Realization Calculus

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1 \section{Split Hilbert obstructions and nonlinear correction}\label{sec:split_obstruction}
2 
3 The quotient obstruction has a concrete orthogonal representative when the correction map acts between Hilbert spaces and has closed range. The Hilbert and Banach space results below are analytic extensions of the finite calculus. They use the stated analytic hypotheses and lie outside the finite formal contract.
4 
5 \begin{theorem}[Orthogonal obstruction projector]\label{thm:orthogonal_obstruction}
6 Let $D:H_1\to H_2$ be a bounded operator between Hilbert spaces with closed range.  Let $D^\dagger$ be its Moore--Penrose inverse.  Then
7 \[
8 P_{\mathrm{ob}}=I_{H_2}-DD^\dagger
9 \]
10 is the orthogonal projector onto
11 \[
12 (\im D)^\perp=\Ker D^*.
13 \]
14 For $r\in H_2$, the following are equivalent:
15 \begin{enumerate}
16 \item $Du=-r$ is solvable
17 \item $[r]=0$ in $\coker D$
18 \item $P_{\mathrm{ob}}r=0$.
19 \end{enumerate}
20 When these conditions hold, $-D^\dagger r$ is the least norm correction.  For every $r$,
21 \[
22 \dist(r,\im D)=\|P_{\mathrm{ob}}r\|.
23 \]
24 \end{theorem}
25 
26 \begin{proof}
27 Closed range gives the orthogonal decomposition
28 \[
29 H_2=\im D\oplus\Ker D^*.
30 \]
31 The Moore--Penrose identities show that $DD^\dagger$ is the orthogonal projector onto $\im D$.  The stated equivalences and distance formula follow from this decomposition.  The least norm property is the standard characterization of $D^\dagger r$.
32 \end{proof}
33 
34 Once the defect lies in the range, nonlinear correction still requires control of the nonlinear term. The next theorem gives that control through a bounded right inverse and a contraction estimate.
35 
36 \begin{theorem}[Right inverse correction theorem]\label{thm:right_inverse_correction}
37 Let $D:X\to Y$ be a bounded linear map between Banach spaces.  Assume $Z=\im D$ is closed and there is a bounded linear right inverse
38 \[
39 R:Z\longrightarrow X,
40 \qquad
41 DR=I_Z.
42 \]
43 Put $X_R=R(Z)$.  Let
44 \[
45 N:\overline B_\rho(0)\cap X_R\longrightarrow Z
46 \]
47 be $L$-Lipschitz.  Let $y\in Z$ and put
48 \[
49 q=\|R\|L.
50 \]
51 Assume $q<1$ and
52 \[
53 \|R(y-N(0))\|
54 \le
55 (1-q)\rho.
56 \]
57 Then the equation
58 \[
59 Dx+N(x)=y
60 \]
61 has a unique solution in $\overline B_\rho(0)\cap X_R$.
62 \end{theorem}
63 
64 \begin{proof}
65 The space $X_R$ is closed.  Indeed, if $Rz_n\to x$, then $z_n=DRz_n\to Dx$, so $x=R(Dx)$.  Define
66 \[
67 \Psi(x)=R(y-N(x)).
68 \]
69 This map takes $X_R$ to itself.  For $x\in\overline B_\rho(0)\cap X_R$,
70 \[
71 \begin{aligned}
72 \|\Psi(x)\|
73 &\le
74 \|R(y-N(0))\|
75 +
76 \|R\|\,\|N(x)-N(0)\|\\
77 &\le
78 (1-q)\rho+q\rho
79 =\rho.
80 \end{aligned}
81 \]
82 Also,
83 \[
84 \|\Psi(x)-\Psi(x')\|
85 \le q\|x-x'\|.
86 \]
87 The contraction mapping theorem gives a unique fixed point in the stated set.  Applying $D$ to the fixed point identity gives the equation.  Conversely, a solution in $X_R$ equals $R(Dx)$ and is therefore a fixed point of $\Psi$.
88 \end{proof}
89 
90 \begin{keyidea}[title={The hypotheses behind correction}]
91 Closed range gives the obstruction an orthogonal representative in the Hilbert setting. A bounded right inverse chooses a correction slice. In the nonlinear theorem, the target and the nonlinear term both lie in the range, and the Lipschitz bound makes the correction map a contraction on the chosen ball. The uniqueness statement concerns the chosen ball within that correction slice.
92 \end{keyidea}
93 
94 \begin{corollary}[Stability with respect to the target]\label{cor:correction_stability}
95 Under the hypotheses of \cref{thm:right_inverse_correction}, let $x$ and $x'$ be the fixed points associated with targets $y$ and $y'$.  If both fixed point maps preserve the same ball, then
96 \[
97 \|x-x'\|
98 \le
99 \frac{\|R\|}{1-q}\|y-y'\|.
100 \]
101 \end{corollary}
102 
103 \begin{proof}
104 Subtract the two fixed-point equations and use the Lipschitz bound for $N$:
105 \[
106 \|x-x'\|
107 \le
108 \|R\|\|y-y'\|+q\|x-x'\|.
109 \]
110 \end{proof}
111 
112 Lyapunov--Schmidt reduction separates the equation into two parts. The implicit function theorem solves the range component, leaving an equation in the obstruction space. Standard accounts of this reduction appear in \cite{Kielhoefer2012,Zeidler1986}.
113 
114 \begin{theorem}[Local Lyapunov--Schmidt reduction]\label{thm:lyapunov_schmidt}
115 Let $\mathcal U\subseteq X$ be an open neighborhood of zero and let $F:\mathcal U\to Y$ be a $C^r$ map between Banach spaces with $r\ge1$, $F(0)=0$, and $L=DF(0)$.  Assume there are closed decompositions
116 \[
117 X=K\oplus X_1,
118 \qquad
119 Y=R\oplus O,
120 \]
121 where $K=\Ker L$, $R=\im L$, and
122 \[
123 L_1=L|_{X_1}:X_1\longrightarrow R
124 \]
125 is a Banach space isomorphism.  Let $\pi_R$ and $\pi_O$ be the two projections on $Y$.  Then there are neighborhoods $U_K\subseteq K$ and $U_1\subseteq X_1$ of zero and a unique $C^r$ map
126 \[
127 \psi:U_K\longrightarrow U_1
128 \]
129 with $\psi(0)=0$ such that
130 \[
131 \pi_RF(k+\psi(k))=0
132 \]
133 for every $k\in U_K$.  After the neighborhoods are made smaller if needed, the solutions of $F(x)=0$ near zero are exactly the points
134 \[
135 x=k+\psi(k)
136 \]
137 for which the reduced obstruction map
138 \[
139 \kappa(k)=\pi_OF(k+\psi(k))
140 \]
141 vanishes.  At the base point,
142 \[
143 D\psi(0)=0,
144 \qquad
145 D\kappa(0)=0.
146 \]
147 \end{theorem}
148 
149 \begin{proof}
150 Define
151 \[
152 G(k,x_1)=\pi_RF(k+x_1).
153 \]
154 The derivative of $G$ in the $X_1$ variable at $(0,0)$ is $L_1$, which is invertible.  The implicit function theorem gives the unique map $\psi$.  Every nearby $x$ has a unique decomposition $x=k+x_1$.  The range equation forces $x_1=\psi(k)$, and the remaining equation is precisely $\kappa(k)=0$.
