Paper VI: Transfer Semigroups: Stochastic, Bayesian, Fractional, and PDE Realizations

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1 \begingroup
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5 \vpage{1}{Three ways to reduce a model}{PAPER VI / A WORKED VISUAL ATLAS}{Traditional Part I: Sections 1--3, 8, and 13. Read the three operations separately.}
6 This guide follows small objects that can be computed completely. A four-state chain connects packet transfer, Markov aggregation, and Bayesian conditioning. A two-state coupled evolution exposes memory. A three-node Dirichlet problem makes the Green-operator normalization visible. These models are related by the paper's language, not by pretending that their state spaces are identical.
7 \vfig{three_views}
8 \platecaption{Left and center use one probability vector and its two block masses. The right panel is a separate retained/hidden model, developed on \pagerefV{22}. Every curve is drawn from an explicit formula.}
9 \head{Coordinates retain; aggregation forgets; elimination remembers}
10 Keeping every component $P_a x$ is an exact change of representation when $\sum_aP_a=I$. Reconstruct by summing the components. In contrast, replacing $(p_0,p_1,p_2,p_3)$ by $(p_0+p_1,p_2+p_3)$ forgets how mass was distributed inside either pair. That information can matter for later observations.
11 
12 Eliminating a hidden state is different again. Its coordinates disappear from the displayed state, but the exact retained equation acquires a history integral. Simply deleting the coupling produces the dashed curve above, not the original evolution.
13 \begin{center}\small
14 \begin{tabularx}{\linewidth}{@{}lYl@{}}\toprule
15 Question & Worked route & Pages\\\midrule
16 When is coarse evolution exact? & Aggregate columns, then test intertwining. & \pagerefV{4}--\pagerefV{6}\\
17 When is a coarse posterior exact? & Compare two hidden shapes with the same masses. & \pagerefV{9}--\pagerefV{13}\\
18 What changes under a fractional power? & Keep modes, change spectral rates. & \pagerefV{15}--\pagerefV{17}\\
19 What returns after elimination? & Compute the self-energy and memory kernel. & \pagerefV{21}--\pagerefV{25}\\\bottomrule
20 \end{tabularx}\end{center}
21 \takeaway{The paper supplies exact identities and conditional bounds. Whether a particular coarse model satisfies their hypotheses is a separate calculation.}
22 
23 \vpage{2}{A dictionary with orientations}{READ BEFORE MULTIPLYING / CONVENTIONS}{Traditional Part I: Sections 2, 4, 6, 7, and 12. Column probabilities and backward observables.}
24 An apparent disagreement between two transfer formulas often comes from acting on different objects. Throughout this guide, probabilities are columns and observables are functions written as columns. Their pairing is $f^{\mathsf T}p$. The transpose distinguishes the forward action on distributions from the backward action on observations.
25 \begin{center}\small
26 \begin{tabularx}{\linewidth}{@{}lYY@{}}\toprule
27 Object & Exact meaning & What to check\\\midrule
28 $p$ & A nonnegative column with $\ones^{\mathsf T}p=1$. & It is not a vector of function values.\\
29 $Q$ & Continuous Markov generator. & Off-diagonal entries are nonnegative and columns sum to zero.\\
30 $K$ & One discrete stochastic step. & $\ones^{\mathsf T}K=\ones^{\mathsf T}$ and $K_{ij}\ge0$.\\
31 $K^{\mathsf T}f$ & Expected next observable given the current state. & The same dynamics, with the opposite action on the dual object.\\
32 $P_a$ & A coordinate packet projector. & All packets together reconstruct the full state.\\
33 $C$ & Sums fine probability masses into coarse blocks. & Usually rectangular and not invertible.\\
34 $E_\pi$ & Conditional expectation of observables. & A weighted average, self-adjoint for the $\pi$ inner product.\\
35 $\mathcal B_\lambda$ & $\operatorname{diag}(\lambda)p/(\lambda^{\mathsf T}p)$. & Nonnegative likelihood and positive total evidence.\\
36 $L$ & A nonnegative self-adjoint operator. & Its evolution is $e^{-tL}$, not $e^{tL}$.\\
37 $R(z)$ & $(zI-A)^{-1}$. & $z$ belongs to the resolvent set.\\
38 $\mathcal K(t)$ & The memory kernel after hidden-state elimination. & It is not the transition matrix $K$.\\\bottomrule
39 \end{tabularx}\end{center}
40 \head{One convention test}
41 For
42 \[
43 K_2=\begin{pmatrix}3/4&1/4\\1/4&3/4\end{pmatrix},
44 \qquad p=e_0,\qquad f=(2,1)^{\mathsf T},
45 \]
46 we obtain $K_2p=(3/4,1/4)^{\mathsf T}$ and
47 \[
48 f^{\mathsf T}K_2p=7/4=(K_2^{\mathsf T}f)_0.
49 \]
50 With departure weights $w_i=e^{-V(i)}$, the backward Feynman--Kac operator is $\operatorname{diag}(w)K_2^{\mathsf T}$. Its forward dual is $K_2\operatorname{diag}(w)$. Interchanging these factors changes when the weight is applied.
51 \takeaway{Always name the object being transported. The transpose and the position of the likelihood or potential are mathematical choices, not typographic details.}
52 
53 \vpage{3}{A clock generated by one matrix}{CONTINUOUS TRANSFER / THE ENTIRE SOLUTION}{Traditional Part I: Theorems 1.2 and 4.1. Finite matrix exponentials and Markov positivity.}
54 Take the generator
55 \[
56 G=\begin{pmatrix}-1&1\\1&-1\end{pmatrix}.
57 \]
58 The total-mass direction $(1,1)$ has eigenvalue zero. The contrast $(1,-1)$ has eigenvalue $-2$. Resolving those two directions gives the full transfer, not only its first-order approximation:
59 \[
60 T(t)=e^{tG}=\frac12\begin{pmatrix}1+e^{-2t}&1-e^{-2t}\\1-e^{-2t}&1+e^{-2t}\end{pmatrix}.
61 \]
62 \vfig{clock}
63 \platecaption{A point seed at state zero approaches the uniform distribution. The displayed times were selected so that every bar label is rational. This is continuous-time evolution, not a simulated trajectory.}
64 \head{Two checks explain the mechanism}
65 The constant mode is unchanged, while each elapsed interval multiplies the contrast by $e^{-2t}$. Consequently two intervals multiply their contrast factors:
66 \[
67 e^{-2s}e^{-2t}=e^{-2(s+t)},\qquad T(s)T(t)=T(s+t).
68 \]
69 For $t\ge0$, the entries are nonnegative and every column sums to one. Differentiation gives $T'(t)=GT(t)$. These three facts separately check the initial condition, the evolution equation, and preservation of probability.
70 
71 At $t=\log 2$, the retained first coordinate is $5/8$. It would instead be $1/2$ under the isolated scalar equation $\dot x=-x$. The discrepancy is the mass that can leave state zero and return. On \pagerefV{22}, we remove the second coordinate correctly and recover exactly the same $5/8$ using a memory integral.
72 \takeaway{The semigroup lives on the complete state. One diagonal entry of a semigroup need not itself satisfy a scalar semigroup law.}
73 
74 \vpage{4}{One evolution, two packet choices}{FOUR-STATE MODEL / COORDINATES AND MODES}{Traditional Part I: Theorem 2.1, Corollary 2.2, and the packet path expansion.}
75 Our main finite model has generator
76 \[
77 Q=\begin{pmatrix}-3&1&2&0\\1&-3&0&2\\2&0&-3&1\\0&2&1&-3\end{pmatrix}.
78 \]
79 The graph shows rates between actual states. The four normalized sign vectors on the right form an orthonormal basis $u_j$. Their projectors $P_j=u_ju_j^{\mathsf T}$ sum to the identity and satisfy $P_iP_j=0$ for $i\ne j$.
80 \vfig{generator_modes}
81 \platecaption{State coordinates have off-diagonal transfers. Spectral packets reduce the generator and evolve independently. The two dashed state groups are the coarse partition used next.}
82 \head{The same evolution has a closed formula}
83 Writing $q=e^{-2t}$ gives
84 \[
85 e^{tQ}=P_0+qP_1+q^2P_2+q^3P_3.
86 \]
87 At $q=1/2$, the point seed $e_0$ becomes $(15,5,9,3)^{\mathsf T}/32$. The polynomial identity $T(q)T(r)=T(qr)$ checks composition exactly, without numerical matrix exponentiation.
88 
89 Any other complete packet family still has exact block convolution. For the coordinate groups A and B,
90 \[
91 P_b e^{(s+t)Q}P_a=\sum_{c\in\{\mathrm A,\mathrm B\}}(P_b e^{sQ}P_c)(P_c e^{tQ}P_a).
92 \]
93 The intermediate packet cannot generally be omitted. Only a reducing family eliminates all cross terms.
94 
95 For later discrete calculations define $K=I+Q/8$. This matrix is stochastic, but \emph{it is not} $e^{Q/8}$. It is the step kernel in the uniformization identity $e^{tQ}=e^{-8t}\sum_{n\ge0}(8t)^nK^n/n!$.
96 \takeaway{Changing packet coordinates does not change the evolution. Replacing continuous time by one discrete step does.}
97 
98 \vpage{5}{When two states are truly equivalent}{EXACT LUMPABILITY / AGGREGATE COLUMNS}{Traditional Part I: Theorem 5.1. The hypothesis is an intertwining identity.}
99 Group states $0,1$ into A and states $2,3$ into B. The aggregation map and the candidate coarse kernel are
100 \[
101 C=\begin{pmatrix}1&1&0&0\\0&0&1&1\end{pmatrix},\qquad
102 \bar K=\begin{pmatrix}3/4&1/4\\1/4&3/4\end{pmatrix}.
103 \]
104 A fine state in A has probability $1/4$ of entering B in one step, regardless of which fine state in A was occupied. The analogous statement holds for B.
105 \vfig{lumpability}
106 \platecaption{The fine columns are visibly different. Their sums over each target packet agree within each source packet. That equality is the reason a two-state description works.}
107 \head{A one-step calculation controls every discrete horizon}
108 The displayed columns give $CK=\bar KC$. Multiplication then yields
109 \[
110 CK^n=\bar K^nC\qquad(n\ge0).
111 \]
112 For example, $e_0$ and $e_1$ have the same coarse initial vector $(1,0)^{\mathsf T}$. Their fine evolutions differ, but their coarse vectors agree at every step. No choice of the within-A conditional distribution is needed for the prediction.
113 
114 The continuous generator also closes:
115 \[
116 CQ=\bar Q C,\qquad
117 \bar Q=\begin{pmatrix}-2&2\\2&-2\end{pmatrix}.
118 \]
119 The Duhamel identity therefore gives $Ce^{tQ}=e^{t\bar Q}C$. This is a second consequence of the corresponding generator identity, not an inference that $K$ is a fixed-time sample of $Q$.
