold supplies an algebraic Weil direction, even if the original effective cycle has a large ambient component.","This also corrects a source-level eigencharacter confusion. The polarization cube has eigenvalue\n\\[\n[2+i]^*(\\theta^3)=125\\,\\theta^3=\\operatorname{Norm}(2+i)^3\\theta^3,\n\\]\nwhile the Weil directions have eigenvalues $-117\\pm44i$. The norm-cubed character therefore identifies the polarization cube, and does not identify the two-dimensional Weil subspace. The September audit recorded this local failure in Section 4.3 of the cited secant-cycle preprint. That specific calculation is not a full audit of the preprint \\cite{SecantPreprintAudit}.","\\section{The seed graph supplies both rational Weil classes}\nAt the seed, take the graph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny","\\section{The seed graph supplies both rational Weil classes}\nAt the seed, take the graph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil direction","\\section{The seed graph supplies both rational Weil classes}\nAt the seed, take the graph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad","\\section{The seed graph supplies both rational Weil classes}\nAt the seed, take the graph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directi","e seed graph supplies both rational Weil classes}\nAt the seed, take the graph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual gr…","raph of\n\\[\n(z_1,z_2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","2,z_3)\\longmapsto\n(z_1,z_2,z_3,\\overline z_1,\\overline z_2,\\overline z_3).\n\\]\nConjugation preserves the $\\Z[i]$ lattice and is holomorphic from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","c from an elliptic factor with complex structure $i$ to a target factor with complex structure $-i$. Thus this map defines an abelian-subvariety graph and is algebraic at $J_0$. On its three source factors the polarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","olarization restricts to weights $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","hts $(2,2,4)$, giving $\\int_Z\\theta^3=3!\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","\\cdot2\\cdot2\\cdot4=96$. Pulling the explicit determinant form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","nt form back to the graph gives the other two periods. In full,\n\\[\n\\left(\\int_Z\\theta^3,\\int_Zu,\\int_Zv\\right)=(96,0,-8).\n\\]\nIts exact Poincaré dual has 64 nonzero basis terms. Pairing that dual against all 924 basis vectors reproduces the graph periods. The projector gives\n\\[\nP_W[Z]=v/4.\n\\]\nThe isogeny $[1+i]$ acts on the Weil directions by\n\\[\n[1+i]^*u=8v,\\qquad[1+i]^*v=-8u.\n\\]\nIt follows that\n\\[\nv=4P_W[Z],\\qquad\nu=-\\tfrac12[1+i]^*P_W[Z].\n\\]\nThus both directions are algebraic at this seed, through rational combinations of actual graph and isogeny correspondences.","The class $v$ is primitive. It need not be the class of a single effective threefold. For a nonzero effective codimension-three cycle on a polarized sixfold, its pairing with $\\theta^3$ is positive. A primitive $v$ has pairing zero. The correct target permits signed rational cycles, or asks for a nonzero projection of an arbitrary effective cycle. Looking for a Chow component whose individual effective members equal $v$ would impose the wrong problem.","itive. It need not be the class of a single effective threefold. For a nonzero effective codimension-three cycle on a polarized sixfold, its pairing with $\\theta^3$ is positive. A primitive $v$ has pairing zero. The correct target permits signed rational cycles, or asks for a nonzero projection of an arbitrary effective cycle. Looking for a Chow component whose individual effective members equal $v$ would impose the wrong problem.","\\section{A representative can obstruct without obstructing its class}\nThe Weil domain has real tangent dimension 18 at the seed. For the selected graph, the first-order graph-preservation constraints have real rank 12 in the nonsplit case with weights $(1,1,3)$. The graph-preserving tangent space therefore has real dimension 6, or complex dimension 3.","The split control with weights $(1,1,1)$ has constraint rank 6, leaving real dimension 12, or complex dimension 6. These ranks were checked both in the exterior calculation and by a separate exact real matrix construction.","The nonsplit graph fails to follow many nearby Weil directions, even though $u,v$ remain Hodge classes. This is a failure of that geometric representative. It does not show that every cycle representing the same rational class fails, and it does not bound all components of a Chow variety.","To prove a Hodge counterexample one would need a global statement that every algebraic threefold has zero pairing with both $u$ and $v$ at some fully mixed member, while those rational Hodge classes remain nonzero. A tangent rank bound for one graph provides no such statement.","\\section{The product locus and replacement factorizations}\nSplit the six directions into a four-dimensional Weil factor $C$ and a two-dimensional factor $B$. The determinant forms factor:\n\\[\n\\Omega_6=\\Omega_C\\wedge\\Omega_B.