155 
156 Differentiate $G(k,\psi(k))=0$ at zero.  Since $L$ vanishes on $K$, one obtains $L_1D\psi(0)=0$, hence $D\psi(0)=0$.  Finally,
157 \[
158 D\kappa(0)=\pi_OL|_K=0.
159 \]
160 \end{proof}
161 
162 
163 \begin{corollary}[Quadratic reduced obstruction]\label{cor:quadratic_reduced_obstruction}
164 Assume the hypotheses of \cref{thm:lyapunov_schmidt} with $r\ge2$.  For $k_1,k_2\in K$,
165 \[
166 D^2\kappa(0)[k_1,k_2]
167 =
168 \pi_O D^2F(0)[k_1,k_2].
169 \]
170 \end{corollary}
171 
172 \begin{proof}
173 Differentiate $\kappa(k)=\pi_OF(k+\psi(k))$ twice.  The terms containing $D\psi(0)$ vanish.  The term containing $D^2\psi(0)$ is multiplied by $\pi_OL$, which is zero because $L(X)=R$.
174 \end{proof}
175 
176 \begin{example}[A purely quadratic obstruction]\label{ex:pure_quadratic_obstruction}
177 Let
178 \[
179 F(x,y)=(y,x^2)
180 \]
181 on $\mathbb R^2$.  At the origin,
182 \[
183 L(h,k)=(k,0).
184 \]
185 The kernel is the $x$ axis, the range is the first output axis, and the cokernel is the second output axis.  The range equation forces $y=0$.  The reduced obstruction map is
186 \[
187 \kappa(x)=x^2.
188 \]
189 The linear obstruction vanishes for every kernel direction, but the nonlinear equation still has only the solution $x=0$ near the origin.
190 \end{example}
191 
192 \begin{remark}\label{rem:higher_obstruction}
193 The reduced map $\kappa$ begins at quadratic order.  This is why vanishing of the first-order cokernel class need not settle a nonlinear problem.  Higher terms can produce a genuine obstruction on the kernel directions.
194 \end{remark}
195 
196 \section{Filtered lifting}\label{sec:filtered_lifting}
197 
198 Let $X$ and $Y$ be modules with decreasing filtrations
199 \[
200 X=F^0X\supseteq F^1X\supseteq\cdots,
201 \qquad
202 Y=F^0Y\supseteq F^1Y\supseteq\cdots.
203 \]
204 Assume both are separated and complete for their filtration topologies.  Let $\Phi:X\to Y$ be continuous and filtration congruence preserving in the sense that
205 \[
206 h\in F^mX
207 \quad\Longrightarrow\quad
208 \Phi(x+h)-\Phi(x)\in F^mY
209 \]
210 for every $x\in X$ and every $m$.  This condition makes the statement $\Phi(x)\in F^mY$ depend only on the class of $x$ modulo $F^mX$.
211 
212 Suppose $x_m$ satisfies
213 \[
214 \Phi(x_m)\in F^mY.
215 \]
216 Assume that corrections $h\in F^mX$ obey a first-order congruence
217 \[
218 \Phi(x_m+h)
219 \equiv
220 \Phi(x_m)+L_mh
221 \pmod{F^{m+1}Y},
222 \]
223 where $L_m:F^mX\to F^mY$ is a module homomorphism satisfying
224 \[
225 L_m(F^{m+1}X)\subseteq F^{m+1}Y
226 \]
227 and therefore induces
228 \[
229 \overline L_m:
230 F^mX/F^{m+1}X
231 \longrightarrow
232 F^mY/F^{m+1}Y.
233 \]
234 
235 
236 At depth $m$, only the associated graded correction is visible. The cokernel below has finite length whenever the target graded piece has finite length, even if the completed problem is infinite. Vanishing of the depth obstruction decides whether the current approximation can be improved by one level. When the graded pieces are finite, the branching count below applies. Finite length alone does not imply finite cardinality.
237 
238 \begin{definition}[Depth obstruction]\label{def:depth_obstruction}
239 The depth $m$ obstruction of $x_m$ is
240 \[
241 \Ob_m(x_m)
242 =
243 [\Phi(x_m)]
244 \in
245 \coker\overline L_m,
246 \]
247 where $\Phi(x_m)$ is first viewed in $F^mY/F^{m+1}Y$.
248 \end{definition}
249 
250 \begin{theorem}[One-step lifting criterion]\label{thm:one_step_lifting}
251 The approximation $x_m$ has a correction $h_m\in F^mX$ such that
252 \[
253 \Phi(x_m+h_m)\in F^{m+1}Y
254 \]
255 if and only if
256 \[
257 \Ob_m(x_m)=0.
258 \]
259 When a correction exists, the set of its classes in
260 \[
261 F^mX/F^{m+1}X
262 \]
263 is a coset of $\Ker\overline L_m$.  If the graded pieces are vector spaces, this coset is an affine space.
264 \end{theorem}
265 
266 \begin{proof}
267 The next depth condition is equivalent to
268 \[
269 \overline L_m[h_m]
270 =-[\Phi(x_m)]
271 \]
272 in $F^mY/F^{m+1}Y$.  This is solvable exactly when the displayed defect has zero cokernel class.  If $h_m$ and $h_m'$ are two solutions, their difference lies in $\Ker\overline L_m$.  Adding an element of the kernel to one solution gives another, so the solution set is a coset of the kernel.
273 \end{proof}
274 
275 \begin{corollary}[Finite branching count]\label{cor:finite_branch_count}
276 Suppose the graded pieces are finite vector spaces over a finite field.  If the depth obstruction vanishes, then the number of correction classes at depth $m$ is
277 \[
278 |\Ker\overline L_m|.
279 \]
280 If the obstruction does not vanish, the number is zero.
281 \end{corollary}
282 
283 \begin{proof}
284 A nonempty affine space over a finite vector space has the same number of elements as that vector space.
285 \end{proof}
286 
287 
288 \begin{lemma}[Well-defined lift classes]\label{lem:well_defined_lift_classes}
289 Under the filtration congruence assumption, the condition
290 \[
291 \Phi(x)\in F^mY
292 \]
293 depends only on the class of $x$ in $X/F^mX$.  Fix a representative $x_m$ of a depth $m$ class.  The map
294 \[
295 [h]\longmapsto[x_m+h]_{m+1}
296 \]
297 identifies correction classes $[h]\in F^mX/F^{m+1}X$ with the classes at depth $m+1$ that reduce to $[x_m]_m$.  Under this identification, the solutions of the associated graded correction equation in \cref{thm:one_step_lifting} are exactly the children whose defect lies in $F^{m+1}Y$.  A different representative of the same parent gives a translated parametrization of the same set of children.
298 \end{lemma}
299 
300 \begin{proof}
301 If $x'-x\in F^mX$, then filtration congruence preservation gives
302 \[
303 \Phi(x')-\Phi(x)\in F^mY.
304 \]
305 Thus one defect lies in $F^mY$ exactly when the other does.  If $h-h'\in F^{m+1}X$, then $x_m+h$ and $x_m+h'$ define the same depth $m+1$ class.  Conversely, two depth $m+1$ classes reducing to $[x_m]_m$ differ from $x_m$ by elements of $F^mX$, and they are equal precisely when those elements agree modulo $F^{m+1}X$.  The first-order congruence shows that such a child has defect in $F^{m+1}Y$ exactly when its correction class solves the associated graded equation.  Replacing $x_m$ by another representative translates the correction coordinate by an element of $F^mX/F^{m+1}X$ and leaves the child classes themselves unchanged.
