Paper XII: Patching Support, Hecke Packets, and Conditional Automorphic Completion

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1 % ---------------------------------------------------------------------------
2 \vpage{1}{Seeing the objects}{A STANDALONE READER'S COMPANION}{Patching Support, Hecke Packets, and Conditional Automorphic Completion\quad \href{p12-route:visual-1}{Main statement 12.6}}
3 {\sffamily\small Komorebi P.\quad\textbullet\quad 10 October 2026}\par
4 The paper repeatedly asks three different questions: does an object exist, where is a supplied object nonzero, and when does that object have the desired arithmetic meaning? This guide makes those distinctions tangible through examples that can be followed by hand.
5 
6 \begin{insight}
7 \begin{tabularx}{\linewidth}{@{}p{.18\linewidth}Y@{}}
8 \textbf{Existence} & Is there a represented deformation problem, a nonzero carrier, or a compatible family to start with?\\[3pt]
9 \textbf{Support} & At which prime ideals does the supplied module or complex survive?\\[3pt]
10 \textbf{Realization} & Does the surviving eigensystem come from a classical automorphic object with the required attached representation?
11 \end{tabularx}
12 \end{insight}
13 
14 \head{Three models, rather than one overloaded analogy}
15 \begin{tabularx}{\linewidth}{@{}p{.20\linewidth}Y>{\raggedright\arraybackslash}p{.15\linewidth}@{}}\toprule
16 \textbf{Model} & \textbf{What it lets you see} & \textbf{Lessons}\\\midrule
17 A formal thickening & Conditions on power series, infinitesimal points, obstructed lifts, and compatible precision levels. & \pagerefV{3}--\pagerefV{5}\\
18 A presented module & Matrix rank, support, Fitting ideals, changes of coordinates, and nonreduced structure. & \pagerefV{6}--\pagerefV{8}\\
19 A finite Artin algebra & Generalized eigenspaces, nilpotent directions, canonical packets, and empty packet modules. & \pagerefV{12}--\pagerefV{14}\\\bottomrule
20 \end{tabularx}
21 
22 The models are connected by the paper's theorems, but they are not the same ring. A polynomial coordinate plane is not a complete local spectrum, and a local branch is not an Artin product factor. Explicitly changing models prevents a picture from silently changing the hypotheses.
23 
24 \head{A route through the guide}
25 Use the object atlas on \pagerefV{2} when a symbol becomes unfamiliar. The local model starts on \pagerefV{3}; the matrix model on \pagerefV{6}. Lessons \pagerefV{9}--\pagerefV{11} explain what quotients, patching, and derived fibers preserve. Lessons \pagerefV{15}--\pagerefV{16} separate finite occupancy from genuine infinite compatibility. The exact arithmetic boundary is on \pagerefV{17}.
26 
27 \head{How to read a visual claim}
28 Each lesson gives an explicit object, a calculation, and a source reference. Coordinate sketches display only the stated real points. Coefficient arrays and finite specialization diagrams describe the stated algebraic data exactly. Line thickness never stands for nilpotent multiplicity.
29 
30 \takeaway{These are explanations of the formal release, not additional assumptions or new Lean certificates. The accompanying scripts check the explicit finite calculations. They do not establish universal arithmetic existence.}
31 
32 % ---------------------------------------------------------------------------
33 \vpage{2}{An atlas of the objects}{ORIENTATION}{Main text: Sections 2, 3, 6, 7, 8 and 12\quad \href{p12-route:visual-2}{Main statement 3.3}}
34 A \emph{ring} describes possible points. A \emph{module over the ring} describes what is carried at those points. A nonempty space of possibilities does not imply a nonzero carrier at every point.
35 
36 {\small
37 \begin{tabularx}{\linewidth}{@{}p{.20\linewidth}Yp{.22\linewidth}@{}}\toprule
38 \textbf{Symbol} & \textbf{Object and role} & \textbf{Keep separate}\\\midrule
39 $\OO,\ k$ & Complete coefficient ring and its residue field. The formal teaching model uses $\OO=\ZZ_5$, $k=\FF_5$. & Coefficients versus variables.\\
40 $R_i,\ I_i$ & A supplied local ring and the ideal imposing a declared local condition. & A quotient condition is not an idempotent.\\
41 $R^{\mathrm{loc}}$ & Completed tensor product of the local rings over $\OO$. & Completion uses the declared adic topology.\\
42 $J$ & Closed ideal generated by the images of all $I_i$ in $R^{\mathrm{loc}}$. & Closure, not just a finite list of tests.\\
43 $R^{\mathrm{loc},\tau}$ & Completed tensor of the local quotients; identified with $R^{\mathrm{loc}}/J$. & A constructed comparison, not assumed notation.\\
44 $R^\square$ & Supplied global ring receiving the local structural maps. & Representability is arithmetic input.\\
45 $R^\tau$ & Completed global base change imposing the simultaneous local conditions. & Does not create a missing residual point.\\
46 $R_\infty,\ M_\infty$ & Patched ring and its module; a perfect complex is used in the derived version. & Freeness over an auxiliary ring is weaker than full support.\\
47 $R,\ M$ & Finite-level ring and carrier after the specified quotient and control isomorphism. & The module may be zero on a component.\\
48 $T,\ H$ & Hecke algebra and its ordinary or derived carrier. & Ring comparison does not imply classicality.\\
49 $e_{\m}H$ & Packet selected by a canonical Artin idempotent. & A generalized packet need not be an eigenspace.\\
50 $Y_m$ & Stable image at level $m$ of an inverse system. & Nonzero levels alone do not give a nonzero limit.\\\bottomrule
51 \end{tabularx}}
52 
53 \head{Why arrows reverse when points appear}
54 A ring map $f:R\to T$ sends a prime of $T$ to its inverse image in $R$:
55 \[
56 f^*:\Spec T\longrightarrow\Spec R,\qquad \mathfrak q\longmapsto f^{-1}(\mathfrak q).
57 \]
58 Similarly, a point $y:T\to E'$ gives the point $y\circ f:R\to E'$. The equality $y\circ f=x$ is an actual requirement in the final theorem, not a consequence of drawing a connecting arrow.
59 
60 \head{Two kinds of point}
61 Local quotient factorization is stated for continuous local coefficient-algebra maps to deformation test rings. The later field-valued point $x:R\to E'$ uses its kernel and scalar fiber. A characteristic-zero field point is generally not a local ring map: the coefficient uniformizer becomes a unit. These two uses of ``point'' must retain their own categories.
62 
63 % ---------------------------------------------------------------------------
64 \vpage{3}{Local conditions remove directions}{MODEL A / A FORMAL THICKENING}{Main text: Theorem 3.1 and Corollary 3.2\quad \href{p12-route:visual-3}{Main statement 3.1}}
65 Fix $\OO=\ZZ_5$. Take two independent local variables and impose two different kinds of condition:
66 \[
67 R_1=\OO[[x]],\quad I_1=(x^2),\qquad
68 R_2=\OO[[y]],\quad I_2=(y).
69 \]
70 Their completed local tensor is $A=\OO[[x,y]]$. The closed generated ideal is $J=(x^2,y)$, and both routes through Theorem 3.1 give
71 \[
72 (R_1/I_1)\whot_{\OO}(R_2/I_2)
73 \cong A/J
74 \cong \OO[\varepsilon]/(\varepsilon^2).
75 \]
76 
77 \vfig[.96]{monomial_quotient}
78 \captiontext{An exact finite window of monomials over $\OO$. Crosses mean that the monomial is zero in the quotient; they do not represent a small numerical coefficient. All terms outside the window satisfy the same divisibility rule.}
79 
80 \head{The calculation behind the image}
81 For $f=\sum_{i,j\ge0}a_{ij}x^iy^j$, reduction gives
82 \[
83 f\longmapsto a_{00}+a_{10}\varepsilon.
84 \]
85 Every term involving $y$ vanishes. Every term involving $x^2$ vanishes. The class of $x$ does \emph{not} vanish: it becomes a nonzero element whose square is zero. The quotient is a free $\OO$-module with basis $1,\varepsilon$.
86 
87 \head{Why a point picture alone would miss this}
88 After passing to the special fiber, $k[\varepsilon]/(\varepsilon^2)$ has the same underlying one-point spectrum as $k$. The extra direction is visible in the algebra and in maps from test rings, not as a second point. It is also not a second component or a second Artin packet.
89 
90 \takeaway{The condition $y=0$ deletes a variable. The condition $x^2=0$ retains a first-order nilpotent direction. Both are quotient conditions, but their effects are different.}
91 
92 % ---------------------------------------------------------------------------
93 \vpage{4}{A point is a test of the equations}{MODEL A / FACTORIZATION AND OBSTRUCTION}{Main text: Proposition 3.3; this explicit example also exhibits a failed infinitesimal lift\quad \href{p12-route:visual-4}{Main statement 3.3}}
94 Let $R^\square=\OO[[u,v]]$ and send $x\mapsto u$, $y\mapsto v-u$. The local conditions become
95 \[
96 u^2=0,\qquad v-u=0,
97 \quad\text{so}\quad
98 R^\tau=\OO[[u,v]]/(u^2,v-u)\cong\OO[\varepsilon]/(\varepsilon^2).
99 \]
100 Probe this ring using $D_3=\FF_5[\delta]/(\delta^3)$, with the fixed coefficient map $\ZZ_5\to\FF_5$. A continuous local point is specified by
101 \[
102 u\longmapsto a\delta+b\delta^2,\qquad
103 v\longmapsto c\delta+d\delta^2,
104 \qquad a,b,c,d\in\FF_5.