\n\\]\nWriting their real and imaginary parts as $c_R,c_I,b_R,b_I$ gives\n\\[\nu=c_Rb_R-c_Ib_I,\\qquad\nv=c_Rb_I+c_Ib_R.\n\\]\nOn the product locus, the degree-four Weil classes on $C$ are algebraic by the checked fourfold theorem, and the degree-two classes on $B$ are divisors by Lefschetz $(1,1)$. Their products give the degree-six classes. This locus has complex dimension five. The product argument does not establish the fully mixed sixfold statement \\cite{MarkmanWeil,MarkmanICM2026}.","rections into a four-dimensional Weil factor $C$ and a two-dimensional factor $B$. The determinant forms factor:\n\\[\n\\Omega_6=\\Omega_C\\wedge\\Omega_B.\n\\]\nWriting their real and imaginary parts as $c_R,c_I,b_R,b_I$ gives\n\\[\nu=c_Rb_R-c_Ib_I,\\qquad\nv=c_Rb_I+c_Ib_R.\n\\]\nOn the product locus, the degree-four Weil classes on $C$ are algebraic by the checked fourfold theorem, and the degree-two classes on $B$ are divisors by Lefschetz $(1,1)$. Their products give the degree-six classes. This locus has complex dimension five. The product argument does not establish the fully mixed sixfold statement \\cite{MarkmanWeil,MarkmanICM2026}.","al Weil factor $C$ and a two-dimensional factor $B$. The determinant forms factor:\n\\[\n\\Omega_6=\\Omega_C\\wedge\\Omega_B.\n\\]\nWriting their real and imaginary parts as $c_R,c_I,b_R,b_I$ gives\n\\[\nu=c_Rb_R-c_Ib_I,\\qquad\nv=c_Rb_I+c_Ib_R.\n\\]\nOn the product locus, the degree-four Weil classes on $C$ are algebraic by the checked fourfold theorem, and the degree-two classes on $B$ are divisors by Lefschetz $(1,1)$. Their products give the degree-six classes. This locus has complex dimension five. The product argument does not establish the fully mixed sixfold statement \\cite{MarkmanWeil,MarkmanICM2026}.","The selected mixed tangent direction $B_{13}=1$ breaks the original factorization infinitesimally: the old surface factor loses type $(1,1)$ and the old fourfold factor loses type $(2,2)$. Nevertheless the full $u,v$ remain type $(3,3)$. At first this looked like a promising obstruction.","The exact audit then found replacement divisor factors using the pairs\n\\[\n(1,6),\\qquad(2,5),\\qquad(3,4).\n\\]\nThese replacement factors remain of divisor type along the selected tangent. With\n\\[\nX=-\\tfrac12(\\Delta J)J_0,\\qquad\nJ(t)=e^{tX}J_0e^{-tX},\n\\]\nthe pair blocks are preserved by the integrating generator. Thus the alternate factorization defeats the tested counterexample path. It proves no universal result for every fully mixed point, and it invalidates the inference that failure of the original factorization establishes nonalgebraicity.","\\section{The nilpotent does not construct a degeneration}\nThe exact computation also produced a nilpotent in the unitary Lie algebra, with block\n\\[\ni\\begin{pmatrix}1&-1\\\\1&-1\\end{pmatrix}\n\\]\non each of the pairs $(e_1,e_4)$ and $(e_2,e_5)$. It has square zero, rank two over $K$ and rank four over $\\Q$. The remaining Hermitian plane is $\\operatorname{diag}(1,-3)$.","1&-1\\end{pmatrix}\n\\]\non each of the pairs $(e_1,e_4)$ and $(e_2,e_5)$. It has square zero, rank two over $K$ and rank four over $\\Q$. The remaining Hermitian plane is $\\operatorname{diag}(1,-3)$.","Modulo 3, the equation $x^2+y^2=0$ has only the solution $(0,0)$. This is a finite anisotropy control for the residual discriminant. It does not construct a semistable algebraic family with this nilpotent as monodromy. An abstract rational nilpotent satisfying Lie-algebra identities is weaker than a geometric degeneration with the required integral, polarization and admissibility data.","\\section{The remaining question}\nFor a fully mixed member of this discriminant-$-3$ Weil family, does there exist an algebraic threefold $Z$ such that\n\\[\n\\left(\\int_Zu,\\int_Zv\\right)\\ne(0,0)?\n\\]\nOne such nonzero period, followed by the algebraic projector and the $[1+i]$ action, would generate both Weil directions. The tested graph is one source at the seed. The product and replacement factorization give additional sources on special loci. Exhausting every fully mixed member is a separate task.","The checked Markman results cover abelian fourfolds of Weil type and the stated split or discriminant-$-1$ sixfold cases. They are not silently extended to the nonsplit discriminant-$-3$ sixfold considered here. The source contracts retain these hypotheses explicitly."]}