\documentclass[11pt]{article} \usepackage[a4paper,margin=24mm]{geometry} \usepackage[T1]{fontenc} \usepackage{lmodern,microtype} \usepackage{amsmath,amssymb,amsfonts,amsthm,mathtools,bm,mathrsfs} \usepackage{esint} \usepackage{enumitem} \usepackage{array,booktabs,longtable} \usepackage{tikz,float} \usetikzlibrary{arrows.meta,calc,patterns} \usepackage{hyperref} \usepackage{aliascnt} \usepackage[nameinlink,capitalise]{cleveref} \allowdisplaybreaks \setlength{\parskip}{1pt} \setlength{\parindent}{15pt} % Theorem-like environments share one counter, numbered within sections. \newtheorem{theorem}{Theorem}[section] \newaliascnt{proposition}{theorem} \newtheorem{proposition}[proposition]{Proposition} \aliascntresetthe{proposition} \newaliascnt{lemma}{theorem} \newtheorem{lemma}[lemma]{Lemma} \aliascntresetthe{lemma} \newaliascnt{corollary}{theorem} \newtheorem{corollary}[corollary]{Corollary} \aliascntresetthe{corollary} \newaliascnt{claim}{theorem} \newtheorem{claim}[claim]{Claim} \aliascntresetthe{claim} \theoremstyle{definition} \newaliascnt{definition}{theorem} \newtheorem{definition}[definition]{Definition} \aliascntresetthe{definition} \theoremstyle{remark} \newaliascnt{remark}{theorem} \newtheorem{remark}[remark]{Remark} \aliascntresetthe{remark} \numberwithin{equation}{section} \crefname{theorem}{Theorem}{Theorems} \Crefname{theorem}{Theorem}{Theorems} \crefname{proposition}{Proposition}{Propositions} \Crefname{proposition}{Proposition}{Propositions} \crefname{lemma}{Lemma}{Lemmas} \Crefname{lemma}{Lemma}{Lemmas} \crefname{corollary}{Corollary}{Corollaries} \Crefname{corollary}{Corollary}{Corollaries} \crefname{definition}{Definition}{Definitions} \Crefname{definition}{Definition}{Definitions} \crefname{claim}{Claim}{Claims} \crefname{remark}{Remark}{Remarks} \newcommand{\T}{\mathbb T} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\Z}{\mathbb Z} \newcommand{\E}{\mathbb E} \newcommand{\Pp}{\mathbb P} \newcommand{\F}{\mathcal F} \newcommand{\Q}{\mathcal Q} \newcommand{\pr}{\mathcal P} \newcommand{\tr}{\operatorname{tr}} \newcommand{\divv}{\operatorname{div}} \newcommand{\dist}{\operatorname{dist}} \newcommand{\diam}{\operatorname{diam}} \newcommand{\supp}{\operatorname{supp}} \newcommand{\Lip}{\operatorname{Lip}} \newcommand{\one}{\mathbf 1} \newcommand{\pre}{\preceq} \newcommand{\ps}{\psi} \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\dd}{\,\mathrm d} \newcommand{\wideD}{\widehat D} \newcommand{\Id}{\mathrm{Id}} \newcommand{\Var}{\operatorname{Var}} \newcommand{\HS}{\mathrm{HS}} \newcommand{\op}{\mathrm{op}} \newcommand{\cF}{\mathcal F} \newcommand{\cT}{\mathcal T} \newcommand{\cM}{\mathcal M} \newcommand{\cP}{\mathcal P} \newcommand{\cB}{\mathcal B} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\pa}[1]{\left(#1\right)} \newcommand{\br}[1]{\left[#1\right]} \newcommand{\ip}[2]{\left\langle #1,#2\right\rangle} \renewcommand{\fint}{\mathop{\rlap{-}\!\int}} \newcommand{\circleavg}[1]{\frac1{2\pi}\int_0^{2\pi}#1\,\mathrm d\theta} \title{The Ultimate Matching Theorem} \hypersetup{pdfauthor={Michael Goldman, Francesco Grotto, Dario Trevisan}} \author{% Michael Goldman\textsuperscript{1}\qquad Francesco Grotto\textsuperscript{2}\qquad Dario Trevisan\textsuperscript{2}\\[0.7em] \small\textsuperscript{1}LAMA, CNRS, Universit\'e Gustave Eiffel, Universit\'e Paris-Est Cr\'eteil,\\ \small 5 boulevard Descartes, 77454 Marne-la-Vall\'ee, France\\[-0.1em] \small\texttt{michael.goldman@cnrs.fr}\\[0.65em] \small\textsuperscript{2}Universit\`a di Pisa, Dipartimento di Matematica,\\ \small Largo Bruno Pontecorvo 5, 56127 Pisa, Italia\\[-0.1em] \small\texttt{francesco.grotto@unipi.it}\quad \texttt{dario.trevisan@unipi.it} } \date{\today} \begin{document} \maketitle \begin{abstract} Let $X_1,\dots,X_N$ be independent uniform points in the unit square, matched either to an evenly spread deterministic configuration or to a second independent sample. For exponents $\alpha_1,\alpha_2>2$ with $1/\alpha_1+1/\alpha_2=1/2$, we prove that with probability tending to one a single permutation controls the $j$th coordinate displacements in the exponential Orlicz norm of exponent $\alpha_j$, at the scale $\sqrt{\log N/N}$, for both $j=1,2$. This settles Talagrand's Ultimate Matching Conjecture for all pairs of finite critical exponents. The proof constructs, on the flat torus, a current whose divergence is Gaussian white noise projected on a dyadic grid and whose coordinates have the critical anisotropic exponential Orlicz bounds, and transfers it to matching through a primal optimal-transport reduction. At the hard endpoints $(2,\infty)$ and $(\infty,2)$ the conjecture fails: a Brownian local-time construction shows that the normalized optimal matching cost diverges in probability. This file was mostly generated by LLMs as a result of interactions with us to try and understand the 'monolithic' versions of the proofs which may be found in the \textbf{associated GitHub repository}~\cite{umc:repository}. A digested version of the proofs will be made available soon. \end{abstract} \setcounter{tocdepth}{1} \tableofcontents \section{Introduction}\label{umc:integrated-overview} \subsection{The matching problem} Let $X_1,\ldots,X_N$ be independent uniform points in the unit square $D=[0,1]^2$. Talagrand's \emph{Ultimate Matching Conjecture}, posed as Research Problem~17.1.2 (equation~(17.3)) of \cite{umc:Talagrand2021}, asks for a permutation matching this random sample to \emph{evenly spread} deterministic points $Y_1,\ldots,Y_N$ with simultaneous exponential tail bounds in both coordinates. Here ``evenly spread'' has Talagrand's meaning: $D$ admits a partition into $N$ rectangles of area $1/N$, with one point $Y_i$ in each rectangle $R_i$, such that $R_i\subset B(Y_i,10/\sqrt N)$. We also consider the matching of two independent uniform samples, which satisfies the same conclusion. Our first result, which we call the Ultimate Matching Theorem, is the following. \begin{theorem}\label{umc:thm:main} Let $\alpha_1,\alpha_2$ be fixed finite exponents with \[ \alpha_1,\alpha_2>2, \qquad \frac{1}{\alpha_1}+\frac{1}{\alpha_2}=\frac12. \] There exists a constant $L=L(\alpha_1,\alpha_2)>0$, independent of $N$ and of the evenly spread configuration, such that the following holds with probability tending to one as $N\to\infty$: for \emph{every fixed evenly spread deterministic} $(Y_i)_{i=1}^N$, or when $(Y_i)_{i=1}^N$ is a second independent uniform sample, \emph{one and the same} permutation $\pi\in S_N$ satisfies, for both coordinates $j=1,2$, \begin{equation}\label{umc:eq:combined-main} \sum_{i=1}^N \exp\!\left[ \left( \frac{\sqrt N\,\bigl|X_i^j-Y_{\pi(i)}^j\bigr|} {L\sqrt{\log N}} \right)^{\alpha_j} \right] \le 2N,\qquad j=1,2. \end{equation} \end{theorem} The symmetric quartic case $(4,4)$ is included. The proof is completed in Corollaries~\ref{cor:big-picture} and~\ref{red:cor:evenly-spread}, which combine the Gaussian flow estimate of Theorem~\ref{thm:gauss} with the reduction of Part~\ref{umc:partII}. \paragraph{Finite subcritical exponents.} The theorem also gives \eqref{umc:eq:combined-main} for every fixed pair $0<\alpha_1,\alpha_2<\infty$ with $\alpha_1^{-1}+\alpha_2^{-1}>1/2$, after changing the constant $L$. Indeed, set \[ c=2(\alpha_1^{-1}+\alpha_2^{-1})>1,\qquad r_i=c\alpha_i. \] Then $r_i>\alpha_i$, $r_i>2$, and $r_1^{-1}+r_2^{-1}=1/2$, so the critical theorem applies with exponents $r_1,r_2$. For the resulting permutation, write $t_{i,j}=\sqrt N\,|X_i^j-Y_{\pi(i)}^j|/(L\sqrt{\log N})$. Since $t^{\alpha_j}\le1+t^{r_j}$ for $t\ge0$, Jensen's inequality with a fixed $A\ge1$ gives \[ \frac1N\sum_{i=1}^N e^{(t_{i,j}/A)^{\alpha_j}} \le\left(\frac1N\sum_{i=1}^N e^{t_{i,j}^{\alpha_j}}\right)^{A^{-\alpha_j}} \le(2e)^{A^{-\alpha_j}}\le2 \] when $A^{\alpha_j}\ge\log(2e)/\log2$ for both $j=1,2$. Thus the same permutation works at the same scale, with $L$ replaced by $AL$; the probability of success is unchanged. \paragraph{The hard endpoints.} The coordinate endpoints $(2,\infty)$ and $(\infty,2)$ behave differently. Theorem~\ref{neg:thm:main} in Part~\ref{umc:partIII} shows that there the optimal matching cost, divided by $\sqrt{\log N/N}$, diverges in probability, uniformly over deterministic reference configurations. The Gaussian construction of Part~\ref{umc:partI} applies only to finite exponents, and the endpoint obstruction is of a different nature. These coordinatewise endpoints should not be confused with the usual scalar $L^\infty$ matching problem or with an endpoint of a power-cost problem. In the endpoint cases it is in fact possible to determine the exact first order asymptotic (in probability) behavior of the cost: for its (undigested, AI assisted) derivation we refer once again to the associated GitHub repository~\cite{umc:repository}, since here we are mainly interested in the failure of Talagrand's conjecture. \subsection{Previous work} \paragraph{The dual route.} The foundations of planar random matching go back to Ajtai, Koml\'os and Tusn\'ady \cite{umc:AKT84}. Talagrand's work \cite{umc:Talagrand92,umc:Talagrand2005,umc:Talagrand2021} identified the multiscale nature of the problem and led to Research Problem~17.1.2 in the 2021 monograph. The guiding method is \emph{dual}: one estimates transport or discrepancy functionals against classes of test functions, and controls the resulting random suprema through chaining, majorizing measures and geometric decompositions. This approach has produced sharp matching estimates in several regimes. In dimension three, Chapter~18 of \cite{umc:Talagrand2021} proves a sharp matching theorem between an independent uniform sample and an evenly spread deterministic configuration, including coordinate-dependent exponential costs and hard coordinate constraints (Theorem~18.1.1 and Exercise~18.1.4); Talagrand observes that the extension to dimensions $d>3$ requires no new ideas. Above dimension two, irregularities at different scales cannot accumulate in the same way, whereas in the plane their cumulative effect produces the critical logarithmic factor. The remaining difficulty is therefore specifically two-dimensional. The dual direction continues to develop: Cheng, Otto, Palmieri and Wachtel \cite{umc:COPW26} study min--max matching through a geometrically linear approximation of a dual formulation, and Ouimet \cite{umc:Ouimet26} proposes a solution to Problem~5.3 in Talagrand's \emph{My favorite problems} \cite{TalagrandProblems}, controlling simultaneously total distance and maximal displacement through electrical flows and planar Lipschitz truncation of lattice circulations. \paragraph{The primal route.} Our proof follows a complementary tradition. Rather than estimating the dual supremum, we construct an actual \emph{primal} transport current carrying the discrepancy of the random measures. This approach was brought into focus by the statistical-physics prediction of Caracciolo, Lucibello, Parisi and Sicuro \cite{umc:CLPS14}: at leading order in quadratic bipartite matching, the displacement should behave like the gradient of a Poisson potential driven by empirical-density fluctuations, which is the linearization of the Monge--Amp\`ere equation. Ambrosio, Stra and Trevisan \cite{umc:AST19} established the leading two-dimensional quadratic asymptotics rigorously, and finer results on the cost and the optimal map followed \cite{umc:AG19,umc:AGTmap19}. In parallel, the variational regularity theory of optimal transport, through the harmonic approximation of Goldman and Otto \cite{umc:GO20} and its quantitative versions by Goldman, Huesmann and Otto \cite{umc:GHO21,umc:GHO23}, quantified the approximation of displacements by Poisson fluxes; related methods treat nonuniform planar domains \cite{umc:AGT22}. These results make it natural to study random matching through the Eulerian continuity equation and a compatible vector current. For nonquadratic costs the corresponding equation is nonlinear. Ambrosio, Vitillaro and Trevisan \cite{umc:AVT24} obtained sharp estimates for power-law costs, and Armegioiu, Goldman, Grotto and Trevisan \cite{umc:AGGT26} related finite power-cost asymptotics on the torus to a Gaussian variational-martingale problem and a parabolic Monge--Amp\`ere Bellman equation. In this viewpoint successive logarithmic scales are accumulated through stochastic control; it directly motivates the construction of Part~\ref{umc:partI}. The power-cost results do not, by themselves, give the coordinatewise exponential Orlicz bounds proved here. \subsection{Strategy of the proof} \paragraph{Critical Gaussian currents.} The main analytic task is to solve a divergence equation for mean-zero Gaussian white noise projected onto dyadic squares of mesh $h=2^{-m}$ on the flat two-dimensional torus. With $\psi_r(t)=e^{|t|^r}-1$ and $\|\cdot\|_{\psi_r}$ the associated Luxemburg norm, Part~\ref{umc:partI} constructs, with probability bounded below uniformly in $m$, a periodic current $J$ with \[ \operatorname{div}J=W_h, \qquad \max_{i=1,2}\|J_i\|_{\psi_{r_i}}\le C\sqrt m, \] for every finite critical pair $(r_1,r_2)=(\alpha_1,\alpha_2)$ (Theorem~\ref{thm:gauss}). The difficulty is not the existence of an inverse of the divergence, but the selection of one whose two coordinate tails are simultaneously of the critical size. At a single point, a Bellman equation suggests an inverse-Hessian covariance for a Gaussian martingale, and a supersolution evaluated backward in Bellman time formally gives a supermartingale. The pointwise choice, however, need not be compatible across space. Part~\ref{umc:partI} therefore reveals the Gaussian increments through predictable local divergence corrections on anisotropic Whitney cells. An adaptive metric follows the locally preferred covariance, and a discounted memory records spatial oscillations of the accumulated current. A simultaneous induction on the metric and the memory yields a contracting Lyapunov quantity, whose negative drift pays for the variance of the critical exponential tilt. A terminal Gaussian correction then transfers the current from the adaptive partition to the fixed dyadic grid. \paragraph{From currents to matching.} Part~\ref{umc:partII} supplies the reductions, independently of the adaptive construction. The minimum-flow gauge is uniformly Lipschitz in the dyadic Gaussian coefficients, so a fixed positive success probability upgrades to high probability by Gaussian concentration. An exact four-sheet folding transfers the torus estimate to the square. Dyadic regularization, a comparison between transport and static currents, and an interval-encoded KMT coupling relate the Gaussian source to one empirical histogram against Lebesgue measure, and then to two independent histograms. Talagrand's evenly spread points are a further $O(N^{-1/2})$ transport from Lebesgue measure, and a Birkhoff decomposition yields one permutation with both coordinate bounds. These steps give \eqref{umc:eq:combined-main} in both formulations. \paragraph{The endpoint obstruction.} At the endpoint $(2,\infty)$ the reduction of Part~\ref{umc:partII} still applies, so it suffices to show that the square Gaussian minimum-flow gauge is not of order $\sqrt m$. Part~\ref{umc:partIII} builds random dual test potentials of bounded cost whose expected pairing with the projected noise is too large. The input is a planar Brownian martingale with bounded terminal anisotropic Orlicz cost but arbitrarily large expected integrated square-root determinant, obtained from short Brownian local-time stages. An adaptive tree of piecewise affine Gaussian bubbles converts this determinant into a positive Gaussian pairing. \subsection{Organization} Part~\ref{umc:partI} (Sections~\ref{sec:introduction}--\ref{sec:terminal}) proves the Gaussian flow estimate on the torus; the geometric estimates for its adaptive construction are collected in Appendix~\ref{app:geometry}. Part~\ref{umc:partII} (Sections~\ref{sec:interface}--\ref{red:sec:consequences}) reduces matching to the square Gaussian flow problem, upgrades fixed-probability bounds to high probability, transfers the torus estimate to the square, and completes the proof of Theorem~\ref{umc:thm:main}. Part~\ref{umc:partIII} (Sections~\ref{neg:sec:introduction}--\ref{neg:sec:conclusion}) proves the failure at the hard endpoints. \clearpage \part{Critical Gaussian Orlicz flows on the torus}\label{umc:partI} In this part, $W_h$ is the projection of mean-zero Gaussian white noise on the flat torus $\T^2$ onto the dyadic square grid of mesh $h=2^{-m}$. For fixed $22$ on the critical line \begin{equation}\label{eq:crit} a_i=\frac1{r_i},\qquad a_1+a_2=\frac12. \end{equation} \phantomsection\label{not:astar} We also set $a_*:=\max_i a_i<1/2$. For every mean-zero source $f$, define the minimum-flow gauge \begin{equation}\label{eq:gauge} \mathfrak F_{\boldsymbol r}(f) =\inf_{\divv J=f}\max_{i=1,2}\|J_i\|_{\psi_{r_i}}, \end{equation} where $\boldsymbol r=(r_1,r_2)$ and the divergence constraint is understood distributionally for periodic currents. The problem is to control this gauge for $W_h$. \begin{theorem}\label{thm:gauss} For every fixed finite $r_1,r_2>2$ with $1/r_1+1/r_2=1/2$, there exist constants $C<\infty$, $\varepsilon_0>0$ and $m_0<\infty$, depending only on the exponents, such that for $m\ge m_0$, \begin{equation}\label{eq:target} \Pp\left(\mathfrak F_{\boldsymbol r}(W_h)\le C\sqrt m\right) >\varepsilon_0. \end{equation} \end{theorem} \begin{remark}\label{rem:concentration-upgrade} The fixed-positive-probability estimate is the conclusion proved in this part. As shown in Part~\ref{umc:partII} (Subsection~\ref{red:subsec:gaussian-concentration}), the fixed-grid minimum-flow gauge is Lipschitz in the Gaussian coefficients with a constant independent of $h$. Gaussian concentration therefore upgrades \eqref{eq:target} to $\Pp(\mathfrak F_{\boldsymbol r}(W_h)>C'\sqrt m) \le 2e^{-cm}$ for suitable $C',c>0$. This improvement requires no further adaptive-flow estimates and is not used below. \end{remark} \subsection{The quartic case} Consider first $r_1=r_2=4$. The proof controls the exponential cost of the accumulated current through the phase \[ H(y)=(1+y_1^2)^2+(1+y_2^2)^2. \] Here $y\in\R^2$ is a current value: at a spatial point $x$ and reveal generation $k$, it will be $y=U_k(x)$. It is not a spatial coordinate. The quadratic terms in $H$ give nondegenerate curvature near the origin, while its growth at infinity is quartic. The choice of the next covariance is suggested by the curvature of $e^H$. Since $D^2H\preceq C H^{1/2}I$, the matrix \[ B(y)=H(y)^{1/2}I+\nabla H(y)\nabla H(y)^T \] controls both the Hessian term and the square of the linear variation of the phase. Among positive definite matrices of determinant one, the choice $C(y)=\sqrt{\det B(y)}B(y)^{-1}$ minimizes $\tr(B(y)G)$; the calculation is given in Section~\ref{sec:selector}. For example, at $y=(R,0)$ with $R\to\infty$, \[ B(y)\asymp\operatorname{diag}(R^6,R^2),\qquad C(y)\asymp\operatorname{diag}(R^{-2},R^2). \] The corresponding phase cost is $\tr(BC)\asymp R^4\asymp H$, whereas an isotropic choice gives $\tr B\asymp R^6$. Thus the preferred covariance reduces the variance in the direction where the current is already large. This pointwise choice cannot be imposed independently at every spatial point. The local divergence equation must be solved on a whole cell, in a chart chosen before the next noise is revealed. The construction therefore uses one covariance-control matrix $G$ per cell and moves it towards $C$ by bounded updates. A memory of old spatial oscillations controls the error between the state at the cell representative and the state at other points. The next subsection explains this simultaneous metric--memory induction. For general fixed finite critical exponents, the same construction uses \begin{equation}\label{eq:phaseH} F_i(t)=(1+t^2)^{r_i/2},\qquad H(y)=F_1(y_1)+F_2(y_2),\qquad v(y)=\nabla H(y). \end{equation} \subsection{Structure of the proof} The noise is revealed through a finite predictable sequence of Gaussian increments. At each stage a local divergence correction adds the newly revealed noise to the current, with a covariance chosen from the old history on each active cell. The aim is to control a spatial exponential modular of the accumulated current. The phase is $H$ from \eqref{eq:phaseH}, and the preferred covariance is selected by the regularized Hessian proxy in \eqref{eq:B}. No continuous-time Bellman equation is solved in the construction; its formal relation to the discrete argument is explained in Subsection~\ref{subsec:bellman-motivation}. The inverse-Hessian rule does not by itself solve the spatial problem. The preferred covariance $C(U_k(x))$ depends on $x$, whereas an inverse of the divergence must be constructed coherently on a whole region. Moreover, the covariance must be chosen \emph{before} the next Gaussian variables are revealed. The construction therefore follows the pointwise Bellman covariance with a predictable metric and records the price of not following it exactly. This is done by a simultaneous induction on the routing metric and on a memory of spatial oscillations. Section~\ref{sec:innovation} reveals the source through a finite predictable tree. Section~\ref{sec:geom} provides a local divergence inverse in intrinsic Whitney coordinates and a normalized Gaussian increment with uniform conditional variance and H\"older envelopes $E_{k+1}$. These bounds do not depend on the eccentricity of the physical chart. We use the scales \begin{equation}\label{eq:parameters} s=hm^3,\qquad \delta=\frac{\varepsilon^2}{m},\qquad \lambda_k=e^{-\Lambda k\delta}, \end{equation} where $\varepsilon>0$ is a small fixed step parameter and $\Lambda>0$ is a sufficiently large fixed damping constant. The cumulative normalized current satisfies $\divv U_k=\sqrt\delta\,P_{\pr_k}W$; its unscaled counterpart is $U_k/\sqrt\delta$. The grid parameter $m$ is fixed throughout the construction; $k$ counts reveals and does not specify a common spatial size of the cells. \phantomsection\label{not:depthbound} The depth of any actual revealed chain is at most $J_0\le C_{\rm dep}m$, for a fixed geometric constant $C_{\rm dep}$. \emph{Metric lag.