306 \end{proof}
307 
308 \begin{definition}[Lift tree]\label{def:lift_tree}
309 Fix an initial approximation.  A vertex at depth $m$ is a class $[x]_m\in X/F^mX$ with $\Phi(x)\in F^mY$ that reduces to the fixed initial class.  A depth $m+1$ vertex is joined to its reduction in $X/F^mX$.  By \cref{lem:well_defined_lift_classes}, this rooted graph is independent of the chosen representatives.  It is called the lift tree.
310 \end{definition}
311 
312 \begin{theorem}[Finitely branching lift tree]\label{thm:finitely_branching_lift_tree}
313 Assume every level of the lift tree is nonempty and every vertex has only finitely many children.  Then the tree contains an infinite branch.  If $X$ and $Y$ are separated and complete and $\Phi$ is continuous, that branch determines an exact solution of $\Phi(x)=0$.
314 \end{theorem}
315 
316 \begin{proof}
317 The first assertion is K\"onig's infinity argument.  The root has finitely many children.  Since vertices occur at arbitrarily large depths, at least one child has descendants at arbitrarily large depths.  Choose such a child and repeat the argument at every chosen vertex.  This gives one compatible vertex at each depth.
318 
319 Choose representatives recursively.  Once $x_m$ represents the selected depth $m$ vertex, choose a representative $x_{m+1}$ of its selected child with
320 \[
321 x_{m+1}-x_m\in F^mX.
322 \]
323 The resulting sequence is Cauchy for the filtration topology.  Each representative satisfies $\Phi(x_m)\in F^mY$.  The conclusion now follows from \cref{thm:complete_filtered_lifting}.
324 \end{proof}
325 
326 \begin{corollary}[Unique lift branch]\label{cor:unique_lift_branch}
327 Suppose an initial approximation is fixed, every depth obstruction vanishes along its descendants, and every map $\overline L_m$ has zero kernel.  Then at most one compatible lift branch extends the initial approximation.
328 \end{corollary}
329 
330 \begin{proof}
331 By \cref{thm:one_step_lifting}, the correction classes at each depth form a coset of $\Ker\overline L_m$.  A zero kernel leaves at most one correction class, so induction gives at most one branch.
332 \end{proof}
333 
334 \begin{theorem}[Complete filtered lifting]\label{thm:complete_filtered_lifting}
335 Assume there is a compatible sequence
336 \[
337 x_{m+1}\equiv x_m\pmod{F^mX}
338 \]
339 with
340 \[
341 \Phi(x_m)\in F^mY
342 \]
343 for every $m$.  Then $x_m$ converges to an element $x_\infty\in X$ and
344 \[
345 \Phi(x_\infty)=0.
346 \]
347 In particular, if the obstruction vanishes at every depth and compatible corrections are chosen, an exact solution exists.
348 \end{theorem}
349 
350 \begin{proof}
351 For every fixed $r$, the congruences imply
352 \[
353 x_n-x_m\in F^rX
354 \]
355 whenever $n\ge m\ge r$.  Thus $(x_m)$ is Cauchy for the filtration topology.  Completeness gives a limit $x_\infty$.
356 
357 Fix $r$.  For every $m\ge r$, one has $\Phi(x_m)\in F^mY\subseteq F^rY$.  The submodule $F^rY$ is open and therefore closed in the filtration topology.  Continuity of $\Phi$ gives $\Phi(x_m)\to\Phi(x_\infty)$, so $\Phi(x_\infty)\in F^rY$.  Since this holds for every $r$ and $Y$ is separated,
358 \[
359 \Phi(x_\infty)\in\bigcap_rF^rY=0.
360 \]
361 \end{proof}
362 
363 \begin{theorem}[Naturality of filtered obstructions]\label{thm:filtered_obstruction_naturality}
364 Suppose a morphism of filtered lifting problems sends $x_m$ to $x_m'$, intertwines $L_m$ with $L_m'$, and sends $\Phi(x_m)$ to $\Phi'(x_m')$ on the associated graded pieces.  Then the induced cokernel map sends
365 \[
366 \Ob_m(x_m)
367 \quad\text{to}\quad
368 \Ob_m'(x_m').
369 \]
370 \end{theorem}
371 
372 \begin{proof}
373 Apply \cref{thm:obstruction_naturality} to the induced commuting square on the depth $m$ graded pieces.
374 \end{proof}
375 
376 \begin{example}[Three lifting outcomes]\label{ex:three_lifting_outcomes}
377 The scalar equation $x^2-2=0$ at $x=3$ modulo $7$ has invertible derivative.  It has one correction modulo $49$ and gives $x=10$ modulo $49$.
378 
379 For $x^2-3=0$ at $x=0$ modulo $3$, the derivative is zero and the defect class is nonzero.  No correction exists.
380 
381 For $x^2=0$ at $x=0$ modulo $3$, the derivative and defect both vanish.  Every correction digit lies in the kernel, so the lift branches.
382 \end{example}
383 
384 \begin{proposition}[Maurer Cartan lifting obstruction]\label{prop:maurer_cartan_lifting}
385 Let $\Lcal$ be a complete filtered differential graded Lie algebra over a ring in which two is invertible.  Assume
386 \[
387 dF^i\Lcal\subseteq F^i\Lcal,
388 \qquad
389 [F^i\Lcal,F^j\Lcal]\subseteq F^{i+j}\Lcal.
390 \]
391 For a degree one element $\eta$, put
392 \[
393 \Omega(\eta)=d\eta+\frac12[\eta,\eta].
394 \]
395 Suppose $\Omega(\eta)\in F^m\Lcal^2$ with $m\ge1$.  For every $h\in F^m\Lcal^1$,
396 \[
397 \Omega(\eta+h)
398 \equiv
399 \Omega(\eta)+d_\eta h
400 \pmod{F^{m+1}\Lcal^2},
401 \qquad
402 d_\eta=d+[\eta,\mathord\cdot].
403 \]
404 The obstruction to improving $\eta$ by one depth is therefore the class of $\Omega(\eta)$ in the cokernel of
405 \[
406 \gr^m(d_\eta):
407 \gr^m\Lcal^1
408 \longrightarrow
409 \gr^m\Lcal^2.
410 \]
411 \end{proposition}
412 
413 \begin{proof}
414 Direct expansion gives
415 \[
416 \Omega(\eta+h)
417 =
418 \Omega(\eta)+dh+[\eta,h]+\frac12[h,h].
419 \]
420 The filtration rule places $[h,h]$ in $F^{2m}\Lcal^2$, which is contained in $F^{m+1}\Lcal^2$ because $m\ge1$.  The congruence follows.  The last statement is \cref{thm:one_step_lifting} applied to the induced graded correction map.
421 \end{proof}
422 
423 \section{Filtered obstruction complexes}\label{sec:obstruction_spectral_sequence}
424 
425 A direct packet decomposition splits a complex into independent pieces, so cohomology splits packet by packet. A filtration retains relations between depths. Its associated graded cohomology can contain classes that disappear on later pages of a spectral sequence.
426 
427 There are therefore two ways for fine obstruction data to give zero total obstruction. Each fine class may already vanish, or later differentials may remove nonzero classes through compatibility between depths. Only the first mechanism occurs in a direct sum of subcomplexes. A filtration can exhibit both.