120 
121 Aggregation is still not invertible. The theorem says the discarded details are irrelevant for \emph{these predictions}. It does not say they are irrelevant to every later likelihood. The observation counterexample on \pagerefV{10} uses this exact same partition.
122 \takeaway{A partition is not automatically a valid Markov model. The aggregate transition columns must be constant within each source block.}
123 
124 \vpage{6}{A broken lump and its error budget}{DEFECTS / A GUARANTEE WITHOUT A FIT}{Traditional Part I: Theorems 3.1, 3.3, and 5.2. Continuous and discrete defect identities.}
125 Modify only column zero of $K$: move $1/16$ of its mass from target state zero to target state two. The perturbed matrix $K_\varepsilon$ remains stochastic, but its aggregate column now differs from column one. Relative to the unchanged coarse kernel,
126 \[
127 \Delta=CK_\varepsilon-\bar KC
128 =\begin{pmatrix}-1/16&0&0&0\\1/16&0&0&0\end{pmatrix},
129 \qquad \|\Delta\|_1=1/8.
130 \]
131 \vfig{defect_budget}
132 \platecaption{The solid values are computed with rational matrix powers. The dashed certificate uses only stochasticity and the exact one-step defect. It need not be sharp.}
133 \head{Telescoping tells us where every error originates}
134 For $n\ge0$,
135 \[
136 CK_\varepsilon^n-\bar K^nC
137 =\sum_{j=0}^{n-1}\bar K^{n-1-j}\Delta K_\varepsilon^j.
138 \]
139 Each summand propagates one defect between an earlier fine evolution and a later coarse evolution. Column-stochastic matrices have induced $\ell^1$ norm one, so the norm is at most $n/8$. The empty sum at $n=0$ is zero.
140 
141 The continuous counterpart is equally explicit. With fine generator $-1$, coarse generator $-2$, and $C=1$, the generator defect is $1$ and
142 \[
143 e^{-t}-e^{-2t}=\int_0^t e^{-2(t-r)}e^{-r}\,dr.
144 \]
145 Its resolvent version is $1/(z+1)-1/(z+2)=1/((z+1)(z+2))$ for $z>0$.
146 \takeaway{A small defect does not disappear. The identity specifies how it accumulates, and the norm assumptions turn that identity into a finite-horizon certificate.}
147 
148 \vpage{7}{Averaging is not summing}{OBSERVABLES / WEIGHTED ORTHOGONALITY}{Traditional Part I: Theorem 6.1. Strictly positive weights are part of the setting.}
149 For this observable example, choose the fixed probability weights $\pi=(1,2,3,4)/10$. They are not being identified with the stationary law of the earlier symmetric chain. On the same two-block partition, conditional expectation is
150 \[
151 E=\begin{pmatrix}
152 1/3&2/3&0&0\\1/3&2/3&0&0\\0&0&3/7&4/7\\0&0&3/7&4/7
153 \end{pmatrix}.
154 \]
155 It replaces function values by their \emph{weighted block average}. In particular, $f=(0,3,2,5)^{\mathsf T}$ becomes $Ef=(2,2,26/7,26/7)^{\mathsf T}$.
156 \vfig{weighted_projection}
157 \platecaption{The bars are function values, not state probabilities. Their total need not be one. The block masses $3/10$ and $7/10$ determine the denominators in the averaging formula.}
158 \head{Which inner product makes this an orthogonal projection?}
159 Put $W=\operatorname{diag}(\pi)$. Direct multiplication checks
160 \[
161 E^2=E,\qquad WE=E^{\mathsf T}W,\qquad E\ones=\ones.
162 \]
163 The middle equation is weighted self-adjointness. Ordinary symmetry fails: $E_{01}=2/3$ whereas $E_{10}=1/3$. Testing $E=E^{\mathsf T}$ would therefore reject a correct conditional-expectation projector.
164 
165 The range consists exactly of functions constant on each block. The error $f-Ef$ has weighted mean zero inside every block, so it is weighted-orthogonal to every block-constant function. This is the finite projection interpretation of conditional expectation.
166 
167 In contrast, $C$ acts on probability columns by summing masses. The shapes of $C$ and $E$ differ, and their domains have different interpretations. The notation ``coarsening'' alone does not specify which operation is intended.
168 \takeaway{An exact weighted projector can be oblique in the unweighted Euclidean picture. Keep the measure and the inner product in the model.}
169 
170 \vpage{8}{Where the variance goes}{MARTINGALE PACKETS / THREE EXACT PIECES}{Traditional Part I: Theorem 6.2. Nested weighted projections and their differences.}
171 Keep the weights and observable from \pagerefV{7}. Let $E_0f$ be the constant global mean, let $E_1=E$ average within A and B, and let $E_2=I$ retain every value. Define
172 \[
173 D_0=E_0,\qquad D_1=E_1-E_0,\qquad D_2=I-E_1.
174 \]
175 Then $D_0f$ is the mean, $D_1f$ describes the contrast between block means, and $D_2f$ is the fluctuation left inside each block.
176 \vfig{variance}
177 \platecaption{The strip lengths are proportional to the three exact weighted energies. The between-block component is small but not zero. The vectors below specify the corresponding directions, not only their sizes.}
178 \head{The energy identity follows from actual orthogonality}
179 The tower identities give $E_0E_1=E_1E_0=E_0$. Together with weighted self-adjointness, these imply
180 \[
181 D_iD_j=0\ (i\ne j),\qquad \sum_{j=0}^2 D_j=I.
182 \]
183 For the chosen $f$, the total weighted energy is $\|f\|_\pi^2=13$ and the mean is $16/5$. Consequently
184 \[
185 \operatorname{Var}_\pi(f)
186 =13-\left(\frac{16}{5}\right)^2
187 =\frac{69}{25}
188 =\frac{108}{175}+\frac{15}{7}.
189 \]
190 The first term is the energy of the between-block contrast, and the second is the weighted within-block variance. This is a decomposition of one observable in one fixed weighted space.
191 
192 Changing the probability weights between levels would change the inner products and the averaging operators. The displayed orthogonal decomposition would no longer follow merely from nesting the sets. Likewise, choosing unrelated projections can preserve the telescope $\sum D_j=I$ while losing orthogonality and the energy sum.
193 \takeaway{The differences between nested conditional expectations are not just bookkeeping. They are orthogonal, separately measurable parts of the original signal.}
194 
195 \vpage{9}{Evidence changes shape and mass}{BAYESIAN TRANSFER / AN EXACT POSTERIOR}{Traditional Part I: Theorems 7.1--7.3. Normalization, packet reconstruction, and positive evidence.}
196 Start with $p=(1,1,1,1)^{\mathsf T}/4$ and likelihood weights $\lambda=(4,1,1,1)^{\mathsf T}$. The unnormalized vector is $(1,1/4,1/4,1/4)^{\mathsf T}$, its evidence is $Z=7/4$, and
197 \[
198 \mathcal B_\lambda(p)=\frac{(4,1,1,1)^{\mathsf T}}7.
199 \]
200 Likelihood weights may be rescaled, so an evidence larger than one is not a probability claim. Multiplying all likelihoods by one common positive factor leaves this posterior unchanged.
201 \vfig{evidence}
202 \platecaption{The A packet becomes heavier and its internal shape changes. The B packet retains an equal internal split but loses total mass. These are different effects of the same likelihood.}
203 \head{Reconstruct from packet mass times packet shape}
204 The unnormalized packet evidences are $Z_{\mathrm A}=5/4$ and $Z_{\mathrm B}=1/2$. The resulting masses and conditional shapes are
205 \[
206 m_{\mathrm A}=5/7,\quad q_{\mathrm A}=(4/5,1/5,0,0)^{\mathsf T},
207 \qquad m_{\mathrm B}=2/7,\quad q_{\mathrm B}=(0,0,1/2,1/2)^{\mathsf T}.
208 \]
209 Thus $m_{\mathrm A}q_{\mathrm A}+m_{\mathrm B}q_{\mathrm B}$ is exactly the displayed posterior. Keeping only the masses would not reconstruct the fine state.
210 
211 There is no need to divide by a zero packet evidence. For likelihood $(0,0,1,1)$, packet A has mass zero and contributes the zero vector. Only the occupied B packet has a conditional probability vector. If the \emph{total} evidence is zero, however, the Bayesian update is not defined and the implementation rejects the input.
212 \takeaway{Bayesian reconstruction uses two kinds of data: how much posterior mass a packet carries and how that mass is distributed inside the packet.}
213 
214 \vpage{10}{Two priors that fool the coarse model}{SUFFICIENCY / A COMPUTABLE FAILURE WITNESS}{Traditional Part I: Theorem 8.2 and the one-packet exception.}
215 Keep the same likelihood $\lambda=(4,1,1,1)^{\mathsf T}$ and partition A$=\{0,1\}$, B$=\{2,3\}$. Compare
216 \[
217 p^{(a)}=(1/2,0,1/2,0)^{\mathsf T},\qquad
218 p^{(b)}=(0,1/2,1/2,0)^{\mathsf T}.
219 \]
220 Both have coarse prior $(1/2,1/2)^{\mathsf T}$. A procedure given only that two-entry vector cannot distinguish them.
221 \vfig{insufficiency}
222 \platecaption{After the identical fine likelihood is applied, the coarse posteriors are $(4/5,1/5)$ and $(1/2,1/2)$. A single deterministic update of the common coarse prior cannot produce both.}
223 \head{Why this witnesses the necessity of blockwise constancy}
224 The A evidence for $p^{(a)}$ is $2$, whereas it is $1/2$ for $p^{(b)}$. Their B evidence is $1/2$ in both cases. The posterior odds therefore distinguish the hidden distribution inside A.
225 
226 For strictly positive fine and coarse likelihoods, and at least two coarse blocks, the manuscript proves the all-prior criterion
227 \[
228 C\mathcal B_\lambda(p)=\mathcal B_{\bar\lambda}(Cp)\quad\hbox{for every }p
229 \quad\Longleftrightarrow\quad
230 \lambda_i=c\bar\lambda_a\quad(i\in I_a)
231 \]
232 for one global $c>0$. The global scale is invisible after normalization. A different arbitrary scale for each packet is not invisible: it changes posterior odds.
233 
234 If all states are merged into a single packet, every normalized posterior coarsens to $(1)$. That degenerate commutation holds for every likelihood and cannot identify any within-packet information. It is explicitly excluded from the nontrivial criterion.
235 \takeaway{A prior-dependent effective likelihood can always summarize the update. A fixed likelihood that works for every prior is a much stronger requirement.}
236 
237 \vpage{11}{The exact information-loss vector}{BAYESIAN DEFECT / MORE THAN AN ERROR BAR}{Traditional Part I: Theorems 8.1 and 8.4, and Corollary 8.6.}
238 For a prior $p$, let $P=Cp$ and let $r_a(p)$ be the conditional average of the fine likelihood inside packet $a$. For the uniform prior and $\lambda=(4,1,1,1)$,
239 \[
240 P=(1/2,1/2)^{\mathsf T},\qquad r=(5/2,1)^{\mathsf T}.
241 \]
242 The exact coarse posterior is $\mathcal B_r(P)=(5/7,2/7)^{\mathsf T}$. Suppose instead that the declared coarse likelihood is $\bar\lambda=(1,1)$.