} On a parent $Q$, choose an old-measurable representative $x_Q$. Section~\ref{sec:selector} selects an updated determinant-one metric $G_Q^+$ by clipping the old metric $G_k$ towards $C(U_k(x_Q))$. The clipping stays within a fixed factor of $G_k$ and contracts, at $x_Q$, the lag \[ L_k(x)^2=\operatorname{tr}\!\left(G_k(x) C(U_k(x))^{-1}\right). \] If $U_k$ were constant on $Q$, this would be a contraction throughout $Q$. But $U_k$ varies in space while the metric is chosen once on $Q$, and the difference $U_k(x)-U_k(x_Q)$ is exactly what prevents applying the pointwise Bellman optimization independently at every location. \emph{Memory.} Each old increment is H\"older in the intrinsic metric in which it was created, and descendants shrink in these birth metrics, even after the bounded changes of metric between refinements. Hence its oscillation across a much later descendant is reduced by a factor $r_{\rm osc}^{k-j}$. The old diagonal metric caps and relative comparisons of $H(U_j)$ then give, schematically for $x,z$ in the current parent, \[ |U_{k,i}(x)-U_{k,i}(z)| \lesssim\sqrt{\delta M}\,H(U_k(x))^{a_i} \sum_{j\le k}r_{\rm osc}^{k-j}E_j(x). \] Along the actual reveals of each branch we retain the discounted square function \[ T_k=\sum_{j\le k}\rho_1^{k-j}E_j^2,\qquad S_k=L_k^2+T_k,\qquad r_{\rm osc}^2<\rho_1<1. \] Weighted Cauchy--Schwarz bounds the ancestor sum by $C\sqrt{T_k}$. Thus $T_k$ pays for the oscillation of the old field between the representative, where the metric is selected, and the point where the estimate is used. \emph{Simultaneous induction.} Section~\ref{sec:bootstrap} closes three estimates together: the coordinatewise caps $G_{k,ii}\le M H(U_k)^{2a_i}$, with a fixed large constant $M$; small relative changes of the phase $H(U_k)$; and the contraction of $S_k$. The \emph{old} memory estimate controls $U_k(x)-U_k(x_Q)$, so the representative-based metric update preserves the diagonal caps with a strict margin. The \emph{fresh} Gaussian envelope gives $|V_{k+1,i}|\lesssim\sqrt\delta\,H(U_k)^{a_i}E_{k+1}$, which preserves the caps and the phase comparison at the new state. Only then do we use the old stopped exponential modular, together with the lower area bound for active cells, to obtain $H(U_k)\lesssim m$ pointwise. Consequently $\sqrt{\delta H(U_k)}\lesssim\varepsilon$. This small factor makes the metric update stable under spatial variation of the state. After the representative contraction and a comparison of the lag along the old-to-new state segment, which uses only old-state information, one obtains \[ L_{k+1}^2\le q_L L_k^2+C+C\varepsilon^2T_k +CE_{k+1}^2,\qquad q_L<1. \] On an actual reveal, $T_{k+1}=\rho_1T_k+E_{k+1}^2$. Choosing $\rho_1$ and $\varepsilon$ with strict margins, the two inequalities combine into the renewal estimate \begin{equation}\label{eq:intro-renewal} \boxed{\quad S_{k+1}\le q_0S_k+C(1+E_{k+1}^2), \qquad q_0<1.\quad} \end{equation} The old memory justifies the new metric, and the contraction of the new metric pays for retaining the memory. This is why the two are propagated simultaneously. \emph{Exponential estimate.} The renewal bound \eqref{eq:intro-renewal} supplies the negative drift that turns the exponential Bellman trial into a spatial supermartingale. Write $H_k=H(U_k)$. The phase used in Section~\ref{sec:phase} is \phantomsection\label{not:lyapunovfirst} \[ \Phi_k=\lambda_k H_k(1+\kappa\delta S_k), \] where $\kappa>0$ is fixed and the clock $\lambda_k$ decreases slightly with each reveal. Under a fresh reveal, the linear variation of $H_k$ is Gaussian. At criticality its conditional variance is bounded, up to constants, by $\lambda_k\delta H_kS_k$: the factor $S_k$ measures how far the actual covariance is from the pointwise Bellman covariance. Exponentiating this linear term produces a positive drift of that size, and the term $\kappa\delta H_kS_k$ is added to the exponent to compensate it. On a good reveal, $H_{k+1}\le(1+\xi)H_k$ for a small fixed $\xi$, and \eqref{eq:intro-renewal} gives \[ H_{k+1}S_{k+1}-H_kS_k \le -d_0 H_kS_k+C H_k(1+E_{k+1}^2), \qquad d_0=1-q_0(1+\xi)>0. \] Thus the additional phase contributes $-\kappa d_0\lambda_{k+1}\delta H_kS_k$, which dominates the conditional Gaussian variance cost once $\kappa$ is large. The remaining errors have size $\lambda_{k+1}\delta H_k(1+E_{k+1}^2)$. The square of the envelope has a uniform conditional exponential moment, and its coefficient satisfies $\delta H_k\lesssim\varepsilon^2$ by the old modular bound $H_k\lesssim m$. The constant part is paid for by the decrease $\lambda_k-\lambda_{k+1}\asymp\lambda_{k+1}\delta$. The order of these estimates matters. We bound the product $H_{k+1}S_{k+1}$ by relative phase control, rather than expanding the new lag inside the phase. Consequently the only linear Gaussian term is the variation of $H_k$, whose coefficient depends on the old state and on the metric selected \emph{before} the noise is revealed. Its conditional moment-generating function can therefore be evaluated directly, and no independence between the Gaussian increment and its H\"older envelope is needed. This gives a one-step conditional estimate for the spatial exponential modular. To iterate it, Section~\ref{sec:phase} uses a nonnegative killed-and-frozen numerical supermartingale. On histories that pass every envelope and modular test it is the spatial integral of $e^{\Phi_k}$; on an envelope failure it is set to zero. Overshoot stopping and the separately controlled probability of an envelope failure then give a bounded spatial modular with a fixed positive probability. Terminal branches are frozen without artificial memory discounts. In this sense the construction plays the role of the formal Bellman supermartingale $u(T-t,X_t)$ from \eqref{eq:intro-bellman-supermart}, with the metric--memory mechanism paying for spatial incompatibility. \emph{Terminal correction.} Section~\ref{sec:terminal} transfers the divergence from the adaptive partition to the fixed-grid source $W_h$ in physical coordinates. Combining that correction with the successful leaf current gives \eqref{eq:target}. The geometric estimates used in the construction are collected in Appendix~\ref{app:geometry}. \subsection{Bellman motivation}\label{subsec:bellman-motivation} The starting analogy is the power-energy problem \[ \inf_{\divv J=W_h}\int_{\T^2}|J(x)|^p\,\dd x, \qquad 10$ and an initial current state $X_0=y$. Formally let the one-point current evolve as \[ \dd X_t=\frac1{\sqrt2}C_u(T-t,X_t)^{1/2} \,\dd\mathbf B_t,\qquad 0\le t\le T, \] where $\mathbf B_t$ is a standard planar Brownian motion modeling the newly revealed Gaussian innovations after the transverse control. Thus $X_t$ is \emph{the current accumulated at the fixed spatial point $x$}, not the position of a particle or a spatial current on the whole torus. Its conditional covariance per unit time is $\frac12C_u(T-t,X_t)$. The factor $1/2$ is exactly the time normalization in \eqref{eq:intro-ma}. Since $\frac14\operatorname{tr}(C_uD_y^2u) =\frac12\sqrt{\det D_y^2u}$, It\^o's formula gives \begin{equation}\label{eq:intro-bellman-supermart} \dd\,u(T-t,X_t) =\left[\frac14\operatorname{tr}(C_uD_y^2u) -\partial_tu\right](T-t,X_t)\dd t +\dd M_t\ \le\ \dd M_t \end{equation} whenever $u$ is a Bellman supersolution. With equality in the Bellman equation the drift vanishes. In the power case, $u_p(0,X_T)=|X_T|^p$, so the supermartingale property yields the one-point terminal-cost bound $\E|X_T|^p\le u_p(T,y)$ (after localization and the necessary integrability justification). The reversal of Bellman time is essential: the PDE evolves in the remaining-time variable, while $X_t$ accumulates newly revealed noise. This calculation is formal (it requires localization and a well-defined feedback martingale), but it identifies the goal: a supermartingale built from an exponential trial evaluated along the evolving current. For the Orlicz problem, use the phase $H$ from \eqref{eq:phaseH} and the critical weights $a_i=1/r_i$. \phantomsection\label{not:bellmanclock} Try $u(t,y)=\exp\{\zeta_{\rm B}(t)H(y)\}$, with $\zeta_{\rm B}>0$ an auxiliary increasing Bellman-time coefficient. \phantomsection\label{not:bellmansigma} To compare with the subcritical exponents, put $\sigma=a_1+a_2-\frac12\ge0$, so $\sigma=0$ on the critical line. Differentiation and the $2\times2$ rank-one determinant identity yield \begin{equation}\label{eq:intro-bellman-determinant} e^{-\zeta_{\rm B}H}\sqrt{\det D_y^2u} \le C(\zeta_{\rm B}+\zeta_{\rm B}^{3/2})H^{1-\sigma}. \end{equation} As $\partial_tu=\zeta_{\rm B}'Hu$, choosing $\zeta_{\rm B}'\ge C(\zeta_{\rm B}+\zeta_{\rm B}^{3/2})$ produces a supersolution for a prescribed finite interval, provided the initial coefficient is sufficiently small. In particular, the exponential Bellman supersolution persists at criticality, including the symmetric quartic example $r_1=r_2=4$. The subcritical advantage is instead the large-amplitude power reserve $H^{-\sigma}$, which disappears when $\sigma=0$. The one-point limiting cost is heuristically finite, but this discards all the geometry. A pointwise covariance optimization does not construct a \emph{spatially compatible} current $x\mapsto J(x)$ driven by the same white noise and satisfying $\divv J=W_h$. At polynomial order the reconstruction errors can be controlled by Wiener-chaos estimates as in \cite{umc:AGGT26}. At critical exponential growth the missing power reserve means that the one-point supermartingale cannot by itself absorb the errors of spatial reconstruction. The solution sought in this part is therefore a \emph{spatial} analogue of \eqref{eq:intro-bellman-supermart}, coupled to an adaptive routing metric and a memory of spatial oscillations. \subsection{Notation}\label{subsec:notation} The notation used throughout Part~\ref{umc:partI} is collected in Table~\ref{tab:notation}. All references point to the first definition of the corresponding object. {\small \setlength{\tabcolsep}{4pt} \renewcommand{\arraystretch}{1.12} \begin{longtable}{@{}p{0.285\linewidth}p{0.535\linewidth}p{0.14\linewidth}@{}} \caption{Principal notation and location of first definitions.}\label{tab:notation}\\ \toprule Symbol & Meaning & Definition \\ \midrule \endfirsthead \toprule Symbol & Meaning & Definition \\ \midrule \endhead \bottomrule \endfoot $W$ & Mean-zero white noise on $\T^2$. & \eqref{eq:white}\\ $P_h,\,W_h$ & Averaging on the fixed dyadic $h$-grid; $W_h=P_hW$. & p.~\pageref{not:fixedgrid}\\ $\boldsymbol r=(r_1,r_2),\,a_i$ & Orlicz exponents and critical weights $a_i=1/r_i$. & \eqref{eq:crit}\\ $a_*$ & Larger critical weight $\max_i a_i<1/2$. & p.~\pageref{not:astar}\\ $\psi_r,\,\|\cdot\|_{\psi_r}$ & Exponential Young function and Luxemburg norm. & p.~\pageref{not:orlicz}\\ $\mathfrak F_{\boldsymbol r}$ & Minimum coordinatewise Orlicz flow gauge. & \eqref{eq:gauge}\\ $h,\,s,\,\delta,\,\varepsilon$ & Fixed-grid mesh, terminal intrinsic scale, time step $\delta=\varepsilon^2/m$, and fixed small $\varepsilon$. & \eqref{eq:parameters}\\ $J_0,\,C_{\rm dep}$ & Deterministic bound $J_0\le C_{\rm dep}m$ for actual reveal depth. & p.~\pageref{not:depthbound}\\ $\lambda_k,\,\Lambda$ & Deterministic damping clock and its fixed rate. & \eqref{eq:parameters}\\ $\pr_k$ & Nested partition after the $k$th reveal. & p.~\pageref{not:predictabletree}\\ $\F_k$ & Sigma-field of geometry and masses revealed through stage $k$. & p.~\pageref{not:revealedfiltration}\\ $f_k,\,\Delta f_k$ & Adaptive projection of $W$ and its next Gaussian increment. & p.~\pageref{not:adaptiveprojection}\\ $Q_k(x),\,M_k(x),\,v_{k+1}(x)$ & Cell containing $x$, its Gaussian average, and next conditional variance. & \eqref{eq:cellmartnotation}\\ $G_Q^+,\,A_Q,\,\widehat Q$ & Predictable routing metric, its square root, and the intrinsic parent chart. & \eqref{eq:intrinsicchart}\\ $K,\,c_0,\,D$ & Fixed adjacent-metric distortion, inball ratio, and current intrinsic parent diameter. & p.~\pageref{not:Kconstant}\\ $\theta,\,D_0$ & Whitney size cap and intrinsic terminal stopping threshold $D\le D_0s$. & p.~\pageref{not:whitneyparameters}\\ $\gamma,\,A,\,C_A$ & Whitney boundary-clearance, inball-displacement and receiver-size constants. & Lemma~\ref{lem:whitney}\\ $\vartheta$ & Child size contraction across an actual birth-chart refinement, distinct from $\theta$. & p.~\pageref{not:whitneyparameters}\\ $z,\,\widehat S$ & Intrinsic coordinates and transformed child. & \eqref{eq:intrinsicchart}\\ $d_{G_Q^+}$ & Distance in the predictable intrinsic metric. & p.~\pageref{not:intrinsicdistance}\\ $\ell_S$ & Intrinsic scale of a Whitney child. & \eqref{eq:whitney}\\ $\alpha,\,\beta$ & H\"older exponents: kernel $\alpha$ and field $\beta$. & p.~\pageref{not:holderparameters}\\ $c_*$ & Fixed positive floor $|S|\ge c_*s^2$ for revealed-cell areas. & Lemma~\ref{lem:randommax}\\ $\widehat K_Q,\,\widehat{\mathcal B}_Q$ & Ray kernel and local divergence inverse in intrinsic coordinates. & \eqref{eq:ray}\\ $K_Q^{\rm phys},\,\mathcal B_Q^{\rm phys}$ & Corresponding physical Piola kernel and inverse. & \eqref{eq:piolakernel}\\ $\widehat P$ & Averaging onto the intrinsic Whitney child partition. & p.~\pageref{not:childprojector}\\ $H(y),\,v(y)$ & Anisotropic exponential phase and its gradient. & \eqref{eq:phaseH}\\ $\zeta_{\rm B}(t),\,\sigma$ & Bellman time coefficient (distinct from $\lambda_k$) and subcritical reserve $\sigma=a_1+a_2-1/2$. & pp.~\pageref{not:bellmanclock}--\pageref{not:bellmansigma}\\ $B(y),\,C(y)$ & Bellman matrix and determinant-one target metric. & \eqref{eq:B}\\ $\mathcal T_C$ & Spectral metric clipping toward $C$, with relative distortion bounded by $K$. & Lemma~\ref{lem:selector}\\ $L(y,G)$ & Lag between actual and target metrics. & \eqref{eq:lag}\\ $G_k,\,G_k^+$ & Birth metric and predictable updated metric at a split. & p.~\pageref{not:birthmetric}\\ $M$ & Fixed coefficient in the diagonal cap $G_{ii}\le M H(U)^{2a_i}$. & Lemma~\ref{lem:selector}\\ $\eta$ & Relative change allowed for $H(U_k)$ at an actual reveal. & p.~\pageref{not:relativephase}\\ $r_{\rm osc}$ & Effective ancestor spatial-oscillation decay factor. & Lemma~\ref{lem:actualphase}\\ $q_d,\,q_L,\,q_0$ & Diagonal-cap, lag and combined-memory contraction factors. & pp.~\pageref{not:qd},\pageref{not:qL}\\ $\xi$ & Realized relative phase-change bound $|H'-H|\le\xi H$, controlled by $\varepsilon$. & p.~\pageref{not:xi}\\ $Z_{k+1}$ & Intrinsic normalized Gaussian increment field. & p.~\pageref{not:intrinsicgauss}\\ $V_{k+1},\,U_k$ & Physical increment and cumulative divergence current. & \eqref{eq:update}\\ $E_{k+1,S}$ & Gaussian supremum--Hölder envelope on a child. & \eqref{eq:envelope}\\ $T_k,\,S_k$ & Discounted spatial memory and combined lag--memory energy. & \eqref{eq:memory}\\ $\rho_1$ & Fixed memory discount; satisfies $r_{\rm osc}^2<\rho_10$ depending only on $c_*$ such that for every $b>0$, \begin{equation}\label{eq:randommax} \Pp\!\left(\max_{0\le k\le J}\max_{Q\in\pr_k} |M(Q)|>\frac b s\right) \le Cs^{-2}\exp(-cb^2). \end{equation} In particular, if $s=2^{-m}m^3$, then \begin{equation}\label{eq:randommaxwhp} \max_{k,Q}\frac{|W(Q)|}{|Q|}\le C\frac{\sqrt m}{s} \quad\text{with probability }1-o(1). \end{equation} \end{lemma} \begin{proof} Fix a deterministic $x$ for which \eqref{eq:cellmartnotation} is well defined at every stage. Conditional on $\F_k$, the chosen subdivision of $Q_k(x)$ is fixed, and the centered child bridge is Gaussian. Hence $M_k(x)$ is a martingale with conditional increment variance $v_{k+1}(x)\ge0$. The predictable variance telescopes \emph{pathwise}: \[ \sum_{j=0}^{k-1}v_{j+1}(x) =|Q_k(x)|^{-1}-|Q_0(x)|^{-1}\le(c_*s^2)^{-1}. \] For each fixed $u\in\R$, the exponential \[ \exp\Big\{u(M_k(x)-M_0(x))- \tfrac12u^2\sum_{j0$ and then with $-u$, and optimize at $u\asymp bs$ to obtain \begin{equation}\label{eq:pointxmax} \Pp\bigl(\max_{k\le J}|M_k(x)|>b/s\bigr) \le 2\exp(-c b^2). \end{equation} At the root $M_0(x)=W(\T^2)=0$. If \emph{any} revealed cell at any time has a large average, then the spatial set of $x$ for which the running maximum is large contains the entire cell and hence has measure at least $c_*s^2$. Consequently, for the event $B$ on the left of \eqref{eq:randommax}, \[ \one_B\le (c_*s^2)^{-1}\int_{\T^2} \one_{\{\max_k|M_k(x)|>b/s\}}\dd x. \] Take expectations and use \eqref{eq:pointxmax}. For $b=A\sqrt m$, $s^{-2}=2^{2m}m^{-6}$, and $A$ large enough that $cA^2>2\log 2$, the upper bound is $o(1)$. \end{proof} No independence between the random partition and $W$ is assumed. \section{Local divergence inverses and Whitney partitions}\label{sec:geom} At each refinement step the new current is obtained by inverting the divergence on an active parent cell and projecting onto the children of that cell. This section constructs the two geometric ingredients. The first is a ray inverse of the divergence, defined in an intrinsic chart attached to the predictable routing metric and transported to physical coordinates by a Piola transform. The second is a finite Whitney partition of the parent in the same chart, in which clipped boundary pieces are attached to full receiver squares. The section ends with the projected-kernel bounds of Lemma~\ref{lem:pk}, which are the geometric input for the covariance estimates of Section~\ref{sec:selector}. Proofs of some of the geometric statements are collected in Appendix~\ref{app:geometry}. \paragraph{Physical and intrinsic coordinates.