\section{A sharp coherence--transport resource law}\label{coupling}
We now represent the geometric repair residual by an actual finite graded operator and determine the quantitative regularity needed to control its long-time moment. The parameter family is fixed, while its compensator may depend on the observation horizon.

\subsection{An explicit graded representation of the residual}
Take $n=3$ and the flat unit-volume torus $F=\R^5/\mathbb Z^5$. Put
\[
 \eta_1=dx_1\wedge dx_2,\quad \eta_2=dx_3\wedge dx_4,\quad
 \eta_3=-dx_1\wedge dx_3,\quad \alpha=dx_2\wedge dx_4\wedge dx_5.
\]
For $p=(x,y)$ define
\begin{equation}\label{couple-geometry}
 \Omega_p=x^2\eta_1+y^2\eta_2+\tfrac12xy^2\eta_3,
 \qquad \alpha\wedge\partial_y\Omega_p=xy\vol.
\end{equation}
For a coefficient vector $g(p)\in\R^3$, let
\[
 \gamma_p=\sum_{j=1}^3g_j(p)*\eta_j,\qquad
 \mathbf a(p)=(x^2,y^2,xy^2/2),\qquad
 r_g(p)=xy-\mathbf a(p)\cdot g(p).
\]
Then $C_p=\theta\wedge\gamma_p\otimes c$ has norm $\norm{C_p}_{L^2}=\norm{g(p)}$, and its normalized mixed-identity residual is $r_g(p)\vol\otimes c$. Torus averaging commutes with the leaf differential and contracts parameter H\"older seminorms and leafwise norms. Extracting these three constant coefficients therefore retains the horizontal mean residual of every admissible general correction. Conversely the displayed invariant family realizes every $g$ isometrically. Thus the optimum of the full uniform leafwise $L^2$ repair problem, with these norm and H\"older budgets, equals the scalar residual optimum below. The finite representation is built from this canonical invariant reduction; it is not the full operator on nonconstant leaf modes.

Use generators $a^1,\ldots,a^5,\theta,u,w$ of degree one; $b^1,\ldots,b^5,b,c^0,c^1$ of degree three; and $\nu_x,\nu_y$ of degree two. Let $\mathcal A$ be their free graded-commutative real algebra. Write $\Omega_p(a)$ for substitution of the $a^i$ into the two-form, and set
\[
 \Omega_p(a,b^i)=\sum_{i<j}(\Omega_p)_{ij}(a^i b^j-a^j b^i).
\]
At each fixed parameter define an odd degree-one derivation $Q=Q_{p,g}$ by
\begin{align}\label{couple-Q}
 Q\theta&=-\Omega_p(a),& Q\nu_y&=\alpha(a),&Q\nu_x&=0,\notag\\
 Qb&=-\Omega_p(a,b^i)-\nu_x(\partial_x\Omega_p)(a)
       -\nu_y(\partial_y\Omega_p)(a)-\theta\gamma_p(a),\notag\\
 Qw&=-uw,&Qc^1&=-uc^1+wc^0,
\end{align}
with zero image on the other generators. These are source-form graded tangent equations with an independent affine factor. Their role here is an explicit representation of the declared all-pairs normalized repair equation; we do not identify that convention with the source's printed cyclic-anchor condition.

\begin{lemma}\label{couple-square}
Writing $V=a^1a^2a^3a^4a^5$, the derivation satisfies
\begin{equation}\label{couple-curvature}
 Q^2=-r_g(p)V\partial_b,\qquad Q^4=0,\qquad
 Q^3(bw)=-r_g(p)Vuw.
\end{equation}
\end{lemma}
\begin{proof}
On $b$, the square is
$-\alpha\wedge\partial_y\Omega_p+\Omega_p\wedge\gamma_p=-r_g(p)\vol$.
Every other generator has square zero, including $c^1$ by cancellation of its two affine terms. The square of an odd derivation is an even derivation, so the generator computation proves the first identity globally. Squaring it gives zero since $V^2=0$ and $\partial_bV=0$. Finally $QV=0$ and $Qw=-uw$ give the displayed cube.
\end{proof}