243 \vfig{loss_vector}
244 \platecaption{The posterior error lies along the mass-preserving direction of the two-state simplex. Its entries are opposite because both endpoints have total mass one.}
245 \head{Compute the factorization, including the null direction}
246 Here $\bar b=\mathcal B_{\bar\lambda}(P)=P$, $\delta=r-\bar\lambda=(3/2,0)^{\mathsf T}$, and $Z_r=7/4$. Hence
247 \[
248 \frac1{Z_r}\left(\diag(P)-\bar bP^{\mathsf T}\right)\delta
249 =\frac4{7}\begin{pmatrix}1/4&-1/4\\-1/4&1/4\end{pmatrix}
250 \begin{pmatrix}3/2\\0\end{pmatrix}
251 =\begin{pmatrix}3/14\\-3/14\end{pmatrix}.
252 \]
253 This is the actual fine-to-coarse discrepancy, not a fitted approximation. The matrix kills $\bar\lambda$, recording the common-scale null direction. Its $\ell^1$ error is $3/7$.
254 
255 For a family of likelihoods, states may be grouped by equality of their \emph{whole likelihood profile}. The family $(3,3,1,1)$ and $(1,2,1,2)$ separates all four states even though either likelihood alone permits a coarser grouping. The common refinement is therefore the discrete partition. The theorem's one-packet exception still remains.
256 \takeaway{The defect factors through exactly the conditional evidence information missing from the declared coarse model. A new observation family can require a finer partition.}
257 
258 \vpage{12}{A nonlinear semigroup of odds}{BAYESIAN FLOW / COMPOSITION WITHOUT LINEARITY}{Traditional Part I: Theorems 7.1 and 9.1. Likelihood composition and the replicator field.}
259 Let the score vector be $g=(1,0)$ and start at $(1/2,1/2)$. The likelihood at time $t$ is $(e^t,1)$, so the first posterior coordinate is
260 \[
261 p_0(t)=\frac{e^t}{1+e^t},\qquad p_1(t)=\frac1{1+e^t}.
262 \]
263 The odds $p_0/p_1$ grow exponentially, while the probabilities remain inside the simplex. At $t=\log 2$ the posterior is exactly $(2/3,1/3)$.
264 \vfig{odds}
265 \platecaption{The solid and dashed curves sum to one. The curvature is caused by the shared evidence denominator, not by a nonlinear likelihood multiplication law.}
266 \head{Two properties that must not be conflated}
267 If $a>0$ multiplies the odds, the first-coordinate map is
268 \[
269 \phi_a(x)=\frac{ax}{1+(a-1)x}.
270 \]
271 Direct algebra gives $\phi_a(\phi_b(x))=\phi_{ab}(x)$. Thus $a=e^t$ gives a nonlinear semigroup. Nevertheless, $\phi_3(1/2)=3/4$, while the mean of $\phi_3(0)$ and $\phi_3(1)$ is $1/2$. Superposition fails.
272 
273 Differentiation of the normalized flow yields the centered score equation
274 \[
275 \dot p_i=\bigl(g_i-\langle g,p\rangle\bigr)p_i.
276 \]
277 For the displayed model, $\dot p_0=p_0(1-p_0)$. The negative average-score term is what keeps the total derivative zero. Omitting the derivative of the normalizer would not preserve mass.
278 
279 Likelihood composition also preserves the evidence identity: the first evidence times the second conditional evidence equals the combined evidence. Normalization cancels in the posterior, but that factor still matters when comparing data models.
280 \takeaway{A semigroup is a composition law in time. It does not imply that the maps are linear or that they preserve mixtures.}
281 
282 \vpage{13}{Prediction and observation do not commute}{HIDDEN MARKOV FILTERING / TIME ORDER}{Traditional Part I: Proposition 9.3. Linear prediction followed by a positive-evidence update.}
283 Use the two-state kernel $K_2$ from \pagerefV{2}, the initial distribution $e_0$, and likelihood $(1,1/2)$. Predicting first gives $(3/4,1/4)$. Its likelihood-weighted vector is $(3/4,1/8)$, with evidence $7/8$, so the posterior is $(6/7,1/7)$.
284 
285 Updating first leaves $e_0$ unchanged. Predicting afterward gives $(3/4,1/4)$ instead. The order of operations expresses which time the observation concerns.
286 \vfig{filter_order}
287 \platecaption{Both routes produce valid probability vectors. Their difference is therefore not a normalization bug. It is a different conditioning problem.}
288 \head{Keep an unnormalized path alongside the filter}
289 For a sequence of observations, let $D_r=\diag(\lambda^{(r)})$. The forward recursion is
290 \[
291 \widehat p_r=Kp_{r-1},\qquad p_r=D_r\widehat p_r/Z_r.
292 \]
293 Its unnormalized companion is
294 \[
295 v_r=D_rK\cdots D_1Kp_0,
296 \qquad p_r=\frac{v_r}{\ones^{\mathsf T}v_r},
297 \qquad \ones^{\mathsf T}v_r=\prod_{j=1}^r Z_j.
298 \]
299 The product of evidences records the normalization lost at each individual step. The four-state computation on \pagerefV{26} checks both paths independently, including their coarse versions.
300 
301 Here $D_rK$ means an observation after prediction. In the Feynman--Kac example on \pagerefV{20}, departure-state weighting has forward operator $K D_w$. Both conventions are legitimate once their meanings are fixed, but they cannot be substituted for one another.
302 \takeaway{Write the chronological sequence before multiplying matrices. A product can be normalized correctly and still answer the wrong time-ordered question.}
303 
304 \vpage{14}{Conjugacy tracks exponents}{BETA MODEL / POSTERIOR AND EVIDENCE}{Traditional Part I: Example 9.2. A finite-parameter conjugate family, not a universal posterior model.}
305 For a small exact example, take the $\mathrm{Beta}(2,3)$ density
306 \[
307 \rho(\theta)=12\theta(1-\theta)^2,\qquad0\le\theta\le1.
308 \]
309 Observe a specified ordered sequence with two successes and one failure. Its likelihood is $\theta^2(1-\theta)$. Multiplication adds exponents, so the posterior belongs to $\mathrm{Beta}(4,4)$ with density $140\theta^3(1-\theta)^3$.
310 \vfig{beta}
311 \platecaption{The dashed prior is asymmetric. The solid posterior is symmetric with mean $1/2$. Polynomial integration verifies the normalizers exactly; the curves are sampled drawings of those polynomials.}
312 \head{The observation convention changes the evidence}
313 The integral of the unnormalized posterior is
314 \[
315 \int_0^1 12\theta^3(1-\theta)^3\,d\theta=\frac3{35}.
316 \]
317 Dividing by $3/35$ gives the displayed normalized density. If the observation is instead the \emph{count} of two successes in three trials, there are $\binom32=3$ compatible ordered sequences. The count likelihood is therefore $3\theta^2(1-\theta)$ and its evidence is $9/35$.
318 
319 The extra factor does not depend on $\theta$, so it cancels from the posterior. The two experiments thus have the same posterior distribution but different data probabilities. Evidence is not determined by the final normalized density alone.
320 
321 The exact check also integrates both densities to one and integrates $\theta$ against the posterior to obtain $1/2$. This demonstrates one conjugate family in elementary terms. It does not show that arbitrary likelihoods preserve that family or that all posterior information can be compressed into two exponents.
322 \takeaway{Conjugacy gives a closed parameter update when a specified likelihood respects a specified family. Preserve the evidence as well as the normalized posterior.}
323 
324 \vpage{15}{The modes stay; their clocks change}{FRACTIONAL DIFFUSION / NEW COUPLINGS}{Traditional Part I: Theorem 10.1. A positive self-adjoint finite model and its spectral powers.}
325 Return to the four-state generator on \pagerefV{4} and put $L=-Q$. It is real symmetric and nonnegative, with orthogonal projectors $P_0,P_1,P_2,P_3$ and eigenvalues $0,2,4,6$. Its square root is
326 \[
327 L^{1/2}=\sqrt2P_1+2P_2+\sqrt6P_3,
328 \qquad (L^{1/2})^2=L.
329 \]
330 Both heat evolutions retain these modes. Ordinary diffusion multiplies mode $j$ by $e^{-2jt}$; square-root diffusion multiplies it by $e^{-\sqrt{2j}\,t}$. The constant mode is unchanged in both.
331 \vfig{fractional}
332 \platecaption{The highest mode decays at rates $6$ and $\sqrt6$. The graph beside the curves shows the additional couplings in the fractional operator. Curves display the explicit formulas, not an empirical fit.}
333 \head{Fractional does not mean entrywise}
334 Although $L_{03}=0$, its spectral square root satisfies
335 \[
336 (L^{1/2})_{03}=\frac{\sqrt6-\sqrt2-2}{4}<0.
337 \]
338 Thus $-L^{1/2}$ permits a direct transition between states $0$ and $3$, which were not neighbors for $Q$. Taking the square root of each matrix entry would neither produce this value nor square back to $L$.
339 
340 The exact block aggregation remains compatible in this example:
341 \[
342 C L^{1/2}=\mat{1&-1\\-1&1}C.
343 \]
344 The coarse matrix is the positive square root of $-Q_c$. This is a concrete naturality identity, checked on the full matrices. It is not a claim that an arbitrary partition commutes with every fractional construction.
345 
346 The computation uses a nonnegative self-adjoint operator and a specified spectral branch. General nonnormal matrices, unbounded domains, and positivity of other fractional evolutions need their own hypotheses.
347 \takeaway{Fractional powers alter modal decay rates. Reassembling those modes can introduce long-range spatial interactions even when the original generator was sparse.}
348 
349 \vpage{16}{A gap changes the error scale}{FRACTIONAL DEFECTS / NEAR ZERO VERSUS AWAY FROM ZERO}{Traditional Part I: Theorem 10.3 and Corollary 10.4. Separate generator error from fractional error.}
350 The function $a\mapsto\sqrt a$ changes most rapidly near zero. A scalar model already shows why a spectral gap appears in fractional-defect estimates:
351 \[
352 |\sqrt\varepsilon-\sqrt0|=\sqrt\varepsilon,
353 \qquad
354 \sqrt{1+\varepsilon}-1
355 =\frac{\varepsilon}{\sqrt{1+\varepsilon}+1}\le\frac\varepsilon2.