105 \]
106 It factors through $R^\tau$ precisely when
107 \[
108 a^2=0,\quad c=a,\quad d=b
109 \quad\Longleftrightarrow\quad a=c=0,\ d=b.
110 \]
111 
112 \begin{minipage}[t]{.48\linewidth}\centering
113 {\sffamily\small\bfseries First-order coefficients}\par
114 \vfig[.68]{points_slopes}
115 \captiontext{Only the slope pair $(a,c)=(0,0)$ survives. Each square initially has 25 choices of $(b,d)$.}
116 \end{minipage}\hfill
117 \begin{minipage}[t]{.48\linewidth}\centering
118 {\sffamily\small\bfseries Remaining second-order coefficients}\par
119 \vfig[.68]{points_curvatures}
120 \captiontext{Once $a=c=0$, the five diagonal choices $d=b$ give all accepted points.}
121 \end{minipage}
122 
123 \head{An obstruction you can calculate}
124 There are $5^4=625$ candidate $D_3$-points and exactly five accepted points. Over $D_2=\FF_5[\delta]/(\delta^2)$, however, the five maps $u=v=a\delta$ all satisfy the equations. Reduction of \emph{every} accepted $D_3$-point gives the zero $D_2$-point. The four nonzero tangent directions therefore do not lift to order three.
125 
126 \takeaway{A point satisfies equations in a target ring. A tangent direction can exist while its next-order lift fails. The factorization theorem identifies the condition; it does not promise that every point lifts.}
127 
128 % ---------------------------------------------------------------------------
129 \vpage{5}{Completion retains every precision}{MODEL A / COMPATIBLE COEFFICIENTS}{Main text: Theorem 3.1; reconstruction and finite occupancy in Section 10\quad \href{p12-route:visual-5}{Main statement 10.5}}
130 For $B=\ZZ_5[\varepsilon]/(\varepsilon^2)$, consider the finite quotient rings
131 \[
132 B_n=(\ZZ/5^n\ZZ)[\varepsilon]/(\varepsilon^2),
133 \qquad |B_n|=5^{2n},\qquad B\cong\varprojlim_n B_n.
134 \]
135 A point of the limit is a \emph{single compatible choice} at every level, not unrelated choices from the nonempty finite rings.
136 
137 \head{Two explicit 5-adic numbers}
138 Choose the canonical digits
139 \[
140 \alpha_j=(j^2+1)\bmod5,\qquad \beta_j=(2j+1)\bmod5,
141 \]
142 with values in $\{0,1,2,3,4\}$. Set $a=\sum_{j\ge0}\alpha_j5^j$ and $b=\sum_{j\ge0}\beta_j5^j$. The element $a+\varepsilon b$ is determined by the following coefficient array.
143 
144 \vfig[.87]{adic_digits}
145 \captiontext{Cutting before column $5^n$ produces reduction modulo $5^n$. The digit formulas define every column, including those not drawn.}
146 
147 \begin{center}\small
148 \begin{tabular}{@{}lrrrrr@{}}\toprule
149 $n$ & 1 & 2 & 3 & 4 & 5\\\midrule
150 $5^n$ & 5 & 25 & 125 & 625 & 3125\\
151 $a\bmod5^n$ & 1 & 11 & 11 & 11 & 1261\\
152 $b\bmod5^n$ & 1 & 16 & 16 & 266 & 2766\\\bottomrule
153 \end{tabular}
154 \end{center}
155 Every longer prefix reduces to the previous one; the digit formulas specify all later coefficients.
156 
157 \head{Where the topology enters}
158 The $5$-adic and $(5,\varepsilon)$-adic topologies on $B$ agree: for every $n$,
159 \[
160 (5,\varepsilon)^{n+1}\subseteq 5^nB\subseteq(5,\varepsilon)^n.
161 \]
162 For the general quotient theorem, the condition ideal must be closed. A continuous map to a Hausdorff target has closed kernel: vanishing on the original generators therefore kills their closure.
163 
164 
165 
166 % ---------------------------------------------------------------------------
167 \vpage{6}{A matrix becomes a support locus}{MODEL B / FITTING STRATA}{Main text: Theorems 6.2, 6.3 and 6.4\quad \href{p12-route:visual-6}{Main statement 6.3}}
168 Now change models. Let $R=k[x,y]$ for a field $k$, and present a module by
169 \[
170 R^2\xrightarrow{A}R^2\longrightarrow M\longrightarrow0,
171 \qquad
172 A=\begin{pmatrix}x&0\\0&y\end{pmatrix}.
173 \]
174 The relations are $xe_1=0$ and $ye_2=0$, so $M\cong R/(x)\oplus R/(y)$.
175 
176 \begin{minipage}[c]{.48\linewidth}
177 \vfig[.98]{support_axes}
178 \end{minipage}\hfill
179 \begin{minipage}[c]{.48\linewidth}
180 {\sffamily\bfseries The minor calculation}\par
181 \[
182 \begin{aligned}
183 \Fitt_0(M)&=(xy),\\
184 \Fitt_1(M)&=(x,y),\\
185 \Fitt_j(M)&=R\quad(j\ge2).
186 \end{aligned}
187 \]
188 At a prime $\p$, reduce the matrix to its residue field:
189 \[
190 \mu_{\p}(M)=2-\rank A(\p).
191 \]
192 Two independent relations disappear at the origin. Only one disappears along either punctured axis.
193 \end{minipage}
194 \captiontext{For the drawing, take $k=\RR$. It shows real closed points, not every point of $\Spec R$. The displayed ideals and rank criterion describe the full algebraic loci over any field.}
195 
196 \begin{center}\small
197 \begin{tabular}{@{}llll@{}}\toprule
198 Position & Matrix rank & Fiber dimension & What survives\\\midrule
199 $x\ne0$, $y\ne0$ & 2 & 0 & neither generator\\
200 $x=0$, $y\ne0$ & 1 & 1 & the $e_1$ direction\\
201 $x\ne0$, $y=0$ & 1 & 1 & the $e_2$ direction\\
202 $x=y=0$ & 0 & 2 & both directions\\\bottomrule
203 \end{tabular}
204 \end{center}
205 Thus $\Supp(M)=V(xy)=V(x)\cup V(y)$, whereas the locus where the fiber needs at least two generators is $V(x,y)$. The latter is a \emph{higher} Fitting stratum. For $r>2$, the locus $\mu_{\p}(M)\ge r$ is empty, corresponding to the unit Fitting ideal.
206 
207 \takeaway{Support asks whether any direction survives. Higher Fitting strata ask how many generators the residue-field fiber needs. This is not an automorphic multiplicity formula.}
208 
209 % ---------------------------------------------------------------------------
210 \vpage{7}{The same support can hide more algebra}{MODEL B / NILPOTENT STRUCTURE}{Main text: Theorems 6.3--6.4; distinguish an ideal from its vanishing set\quad \href{p12-route:visual-7}{Main statement 6.4}}
211 Compare the previous module with a slightly different presentation:
212 \[
213 M=\coker\begin{pmatrix}x&0\\0&y\end{pmatrix},
214 \qquad
215 M'=\coker\begin{pmatrix}x^2&0\\0&y\end{pmatrix}.
216 \]
217 Their support sets coincide. Their residue-field fiber dimensions also coincide at every prime. But their Fitting ideals do not.