This is an honest finite construction. In the torus-invariant graded tangent algebra the transverse parameters obey $Qp=0$; their evaluation ideals are invariant. One first freezes $p$, as in \eqref{couple-Q}. All remaining generators then have positive degree. The ideal of monomials of weighted degree greater than an integer $L$ is $Q$-invariant, and its quotient $\mathcal A_{\le L}$ is finite-dimensional. For $L\ge7$, the cube witness in \eqref{couple-curvature} survives. Give its monomial basis the Euclidean inner product. On a fixed compact parameter set with $\norm g\le B$, constants independent of $p,g$ satisfy
\[
 1\le\norm Q\le M,
 \qquad |r_g|\le\norm{Q^2}\le M_2|r_g|,
 \qquad |r_g|\le\norm{Q^3}\le M_3|r_g|.
\]
The unit monomial $\nu_y$ gives the first lower bound, and $b,bw$ give the others.

Let $\mathcal P_N$ be the ordered product of independent equiprobable branches $I+Q$ and $I-Q$. For $G_N=\mathbb E(\mathcal P_N^*\mathcal P_N)$, conditioning yields
\begin{equation}\label{couple-moment}
 G_N=I+NQ^*Q+\binom N2(Q^2)^*Q^2+\binom N3(Q^3)^*Q^3,
 \qquad \norm{G_N}\asymp N+r_g(p)^2N^3.
\end{equation}
The comparison is uniform for $N\ge1$ under the fixed bounds above: every summand is positive semidefinite, and the operator bounds prove both directions, adjusting constants for finitely many small $N$. For $\Phi(H)=Q^*HQ$, one has $\Phi^4=0$, hence $\rho(\Phi)=0$ and $\rho(\Id+\Phi)=1$ throughout the transition.

\subsection{The exact exponent of the regularity resource}
Fix $0<R\le1$, the disk $D_R=\{p:\norm p\le R\}$, a coefficient norm budget $B\ge1/\sqrt2$, and a H\"older exponent $0<\sigma\le1$. Let
\begin{align*}
 \mathcal G_{B,K,\sigma}&=\{g:D_R\to\R^3:\ 
 \sup\norm g\le B,\ \norm{g(p)-g(q)}\le K\norm{p-q}^{\sigma}\},\\
 e_{K,\sigma}&=\inf_{g\in\mathcal G_{B,K,\sigma}}\sup_{p\in D_R}|r_g(p)|.
\end{align*}
\begin{theorem}[Sharp regularity--horizon tradeoff]\label{couple-resource}
For sufficiently large $K$, with constants depending only on $R,B,\sigma$,
\begin{equation}\label{couple-error}
 e_{K,\sigma}\asymp K^{-2/\sigma}.
\end{equation}
For the fixed quotient $L\ge7$, the optimal normalized worst-parameter moment satisfies, uniformly for $N\ge1$,
\begin{equation}\label{couple-minimax}
 \inf_{g\in\mathcal G_{B,K,\sigma}}\sup_{p\in D_R}
 \frac{\norm{G_N(Q_{p,g})}}N
 \asymp1+N^2K^{-4/\sigma}.
\end{equation}
Thus, as $N\to\infty$, horizon-dependent implementations retain uniformly linear energy precisely when their available H\"older budget is at least of order $N^{\sigma/2}$. In particular the necessary and sufficient Lipschitz scale is $\sqrt N$.
\end{theorem}
\begin{proof}
At $p_+=(t,t)$ and $p_-=(t,-t)$ the curvature rows coincide:
$\mathbf a(p_\pm)=(t^2,t^2,t^3/2)$, while the targets are $\pm t^2$. Consequently
\[
 \max\{|r_g(p_+)|,|r_g(p_-)|\}
 \ge t^2\left[1-\tfrac12\sqrt{2+t^2/4}\,K(2|t|)^\sigma\right].
\]
Choose $t$ a sufficiently small fixed multiple of $K^{-1/\sigma}$; it lies in the disk for large $K$. This proves the lower bound in \eqref{couple-error}. For $\sigma=1$, $t=4/(9K)$ gives the explicit bound $16/(243K^2)$ whenever $K\ge1/R$.