356 \]
357 The same input difference has a square-root scale at zero and a linear bound above a fixed positive lower bound.
358 \vfig{gap}
359 \platecaption{The upper curve compares $0$ with $\varepsilon$. The lower curve compares $1$ with $1+\varepsilon$. The dotted line $\varepsilon/2$ bounds the latter. All plotted values come from the displayed scalar formulas.}
360 \head{Read an operator bound with its hypotheses}
361 For positive self-adjoint finite matrices $A,B\ge mI$ and $0<\alpha<1$, the manuscript's gapped estimate has the form
362 \[
363 \norm{CA^\alpha-B^\alpha C}
364 \le \alpha m^{\alpha-1}\norm{CA-BC}.
365 \]
366 It concerns an intertwining defect, not just two eigenvalue lists. With $C=1$, $A=4$, $B=9$, $m=4$, and $\alpha=1/2$, the generator defect has size $5$, the fractional defect has size $1$, and the bound is $5/4$.
367 
368 The common zero mode of a Markov generator cannot simply be included in a positive-gap estimate. One must either work on an appropriate gapped subspace, handle the zero mode separately, or use the gapless estimate. In particular, a good estimate on each finite level need not have a constant uniform in the refinement level.
369 
370 On \pagerefV{27}, the sequences $1/n^2$ and $1/n$ turn this change in error scale into a concrete distinction between summable and nonsummable defects.
371 \takeaway{A fixed gap is quantitative information. It prevents the derivative of the fractional function from becoming singular along the comparison.}
372 
373 \vpage{17}{A random diffusion clock}{SUBORDINATION / A DISTRIBUTION OF ORDINARY TIMES}{Traditional Part I: Theorem 10.2. The Poisson subordination formula for the finite positive model.}
374 Square-root evolution can be written as an average of ordinary diffusion over a positive clock variable $s$:
375 \[
376 e^{-t\sqrt L}=\int_0^\infty \eta_t(s)e^{-sL}\,ds,
377 \qquad
378 \eta_t(s)=\frac{t}{2\sqrt\pi}s^{-3/2}e^{-t^2/(4s)},\quad t>0.
379 \]
380 Here $t$ is the external time; $s$ is the ordinary diffusion time being mixed. They are not interchangeable variables.
381 \vfig{random_clock}
382 \platecaption{The density $\eta_1$ peaks at $s=1/6$ but has a long right tail. The drawing ends at $s=2$; it does not claim that the missing tail has zero mass.}
383 \head{Check the normalizer before interpreting a density}
384 Under $s=t^2/(4u^2)$, reversing the integration limits gives
385 \[
386 \eta_t(s)\,|ds|=\frac2{\sqrt\pi}e^{-u^2}\,du.
387 \]
388 Its integral over $u>0$ is one, so $\eta_t$ is a probability density. On each eigenmode of $L$, the operator formula reduces to the scalar Laplace identity
389 \[
390 \int_0^\infty\eta_t(s)e^{-\lambda s}\,ds=e^{-t\sqrt\lambda}.
391 \]
392 The manuscript supplies that scalar identity and its spectral assembly. The appendix separately checks the change of variables and normalization; sampled curves are not used to certify an improper integral.
393 
394 A normalized density need not have a finite mean. For large $s$, the exponential factor tends to one and $\eta_t(s)$ is proportional to $s^{-3/2}$. Consequently $s\eta_t(s)$ has a nonintegrable $s^{-1/2}$ tail. The clock has infinite mean even though its total mass is one.
395 \takeaway{Subordination is a weighted mixture of diffusion times. Replacing the clock distribution by a single average time is not justified, and here that average is infinite.}
396 
397 \vpage{18}{A Green kernel is an inverse with units}{DIRICHLET GRID / THE MISSING FACTOR OF h}{Traditional Part I: Theorem 11.1. Exact nodal Green reconstruction for this one-dimensional stencil.}
398 Take three interior nodes $x_i=i/4$ on $[0,1]$, with zero boundary values and $h=1/4$. The discrete negative second derivative is
399 \[
400 L_h=\frac1{h^2}\mat{2&-1&0\\-1&2&-1\\0&-1&2}.
401 \]
402 For the continuous Dirichlet problem, the Green function is $G(x,y)=\min(x,y)-xy$. At these nodes the exact comparison is
403 \[
404 L_h^{-1}=h\,[G(x_i,x_j)]_{i,j=1}^3
405 =\frac1{64}\mat{3&2&1\\2&4&2\\1&2&3}.
406 \]
407 \vfig{green}
408 \platecaption{The tent is the central column of the continuous Green kernel. It has nodal values $(1,2,1)/8$; its discrete inverse column is $(1,2,1)/32$, smaller by the factor $h$.}
409 \head{A forcing vector is not a quadrature-weighted measure}
410 For sampled forcing $f_j$, the finite solution is
411 \[
412 u_i=\sum_j(L_h^{-1})_{ij}f_j
413 =h\sum_jG(x_i,x_j)f_j.
414 \]
415 The factor $h$ converts samples into the quadrature weights appropriate to this formula. Dropping it multiplies the computed solution by four on the displayed grid.
416 
417 A discrete unit coordinate vector and a discretized unit point mass use different normalizations. For $f=e_2$, the result is the second inverse column. For $f=e_2/h$, it is the unscaled tent sampled at the nodes. Either convention can be useful, but it must be specified before comparing answers.
418 
419 Exact checks multiply $L_h$ by the proposed inverse at $1,3,7,$ and $15$ interior nodes. The general identity is the manuscript theorem. These finite checks do not extend it to arbitrary meshes, dimensions, or boundary conditions.
420 \takeaway{An inverse kernel carries a measure convention. A correct-looking tent can give a wrong solution if the grid weight is omitted.}
421 
422 \vpage{19}{Convergence with a nonzero exact error}{FINITE DIFFERENCES / A QUARTIC TEST PROBLEM}{Traditional Part I: Theorems 11.1--11.2. Exact nodal inversion and the scale of low-mode errors.}
423 Use $u(x)=x-x^4$, which vanishes at both endpoints and solves $-u''=12x^2$. Its centered second difference is not exact:
424 \[
425 L_hu|_{\rm nodes}=12x_i^2+2h^2.
426 \]
427 Since the stencil differentiates the quadratic $x(1-x)$ exactly, the solution $u_h$ to the sampled forcing $12x_i^2$ satisfies
428 \[
429 \boxed{u_h(x_i)-u(x_i)=-h^2x_i(1-x_i).}
430 \]
431 This gives both the sign and size of the error, rather than merely a log--log slope.
432 \vfig{mesh}
433 \platecaption{The three curves show magnitudes of the exact nodal errors for $h=1/4,1/8,1/16$. Lines between grid points are visual guides, not an additional continuous numerical solution. The error is negative on every interior node.}
434 \head{One refinement divides this error by four}
435 All three grids include $x=1/2$, where the maximum absolute nodal error is $h^2/4$. It equals $1/64$, $1/256$, and $1/1024$, respectively. Exact rational matrix inversion verifies these values independently of the closed error formula.
436 
437 The manuscript also compares the discrete eigenvalues
438 \[
439 \lambda_{k,h}=4h^{-2}\sin^2(k\pi h/2)
440 \]
441 with $(k\pi)^2$, with error at most $(k\pi)^4h^2/12$. For a fixed mode $k$, the bound tends to zero quadratically. It does not claim a uniform small error for modes whose index grows like $1/h$.
442 
443 The nodal Green identity is exact, while the discretization of the differential equation still has a truncation error. These statements concern different comparisons and do not contradict one another.
444 \takeaway{An exact inverse of the discrete operator does not make that operator equal to the differential operator. Keep algebraic inversion and continuum approximation separate.}
445 
446 \vpage{20}{Eight paths, one weighted expectation}{FEYNMAN--KAC / DEPARTURE-STATE WEIGHTS}{Traditional Part I: Theorem 12.1. A finite path sum with its time convention displayed.}
447 Let $K_2=\frac14\mat{3&1\\1&3}$, start at state $0$, use weights $w=(1,1/2)$, and terminal observable $f=(2,1)$. For three transitions, there are eight possible state sequences after the fixed initial state. We evaluate
448 \[
449 \EE_0\left[\prod_{r=0}^{2}w(X_r)\,f(X_3)\right].
450 \]
451 Each path contributes its transition probability times its three departure weights times its terminal value.
452 \vfig{paths}
453 \platecaption{Every possible path is included. The probability column sums to one. The contribution column sums to $311/256$, agreeing exactly with the transfer calculation.}
454 \head{Check which side of the transition gets weighted}
455 The backward operator on observables is $D_w K_2^{\mathsf T}$. Therefore the expectation is the zeroth coordinate of
456 \[
457 (D_wK_2^{\mathsf T})^3f.
458 \]
459 Its forward dual is $K_2D_w$: weight the mass at the departing state, then move it. Weighting arrival states instead gives $D_wK_2$, a different expression. The exact tests deliberately check that the two conventions do not agree in this example.
460 
461 The answer $311/256$ exceeds one, but it is not a probability. The terminal observable can equal two. With $0\le w_i\le1$, the weighted expectation is bounded by the maximum terminal value, not necessarily by one.
462 
463 This calculation certifies a finite three-step identity for a stated model. The continuous-time theorem uses the corresponding infinitesimal generator and potential, with the direction of action kept consistent.
464 \takeaway{A Feynman--Kac weight changes the value assigned to each path. Its location relative to a transition is part of the model, not a typographical choice.}
465 
466 \vpage{21}{Elimination makes a frequency-dependent term}{SCHUR--FESHBACH / RETAINED RESOLVENT}{Traditional Part I: Theorem 13.1. A two-state Schur complement with the inverse checked directly.}
467 The two-state evolution from \pagerefV{3} has generator
468 \[
469 A=\mat{-1&1\\1&-1}.
470 \]
471 Retain the first coordinate and eliminate the second. For $z$ away from the relevant poles, the retained entry of $(zI-A)^{-1}$ is
472 \[
473 R_x(z)=\left[z+1-\frac1{z+1}\right]^{-1}
474 =\frac{z+1}{z(z+2)}.
475 \]
476 The term $\Sigma(z)=1/(z+1)$ is the hidden coordinate's self-energy.
477 \vfig{resolvent}
478 \platecaption{The full inverse gives a rational response with two poles, at $0$ and $-2$. Its partial fractions are $R_x(z)=\frac12z^{-1}+\frac12(z+2)^{-1}$. The hidden intermediate inverse has its own excluded point $z=-1$.}
479 \head{An eliminated variable has not disappeared}
480 If the coupling to the second coordinate were simply deleted, the retained resolvent would be $(z+1)^{-1}$. That has a different pole and a different long-time limit. The correction is frequency dependent, so it cannot be represented by one constant scalar generator on the retained coordinate.
481 
482 The Schur expression first requires $z+1\ne0$, as well as invertibility of the full matrix. Its simplified rational form has a removable intermediate singularity at $z=-1$, where the retained entry of the full inverse is well-defined and equals zero. Simplification does not retroactively make the eliminated block invertible there.