218 
219 \begin{center}
220 \begin{tabular}{@{}lll@{}}\toprule
221  & $M$ & $M'$\\\midrule
222 Zeroth Fitting ideal & $(xy)$ & $(x^2y)$\\
223 First Fitting ideal & $(x,y)$ & $(x^2,y)$\\
224 Support as a set & $V(x)\cup V(y)$ & $V(x)\cup V(y)$\\
225 Fiber dimension at $(x,y)$ & 2 & 2\\\bottomrule
226 \end{tabular}
227 \end{center}
228 
229 \head{Magnify transversely, rather than draw a thicker line}
230 Take the slice $y=1$. The second summand vanishes because $y$ becomes a unit. What remains is
231 \[
232 M|_{y=1}\cong k[x]/(x),\qquad
233 M'|_{y=1}\cong k[x]/(x^2).
234 \]
235 The first has basis $1$; the second has basis $1,x$. Multiplication by $x$ distinguishes them exactly:
236 
237 \begin{center}
238 \begin{tikzpicture}[x=1cm,y=1cm,font=\small,>=Stealth]
239 \node[font=\sffamily\bfseries] at (1.3,1.8) {reduced slice};
240 \node at (.2,.85) {$1$};\node at (2.4,.85) {$0$};
241 \draw[->] (.5,.85)--(2.1,.85) node[midway,above] {$\times x$};
242 \node[font=\sffamily\bfseries] at (7.2,1.8) {nilpotent slice};
243 \node at (5,.85) {$1$};\node at (7.2,.85) {$x$};\node at (9.4,.85) {$0$};
244 \draw[->] (5.3,.85)--(6.9,.85) node[midway,above] {$\times x$};
245 \draw[->] (7.5,.85)--(9.1,.85) node[midway,above] {$\times x$};
246 \node at (1.3,-.05) {$[x]=[0]$};
247 \node at (7.2,-.05) {$[x]=\begin{pmatrix}0&0\\1&0\end{pmatrix}$};
248 \end{tikzpicture}
249 \end{center}
250 
251 \head{A test ring detects the higher-stratum thickening}
252 Let $D=k[\delta]/(\delta^2)$ and send $x\mapsto\delta$, $y\mapsto0$. The ideal $(x^2,y)$ maps to zero, but $(x,y)$ does not. Therefore this $D$-point lies in the scheme defined by the first Fitting ideal of $M'$, but not in the corresponding scheme for $M$.
253 
254 No extra ordinary point has appeared. What changed is the behavior on nilpotent test rings. This is the same distinction encountered in the formal model on \pagerefV{3}.
255 
256 \takeaway{Taking $V(I)$ forgets nilpotent thickness. Fitting \emph{ideals} can retain information that both the support set and the fiber-dimension function fail to distinguish.}
257 
258 % ---------------------------------------------------------------------------
259 \vpage{8}{Base change is an algebraic pullback}{MODEL B / SPECIALIZATION AND TRANSPORT}{Main text: Proposition 6.5 and Theorem 7.1\quad \href{p12-route:visual-8}{Main statement 6.5}}
260 Return to $M=\coker\operatorname{diag}(x,y)$ over $R=k[x,y]$. Pull back along the diagonal $R\to k[t]$, $x\mapsto t$, $y\mapsto t$. The base-changed presentation is literal:
261 \[
262 M\otimes_R k[t]\cong\coker\begin{pmatrix}t&0\\0&t\end{pmatrix}
263 \cong\bigl(k[t]/(t)\bigr)^2.
264 \]
265 
266 \begin{minipage}[c]{.44\linewidth}
267 \vfig[.94]{diagonal_slice}
268 \end{minipage}\hfill
269 \begin{minipage}[c]{.51\linewidth}
270 \head{Both relations vanish together}
271 \[
272 (xy)\longmapsto(t^2),\qquad
273 (x,y)\longmapsto(t).
274 \]
275 The support is the origin, but the origin has fiber dimension two. Replacing $(t^2)$ by its radical $(t)$ would lose the zeroth Fitting ideal.
276 
277 This map is not flat: it is the quotient by $x-y$. The Fitting base-change theorem still applies; it does not require flatness.
278 \end{minipage}
279 \captiontext{A real-point shadow of the diagonal slice. The double algebraic contribution is certified by the minor $t^2$, not by the visual thickness of the intersection point.}
280 
281 \head{An isomorphism changes coordinates, not occupancy}
282 Over a field of characteristic other than two, the isomorphism
283 \[
284 k[x,y]\xrightarrow{\sim}k[a,b],\qquad x\mapsto a+b,\quad y\mapsto a-b
285 \]
286 sends the Fitting ideals to $(a^2-b^2)$ and $(a,b)$. The two axes become the two lines $a=b$ and $a=-b$. No previously missing module vector is created by this relabeling.
287 
288 \head{An extra presentation generator changes nothing}
289 Replacing $\operatorname{diag}(x,y)$ by $\operatorname{diag}(x,y,1)$ adds a generator killed by the relation $1$. The new maximal minor is $xy$, and the new next-size minors generate $(xy,x,y)=(x,y)$. The module and its intrinsic Fitting ideals are unchanged. This is a small visible instance of presentation independence of Fitting ideals.
290 
291 \takeaway{Specialization substitutes into the actual relation matrix. An isomorphism transports all the data. Neither operation may be replaced by an informal edit of the support picture.}
292 
293 % ---------------------------------------------------------------------------
294 \vpage{9}{A branch is not an Artin packet}{LOCAL GEOMETRY / AN EXACT FINITE SPECTRUM}{Main text: Proposition 3.4 and Proposition 7.3\quad \href{p12-route:visual-9}{Main statement 7.3}}
295 Consider the complete local node
296 \[
297 R=k[[x,y]]/(xy),\qquad T=R/(x)\cong k[[y]].
298 \]
299 Every prime of $R$ contains $x$ or $y$. Since $k[[x]]$ and $k[[y]]$ are discrete valuation rings, the following diagram lists \emph{all three} prime ideals of $R$ and their specialization relations.
300 
301 \vfig[.97]{node_spectrum}
302 
303 \head{What the quotient removes}
304 As an $R$-module, $T$ has annihilator $(x)$, hence
305 \[
306 \Supp_R(T)=V(x)=\{(x),(x,y)\}.
307 \]
308 The other generic point $(y)$ is missing because it does not contain $x$. The quotient has not created a disconnected projector summand inside the original local ring. It has changed the ring by imposing an equation.
309 
310 \head{Why no idempotent selects just this branch}
311 In a commutative local ring, $e^2=e$ implies $e=0$ or $e=1$. Indeed, $e+(1-e)=1$ forces at least one of $e,1-e$ to be a unit; canceling in $e(1-e)=0$ gives the conclusion. Thus this node has no nontrivial idempotent even though it has two irreducible components.
312 
313 By contrast, a product of two Artin local rings has two disjoint maximal points, with canonical idempotents $(1,0)$ and $(0,1)$. Those idempotents act on \emph{any supplied module over the product}. Their two ranges are the packets studied on \pagerefV{12}.
314 
315 \takeaway{Intersecting local components are not product factors. Use quotient ideals and support for the node; use canonical idempotents only where an actual product decomposition exists.}
316 
317 % ---------------------------------------------------------------------------
318 \vpage{10}{Free over what? Supported where?}{PATCHING / A COUNTEREXAMPLE WITH A VISIBLE SPECIAL FIBER}{Main text: Theorems 6.7, 6.9 and 6.11; Example 6.13\quad \href{p12-route:visual-10}{Main statement 6.11}}
319 Let
320 \[
321 S=k[[t]],\qquad R_\infty=S[[x]]/(x(x-t)),\qquad
322 M_\infty=R_\infty/(x)\cong S.
323 \]
324 The module is free of rank one over $S$. Yet over $R_\infty$ its support is only $V(x)$; the generic point $(x-t)$ of the other branch is not occupied.
325 
326 \begin{minipage}[c]{.47\linewidth}
327 \vfig[.94]{patching_branches}
328 \end{minipage}\hfill
329 \begin{minipage}[c]{.49\linewidth}
330 \head{Remove the auxiliary variable}
331 Set $t=0$. Then
332 \[
333 R=R_\infty/(t)\cong k[[x]]/(x^2),
334 \]
335 \[
336 M=M_\infty/tM_\infty\cong k.
337 \]
338 The two branches meet in a nonreduced special fiber. The relation becomes $x^2=0$, not $x=0$.
339 \end{minipage}
340 \captiontext{The drawing is a real algebraic shadow of $x(x-t)=0$. It is not the complete local spectrum. The special fiber's nilpotent structure is recorded by its quotient algebra, not by the point drawn at the intersection.}
341 
342 \head{Full support need not mean faithful action}
343 The ring $R$ has one prime, $(x)$. The nonzero module $M=k$ therefore has full support. But $x$ annihilates $M$, so the action is not faithful. It is \emph{nearly faithful}: the annihilator $(x)$ is nilpotent. This distinction disappears over a reduced ring.
344 
345 \head{The actual descent theorem is one-way}
346 For a finite module over a Noetherian ring,
347 \[
348 \Supp_{R_\infty/(t)}(M_\infty/tM_\infty)
349 =\Supp_{R_\infty}(M_\infty)\cap V(t).
350 \]
351 If the initial support is full, the resulting support is full on the quotient. Our example shows that full support at the quotient can occur even when initial support was not full. It also shows why freeness over $S$ alone is not the required patching hypothesis.
352 
353 \takeaway{A statement about freeness must name the base ring. A statement about occupancy must name the spectrum. Switching between them without a theorem loses the missing component.}
354 
355 % ---------------------------------------------------------------------------
356 \vpage{11}{A vanishing differential changes the fiber}{DERIVED SUPPORT / AN ACTUAL TWO-TERM COMPLEX}{Main text: Definition 6.14 and Theorem 6.16\quad \href{p12-route:visual-11}{Main statement 6.16}}
357 Take $R=k[t]$ and the perfect complex
358 \[
359 C=\bigl[R\xrightarrow{\;t\;}R\bigr]
360 \qquad\text{in cohomological degrees }0,1.