428 
429 \begin{standardfact}[Spectral sequence of a finite filtered complex]\label{fact:filtered_spectral_sequence}
430 Let $C^\bullet$ be a cochain complex with a finite decreasing filtration by subcomplexes
431 \[
432 C^\bullet=F^0C^\bullet\supseteq F^1C^\bullet\supseteq\cdots\supseteq F^{N+1}C^\bullet=0.
433 \]
434 There is a natural spectral sequence with
435 \[
436 E_0^{p,q}=\gr^pC^{p+q},
437 \qquad
438 E_1^{p,q}=H^{p+q}(\gr^pC^\bullet),
439 \]
440 which converges to the associated graded object of $H^{p+q}(C^\bullet)$ with its induced filtration.  A filtered cochain map induces a morphism of spectral sequences.
441 \end{standardfact}
442 
443 This is the standard construction for a filtered complex, in the form given in \cite{Weibel1994,McCleary2001}. When $C^\bullet$ is a correction or deformation complex, the groups on its pages can be read as graded obstruction candidates. For an arbitrary filtered complex, they are cohomology classes of the associated graded problem.
444 
445 \begin{interpretation}[title={Lifting and survival answer different questions}]
446 A depth obstruction tests whether a chosen approximation admits one further correction. A spectral sequence follows cohomology classes of a filtered complex through later differentials. Turning successive corrections into an exact solution also uses compatible choices, completeness, continuity, and separatedness.
447 \end{interpretation}
448 
449 \begin{proposition}[Fine obstruction survival]\label{prop:fine_obstruction_survival}
450 A class on the first page contributes to the total obstruction only if it is represented by a permanent cycle, so it is neither killed nor identified as a boundary on any later page.  If the spectral sequence degenerates at the first page, then $H^n(C^\bullet)=0$ exactly when $E_1^{p,n-p}=0$ for every $p$.
451 \end{proposition}
452 
453 \begin{proof}
454 Convergence identifies the last page with the associated graded object of the total cohomology.  A first page class contributes to that object exactly when it determines a nonzero class on the last page.  If the sequence degenerates at the first page, the associated graded pieces of $H^n(C^\bullet)$ are the groups $E_1^{p,n-p}$.  A finite filtered module is zero exactly when all its associated graded pieces are zero.
455 \end{proof}
456 
457 \begin{corollary}[Direct packet complexes]\label{cor:direct_packet_no_cancellation}
458 Suppose
459 \[
460 C^\bullet=\bigoplus_{\alpha\in I}C_\alpha^\bullet
461 \]
462 is a finite direct sum of subcomplexes.  Then
463 \[
464 H^n(C^\bullet)
465 \cong
466 \bigoplus_{\alpha\in I}H^n(C_\alpha^\bullet).
467 \]
468 A total cohomology class vanishes if and only if every packet class vanishes.
469 \end{corollary}
470 
471 \begin{proof}
472 The differential acts independently on the summands.  Kernels and images therefore split as finite direct sums, and so does their quotient.
473 \end{proof}
474 
475 \begin{example}[A fine class killed at the next page]\label{ex:spectral_cancellation}
476 Let $C^0=\bbk x$ and $C^1=\bbk y$, with differential $d(x)=y$.  Give $C^0$ the filtration
477 \[
478 F^0C^0=\bbk x,
479 \qquad
480 F^1C^0=0,
481 \]
482 and give $C^1$ the filtration
483 \[
484 F^0C^1=F^1C^1=\bbk y,
485 \qquad
486 F^2C^1=0.
487 \]
488 The associated graded differential is zero in the relevant positions, so both $x$ and $y$ give nonzero classes on the next page.  The next differential sends the class of $x$ to the class of $y$.  The total complex is acyclic.  Thus nonzero fine obstruction classes can cancel through a later refinement differential.
489 \end{example}
490 
491 \begin{remark}\label{rem:spectral_sequence_meaning}
492 The spectral sequence does not guarantee that a total obstruction vanishes.  It records which fine classes survive and which are removed by compatibility across depths.
493 \end{remark}
494 
495 \section{Nonlinear elimination}\label{sec:nonlinear_elimination}
496 
497 Let $U,V,Y_P,Y_Q$ be Banach spaces. On an open neighborhood of $(u_0,v_0)\in U\times V$, split the variables and equations as
498 \[
499 x=(u,v),
500 \qquad
501 F(u,v)=
502 \begin{pmatrix}
503 F_P(u,v)\\
504 F_Q(u,v)
505 \end{pmatrix}.
506 \]
507 Assume $F$ takes values in $Y_P\times Y_Q$, is twice continuously Fr\'echet differentiable near $(u_0,v_0)$, satisfies $F_Q(u_0,v_0)=0$, and
508 \[
509 A=D_vF_Q(u_0,v_0)
510 \]
511 is a Banach space isomorphism from $V$ to $Y_Q$. The implicit function theorem gives a local map $v=\varphi(u)$, with $\varphi(u_0)=v_0$, satisfying
512 \[
513 F_Q(u,\varphi(u))=0.
514 \]
515 Define the reduced equation
516 \[
517 R(u)=F_P(u,\varphi(u)).
518 \]
519 
520 \begin{theorem}[Exact local elimination]\label{thm:exact_local_elimination}
521 After the neighborhoods are chosen small enough, the equation
522 \[
523 F(u,v)=0
524 \]
525 is equivalent to
526 \[
527 v=\varphi(u)
528 \qquad\text{and}\qquad
529 R(u)=0.
530 \]
531 Thus eliminating the hidden variable loses no local solutions.
532 \end{theorem}
533 
534 \begin{proof}
535 The hidden equation $F_Q(u,v)=0$ has the unique local solution $v=\varphi(u)$ by the implicit function theorem.  After this identity is imposed, the remaining component of $F$ is exactly $R(u)$.
536 \end{proof}
537 
538 \begin{theorem}[Nonlinear Schur derivative]\label{thm:schur_derivative}
539 At $(u_0,v_0)$,
540 \[
541 DR
542 =
543 D_uF_P
544 -
545 D_vF_P A^{-1}D_uF_Q.
546 \]
547 \end{theorem}
548 
549 \begin{proof}
550 Differentiate $F_Q(u,\varphi(u))=0$:
551 \[
552 D_uF_Q+A D\varphi=0.
553 \]
554 Thus
555 \[
556 D\varphi=-A^{-1}D_uF_Q.
557 \]
558 Substitute this into
559 \[
560 DR=D_uF_P+D_vF_PD\varphi.
561 \]
562 \end{proof}
563 
564 \begin{interpretation}[title={The hidden variable still contributes}]
565 The reduced derivative includes the response of the hidden variable to a retained variation. That response travels through the inverse hidden derivative and then back into the retained equation. The Schur term records this contribution. Its use depends on the hidden derivative being invertible, as required in the local elimination setup.
566 \end{interpretation}
567 
568 The second derivative records how this response enters the first nonlinear feedback from the hidden variables.
569 
570 \begin{theorem}[Second Schur derivative]\label{thm:schur_second_derivative}
571 For retained directions $h,k$, set
572 \[
573 \widehat h=(h,D\varphi h),
574 \qquad
575 \widehat k=(k,D\varphi k).
576 \]
577 Then
578 \[
579 D^2\varphi[h,k]
580 =
581 -A^{-1}D^2F_Q[\widehat h,\widehat k]
582 \]
583 and
584 \[
585 D^2R[h,k]
586 =
587 D^2F_P[\widehat h,\widehat k]
588 -
589 D_vF_P A^{-1}D^2F_Q[\widehat h,\widehat k].
590 \]
591 All derivatives on the right are evaluated at $(u_0,v_0)$.
592 \end{theorem}
593 
594 \begin{proof}
595 Differentiate the identity $F_Q(u,\varphi(u))=0$ twice.  The chain rule gives
596 \[
597 D^2F_Q[\widehat h,\widehat k]
598 +A D^2\varphi[h,k]
599 =0.