483 
484 For $z>0$, the Laplace transform of $(1+e^{-2t})/2$ is exactly $R_x(z)$. The next page derives the time-domain equation producing that response.
485 \takeaway{The Schur complement is exact elimination, not independent dynamics on a smaller state space. Hidden evolution returns as a $z$-dependent self-energy.}
486 
487 \vpage{22}{Hidden state returns as memory}{EXACT RETAINED DYNAMICS / NOT A SCALAR SEMIGROUP}{Traditional Part I: Theorem 13.2. The retained-state memory identity, including hidden initial data.}
488 Write the same system as
489 \[
490 x'=-x+y,\qquad y'=x-y.
491 \]
492 Solving the second equation first gives
493 \[
494 y(t)=e^{-t}y(0)+\int_0^t e^{-(t-r)}x(r)\,dr.
495 \]
496 Substitution is exact:
497 \[
498 x'(t)=-x(t)+e^{-t}y(0)+\int_0^t e^{-(t-r)}x(r)\,dr.
499 \]
500 The hidden initial state produces a separate forcing term. It must be retained even when the memory kernel is known.
501 \vfig{memory}
502 \platecaption{For $(x(0),y(0))=(1,0)$, the exact retained coordinate is $(1+e^{-2t})/2$. The dashed curve $e^{-t}$ solves the equation obtained by deleting the hidden coordinate rather than eliminating it.}
503 \head{Projection can destroy the semigroup law}
504 Let $R(t)=(1+e^{-2t})/2$, the response to this fixed initial preparation. At $t=\log2$,
505 \[
506 R(t)=\frac58,\qquad R(2t)=\frac{17}{32},
507 \qquad R(2t)-R(t)^2=\frac9{64}.
508 \]
509 The full two-state propagator is a semigroup, but this retained scalar response is not. Restarting it as a scalar model silently resets the hidden coordinate, which generally carries information from the first interval.
510 
511 The appendix verifies the full differential equation, initial condition, integral memory equation, and Laplace resolvent in this model. Its kernel is $e^{-t}$, whose transform is the self-energy on \pagerefV{21}.
512 \takeaway{Exact reduction retains a record of the eliminated state through convolution and initial forcing. A memoryless approximation must justify the information it removes.}
513 
514 \vpage{23}{A repeated pole remembers a Jordan chain}{NONDIAGONAL HIDDEN DYNAMICS / POLYNOMIAL EXPONENTIALS}{Traditional Part I: Theorem 13.2 and Section 15. Functional calculus must retain nilpotent data.}
515 Now use a two-dimensional hidden block
516 \[
517 J=-2I+N,\qquad N=\mat{0&1\\0&0},\qquad N^2=0.
518 \]
519 Choose the entrance column $V=(0,1)^{\mathsf T}$ and exit row $U=(1,0)$. The hidden exponential is not $e^{-2t}I$:
520 \[
521 e^{tJ}=e^{-2t}(I+tN),
522 \qquad K_{\rm mem}(t)=Ue^{tJ}V=t e^{-2t}.
523 \]
524 The nilpotent link carries an input from the second hidden coordinate to the first.
525 \vfig{jordan}
526 \platecaption{The memory kernel starts at zero, becomes positive, and peaks at $t=1/2$. A rule that retained the eigenvalue $-2$ but discarded the Jordan chain would predict the zero kernel for these entrance and exit maps.}
527 \head{The transfer function sees pole order}
528 The exact transform is
529 \[
530 U(zI-J)^{-1}V=\frac1{(z+2)^2}
531 =\int_0^\infty t e^{-(z+2)t}\,dt,
532 \qquad \operatorname{Re}z>-2.
533 \]
534 The repeated pole and the polynomial factor $t$ are the same information in two domains. For a longer nilpotent chain, higher powers of $t$ occur with factorial denominators.
535 
536 A zero value $K_{\rm mem}(0)$ does not make the entire kernel zero. Nor do stable eigenvalues alone specify every transient contribution: entrance maps, exit maps, and nilpotent structure matter. Here the full kernel is nevertheless elementary, so its peak and transform can be checked without numerical fitting.
537 
538 This page is an explicit finite realization. It does not show that an arbitrary continuum memory kernel admits a finite-dimensional representation of this form.
539 \takeaway{Repeated eigenvalues may carry directional information through Jordan chains. Keep the finite jet of the functional calculus, not just its values on the spectrum.}
540 
541 \vpage{24}{A zero initial hazard can still lead to certain exit}{PHASE-TYPE SURVIVAL / TWO SUCCESSIVE STAGES}{Traditional Part I: Theorem 14.1. Survival, density, hazard, and moments have distinct meanings.}
542 Let a transient process move through two successive rate-one stages, then leave the system. With column probabilities, the transient generator and starting vector are
543 \[
544 T=\mat{-1&0\\1&-1},\qquad p(0)=e_0.
545 \]
546 Thus $p(t)=e^{-t}(1,t)^{\mathsf T}$. Its total remaining mass, exit density, and hazard are
547 \[
548 S(t)=(1+t)e^{-t},\qquad f(t)=t e^{-t},
549 \qquad h(t)=\frac{t}{1+t}.
550 \]
551 The initial hazard is zero because an exit needs two stages, not one.
552 \vfig{survival}
553 \platecaption{Survival falls from one to zero, while absorbed mass $1-S$ rises to one. The dotted hazard is an instantaneous rate, not a probability, even though it shares the displayed numerical scale. Time is measured in inverse stage-rate units.}
554 \head{Missing transient mass is absorbed mass}
555 The column sums of $T$ are $0$ and $-1$. The exit-rate row is therefore $(0,1)$, and $f(t)=(0,1)p(t)$. Adding an absorbing state restores a three-state probability vector with total mass one.
556 
557 Exact integration gives
558 \[
559 \int_0^\infty f(t)\,dt=1,\qquad
560 \EE[\tau]=2,\qquad \EE[\tau^2]=6,
561 \qquad \Var(\tau)=2.
562 \]
563 Equivalently, the mean is $\ones^{\mathsf T}(-T)^{-1}e_0$. A single exponential waiting time with mean two would have constant hazard $1/2$ and variance four, so matching the mean alone would not reproduce this distribution.
564 
565 The hazard formula is used where $S(t)>0$. The model has that property for every finite $t\ge0$. No division by an already vanished survival mass is required.
566 \takeaway{Survival tracks mass, density tracks exits per unit time, and hazard conditions on survival. Equal means or equal initial rates do not identify a waiting-time law.}
567 
568 \vpage{25}{Naturality also transports derivatives}{HOLOMORPHIC CALCULUS / A RECTANGULAR COMPARISON}{Traditional Part I: Theorem 15.1. Exact intertwining passes to the specified functional calculus.}
569 Consider a fine operator, a coarse operator, and a rectangular map
570 \[
571 A_f=\mat{2&1&0\\0&2&0\\0&0&9},\qquad
572 A_c=\mat{2&1\\0&2},\qquad
573 C_r=\mat{1&0&0\\0&1&0}.
574 \]
575 Direct multiplication gives $C_rA_f=A_cC_r$. For $F(z)=1/(1+z)$, which is defined at both spectra, compute the matrices by inversion:
576 \[
577 F(A_f)=\mat{1/3&-1/9&0\\0&1/3&0\\0&0&1/10}.
578 \]
579 \vfig{naturality}
580 \platecaption{The discarded eigenvalue $9$ contributes $1/10$ only on the fine side. On the retained Jordan block, the off-diagonal entry is $F'(2)=-1/9$, not zero. The rectangular comparison preserves that entry.}
581 \head{Why spectral values are not enough}
582 On the retained block $A_c=2I+N$, with $N^2=0$, one has
583 \[
584 F(A_c)=F(2)I+F'(2)N=\tfrac13I-\tfrac19N.
585 \]
586 Hence $C_rF(A_f)=F(A_c)C_r$. Replacing the block by $F(2)I$ would lose the nonzero derivative term and no longer give the actual inverse of $I+A_c$.
587 
588 For polynomials, naturality follows by transporting every operator power. The manuscript's finite holomorphic statement uses finite Taylor data at each primary block and the corresponding nilpotent orders. This example verifies a rational instance with an actual inverse rather than asserting that arbitrary scalar substitutions are meaningful.
589 
590 The same mechanism explains why the Jordan memory kernel on \pagerefV{23} contains $t e^{-2t}$. Exponentiation also has nonzero derivatives at repeated spectral values.
591 \takeaway{A functional calculus acts on operators, including their nilpotent parts. Exact intertwining transports that full action, not merely a list of scalar eigenvalue images.}
592 
593 \vpage{26}{One complete coarse filter}{END-TO-END MODEL / TWO OBSERVATIONS}{Traditional Part I: Proposition 9.3 with Theorems 5.1 and 8.2. Markov closure and likelihood sufficiency are separate.}
594 Use the four-state kernel $K$ and aggregation $C$ from \pagerefV{5}, starting at $p_0=e_0$. Observe two likelihood vectors constant inside both packets:
595 \[
596 \lambda^{(1)}=(3/4,3/4,1/4,1/4),\qquad
597 \lambda^{(2)}=(1/2,1/2,1,1).
598 \]
599 Predict, then update at each observation. The fine calculation gives
600 \[
601 p_1=\tfrac1{20}(15,3,2,0)^{\mathsf T},\qquad
602 p_2=\tfrac1{104}(41,15,40,8)^{\mathsf T}.
603 \]
604 \vfig{integrated_filter}
605 \platecaption{Within-packet probabilities change during the two observations, but the packet totals agree exactly with an independent two-state filter. The second coarse posterior is $(7/13,6/13)$.}
606 \head{Check the normalizers, not only the final vector}
607 The first and second predictive evidences are
608 \[
609 Z_1=\frac58,\qquad Z_2=\frac{13}{20},
610 \qquad Z_1Z_2=\frac{13}{32}.
611 \]
612 The unnormalized product $D_2KD_1Kp_0$ has mass $13/32$ and normalizes to $p_2$. Because the likelihood entries here lie in $[0,1]$, they can serve as probabilities of the specified observations. This makes the product their joint evidence in the stated model.
613 
614 Two distinct equalities justify the coarse computation. First, $CK=K_cC$ transports prediction. Second, the likelihood is constant on each packet, so $CD_r=\overline D_rC$ transports observation weighting. Their composition preserves both the posterior packet totals and the evidence factors.
615 
616 The coarse filter does not recover the within-packet shape. If a later observation distinguishes states inside a packet, those missing probabilities can become necessary, as the counterexample on \pagerefV{10} shows.