361 \]
362 Specialization at $t=a$ gives $C_a=[k\xrightarrow{a}k]$. When $a\ne0$, the differential is invertible and the complex is acyclic. When $a=0$, the differential is zero and both cohomology groups survive.
363 
364 \vfig[.81]{derived_fibers}
365 \captiontext{A few fibers over $k=\QQ$ are displayed. The statement for every $a\ne0$ follows from invertibility of the scalar $a$; it is not inferred from the sample points.}
366 
367 \head{Compute before and after specialization}
368 Before specialization,
369 \[
370 H^0(C)=0,\qquad H^1(C)=k[t]/(t).
371 \]
372 After \emph{derived} specialization at zero,
373 \[
374 C\otimes_R^{\mathbf L}k=[k\xrightarrow{0}k],
375 \quad H^0\cong k,\quad H^1\cong k.
376 \]
377 If one retained only $H^1(C)$ and took its ordinary tensor product with $k$, one would see a single $k$ and miss the extra cohomology in degree zero. The projective complex keeps the information needed for specialization.
378 
379 \head{The support is still a precise locus}
380 At any prime not containing $t$, the residue-field differential is invertible. At the prime $(t)$ it is zero. Therefore
381 \[
382 \Supp_R(C)=V(t).
383 \]
384 Theorem 6.16 identifies support after derived quotient with the intersection of original support and the quotient locus.
385 
386 \takeaway{Derived nonvanishing means that some cohomology group survives after derived specialization. It neither specifies a classical automorphic class nor eliminates the need for a realization theorem.}
387 
388 % ---------------------------------------------------------------------------
389 \vpage{12}{A packet contains an entire Jordan chain}{MODEL C / A NONSEMISIMPLE HECKE ALGEBRA}{Main text: Theorem 8.2, Corollary 8.3 and Example 8.6\quad \href{p12-route:visual-12}{Main statement 8.2}}
390 Work over $\QQ$ and let
391 \[
392 T=\QQ[z]\big/\bigl((z-1)^2(z+1)^3\bigr),\qquad V=T.
393 \]
394 The Chinese remainder theorem gives
395 \[
396 T\cong \underbrace{\QQ[\varepsilon]/(\varepsilon^2)}_{T_+,\ z=1+\varepsilon}
397 \times
398 \underbrace{\QQ[\eta]/(\eta^3)}_{T_-,\ z=-1+\eta}.
399 \]
400 Multiplication by $z$ is an operator $H$. In the ordered local bases $1,\varepsilon$ and $1,\eta,\eta^2$, its matrix is
401 \[
402 H=\left(\begin{array}{cc|ccc}
403 1&0&0&0&0\\1&1&0&0&0\\\hline
404 0&0&-1&0&0\\0&0&1&-1&0\\0&0&0&1&-1
405 \end{array}\right).
406 \]
407 
408 \vfig[.98]{jordan_packets}
409 \captiontext{These arrows are the actual nilpotent actions $H-1$ and $H+1$. Each nonzero displayed basis vector belongs to the indicated packet. Only the last nonzero vector in each chain is an ordinary eigenvector.}
410 
411 \head{The projectors can be written explicitly}
412 Modulo $(z-1)^2(z+1)^3$, put
413 \[
414 e_+(z)=\frac{(5-3z)(z+1)^3}{16},\qquad e_-(z)=1-e_+(z).
415 \]
416 Then $e_+\equiv1\pmod{(z-1)^2}$ and $e_+\equiv0\pmod{(z+1)^3}$. Consequently
417 \[
418 e_+(H)=\operatorname{diag}(1,1,0,0,0),\qquad
419  e_-(H)=\operatorname{diag}(0,0,1,1,1).
420 \]
421 The packet dimensions are two and three. The ordinary eigenspaces each have dimension one; the generalized eigenspaces have dimensions two and three.
422 
423 \takeaway{Canonical idempotents isolate the full primary pieces, including their nilpotent structure. Semisimplicity is not required, and replacing a packet by an ordinary eigenspace can discard most of it.}
424 
425 % ---------------------------------------------------------------------------
426 \vpage{13}{A possible packet can still be empty}{MODEL C / OCCUPANCY IS MODULE DATA}{Main text: Theorem 7.4, Theorem 8.2 and Corollary 8.4\quad \href{p12-route:visual-13}{Main statement 7.4}}
427 Keep the algebra $T=T_+\times T_-$ from \pagerefV{12}, but compare two different $T$-modules:
428 \[
429 V=T,\qquad W=T_+,
430 \]
431 where $T$ acts on $W$ through projection to its first factor. The algebra still has two maximal ideals, but only one of them is in $\Supp_T(W)$. Here $T$ is a supplied algebra; its action on $W$ is \emph{not} assumed faithful.
432 
433 \begin{center}
434 \begin{tikzpicture}[x=1cm,y=1cm,font=\small]
435 \node[anchor=east] at (-.25,1.65) {$V=T$};\node[anchor=east] at (-.25,.55) {$W=T_+$};
436 \foreach \j in {0,...,4}{\draw[fill=black!15,draw=black!35] (\j*.95,1.3) rectangle ++(.75,.7);}
437 \foreach \j in {0,1}{\draw[fill=black!15,draw=black!35] (\j*.95,.2) rectangle ++(.75,.7);}
438 \foreach \j in {2,3,4}{\draw[black!30] (\j*.95,.2) rectangle ++(.75,.7);\node[text=black!40] at (\j*.95+.375,.55) {$\times$};}
439 \draw[dashed] (1.80,-.15)--(1.80,2.35);
440 \node[font=\footnotesize] at (.85,2.48) {$T_+$: 2 slots};
441 \node[font=\footnotesize] at (3.25,2.48) {$T_-$: 3 slots};
442 \node[font=\footnotesize,anchor=west] at (5.15,1.65) {dimensions $2+3$};
443 \node[font=\footnotesize,anchor=west] at (5.15,.55) {dimensions $2+0$};
444 \end{tikzpicture}
445 \end{center}
446 \captiontext{Squares refer to the local bases of the regular module. Crossed reference slots are absent from $W$; they are not additional basis vectors of $W$.}
447 
448 \begin{center}
449 \begin{tabular}{@{}lcc@{}}\toprule
450  & regular module $V$ & first-factor module $W$\\\midrule
451 $\dim e_+(-)$ & 2 & 2\\
452 $\dim e_-(-)$ & 3 & 0\\
453 Support & both maximal points & only the $+$ point\\
454 Trace of $e_-$ & 3 & 0\\\bottomrule
455 \end{tabular}
456 \end{center}
457 
458 \head{A generalized packet is not its residue-field fiber}
459 At the point $z=1$, the full local packet $e_+V$ has dimension two, but its scalar fiber is
460 \[
461 V\otimes_{T,\,z\mapsto1}\QQ
462 \cong T_+/(\varepsilon)\cong\QQ.
463 \]
464 The packet has the basis $1,\varepsilon$. Its scalar fiber kills $\varepsilon$. The ordinary eigenspace is instead the kernel of multiplication by $\varepsilon$, spanned by $\varepsilon$. In this example the last two spaces have the same dimension, but they are obtained by different operations.
465 
466 For $W$, the fiber at $z=-1$ is zero. At $z=1$ it is nonzero. The field-fiber criterion detects exactly the support of the supplied module.
467 
468 \takeaway{A ring factor describes an available packet location. A nonzero localized module certifies occupancy there. A field fiber then specializes that module; it is not the whole generalized packet.}
469 
470 % ---------------------------------------------------------------------------
471 \vpage{14}{At two, division is the obstruction}{A COEFFICIENT-PRIME BOUNDARY}{Main text: Section 5, Remark 8.5 and Section 11\quad \href{p12-route:visual-14}{Main statement 11.1}}
472 Let $C_2=\langle g:g^2=1\rangle$. Over $\QQ_2$, the two familiar sign projectors are
473 \[
474 e_+=\frac{1+g}{2},\qquad e_-=\frac{1-g}{2},
475 \qquad \QQ_2[C_2]\cong\QQ_2\times\QQ_2.
476 \]
477 They do not belong to $\ZZ_2[C_2]$. Therefore one cannot first use them integrally and then reduce modulo two.
478 
479 \head{Compute the actual residual algebra}
480 Over $\FF_2$, put $\delta=g-1=g+1$. Then $\delta^2=g^2-2g+1=0$ and
481 \[
482 \FF_2[C_2]\cong\FF_2[\delta]/(\delta^2).
483 \]
484 The two sign eigenvalues have merged. The ring is local and retains a nilpotent direction instead of two residual projector pieces.