} The current $U_k$, the two coordinate directions, and the phase $H(U_k)$ always live in the original physical coordinates on $\T^2$. The local inverse of divergence, however, is constructed in a chart chosen using the \emph{old} Gaussian history. Fix an active physical parent $Q$, regarded as a convex set in one chosen lift of the torus, and let $G_Q^+$ be its new predictable routing metric. Set \begin{equation}\label{eq:intrinsicchart} A_Q=(G_Q^+)^{1/2}=A_Q^T,\quad \det A_Q=1,\quad z=A_Q^{-1}(x-b_Q),\quad \widehat Q=A_Q^{-1}(Q-b_Q),\quad \widehat S=A_Q^{-1}(S-b_Q). \end{equation} Here $b_Q$ is an old-measurable choice of origin, and all displayed quantities are local to this parent. In particular, $A_Q$ is generally not an automorphism of the torus. \phantomsection\label{not:intrinsicdistance} The intrinsic distance is \[ d_{G_Q^+}(x,y)=|A_Q^{-1}(x-y)|,\qquad d_{G_Q^+}(x,\partial Q)=\dist(A_Q^{-1}(x-b_Q),\partial\widehat Q). \] The physical child $S=b_Q+A_Q\widehat S$ has $|S|=|\widehat S|$, but need not be a Euclidean square (Figure~\ref{fig:intrinsic-physical}). All local radii, Whitney lengths, H\"older seminorms and point-source kernel estimates below are in the \emph{intrinsic} $z$-chart. There is no uniform Euclidean physical kernel bound independent of the eccentricity of $A_Q$. The transformed parent $\widehat Q$ is a uniformly shape-regular convex parallelogram: before updating the metric the active parent is a square in its birth chart, and successive metrics have a uniformly bounded relative Loewner distortion. Let $D=\diam(\widehat Q)$ and choose an inball $B(z_0,c_0D)\subset\widehat Q$. \phantomsection\label{not:Kconstant} Here $K>1$ denotes the fixed bound on distortion between a cell's birth metric $G$ and the updated metric $G^+$: $K^{-1}G\preceq G^+\preceq KG$ (see \eqref{eq:clipping}). The inball ratio $c_0=c_0(K)>0$ is independent of the physical eccentricity of the chart. \begin{figure}[tbp] \centering \begin{tikzpicture}[>=Latex, font=\footnotesize, line join=round] % intrinsic chart \begin{scope}[shift={(0.5,0)}] \fill[black!3] (0,0.35)--(3.0,0.35)--(3.38,2.58)--(0.38,2.58)--cycle; \draw[line width=.65pt] (0,0.35)--(3.0,0.35)--(3.38,2.58)--(0.38,2.58)--cycle; \draw[densely dashed,gray] (2.28,1.60) circle (.45); \draw[fill=black!14, line width=.7pt] (.70,.70) rectangle (1.39,1.39); \node[anchor=north] at (1.65,.15) {$\widehat Q\subset\R^2_z$}; \node at (1.045,1.045) {$\widehat S$}; \node at (2.28,1.60) {$\scriptstyle z_0$}; \end{scope} % affine arrow between panels \draw[->,line width=.7pt] (4.25,1.45)--(5.55,1.45); \node[align=center] at (4.9,1.92) {$x=b_Q+A_Qz$}; \node[align=center] at (4.9,.91) {$\det A_Q=1$}; % physical chart, affine image under an area preserving shear/stretch \begin{scope}[shift={(6.0,.1)},x={(1.18cm,0cm)},y={(.44cm,.84746cm)}] \fill[black!3] (0,.35)--(3,.35)--(3.38,2.58)--(.38,2.58)--cycle; \draw[line width=.65pt] (0,.35)--(3,.35)--(3.38,2.58)--(.38,2.58)--cycle; \draw[fill=black!14,line width=.7pt] (.70,.70)--(1.39,.70)--(1.39,1.39)--(.70,1.39)--cycle; \draw[densely dashed,gray] (2.28,1.60) circle (.45); \node at (1.045,1.045) {$S$}; \end{scope} \node at (8.47,.24) {$Q\subset\R_x^2$}; \end{tikzpicture} \caption{The predictable intrinsic chart. Whitney squares and kernel estimates are defined on $\widehat Q$; their physical images are generally parallelograms. The local affine map preserves area, but not Euclidean lengths. The dashed intrinsic ball becomes an ellipse in physical coordinates.} \label{fig:intrinsic-physical} \end{figure} Take a fixed nonnegative smooth probability bump $\omega$ supported in $B(0,c_0/4)$ and define \[ \widehat\omega_Q(w)=D^{-2}\omega((w-z_0)/D),\qquad \int_{\widehat Q}\widehat\omega_Q=1,\quad \|D^j\widehat\omega_Q\|_\infty\le C_jD^{-2-j}\quad(j=0,1). \] Define the \emph{intrinsic} ray inverse by \begin{equation}\label{eq:ray} \widehat K_Q(z,w)=(z-w)\int_1^\infty t\widehat\omega_Q(w+t(z-w))\,dt, \qquad \widehat{\mathcal B}_Q\widehat f(z) =\int_{\widehat Q}\widehat K_Q(z,w)\widehat f(w)\,dw. \end{equation} For mean-zero $\widehat f$, integration along the ray proves $\divv_z\widehat{\mathcal B}_Q\widehat f=\widehat f$ and zero normal trace on $\partial\widehat Q$. The \emph{physical} Piola kernel and inverse are \begin{equation}\label{eq:piolakernel} K_Q^{\rm phys}(x,y)=A_Q\widehat K_Q(z,w),\qquad \mathcal B_Q^{\rm phys}f(x)=A_Q\widehat{\mathcal B}_Q\widehat f(z), \quad \widehat f(w)=f(b_Q+A_Qw), \end{equation} where $z=A_Q^{-1}(x-b_Q)$ and $w=A_Q^{-1}(y-b_Q)$. Because $\det A_Q=1$, no Jacobian factor is missing. A change of variables and the Piola identity give \begin{equation}\label{eq:weakray} \int_Q\mathcal B_Q^{\rm phys}f\cdot\nabla_x\phi\,dx =\int_{\widehat Q}\widehat{\mathcal B}_Q\widehat f \cdot\nabla_z\widehat\phi\,dz =-\int_Qf\phi\,dx,\qquad \int_Q f=0. \end{equation} The zero extension across the physical boundary has no flux. The inverse acts on the convex parent, not on its potentially disconnected terminal children. \paragraph{Intrinsic kernel bounds.} The scaled bump gives, for $z\ne w$, \begin{equation}\label{eq:singular} |\widehat K_Q(z,w)|\le\frac{C}{|z-w|},\qquad |D_z\widehat K_Q(z,w)|+|D_w\widehat K_Q(z,w)| \le\frac{C}{|z-w|^2}. \end{equation} Furthermore, \begin{equation}\label{eq:support} \widehat K_Q(z,w)=0\quad\hbox{if}\quad |z-w|>C\dist(z,\partial\widehat Q). \end{equation} Both properties are proved in Lemma~\ref{lem:app-ray} (Appendix~\ref{app:kernelchain}): \eqref{eq:singular} follows by differentiating the normalized bump under the integral, and \eqref{eq:support} from convexity, since a ray point $w+t(z-w)$ in the central bump forces $\dist(z,\partial\widehat Q)\ge cD/t$ while $|z-w|\le D/t$. In physical variables the uniform bounds read \[ |A_Q^{-1}K_Q^{\rm phys}(x,y)|\le\frac{C}{d_{G_Q^+}(x,y)}, \quad |A_Q^{-1}D_xK_Q^{\rm phys}(x,y)A_Q| +|A_Q^{-1}D_yK_Q^{\rm phys}(x,y)A_Q| \le\frac{C}{d_{G_Q^+}(x,y)^2}. \] The support condition is likewise expressed in $d_{G_Q^+}$. For example $A_Q=\operatorname{diag}(L,L^{-1})$ makes the raw physical kernel as large as $L$ times its intrinsic counterpart, and the bound $|K_Q^{\rm phys}(x,y)|\lesssim|x-y|^{-1}$ can fail by a factor of order $L^2$. Determinant one preserves area, not Euclidean lengths. \subsection{Whitney partitions in intrinsic coordinates} Truncating a Whitney decomposition and completing it by boundary sets of diameter $O(s)$ and area comparable to $s^2$ is the geometric construction of \cite[Lemma~5.1 and Appendix~A]{umc:AGT22}, also recalled in \cite[Section~2.3, Lemma~2.1]{GT-combinatorial}. Here it is carried out inside each predictably chosen intrinsic parent. The explicit receiver construction below also gives the child-size cap and the constants, uniform under successive changes of chart, which the adaptive construction requires. \phantomsection\label{not:whitneyparameters} The constants $\theta\in(0,1)$ and $D_0>0$ denote, respectively, the upper fraction of a parent's intrinsic diameter allowed for a Whitney square and the terminal stopping threshold $D\le D_0s$. The distinct contraction parameter $\vartheta\in(0,1)$ bounds the \emph{birth-chart} size ratio at an actual refinement: $\ell_{j+1}\le\vartheta\ell_j$, with one corresponding diameter contraction for a newly terminal union. It is achieved by choosing $\theta$ small and $D_0$ large after $K$ is fixed; Lemma~\ref{lem:birthchart} gives the precise constants. \begin{lemma}\label{lem:whitney} Fix a shape parameter $c_0>0$ and a sufficiently small fixed $\theta>0$. There are $\gamma,D_0,C$ depending only on $c_0,\theta$ so that, whenever $D>D_0s$, the \emph{intrinsic} parallelogram $\widehat Q$ admits a disjoint measurable partition into children $\widehat S$ with scales $\ell_S\ge s$: \begin{equation}\label{eq:whitney} |\widehat S|\asymp \ell_S^2,\quad \diam(\widehat S)\le C\ell_S,\quad \ell_S\asymp\max\{s,\dist(z,\partial\widehat Q)\} \quad(z\in\widehat S). \end{equation} Nonterminal children are \emph{intrinsic} dyadic squares with $\ell_S\le\theta D$; terminal children have $\ell_S=s$, area at least $s^2$ and intrinsic diameter at most $Cs$. After choosing $D_0$ sufficiently large, the maximal intrinsic child diameter is a fixed small fraction of $D$. The physical partition consists of $S=b_Q+A_Q\widehat S$. If $D\le D_0s$, the parent is left terminal. \end{lemma} Since $\det A_Q=1$, the physical children have the same areas as the intrinsic ones, but not in general the same Euclidean diameters. \begin{proof} Throughout this proof write $Q,S,R,x$ for \emph{intrinsic} subsets and points of $\widehat Q$; they are not physical coordinates. Use a dyadic lattice in this predictable intrinsic chart, with smallest squares of side $s$. Take \emph{all inclusion-maximal} dyadic squares $R\subset Q$ with side $\ell_R\in[2s,\theta D]$ and $\dist(R,\partial Q)\ge\gamma\ell_R$. Since dyadic squares are nested or disjoint, these maximal squares have disjoint interiors. A maximal square's dyadic parent fails admissibility: it either is not contained in $Q$, violates the distance condition, or exceeds $\theta D$. These three alternatives imply $\dist(x,\partial Q)\lesssim_{\gamma,\theta}\ell_R$ for $x\in R$. The reverse inequality follows from admissibility. More explicitly, for a square of side $\ell$, if its $2\ell$ parent is not contained in $Q$, every point of the child is within $2\sqrt2\ell$ of the boundary; if the parent fails the distance requirement, the same bound is $(2\gamma+2\sqrt2)\ell$; if it exceeds the cap, then $D<2\ell/\theta$. This proves the upper Whitney comparison, with constants fixed before the Gaussian experiment, so the nonterminal squares satisfy \eqref{eq:whitney}. Every remaining point of $Q$ belongs to an uncovered $s$-lattice square. If this $s$-square is wholly inside $Q$, retain it as a full uncovered square. Its $2s$-ancestor cannot be admissible: otherwise that ancestor would lie in some inclusion-maximal admissible square and would cover the $s$-square. The cap cannot fail since $\theta D>4s$. If the ancestor fails containment or boundary clearance, every point of the uncovered $s$-square lies within $(2\gamma+3\sqrt2)s$ of $\partial Q$. In particular every full uncovered square is within $Cs$ of the boundary of $Q$. It remains to assign a \emph{clipped} piece $P=Q\cap R$, where an $s$-square $R$ meets $\partial Q$, to a full uncovered $s$-square. Choose $x\in P$. Then $\dist(x,\partial Q)\le\sqrt2s$. \phantomsection\label{not:receiverconstants} Let $z_0$ be the center of an inball $B(z_0,c_0D)\subset Q$ and set $x_t=(1-t)x+tz_0$ with $t=A s/D$. Choose a constant $A=A(c_0)$ large enough that $c_0As>5s$, and choose $D_0>2A$ to ensure $t<1/2$. Convexity and the supporting-line characterization of boundary distance give \[ 5s<\dist(x_t,\partial Q)\le(\sqrt2+A)s, \qquad |x_t-x|\le As. \] The $s$-lattice square $R'$ containing $x_t$ is therefore fully inside $Q$. Take $\gamma$ larger than $(A+\sqrt2)/2$ (and enlarge it by an absolute constant if necessary). Then $R'$ cannot lie in an admissible dyadic square of side at least $2s$, because any such square would put $x_t$ at boundary distance at least $2\gamma s$. Hence $R'$ is an uncovered full receiver within $O(s)$ of $P$. Choose one receiver by a deterministic tie-break rule for every clipped $s$-square and join its clipped piece to that receiver. A receiver can get only $O_A(1)$ pieces because their originating lattice squares lie within $O(As)$; the union has area at least $s^2$, diameter $O_A(s)$, and all points have boundary distance $O_A(s)$. Any full uncovered square not receiving a clipped piece is a terminal child by itself. The maximal squares and these terminal unions partition $Q$ modulo a fixed measure-zero boundary convention; see Figure~\ref{fig:whitney-receivers}. Choose $D_0$ only after $A,\gamma,\theta$ and the terminal diameter constant $C_A$ have been fixed, so that $C_As\le\theta D$ whenever $D>D_0s$. The exact old/new-chart side ratio and the metric transport factor are recorded in Lemmas~\ref{lem:birthchart} and \ref{lem:ancestorindexed}. In particular, choose $\theta$ and then $D_0$ so that the birth-side shrink factor $\vartheta$ in those lemmas satisfies $\vartheta\sqrt K<1$. All selections can be made by a deterministic lexicographic rule. The construction terminates in at most $C_K\log(1/s)=O(m)$ steps: every nonterminal square has intrinsic area at most a fixed fraction of its parent's area, and every nonempty cell has area at least $c s^2$. No lower bound on the area of an individual clipped fragment is needed. \end{proof} \begin{figure}[tbp] \centering \begin{tikzpicture}[>=Latex,font=\footnotesize,line join=round] \def\Pleft{1.0} \def\Pright{9.1} % convex oblique parent; draw a clipped intrinsic s grid \begin{scope} \clip (1.0,.55)--(9.05,.55)--(7.67,4.45)--(-.38,4.45)--cycle; \fill[black!3] (-.5,.5) rectangle (9.2,4.6); \draw[step=.46cm,very thin,black!22] (-.5,.3) grid (9.35,4.6); \fill[black!12] (2.26,2.14) rectangle (3.18,3.06); \draw[line width=.7pt] (2.26,2.14) rectangle (3.18,3.06); % clipped boundary piece of a small lattice square \fill[black!37] (8.70,.76) rectangle (9.16,1.22); % an uncovered full receiver (schematic size s) \fill[black!22] (7.32,1.22) rectangle (7.78,1.68); \draw[line width=.85pt] (7.32,1.22) rectangle (7.78,1.68); \end{scope} \draw[line width=.85pt] (1.0,.55)--(9.05,.55)--(7.67,4.45)--(-.38,4.45)--cycle; \draw[dashed,black!65] (4.2,2.45) circle (.85); \fill (4.2,2.45) circle (.035); \node[anchor=south west] at (4.25,2.49) {$z_0$}; % extrapolation of the ray from clipped piece towards inball center \coordinate (x) at (8.84,.93); \coordinate (xt) at (7.68,1.31); \draw[->,line width=.85pt] (x)--(xt); \draw[densely dashed,gray] (xt)--(4.2,2.45); \fill (x) circle (.042); \fill (xt) circle (.042); \node[anchor=south west] at (8.88,.95) {$x$}; \node[anchor=south] at (7.55,1.75) {$x_t$}; \draw[->,thin] (2.72,3.57)--(2.72,3.09); \node[anchor=south] at (2.72,3.59) {interior dyadic square}; \draw[->,thin] (9.15,1.92)--(8.91,1.17); \node[anchor=west,align=left] at (9.13,2.01) {clipped\\piece}; \draw[->,thin] (6.59,.32)--(7.35,1.22); \node[anchor=north,align=center] at (6.20,.27) {full $s$-square receiver}; \end{tikzpicture} \caption{Clipped-square receivers in an intrinsic chart (schematic, not to scale). A boundary $s$-square may have an arbitrarily small intersection with $\widehat Q$. Moving $x$ toward the inball center gives $x_t$, contained in an uncovered full $s$-square. The clipped fragment is assigned to that receiver; only the resulting union is a terminal child and has area at least $s^2$.} \label{fig:whitney-receivers} \end{figure} The half-open boundary convention and the finite receiver assignment are spelled out in Lemma~\ref{lem:exactcover}. \subsection{Boundary receivers and the choice of constants} We fix the constants in the following order. Fix an inball ratio $c_0$. First take $A>5/c_0$ and $\gamma>(A+\sqrt2)/2$. Next choose $\theta>0$ small enough for subsequent metric contraction. Finally take $D_0$ so large that $D_0>2A$, $\theta D_0>4$, and every $O_A(s)$ terminal union has diameter at most a prescribed small fraction of $D$ in the current chart. The constants in the Whitney bounds may depend on $c_0,A,\gamma,\theta$ but \emph{not} on $D/s$, the lattice translation, or the physical eccentricity of a determinant-one chart. The two kinds of children behave differently. A maximal admissible square has its full area $\ell^2$, whereas a clipped mesh piece may have arbitrarily small area. The receiving full $s$-square provides the area floor for the union, and only this union is a child of the partition. At an acute corner the receiving square may lie along a different face, but the inball displacement gives bounded distance and bounded multiplicity without any angle-dependent connectivity hypothesis. The operator $\mathcal B_Q$ is applied to the convex parent, not to a terminal union. Only the area, diameter and Whitney bounds of the children enter the projected kernel estimates, so neither connectedness nor a John constant for terminal unions is used. \paragraph{Lattice conventions.} All squares are half-open after applying one fixed lexicographic boundary convention. The $s$-lattice belongs to the predictable intrinsic chart, not to the fixed physical grid; it is translated according to a preassigned measurable rule. At exact lattice contacts we choose one closed representative for the support test and assign measure-zero interfaces arbitrarily. No Gaussian sample is consulted when making any of these choices. A parent with $D\le D_0s$ is terminal, and no inverse is applied on its children. Lemma~\ref{lem:whitney} is thus a finite, measurable construction rather than a limiting Whitney decomposition. \paragraph{Notation for the projected kernels.} \phantomsection\label{not:holderparameters} We use $0<\alpha<1$ for the projected-kernel H\"older exponent. The Gaussian field and its spatial memory use a fixed smaller exponent $0<\beta<\alpha$; constants may depend on both exponents. Here $G_Q^+$ is the updated physical covariance metric, $A_Q=(G_Q^+)^{1/2}$, $z=A_Q^{-1}(x-b_Q)$ and $\widehat S=A_Q^{-1}(S-b_Q)$. The length $\ell_S$ and every distance in Lemma~\ref{lem:pk} are measured in this updated intrinsic chart. The averaging operator $\widehat P$ is intrinsic; a physical vector is recovered only after multiplying by $A_Q$. The fixed physical distance of the torus occurs only in Section~\ref{sec:terminal}. The cell-averaged singular-kernel estimates are related to the near- and far-cell bounds in \cite[Section~2.4, Lemma~2.6]{GT-combinatorial} and to the covariance calculation for the histogram correction in \cite[Section~4, Step~3]{umc:AGT22}. In the present argument, the boundary-distance support \eqref{eq:support} and the exact Gaussian covariance identity \eqref{eq:gramZ} give a pointwise conditional variance bound uniform in the intrinsic scale. The projected-kernel bounds needed for this are the following. \begin{lemma}\label{lem:pk} \phantomsection\label{not:childprojector} Let $\widehat P$ average over the children $\widehat S$ in the intrinsic chart $\widehat Q$, and let $z,z'$ belong to one $\widehat S$ of intrinsic scale $\ell$. For every fixed $0<\alpha<1$, \begin{align}\label{eq:pk} \|\widehat P\widehat K_Q(z,\cdot)\|_2&\le C,& \|\widehat P\widehat K_Q(z,\cdot)\|_\infty&\le C\ell^{-1},\\ \|\widehat P(\widehat K_Q(z,\cdot)-\widehat K_Q(z',\cdot))\|_2 &\le C(|z-z'|/\ell)^\alpha,& \|\widehat P(\widehat K_Q(z,\cdot)-\widehat K_Q(z',\cdot))\|_\infty &\le C\ell^{-1}(|z-z'|/\ell)^\alpha.