600 \]
601 This proves the first formula.  Differentiate $R(u)=F_P(u,\varphi(u))$ twice and substitute the first formula.
602 \end{proof}
603 
604 \begin{corollary}[Packet jets of the reduced equation]\label{cor:reduced_packet_jets}
605 If the retained and hidden spaces carry packet systems, then the first and second packet jets of $R$ are obtained by projecting the formulas in \cref{thm:schur_derivative,thm:schur_second_derivative}.  Hidden packet feedback is carried by the factors involving $A^{-1}$.
606 \end{corollary}
607 
608 \begin{proof}
609 Apply the packet projectors to the input and output slots of the two derivative formulas.
610 \end{proof}
611 
612 
613 \begin{example}[Two retained variables and one hidden variable]\label{ex:finite_schur}
614 Suppose the derivative of $F$ at a solution has block form
615 \[
616 DF=
617 \begin{pmatrix}
618 A&B\\
619 C&D
620 \end{pmatrix},
621 \]
622 where $D$ is invertible.  Eliminating the hidden variable replaces the retained derivative $A$ by
623 \[
624 A-BD^{-1}C.
625 \]
626 The feedback term is exact.  Keeping only $A$ would ignore the path from retained variables into the hidden variable and back.
627 \end{example}
628 
629 \begin{remark}\label{rem:elimination_vs_obstruction}
630 Elimination and obstruction reduction address different situations.  The Schur formula assumes that the hidden derivative is invertible.  Lyapunov--Schmidt reduction allows a nontrivial kernel and cokernel, then leaves a reduced obstruction equation on the kernel directions.
631 \end{remark}
632 
633 \section{Finite transport and holonomy}\label{sec:finite_holonomy}
634 
635 Let $\Gamma=(V,E)$ be a finite connected graph and fix a vector space $X$.  At every vertex take one labeled copy of $X$.  Every oriented edge $u\to v$ carries an invertible transport
636 \[
637 T_{vu}\in\operatorname{GL}(X),
638 \qquad
639 T_{uv}=T_{vu}^{-1}.
640 \]
641 Equivalently, one may begin with separate but isomorphic vertex fibers and choose one frame in each fiber.  The matrices below are then the transports in those frames.  For a path $\gamma=(v_0,\ldots,v_r)$, set
642 \[
643 T_\gamma
644 =
645 T_{v_rv_{r-1}}\cdots T_{v_1v_0}.
646 \]
647 For a loop, this product is its holonomy. In the based loop group, $\gamma_2\gamma_1$ means first traverse $\gamma_1$ and then $\gamma_2$, so $T_{\gamma_2\gamma_1}=T_{\gamma_2}T_{\gamma_1}$.  Standard graph terminology may be found in \cite{GrossTucker1987}.
648 
649 An invertible map on each edge supplies the local transport data. A global frame must make these maps agree around every cycle. A spanning tree has no cycles, so a gauge can make every tree transport the identity. Each remaining edge closes a loop and retains that loop's holonomy.
650 
651 \begin{definition}[Gauge change]\label{def:gauge_change}
652 A gauge is a family $G_v\in\operatorname{GL}(X)$.  It changes the edge transports to
653 \[
654 T'_{vu}=G_vT_{vu}G_u^{-1}.
655 \]
656 \end{definition}
657 
658 \begin{proposition}[Gauge covariance]\label{prop:gauge_covariance}
659 For a loop $\gamma$ based at $v_0$,
660 \[
661 \Hol'(\gamma)
662 =
663 G_{v_0}\Hol(\gamma)G_{v_0}^{-1}.
664 \]
665 Thus the conjugacy class of each loop holonomy is gauge invariant.
666 \end{proposition}
667 
668 \begin{proof}
669 Every intermediate gauge factor cancels with its inverse.  The initial and terminal factors remain at the base vertex.
670 \end{proof}
671 
672 \begin{theorem}[Spanning tree normalization]\label{thm:tree_normalization}
673 Choose a root and a spanning tree $\Tcal$ of $\Gamma$.  There is a gauge in which every tree edge transport is the identity.  In that gauge, the transport on each edge outside the tree is the holonomy of its fundamental cycle, with the orientation used to define that cycle.
674 \end{theorem}
675 
676 \begin{proof}
677 Set the gauge at the root to the identity.  Propagate gauges along the unique tree paths so that every tree edge becomes the identity.  An edge outside the tree closes one fundamental cycle.  Since every tree factor is now the identity, the remaining edge transport is the cycle product.
678 \end{proof}
679 
680 \begin{theorem}[Cycle gluing criterion]\label{thm:cycle_gluing}
681 There are vertex coordinates in which every edge transport is the identity if and only if every fundamental cycle holonomy is the identity.
682 \end{theorem}
683 
684 \begin{proof}
685 If all edge transports are identity maps, every loop product is the identity.  Conversely, use \cref{thm:tree_normalization}.  Identity fundamental holonomy forces every edge outside the tree to become the identity as well.
686 \end{proof}
687 
688 \begin{corollary}[Number of independent cycle tests]\label{cor:cycle_number}
689 For a connected graph with $|V|$ vertices and $|E|$ unoriented edges, a spanning tree leaves
690 \[
691 |E|-|V|+1
692 \]
693 fundamental cycle holonomies.
694 \end{corollary}
695 
696 \begin{proof}
697 A spanning tree has $|V|-1$ edges.  Every remaining edge creates one fundamental cycle.
698 \end{proof}
699 
700 
701 \subsection{Intertwining two transport systems}
702 
703 Let $A$ be a transport system on copies of a fixed vector space $V_0$, and let $B$ be a transport system on copies of a fixed vector space $W_0$.  Fix a root $r$ and an isomorphism
704 \[
705 U_r:V_0\longrightarrow W_0.
706 \]
707 
708 \begin{theorem}[Fundamental holonomy intertwining criterion]\label{thm:holonomy_intertwining}
709 The following conditions are equivalent.
710 \begin{enumerate}
711 \item There is a family of isomorphisms $U_v:V_0\to W_0$ with the prescribed root value $U_r$ and satisfying
712 \[
713 U_vA_{vu}=B_{vu}U_u
714 \]
715 for every oriented edge $u\to v$.
716 \item For every fundamental loop $\gamma$ based at $r$,
717 \[
718 U_r\Hol_A(\gamma)=\Hol_B(\gamma)U_r.
719 \]
720 \item The same identity holds for every loop based at $r$.
721 \end{enumerate}
722 When these conditions hold, the family $(U_v)$ is uniquely determined by $U_r$.
723 \end{theorem}
724 
725 \begin{proof}
726 If the edge intertwiners exist, multiply their equations around a loop.  Every intermediate vertex map cancels.  This proves that the first condition implies the third, and the third implies the second.
727 
728 Conversely, choose a spanning tree.  Starting from $U_r$, propagate $U_v$ along the unique tree path by the rule
729 \[
730 U_v=B_{vu}U_uA_{vu}^{-1}
731 \]
732 on every tree edge.  This is the only possible extension satisfying the tree edge equations.  An edge outside the tree closes one fundamental loop.  Its intertwining equation holds exactly when $U_r$ intertwines the two corresponding fundamental holonomies.  The second condition therefore supplies every remaining edge equation.  Uniqueness follows from propagation along the tree.