485 
486 \begin{minipage}[c]{.48\linewidth}
487 \centering
488 \begin{tabular}{@{}ccc@{}}\toprule
489 Element & Square & Idempotent?\\\midrule
490 $0$ & $0$ & yes\\
491 $1$ & $1$ & yes\\
492 $\delta$ & $0$ & no\\
493 $1+\delta$ & $1$ & no\\\bottomrule
494 \end{tabular}
495 \end{minipage}\hfill
496 \begin{minipage}[c]{.47\linewidth}
497 In the basis $1,\delta$, multiplication by $g=1+\delta$ is
498 \[
499 G=\begin{pmatrix}1&0\\1&1\end{pmatrix},\qquad
500 G^2=I\quad\text{over }\FF_2.
501 \]
502 It is not the identity: $G-I$ is nonzero and square-zero.
503 \end{minipage}
504 
505 \head{A trace can vanish while a packet is occupied}
506 The identity projector on $\FF_2^2$ has rank two and trace $2=0$. Hence nonzero rank and nonzero trace are not equivalent in this characteristic. The localization and rank criteria still detect the nonzero space. The characteristic-zero qualification in Corollary 8.4 is essential.
507 
508 \head{Recovery without averaging}
509 The augmentation map sends $g$ to $1$ and has kernel $(g-1)$. Thus
510 \[
511 \ZZ_2[C_2]/(g-1)\cong\ZZ_2,
512 \qquad
513 \bigl(\ZZ_2[C_2]^r\bigr)/(g-1)\ZZ_2[C_2]^r\cong\ZZ_2^r.
514 \]
515 These quotient identities require no division by two. They explain why augmentation and coinvariants remain usable even when sign-projector formulas do not.
516 
517 \takeaway{Use the algebra available at the coefficient prime. Augmentation recovery is valid here; it still does not prove full deformation support or supply a universal automorphy theorem at two.}
518 
519 % ---------------------------------------------------------------------------
520 \vpage{15}{Nonzero at every level is not enough}{INVERSE SYSTEMS / THE STABLE CORE}{Main text: Definition 8.9, Lemma 8.10 and Theorem 8.11\quad \href{p12-route:visual-15}{Main statement 8.11}}
521 Let every level be $V_m=k^3$. For a transition matrix $F$, compatibility means $v_m=Fv_{m+1}$. The directions visible at level $m$ after looking $h$ levels ahead form $\im(F^h)$.
522 
523 \begin{minipage}[t]{.48\linewidth}\centering
524 {\sffamily\bfseries A surviving direction}\par
525 \[
526 P=\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix},\qquad P^2=P.
527 \]
528 \vfig[.45]{core_projection}
529 \captiontext{At every positive horizon, precisely the $e_1$ direction remains visible.}
530 \end{minipage}\hfill
531 \begin{minipage}[t]{.48\linewidth}\centering
532 {\sffamily\bfseries Every direction eventually dies}\par
533 \[
534 N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\qquad N^3=0.
535 \]
536 \vfig[.45]{core_nilpotent}
537 \captiontext{Image dimensions are $3,2,1,0,0,\ldots$, although every level still has dimension three.}
538 \end{minipage}
539 
540 \head{Now compute the limits, not just their level dimensions}
541 For the $P$-system, the limit consists of the constant compatible sequences $(a e_1,a e_1,\ldots)$, so it is isomorphic to $k$. For the $N$-system, every compatible sequence satisfies
542 \[
543 v_m=N^3v_{m+3}=0\quad\text{for every }m,
544 \]
545 and the limit is zero. This is an all-level argument from the identity $N^3=0$, not a finite experimental guess.
546 
547 \head{What the stable core measures}
548 In a general inverse system of finite-dimensional spaces,
549 \[
550 Y_m=\bigcap_{n\ge m}\im(f_{n,m})
551 \]
552 contains exactly the vectors visible from arbitrarily remote levels. The paper proves that the transition maps restrict to surjections $Y_{m+1}\twoheadrightarrow Y_m$. A prescribed nonzero vector in $Y_1$ can then be lifted recursively to a coherent nonzero sequence.
553 
554 \takeaway{The stable core tests persistence through the transition maps. Counting nonzero spaces at separate levels does not test persistence.}
555 
556 % ---------------------------------------------------------------------------
557 \vpage{16}{Finite dimension prevents tail escape}{TWO DIFFERENT COMPACTNESS MECHANISMS}{Main text: Theorem 8.7, Corollary 8.8 and Example 8.13\quad \href{p12-route:visual-16}{Main statement 8.7}}
558 Nested nonzero subspaces inside \emph{one fixed finite-dimensional space} behave differently from nested tails inside an infinite-dimensional space.
559 
560 \begin{minipage}[t]{.44\linewidth}\centering
561 {\sffamily\bfseries Fixed carrier $k^5$}\par\vspace{4pt}
562 \vfig[.75]{nested_finite}
563 \captiontext{The projectors retain initial basis directions with dimensions $5,4,4,2,1,1,\ldots$. In this explicitly defined example the projector is constant from level five onward.}
564 \end{minipage}\hfill
565 \begin{minipage}[t]{.52\linewidth}\centering
566 {\sffamily\bfseries Infinite carrier $\ell^2(\mathbb N)$}\par\vspace{4pt}
567 \vfig[.91]{infinite_tail}
568 \captiontext{Row $m$ retains all basis vectors $e_j$ with $j\ge m$. Every row continues infinitely to the right. The display is a finite window, not a truncation of the actual carrier.}
569 \end{minipage}
570 
571 \head{Why the finite chain eventually stops changing}
572 Suppose idempotents satisfy
573 \[
574 e_{m+1}e_m=e_me_{m+1}=e_{m+1}.
575 \]
576 Their ranges are nested. In dimension $d$, nonzero ranges allow at most $d-1$ strict dimension drops. There may be long gaps between drops, but only finitely many drops. The chain therefore becomes constant and has nonzero intersection.
577 
578 \begin{insight}
579 A single repeated dimension does \emph{not} certify final stabilization. The finite example deliberately has a plateau $4,4$ before another drop. Finite dimension bounds the number of strict drops, not the time of the last one.
580 \end{insight}
581 
582 \head{Why every infinite tail can be nonzero while their intersection vanishes}
583 Let $P_m$ project onto the closed span of $e_m,e_{m+1},\ldots$. Each range is nonzero. But a vector belonging to every range has its $j$-th coordinate zero once $m>j$. Every coordinate is therefore zero:
584 \[
585 \bigcap_{m\ge1}P_m\ell^2(\mathbb N)=\{0\}.
586 \]
587 
588 \takeaway{Fixed finite-carrier stabilization, stable images in an inverse system, and infinite Hilbert-space tails are different situations. Their hypotheses cannot be exchanged.}
589 
590 % ---------------------------------------------------------------------------
591 \vpage{17}{Where the arithmetic still enters}{THE CONDITIONAL THEOREM / INPUTS AND CONSEQUENCES}{Main text: revised Theorem 12.6 and Section 13\quad \href{p12-route:visual-17}{Main statement 12.6}}
592 The final theorem is a composition of proved implications with explicitly supplied data. Use a single coherent Frobenius normalization in every polynomial comparison. These are not competing labels: an implication may be internally proved while its arithmetic premises remain external.
593 
594 {\small
595 \begin{tabularx}{\linewidth}{@{}p{.13\linewidth}YY@{}}\toprule
596 \textbf{Gate} & \textbf{What must be supplied} & \textbf{What the formal deduction gives}\\\midrule
597 Residual & Residual automorphic occupancy and a represented deformation problem. & The rings, universal representation, and starting carrier are available as inputs.\\
598 Patching & An applicable patching package: full patched support, quotient, and control isomorphism. & Full support descends to the finite-level module or perfect complex.\\
599 Comparison & A ring map $f:R\to T$ with the stated support comparison, or the suitable isomorphism and carrier identification. & Support is transported at compatible points.\\
600 Fiber & A Hecke point $y:T\to E'$ over the same field, with $y\circ f=x$. & A nonzero scalar fiber and an actual eigensystem; in the derived case, one survives in a cohomological degree.\\
601 Realization & Classicality for that eigensystem, an attached representation, continuity, equality of the universal specialization and attached Frobenius polynomials on a dense union of conjugacy classes, and semisimplification/Brauer--Nesbitt inputs. & The specialization is identified after semisimplification with the attached representation.\\\bottomrule
602 \end{tabularx}}
603 
604 \head{Support alone does not supply a point over the same field}
605 The inclusion $\QQ\to\QQ(i)$ preserves full support of the nonzero scalar-extended carrier. But no $\QQ$-algebra map $\QQ(i)\to\QQ$ exists: the image of $i$ would be a rational number $a$ with $a^2=-1$. Thus one cannot infer the required Hecke point $y$ merely from a support statement.
606 
607 \head{Propagation needs two existing families and the same automorphic object}
608 After producing an automorphic object $\pi$, retain \emph{that same} $\pi$. The propagation statement assumes both the original compatible family and the family attached to $\pi$ already exist, with the common Frobenius polynomial data and the stated comparison hypotheses. It then identifies corresponding members. It does not construct a missing family.