617 \takeaway{Exact filtering at a coarse level needs both dynamical lumpability and observation sufficiency. Agreement under one step does not excuse testing the other.}
618 
619 \vpage{27}{Small at each level is not summable across levels}{REFINEMENT LIMITS / WHAT THE FINITE MODELS DO NOT PROVE}{Traditional Part I: Corollary 10.5 and Section 17. Summability and uniform control are additional data.}
620 For each $n\ge1$, compare the two scalar operators $A_n=0$ and $B_n=1/n^2$ through the identity. Their ordinary defects have sizes $1/n^2$, while their square-root defects have sizes $1/n$:
621 \[
622 \sum_n|A_n-B_n|<\infty,
623 \qquad
624 \sum_n|A_n^{1/2}-B_n^{1/2}|=\infty.
625 \]
626 These are independent scalar comparisons, not a constructed inverse system. They isolate an error-scale obstruction without claiming a coherent refinement tower.
627 \vfig{summability}
628 \platecaption{Partial sums through $n=16$ illustrate the contrast. The convergence and divergence statements follow from the displayed series, not from the finite plot. A common positive gap gives the third, summable upper bound.}
629 \head{A uniform hypothesis can change the conclusion}
630 Replace the pair by $A_n=1$ and $B_n=1+1/n^2$. Then
631 \[
632 \left|B_n^{1/2}-A_n^{1/2}\right|
633 =\frac{1/n^2}{\sqrt{1+1/n^2}+1}
634 \le\frac1{2n^2}.
635 \]
636 This is the elementary form of the uniform-gap mechanism on \pagerefV{16}. Without such information, transferring a summable generator-defect budget through a fractional function can fail.
637 
638 The formal release keeps three original research questions separate: realization of broader transfer semigroups, uniform stochastic-refinement control, and latent packet closure. The exact finite examples here do not settle their existence, domain, topology, or uniformity requirements.
639 
640 Likewise, exact hidden-state elimination only yields the finite memory models that were supplied. It does not establish a finite latent realization for every memory law. A complete list of matrix identities can still leave an infinite-dimensional approximation problem untouched.
641 \takeaway{A finite example can expose a missing hypothesis. It cannot supply a projective-limit construction, a uniform constant, or a continuum realization merely by passing at several levels.}
642 
643 \vpage{28}{How to use the evidence}{READER'S GUIDE / CALCULATION, DRAWING, AND FORMAL CONTRACT}{Source release: Paper VI v0.5.0, Checked Fifth Block. Inherited evidence: September 2026.}
644 The first 28 pages retain the worked mathematical atlas from the September release. Traditional Part I carries the corresponding core manuscript, including its figures and formal-status appendix; its theorem numbers are the references beneath these titles. The standalone guide uses internal visual-page links. The added pages beginning at V--29 address the strengthened branch in Traditional Part II.
645 
646 The chosen release records all 33 scoped results complete, with 341 supporting theorem declarations in 47 proof modules and a saved audit of 782 namespace declarations. Those are inherited verification records, not a new Lean run and not counts of historically novel results. The three original conjectures retain their stated status.
647 \head{Read each kind of evidence at its actual strength}
648 \begin{tabularx}{\linewidth}{@{}>{\raggedright\arraybackslash}p{1.22in}YY@{}}
649 \toprule
650 Evidence & What is checked here & What it does not establish\\
651 \midrule
652 Exact matrices & Stochasticity, intertwining, projector identities, inverse formulas, and specific defects. & The same properties for arbitrary partitions or unbounded operators.\\
653 Rational algebra & Posterior/evidence identities, polynomial integrals, nodal errors, and complete finite path sums. & An unstated observation model or a continuum convergence theorem.\\
654 Symbolic formulas & The displayed exponentials, Laplace transforms, derivatives, and finite spectral powers under stated conditions. & A general-purpose formal proof of every analytic identity.\\
655 Vector drawings & Reproducible geometry and sampled curves from the explicit models. & Numerical quadrature certificates or an exhaustive infinite-time experiment.\\
656 Saved Lean records & The original contract, source files, dependency pins, and earlier verification logs. & Fresh kernel verification of this editorial addition.\\
657 \bottomrule
658 \end{tabularx}
659 \head{Reproduce a calculation before adapting it}
660 The visual-appendix tree contains the mathematical models, drawing generator, and exact-example verifier. Its README records their paths and execution commands. The inherited September verifier records 204 exact or symbolic assertions and 11 deliberately false controls rejected. The monograph reproduces that symbolic suite with the available SymPy and mpmath environment. Its current execution record is identified on V--42. Its JSON record names every check and its finite scope.
661 
662 The finite families are explicit: three-step Feynman--Kac paths are fully enumerated, selected rational priors are checked exhaustively within their stated denominator bound, and Green inverses are checked on four named grids. No finite count is used as a substitute for a general theorem.
663 
664 To change a model, specify the state convention, partition, likelihood meaning, topology, and domain first. Then rerun the calculations and inspect the regenerated drawings. A preserved formula with changed assumptions is not automatically a preserved result.
665 \takeaway{Use the guide to see why a hypothesis matters and to compute a small case. Use the manuscript and its formal map for the scope of the general claim.}
666 
667 \vpage{29}{Acyclic paths stop after finitely many transfers}{DIRECTED TRANSFER / A FINITE DYSON EXAMPLE}{Traditional Part II: Theorem E2.8 and Corollary E2.9. Acyclicity concerns the transfer graph.}
668 Let three scalar packets form the directed chain $0\to1\to2$, with transfer coefficients $2$ and $3$. The generator and its powers are
669 \[
670 A=\mat{0&0&0\\2&0&0\\0&3&0},\qquad
671 A^2=\mat{0&0&0\\0&0&0\\6&0&0},\qquad A^3=0.
672 \]
673 Therefore the exponential is the exact polynomial
674 \[
675 e^{tA}=I+tA+\frac{t^2}{2}A^2.
676 \]
677 \vfig{ext_paths}
678 \platecaption{Only the empty path, one-edge paths, and the two-edge path exist. The coefficient $3t^2$ includes both the edge product $2\cdot3$ and the ordered-time simplex area $t^2/2$.}
679 \head{Path length counts transfers, not elapsed time}
680 Starting at $e_0$, the complete state is $(1,2t,3t^2)^{\mathsf T}$. No fourth packet or repeated cycle can contribute. Direct differentiation gives $x'(t)=Ax(t)$, including the relation $x_2'(t)=3x_1(t)$.
681 
682 If nonzero diagonal generators are added, a path acquires an integral of the within-packet evolutions between its transfer times. Acyclicity still bounds the number of transfers. It does not make the diagonal exponential factors polynomial in time.
683 
684 This is a linear transfer example. The columns of $A$ do not sum to zero; the total of the displayed state is $1+2t+3t^2$. It is consequently not a Markov probability model. Positivity of off-diagonal coefficients alone does not imply conservation of probability.
685 \takeaway{An acyclic packet graph terminates the transfer expansion. Nilpotence makes this particular full exponential polynomial; the two statements have different hypotheses.}
686 
687 \vpage{30}{Convergence does not close the dynamics}{REFINEMENT / A PERSISTENT RANGE EXCURSION}{Traditional Part II: Theorem E4.8 and Example E4.10. Summability and closure are different tests.}
688 Let the coarse space be $\RR$ and every finer space be $\RR^2$. The first connecting maps are $C(x,y)=x$ and $Rx=(x,0)$; all later maps are identities. Let the coarse evolution be the identity and all fine evolutions be rotations:
689 \[
690 T(t)=\mat{\cos t&-\sin t\\\sin t&\cos t},\qquad
691 U_n(t)=CT(t)R=\cos t\quad(n\ge1).
692 \]
693 The reconstructed sequence is already constant above its first level. Only one adjacent defect is nonzero, so its defect sum is finite on every bounded time interval.
694 \vfig{ext_closure}
695 \platecaption{The full state returns to the coarse line with negative sign after two quarter turns. Resetting the hidden coordinate after the first quarter turn loses the returning contribution.}
696 \head{A limit can fail the semigroup identity}
697 At $s=t=\pi/2$,
698 \[
699 U(s)U(t)=0,\qquad U(s+t)=-1.
700 \]
701 Write $P=RC=\diag(1,0)$. The exact algebraic difference is
702 \[
703 U(s)U(t)-U(s+t)=CT(s)(P-I)T(t)R.
704 \]
705 Its value is $1$ at these times, matching the missing hidden excursion. This is a failure of closure, not a failure of convergence.
706 
707 The strengthened closure theorem asks for the corresponding excursion norm to tend to zero with refinement, uniformly on bounded time intervals. Generator control of range leakage supplies a sufficient condition in its stated contractive setting. The present rotation example never meets that vanishing condition.
708 \takeaway{A convergent sequence of reconstructed evolutions need not converge to a semigroup. The limit must also forget no excursion that can later return.}
709 
710 \vpage{31}{A tail bound states what refinement buys}{SUMMABLE DEFECTS / A SCALAR COMPLETION}{Traditional Part II: Theorem E4.5 and Corollary E4.7. This example has identity connecting maps.}
711 Take $X_n=\RR$, $C_n=R_n=1$, and
712 \[
713 a_n=1-2^{-n},\qquad A_n=-a_n,\qquad T_n(t)=e^{-a_nt}\quad(n\ge0).
714 \]
715 Each semigroup is contractive for $t\ge0$. The exact adjacent generator-defect magnitude is
716 \[
717 |A_{n+1}-A_n|=2^{-(n+1)},\qquad
718 \sum_{j=n}^{\infty}|A_{j+1}-A_j|=2^{-n}.
719 \]
720 \vfig{ext_tails}
721 \platecaption{The bars show the certified tail $2^{-n}$ at horizon $T=1$. They are bounds on a whole time interval, not errors sampled at the right endpoint.}
722 \head{The estimate follows before any graph is drawn}
723 For $0\le t\le T$, the derivative of $a\mapsto e^{-at}$ has absolute value at most $T$ when $a\ge0$. Thus
724 \[
725 \sup_{0\le t\le T}|T_n(t)-e^{-t}|\le T2^{-n}.
726 \]
727 The limit is $e^{-t}$, which does have the semigroup law. Here every reconstruction is the identity, so the hidden-range excursion is identically zero. That closure fact is additional to the tail estimate.
728 
729 In a general tower, the weighted estimate includes norms of composed connecting and reconstruction maps. Dropping those weights without contraction or another uniform bound can change a convergent error budget into an invalid one. Even a small defect at each individual level does not control the sum over all levels.
730 \takeaway{A summable tail gives an explicit stopping rule for a fixed horizon. It does not remove weights, supply uniformity on all time, or prove closure without its own hypotheses.}
731 
732 \vpage{32}{The obstruction lives in the forgotten directions}{MARKOV DESCENT / TEST THE AGGREGATION KERNEL}{Traditional Part II: Theorem E5.3 and Corollary E5.4. A quotient criterion with a concrete witness.}
733 Return to the four-state model of V--4 and the perturbed stochastic matrix $K_\varepsilon$ of V--6. Let
734 \[
735 C=\mat{1&1&0&0\\0&0&1&1},\qquad w=e_0-e_1.
736 \]
737 The vector $w$ changes the fine distribution inside the first packet without changing its mass: $Cw=0$. Yet
738 \[
739 CK_\varepsilon w=\mat{-1/16\\1/16}\ne0.
740 \]
741 \vfig{ext_obstruction}
742 \platecaption{The coarse input cannot see $w$. The next coarse output can. Therefore no linear coarse operator can reproduce this fine step for every input.}
743 \head{Why one witness rules out every coarse matrix}
744 If some $\bar K$ satisfied $CK_\varepsilon=\bar KC$, then $CK_\varepsilon w=\bar KCw=0$, contradicting the displayed vector. The failure is not repaired by fitting a different coarse transition matrix to selected priors.