\nonumber \end{align} The same inequalities hold after subtracting the parent-mean projection. They are uniform in the intrinsic chart. \end{lemma} Lemma~\ref{lem:pk} concerns the intrinsic kernel only; the passage to physical vectors is discussed after \eqref{eq:projectedpiola}. \begin{proof} In this proof, for compactness, write $Q,S,T,x,y$ and $K_Q,P$ for the intrinsic objects $\widehat Q,\widehat S,\widehat T,z,w$ and $\widehat K_Q,\widehat P$. Set $d_Q(x)=\dist(x,\partial Q)$. The Whitney bounds give $d_Q(x)\le C_W\ell$ for $x\in S$; by the support restriction, $K_Q(x,\cdot)$ is supported in $B(x,C\ell)$. If a child $T$ meets this ball, choose $y\in T\cap B(x,C\ell)$. The distance-to-boundary function is $1$-Lipschitz, so $\ell_T\le C_W\max\{s,d_Q(y)\}\le C\ell$. Thus $\diam T\le C\ell$ and the union of projected-support children lies in $B(x,C'\ell)$ and has area at most $C\ell^2$. For the one-point $L^\infty$ bound we separate near and far children. Choose $c_1>0$ smaller than the inverse of $10$ times every fixed Whitney comparison constant. For any child $T$ meeting $B(x,c_1\ell)$, \begin{equation}\label{eq:nearchild} |T|\ge c\ell^2. \end{equation} Indeed, if $\ell=s$ this is the global area floor $|T|\ge s^2$. If $\ell>s$, then $S$ is a nonterminal square and admissibility implies $d_Q(x)\ge\gamma\ell$; the $1$-Lipschitz property gives $d_Q(y)\ge\gamma\ell/2$ for $y\in B(x,c_1\ell)$. The pointwise Whitney comparison on $T$ implies $\ell_T\ge c\ell$, and $|T|\ge c\ell_T^2$. For these near children, \[ \left|\fint_T K_Q(x,y)\,dy\right| \le \frac{C}{|T|}\int_{B(x,C'\ell)}\frac{dy}{|x-y|} \le \frac{C\ell}{\ell^2}=\frac{C}{\ell}. \] On each other projected-support child $T$, all $y\in T$ satisfy $|x-y|\ge c_1\ell$, so the same bound follows pointwise from $|K_Q(x,y)|\le C/|x-y|$. Multiplication by the square root of the $O(\ell^2)$ projected-support area proves the $L^2$ estimate. For the two-point estimates put $d=|x-x'|$. If $d\ge c_1\ell/4$, apply the two one-point estimates and enlarge the constant. Suppose $01$, the determinant-one spectral clipping \[ \mathcal T_C(G)=G^{1/2} \operatorname{clip}_{[K^{-1},K]}\bigl(G^{-1/2}CG^{-1/2}\bigr)G^{1/2} \] satisfies, with $G^+=\mathcal T_C(G)$, \begin{equation}\label{eq:clipping} K^{-1}G\preceq G^+\preceq KG,\quad G^+\preceq K^{-1}G+C,\quad L(y,G^+)\le K^{-1/2}L(y,G)+C_K. \end{equation} \end{lemma} \begin{proof} Write $D_i=H^{1-2a_i}$. The rank-one determinant formula gives \[ \det B=D_1D_2\left(1+\sum_i\frac{v_i^2}{D_i}\right) =H\left(1+\sum_i v_i^2H^{-1+2a_i}\right)\asymp H^2, \] where the last comparison follows by examining the dominant summand $F_i(y_i)\ge H/2$. The formula for the inverse of a $2\times2$ positive matrix gives $C_{ii}=B_{jj}/\sqrt{\det B}\lesssim H^{2a_i}$. Also \[ C^{-1}=\frac{B}{\sqrt{\det B}} =\chi\operatorname{diag}(H^{-2a_i})+ww^T, \quad \chi=\frac H{\sqrt{\det B}}\asymp1, \quad w=\frac v{(\det B)^{1/4}}. \] One has $|\partial_i\chi|\lesssim H^{-a_i}$ and $|\partial_i w_j|\lesssim H^{1/2-a_i-a_j}$. Represent $L$ as the Euclidean norm of the direct sum of $\sqrt\chi\,G_{ii}^{1/2}H^{-a_i}$ and $G^{1/2}w$. Under the diagonal caps, differentiation of this vector proves \eqref{eq:lagder} without dividing by a possibly small $L$. The two eigenvalues of $G^{-1/2}CG^{-1/2}$ are reciprocal, and clipping both to $[K^{-1},K]$ preserves their product. The inequality $\operatorname{clip}(t)\le K^{-1}+t$ yields the Loewner bound. Along the common eigendirections the logarithmic mismatch is shortened by $\log K$. Separating the regimes of large and bounded log-mismatch, this gives $L(y,G^+)\le K^{-1/2}L(y,G)+C_K$. \end{proof} \section{The adaptive construction and the bootstrap}\label{sec:bootstrap} We now build the adaptive field $U_k$ generation by generation and prove that it stays in the regime where the estimates of Sections~\ref{sec:geom} and~\ref{sec:selector} apply. At each active parent the metric of the next increment is obtained by clipping the current metric toward the target $C$ of Section~\ref{sec:selector}; the fresh Gaussian increment is then pushed forward by this metric. The main result, Proposition~\ref{prop:bootstrap}, shows by induction that, as long as no envelope is too large and the modular has not overshot, the diagonal caps, the relative phase comparisons and a contraction of the lag-plus-memory functional $S_k$ hold simultaneously. These are the inputs of the exponential moment estimates of Section~\ref{sec:phase}. Throughout, every stage $k\to k+1$ has the same order. First the geometry is chosen from the old history: the representative, the metric $G_Q^+$, the intrinsic parent and its Whitney partition, and the stopping decisions are $\F_k$-measurable. Only then are the new Gaussian child masses revealed. Every quantity called old, predictable or old-measurable below is fixed before the noise of the current stage. \phantomsection\label{not:birthmetric} Initialize $U_0=0,G_0=I,\pr_0=\{Q_0\}$. At an active physical parent $Q$ in a chosen lift, choose a representative $x_Q$ by a fixed old-measurable rule, and set $C_Q=C(U_k(x_Q))$, $G_Q^+=\mathcal T_{C_Q}(G_k)$ and $A_Q=(G_Q^+)^{1/2}$. This matrix, the intrinsic parent $\widehat Q$, its bump and its Whitney partition are all fixed before the new child masses are revealed. After revelation let $\Delta f_k=(P_{\pr_{k+1}}-P_{\pr_k})W$ be the physical, parentwise mean-zero step density and put \phantomsection\label{not:intrinsicgauss} \[ \widehat{\Delta f}_k(w)=\Delta f_k(b_Q+A_Qw),\qquad Z_{k+1}(z)=\int_{\widehat Q}\widehat K_Q(z,w) \widehat{\Delta f}_k(w)\,dw. \] Thus $Z_{k+1}$ is an intrinsically normalized Gaussian vector field. The physical increment is its Piola push-forward: \begin{equation}\label{eq:update} V_{k+1}(x)=\sqrt\delta\,A_Q Z_{k+1}\bigl(A_Q^{-1}(x-b_Q)\bigr),\qquad U_{k+1}=U_k+V_{k+1},\qquad \divv_x U_{k+1}=\sqrt\delta\,f_{k+1}. \end{equation} Every such increment has zero normal trace on its physical parent, so its zero extension is globally periodic. In particular the local matrix $A_Q\in\mathrm{SL}(2,\mathbb R)$ is never used as a global coordinate change on $\T^2$. \begin{lemma} \label{lem:piolaGaussian} Conditional on the old filtration and on the predictably selected chart, $Z_{k+1}(z)$ is a centered Gaussian vector with covariance matrix \begin{equation}\label{eq:gramZ} \operatorname{Cov}_k\bigl(Z_{k+1}(z)\bigr)_{ab} =\int_{\widehat Q} \bigl[(\widehat P-\widehat P_{\widehat Q}) \widehat K_{Q,a}(z,\cdot)\bigr](w) \bigl[(\widehat P-\widehat P_{\widehat Q}) \widehat K_{Q,b}(z,\cdot)\bigr](w)\,dw. \end{equation} In particular, \begin{equation}\label{eq:covphysical} \operatorname{Cov}_k Z_{k+1}(z)\preceq CI, \qquad \operatorname{Cov}_k V_{k+1}(x)\preceq C\delta G_Q^+, \qquad \divv_x V_{k+1}=\sqrt\delta\,\Delta f_k. \end{equation} \end{lemma} \begin{proof} Fix the old history. The chart $A_Q$ and the partition are then deterministic. Pulling back a white-noise test $g$ by $x=b_Q+A_Qw$ preserves its $L^2$ inner product: the Jacobian is $\det A_Q=1$. The conditional increment $\Delta f_k$ is precisely the orthogonal Gaussian projection onto the span of child-constant functions with their parent constant removed. Pairing it with the two components of $\widehat K_Q(z,\cdot)$ gives the Gram matrix \eqref{eq:gramZ}. Lemma~\ref{lem:pk} provides the intrinsic $L^2$ bound, so the Gram matrix is $\preceq CI$. The covariance identity $\operatorname{Cov}_k(V)=\delta A_Q\operatorname{Cov}_k(Z)A_Q^T$ then proves the physical bound, since $A_QA_Q^T=G_Q^+$. For the divergence, $D_x V=\sqrt\delta\,A_Q(D_zZ)A_Q^{-1}$, so $\operatorname{tr}D_xV=\sqrt\delta\operatorname{tr}D_zZ =\sqrt\delta\widehat{\Delta f}_k$. Equivalently, the weak identity \eqref{eq:weakray} proves the result without pointwise differentiation. The Piola normal-flux identity preserves the no-flux boundary condition. \end{proof} The physical field does not, in general, have covariance bounded by $C\delta I$; its covariance is controlled only through the anisotropic metric $G_Q^+$. The next lemma gives uniform sub-Gaussian control of the normalized field $Z_{k+1}$ and of its H\"older seminorm on each child. \begin{lemma}\label{lem:envelope} Fix $0<\beta<\alpha<1$. At generation $k$, condition on $\F_k$. Let $Q$ be an active parent, let its intrinsic chart and Whitney children be selected $\F_k$-measurably as above, and let $Z_{k+1}$ be the centered intrinsic field in \eqref{eq:update}. For each newly revealed child $S=b_Q+A_Q\widehat S$ of intrinsic scale $\ell_S$, choose the continuous version of $Z_{k+1}$ on $\overline{\widehat S}$ and set \begin{equation}\label{eq:envelope} E_{k+1,S}:= \sup_{z\in\overline{\widehat S}}|Z_{k+1}(z)| +\ell_S^\beta[Z_{k+1}]_{C^\beta(\overline{\widehat S})}. \end{equation} There are deterministic $c,C>0$, depending only on $\alpha,\beta$ and the uniform intrinsic Whitney and kernel constants, such that, for every $t\ge0$, \begin{equation}\label{eq:envelopetail} \Pp(E_{k+1,S}>t\mid\F_k)\le Ce^{-ct^2}, \qquad \E_k e^{cE_{k+1,S}^2}\le C, \end{equation} possibly after reducing $c$. These estimates hold for each $\F_k$-measurably selected child and are uniform in $k,m$, its intrinsic scale, and the eccentricity of $A_Q$. The H\"older seminorm in \eqref{eq:envelope} is computed with the intrinsic $z$-distance; its physical pullback uses $d_{G_Q^+}(x,x')=|A_Q^{-1}(x-x')|$. If $J_0\le C_{\rm dep}m$ is the deterministic reveal-depth bound, $s=2^{-m}m^3$, and each child has area at least $c_*s^2$, then for a sufficiently large fixed \emph{global envelope cutoff} $R>0$, \begin{equation}\label{eq:globalcut} \Pp\!\left(\exists\text{ an actually attempted reveal }(k+1,S): E_{k+1,S}>R\sqrt m\right) \le C J_0s^{-2}e^{-cR^2m} \le e^{-c_Rm} \end{equation} for all sufficiently large $m$, with $c_R>0$. \end{lemma} \begin{proof} We work conditionally on $\F_k$ and write $\Pp_k(\cdot)=\Pp(\cdot\mid\F_k)$; in particular the chart, $\widehat Q$, the averaging projector and the child $\widehat S$ are deterministic. By the Gram formula \eqref{eq:gramZ} and the projected evaluations \eqref{eq:pk}, \begin{equation}\label{eq:envelopecanonical} \sup_{z\in\widehat S}\E_k|Z_{k+1}(z)|^2\le C_0, \qquad \E_k|Z_{k+1}(z)-Z_{k+1}(z')|^2 \le C_0\left(\frac{|z-z'|}{\ell_S}\right)^{2\alpha} \quad(z,z'\in\widehat S). \end{equation} The same bounds extend to the closure by continuity. Both components are centered Gaussian. The constants do not see the Piola matrix, because \eqref{eq:envelopecanonical} concerns the normalized field $Z_{k+1}$. We bound $E_{k+1,S}$ by a chaining argument. Set $\ell=\ell_S$ and fix an exponent $\gamma$ with $\beta<\gamma<\alpha$. The intrinsic diameter bound $\operatorname{diam}(\widehat S)\le C_W\ell$ allows us to construct, for every integer $j\ge0$, a finite $2^{-j}\ell$-net $N_j\subset\overline{\widehat S}$ with $|N_j|\le C2^{2j}$: take a fixed countable dense set in the child, run a deterministic greedy procedure with a fixed order, and retain the finitely many selected points. Since the intrinsic chart is predictable, the nets are measurable with respect to the old history. Each point $z\in\overline{\widehat S}$ has a representative $z_j\in N_j$ at distance at most $C2^{-j}\ell$. The nets need not be nested, nor need $\widehat S$ be connected. At level zero test the two Gaussian coordinates at all points of $N_0$. At every level $j\ge0$ test all coordinate differences between pairs of points in $N_j\cup N_{j+1}$ whose separation is at most $C_1 2^{-j}\ell$, where $C_1$ is chosen to cover both consecutive-level approximations and pairs of nearest representatives of nearby points. There are at most $C2^{4j}$ such tests. By \eqref{eq:envelopecanonical}, each tested difference has conditional standard deviation at most $C2^{-\alpha j}$. For $A\ge1$ and $u\ge0$, let $\mathcal G_u$ be the event on which every level-zero coordinate has absolute value at most $A(1+u)$ and every tested difference at level $j$ has absolute value at most $A(1+u)2^{-\gamma j}$. The scalar Gaussian tail bound and a union bound give \begin{align}\label{eq:chainingfailure} \Pp_k(\mathcal G_u^c) &\le Ce^{-cA^2(1+u)^2} +C\sum_{j\ge0}2^{4j} \exp\!\left[-cA^2(1+u)^2 2^{2(\alpha-\gamma)j}\right] \le C'e^{-c'u^2}. \end{align} Here $A$ is fixed sufficiently large. The series converges uniformly: since $\alpha-\gamma>0$, its exponential factor dominates $2^{4j}$, and $(1+u)^2\ge1+u^2$ separates the $u$-tail. The increment threshold is $2^{-\gamma j}$, without a factor growing with $j$; this is what yields a genuine $C^\beta$ bound. On $\mathcal G_u$, telescope the values at the representatives $z_j$ of a point $z$ from level zero to arbitrarily fine levels. Consecutive representatives are $O(2^{-j}\ell)$ apart, so every increment occurs among the tested pairs. The convergent geometric series $\sum_{j\ge0}2^{-\gamma j}$ gives \[ \sup_{\widehat S}|Z_{k+1}|\le C A(1+u). \] For $z,z'$ with $d=|z-z'|\le\ell$, choose $j$ so that $2^{-j}\ell\asymp d$. Their level-$j$ representatives are $O(2^{-j}\ell)$ apart, hence their difference is one of the tested pairs; the two finer telescopes cost at most $CA(1+u)2^{-\gamma j}$. Consequently, \[ |Z_{k+1}(z)-Z_{k+1}(z')| \le CA(1+u)(d/\ell)^\gamma \le CA(1+u)(d/\ell)^\beta. \] For $d>\ell$ the supremum bound gives the same $C^\beta$ estimate. Taking suprema over a fixed countable dense set and extending to the continuous version makes $E_{k+1,S}$ measurable. We conclude from \eqref{eq:chainingfailure} that $\Pp_k(E_{k+1,S}>C A(1+u))\le C'e^{-c'u^2}$, which, after adjusting constants, proves the first inequality in \eqref{eq:envelopetail}. Tail integration gives, for $c$ smaller than the tail exponent, \[ \E_k e^{cE_{k+1,S}^2} =1+\int_0^\infty 2ct e^{ct^2} \Pp_k(E_{k+1,S}>t)\,dt\le C. \] Finally, the area floor implies that at each generation there are at most $(c_*s^2)^{-1}$ children, including terminal unions. Because the next subdivision is $\F_k$-measurable, the conditional bound in \eqref{eq:envelopetail} can be summed over all children selected for the next actual reveal, and then averaged over the past. A further union bound over the $J_0$ generations gives the first inequality of \eqref{eq:globalcut}. The envelopes on distinct children need not be independent for this argument. Since $s=2^{-m}m^3$ and $J_0\le C_{\rm dep}m$, \[ C J_0s^{-2}e^{-cR^2m} \le C' m^{-5}\exp\bigl[-(cR^2-2\log2)m\bigr]. \] Choose fixed $R$ with $cR^2>2\log2$. This proves the second inequality in \eqref{eq:globalcut}. \end{proof} \paragraph{Local estimates used in the induction.} Conditionally on its old history, every active parent satisfies the following, by Lemmas~\ref{lem:whitney}, \ref{lem:pk}, \ref{lem:piolaGaussian} and~\ref{lem:envelope}. The chart and the entire child partition are predictable, and every child has area at least $c_*s^2$. Active children contract in birth charts, while terminal children have intrinsic diameter $O(s)$; the contraction is quantified in Lemma~\ref{lem:birthchart}, and its use along an entire ancestry is recorded in Lemma~\ref{lem:actualphase}. The increment $V_{k+1}=\sqrt\delta A_QZ_{k+1}$ has zero normal trace on its parent and divergence $\sqrt\delta\,\Delta f_k$. For $x=b_Q+A_Qz$ in a child $S$, \[ \operatorname{Cov}_k V_{k+1}(x)\preceq C\delta G_Q^+, \qquad |V_{k+1,i}(x)|\le \sqrt\delta\sqrt{(G_Q^+)_{ii}}\,E_{k+1,S}, \qquad \E_k e^{cE_{k+1,S}^2}\le C, \] and the same envelope controls the intrinsic $C^\beta$ seminorm. The constants are independent of the physical eccentricity, whose effect is confined to the factors involving $G_Q^+$. \paragraph{Lag and memory.} For each $x$, let $E_j(x)$ be the envelope associated to its stage-$j$ child, or zero on an inactive generation. Fix a deterministic memory discount $0<\rho_1<1$; it is chosen in Section~\ref{subsec:parameters} so that the square of the geometric ancestor-decay factor ($r_{\rm osc}$ in Lemma~\ref{lem:actualphase}) is strictly smaller than $\rho_1$. The subscript in $\rho_1$ is part of its name, not a generation index. Set \begin{equation}\label{eq:memory} T_0=0,\quad T_{k+1}=\rho_1T_k+E_{k+1}^2,\quad L_k(x)=L(U_k(x),G_k(x)),\quad S_k=L_k^2+T_k. \end{equation} Both terms of $S_k$ are nonnegative. The memory $T_k$ records old oscillations with an exponentially discounted weight, and $L_k^2$ measures how far the current routing metric lags behind the instantaneous Bellman metric. The recurrence for $T_k$ applies only at actual reveals: on a frozen leaf, $T_k$, $L_k$ and $S_k$ remain unchanged, so padding the filtration does not discount the past. \paragraph{Stopping rules.} Fix the deterministic modular stop threshold $T_*>0$. The construction uses exactly two stopping rules, both checked before the next subdivision is selected. If some previously attempted child envelope exceeded $R\sqrt m$, the history has been killed. If the old modular \phantomsection\label{not:modularfirst} $\mathcal Z_k:=\int_{\T^2}e^{\Phi_k}$ (see \eqref{eq:Phi}) satisfies $\mathcal Z_k\ge T_*$, the history has overshot. Otherwise the next subdivision and metric update are chosen, the new child masses are revealed, and their envelopes are computed. If a new envelope exceeds $R\sqrt m$, we kill the scalar stopped supermartingale. If an overshoot occurs, we retain the attained modular value and freeze further geometric refinement. The diagonal caps and the relative phase comparisons are not tested by the construction. On every history that has been neither killed nor stopped by overshoot they are consequences of Proposition~\ref{prop:bootstrap}. This separates a random failure event from the field estimates valid on successful histories. \paragraph{Birth and reveal convention.} A stage-$j$ square $Q_j$ is born with metric $G_j$, which is also the physical covariance metric of the increment $V_j=\sqrt\delta\,G_j^{1/2}Z_j$. On an active parent $Q_j$ the predictable next metric is $G_j^+$, and a newly created child is born with $G_{j+1}=G_j^+$. A child created as a clipped terminal union receives that increment once and then freezes. A square stopped directly at the threshold is not updated and reveals no further Gaussian increment. At padded times, the exponent of the ancestor contraction is the number of genuine reveals, not the calendar difference. We use essential suprema across cell boundaries and pointwise continuous versions on each cell interior. \subsection{Auxiliary estimates} \phantomsection\label{not:relativephase} Throughout the induction, $0<\eta<1$ is the fixed relative phase tolerance per actual Gaussian reveal: $(1-\eta)H(U_{j-1})\le H(U_j)\le(1+\eta)H(U_{j-1})$. It is chosen small before the Gaussian step size $\varepsilon$. There is no new phase comparison on a frozen stage. The two lemmas below are the deterministic inputs of the induction. Lemma~\ref{lem:actualphase} bounds the oscillation of $U_b$ on a cell by the memory $T_b$, summing only over actual reveals and not over padded calendar times. Lemma~\ref{lem:lagsegments} identifies short anisotropic segments along which the lag may be differentiated using only the old diagonal cap. Neither follows from the local ray-kernel estimate alone. \begin{lemma}\label{lem:actualphase} Let $Q_0\supset\cdots\supset Q_b$ be the actual ancestral chain of a point, with exactly one new Gaussian increment $V_j$ revealed when $Q_j$ is born, $1\le j\le b$. The birth metric $G_j$ and intrinsic length $\ell_j$ belong to $Q_j$, and $E_j(Q_j)$ is the intrinsic $C^\beta$ envelope of the normalized field $Z_j$, before the Piola factor $G_j^{1/2}$. Suppose the already certified caps and phase inequalities hold at each reveal: \[ (G_j)_{ii}\le M H(U_j)^{2a_i},\qquad (1-\eta)H(U_{j-1})\le H(U_j)\le(1+\eta)H(U_{j-1}). \] If the actual child contractions and metric comparisons of Lemma~\ref{lem:ancestorindexed} hold, define the effective ancestor-oscillation ratio $r_{\rm osc}$ by \[ r_{\rm osc}=(\vartheta\sqrt K)^\beta(1-\eta)^{-a_*}, \qquad r_{\rm osc}^2<\rho_1<1. \] Then for $x,z$ in the same stage-$b$ cell (whether this cell is active, newly terminal, or directly stopped), \begin{equation}\label{eq:actualphasetransfer} |U_{b,i}(x)-U_{b,i}(z)| \le \frac{C\sqrt{\delta M}}{\sqrt{1-r_{\rm osc}^2/\rho_1}} H(U_b(x))^{a_i}\sqrt{T_b(x)}, \qquad T_b(x)=\sum_{j=1}^{b}\rho_1^{b-j}E_j(Q_j(x))^2. \end{equation} The sum is over actual revelations only. After a cell becomes terminal, $U$, $G$ and $T$ are frozen there and no extra contraction factor is generated by padding the filtration to the deterministic horizon. \end{lemma} \begin{proof} We bound each revealed increment separately and sum. For $1\le j\le b$, the pair $x,z$ lies in the actual stage-$j$ child on which $V_j=\sqrt\delta G_j^{1/2}Z_j$ was constructed. The row bound for the Piola transform and the intrinsic envelope give \[ |V_{j,i}(x)-V_{j,i}(z)| \le\sqrt\delta\sqrt{(G_j)_{ii}}\,E_j(Q_j) \left(\frac{d_{G_j}(x,z)}{\ell_j}\right)^\beta. \] The $j$-th cap and the inequalities along the actual chain imply \[ \sqrt{(G_j)_{ii}} \le \sqrt M(1-\eta)^{-a_i(b-j)}H(U_b(x))^{a_i}. \] By Lemma~\ref{lem:ancestorindexed}, with a final terminal transition counted only when it was actually revealed, $d_{G_j}(x,z)/\ell_j\le C(\vartheta\sqrt K)^{b-j}$. Since $U_0$ is spatially constant, summing the revealed increments and using $a_i\le a_*$ yields \[ |U_{b,i}(x)-U_{b,i}(z)|\le C\sqrt{\delta M}H(U_b(x))^{a_i} \sum_{j=1}^{b}r_{\rm osc}^{b-j}E_j(Q_j(x)). \] The weighted Cauchy--Schwarz inequality, applied over actual reveals only, gives \[ \sum_{j=1}^{b}r_{\rm osc}^{b-j}E_j \le\left(\sum_{n\ge0}(r_{\rm osc}^2/\rho_1)^n\right)^{1/2} \left(\sum_{j=1}^{b}\rho_1^{b-j}E_j^2\right)^{1/2}. \] For a directly stopped cell $Q_b$ there is no subsequent increment; for a newly terminal union the $b$-th increment is included once. After termination the field does not change and the inequality is retained without further discounting. \end{proof} \begin{lemma}\label{lem:lagsegments} Fix the finite critical exponents and $K,M<\infty$. There are $b_0>0$ and $C=C(K,M,\boldsymbol r)$ such that the following holds. Let $y\in\R^2$, let $G$ be positive definite with $\det G=1$ and $G_{ii}\le M H(y)^{2a_i}$, and let $\widetilde G$ satisfy $0\prec\widetilde G\preceq K G$ and $\det\widetilde G=1$. Suppose $y^{(1)},y^{(2)}$ belong to the anisotropic rectangle \[ \mathcal R_b(y)=\{w:|w_i-y_i|\le bH(y)^{a_i},\ i=1,2\}, \qquad 01$ and choose the constants in the order of Section~\ref{subsec:parameters}. Then the following holds for every $k\ge0$. \begin{enumerate}[label=(\roman*),leftmargin=2.7em] \item On $\mathcal H_k$, before the stage-$(k+1)$ revelation, every active parent $Q$ and almost every $x\in Q$ satisfy \begin{equation}\label{eq:Hcap} H(U_k(x))\le C_{T_*}m, \end{equation} and the predictable metric $G^+=G_Q^+$ satisfies \eqref{eq:margin} below. \item On $\mathcal H_k\cap\mathcal G_{k+1}$, simultaneously for all new children and almost every point (and for the chosen continuous representatives inside each child), \begin{align}\label{eq:threecaps} (G_{k+1})_{ii}(x)&\le M H(U_{k+1}(x))^{2a_i},\\ (1-\eta)H(U_k(x))&\le H(U_{k+1}(x)) \le(1+\eta)H(U_k(x)),\nonumber\\ S_{k+1}(x)&\le q_0S_k(x)+C(1+E_{k+1}(x)^2).