609 
610 \takeaway{The algebra locates and transports supplied nonzero data. Arithmetic inputs establish occurrence, classicality, and attachment. A visualization must not turn one of those inputs into an automatic consequence.}
611 
612 % ---------------------------------------------------------------------------
613 \vpage{18}{Return to the proofs with a concrete object}{READING ROUTE / SOURCE CROSSWALK / CHECKS}{Reading guide to Paper XII; formal source v0.17.0 and evidence v0.17.1\quad \href{p12-route:visual-18}{Main statement 3.1}}
614 Use the example on the left to keep the relevant object in view while reading the exact statement on the right. The guide is explanatory; the main text supplies the theorem hypotheses and the proof contract.
615 
616 {\small
617 \begin{tabularx}{\linewidth}{@{}p{.12\linewidth}Yp{.30\linewidth}@{}}\toprule
618 \textbf{Guide} & \textbf{Question or calculation} & \textbf{Paper XII reference}\\\midrule
619 V--3 & Quotient monomials before or after completed combination & Theorem 3.1; Corollary 3.2\\
620 V--4 & A test point factors exactly when its local equations vanish & Proposition 3.3\\
621 V--5 & Compatible precision data and the closed-kernel argument & Section 3; Section 10\\
622 V--6--7 & Minors, fiber dimensions, support, and nilpotent structure & Theorems 6.3--6.4\\
623 V--8 & Base change and transport of intrinsic Fitting ideals & Proposition 6.5; Theorem 7.1\\
624 V--9 & A quotient loses a generic point without creating a projector & Propositions 3.4 and 7.3\\
625 V--10 & Auxiliary freeness versus full deformation support & Theorems 6.7, 6.9, 6.11; Example 6.13\\
626 V--11 & A two-term projective complex and its derived fibers & Definition 6.14; Theorem 6.16\\
627 V--12--13 & Whole primary packets, localization, and scalar fibers & Theorems 7.4, 8.2; Corollaries 8.3--8.4\\
628 V--14 & Sign-projector failure and augmentation recovery & Section 5; Remark 8.5; Section 11\\
629 V--15--16 & Stable images, finite stabilization, and tail escape & Section 8, items 8.7--8.13\\
630 V--17 & Compatible Hecke points and conditional realization & Theorem 12.6; Section 13\\\bottomrule
631 \end{tabularx}}
632 
633 \head{Three short proof-reading tests}
634 \textbf{Can a module have full support but fail to be faithful?} Yes. On \pagerefV{10}, the module $k$ over $k[[x]]/(x^2)$ has a nonzero nilpotent annihilator and full support.
635 
636 \textbf{Do equal support sets force equal Fitting ideals?} No. On \pagerefV{7}, $(xy)$ and $(x^2y)$ have the same vanishing set but different infinitesimal behavior.
637 
638 \textbf{Do nonzero finite levels force a nonzero inverse limit?} No. On \pagerefV{15}, $N^3=0$ forces every compatible sequence to vanish, although every level is $k^3$.
639 
640 \head{Provenance and reproducibility}
641 The source is \emph{Patching Support, Hecke Packets, and Conditional Automorphic Completion}, Komorebi P., revised formalization v0.17.0. The theorem references above are internal references to that manuscript. The existing 49-claim revised formal contract and its inherited original-wording boundaries are unchanged.
642 
643 The accompanying Python script performs \textbf{46 exact checks}: exhaustive finite test-ring enumeration, symbolic minors and CRT identities, exact matrix calculations, and the identities behind the stable-core examples. All drawings are reproducible vector graphics. Further worked details appear on \pagerefV{19}--\pagerefV{25}; the inherited revised/original claim boundary is on \pagerefV{26}. The illustrative examples are not newly certified in Lean by this guide; its finite checks are reported separately from the source release. Infinite claims are supported by the arguments given here, not by extrapolating a finite image.
644 
645 % ---------------------------------------------------------------------------
646 \setlength{\parskip}{5pt}
647 \vpage{19}{Complete the actual topology}{TOPOLOGICAL ALGEBRA / THE COMPARISON MAP}{Main text: Theorem 3.1, Corollary 3.2 and Proposition 3.3\quad \href{p12-route:visual-19}{Main statement 3.1}}
648 The hat in $\whot$ names an actual completion. For the finite family of local rings, start with the ordinary tensor and its ideal of definition:
649 \[
650  S=\bigotimes_{i=1}^s R_i,\qquad
651  \mathfrak a=\m_{\OO}S+\sum_i\m_{R_i}S,\qquad L=\sum_i I_iS.
652 \]
653 The comparison uses the $\mathfrak a$-adic additive uniformity on $S$, the image ideal topology on $S/L$, and the closed extended ideal $J=\overline{L\widehat S}$.
654 
655 \vfig[.82]{completion_square}
656 \captiontext{The square names the actual quotient and completion maps. The lower identification is bicontinuous and preserves the coefficient map and every canonical factor map. The square does not assert that completion leaves $S$ unchanged.}
657 
658 \head{The explicit model revisited}
659 For $R_1=\ZZ_5[[x]]$ and $R_2=\ZZ_5[[y]]$, the completion of their ordinary tensor in this topology is $\ZZ_5[[x,y]]$. With $L$ generated by $x^2,y$, the lower quotient is $\ZZ_5[\varepsilon]/(\varepsilon^2)$ with its $(5,\varepsilon)$-adic topology. The containment on \pagerefV{5} identifies this with its $5$-adic topology.
660 
661 \head{Boundary cases are part of the contract}
662 \begin{center}\small
663 \begin{tabularx}{\linewidth}{@{}p{.22\linewidth}Y@{}}\toprule
664 Case & Actual object and consequence\\\midrule
665 Empty family & $S=\OO$ and $\mathfrak a=\m_\OO$. The scalar adic topology remains; an empty sum of factor ideals does not erase it.\\
666 All $I_i=0$ & $J=0$; the comparison returns the completed tensor itself.\\
667 Some $I_i=R_i$ & $J=\widehat S$ and both sides are the zero ring. This comparison does not give a local ring with residue field $k$.\\\bottomrule
668 \end{tabularx}
669 \end{center}
670 
671 For global base change, the structural map $A\to B$ belongs to the continuous local coefficient-algebra setting. Then $B\whot_A(A/J)\cong B/\overline{JB}$. For a continuous algebra point to a complete Hausdorff target, closedness of its kernel gives the factorization test. Locality of a point, when supplied, passes to its unique quotient factor.
672 
673 \takeaway{Closure and topology are structural data. A list of equations in a finite model illustrates the quotient but does not replace the actual completed comparison.}
674 
675 % ---------------------------------------------------------------------------
676 \vpage{20}{A finite adequacy calculation}{RESIDUAL IMAGE / ONE EXPLICIT CERTIFICATE}{Main text: Theorem 4.2; label \texttt{thm:s3\_adequacy}. Adequacy convention: external Standard Fact 4.1.\quad \href{p12-route:visual-20}{Main statement 4.2}}
677 Over $k=\FF_7$, the standard representation of $S_3$ has matrices
678 \[
679  s=\begin{pmatrix}6&1\\0&1\end{pmatrix},\qquad
680  t=\begin{pmatrix}0&6\\1&6\end{pmatrix}.
681 \]
682 All entries and calculations on this page are modulo seven. Direct multiplication gives $s^2=t^3=I$ and $sts=t^{-1}$. Their eigenvalues are $\{-1,1\}$ and $\{2,4\}$. Distinct eigenvalues allow an exact projector formula
683 \[
684 P_{D,\alpha}=(\alpha-\beta)^{-1}(D-\beta I).
685 \]
686 
687 \vfig[.92]{adequacy_pairings}
688 \captiontext{Each row displays the supplied matrix, an actual canonical eigenprojector and its nonzero trace pairing. It detects that stated adjoint constituent; it does not create a global automorphic carrier.}
689 
690 Write $W(a,b,c)=\left(\begin{smallmatrix}a&b\\c&-a\end{smallmatrix}\right)$. The manuscript exhibits $Z=W(2,3,4)$, $A=W(1,1,2)$ and $B=W(1,5,6)$; their coordinate determinant is $3$. Thus they are a basis of the trace-zero adjoint. Conjugation splits this space as $\operatorname{sgn}\oplus V$. In the drawing the standard witness is $W(6,4,3)=3A+3B$.
691 
692 \head{Why the cohomology argument is uniform in the module}
693 Since $|S_3|=6$ is a unit in $\FF_7$, a cocycle $f:S_3\to M$ is a coboundary: set $u=-6^{-1}\sum_h f(h)$ and sum $f(gh)=g f(h)+f(g)$ to obtain $gu-u=f(g)$. This proves $H^1(S_3,M)=0$ for every $\FF_7[S_3]$-module $M$. The printed invariant calculation separately proves $H^0(S_3,\operatorname{ad}^0V)=0$.
694 
695 \head{What makes this a finite certificate}
696 The manuscript also checks absolute irreducibility and projector detection on every irreducible adjoint submodule. In this example $7\nmid2$, and every group element's eigenvalues lie in $\FF_7$, as required by the stated convention. The two displayed pairings alone are a worked part of that complete certificate.