745 
746 More generally, for a surjective aggregation map $C$, a linear map $M$ descends exactly if $CM$ vanishes on $\ker C$. The descended map is defined by $\bar M(Cx)=CMx$; the kernel condition makes this independent of the representative $x$.
747 
748 For a deterministic partition of a finite stochastic system, differences $e_i-e_j$ within the same source block span the forgotten directions. Testing them gives the familiar equality of aggregate columns. Nonnegativity and column sums of the descended stochastic matrix then follow from the fine stochastic matrix and the partition structure.
749 \takeaway{The canonical obstruction is the action on information that aggregation discards. A nonzero witness in that kernel rules out exact descent at once.}
750 
751 \vpage{33}{Gaussian packets have separate clocks}{ORNSTEIN--UHLENBECK / HERMITE ENERGY}{Traditional Part II: Theorem E7.1 and Corollaries E7.2, E7.4, and E7.5. Standard Gaussian measure.}
752 In $L^2(\gamma)$ for the standard Gaussian law on $\RR$, use the probabilists' Hermite polynomials
753 \[
754 H_1(x)=x,\qquad H_2(x)=x^2-1,\qquad
755 H_4(x)=x^4-6x^2+3.
756 \]
757 The Ornstein--Uhlenbeck generator is $\mathcal Lf=f''-xf'$. Direct polynomial differentiation gives $\mathcal LH_n=-nH_n$ for these three polynomials. Their squared norms are $n!$ and different Hermite degrees are orthogonal.
758 \vfig{ext_hermite}
759 \platecaption{For $f=H_1+H_2$, the two bars are its separate squared-energy contributions. At $t=\log2$, their sum is $3/8$, below the spectral-gap bound $3/4$.}
760 \head{Reconstruction preserves the modes; time changes their weights}
761 The semigroup sends $f=H_1+H_2$ to
762 \[
763 T(t)f=e^{-t}H_1+e^{-2t}H_2,\qquad
764 \|T(t)f\|_2^2=e^{-2t}+2e^{-4t}.
765 \]
766 Its mean is zero. The Gaussian gap bounds this by $e^{-2t}\|f\|_2^2=3e^{-2t}$. It need not be an equality when multiple degrees are present.
767 
768 For the nonnegative operator $-\mathcal L$, a square-root evolution replaces rate $n$ by $\sqrt n$. On $g=H_1+H_4$, the coefficients at $t=\log2$ are $1/2$ and $1/4$, whereas ordinary OU evolution gives $1/2$ and $1/16$. The eigenfunctions agree; the rates differ.
769 
770 These calculations concern polynomial elements of Gaussian $L^2$. The general expansion and generator-domain statements belong to the traditional proof; a finite list of Hermite checks does not establish completeness of the basis.
771 \takeaway{Orthogonal packets make the energy accounting exact. A fractional generator alters the spectral clock while keeping the same Gaussian modes.}
772 
773 \vpage{34}{Several observations determine one sufficient partition}{FIXED LIKELIHOODS / JOINT PROFILES}{Traditional Part II: Theorem E9.3. A finite family of strictly positive likelihoods.}
774 On four states, consider two likelihoods
775 \[
776 \lambda^{(1)}=(2,2,1,1),\qquad
777 \lambda^{(2)}=(1,1,3,1).
778 \]
779 Two states may share a packet in the coarsest partition on which every listed likelihood is constant exactly when their entire likelihood profiles agree. The profiles are
780 \[
781 (2,1),\ (2,1),\ (1,3),\ (1,1),
782 \]
783 so the partition is $\{0,1\},\{2\},\{3\}$.
784 \vfig{ext_profiles}
785 \platecaption{The first observation alone allows the last two states to be merged. The second observation distinguishes them. The shared partition must respect both likelihoods.}
786 \head{A complete two-stage calculation}
787 Start from the uniform prior. The coarse prior is $(1/2,1/4,1/4)$. After the first likelihood, it becomes $(2/3,1/6,1/6)$; after the second, it is
788 \[
789 (1/2,3/8,1/8).
790 \]
791 The direct fine update multiplies the likelihoods componentwise, giving weights $(2,2,3,1)$ and posterior $(2,2,3,1)/8$. Its packet masses are precisely the displayed coarse posterior.
792 
793 The individual evidences are $Z_1=3/2$ and $Z_2=4/3$, and their product is $2$, the direct evidence of the product likelihood. These are relative likelihood weights, so evidence larger than one is allowed.
794 
795 This partition supplies fixed coarse likelihoods that work for every prior. A single-packet aggregation always gives posterior $(1)$; that mass-only commutation does not identify likelihood profiles. A prior-dependent conditional average can summarize one update on a coarser partition, but requires the within-packet information that a fixed sufficient rule avoids.
796 \takeaway{Equal full likelihood profiles determine the coarsest partition on which every listed likelihood is constant. This partition admits fixed sufficient update rules.}
797 
798 \vpage{35}{Information loss has an exact location}{RELATIVE ENTROPY / COARSE AND CONDITIONAL TERMS}{Traditional Part II: Theorem E9.8 and Corollary E9.9. Positive reference masses in this example.}
799 Take
800 \[
801 p=(1/2,1/4,1/8,1/8),\qquad q=(1/4,1/4,1/4,1/4)
802 \]
803 and group the first two and last two states. Their coarse laws are $P=(3/4,1/4)$ and $Q=(1/2,1/2)$. Define $D(p\Vert q)=\sum_i p_i\log(p_i/q_i)$ with natural logarithms.
804 \vfig{ext_information}
805 \platecaption{The bar lengths use the displayed exact formulas evaluated for illustration. The split is an identity: coarse information plus posterior-weighted conditional information. It is not a fitted allocation.}
806 \head{The chain rule identifies exactly what aggregation removes}
807 Inside packet A, the conditional laws are $(2/3,1/3)$ and $(1/2,1/2)$; inside B they agree. Hence
808 \[
809 \begin{aligned}
810 D(p\Vert q)&=\tfrac14\log2,\\
811 D(P\Vert Q)&=\tfrac34\log3-\log2,\\
812 \sum_aP_aD(p(\cdot\mid a)\Vert q(\cdot\mid a))
813 &=\tfrac54\log2-\tfrac34\log3.
814 \end{aligned}
815 \]
816 Adding the last two lines recovers the first. Every conditional term is nonnegative, so aggregation cannot increase relative entropy. Equality holds precisely when the occupied packets have the same conditional laws, under the stated support convention.
817 
818 The weighting is by the packet mass $P_a$ from the first argument of $D(p\Vert q)$. Replacing it by $Q_a$ generally breaks the identity. Likewise, forgetting zero-mass conventions or allowing a positive $p_i$ over a zero $q_i$ changes which expressions are finite.
819 \takeaway{Coarse relative entropy measures distinguishability that survives aggregation. The conditional term records exactly the distinguishability removed inside the packets.}
820 
821 \vpage{36}{The score rises because it has variance}{BAYESIAN FLOW / EVIDENCE CUMULANTS}{Traditional Part II: Theorems E10.1 and E10.2. Exponential tilting of a fixed finite prior.}
822 Start with equal prior masses on scores $u=(0,1,2)$ and likelihood $e^{tu}$. Put $q=e^t$. The evidence and tilted distribution are
823 \[
824 Z(t)=\frac{1+q+q^2}{3},\qquad
825 p(t)=\frac{(1,q,q^2)}{1+q+q^2}.
826 \]
827 At $t=\log2$, they are $Z=7/3$ and $p=(1,2,4)/7$.
828 \vfig{ext_cumulants}
829 \platecaption{The score values stay fixed. Their probabilities change. The rising mean score is a property of this tilted family, not an increase in the individual scores.}
830 \head{Two derivatives have different statistical meanings}
831 The first derivative of log evidence is the current mean score, and the second is its variance:
832 \[
833 \frac{d}{dt}\log Z(t)=\EE_{p(t)}u,
834 \qquad
835 \frac{d^2}{dt^2}\log Z(t)=\Var_{p(t)}(u)\ge0.
836 \]
837 At $q=2$ the exact values are
838 \[
839 \EE u=\frac{10}{7},\qquad
840 \EE u^2=\frac{18}{7},\qquad
841 \Var(u)=\frac{26}{49}.
842 \]
843 Thus the derivative of the mean is $26/49$ at this time. The flow equation $p_i'=(u_i-\EE u)p_i$ gives derivatives $(-10,-6,16)/49$, which sum to zero.
844 
845 As $t\to\infty$, mass concentrates on the largest score present in the initial support. A score with zero initial prior mass is never created by multiplication. If several occupied states tie for the maximum, their relative prior proportions remain unchanged within that maximal packet.
846 \takeaway{Log evidence is a cumulant generator for the tilted law. Its curvature is variance, so the mean score is monotone even though the flow is nonlinear.}
847 
848 \vpage{37}{Keep blocks by their certified contribution}{MEMORY COMPRESSION / AN OPTIMIZATION WITH A DEFINED COST}{Traditional Part II: Theorem E15.1 and Corollary E15.2. Stable primary blocks; block count is the budget.}
849 Consider the scalar memory kernel
850 \[
851 \mathcal K(t)=\tfrac14e^{-t}+(3+2t)e^{-2t}+\tfrac13e^{-3t}.
852 \]
853 It has three primary blocks. The middle block carries a Jordan term and must be retained or omitted as a whole for this selection rule. Their certified integrated weights are
854 \[
855 w_1=\frac14,\qquad w_2=\frac32+\frac{2}{2^2}=2,
856 \qquad w_3=\frac{1/3}{3}=\frac19.
857 \]
858 \vfig{ext_selection}
859 \platecaption{With one primary block retained, choose the middle block and certify omitted weight $13/36$. With two retained, add the rate-one block and the remaining certificate is $1/9$.}
860 \head{The slowest eigenvalue is not necessarily the largest effect}
861 The rate-two block has the greatest weight because its couplings and nilpotent contribution dominate. Retaining the slowest rate alone would omit weight $2+1/9$, a much larger certificate.
862 
863 The descending order is $2,1/4,1/9$. A target tolerance $1/8$ needs two blocks according to this bound: one block leaves $13/36>1/8$, while two leave $1/9\le1/8$. Every term in this particular scalar kernel is nonnegative, so its omitted integral equals the certificate.