\nonumber \end{align} On an inactive cell, $U$, $G$ and $S$ are unchanged, so the first two bounds persist and $S_{k+1}=S_k$. \end{enumerate} In particular, $\mathcal H_k\cap\mathcal G_{k+1}\cap\{\mathcal Z_{k+1}0$. Moreover, with $q_d<(1-\eta)^{2a_*}$ (Section~\ref{subsec:parameters}), \eqref{eq:margin} and this phase comparison preserve the diagonal cap $(G_{k+1})_{ii}\le M H(U_{k+1})^{2a_i}$. This proves the first two lines of \eqref{eq:threecaps} for all new children simultaneously. \textbf{Step 4: the old modular gives a pointwise phase cap.} On an active parent $Q$, the oscillation bound \eqref{eq:osc} of Step~1 makes $H(U_k)$ comparable throughout $Q$. Every active $Q$ has area at least $cs^2$. By (H3), $\mathcal Z_k\le T_*$, and $\lambda_k\ge\lambda_{J_0}\ge1/2$ by the choice of $\varepsilon$ after $\Lambda$ and $C_{\rm dep}$ (Section~\ref{subsec:parameters}). Integration over $Q$ gives \[ T_*\ge\int_Qe^{\lambda_kH(U_k)} \ge c|Q|e^{cH(U_k(x))}, \] so $H(U_k(x))\le C\bigl(1+\log(T_*/s^2)\bigr)\le C_{T_*}m$. This proves \eqref{eq:Hcap}, which is $\F_k$-measurable. \textbf{Step 5: contraction of the lag and memory.} Set $y=U_k(x)$, $y_Q=U_k(x_Q)$ and $y'=U_{k+1}(x)$. Steps~1 and~3 establish $|y_{Q,i}-y_i|+|y'_i-y_i|\le bH(y)^{a_i}$ with $b\le C\varepsilon R\sqrt M$. Both metrics $G_k$ and $G^+$ satisfy the old diagonal cap, with $G^+\preceq KG_k$ and $\det G^+=1$. Once $\varepsilon$ is chosen so that $b\le b_0$, Lemma~\ref{lem:lagsegments} applies with $y^{(1)}=y_Q$ and $y^{(2)}=y'$. It gives the cap throughout the segments $[y,y_Q]$ and $[y,y']$, which permits integration of \eqref{eq:lagder} along them; no conclusion of the lag update is used. At the representative the clipping contracts the lag, $L(y_Q,G^+)\le K^{-1/2}L(y_Q,G_k)+C_K$ by \eqref{eq:clipping}. Together with the segment estimates, this gives \[ L_{k+1}(x) \le K^{-1/2}L_k(x)+C_K +C_M\sqrt{\delta H(U_k(x))} \bigl(\sqrt{T_k(x)}+E_{k+1}(x)\bigr). \] By Step~4, $\sqrt{\delta H}\le C_{T_*}\varepsilon$. \phantomsection\label{not:qL} Fix the lag-contraction factor $q_L>K^{-1}$ and a Young slack $\tau>0$ with $(1+\tau)K^{-1}\le q_L$. Squaring with this slack gives \[ L_{k+1}^2\le q_LL_k^2+C_\tau+C_\tau\varepsilon^2T_k +C_\tau\varepsilon^2E_{k+1}^2. \] Since $T_{k+1}=\rho_1T_k+E_{k+1}^2$, choose $\max\{q_L,\rho_1\}1$, $0<\beta<\alpha<1$ & Metric distortion and H\"older exponents; arbitrary.\\ $c_0$; $A$, $\gamma$ & Inball ratio of intrinsic parents, depending on $K$; then the Whitney displacement and clearance constants, depending only on $c_0$ (Lemma~\ref{lem:whitney}).\\ $q_d,\,q_L\in(K^{-1},1)$; $\tau$ & Cap and lag contraction factors; Young slack $\tau>0$ with $(1+\tau)K^{-1}\le q_L$ (Step~5 of Proposition~\ref{prop:bootstrap}).\\ $\eta$, $q_0$ & $q_d<(1-\eta)^{2a_*}$, and $q_0$ with $q_L\mathcal Z_0$, large enough for a positive success probability (Proposition~\ref{prop:stopped}).\\ $R$ & $cR^2>2\log2$, with $c$ the envelope tail constant (Lemma~\ref{lem:envelope}).\\ $\varepsilon$ & Last: $b\le b_0$ in Lemma~\ref{lem:lagsegments}; $\xi<\eta$; absorption of the representative error in \eqref{eq:margin} and of $\varepsilon^2T_k$ in Step~5; $\lambda_{J_0}\ge1/2$, i.e.\ $\Lambda C_{\rm dep}\varepsilon^2\le\log2$; the smallness conditions of Section~\ref{sec:phase}; the terminal corrections of Section~\ref{sec:terminal}.\\ \bottomrule \end{tabular} \caption{Order of the parameter choices in Part~\ref{umc:partI}.} \label{tab:parameters} \end{table} \noindent In particular, \begin{equation}\label{eq:strictschedule} r_{\rm osc}^2<\rho_10, \qquad \ell=\lambda\sqrt\delta\,A_Q^Tv(y) =\lambda\sqrt\delta\,(G_Q^+)^{1/2}v(y). \] In particular, $\ell$ is measurable \emph{before} the fresh Gaussian variables are revealed. \end{lemma} \begin{proof} Taylor's formula and $F_i''(t)\lesssim H^{1-2a_i}$ along the short segment give \begin{equation}\label{eq:taylor} H'-H\le v(y)\cdot V_{k+1}+A\delta H E^2, \qquad |H'-H|\le\xi H. \end{equation} By the memory inequality $S'\le q_0S+A(1+E^2)$ and positivity, \[ H'S'-HS \le (1+\xi)H\bigl(q_0S+A(1+E^2)\bigr)-HS =-d_0HS+A(1+\xi)H(1+E^2). \] Choose $\xi$ so that $q_0(1+\xi)<1$. Now use the exact algebraic identity \[ \Phi_{k+1}-\Phi_k =-(\lambda_k-\lambda)H(1+\kappa\delta S) +\lambda(H'-H)+\kappa\lambda\delta(H'S'-HS). \] The clock gives $\lambda_k-\lambda\ge\Lambda\lambda\delta$. Insert the two preceding bounds and $V_{k+1}=\sqrt\delta A_QZ_{k+1}$. This proves \eqref{eq:path}. \end{proof} \begin{lemma}\label{lem:joint} Suppose, conditionally on $\F_k$, $Z$ is a centered Gaussian vector with covariance at most $\sigma^2I$ and an accompanying (not necessarily independent) envelope satisfies $\E_ke^{cE^2}\le C$. For $0\le u\le u_0C_K$ and then $\Lambda>C_\kappa$ makes the right-hand side nonpositive. \end{proof} Because the tilt $\ell$ is predictable and the fresh field enters linearly, the estimate involves no $\delta H^2$ term and no fourth cumulant. \subsection{Stopping rule} The one-step estimate is applied up to a stopping time. We define the numerical process on all outcomes of each attempted reveal. Let $J_0$ be a fixed upper bound on active depth, and let $\F_k$ be the completed sigma-field generated by the geometry and all Gaussian cell masses revealed through stage $k$; as described in Section~\ref{sec:bootstrap}, the entire next subdivision is $\F_k$-measurable. We pad terminal branches to the deterministic horizon $J_0$ by doing nothing. The old good event is $A_k:=\mathcal H_k\in\F_k$, defined before Proposition~\ref{prop:bootstrap}: previous envelopes were acceptable and the last attained modular is strictly below $T_*$; the caps and relative-phase bounds then hold by that proposition. At stage $k+1$ write \[ B_{k+1}=\mathcal G_{k+1} =\{\text{every newly revealed child envelope is }\le R\sqrt m\} \in\F_{k+1}, \] where no new child is assigned an envelope on a frozen branch. Define a nonnegative adapted scalar process $\widehat{\mathcal Z}_k$, the killed-and-frozen modular, as follows: \begin{enumerate}[label=(\roman*),leftmargin=2.7em] \item Initially $\widehat{\mathcal Z}_0=\mathcal Z_0$. \item Before failure or overshoot, attempt the next geometric refinement. If $B_{k+1}$ fails, put $\widehat{\mathcal Z}_{k+1}=0$ and thereafter keep the numerical process zero. \item If $B_{k+1}$ holds, use the actual new modular $\widehat{\mathcal Z}_{k+1}=\mathcal Z_{k+1}$. If it is at least $T_*$, freeze the geometric field and its memory for all later stages, allowing only $\lambda$ to decrease. \item On a branch that became terminal for geometric reasons, freeze $(U,S)$ on that branch from then on, while the clock continues to decrease. \end{enumerate} \begin{proposition} \label{prop:stopped} The process $\widehat{\mathcal Z}_k$ is a nonnegative supermartingale and \begin{equation}\label{eq:overshoot} \Pp\bigl(\exists j\le J_0:\,\widehat{\mathcal Z}_j\ge T_*\bigr) \le \frac{\mathcal Z_0}{T_*}. \end{equation} The global envelope-failure probability is $o(1)$. Consequently, for large fixed $T_*$, there is a constant $p_*>0$ independent of $m$ such that with probability at least $p_*$ all leaves terminate and the leaf current satisfies \begin{equation}\label{eq:leafbound} \divv J_{\mathrm{leaf}}=P_{\pr_*}W, \qquad \max_i\norm{J_{\mathrm{leaf},i}}_{\psi_{r_i}} \le C\delta^{-1/2}\le C_{\boldsymbol r}\sqrt m. \end{equation} \end{proposition} \begin{proof} We first prove the supermartingale property, case by case. Suppose the past has neither failed nor overshot. For each active $x$, Propositions~\ref{prop:bootstrap} and~\ref{prop:mgf} give \[ \E_k\left[\one_{B_{k+1}}e^{\Phi_{k+1}(x)}\right] \le\E_k\left[\one_{\{E_{k+1}(x)\le R\sqrt m\}} e^{\Phi_{k+1}(x)}\right] \le e^{\Phi_k(x)}. \] The first inequality holds because the global envelope event is contained in the local one. On an inactive branch the field and memory are frozen and the clock decreases, so the integrand can only decrease and the same inequality holds. Integrating over $x$ and using conditional Tonelli, we obtain \[ \E_k[\one_{B_{k+1}}\mathcal Z_{k+1}]\le\mathcal Z_k. \] Since the definition sets $\widehat{\mathcal Z}_{k+1}=0$ on $B_{k+1}^c$, the left side is exactly the conditional expectation of the numerical process at the new stage, until the first overshoot. After the overshoot there is no further geometric evolution, and $\lambda_{k+1}\le\lambda_k$ ensures $\widehat{\mathcal Z}_{k+1}\le \widehat{\mathcal Z}_k$ pathwise. After a failure the process is zero. No conditioning on the success of a future reveal is involved. The initial values are deterministic: $U_0=0$, $H(U_0)=2$, $G_0=I$ and $T_0=0$; hence $\mathcal Z_0=\exp(2(1+\kappa\delta L(0,I)^2))=O(1)$. For $\tau=\inf\{k:\widehat{\mathcal Z}_k\ge T_*\}$, Ville's inequality yields \eqref{eq:overshoot} for the killed process. Independently of this stopping argument, Lemma~\ref{lem:envelope} bounds the probability of any actual attempted envelope failure by $o(1)$, using conditional tails and the deterministic bound on the number of attempted children. Write $\mathcal E_m$ for the event of any attempted envelope failure and $\mathcal O_m$ for the first overshoot along an all-envelope-good history. The successful completion event satisfies \[ \mathbb P(A_m)\ge1-\mathbb P(\mathcal O_m) -\mathbb P(\mathcal E_m) \ge1-\mathcal Z_0/T_*-o(1). \] Thus $T_*$ may be fixed sufficiently large so that this lower bound exceeds some $p_*>0$. On this good completion event the final positive modular is at most $T_*$. Since $\lambda_{J_0}\ge1/2$ (after choosing $\varepsilon$ small), $\Phi_{J_0}\ge\frac12 H(U_{J_0})$. The fixed exponential modular bound implies the coordinate Luxemburg bounds for $U_{J_0}$ after an absolute dilation, and dividing by $\sqrt\delta$ proves \eqref{eq:leafbound}. \end{proof} The threshold in \eqref{eq:overshoot} is tested on the killed process $\widehat{\mathcal Z}_k$, so Ville's bound concerns a first overshoot along a history on which all envelopes are good. On a killed history the process is not the modular of an admissible current. The actual-reveal geometry, including direct stops, terminal unions and padded times, is recorded in Lemma~\ref{lem:actualreveal}. The next section corrects the adaptive current from $P_{\pr_*}W$ to the fixed-grid source $W_h$. \section{The terminal Gaussian correction}\label{sec:terminal} The adaptive field from Proposition~\ref{prop:stopped} solves the cell-averaged source $PW$, where $P$ averages over the terminal leaves of a random partition. The target source is $W_h$, obtained by averaging over the fixed $h$-grid. In this section we construct a current with divergence $W_h-PW$ and show that it is negligible in the Orlicz norms under consideration; this completes the proof of Theorem~\ref{thm:gauss}. The difference $W_h-PW$ splits exactly into two pieces. The first is built from the revealed averages $PW$ and has zero mean on each fixed $h$-square, so it is inverted square by square. The second is the fixed-grid average of the unrevealed residual $W-PW$; conditionally on the terminal history it is Gaussian, and we invert it by convolution with the periodic Green kernel. On the torus this is a periodic convolution, with no physical boundary data. \subsection{The exact splitting} On a successful history, every terminal leaf $S$ has area at least $cs^2$ and diameter $O(s)$ in its birth metric $G_S$. This includes both the clipped unions born at a split and the active squares stopped without another split; by Lemma~\ref{lem:birthchart}, the latter have intrinsic diameter at most $C D_0s$. Write $A_S=G_S^{1/2}$ and $S=b_S+A_S\widehat S$ in its chosen lift. The diagonal cap holds in physical coordinate directions, so at the final good stage $H(U)\le C_{T_*}m$ gives \[ \|A_S\|_{\rm op}^2\le\operatorname{tr}G_S \le M\sum_{i=1}^2H(U)^{2a_i}\le C_{M,T_*}m^{2a_*}. \] We now pass from the local intrinsic geometry to the fixed physical flat-torus metric, in which \begin{equation}\label{eq:Dstar} \diam_{\rm physical}(S)\le D_*:=Csm^{a_*}, \qquad a_*<\frac12,\qquad D_*/h=Cm^{3+a_*}. \end{equation} This bound uses the cap: with $Cs$ in place of $D_*$ it would fail to hold uniformly in the eccentricity. \begin{lemma} \label{lem:terminalhistory} On the successful first-failure event, every terminal leaf $S$, including a directly stopped square and a clipped union born at the last reveal, satisfies \[ |S|\ge cs^2,\qquad \operatorname{diam}_{\rm physical}S\le Csm^{a_*}. \] The terminal partition $\pr_*$ and the success event belong to the finite terminal revealed sigma field $\F_*$. For every fixed deterministic $g\in L^2(\T^2)$, conditional on $\F_*$, \[ \operatorname{Var}\bigl((W-P_{\pr_*}W)(g)\mid\F_*\bigr) =\sum_{S\in\pr_*}\int_S|g-g_S|^2. \] \end{lemma} \begin{proof} There are two kinds of terminal leaves. A newly terminal union has $G_S$-intrinsic diameter at most $C_As$, and a square stopped without another reveal has birth-metric diameter at most $\sqrt K D_0s$. In both cases the receiver construction gives the area floor. On successful completion the final modular is at most $T_*$ and $\lambda_{J_0}\ge1/2$. Lemma~\ref{lem:actualphase}, applied at the last actual reveal of $S$, gives throughout $S$ the same phase comparability that holds on an active cell. Since $U$ is frozen afterwards, the comparison remains valid at time $J_0$. Integrating the final modular over $S$ then gives $H(U|_S)\le C_{T_*}m$, also when $S$ was created at the final reveal. The newly inherited diagonal cap, or the old cap for a directly stopped square, yields $\|G_S^{1/2}\|_{\rm op}\le C m^{a_*}$. Multiplying by the intrinsic diameter proves the physical bound. The algorithm makes all geometric choices before the corresponding fresh Gaussian child contrasts are revealed, and it is padded to a finite deterministic horizon. Since stopping and the success event depend only on revealed contrasts, they are $\F_*$-measurable. The residual white noise in the orthogonal complement of the finally revealed cell indicators is conditionally Gaussian. Testing it against a deterministic $g$ and using the conditional orthogonal projection gives the displayed variance. \end{proof} From now on all distances and derivatives are physical, not intrinsic; in particular the periodic Green kernel is never pulled back by an adaptive $A_S$. Write $P=P_{\pr_*}$. The identity \begin{equation}\label{eq:decomp} W_h-PW=(P_h-I)PW+P_h(W-PW)=:g^{(1)}+g^{(2)} \end{equation} holds exactly, as an identity of centered distributions. \subsection{Correction of the revealed part} \phantomsection\label{not:terminalcurrents} The first source $g^{(1)}$ has zero integral on each fixed $h$-square. Applying the ray inverse on these squares gives a periodic current $J^{(1)}$ with \[ \divv J^{(1)}=g^{(1)},\qquad \norm{J^{(1)}}_\infty\le Ch\norm{PW}_\infty. \] Lemma~\ref{lem:randommax} controls the random final cells directly and gives, with probability $1-o(1)$, \begin{equation}\label{eq:smallfirst} \norm{PW}_\infty\le C\sqrt m/s,\qquad \norm{J^{(1)}}_\infty\le C\sqrt m/m^3=o(1). \end{equation} \subsection{Periodic kernel estimates for the unrevealed bridge} \phantomsection\label{not:periodicgreen} Let $G_{\T}$ be the mean-zero Green function of the torus Laplacian in the fixed physical coordinates, and choose the periodic vector kernel $K_{\T}(x,y)=\nabla_xG_{\T}(x-y)$ with the sign convention \[ \divv_xK_{\T}(x,y)=\delta_y-1. \] The standard local expansion of the periodic Green function and smoothness away from the diagonal give \begin{equation}\label{eq:periodicK} |K_{\T}(x,y)|\le C(1+d_{\T}(x,y)^{-1}),\qquad |\nabla_yK_{\T}(x,y)|\le C(1+d_{\T}(x,y)^{-2}). \end{equation} In the source variable let $K_{x,h}=P_h^{(y)}K_{\T}(x,\cdot)$, a piecewise constant function with jumps along the fixed $h$-grid, and define \[ J^{(2)}(x)=(W-PW)(K_{x,h}). \] Its divergence is $g^{(2)}$, with the vector notation understood coordinatewise. Lemma~\ref{lem:terminalhistory}, applied to the deterministic function $K_{x,h}$, gives the exact conditional variance \begin{equation}\label{eq:terminalvar} \sigma_*^2(x) :=\operatorname{Var}\bigl(J^{(2)}(x)\mid\F_*\bigr) =\sum_{Q\in\pr_*}\int_Q |K_{x,h}-(K_{x,h})_Q|^2\dd y. \end{equation} The next lemma collects the kernel bounds used to estimate this variance and the spatial modulus of $J^{(2)}$. \begin{lemma} \label{lem:periodic} For $h$ sufficiently small, uniformly in $x,y\in\T^2$, \begin{equation}\label{eq:gridreg} |K_{x,h}(y)|\le \frac{C}{h+d_{\T}(x,y)}. \end{equation} If $y,z$ lie in a common physical set of Euclidean torus diameter at most $D_*$, with $r=d_{\T}(x,y)\ge4(D_*+2h)$, then \begin{equation}\label{eq:faravg} |K_{x,h}(y)-K_{x,h}(z)| \le C\frac{D_*+h}{r^2}. \end{equation} Finally, for $|u|\le1/4$, \begin{equation}\label{eq:periodictrans} \norm{K_{\T}(\cdot+u,\cdot)-K_{\T}(\cdot,\cdot)}_{L^1(\T^2)} \le C|u|(1+|\log|u||). \end{equation} \end{lemma} \begin{proof} Fix a periodic lift at distance less than $1/4$ from the pole. When the $h$-square containing $y$ is at distance at least $2h$ from $x$, average \eqref{eq:periodicK} over that square and use $|x-z|\asymp d_{\T}(x,y)$. When the square is close to the pole, the integral of $1/|x-z|$ over an $h$-square is $O(h)$, and division by the area $h^2$ gives $O(h^{-1})$. The smooth periodic remainder contributes $O(1)$, which proves \eqref{eq:gridreg}. For the second claim, let $R_y,R_z$ be the fixed $h$-squares containing $y,z$. If $\zeta\in R_y$ and $\zeta'\in R_z$, then $|\zeta-\zeta'|\le D_*+2\sqrt2h$. Every point of the segment $[\zeta,\zeta']$ lies within $\sqrt2h+D_*+2\sqrt2h\le\frac{3\sqrt2}{2}(D_*+2h)$ of $y$, and $\frac{3\sqrt2}{2}(D_*+2h)\le\frac{3\sqrt2}{8}r<\frac34r$ because $r\ge4(D_*+2h)$. Hence the whole segment stays at distance at least $r/4$ from the pole. We apply the derivative bound for the original kernel $K_{\T}$ in \eqref{eq:periodicK} along this segment and average twice; this proves \eqref{eq:faravg}. The step function $K_{x,h}$ itself is never differentiated. For \eqref{eq:periodictrans}, split the torus into $B(x,2|u|)$ and its complement. In the ball, integrating the two $1/r$ singularities gives $C|u|$. Outside, the mean-value theorem gives the pointwise bound $C|u|/r^2$, whose integral in two dimensions is $C|u|(1+|\log|u||)$. \end{proof} \begin{proposition} \label{prop:termvar} Under the leaf diameter bound \eqref{eq:Dstar}, \begin{equation}\label{eq:logvar} \sup_{x\in\T^2}\sigma_*^2(x) \le C\left[1+\log_+\left(\frac{D_*}{h}\right)\right] \le C\log m. \end{equation} \end{proposition} The estimate is deterministic once the successful terminal geometry is fixed, and it uses nothing about how the tree was selected. \begin{proof} Put $R_*=8(D_*+2h)$ and split the leaves into those meeting $B(x,R_*)$ and the remaining ones. Any near leaf lies inside $B(x,R_*+D_*)\subset B(x,CR_*)$. Subtracting a leaf average only decreases the $L^2$ norm, so by \eqref{eq:gridreg} \[ \sum_{Q\,\mathrm{near}}\int_Q|K_{x,h}-(K_{x,h})_Q|^2 \le C\int_{B(x,CR_*)}\frac{\dd y}{(h+d_{\T}(x,y))^2} \le C\bigl[1+\log_+(D_*/h)\bigr]. \] For a far leaf $Q$ and $y,z\in Q$ we apply \eqref{eq:faravg}. Since the leaf is much smaller than its distance to the pole, this gives \[ \int_Q|K_{x,h}-(K_{x,h})_Q|^2 \le C|Q|\frac{(D_*+h)^2}{r_Q^4}, \quad r_Q=\dist_{\T}(x,Q). \] Summing over far leaves and comparing with the integral in the two-dimensional radial variable bounds their contribution