697 
698 \takeaway{The calculation is exact for $S_3/\FF_7$. Applying an automorphy lifting theorem still requires its actual residual image hypotheses and residual occupancy. No coefficient-prime-two adequacy theorem follows from this picture.}
699 
700 % ---------------------------------------------------------------------------
701 \vpage{21}{Augmentation in two variables}{AUXILIARY ALGEBRA / AN EXACT MONOMIAL BASIS}{Main text: Proposition 5.1, Theorem 5.2 and Theorem 5.4\quad \href{p12-route:visual-21}{Main statement 5.2}}
702 Take $\Delta=C_4\times C_2$ in residue characteristic two. If $g$ and $h$ generate its two factors, put $X=g-1$, $Y=h-1$. Then
703 \[
704 \FF_2[\Delta]\cong\FF_2[X,Y]/(X^4,Y^2).
705 \]
706 The identity $(1+X)^4-1=X^4$ and its $Y$ analogue provide the relations. The eight surviving monomials provide the exact dimension $|\Delta|=8$ and establish the isomorphism after surjectivity.
707 
708 \vfig[.50]{augmentation_grid}
709 \captiontext{All eight basis monomials are displayed. The filled scalar slot survives augmentation; all other displayed monomials lie in the augmentation ideal. This is a basis array, not a spectrum with eight points.}
710 
711 \head{The residual ring is local}
712 The augmentation ideal is $(X,Y)$, its quotient is $\FF_2$, and $(X,Y)^5=0$: a degree-five monomial has $X$-exponent at least four or $Y$-exponent at least two. Every element outside $(X,Y)$ has the form $1+n$ for nilpotent $n$, with inverse the finite geometric series. Thus $(X,Y)$ is the unique maximal ideal, and the only idempotents are $0,1$.
713 
714 \head{A quotient of groups gives a quotient of algebras}
715 Map $C_4\times C_2\to C_2$ by $g\mapsto\bar g$, $h\mapsto1$. The induced algebra map sends $X$ to $\bar X$ with $\bar X^2=0$ and sends $Y$ to zero. It is surjective. Augmentation commutes with this map because every group basis element is sent to another group element of augmentation one.
716 
717 \head{Recover a free carrier, with its base ring named}
718 For $\Lambda=\ZZ_2[\Delta]$ and the right module $M=\Lambda^r$,
719 \[
720 M\otimes_{\Lambda,\epsilon}\ZZ_2
721 \cong M/M\ker\epsilon\cong\ZZ_2^r.
722 \]
723 The quotient applies augmentation to each basis coordinate. This remains valid for nonabelian finite two groups with the stated right-module convention. It establishes the rank over the coefficient ring after coinvariants, without establishing full support over a separate deformation ring.
724 
725 \takeaway{The augmentation quotient keeps the scalar coordinate. It needs no averaging denominator and supplies no missing patching-support hypothesis.}
726 
727 % ---------------------------------------------------------------------------
728 \vpage{22}{A finite forest selects a branch}{FINITE OCCUPANCY / MAPS NEED NOT BE SURJECTIVE}{Main text: Theorem 10.3 and Corollary 10.4\quad \href{p12-route:visual-22}{Main statement 10.3}}
729 A finite nonempty set at every precision, with total reduction maps, can select a compatible branch even when some nodes have no deeper lift. Here is an explicit system:
730 \[
731 X_1=\{r\},\quad X_2=\{a,b\},\quad X_3=\{c,d\},\quad
732 X_n=\{e_n\}\quad(n\ge4).
733 \]
734 Both $a,b$ reduce to $r$; both $c,d$ reduce to $a$; $e_4$ reduces to $c$; and $e_{n+1}$ reduces to $e_n$ for every $n\ge4$.
735 
736 \vfig[.66]{finite_forest}
737 \captiontext{Every displayed edge is an actual reduction map. The formula specifies the infinite continuation. The chosen branch is $r,a,c,e_4,e_5,\ldots$. The nodes $b,d$ are admissible at their own precision and have no successor.}
738 
739 \head{What the compactness argument proves}
740 For general finite levels, the reduction forest is a locally finite rooted tree with vertices at every depth. An infinite branch exists. Surjectivity is sufficient but is not necessary: the map $X_3\to X_2$ above misses $b$, and $X_4\to X_3$ misses $d$.
741 
742 Finite occupancy does not guarantee a branch through a prescribed higher-level node. It also does not prove arithmetic occupancy of the finite sets. Those sets must already have been supplied as admissible realizations of the declared spine and residual root.
743 
744 \head{Why the vector-space counterexample is consistent with this}
745 The nilpotent system on \pagerefV{15} has a compatible sequence: the zero sequence. Its finite levels are nonempty sets because they contain zero. The question there is whether the inverse limit contains a \emph{nonzero} vector, which needs stable-core data. Here the question is nonemptiness of an inverse limit of admissible finite sets. Different conclusions require different hypotheses.
746 
747 \takeaway{Finite set compactness chooses a branch among existing admissible nodes. It neither extends every node nor produces an arithmetic node absent from the finite forest.}
748 
749 % ---------------------------------------------------------------------------
750 \vpage{23}{Classes are not matrix frames}{COMPANION COMPLETION / ACTUAL REPRESENTATIVES}{Main text: revised Theorem 10.5 and its preceding admissibility hypotheses\quad \href{p12-route:visual-23}{Main statement 10.5}}
751 Let $\Gamma=C_2$ with the discrete topology, and represent its generator by $D=\operatorname{diag}(1,-1)$ over $\ZZ_5$. At precision $n$, its conjugacy class is an admissible class in $\GL_2(\ZZ/5^n\ZZ)$. These classes are compatible. Arbitrary representatives need not be.
752 
753 \vfig[.70]{frames_example}
754 \captiontext{The upper representatives are conjugate to the same diagonal involution, but their reductions are unequal matrices. The lower choices use $D$ consistently at both precisions. All entries displayed are canonical residues.}
755 
756 \head{Calculate the mismatch}
757 For $Q=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$,
758 \[
759 QDQ^{-1}=\begin{pmatrix}1&-2\\0&-1\end{pmatrix}
760 \equiv\begin{pmatrix}1&23\\0&24\end{pmatrix}\pmod{25}.
761 \]
762 It reduces to $\left(\begin{smallmatrix}1&3\\0&4\end{smallmatrix}\right)$ modulo five, not the chosen diagonal matrix $\left(\begin{smallmatrix}1&0\\0&4\end{smallmatrix}\right)$. Nevertheless, their residual conjugacy classes agree.
763 
764 \head{The proof chooses representatives through a finite tree}
765 A finite coefficient quotient has a finite group of invertible basis matrices. Each of the finitely many admissible classes therefore has only finitely many representatives. Take \emph{all} actual continuous representatives of the admissible classes at each level. Reduction is a total map between these finite nonempty sets because admissibility is preserved under reduction. A branch then gives strictly compatible matrices. Completeness reconstructs entries and inverse entries over $\OO_\lambda$; finite-level continuity gives continuity in the ideal-adic topology.
766 
767 This argument does not assume that a chosen conjugating matrix lifts, nor that every representative has an admissible successor. The theorem retains the declared residual \emph{class}. It specifies neither a prescribed residual frame nor uniqueness up to integral conjugacy.
768 
769 \takeaway{Compatibility of classes is the input. Strictly compatible continuous representatives are a constructed output, obtained with the finite-frame and coefficient hypotheses in the revised statement.}
770 
771 % ---------------------------------------------------------------------------
772 \vpage{24}{Equations survive every precision}{ADIC EFFECTIVITY / DECLARED POLYNOMIAL CONDITIONS}{Main text: revised Theorem 10.5 and Corollary 3.2\quad \href{p12-route:visual-24}{Main statement 10.5}}
773 The preceding involution gives an exact polynomial condition. For a matrix $A=(a_{ij})$, require all entries of $A^2-I$ to vanish. In
774 \[
775 B=\ZZ_5[a_{11},a_{12},a_{21},a_{22}],
776 \]
777 the ideal $J$ is generated by
778 \[
779 \begin{aligned}
780 a_{11}^2+a_{12}a_{21}-1,&\qquad a_{12}(a_{11}+a_{22}),\\
781 a_{21}(a_{11}+a_{22}),&\qquad a_{21}a_{12}+a_{22}^2-1.
782 \end{aligned}
783 \]
784 For $D$ and every conjugate $QDQ^{-1}$ these equations vanish. The condition is valid in every admissible frame. By contrast, the equation $a_{12}=0$ holds for $D$ but fails for the conjugate displayed on \pagerefV{23}.
785 
786 \head{The reconstruction argument is coefficientwise}
787 Let $A_n$ be compatible matrices modulo $5^n$, each satisfying these four equations. Completeness reconstructs $A\in M_2(\ZZ_5)$. For each polynomial $f\in J$, evaluation commutes with reduction, so
788 \[
789 f(A)\in\bigcap_{n\ge1}5^n\ZZ_5=\{0\}.
790 \]
791 Separatedness supplies the last equality. Thus the actual evaluation map $B\to\ZZ_5$ kills $J$ and factors uniquely through $B/J$. If inverse matrices were reconstructed as well, this is an actual representation in $\GL_2(\ZZ_5)$.