For the matching construction put $S=\norm{\mathbf a}^2$ and, for $\delta>0$, define the smooth family
\[
 g_\delta(p)=\frac{xy\,\mathbf a(p)}{S(p)+\delta^4},\qquad
 r_{g_\delta}(p)=\frac{xy\,\delta^4}{S(p)+\delta^4}.
\]
Since $|xy|\le\sqrt{S/2}$,
\[
 \sup\norm{g_\delta}\le1/\sqrt2,
 \qquad \sup|r_{g_\delta}|\le\delta^2/(2\sqrt2).
\]
Writing $\rho=\norm p$, the bounds $\norm{\mathbf a}\le(\sqrt5/2)\rho^2$, $\norm{D\mathbf a}\le3\rho$, $\norm{D(xy)}=\rho$, and $S\ge\rho^4/2$ give
\[
 \norm{Dg_\delta}\le(\sqrt5+3)\frac{\rho^3}{\rho^4+\delta^4}
       +15\frac{\rho^7}{(\rho^4+\delta^4)^2}\le\frac{21}{\delta}.
\]
The last step uses $z^3/(1+z^4)\le1$ and $z^7/(1+z^4)^2\le1$ for $z\ge0$. Thus $\operatorname{Lip}(g_\delta)\le21/\delta$ on $D_R$ for $0<\delta\le1$.
A map bounded by $B_0=1/\sqrt2$ and Lipschitz with constant $A$ has H\"older seminorm at most $(2B_0)^{1-\sigma}A^\sigma$, by interpolation between the bounds $2B_0$ and $A\norm{p-q}$. Thus $[g_\delta]_{C^{0,\sigma}}\le C_\sigma\delta^{-\sigma}$. Taking $\delta=(C_\sigma/K)^{1/\sigma}$ proves the upper bound. Finally the uniform comparison \eqref{couple-moment} permits taking the supremum and infimum, giving \eqref{couple-minimax}.
\end{proof}

The norm threshold is sharp: if fixed $B<1/\sqrt2$, a sufficiently small nonzero diagonal point has unavoidable residual at least
$t^2[1-B\sqrt{2+t^2/4}]>0$. Its worst-parameter energy has cubic order regardless of $K$. Also, mere continuity has no uniform quantitative replacement for \eqref{couple-error}: a horizon-dependent continuous implementation can place an arbitrarily narrow transition layer near the origin. The H\"older bound is the resource that makes the minimax statement meaningful.

\subsection{Genuinely noncommuting transport with the same relative law}
\begin{corollary}\label{couple-noncommuting}
For every fixed integer $m\ge1$ there is a finite realization with genuinely noncommuting centered increments and
\[
 \norm{G_N}\asymp N^{m+1}+r_g(p)^2N^{m+3}.
\]
Its optimal moment divided by $N^{m+1}$ still has order $1+N^2K^{-4/\sigma}$.
\end{corollary}
\begin{proof}
Adjoin $m$ independent degree-one affine pairs $u_j,w_j$ and odd derivations $S_jw_j=-u_jw_j$, zero on all other generators. Set $U=Q$. Independent odd factors anticommute, so $US_j=-S_jU$ and $S_iS_j=-S_jS_i$. They fail to commute as ordinary matrices: $US_j(\nu_yw_j)$ is a nonzero monomial. Choose cutoff $L\ge7+2m$ and branches $I\pm U,I\pm S_1,\ldots,I\pm S_m$, each with probability $1/[2(m+1)]$.

Their conjugation maps commute because the signs cancel on both sides. Since $S_j^2=0$, writing $S_J$ for the ordered product over a subset $J$ gives the exact expansion
\[
 \Phi^k(I)=\frac1{(m+1)^k}
 \sum_{\substack{J\subseteq\{1,\ldots,m\}\\|J|\le k}}
 \frac{k!}{(k-|J|)!}
 (U^{k-|J|}S_J)^*U^{k-|J|}S_J.
\]
Terms with $k-|J|\ge4$ vanish. The baseline word $US_1\cdots S_m$ has a unit monomial witness $\nu_y\prod_jw_j$, and the longest word $U^3S_1\cdots S_m$ has witness $bw\prod_jw_j$ with coefficient of magnitude $|r_g|$ and target degree $7+2m$. All words of length above $m+1$ have norm $O(|r_g|)$. Positivity in the binomial moment expansion now gives the stated bounds; \eqref{couple-error} gives the minimax conclusion. In particular $\Phi^{m+4}=0$ and all moment-map spectral radii remain unchanged.
\end{proof}