by \[ C(D_*+h)^2\int_{R_*/2}^{1}r^{-3}\dd r\le C. \] This proves the first inequality in \eqref{eq:logvar}, and the second follows from \eqref{eq:Dstar}. Borderline cells are avoided by increasing the universal radius constants, which does not change the scaling. \end{proof} \subsection{Orlicz and supremum estimates} We now turn the variance bound into estimates for $J^{(2)}$: a $\psi_2$ bound with fixed positive conditional probability, and an $L^\infty$ bound with probability $1-o(1)$. Interpolating between the two gives the $\psi_{r_i}$ bounds. Write $A_m\in\mathcal F_*$ for the event that the adaptive construction completes without numerical killing or modular overshoot, so that $\mathbb P(A_m)\ge p_*>0$ by Proposition~\ref{prop:stopped}. On $A_m$, conditionally on $\F_*$, each coordinate $J^{(2)}_i(x)$ is centered Gaussian by Lemma~\ref{lem:terminalhistory}, and its variance is at most $C_0\log m$ by \eqref{eq:logvar}, with a fixed $C_0$. We first bound the quadratic exponential modular. For a centered Gaussian variable $X$ of variance $\sigma^2$ and $t^2>2\sigma^2$, \[ \E e^{X^2/t^2}=\Bigl(1-\frac{2\sigma^2}{t^2}\Bigr)^{-1/2}. \] With $A_1^2=8C_0/3$ and $t^2=A_1^2\log m$ one has $2\sigma^2/t^2\le3/4$, hence \[ \E\left[\exp\left(\frac{|J^{(2)}_i(x)|^2} {A_1^2\log m}\right)\Big|\F_*\right]\le2 \qquad(x\in\T^2,\ i=1,2). \] By conditional Tonelli the spatial integral has conditional expectation at most $2$. Markov's inequality at level $16$, applied to each of the two coordinates, shows that with conditional probability at least $1-2\cdot\frac2{16}=\frac34$, \[ \int_{\T^2}\exp\left(\frac{|J^{(2)}_i|^2}{A_1^2\log m}\right)\le16, \qquad i=1,2. \] On this event, Jensen's inequality for the concave map $t\mapsto t^{1/4}$ on the unit-area torus gives $\int_{\T^2}\exp\bigl(|J^{(2)}_i|^2/(4A_1^2\log m)\bigr)\le16^{1/4}=2$, that is, $\int_{\T^2}\psi_2\bigl(J^{(2)}_i/(2A_1\sqrt{\log m})\bigr)\le1$. Consequently the event \[ C_m=\Bigl\{\max_{i=1,2}\|J_i^{(2)}\|_{\psi_2} \le 2A_1\sqrt{\log m}\Bigr\} \] satisfies \begin{equation}\label{eq:ps2corr} \mathbb P(C_m\mid\mathcal F_*)\ge\tfrac34 \qquad\text{on }A_m. \end{equation} The $L^\infty$ bound requires a separate high-probability argument, since quadratic exponential integrability in space alone does not control the supremum. Recall that $g^{(2)}=W_h-P_hPW$, and let $D_m$ be the intersection of the fixed-$h$ Gaussian maximum event and the predictable adaptive-cell maximum event. Since $s\gg h$, on $D_m$ \[ \norm{g^{(2)}}_\infty\le C\sqrt m/h. \] The fixed-grid event follows from the ordinary union bound over $h^{-2}$ deterministic Gaussians, and the adaptive event is \eqref{eq:randommaxwhp}. Hence $\mathbb P(D_m^c)=o(1)$; this bound is unconditional. Since $J^{(2)}=K_{\T}*g^{(2)}$, the translation estimate \eqref{eq:periodictrans} gives on $D_m$ the deterministic modulus \[ |J^{(2)}(x)-J^{(2)}(z)| \le C\frac{\sqrt m}{h}\,d_{\T}(x,z) \bigl(1+|\log d_{\T}(x,z)|\bigr). \] Fix a deterministic torus net $\mathcal N_m$ of mesh $q_m=2^{-Lm}$, where $L>3$ is fixed. It has at most $C2^{2Lm}$ points. Since $h=2^{-m}$, the modulus above is $O(m^{3/2}2^{-(L-1)m})=o(1)$ at distance $q_m$. Let $N_m$ be the event that $|J^{(2)}(y)|\le A_0\sqrt{m\log m}$ for every $y\in\mathcal N_m$. The conditional Gaussian tails at variance level $C_0\log m$ and a union bound over the net give, on $A_m$, \[ \mathbb P(N_m^c\mid\mathcal F_*) \le C2^{2Lm}e^{-A_0^2m/(2C_0)}=o(1) \] once $A_0^2>4C_0L\log2$. This estimate uses neither the fixed-grid nor the adaptive-cell maximum event. On $D_m\cap N_m$ the modulus extends the net bound to \begin{equation}\label{eq:supcorr} \norm{J^{(2)}}_\infty\le C\sqrt{m\log m}. \end{equation} For a scalar function $f$ and fixed $r>2$, the elementary inequality \[ \left(\frac{|f|}{A^{1-2/r}B^{2/r}}\right)^r \le \left(\frac{|f|}{B}\right)^2 \quad\text{whenever}\quad \norm f_\infty\le A \] shows, after a fixed dilation, that \[ \norm f_{\psi_r}\le C_r \norm f_\infty^{1-2/r}\norm f_{\psi_2}^{2/r}. \] Consequently, on $C_m\cap D_m\cap N_m$, \begin{equation}\label{eq:interp} \norm{J^{(2)}_i}_{\psi_{r_i}} \le C_{r_i}m^{1/2-1/r_i}\sqrt{\log m} =o(\sqrt m)\qquad (r_i<\infty). \end{equation} The gain over $\sqrt m$ comes from the finiteness of $r_i$ and is not uniform as $r_i\to\infty$. \subsection{Joint probability estimate} Since $A_m\in\mathcal F_*$, \eqref{eq:ps2corr} gives $\mathbb P(A_m\cap C_m)=\E[\one_{A_m}\mathbb P(C_m\mid\F_*)] \ge\frac34\mathbb P(A_m)$. The remaining two events are removed by subtracting their complements, which requires no conditional independence: \begin{align*} \mathbb P(A_m\cap C_m\cap D_m\cap N_m) &\ge\mathbb P(A_m\cap C_m) -\mathbb P(D_m^c)-\mathbb P(A_m\cap N_m^c)\\ &\ge\tfrac34\,\mathbb P(A_m)-o(1) \ge\tfrac34\,p_*-o(1)>0. \end{align*} On this intersection, put $J=J_{\mathrm{leaf}}+J^{(1)}+J^{(2)}$. The exact splitting \eqref{eq:decomp} gives $\divv J=W_h$, and the three estimates are \[ \max_i\|J_{\mathrm{leaf},i}\|_{\psi_{r_i}}\le C\sqrt m, \qquad \|J^{(1)}\|_\infty\le C m^{-5/2}, \qquad \|J^{(2)}_i\|_{\psi_{r_i}} \le C_{r_i}m^{1/2-1/r_i}\sqrt{\log m}. \] In the quartic case the last bound is $C m^{1/4}\sqrt{\log m}=o(\sqrt m)$, and for every fixed finite pair the same strict improvement follows from \eqref{eq:interp}. The triangle inequality therefore gives, for some fixed $p_{**}>0$, \begin{equation}\label{eq:fixedprob} \mathbb P\bigl(\mathfrak F_{\boldsymbol r}(W_h) \le C\sqrt m\bigr)\ge p_{**} \end{equation} for all sufficiently large $m$. \paragraph{Completion of the proof of Theorem~\ref{thm:gauss}.} The estimate \eqref{eq:fixedprob} is the required fixed-positive-probability statement. Taking any $0<\varepsilon_01. \end{cases} \] The Luxemburg norm associated with $\psi_\infty$ is the $L^\infty$ norm. For the exponent pair $\boldsymbol r$ we use the coordinate Young functions $\psi_{r_1}$ and $\psi_{r_2}$ throughout. For a finite vector $b=(b_i)_{i=1}^N$, define \[ \norm b_{\psi_{r_j},N} :=\inf\set{c>0:\frac1N\sum_{i=1}^N\psi_{r_j}(b_i/c)\le1}. \] For $r_j=\infty$, this is $\max_i |b_i|$. For a permutation $\sigma\in S_N$, write $d_{i,j}^\sigma=\abs{X_i^j-Y_{\sigma(i)}^j}$ and define \begin{equation} \mathfrak M_N^{\boldsymbol r} :=\min_{\sigma\in S_N} \max_{j=1,2}\norm{(d_{i,j}^\sigma)_{i=1}^N}_{\psi_{r_j},N}. \label{red:eq:atomic-aniso} \end{equation} The two-sample form of the Ultimate Matching Conjecture is the following. \begin{definition} \label{red:def:UMC} The conjecture $\mathrm{UMC}_{\boldsymbol r}$ is the assertion that there is a deterministic $C<\infty$ such that \begin{equation} \Pp\pa{\mathfrak M_N^{\boldsymbol r}\le Cs_N}\longrightarrow1. \label{red:eq:UMC} \end{equation} \end{definition} Talagrand's Research Problem~17.1.2 \cite{umc:Talagrand2021} states the same coordinatewise criterion for a random sample and an evenly spread reference configuration. The symmetric quartic case is $\boldsymbol r=(4,4)$; the hard coordinate endpoint is $\boldsymbol r=(2,\infty)$. \begin{lemma} \label{red:lem:isotropic-quartic} For $\boldsymbol r=(4,4)$, the coordinate gauge in \eqref{red:eq:atomic-aniso} is equivalent, up to a factor $\sqrt2$, to the isotropic gauge generated by $e^{|z|^4}-1$. \end{lemma} \begin{proof} The lower bound follows from $|z_j|\le|z|$. Conversely, if both coordinate Luxemburg norms are at most $a$, then $|z|^4\le2(|z_1|^4+|z_2|^4)$ and Cauchy--Schwarz yields \[ \frac1N\sum_i e^{|z_i|^4/(\sqrt2a)^4} \le \prod_{j=1}^2 \left(\frac1N\sum_i e^{|z_{i,j}|^4/a^4}\right)^{1/2} \le2. \] \end{proof} \subsection{Dyadic Gaussian flows} Write $\T^2=(\R/\mathbb Z)^2$. For $\Omega\in\{D,\T^2\}$, $P_h^{\Omega}$ denotes averaging onto the fixed $h$-square grid in $\Omega$, acting on mean-zero functions when appropriate. With mean-zero white noise $W^{\Omega}$ (the isonormal process on $L_0^2(\Omega)$), set \begin{equation}\label{red:eq:compatible-source} W_h^{\Omega}=P_h^{\Omega}W^{\Omega}. \end{equation} On the torus we also write $W_h=W_h^{\T}$ and $\mathfrak F_{\boldsymbol r}=\mathfrak F_{\boldsymbol r}^{\T}$ when there is no ambiguity. Choose a sufficiently large constant $A$ and a dyadic $h_N$ satisfying \begin{equation} As_N\le h_N<2As_N. \label{red:eq:h-choice} \end{equation} The Gaussian source on the square used in the reduction is \begin{equation} W_{h_N}^{D}=P_{h_N}^{D}W^{D}, \label{red:eq:haar-source} \end{equation} with the same white-noise normalization as $W_h$ on the torus. For a vector field $J=(J_1,J_2)$, put \begin{equation} \norm J_{\boldsymbol r} :=\max_{j=1,2}\norm{J_j}_{\psi_{r_j}}. \label{red:eq:product-flow-norm} \end{equation} For a mean-zero distribution $f$, we encode both the divergence equation and the no-flux boundary condition weakly. A vector field $J$ is admissible for $f$ when \begin{equation} -\int_D J\cdot\nabla\varphi\,dx=\langle f,\varphi\rangle \qquad\text{for every }\varphi\in C^\infty(\overline D). \label{red:eq:weak-div-no-flux} \end{equation} For regular $J$, this is exactly $\divv J=f$ in $D$ together with the zero normal trace $J\cdot n=0$ on $\partial D$. We then define \begin{equation} \mathfrak F_{\boldsymbol r}^{D}(f) :=\inf\set{\norm J_{\boldsymbol r}: J\text{ satisfies } \eqref{red:eq:weak-div-no-flux}}. \label{red:eq:aniso-flow} \end{equation} The corresponding Gaussian minimum-flow problem is the following. \begin{definition} \label{red:def:GOMF} The conjecture $\mathrm{GOMF}_{\boldsymbol r}$ asserts that there is a deterministic $C<\infty$ such that \begin{equation} \Pp\pa{\mathfrak F_{\boldsymbol r}^{D}(W_{h_N}^{D}) \le C\sqrt{\log N}}\longrightarrow1. \label{red:eq:GOMF} \end{equation} \end{definition} The main result of this part is the following equivalence. \begin{theorem} \label{red:thm:main} For every fixed pair $\boldsymbol r$ satisfying \eqref{red:eq:critical-pair}, and for $A$ sufficiently large, $\mathrm{UMC}_{\boldsymbol r}$ and $\mathrm{GOMF}_{\boldsymbol r}$ are equivalent. More precisely, the following statements are equivalent: \begin{enumerate}[label=\textnormal{(\roman*)}] \item $\Pp(\mathfrak M_N^{\boldsymbol r}\le Cs_N)\to1$ for some $C$; \item \[ \Pp\pa{\mathfrak F_{\boldsymbol r}^{D} (P_{h_N}^D(\mu_N-\nu_N))\le Cs_N}\to1 \] for some $C$; \item \[ \Pp\pa{\mathfrak F_{\boldsymbol r}^{D}(W_{h_N}^{D}) \le C\sqrt{\log N}}\to1 \] for some $C$. \end{enumerate} All constants are deterministic and independent of $N$. \end{theorem} \section{Preliminary estimates} We collect elementary facts on Orlicz norms and on the minimum-flow gauge. We then show that the gauge of projected white noise concentrates around its median, so that a bound with fixed positive probability implies a bound with high probability. \begin{lemma} \label{red:lem:moment} For every fixed $2\le r<\infty$ on a probability space, \begin{equation} \norm Z_{\psi_r} \asymp_r \sup_{s\ge1}s^{-1/r}\norm Z_{L^s}. \label{red:eq:moment-characterization} \end{equation} \end{lemma} \begin{proof} The forward implication follows by integrating the tail furnished by $\E e^{|Z|^r/a^r}\le2$. Conversely, expand the exponential in moments and use $k!\ge(k/e)^k$. \end{proof} \begin{lemma} \label{red:lem:absorb} Let $\Theta$ be any Young function and $K\ge1$. If $\int\Theta(|f|/r)\le K$, then $\norm f_{L^\Theta}\le Kr$. \end{lemma} \begin{proof} Convexity and $\Theta(0)=0$ give $\Theta(t/K)\le K^{-1}\Theta(t)$. \end{proof} \begin{lemma} \label{red:lem:flow-lipschitz} For mean-zero sources $f,g$, \[ \abs{\mathfrak F_{\boldsymbol r}^{D}(f)- \mathfrak F_{\boldsymbol r}^{D}(g)} \le \mathfrak F_{\boldsymbol r}^{D}(f-g), \qquad \mathfrak F_{\boldsymbol r}^{D}(af)=|a|\mathfrak F_{\boldsymbol r}^{D}(f). \] \end{lemma} \begin{proof} Subtract admissible currents and use the triangle inequality in the product norm \eqref{red:eq:product-flow-norm}. \end{proof} \subsection{Gaussian concentration for dyadic minimum-flow gauges} \label{red:subsec:gaussian-concentration} The argument of this subsection uses neither the matching reduction, nor the transport--flow comparison, nor any construction of the Gaussian current. It applies without change when one of the exponents equals $\infty$. For $\Omega\in\{D,\T^2\}$, let $\mathfrak F_{\boldsymbol r}^{\Omega}$ be the minimum-flow gauge of \eqref{red:eq:aniso-flow}, with the periodic divergence convention in the torus case. If $h=2^{-m}$, write its projected mean-zero white noise in the form \begin{equation}\label{red:eq:gaussian-coefficients} W_h^{\Omega}(Z)|_Q=h^{-1}(Z_Q-\overline Z), \qquad \overline Z=h^2\sum_Q Z_Q, \end{equation} where the $h$-grid coefficients $Z_Q$ are independent standard Gaussian variables. \begin{lemma} \label{red:lem:uniform-gaussian-lipschitz} Fix any $r_i\in[2,\infty]$. There is an absolute constant $C$, independent of $h$, the number of cells and the exponent pair, such that for $\Omega\in\{D,\T^2\}$ and coefficient arrays $Z,Z'$, \begin{equation}\label{red:eq:uniform-gaussian-lipschitz} \left|\mathfrak F_{\boldsymbol r}^{\Omega}(W_h^{\Omega}(Z))- \mathfrak F_{\boldsymbol r}^{\Omega}(W_h^{\Omega}(Z'))\right| \le C\norm{Z-Z'}_{\ell^2}. \end{equation} \end{lemma} \begin{proof} Write $v=Z-Z'$ and $g_Q=h(v_Q-\overline v)$ for the mass of the source difference in a grid square $Q$. For any union $A$ of grid squares, centering and Cauchy--Schwarz give \[ \left|\sum_{Q\in A}g_Q\right| \le 2h\norm v_{\ell^2}\sqrt{\min\{\#A,\#A^c\}}. \] The nearest-neighbor square grid satisfies the relative edge-isoperimetric inequality \[ |\partial A|\ge \sqrt2\,\sqrt{\min\{\#A,\#A^c\}}, \] where $\partial A$ counts internal edges (and also periodic seam edges for $\Omega=\T^2$). Indeed, for the $k\times k$ grid with $k=h^{-1}$ and $\#A\le k^2/2$, the edge-isoperimetric inequality of Bollob\'as and Leader \cite{BollobasLeader1991} gives $|\partial A|\ge\min\{2\sqrt{\#A},k\}\ge\sqrt2\sqrt{\#A}$; the torus graph contains all edges of the square grid, so the same bound holds for $\Omega=\T^2$. Consequently every cut has enough capacity when each internal edge is assigned capacity $C h\norm v_{\ell^2}$ in both directions. Gale's supply--demand theorem \cite{Gale1957} (see also \cite{Hoffman1960}), a consequence of the max-flow/min-cut theorem, then produces a consistent signed edge flux $f_e$ with \[ \sum_{e\,\mathrm{out}(Q)}f_e- \sum_{e\,\mathrm{in}(Q)}f_e=g_Q, \qquad |f_e|\le C h\norm v_{\ell^2}. \] On $D$, take the flux on exterior faces to be zero. It remains to realize these integrated face fluxes as a continuum current. On $Q=[ih,(i+1)h]\times[jh,(j+1)h]$, put $t=(x-ih)/h$ and $u=(y-jh)/h$, and use globally consistent rightward/upward signed fluxes $f_L,f_R,f_B,f_T$. Define \begin{equation}\label{red:eq:face-current} J_1(x,y)=\frac{(1-t)f_L+tf_R}{h},\qquad J_2(x,y)=\frac{(1-u)f_B+uf_T}{h}. \end{equation} The normal components match across common faces (and periodically across torus seams). Their weak divergence on each cell is \[ \divv J=\frac{f_R-f_L+f_T-f_B}{h^2} =h^{-1}(v_Q-\overline v). \] This current satisfies the zero-normal-trace convention on $D$ and $\max_i\norm{J_i}_{L^\infty}\le C\norm v_{\ell^2}$. Every Luxemburg $\psi_r$-norm with $r\ge2$, including $r=\infty$, is bounded by a universal multiple of $L^\infty$ on a space of unit area. The triangle inequality for minimum-flow gauges, \cref{red:lem:flow-lipschitz}, now gives \eqref{red:eq:uniform-gaussian-lipschitz}. \end{proof} \begin{proposition} \label{red:prop:gaussian-concentration} Let $h=2^{-m}$, $\Omega\in\{D,\T^2\}$, and fix $r_1,r_2\in[2,\infty]$. Suppose there are $B,p_0>0$ such that, for all sufficiently large $m$, \[ \Pp\!\left(\mathfrak F_{\boldsymbol r}^{\Omega}(W_h^{\Omega}) \le B\sqrt m\right)\ge p_0. \] Then there are $B',c>0$ such that, for all sufficiently large $m$, \begin{equation}\label{red:eq:gaussian-concentration} \Pp\!\left(\mathfrak F_{\boldsymbol r}^{\Omega}(W_h^{\Omega}) >B'\sqrt m\right)\le2e^{-cm}. \end{equation} \end{proposition} \begin{proof} By \cref{red:lem:uniform-gaussian-lipschitz}, the gauge is an $L$-Lipschitz function of a standard Gaussian vector, with $L$ independent of the dimension. Gaussian concentration around a median $M_m$ shows that $M_m\le B\sqrt m+C(L,p_0)$. Indeed, otherwise the lower tail below $B\sqrt m$ would have probability less than $p_0$. The upper concentration tail beyond a fixed larger multiple $B'\sqrt m$ is then at most $2e^{-cm}$. \end{proof} \begin{corollary} \label{red:cor:positive-probability-equivalence} In the dyadic Gaussian condition (iii) of \cref{red:thm:main}, it is equivalent to require the existence of $B,p_0>0$ such that \[ \Pp\!\left(\mathfrak F_{\boldsymbol r}^{D} (W_{h_N}^{D})\le B\sqrt{\log N}\right)\ge p_0 \] for all sufficiently large $N$. The same conclusion holds for a torus flow with its periodic divergence constraint. \end{corollary} \begin{proof} The high-probability statement implies the positive-probability bound. Conversely, $h_N=2^{-m(N)}$ and $m(N)\asymp\log N$, so \cref{red:prop:gaussian-concentration} applies with a changed constant and yields the Gaussian condition (iii). \end{proof} Thus the Gaussian input to the matching equivalence need only be proved with a fixed positive probability. \section{From point configurations to histograms} We show that, up to an additive error of order $h$, matching the two point clouds is equivalent to transporting their dyadic histograms. This replaces the atomic problem by a transport problem between densities. For a coupling $\pi$ on $D\times D$, define \begin{equation} \mathfrak T_{\boldsymbol r}(\varrho_0,\varrho_1) :=\inf_{\pi\in\Pi(\varrho_0\dd x,\varrho_1\dd x)} \max_{j=1,2}\norm{x_j-y_j}_{L^{\psi_{r_j}}(\pi)}. \label{red:eq:aniso-transport-gauge} \end{equation} Let $P_h^D\mu_N$ be obtained by spreading each atomic mass uniformly on its dyadic cell. \begin{proposition} \label{red:prop:atomic-hist} There is a universal $C$ such that, for every pair of point clouds and every dyadic $h$, \begin{equation} \mathfrak T_{\boldsymbol r}(P_h^D\mu_N,P_h^D\nu_N) \le \mathfrak M_N^{\boldsymbol r}+Ch, \qquad \mathfrak M_N^{\boldsymbol r} \le C\bigl(\mathfrak T_{\boldsymbol r}(P_h^D\mu_N,P_h^D\nu_N)+h\bigr). \label{red:eq:atomic-hist-equivalence} \end{equation} The constant is uniform over the critical exponents, including an $L^\infty$ endpoint. \end{proposition} \begin{proof} Fix a permutation. Couple the two associated cells using the same relative coordinate. Each coordinate displacement increases by at most $2h$, and the first inequality follows from the Luxemburg triangle inequality. Conversely, let $\pi$ couple the histograms. Attach posterior atomic labels inside every occupied cell and define \[ q_{ik}=N\int\lambda_i(x)\mu_k(y)\dd\pi(x,y). \] Then $Q=(q_{ik})$ is doubly stochastic. If the histogram coordinate norms are at most $r$, the fractional labelled coupling has both coordinate norms at most $r+2h$. Decompose $Q$ into permutation matrices. For finite coordinates, the average over this decomposition of the sum of the two coordinate modulars is at most $2$, so one permutation has that sum at most $2$; \cref{red:lem:absorb} absorbs this fixed factor. For an $L^\infty$ coordinate, every positive entry of $Q$ obeys the corresponding displacement bound, and a Birkhoff decomposition can be chosen inside the positive support. Thus the same permutation satisfies the hard coordinate bound. \end{proof} \section{Transport and divergence currents} This section contains the deterministic core of the reduction. For densities bounded above and away from zero, we prove that the anisotropic transport gauge $\mathfrak T_{\boldsymbol r}$ is comparable to the minimum-flow gauge $\mathfrak F_{\boldsymbol r}^D$ of their difference, first for finite exponents and then at the endpoint $(2,\infty)$. The passage from transport to flow requires a bound on the densities along displacement interpolation, which we derive from noncrossing of transport rays and a Jacobian inequality. The domain $D$ is convex, so straight segments remain in $D$. \subsection{Noncrossing of transport rays} \begin{lemma} \label{red:lem:noncrossing} Let $L:\R^2\to\R\cup\{+\infty\}$ be strictly convex on its convex effective domain. Let $\Gamma\subset D\times D$ be $c$-cyclically monotone for $c(x,y)=L(y-x)$. For $00$ because $(1-t)+t\lambda_k>0$, so the absolute value in \eqref{red:eq:jacobian-interpolation} can also be removed. We have \[ \det DT_t =\prod_{k=1}^2((1-t)+t\lambda_k) \ge\br{(1-t)+t\sqrt{\det DT}}^2. \] By \eqref{red:eq:jacobian-map}, now without absolute values, $\det DT=\varrho_0/(\varrho_1\circ T)\ge \underline\varrho/\overline\varrho$, hence $\det DT_t\ge \underline\varrho/\overline\varrho$. Using \eqref{red:eq:jacobian-interpolation}, \[ \varrho_t(T_t(x))=\frac{\varrho_0(x)}{\det DT_t(x)} \le\frac{\overline\varrho^2}{\underline\varrho}. \] At $t=0,1$ the same bound follows directly from $\varrho_0,\varrho_1\le\overline\varrho\le \overline\varrho^2/\underline\varrho$ (the unit-mass bounds imply $\underline\varrho\le1\le\overline\varrho$). \end{proof} \begin{remark} With the opposite convention for the Kantorovich potential, all signs in the intermediate identities reverse simultaneously. The invariant statement is that $DT$ is similar to a symmetric positive semidefinite matrix. \end{remark} \subsection{Boundary smoothing and approximation} The previous lemma requires smooth densities, whereas the histogram densities used later are piecewise constant. The passage between them requires a recovery approximation, not merely lower semicontinuity. We use the following local construction on the square. \begin{lemma} \label{red:lem:local-smoothing} Let $D=[0,1]^2$. There are Markov kernels $K_\tau(x,\dd z)$, $0<\tau<1/4$, with smooth nonnegative densities $k_\tau(x,z)$ such that: \begin{enumerate}[label=\textup{(\roman*)}] \item $K_\tau$ preserves Lebesgue measure and is symmetric: \[ \int_D k_\tau(x,z)\,\dd z=1, \qquad \int_D k_\tau(x,z)\,\dd x=1, \qquad k_\tau(x,z)=k_\tau(z,x). \] \item $k_\tau(x,z)=0$ unless $|x-z|_\infty\le\tau$. \item If $S_\tau\varrho(z)=\int_D k_\tau(x,z)\varrho(x)\,\dd x$, then $S_\tau\varrho$ is smooth, $S_\tau\varrho\to\varrho$ in $L^1(D)$, and \[ \underline\varrho\le\varrho\le\overline\varrho \quad\Longrightarrow\quad \underline\varrho\le S_\tau\varrho\le\overline\varrho. \] \item If $\pi$ couples $\varrho_0\dd x$ and $\varrho_1\dd x$, then \[ \pi_\tau(\dd x'\dd y') :=\int_{D\times D}K_\tau(x,\dd x')K_\tau(y,\dd y')\,\pi(\dd x\dd y) \] couples $S_\tau\varrho_0\dd x$ and $S_\tau\varrho_1\dd x$. Moreover $|(y'-x')-(y-x)|_\infty\le2\tau$ on the support of the lifted coupling. Consequently, for every bounded continuous displacement cost $c(y-x)$, \[ \int c(y'-x')\,\dd\pi_\tau\longrightarrow \int c(y-x)\,\dd\pi. \] If in addition $|y_2-x_2|\le r$ $\pi$-a.s., then $|y'_2-x'_2|\le r+2\tau$ $\pi_\tau$-a.s. \end{enumerate} \end{lemma} \begin{proof} Choose an even $C_c^\infty(\R)$ probability density $\eta_\tau$ supported in $(-\tau,\tau)$. In one dimension define the reflected, two-periodic kernel on $[0,1]$ by the method of images \[ k^{(1)}_\tau(x,z) =\sum_{n\in\mathbb Z}\Bigl[ \eta_\tau(z-x+2n)+\eta_\tau(z+x+2n) \Bigr], \qquad x,z\in[0,1]. \] It is nonnegative, symmetric, smooth, and its integral in either variable is one. Equivalently, it is convolution by $\eta_\tau$ after even $2$-periodic reflection. Since $\tau<1/4$, every nonzero image term has $|x-z|\le\tau$; for the reflected terms near $0$ or $1$, this follows from $x+z\le\tau$ or $2-x-z\le\tau$. Take the product of the two one-dimensional kernels to obtain $k_\tau$ on $D$. The Markov and bistochastic identities imply preservation of mass and the pointwise bounds $\underline\varrho,\overline\varrho$. Standard approximate-identity theory for the even periodic extension gives $L^1$ convergence and smoothness. The formula for $\pi_\tau$ gives the asserted marginals by Fubini. Local support gives $|x'-x|_\infty,|y'-y|_\infty\le\tau$, hence the displacement and hard-support bounds. Uniform continuity of a bounded continuous cost on the compact set $D-D$ gives the final convergence. \end{proof} \begin{corollary} \label{red:cor:transport-recovery} Assume $0<\underline\varrho\le\varrho_i\le\overline\varrho$ and let $\varrho_i^\tau=S_\tau\varrho_i$. If $r>\mathfrak T_{\boldsymbol r}(\varrho_0,\varrho_1)$ and $r'>r$, then for all sufficiently small $\tau$ there is a coupling of $\varrho_0^\tau$ and $\varrho_1^\tau$ whose finite-coordinate modulars at scale $r'$ are bounded by a fixed constant independent of $\tau$. At a hard coordinate, a bound $|y_j-x_j|\le r$ is replaced by $|y_j'-x_j'|\le r'$. The fixed modular constant may be absorbed into a fixed further enlargement of $r'$ by \cref{red:lem:absorb}. \end{corollary} \begin{proof} Start from a coupling witnessing the gauge at scale $r$ and apply \cref{red:lem:local-smoothing}. For finite coordinates the costs at the larger scale $r'$ are bounded continuous functions on $D-D$ and their integrals converge to those of the original coupling, which are at most one at scale $r$ and hence at most one at scale $r'$. For a hard coordinate choose $\tau<(r'-r)/2$. \end{proof} \subsection{Finite critical exponents} For finite $r_1,r_2$, set \[ L_r(z)=\sum_{j=1}^2\psi_{r_j}(|z_j|/r). \] The generalized Benamou--Brenier formula for this convex Lagrangian is standard; see, for example, \cite{Santambrogio,Villani}: \begin{equation} \inf_\pi\int L_r(y-x)\dd\pi =\inf_{\partial_t\varrho_t+\divv J_t=0} \int_0^1\int_D\varrho_tL_r(J_t/\varrho_t)\dd x\dd t. \label{red:eq:BB} \end{equation} \begin{theorem} \label{red:thm:transport-flow-finite} Assume $2\norm J_{\boldsymbol r}$. Since $\varrho_t\ge\underline\varrho$ and the perspective $s\mapsto s\psi_{r_j}(q/s)$ is decreasing, \[ \int_0^1\!\int\varrho_t \psi_{r_j}\pa{\frac{|J_j|}{\varrho_t(a/\underline\varrho)}} \le\underline\varrho\int\psi_{r_j}(|J_j|/a)\le\underline\varrho. \] Thus the total $L_{a/\underline\varrho}$ action is at most $2\underline\varrho\le2$. By \eqref{red:eq:BB} and \cref{red:lem:absorb}, the two displacement norms are at most $2a/\underline\varrho$. \emph{Transport to flow.} Take $r>\mathfrak T_{\boldsymbol r}(\varrho_0,\varrho_1)$ and then fix $r'>r$. Let $\varrho_i^\tau=S_\tau\varrho_i$. By \cref{red:cor:transport-recovery}, for small $\tau$ there is a coupling of the smooth densities $\varrho_0^\tau,\varrho_1^\tau$ whose two coordinate modulars at scale $r'$ are bounded by a universal fixed constant. After a fixed enlargement of $r'$ using \cref{red:lem:absorb}, we may assume that their sum is at most $2+o_\tau(1)$. For $\varepsilon>0$, minimize $L_{r'}(z)+\varepsilon|z|^2$ between the smoothed marginals. The minimizer is induced by a map and \cref{red:lem:density-interpolation} applies. Its interpolation densities obey $\varrho_t^{\tau,\varepsilon}\le\varrho_*=\overline\varrho^2/\underline\varrho$, uniformly in both $\tau$ and $\varepsilon$. If $v_t(T_t(x))=T(x)-x$ and $J_t=\varrho_tv_t$, set $J^{\tau,\varepsilon}=\int_0^1J_t\dd t$ and $a(x)=\int_0^1\varrho_t(x)\dd t\le\varrho_*$. Coordinatewise Jensen gives \[ \varrho_*\psi_{r_j}\pa{\frac{|J_j^{\tau,\varepsilon}|} {\varrho_* r'}} \le\int_0^1\varrho_t\psi_{r_j}(|v_{t,j}|/r')\dd t. \] Hence the currents are bounded, uniformly in $\tau,\varepsilon$, in the corresponding Orlicz spaces at scale $C_{\underline\varrho,\overline\varrho}r'$. First let $\varepsilon\downarrow0$. Then let $\tau\downarrow0$. The superlinear Young functions give uniform integrability of the current coordinates, hence weak compactness in $L^1$; the divergence constraint passes distributionally because $\varrho_i^\tau\to\varrho_i$ in $L^1$. Lower semicontinuity of convex integral functionals gives a limiting current $J$ with $\divv J=\varrho_0-\varrho_1$ and $\norm{J_j}_{L^{\psi_{r_j}}}\le C_{\underline\varrho,\overline\varrho}r'$. Finally let $r'\downarrow r$ and $r\downarrow\mathfrak T_{\boldsymbol r}(\varrho_0,\varrho_1)$. The smoothing step is where \cref{red:cor:transport-recovery} enters: it gives an upper bound for the smoothed transport problem, whereas lower semicontinuity alone gives the inequality in the wrong direction. \end{proof} \subsection{The critical endpoint $(2,\infty)$} Assume, after exchanging coordinates if necessary, that $\boldsymbol r=(2,\infty)$. \begin{theorem} \label{red:thm:transport-flow-endpoint} If $0<\underline\varrho\le\varrho_0,\varrho_1\le\overline\varrho$, then \begin{equation} \mathfrak T_{(2,\infty)}(\varrho_0,\varrho_1) \asymp_{\underline\varrho,\overline\varrho} \mathfrak F_{(2,\infty)}^{D}(\varrho_0-\varrho_1). \label{red:eq:endpoint-transport-flow} \end{equation} \end{theorem} \begin{proof} \emph{Flow to transport.} Let $a>\max(\norm{J_1}_{\exp L^2},\norm{J_2}_\infty)$ and use the linear density path. Then $|J_2/\varrho_t|\le a/\underline\varrho$. By the weak admissibility identity \eqref{red:eq:weak-div-no-flux}, extending $\varrho_t$ and $J$ by zero outside $D$ produces a distributional continuity equation on $\R^2$ with no boundary source. The standard superposition principle for continuity equations, see \cite{AGS}, then produces a dynamic plan concentrated on trajectories in $\overline D$ whose endpoints satisfy $|y_2-x_2|\le a/\underline\varrho$. Jensen along trajectories and the perspective estimate control the first coordinate exponential-square modular at the same scale, up to a fixed factor. \emph{Transport to flow.} Let $r$ exceed the endpoint transport gauge and fix $r'>r$. Choose a coupling $\pi_0$ with \[ |y_2-x_2|\le r\quad\pi_0\text{-a.s.}, \qquad \int\psi_{r_1}(|y_1-x_1|/r)\dd\pi_0\le1. \] Smooth the two marginals and the coupling by \cref{red:lem:local-smoothing}. For all sufficiently small $\tau$ the resulting coupling $\pi_0^\tau$ has smooth marginals $\varrho_i^\tau=S_\tau\varrho_i$, still satisfying $\underline\varrho\le\varrho_i^\tau\le\overline\varrho$, and \[ |y_2-x_2|\le r'\quad\pi_0^\tau\text{-a.s.}, \qquad \int\psi_{r_1}(|y_1-x_1|/r')\dd\pi_0^\tau\le 1+o_\tau(1). \] Thus it is enough to obtain estimates uniform in $\tau$ for these smooth marginals. Use the explicit smooth convex barriers \[ b_k(s)=k\,(|s|-1)_+^4+k^{-1}s^2. \] They are $C^2$, strictly convex, satisfy $\sup_{|s|\le1}b_k(s)\le k^{-1}$, and diverge uniformly on $\{|s|\ge1+\eta\}$ for every $\eta>0$. For fixed $\tau$, minimize the strictly convex smooth costs \[ L_{r',k}(z)=\psi_{r_1}(|z_1|/r')+b_k(z_2/r'). \] The smoothed benchmark gives a cost bounded uniformly in $k$ and $\tau$. For every $k$, \cref{red:lem:density-interpolation} applies to the smooth marginals and gives $\varrho_t^{\tau,k}\le\varrho_*=\overline\varrho^2/\underline\varrho$. Passing first $k\to\infty$ yields a coupling $\pi^\tau$ satisfying \[ |y_2-x_2|\le r'\quad\pi^\tau\text{-a.s.}, \qquad \int\psi_{r_1}(|y_1-x_1|/r')\dd\pi^\tau\le C, \] and the interpolation density bound passes to the limit by weak convergence against nonnegative test functions. Let $\varrho_t^\tau=(e_t)_\#\pi^\tau$ and $J_t^\tau=(e_t)_\#((y-x)\pi^\tau)$. Disintegrating over $e_t$ shows \[ |J_{t,2}^\tau|\le r'\varrho_t^\tau\le r'\varrho_* \quad\text{a.e.} \] Hence $J^\tau=\int_0^1J_t^\tau\dd t$ satisfies $\norm{J_2^\tau}_\infty\le r'\varrho_*$, while the first component is uniformly bounded in $\exp L^2$ by the same perspective/Jensen estimate as in \cref{red:thm:transport-flow-finite}. Extracting a subsequence as $\tau\downarrow0$, use weak compactness in $L^1$ for the first component and weak-* compactness in $L^\infty$ for the second. The divergence equation passes to the limit and convex lower semicontinuity yields a current $J$ with \[ \divv J=\varrho_0-\varrho_1, \qquad \norm{J_1}_{\exp L^2}+\norm{J_2}_\infty\le C_{\underline\varrho,\overline\varrho}r'. \] Finally let $r'\downarrow r$ and then $r\downarrow\mathfrak T_{(2,\infty)}$. \end{proof} The hard constraint enters this proof only in the limit $k\to\infty$; no uniform $L^\infty$ bound is used for the approximating problems before the convex barriers become infinite. \section{Bounds on the histogram densities} The transport--flow comparison of the previous section requires densities bounded above and below. We check that the histograms satisfy this with probability tending to one. \begin{lemma} \label{red:lem:hist-ellipticity} For every fixed $\varepsilon\in(0,1)$, if $A$ in \eqref{red:eq:h-choice} is sufficiently large, then \[ \Pp\pa{1-\varepsilon\le P_{h_N}^D\mu_N, P_{h_N}^D\nu_N\le1+\varepsilon\text{ on }D}\longrightarrow1. \] \end{lemma} \begin{proof} A terminal cell count has mean $\lambda_N=Nh_N^2\asymp A^2\log N$. Chernoff's inequality and a union bound over the $h_N^{-2}=N/\lambda_N$ cells give \[ \Pp\pa{\max_Q|N\mu_N(Q)-\lambda_N|> \varepsilon\lambda_N} \le\frac{2N}{\lambda_N}e^{-c\varepsilon^2\lambda_N}, \] and similarly for $\nu_N$. \end{proof} Combining \cref{red:prop:atomic-hist,red:thm:transport-flow-finite,red:thm:transport-flow-endpoint,red:lem:hist-ellipticity} proves the equivalence of statements (i) and (ii) in \cref{red:thm:main}. \section{Coupling the histograms with dyadic Gaussian noise} It remains to compare the histogram source with the projected Gaussian source. We use the one-dimensional empirical-process approximation of Koml\'os, Major and Tusn\'ady \cite{KMT1975}, in the nonasymptotic form of Bretagnolle and Massart \cite[Theorem~1]{BretagnolleMassart1989}; see also \cite[Theorem~1.8 and equation~(1.1)]{DudleyUCLT}. For the empirical distribution function $F_N$ of independent uniform variables on $[0,1]$, it provides a coupled standard Brownian bridge $B$ such that, with \[ \Delta_N=\sup_{0\le t\le1}|F_N(t)-t-N^{-1/2}B(t)|, \] one has $\Pp\{\Delta_N>(12\log N+x)/N\}\le2e^{-x/6}$ for all $x\ge0$. In particular there are absolute positive constants $C,c$ such that \begin{equation}\label{red:eq:KMT-cdf} \Pp\{\Delta_N>C(\log N+x)/N\}\le Ce^{-cx}, \qquad N\ge2,\quad x>0. \end{equation} We encode the fixed planar partition by intervals and apply this theorem directly. Since we only compare bounds with fixed constants, a correction of size $O(s_N)$ with probability tending to one suffices. \paragraph{A deterministic divergence inverse.} Resolve the fixed $h=2^{-m}$ square grid by alternating vertical and horizontal equal binary splits, for $2m$ generations. Each level-$k$ rectangle has area $2^{-k}$ and aspect ratio between $1/2$ and $2$. For an internal rectangle $Q$ with children $Q^0,Q^1$, put \[ \mathsf h_Q=|Q|^{-1/2}(\one_{Q^0}-\one_{Q^1}). \] For a vertical split $Q=I\times K$, let $v_I$ vanish at both endpoints of $I$ and satisfy $v_I'=\one_{I^0}-\one_{I^1}$. Set $V_Q=|Q|^{-1/2}v_I(x)\one_K(y)e_1$, extended by zero; for horizontal splits use the analogous $e_2$ field. Integration by parts in the active coordinate gives \begin{equation}\label{red:eq:corrector} \int_D V_Q\cdot\nabla\phi=-\int_D\mathsf h_Q\phi, \qquad \norm{V_Q}_\infty\le C, \qquad \phi\in C^1(\overline D). \end{equation} There is no boundary contribution: on faces normal to the active component the primitive vanishes, and on the other faces it is tangential. If $g$ is constant on the terminal squares and $\int_Dg=0$, set $d_Q=\int_{Q^0}g-\int_{Q^1}g$. The tree Haar basis gives \[ g=\sum_Q\frac{d_Q}{\sqrt{|Q|}}\mathsf h_Q, \qquad R=\sum_Q\frac{d_Q}{\sqrt{|Q|}}V_Q. \] Thus $\divv R=g$ with zero normal flux in the weak sense of \eqref{red:eq:corrector}. At almost every point only one rectangle per generation contributes. Consequently \begin{equation}\label{red:eq:tree-discrepancy-inverse} \norm R_\infty\le C\max_Q|d_Q|\sum_{k=0}^{2m-1}2^{k/2} \le \frac Ch\max_Q|d_Q|. \end{equation} All sums are finite. For $m=0$ the mean-zero grid-constant source is zero. \paragraph{The one-cloud coupling.} Order the terminal squares recursively, placing all descendants of $Q^0$ before those of $Q^1$, and assign them consecutive intervals of length $h^2$ in $[0,1]$. Every tree rectangle $Q$ then corresponds to a single interval $I_Q$ of length $|Q|$. Take the uniform sample in \eqref{red:eq:KMT-cdf}. For each sample variable in a leaf interval $I_S$, independently choose a uniform point in $S$. These points are independent and uniform in $D$, and their counts on tree rectangles equal the interval counts. The construction does not condition on the event where the KMT bound holds. Fixed interval endpoints are avoided almost surely. For $I_S=(a_S,b_S]$, define $W_h^X$ on $S$ by \[ \int_S W_h^X=B(b_S)-B(a_S). \] These masses are centered jointly Gaussian and satisfy \[ \operatorname{Cov}\left(\int_S W_h^X,\int_T W_h^X\right) =|S|\one_{S=T}-|S||T|. \] Hence $W_h^X$ has exactly the law of $P_h^DW^D$, including the mean-zero constraint. For $g=P_h^D\mu_N-1-N^{-1/2}W_h^X$, interval increments of the error in \eqref{red:eq:KMT-cdf} give $|\int_Qg|\le2\Delta_N$ and $|d_Q|\le4\Delta_N$. The deterministic inverse therefore supplies a current $R_h^X$ with \begin{equation}\label{red:eq:one-cloud-KMT} \divv R_h^X=P_h^D\mu_N-1-N^{-1/2}W_h^X, \qquad \Pp\!\left\{\norm{R_h^X}_\infty> C\frac{\log N+x}{Nh}\right\}\le Ce^{-cx}. \end{equation} Its normal flux is zero. The joint construction may depend on $N$ and $h$; this is enough for the probability comparisons below. \begin{theorem}\label{red:thm:KMT} If $Nh^2\asymp\log N$, there is a coupling of two independent uniform point clouds and a field $W_h^D$ having exactly the law of $P_h^DW^D$, together with a zero-normal-flux current $Q_h$, such that \begin{equation}\label{red:eq:KMT-div} \divv Q_h=P_h^D(\mu_N-\nu_N)-\sqrt{2/N}\,W_h^D. \end{equation} For a deterministic $C<\infty$, \begin{equation}\label{red:eq:KMT-infty} \Pp\!\left\{\norm{Q_h}_\infty\le C\frac{\log N}{Nh}\right\}\longrightarrow1. \end{equation} In particular $\norm{Q_h}_{\boldsymbol r}\le Cs_N$ with probability tending to one for every critical exponent pair, including the endpoints. \end{theorem} \begin{proof} Use independent copies of \eqref{red:eq:one-cloud-KMT} for the two clouds and put \[ W_h^D=(W_h^X-W_h^Y)/\sqrt2, \qquad Q_h=R_h^X-R_h^Y. \] The Gaussian law and the divergence identity follow directly. Take $x$ to be a sufficiently large fixed multiple of $\log N$ in \eqref{red:eq:one-cloud-KMT} and use a union bound. Since $Nh^2\asymp\log N$, we have $\log N/(Nh)\asymp s_N$. Finally, on the unit-area square, $\norm f_{\psi_r}\le(\log2)^{-1/r}\norm f_\infty$ for $2\le r<\infty$, and $\norm f_{\psi_\infty}=\norm f_\infty$. \end{proof} The correction $Q_h$ and the Gaussian source $W_h^D$ are in general dependent; the comparison below does not use independence. \section{Proof of the reduction}\label{sec:probability-transfer} We now assemble the previous sections into the proof of \cref{red:thm:main}, and then derive the one-cloud version of the matching estimate. \begin{proof}[Proof of \cref{red:thm:main}] By \cref{red:prop:atomic-hist}, atomic matching and histogram transport are equivalent at scale $s_N$, because $h_N\asymp s_N$. On the event of \cref{red:lem:hist-ellipticity}, the transport--flow comparison \cref{red:thm:transport-flow-finite,red:thm:transport-flow-endpoint} identifies histogram transport with the corresponding static minimum-flow gauge, for finite and endpoint exponents respectively. This proves (i)$\Leftrightarrow$(ii). By \cref{red:lem:flow-lipschitz,red:thm:KMT} and the homogeneity of the minimum-flow gauge, \[ \left|\mathfrak F_{\boldsymbol r}^{D} (P_{h_N}^D(\mu_N-\nu_N)) -\sqrt{\frac2N}\,\mathfrak F_{\boldsymbol r}^{D}(W_{h_N}^{D}) \right|\le\norm{Q_{h_N}}_{\boldsymbol r}. \] By \eqref{red:eq:KMT-infty}, the right-hand side is at most $Cs_N$ with probability tending to one for every critical pair. Intersecting this event with a high-probability bound for either flow gauge transfers that bound to the other after changing its constant; no independence is required. Since $\sqrt{2/N}\,\sqrt{\log N}=\sqrt2\,s_N$, this proves (ii)$\Leftrightarrow$(iii). \end{proof} \subsection{One cloud and Talagrand's evenly spread references} The same argument gives the original one-cloud formulation of the Ultimate Matching Conjecture, and not only its two-sample analogue. It uses no additional Gaussian input. \begin{corollary} \label{red:cor:evenly-spread} Fix a critical exponent pair $\boldsymbol r$ as in \eqref{red:eq:critical-pair} and let $h_N$ satisfy \eqref{red:eq:h-choice}. Suppose that for some deterministic $C$, \begin{equation}\label{red:eq:one-cloud-input} \Pp\!