733 \end{proof}
734 
735 \begin{corollary}[Scalar holonomy bridge]\label{cor:scalar_holonomy_bridge}
736 Suppose every vertex space is one dimensional.  Two transport systems are gauge equivalent if and only if their holonomies agree on a fundamental cycle basis.
737 \end{corollary}
738 
739 \begin{proof}
740 Choose a nonzero root isomorphism.  Scalar loop maps commute with it, so \cref{thm:holonomy_intertwining} reduces to equality of the cycle scalars.
741 \end{proof}
742 
743 \begin{theorem}[Gauge classification of transports]\label{thm:gauge_classification}
744 Let $G\subseteq\operatorname{GL}(X)$ be the chosen structure group and fix a root $r$.  Restrict both edge transports and gauge values to $G$.  Gauge equivalence classes of $G$ valued transport systems are naturally identified with
745 \[
746 \Hom(\pi_1(\Gamma,r),G)/G,
747 \]
748 where $G$ acts by conjugation.  After a spanning tree and orientations of the remaining edges are chosen, this set is identified with simultaneous conjugacy classes of
749 \[
750 b_1=|E|-|V|+1
751 \]
752 tuples in $G$.
753 \end{theorem}
754 
755 \begin{proof}
756 A transport system assigns a group element to every oriented edge, with inversion under reversal.  The product along a based loop gives a homomorphism from the free group $\pi_1(\Gamma,r)$ to $G$.  A gauge change conjugates this homomorphism by the gauge value at the root.  Conversely, normalize a transport system on a spanning tree.  The remaining edge values prescribe the images of a free basis of $\pi_1(\Gamma,r)$, and every homomorphism arises in this way.  A further gauge preserving the tree normalization is constant at all vertices and acts by simultaneous conjugation.  The number of non-tree edges is $b_1$.
757 \end{proof}
758 
759 \begin{corollary}[Abelian cohomology class]\label{cor:abelian_holonomy_class}
760 When the structure group $A$ is abelian, gauge classes of transports are naturally
761 \[
762 H^1(\Gamma,A)
763 \cong
764 \Hom(H_1(\Gamma,\mathbb Z),A).
765 \]
766 After a basis of $H_1(\Gamma,\mathbb Z)$ is chosen, this group is identified with $A^{b_1}$.
767 \end{corollary}
768 
769 \begin{proof}
770 An edge assignment is an $A$ valued one cochain.  A gauge change multiplies it by a coboundary.  A graph has no two cells, so every one cochain is a cocycle.  The usual computation of the first cohomology of a connected graph gives $A^{b_1}$.
771 \end{proof}
772 
773 \begin{corollary}[Projective transport]\label{cor:projective_transport}
774 The same normalization and gluing statements hold for transports valued in a projective linear group.  A projective transport is globally trivial exactly when all fundamental holonomies are identity elements in that projective group.
775 \end{corollary}
776 
777 \begin{proof}
778 Apply \cref{thm:tree_normalization,thm:cycle_gluing} with the projective linear group as the structure group.
779 \end{proof}
780 
781 \begin{example}[A square loop]\label{ex:square_loop}
782 A square graph has four vertices and four edges, so it has one fundamental cycle.  After a tree gauge, three edges are identities and the fourth is the full loop holonomy.  The system glues to one global frame exactly when that remaining matrix is the identity.
783 \end{example}
784 
785 \section{Smooth projector transport}\label{sec:kato_transport}
786 
787 Finite graph transport has a continuous counterpart.  Let $H$ be a Hilbert space and let
788 \[
789 \Pcal(t)=(P_\alpha(t))_{\alpha\in I}
790 \]
791 be a complete orthogonal packet system with finitely many labels. Assume that it is continuously differentiable in operator norm for $t\in[0,T]$, where $T\ge0$. Here every $P_\alpha(t)$ is a bounded self-adjoint projection, so $P_\alpha(t)^*=P_\alpha(t)=P_\alpha(t)^2$.
792 
793 Zero projections are permitted in this section and contribute zero to all generator sums. When $T=0$, every partition product is the empty product $I$. Completeness and orthogonality give
794 \[
795 P_\alpha(t)P_\beta(t)=0
796 \quad\text{for }\alpha\ne\beta,
797 \qquad
798 \sum_\alpha P_\alpha(t)=I.
799 \]
800 Define the canonical generator
801 \[
802 K(t)=\sum_\alpha \dot P_\alpha(t)P_\alpha(t).
803 \]
804 
805 \begingroup
806 \setlength{\abovedisplayskip}{6pt plus 2pt minus 2pt}
807 \setlength{\belowdisplayskip}{6pt plus 2pt minus 2pt}
808 \begin{lemma}[Complete packet commutator identity]\label{lem:packet_commutator}
809 For every packet label $\beta$,
810 \[
811 \dot P_\beta(t)=[K(t),P_\beta(t)].
812 \]
813 The generator is also skew-adjoint:
814 \[
815 K(t)^*=-K(t).
816 \]
817 \end{lemma}
818 
819 \begin{proof}
820 Differentiating $P_\alpha^2=P_\alpha$ gives
821 \[
822 P_\alpha\dot P_\alpha P_\alpha=0.
823 \]
824 Differentiating $P_\beta P_\alpha=0$ for $\alpha\ne\beta$ gives
825 \[
826 P_\beta\dot P_\alpha=-\dot P_\beta P_\alpha.
827 \]
828 Hence
829 \[
830 K P_\beta=\dot P_\beta P_\beta
831 \]
832 and
833 \[
834 P_\beta K=-\dot P_\beta(I-P_\beta).
835 \]
836 Their difference is $\dot P_\beta$.  Also,
837 \[
838 K+K^*
839 =
840 \sum_\alpha(\dot P_\alpha P_\alpha+P_\alpha\dot P_\alpha)
841 =
842 \sum_\alpha\dot P_\alpha
843 =0.
844 \]
845 \end{proof}
846 \endgroup
847 
848 \begin{theorem}[Canonical packet transport]\label{thm:kato_transport}
849 Let $U(t)$ solve
850 \[
851 \dot U(t)=K(t)U(t),
852 \qquad
853 U(0)=I.
854 \]
855 Then $U(t)$ is unitary and transports every packet exactly:
856 \[
857 P_\alpha(t)U(t)=U(t)P_\alpha(0)
858 \]
859 for all $\alpha$.
860 \end{theorem}
861 
862 \begin{proof}
863 The continuous bounded generator has an evolution inverse. Indeed, solve $\dot V=-VK$ with $V(0)=I$. Differentiation gives $(VU)^{\prime}=0$, hence $VU=I$. The product $W=UV$ satisfies $\dot W=KW-WK$ and $W(0)=I$. Uniqueness of this bounded linear equation gives $UV=I$. Thus $U$ is invertible.
864 
865 Skew-adjointness of $K(t)$ gives
866 \[
867 \frac{d}{dt}(U(t)^*U(t))=0,
868 \]
869 so $U(t)$ is unitary.  By \cref{lem:packet_commutator},
870 \[
871 \frac{d}{dt}\big(U(t)^{-1}P_\alpha(t)U(t)\big)
872 =
873 U(t)^{-1}\big(\dot P_\alpha-[K,P_\alpha]\big)U(t)
874 =0.