864 
865 In general matrix kernels, cancellations can make the true error smaller. The theorem optimizes the sum of omitted block weights, not every possible approximation error. Moreover, a Jordan block may cost several state dimensions. A fixed budget in total hidden dimension is a different optimization problem from a fixed number of primary blocks.
866 \takeaway{The largest-weight rule is exact for its stated certificate and block-count budget. It is not a claim of globally optimal model reduction.}
867 
868 \vpage{38}{A kernel certificate becomes a trajectory certificate}{VOLTERRA STABILITY / SAME INITIAL DATA AND FORCING}{Traditional Part II: Theorem E15.6. Norms and a finite time horizon are explicit.}
869 Let $N=\mat{0&1\\0&0}$, so $N^2=0$ and $\|N\|_2=1$. On $[0,T]$, compare the memory kernels $K(t)=N$ and $\widetilde K(t)=0$, with identity free semigroup, no forcing, and initial state $x_0=e_1=(0,1)^{\mathsf T}$. The equations are
870 \[
871 x'(t)=\int_0^t N x(r)\,dr,\qquad
872 \widetilde x'(t)=0.
873 \]
874 Their exact solutions are $x(t)=(t^2/2,1)^{\mathsf T}$ and $\widetilde x(t)=(0,1)^{\mathsf T}$.
875 \vfig{ext_trajectory}
876 \platecaption{The first-coordinate error is exactly $t^2/2$. The certificate bounds its maximum on the entire interval. The bound is deliberately shown separately from the trajectory.}
877 \head{Track every constant before applying the estimate}
878 Here $M_T=1$, $B=\sup\|\widetilde x\|=1$, and
879 \[
880 L=\|K\|_{L^1(0,T)}=T,\qquad
881 \varepsilon=\|K-\widetilde K\|_{L^1(0,T)}=T.
882 \]
883 The theorem yields
884 \[
885 \sup_{0\le t\le T}\|x(t)-\widetilde x(t)\|
886 =\frac{T^2}{2}\le T^2e^{T^2}.
887 \]
888 At $T=1/2$, the exact error is $1/8$ and the certificate is $e^{1/4}/4$. Nilpotence makes this example polynomial; it does not remove the time horizon from the general theorem.
889 
890 The two solutions share initial state, free semigroup, and forcing. If those are changed as well, their errors need additional terms. An integrated kernel error alone does not automatically control a trajectory on an infinite horizon.
891 \takeaway{The compression estimate transports an $L^1$ kernel budget into a finite-horizon state bound, using an explicit bound on the comparison trajectory.}
892 
893 \vpage{39}{Two auxiliary states realize a repeated pole}{FINITE REALIZATION / AN EXACT DIFFERENTIAL MODEL}{Traditional Part II: Corollaries E15.4 and E15.5. This is a supplied rational realization.}
894 The kernel $K(t)=te^{-2t}$ from V--23 has transform $1/(z+2)^2$. It is realized by two auxiliary states:
895 \[
896 \begin{aligned}
897 \dot y_1&=-2y_1+y_2,& y_1(0)&=0,\\
898 \dot y_2&=-2y_2+x,& y_2(0)&=0,\\
899 \dot x&=a x+y_1.&
900 \end{aligned}
901 \]
902 With $D=\mat{-2&1\\0&-2}$, $B=(1,0)$, and $C=(0,1)^{\mathsf T}$, the feedback kernel is $Be^{tD}C=te^{-2t}$.
903 \vfig{ext_auxiliary}
904 \platecaption{An input first enters $y_2$, transfers to $y_1$, and then returns to $x$. The two equal decay rates produce a repeated pole through the intervening link.}
905 \head{Elimination recovers the convolution, including its preparation}
906 Variation of constants gives
907 \[
908 y(t)=e^{tD}y(0)+\int_0^t e^{(t-r)D}Cx(r)\,dr,
909 \]
910 so the retained equation is
911 \[
912 \dot x(t)=a x(t)+Be^{tD}y(0)
913 +\int_0^t (t-r)e^{-2(t-r)}x(r)\,dr.
914 \]
915 The convolution-only version is exact for $y(0)=0$. For a nonzero preparation, $Be^{tD}y(0)=e^{-2t}(y_1(0)+t y_2(0))$ is an additional forcing.
916 
917 The highest pole coefficient is nonzero in this example. A hidden Jordan block can become invisible to a transfer function if its entrance or exit coupling annihilates that coefficient. Pole order is therefore bounded by hidden Jordan depth, with equality requiring a visible highest term.
918 \takeaway{A polynomial-exponential kernel has a concrete finite state realization when the matrices are supplied. This does not realize every possible memory law with finitely many states.}
919 
920 \vpage{40}{Survivors have a different mixture of causes}{COMPETING RISKS / RESIDUAL LIFE}{Traditional Part II: Proposition E16.4 and Theorem E16.5. This example is stable and absorbs surely.}
921 Two transient states have no internal transitions. Their exit rates are $1$ and $2$, their initial masses are both $1/2$, and the two states exit by distinct causes. Thus
922 \[
923 A=\diag(-1,-2),\qquad
924 S(t)=\tfrac12e^{-t}+\tfrac12e^{-2t},
925 \]
926 with cause densities $f_1(t)=e^{-t}/2$ and $f_2(t)=e^{-2t}$.
927 \vfig{ext_risks}
928 \platecaption{At $t=\log2$, surviving mass is $3/8$. Cause 1 has accumulated probability $1/4$ and cause 2 has accumulated probability $3/8$; the three pieces sum to one.}
929 \head{Condition on survival before restarting the clock}
930 At this time the transient masses are $(1/4,1/8)$, so the conditional distribution of survivors is $(2/3,1/3)$. Consequently
931 \[
932 \frac{S(t+s)}{S(t)}=\frac23e^{-s}+\frac13e^{-2s},
933 \qquad
934 \EE[\tau-t\mid\tau>t]=\frac23+\frac16=\frac56.
935 \]
936 The original mean lifetime was $\tfrac12\cdot1+\tfrac12\cdot\tfrac12=3/4$. Surviving longer changes the mixture toward the slower state; it does not restart the original mixture.
937 
938 The cause densities at $t=\log2$ both equal $1/4$, so the cause-specific hazards are both $2/3$ and total hazard is $4/3$. Hazards are rates and need not lie in $[0,1]$. Each eventual cause probability is $1/2$, obtained by integrating its density. Stability of the subgenerator is part of that eventual-absorption statement.
939 \takeaway{Residual life restarts the same transient semigroup from the conditional surviving distribution. Cause probabilities, densities, and hazards answer different questions.}
940 
941 \vpage{41}{An operator substitution transports the defect too}{FUNCTIONAL CALCULUS / AN EXACT RATIONAL TEST}{Traditional Part II: Theorem E3.4; compare Traditional Part I, Theorem 15.1, and V--25.}
942 Use the operators from V--25, but perturb the rectangular comparison map:
943 \[
944 A=\mat{2&1&0\\0&2&0\\0&0&9},\quad
945 B=\mat{2&1\\0&2},\quad
946 C_\eta=\mat{1&0&\eta\\0&1&0}.
947 \]
948 The generator defect is $\Delta=C_\eta A-BC_\eta$, with only one nonzero entry, $\Delta_{13}=7\eta$ (matrix indices begin at one).
949 \vfig{ext_substitution}
950 \platecaption{The perturbation lets the discarded eigenvalue enter the comparison. Exact naturality holds at $\eta=0$; a nonzero defect is transformed, not erased, by the rational calculus.}
951 \head{A resolvent identity supplies the sign and both inverse factors}
952 For $F(z)=1/(1+z)$, all required inverses exist, and
953 \[
954 C_\eta F(A)-F(B)C_\eta
955 =-(I+B)^{-1}\Delta(I+A)^{-1}
956 =\mat{0&0&-7\eta/30\\0&0&0}.
957 \]
958 This follows by multiplying the difference by $I+B$ on the left and $I+A$ on the right. The middle numerator is $(I+B)C_\eta-C_\eta(I+A)=-\Delta$.
959 
960 The retained Jordan block still contributes $F'(2)=-1/9$, as on V--25. The new third-column discrepancy compares $F(9)=1/10$ with $F(2)=1/3$, multiplied by $\eta$.
961 
962 The general holomorphic defect formula integrates the corresponding resolvent defect under its contour and spectral assumptions. This rational example verifies an actual matrix inverse identity. It does not remove those analytic hypotheses or establish a calculus for arbitrary unbounded operators.
963 \takeaway{Exact intertwining gives exact naturality. When intertwining fails, the functional calculus has a structured defect whose factors can be computed.}
964 
965 \vpage{42}{What the final visual edition verifies}{EVIDENCE MAP / INHERITED RECORDS AND FRESH EXAMPLES}{The final package retains original sources, new checks, build logs, and the traditional volume.}
966 \vfig{ext_evidence}
967 \platecaption{The three evidence tracks have distinct scopes. The formal contract is inherited; the new exact examples and the present PDF build are fresh editorial checks.}
968 \head{Read the records without adding incompatible counts}
969 The September visual release records 204 exact or symbolic checks and 11 rejected false controls for its 28-page atlas. The scripts and saved report remain available. The monograph reruns the original symbolic suite with SymPy and mpmath. Its recorded inputs, outputs, resource limits, and execution logs accompany the current result.
970 
971 The monograph also reruns the separate standard-library exact checker for V--29 through V--41. It uses rational matrices, finite polynomial algebra, Gaussian moments, and rational coefficients of logarithms. It checks the stated finite examples and rejects deliberately false alternatives. Its report records 120 assertions and 8 rejected false controls. A separate current verification also reruns 182 rational assertions and 4 rejected false controls for selected inherited models. That narrower check overlaps the full symbolic replay and remains separate from its count.
972 
973 The new checks do not numerically certify improper integrals, all time horizons, or every analytic theorem. For example, the integrated memory weights use the stated exponential-integral formula; the checker verifies its coefficients and finite selection problem. Gaussian polynomial checks do not prove Hermite completeness. Rational entropy coefficient identities do not replace the general support-sensitive chain rule.
974 \head{The formal and editorial boundaries stay separate}
975 The inherited Lean record covers 33 scoped results with 341 supporting theorem declarations in 47 proof modules. This guide neither enlarges that denominator nor claims a fresh kernel build. Traditional Part II retains broader written proofs; Part III retains earlier complementary results and the operator-substitution material migrated from Paper IV. The Lean companion maps their relationship to the formal baseline.
976 
977 All 40 plates are editable vector sources: the 26 original plates plus 14 editorial plates. The portable build compiles the shipped geometry without needing symbolic regeneration. The supplied models and Python generators were used to regenerate all forty plates. Geometry samples remain drawing coordinates rather than additional analytic proofs.
978 \takeaway{A final edition can preserve strong earlier evidence while reporting exactly which checks were rerun. Treat each evidence count as a count of its own declared scope.}
979 
980 \endgroup

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