792 
793 \vfig[.52]{polynomial_effectivity}
794 \captiontext{Every finite evaluation kills the declared ideal. Reduction commutes with evaluation, and separatedness gives integral vanishing. The displayed diagram represents this exact polynomial condition.}
795 
796 \head{The scope of the effectivity conclusion}
797 The companion theorem needs integral Frobenius polynomials, finite admissible continuous classes, reduction closure, finite occupancy, and the declared triviality subgroup. A condition in one frame needs a compatible-frame argument. An arbitrary Hodge or Weil--Deligne condition is not automatically a polynomial ideal in matrix entries.
798 
799 Represented local quotient conditions use continuous ring maps and the closed-kernel argument on \pagerefV{19}. These are different supplied objects from the polynomial evaluation model.
800 
801 \takeaway{The precise coefficient and admissibility hypotheses preserve declared equations. Representability of a desired arithmetic condition remains a separate input.}
802 
803 % ---------------------------------------------------------------------------
804 \vpage{25}{Recover multiplicities from traces}{A FINITE SEPARATOR / A KNOWN SEMISIMPLE CARRIER}{Main text: Proposition 10.9\quad \href{p12-route:visual-25}{Main statement 10.9}}
805 Suppose a known semisimple carrier is a sum of two known constituents with multiplicities $m_1,m_2$. Consider the independent trace functions
806 \[
807 t_1(\omega)=1,\qquad t_2(\omega)=\omega,\qquad \omega\in\QQ.
808 \]
809 At the two sample points $\omega=1,-1$, their evaluation matrix is
810 \[
811 A=\begin{pmatrix}1&1\\1&-1\end{pmatrix},\qquad\det A=-2.
812 \]
813 It is invertible over $\QQ$. If the measured trace vector is $(5,1)$, solve
814 \[
815 A\begin{pmatrix}m_1\\m_2\end{pmatrix}=\begin{pmatrix}5\\1\end{pmatrix}
816 \quad\Longrightarrow\quad
817 \begin{pmatrix}m_1\\m_2\end{pmatrix}=\begin{pmatrix}3\\2\end{pmatrix}.
818 \]
819 
820 \vfig[.46]{trace_separator}
821 \captiontext{The vectors $(1,1)$ and $(1,-1)$ are the two independent rows of the exact evaluation matrix. The coefficients $3,2$ are recovered by the displayed linear equations. This illustration uses functions on $\QQ$, not a newly constructed arithmetic Frobenius set.}
822 
823 \head{Why finitely many evaluations suffice in general}
824 Independent functions $t_1,\ldots,t_r$ on $\Omega$ have evaluation vectors spanning $E'^r$. Otherwise an annihilating nonzero linear functional would give a forbidden relation among the functions. Choosing $r$ independent evaluations gives an invertible matrix that recovers coefficients in their known sum.
825 
826 \head{Two checks before using this in arithmetic}
827 The trace functions must be independent on the permitted Frobenius set; the carrier must already be semisimple with the stated constituents. Arbitrary samples supply neither premise. Nonnegative integral multiplicities come from representation data, not every rational linear-system solution.
828 
829 \head{The coefficient field matters again}
830 The displayed integer matrix reduces modulo two to a matrix with determinant zero and two coincident rows. Those residual samples cannot recover the two coefficients. This matches the trace caution on \pagerefV{14}: characteristic-zero trace detection cannot be silently transported to residue characteristic two.
831 
832 \takeaway{A finite separator reconstructs coefficients in supplied independent trace data. It is not a construction of the constituents, their compatible systems, or automorphic realizations.}
833 
834 % ---------------------------------------------------------------------------
835 \vpage{26}{Read the exact claim contract}{PROVENANCE / REVISED SCOPE AND ORIGINAL WORDING}{Source: formal release v0.17.0 manuscript, v0.17.1 status and claim delta; finalized 10 October 2026\quad \href{p12-route:visual-26}{Main statement 12.6}}
836 This guide explains the revised manuscript and retains its inherited verification boundaries. There are two censuses, with different statements as their subjects:
837 
838 \begin{center}
839 \begin{tabular}{@{}lrrr@{}}\toprule
840 Inherited contract & Complete & Partial & Unchecked\\\midrule
841 Revised 49 internal claims & 49 & 0 & 0\\
842 Original wording of those claims & 41 & 7 & 1\\\bottomrule
843 \end{tabular}
844 \end{center}
845 
846 \head{The eight inherited amendments}
847 \begin{center}\footnotesize
848 \begin{tabularx}{\linewidth}{@{}p{.11\linewidth}Y@{}}\toprule
849 Claim & Scope that the revised statement makes explicit\\\midrule
850 6.12 & Each irreducible component is contained in full support; the original containment was reversed.\\
851 9.4 & Hausdorff coefficient topology and continuity for the compared representations.\\
852 9.5 & The realized locus consists of points satisfying those topology and continuity hypotheses.\\
853 9.6 & The surviving cohomological degree belongs to the declared classicality range; comparison inputs are explicit.\\
854 10.5 & Complete separated coefficients, finite continuous admissible classes, integrality, and actual quotient equations in every admissible frame.\\
855 12.2 & Continuous characteristic-zero pointwise comparison under the stated realization input.\\
856 12.4 & Applicability of realization at the selected derived eigensystem, with the comparison inputs exposed.\\
857 12.6 & A compatible Hecke point $y\circ f=x$, actual ordinary or derived carriers, and both existing families for the same automorphic object.\\\bottomrule
858 \end{tabularx}
859 \end{center}
860 
861 In the inherited record, original 6.12 is unchecked; original 10.5 and the other six amended claims are partial. Revised completeness does not certify the stronger original wording. Claims 3.1 and 3.3 required no additional amendment in the v0.17 completion record.
862 
863 \head{Final editorial clarifications}
864 The final companions expose finite $S$, a coherent Frobenius normalization, the local coefficient-test category in 3.3, the $r>n$ Fitting boundary, and both compared Frobenius polynomial identities in ordinary completion. These clarifications are distinct from the eight inherited amendments; no new Lean replay certifies them.
865 
866 \vpage{27}{How to assess the evidence}{VALIDATION / FOUR DISTINCT KINDS OF EVIDENCE}{Inherited source records, current illustrative calculations and final editorial review\quad \href{p12-route:visual-27}{Main statement 12.6}}
867 \head{What this companion adds and what its checks mean}
868 The guide preserves the earlier worked examples. TikZ generates every drawing from exact objects; there are no raster figures. Python checks finite enumerations, minors, projectors and matrices; the text argues uniform conclusions. These checks supply no new Lean certificate or census update. Supporting Markdown (.md) files accompany the source.
869 
870 \takeaway{The source's internal formal implications are distinguished from the arithmetic inputs: representability, residual occupancy, applicable patching, classicality, attached representations and compatible systems, Frobenius density, semisimplification and Brauer--Nesbitt. Universal companions and unrestricted reciprocity remain outside the proved scope.}
871 
872 
873 \head{Inherited labels and current mathematical judgments}
874 The original-status labels record historical verification coverage. The current mathematical audit judges the actual statements. These are different questions:
875 \begin{center}\small
876 \begin{tabularx}{\linewidth}{@{}>{\raggedright\arraybackslash}p{.12\linewidth}>{\raggedright\arraybackslash}p{.20\linewidth}Y@{}}\toprule
877 Original & Inherited label & Current statement-level judgment\\\midrule
878 6.12 & unchecked & The reversed containment is false: in the node of V--9, $M=R$ has full support, which is not contained in either single component.\\
879 10.5 & partial & The unrestricted effectivity conclusion lacks the coefficient, continuity, integrality and frame hypotheses made explicit in the revision.\\\bottomrule
880 \end{tabularx}
881 \end{center}
882 
883 \head{Use each check for its actual subject}
884 \begin{center}\small
885 \begin{tabularx}{\linewidth}{@{}>{\raggedright\arraybackslash}p{.23\linewidth}Y@{}}\toprule
886 Evidence & What its subject is\\\midrule
887 Inherited Lean record & The mapped revised statements under their supplied hypotheses. This guide neither replays that build nor changes the original-wording record.\\
888 Current exact Python checks & Forty-six illustrative assertions: 31 inherited checks and 15 added checks on the printed finite examples and identities.\\
889 Mathematical arguments & Uniform quotient, specialization and all-level conclusions, with their hypotheses retained in the text. A finite diagram is not an infinite proof.\\
890 PDF and rebuild checks & Topic alignment, bounds, vector-only assets, resolved references, and an isolated byte/text/page comparison. These are document checks.\\\bottomrule
891 \end{tabularx}
892 \end{center}
893 
894 The supporting \texttt{README.md} gives portable rebuild commands. The source-coverage Markdown and JSON map all 49 internal claims to this guide's reading entries and exact manuscript labels. The audit records dependency versions, mathematical self-review and complete-page inspection. No finite calculation here verifies residual automorphic existence, classicality, attached compatible systems, or a universal arithmetic theorem.

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