875 \]
876 The expression equals $P_\alpha(0)$ at time zero.
877 \end{proof}
878 
879 \begin{corollary}[One moving subspace]\label{cor:one_moving_subspace}
880 For one orthogonal projection $P(t)$ and its complement, the canonical generator is
881 \[
882 K(t)=[\dot P(t),P(t)].
883 \]
884 The transport in \cref{thm:kato_transport} maps $\im P(0)$ unitarily onto $\im P(t)$.
885 \end{corollary}
886 
887 \begin{proof}
888 Apply the complete packet formula to $P(t)$ and $I-P(t)$, then simplify the sum.
889 \end{proof}
890 
891 
892 \begin{definition}[Direct rotation of nearby packet systems]\label{def:packet_direct_rotation}
893 Let $(P_\alpha)_{\alpha\in I}$ and $(Q_\alpha)_{\alpha\in I}$ be complete orthogonal packet systems on the same Hilbert space.  Put
894 \[
895 S(\Qcal,\Pcal)=\sum_{\alpha\in I}Q_\alpha P_\alpha.
896 \]
897 When $S(\Qcal,\Pcal)$ is invertible, define
898 \[
899 \Ucal(\Qcal,\Pcal)
900 =
901 S(\Qcal,\Pcal)
902 \big(S(\Qcal,\Pcal)^*S(\Qcal,\Pcal)\big)^{-1/2}.
903 \]
904 \end{definition}
905 
906 \begin{proposition}[Exact direct packet rotation]\label{prop:packet_direct_rotation}
907 If
908 \[
909 \max\bigl(\{0\}\cup\{\|P_\alpha-Q_\alpha\|:\alpha\in I\}\bigr)<1,
910 \]
911 then $S(\Qcal,\Pcal)$ is invertible.  The operator $\Ucal(\Qcal,\Pcal)$ is unitary and satisfies
912 \[
913 Q_\alpha\Ucal(\Qcal,\Pcal)
914 =
915 \Ucal(\Qcal,\Pcal)P_\alpha
916 \]
917 for every $\alpha$.
918 \end{proposition}
919 
920 \begin{proof}
921 The identities
922 \[
923 Q_\alpha S(\Qcal,\Pcal)
924 =
925 S(\Qcal,\Pcal)P_\alpha
926 =
927 Q_\alpha P_\alpha
928 \]
929 follow from pairwise orthogonality.  Also
930 \[
931 S^*S
932 =
933 \sum_\alpha P_\alpha Q_\alpha P_\alpha,
934 \qquad
935 SS^*
936 =
937 \sum_\alpha Q_\alpha P_\alpha Q_\alpha.
938 \]
939 Put $\delta=\max(\{0\}\cup\{\|P_\alpha-Q_\alpha\|:\alpha\in I\})$.  On $P_\alpha H$,
940 \[
941 P_\alpha Q_\alpha P_\alpha
942 =
943 P_\alpha-P_\alpha(P_\alpha-Q_\alpha)P_\alpha,
944 \]
945 so this block is bounded below by $1-\delta>0$.  The same argument on $Q_\alpha H$ applies to $Q_\alpha P_\alpha Q_\alpha$.  Thus both $S^*S$ and $SS^*$ are invertible.  It follows that $S$ is injective and surjective.  The polar factor $\Ucal=S(S^*S)^{-1/2}$ is unitary.  Since $S^*S$ commutes with every $P_\alpha$, the intertwining identity for $S$ passes to $\Ucal$.
946 \end{proof}
947 
948 \begin{lemma}[One-step direct rotation estimate]\label{lem:one_step_direct_rotation}
949 Assume $\Pcal(t)$ is twice continuously differentiable in operator norm on $[0,T]$.  There are constants $h_0>0$ and $C>0$ such that, whenever $t,t+h\in[0,T]$ and $|h|\le h_0$,
950 \[
951 \left\|
952 \Ucal(\Pcal(t+h),\Pcal(t))
953 -
954 I
955 -
956 hK(t)
957 \right\|
958 \le
959 Ch^2.
960 \]
961 The constants can be chosen uniformly in $t$.
962 \end{lemma}
963 
964 \begin{proof}
965 Taylor's theorem in the Banach space of bounded operators gives
966 \[
967 P_\alpha(t+h)
968 =
969 P_\alpha(t)+h\dot P_\alpha(t)+E_{\alpha}(t,h),
970 \qquad
971 \|E_{\alpha}(t,h)\|\le C_1h^2,
972 \]
973 uniformly in $t$ and $\alpha$.  Hence
974 \[
975 S(\Pcal(t+h),\Pcal(t))
976 =
977 I+hK(t)+E(t,h),
978 \qquad
979 \|E(t,h)\|\le C_2h^2.
980 \]
981 Since $K(t)^*=-K(t)$,
982 \[
983 S^*S=I+E_2(t,h),
984 \qquad
985 \|E_2(t,h)\|\le C_3h^2.
986 \]
987 For small $h$, the spectrum of $S^*S$ lies in a fixed compact subinterval of $(0,\infty)$.  The functional calculus for $z^{-1/2}$ gives
988 \[
989 (S^*S)^{-1/2}=I+E_3(t,h),
990 \qquad
991 \|E_3(t,h)\|\le C_4h^2.
992 \]
993 Multiplying the last two expansions proves the estimate.
994 \end{proof}
995 
996 \begin{theorem}[Direct rotations converge to packet transport]\label{thm:direct_rotation_convergence}
997 Assume the complete packet family $\Pcal(t)$ is twice continuously differentiable in operator norm on $[0,T]$.  For a partition
998 \[
999 0=t_0<t_1<\cdots<t_N=T
1000 \]
1001 with sufficiently small mesh, define
1002 \[
1003 V_\pi
1004 =
1005 \Ucal(\Pcal(t_N),\Pcal(t_{N-1}))
1006 \cdots
1007 \Ucal(\Pcal(t_1),\Pcal(t_0)).
1008 \]
1009 Then
1010 \[
1011 V_\pi\longrightarrow U(T)
1012 \]
1013 in operator norm, where $U$ is the canonical transport from \cref{thm:kato_transport}.  On a fixed compact interval the error is bounded by a constant times the mesh.
1014 \end{theorem}
1015 
1016 \begin{proof}
1017 Let $U(t,s)$ be the exact propagator generated by $K$.  Since $K$ is continuously differentiable, the integral equation for $U(t+h,t)$ gives a uniform estimate
1018 \[
1019 \|U(t+h,t)-I-hK(t)\|
1020 \le
1021 C_1h^2.
1022 \]
1023 By \cref{lem:one_step_direct_rotation}, the direct rotation $V_j$ from $t_j$ to $t_{j+1}$ therefore satisfies
1024 \[
1025 \|V_j-U(t_{j+1},t_j)\|
1026 \le
1027 C_2h_j^2,
1028 \qquad
1029 h_j=t_{j+1}-t_j.
1030 \]
1031 Both $V_j$ and the exact propagator are unitary.  A telescoping product comparison gives
1032 \[
1033 \|V_\pi-U(T,0)\|
1034 \le
1035 \sum_j\|V_j-U(t_{j+1},t_j)\|
1036 \le
1037 C_2\sum_jh_j^2
1038 \le
1039 C_2T|\pi|.
1040 \]
1041 This proves norm convergence and the stated first-order global bound.
1042 \end{proof}
1043 
1044 \begin{keyidea}[title={Exact packet matching at each sample}]
1045 Each direct rotation maps a sampled packet exactly to the corresponding packet at the next sample. The product therefore preserves packet labels throughout the partition. Convergence compares this discrete product with the canonical continuous transport. The uniform estimate uses twice continuous differentiability in operator norm, as stated in the theorem.
1046 \end{keyidea}
1047 
1048 \begin{figure}[htbp]
1049 \centering
1050 \includegraphics[width=.96\linewidth]{rotating_projector_transport.pdf}
1051 \caption{A rotating rank-one packet and its orthogonal complement. With $u(\theta)=(\cos\theta,\sin\theta)$ and $P(\theta)=u(\theta)u(\theta)^{\mathsf T}$, the canonical generator is $[P',P]=\bigl(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\bigr)$. For a positive step $\delta<\pi/2$, the polar unitary is the usual rotation $R(\delta)$. On each displayed mesh, the direct rotation product equals $R(\pi/3)$ exactly. This special planar example illustrates the construction. The general theorem supplies convergence for families whose step products need not already equal the continuous transport.}
1052 \label{fig:rotating-projector-example}
1053 \end{figure}
1054 \clearpage
1055 
1056 \begin{theorem}[Convergence of sampled transports]\label{thm:sampled_transport_convergence}
1057 Assume $K$ is Lipschitz on $[0,T]$.  For a partition
1058 \[
1059 0=t_0<t_1<\cdots<t_N=T,
1060 \]
1061 define
1062 \[
1063 U_\pi
1064 =
1065 \exp\big((t_N-t_{N-1})K(t_{N-1})\big)
1066 \cdots
1067 \exp\big((t_1-t_0)K(t_0)\big).
1068 \]
1069 As the mesh $|\pi|$ tends to zero,
1070 \[
1071 U_\pi\longrightarrow U(T)
1072 \]
1073 in operator norm.  There is a constant $C$, depending only on $T$, the supremum norm of $K$, and its Lipschitz constant, such that
1074 \[
1075 \|U_\pi-U(T)\|
1076 \le C|\pi|.
1077 \]
1078 \end{theorem}
1079 
1080 \begin{proof}
1081 On each interval, compare the exact propagator with the frozen exponential.  Variation of constants and the Lipschitz bound give a local error bounded by a constant times the square of the interval length.  Every factor is unitary, so multiplication does not amplify norms.  Summing the local errors gives a bound by a constant times
1082 \[
1083 \sum_j(t_{j+1}-t_j)^2
1084 \le
1085 T|\pi|.
1086 \]
1087 \end{proof}
1088 
1089 \Needspace{10\baselineskip}
1090 \begin{corollary}[Continuous holonomy from graph products]\label{cor:continuous_holonomy}
1091 Assume the complete packet family is twice continuously differentiable in operator norm on $[0,T]$. If $P_\alpha(0)=P_\alpha(T)$ for every $\alpha$, then $U(T)$ preserves every initial packet.  Its restriction to each packet is the holonomy of the closed packet path.  The products in \cref{thm:direct_rotation_convergence,thm:sampled_transport_convergence} converge to these holonomies.  The direct rotation product has the added property that every factor maps each sampled packet exactly to its successor.
1092 \end{corollary}
1093 
1094 \begin{proof}
1095 The transport identity in \cref{thm:kato_transport} gives packet preservation at the endpoint.  The two convergence statements follow from \cref{thm:direct_rotation_convergence,thm:sampled_transport_convergence}.
1096 \end{proof}
1097 
1098 \begin{example}[A rotating line]\label{ex:rotating_line}
1099 In $\mathbb R^2$, let
1100 \[
1101 v(t)=(\cos t,\sin t)
1102 \]
1103 and let $P(t)$ be the orthogonal projection onto the line spanned by $v(t)$.  The canonical transport is the ordinary rotation matrix
1104 \[
1105 U(t)=
1106 \begin{pmatrix}
1107 \cos t&-\sin t\\
1108 \sin t&\cos t
1109 \end{pmatrix}.
1110 \]
1111 It carries the initial line and its orthogonal complement to the two moving packets.  Products of short rotations converge to the same transport.
1112 \end{example}
1113 
1114 
1115 \begin{standardfact}[Curvature of a projected connection]\label{fact:projected_curvature}
1116 Let $P$ be a smooth orthogonal projector-valued map on a parameter manifold.  The connection
1117 \[
1118 \nabla=P\,d
1119 \]
1120 on the range bundle has curvature
1121 \[
1122 \Fcal=P(dP\wedge dP)P.
1123 \]
1124 \end{standardfact}
1125 
1126 \begin{proof}
1127 For a section $s$ satisfying $Ps=s$,
1128 \[
1129 \nabla^2s=P\,d(P\,ds)=P(dP)\wedge ds.
1130 \]
1131 Since $ds=(dP)s+Pds$ and $P(dP)P=0$, the surviving term is
1132 \[
1133 P(dP\wedge dP)Ps.\qedhere
1134 \]
1135 \end{proof}
1136 
1137 \section{Realization data and limits of the calculus}\label{sec:scope}
1138 
1139 The smooth transport theorem starts with a declared projector family. Applying it to a physical model requires a construction of that family and, where relevant, a separate continuum limit. Triviality of holonomy is a further geometric question. The curvature formula in \cref{fact:projected_curvature} is classical. Relating it to global holonomy requires geometric information as in \cite{Kato1950,Kato1995,KobayashiNomizu1963}.
1140 
1141 An application begins by specifying the following objects:
1142 \begin{enumerate}
1143 \item an ambient nonlinear equation or continuation problem
1144 \item the allowed correction space
1145 \item a linear correction map or a correction complex
1146 \item the defect whose class is to be tested
1147 \item a packet decomposition or filtration compatible with the correction data
1148 \item transition maps between refinement levels
1149 \item transport maps when local solutions must be glued.
1150 \end{enumerate}
1151 
1152 Once these objects are fixed, the cokernel and cone constructions are functorial by \cref{thm:obstruction_naturality,thm:cone_obstruction}. Packet splitting and descent follow from \cref{thm:packet_obstruction_decomposition,prop:obstruction_quotient_projectors,cor:packet_obstruction_complexes}. Relative loss under coarsening is measured by \cref{thm:relative_obstruction_sequence}.
1153 
1154 Filtered lifting follows from \cref{thm:one_step_lifting,thm:finitely_branching_lift_tree,thm:complete_filtered_lifting}. Gluing and comparison of transport systems follow from \cref{thm:cycle_gluing,thm:holonomy_intertwining}. Smooth packet transport follows from \cref{thm:kato_transport,thm:direct_rotation_convergence}.
1155 
1156 Exact packet coordinates give a representation of the data. Solving the associated problem still requires the correction and lifting hypotheses. The correction map belongs to the problem data, and its cokernel is canonical relative to that choice. The cokernel may be infinite dimensional. In the Hilbert setting, the orthogonal obstruction projector requires closed range.
1157 
1158 Coarse vanishing can erase a nonzero fine obstruction, and higher nonlinear terms can obstruct a problem whose linear class vanishes. A spectral sequence records which classes survive the later differentials. An application of finite graph holonomy or smooth projector transport to a continuum field theory requires the additional realization data just described.
1159 
1160 Whenever a nonlinear continuation problem supplies a correction map, its residual can be recorded in an exact quotient or cohomology object. Under the stated hypotheses, these objects admit packet decomposition, filtered refinement, natural transport, and holonomy tests. They identify the obstruction that a correction or gluing step must address.

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