\left(\mathfrak F_{\boldsymbol r}^D(W_{h_N}^D) \le C\sqrt{\log N}\right)\longrightarrow1. \end{equation} Let $Y_1,\ldots,Y_N$ be any deterministic evenly spread family in $D$: there are rectangles $R_i$ partitioning $D$ up to null sets, with $|R_i|=1/N$, $Y_i\in R_i$, and $R_i\subset B(Y_i,10/\sqrt N)$. Then for some deterministic $C'<\infty$, independent of the chosen family, \begin{equation}\label{red:eq:one-cloud-matching} \Pp\!\left(\min_{\sigma\in S_N}\max_{j=1,2} \bigl\|(X_i^j-Y_{\sigma(i)}^j)_{i=1}^N\bigr\|_{\psi_{r_j},N} \le C'\sqrt{\frac{\log N}{N}}\right)\longrightarrow1. \end{equation} \end{corollary} The conclusion includes a hard coordinate endpoint whenever the corresponding Gaussian hypothesis \eqref{red:eq:one-cloud-input} holds; the corollary does not supply that hypothesis. \begin{proof} Let $\varrho_h=P_{h_N}^D\mu_N$. The one-cloud coupling \eqref{red:eq:one-cloud-KMT} gives \[ \varrho_h-1=N^{-1/2}W_{h_N}^X+\divv R_{h_N}^X, \] where $W_{h_N}^X$ has the law of $W_{h_N}^D$ and $\norm{R_{h_N}^X}_\infty\le Cs_N$ with probability tending to one. The same bound holds in the coordinate Orlicz gauge. Therefore \eqref{red:eq:one-cloud-input} and the flow-gauge triangle inequality give \[ \mathfrak F_{\boldsymbol r}^D(\varrho_h-1) \le N^{-1/2}\mathfrak F_{\boldsymbol r}^D(W_{h_N}^X) +\norm{R_{h_N}^X}_{\boldsymbol r} \le C_1s_N \] with probability tending to one. By \cref{red:lem:hist-ellipticity} applied to the first cloud, $1/2\le\varrho_h\le3/2$ on another event of probability tending to one. Hence the transport--flow comparison, \cref{red:thm:transport-flow-finite,red:thm:transport-flow-endpoint}, yields a coupling from $\varrho_h\dd x$ to $\dd x$ whose two coordinate Luxemburg norms are bounded by $C_2s_N$. There is also a coupling from $\mu_N$ to $\varrho_h\dd x$ which moves each atom only inside its $h_N$-cell, with coordinate displacement at most $h_N$. Finally, the deterministic map which sends $x\in R_i$ to $Y_i$ couples Lebesgue measure to $N^{-1}\sum_i\delta_{Y_i}$, moving every point by at most $10/\sqrt N$. Gluing these three couplings and using the Luxemburg triangle inequality gives a coupling $\gamma$ between the two equal-weight atomic measures for which \[ \max_j\|x_j-y_j\|_{L^{\psi_{r_j}}(\gamma)} \le C_2s_N+h_N+10/\sqrt N\le C_3s_N. \] It remains to replace $\gamma$ by a permutation. The coupling $\gamma$ determines a doubly stochastic matrix, so by Birkhoff's theorem it is a convex combination of matching permutations. At a scale slightly larger than $C_3s_N$, the mean over these permutations of the sum of the two coordinate Young-function modulars is at most $2$. Therefore some single permutation has this sum at most $2$, and a fixed further enlargement of scale makes each coordinate modular at most $1$. For $r_j=\infty$, all positive entries of the doubly stochastic matrix already obey the hard coordinate bound; we select a permutation in that support while controlling the remaining finite-coordinate modular. This yields \eqref{red:eq:one-cloud-matching} with a constant uniform over the evenly spread family. \end{proof} \section{From the torus to the square} \label{sec:fold} The reduction requires a Gaussian source on the square, whereas the flow construction of Part~\ref{umc:partI} yields a periodic current on $\T^2=(\mathbb R/\mathbb Z)^2$. In this section we show that folding the torus onto the square relates the two projected sources exactly, so that a torus bound implies the corresponding square bound (\cref{prop:torus-implies-square}). \subsection{Folding of currents} On $\T^1=[0,1)$ define \[ F_1(u)=2\min\{u,1-u\},\qquad F=(F_1,F_1):\T^2\longrightarrow D=[0,1]^2. \] The notation $F=(F_1,F_1)$ means $F(u_1,u_2)=(F_1(u_1),F_1(u_2))$. Outside the folds $u_i=0,1/2$, the derivative is $DF(u)=2R_u$, where $R_u$ is diagonal with entries $\pm1$. Each $y\in(0,1)^2$ has exactly four preimages. For integrable scalar functions and vector fields define \begin{equation} (\mathscr S_0 f)(y)=\frac14\sum_{F(u)=y}f(u),\qquad (\mathscr S_1 J)(y)=\frac12\sum_{F(u)=y}R_u J(u). \label{eq:unit-fold} \end{equation} These formulas are the pushforward against normalized Lebesgue measure; in particular $F_\#\mathrm du=\mathrm dy$. For mean-zero distributions the scalar pushforward is defined by $\langle\mathscr S_0 f,\varphi\rangle =\langle f,\varphi\circ F\rangle$. \begin{lemma} \label{lem:unit-fold-current} Suppose $J\in L^1(\T^2;\mathbb R^2)$ and $\operatorname{div}_{\T}J=f$ in distributions. Then $J^D=\mathscr S_1J$ satisfies \[ -\int_D J^D\cdot\nabla\varphi\,\dd y =\langle\mathscr S_0f,\varphi\rangle \quad\text{for every }\varphi\in C^1(\overline D). \] In particular, $J^D$ has zero normal trace on $\partial D$. For every Young function $\Theta$ and $j=1,2$, \begin{equation} \int_D\Theta\left(\frac{|J_j^D|}{2a}\right)\dd y \le\int_{\T^2}\Theta\left(\frac{|J_j|}{a}\right)\dd u, \qquad \|J_j^D\|_{L^\Theta(D)}\le2\|J_j\|_{L^\Theta(\T^2)}. \label{eq:unit-fold-norm} \end{equation} The same inequality holds when $L^\Theta=L^\infty$. \end{lemma} \begin{proof} On every inverse sheet $\dd u=\dd y/4$, and the chain rule gives $\nabla(\varphi\circ F)(u)=2R_u\nabla\varphi(F(u))$. Thus \[ -\int_{\T^2}J\cdot\nabla(\varphi\circ F)\,\dd u =-\int_D\left(\frac12\sum_{F(u)=y}R_uJ(u)\right) \cdot\nabla\varphi(y)\,\dd y. \] For an $L^1$ current, testing with the periodic Lipschitz function $\varphi\circ F$ is justified by smooth approximation; in our application the source is a finite-dimensional Gaussian cell function. The distributional divergence identity on the torus gives the first claim. Since the test functions need not vanish on $\partial D$, this identity also shows that the fold creates no boundary flux. Finally $J_j^D=2\cdot(1/4)\sum\varepsilon_{u,j}J_j(u)$, and Jensen's inequality together with $F_\#\mathrm du=\mathrm dy$ proves \eqref{eq:unit-fold-norm}. \end{proof} \subsection{Compatibility with dyadic projections} Recall that $P_h^{\T}$ and $P_h^D$ are the fixed-grid conditional expectations. Restricted to $L_0^2$, their ranges are the mean-zero cellwise-constant subspaces. Accordingly the projected noises $W_h^{\T},W_h^D$ have exactly the meaning specified in \eqref{red:eq:compatible-source}. If $h=2^{-m}$ is a square mesh, take the torus mesh \[ \widetilde h=h/2=2^{-(m+1)}. \] The four folding lines coincide with grid lines at this mesh. Thus every torus $\widetilde h$-square is mapped bijectively onto a square $h$-square, and the following deterministic relation holds up to cell-boundary null sets: \begin{equation} \mathscr S_0P_{\widetilde h}^{\T}f =P_h^{D}\mathscr S_0f. \label{eq:commute-projections} \end{equation} \begin{lemma} \label{lem:noise-fold} Let $W^{\T}$ be centered mean-zero isonormal white noise over $L_0^2(\T^2)$. For $h=2^{-m}$ and $\widetilde h=h/2$, \begin{equation} \mathscr S_0W_{\widetilde h}^{\T} \ \stackrel{\mathrm{law}}=\ P_h^{D}W^D, \label{eq:exact-fold-law} \end{equation} where $W^D$ is standard mean-zero white noise on $D$. \end{lemma} \begin{proof} For deterministic $\phi,\psi\in L_0^2(D)$, the isonormal covariance of the folded field is \begin{align*} \mathbb E\bigl[\langle\mathscr S_0W_{\widetilde h}^{\T}, \phi\rangle\langle\mathscr S_0W_{\widetilde h}^{\T}, \psi\rangle\bigr] &=\langle P_{\widetilde h}^{\T}(\phi\circ F), P_{\widetilde h}^{\T}(\psi\circ F)\rangle_{L^2(\T^2)}\\ &=\langle P_h^{D}\phi,P_h^{D}\psi\rangle_{L^2(D)}. \end{align*} The second equality holds because $F$ is measure preserving and the reflected dyadic mesh is exactly aligned. Both fields are centered Gaussian on the same finite-dimensional cell space, so equality of covariances proves equality in law. \end{proof} \begin{proposition} \label{prop:torus-implies-square} For every pair $\boldsymbol r$ for which the coordinate Orlicz spaces are defined, and every $m\ge1$, \[ \mathbb P\left\{\mathfrak F_{\boldsymbol r}^{D} (P_{2^{-m}}^{D}W^D)\le2t\right\} \ge \mathbb P\left\{\mathfrak F_{\boldsymbol r}^{\T} (P_{2^{-(m+1)}}^{\T}W^{\T})\le t\right\}, \qquad t>0. \] Consequently, a bound $O(\sqrt m)$ with high probability for the unit-torus Gaussian flow implies the identical exact-cell-source square Gaussian estimate, with at most a universal change of constant. \end{proposition} \begin{proof} For any torus sample and any admissible periodic current, use \cref{lem:unit-fold-current} to obtain a no-flux square current for the folded source, at coordinate cost at most twice the torus cost. Take infima (or approximating sequences) and then use \cref{lem:noise-fold} to identify the source law. \end{proof} \begin{remark} A general no-flux square current cannot be pulled back to the torus at unchanged Gaussian source law simply by reversing the four-sheeted averaging: the fold destroys information from three independent complementary sheet modes. The matching--square-flow equivalence is symmetric, but the folding result is used only in the torus-to-square direction. \end{remark} \section{Choice of dyadic mesh} \label{sec:alignment} The torus Gaussian bound is formulated along the meshes $h_m=2^{-m}$ and expressed in $\sqrt m$, whereas the matching reduction uses the mesh $h_N$ of \eqref{red:eq:h-choice}. We check that these two parametrizations describe the same family of dyadic projections, and then combine the previous sections into a matching criterion stated on the torus. In the square matching reduction one chooses a dyadic $h_N$ with $A\sqrt{\log N/N}\le h_N<2A\sqrt{\log N/N}$. Given $N$, write $h_N=2^{-m(N)}$, so that \[ m(N)=\frac{\log N}{2\log2}+O(\log\log N),\qquad \sqrt{m(N)}\asymp\sqrt{\log N}. \] The square-source estimate follows from the torus theorem at the single adjacent depth $m(N)+1$. Conversely, a square-source high-probability bound formulated at every $h_N$ gives a bound along every $m$ by choosing integers $N_m$ with \[ N_m\asymp A^2m4^m, \qquad A\sqrt{\frac{\log N_m}{N_m}}\le2^{-m} <2A\sqrt{\frac{\log N_m}{N_m}}. \] For instance, one may take $N_m=\lceil4A^2(\log 2)m4^m\rceil$: then $A\sqrt{(\log N_m)/N_m}/2^{-m}\to1/\sqrt2$, which lies strictly between $1/2$ and $1$. This reverse calibration concerns the square Gaussian gauges; it does not reverse the torus fold. \begin{corollary} \label{cor:big-picture} Fix $\boldsymbol r$ satisfying \eqref{red:eq:critical-pair}. Suppose that there exist constants $B,p_0>0$ such that, for every sufficiently large $m$, \[ \mathbb P\left\{\mathfrak F_{\boldsymbol r}^{\T} (P_{2^{-m}}^{\T}W^{\T})\le B\sqrt m\right\} \ge p_0. \] Then Talagrand's square matching estimate follows: \[ \mathbb P\left\{\mathfrak M_N^{\boldsymbol r} \le C'\sqrt{\log N/N}\right\}\longrightarrow1 \] for a deterministic $C'<\infty$. \end{corollary} \begin{proof} \Cref{red:prop:gaussian-concentration}, applied on $\T^2$, upgrades the assumed lower probability bound to an exponentially small failure probability at a possibly larger constant. Choose the square mesh $h_N$ as in \cref{sec:alignment} and apply \cref{prop:torus-implies-square} at the adjacent torus mesh $h_N/2$. This gives the square Gaussian gauge estimate with high probability. The three-way matching--histogram--square-Gaussian equivalence \cref{red:thm:main} finishes the proof, since $\sqrt{2/N}\,\sqrt{\log N}=\sqrt2s_N$. \end{proof} Thus the fixed-positive-probability estimate of \cref{thm:gauss} suffices for the finite critical matching estimates. Part~\ref{umc:partIII} shows that this torus criterion cannot hold at the hard endpoint $(2,\infty)$, since its square consequence is false. \section{Consequences and the hard endpoint} \label{red:sec:consequences} We conclude Part~\ref{umc:partII} by recording what the reduction gives for the quartic pair and for the hard endpoint, which is the subject of Part~\ref{umc:partIII}. The reduction of \cref{red:thm:main} applies to every critical exponent pair, including both hard-coordinate endpoints. The different outcomes for finite and infinite exponents concern the Gaussian minimum-flow estimate itself, not the validity of the matching--Gaussian comparison. \paragraph{The quartic case.} For $\boldsymbol r=(4,4)$, the isotropic comparison \cref{red:lem:isotropic-quartic} identifies the matching statement, up to universal changes of constants, with the square Gaussian estimate \[ \mathbb P\!\left\{ \inf_{\substack{\operatorname{div}J=W_{h_N}^{D}\\J\cdot n=0}} \|J\|_{\exp L^4(D)}\le C\sqrt{\log N}\right\} \longrightarrow 1. \] Thus \cref{thm:gauss}, combined with \cref{cor:big-picture}, resolves the finite critical pairs, including the symmetric quartic case. \paragraph{The hard endpoint.} For $\boldsymbol r=(2,\infty)$ the corresponding square gauge is \[ \mathfrak F_{(2,\infty)}^D(W_{h_N}^{D}) =\inf_{\substack{\operatorname{div}J=W_{h_N}^{D}\\J\cdot n=0}} \max\{\|J_1\|_{\exp L^2(D)},\|J_2\|_{\infty}\}. \] The comparison theorem remains applicable, including the $L^\infty$ coordinate: the endpoint KMT correction is of the target order and can be absorbed by changing the fixed constant. However, \cref{neg:thm:Gaussian} in Part~\ref{umc:partIII} shows that, on the square, \[ \frac{\mathfrak F_{(2,\infty)}^D(W_{2^{-m}}^D)}{\sqrt m} \xrightarrow{\mathbb P}+\infty. \] Consequently the hard-endpoint matching assertion is false (\cref{neg:thm:main}). The one-way torus-to-square implication of \cref{prop:torus-implies-square} also rules out a fixed-positive-probability torus upper bound at this endpoint. \paragraph{Dual tests and the cutoff.} The complementary Young function to $\psi_r(t)=e^{|t|^r}-1$ grows, for large arguments, as $q(\log q)^{1/r}$; near zero it grows as $q^{r/(r-1)}$. At $r=2$, the admissible dual gradient cost is therefore comparable to $|q|\sqrt{\log(2+|q|)}$, while the associate coordinate for $L^\infty$ is $L^1$. The lower-bound argument of Part~\ref{umc:partIII} only needs a weak Orlicz--H\"older inequality for smooth potentials (\cref{neg:lem:dual}); it does not use an identification of the full Banach dual of an exponential Orlicz space. Such an identification would require care, because $\psi_r$ does not satisfy the $\Delta_2$ condition at infinity. The dyadic projection is necessary: unregularized planar white noise is almost surely not in $H^{-1}$, so it has no static $L^2$ divergence current. The specific matching mesh $h_N\asymp\sqrt{\log N/N}$, however, is fixed by the occupancy and histogram-regularization requirements of this part, not by the ultraviolet divergence alone. \clearpage \part{Failure of the hard endpoint}\label{umc:partIII} In this part we prove that at the hard coordinate endpoint $(2,\infty)$, and by symmetry at $(\infty,2)$, the optimal matching cost divided by $\sqrt{\log N/N}$ diverges in probability (\cref{neg:thm:main}). The obstruction we show here is qualitative, but a determination of the exact divergence rate of the cost can be found in the GitHub repository~\cite{umc:repository} associated to this file. \section{Statement, mechanism, and Gaussian reduction} \label{neg:sec:introduction} This section states the result, explains the mechanism, and reduces the theorem to a statement about the projected Gaussian source on the square. This part uses its own notation; where an object coincides with one of Part~\ref{umc:partII}, we say so. Let $D=[0,1]^2$, let $(X_i)_{i=1}^N$ and $(Y_i)_{i=1}^N$ be independent uniform samples in $D$, and write \[ s_N=\sqrt{\frac{\log N}{N}},\qquad \psi_2(t)=e^{t^2}-1. \] For a permutation $\pi$, denote by $\Delta_j^\pi$ the $N$-vector of $j$th coordinate displacements $X_i^j-Y_{\pi(i)}^j$, with the normalized Luxemburg norm $\|\cdot\|_{\psi_2,N}$. Put \begin{equation}\label{neg:eq:matching-cost} M_N=\min_{\pi\in S_N} \max\left\{\|\Delta_1^\pi\|_{\psi_2,N}, \max_i|\Delta_2^\pi(i)|\right\}. \end{equation} \begin{theorem} \label{neg:thm:main} For the two independent point clouds, \begin{equation}\label{neg:eq:negative-result} \frac{M_N}{s_N}\xrightarrow[N\to\infty]{\mathbb P}+\infty. \end{equation} The assertion holds after exchanging the coordinate axes. In fact, the following one-cloud conclusion is uniform over \emph{all} deterministic reference configurations $z^{(N)}=(z_i^{(N)})_{i=1}^N\in D^N$, including the evenly spread configurations of \cref{red:cor:evenly-spread}. For every fixed $A<\infty$, \begin{equation}\label{neg:eq:reference-negative} \sup_{z^{(N)}\in D^N}\mathbb P\!\left\{ \min_{\pi\in S_N}\max\left( \|(X_i^1-z_{\pi(i)}^{(N),1})_i\|_{\psi_2,N}, \max_i|X_i^2-z_{\pi(i)}^{(N),2}| \right)\le A s_N\right\}\longrightarrow0. \end{equation} \end{theorem} \paragraph{Idea of the proof.} Part~\ref{umc:partI} establishes \emph{upper} Gaussian-flow bounds for finite exponents, which Part~\ref{umc:partII} transfers to matching. Here we work in the opposite direction: we build random test potentials of bounded dual cost whose expected pairing with the projected noise is too large for any fixed endpoint bound. Three probability spaces are involved. Brownian motion supplies a model martingale; an auxiliary uniform point in the square samples the gradients of a deterministic tree of triangles; finally, Gaussian white noise selects the signs of the bubbles on that tree. The spatial-branch law holds for \emph{every} assignment of signs, whereas Gaussian freshness produces a positive expectation over those signs. The two conditionings must be kept separate. \paragraph{Why the endpoint is different.} The complementary Young function to $\exp(t^2)-1$ grows like $|v|\sqrt{\log(2+|v|)}$. Thus a dual gradient potential $u$ is charged according to \begin{equation}\label{neg:eq:PhiPsi} \Phi(a)=\int_0^{|a|}\sqrt{\log(2+r)}\,\mathrm dr, \qquad \Psi(a,b)=\Phi(a)+|b|. \end{equation} We seek an upper-triangular predictable control for the Brownian martingale $\int_0^1L_t\,\mathrm dB_t$ with bounded terminal $\Psi$-cost but large \begin{equation}\label{neg:eq:detreward} \mathbb E\int_0^1\sqrt{\det L_t}\,\mathrm dt. \end{equation} In an upper-triangular parametrization, $Y_t>0$ defines a change of probability measure, and the ratio $M_t=X_t/Y_t$ is a centered martingale under the $Y_t$-biased law. A sequence of short local-time stages generates a large accumulated reward while keeping $|M_t|$ small when the logarithmic weight on $Y_t$ is large. The terminal cost therefore remains uniformly bounded even as the reward increases. The obstruction is thus structural; it is not the failure of a particular candidate flow. Spatially, a determinant-one chart converts $\sqrt{\det L_t}$ into the $L^2$ size of a bubble. Choosing its sign from fresh Gaussian noise turns this size into a positive expected pairing. Unlike a deterministic test, the adaptive potentials exploit this reward without increasing the convex terminal gradient cost. \paragraph{Plan of the proof.} The argument has four steps: Brownian reward (\cref{neg:lem:Brownian}), spatial realization (\cref{neg:lem:spatial}), a weak dual lower bound (\cref{neg:lem:dual}), and concentration together with the matching transfer (\cref{neg:thm:Gaussian}). The Brownian and spatial steps are new. From Part~\ref{umc:partII} we use the square-flow Lipschitz estimate \cref{red:lem:uniform-gaussian-lipschitz}, the triangle inequality of \cref{red:lem:flow-lipschitz}, the mesh choice \eqref{red:eq:h-choice}, and the probability-level comparisons of \cref{red:prop:atomic-hist}, \cref{red:thm:transport-flow-endpoint}, \cref{red:lem:hist-ellipticity} and \cref{red:thm:KMT}. The argument takes place directly on the square: the torus-to-square folding of \cref{sec:fold} goes in one direction only, and it is not reversed here. \paragraph{The square Gaussian source.} Let $\mathcal W$ be isonormal white noise on $L^2(D)$, with mean-zero restriction $\mathcal W_0$. Let $P_h^D$ be cellwise averaging on the fixed dyadic square grid of mesh $h=2^{-m}$, and put $W_h=P_h^D\mathcal W_0$. Thus $W_h$ is a random \emph{function}, not a continuum white-noise distribution; its pairing with a test $u$ equals $\mathcal W(P_h^Du)$ even for a random $u$ (interpreted through the finite cell coefficients). For a mean-zero distribution $f$, define the square, no-flux minimum-flow gauge \begin{equation}\label{neg:eq:flow-gauge} F_h(f)=\inf_{\substack{J=(J_1,J_2)\\ -\int_DJ\cdot\nabla\varphi=\langle f,\varphi\rangle\ \forall\varphi\in C^\infty(\overline D)}} \max\{\|J_1\|_{L^{\psi_2}(D)},\|J_2\|_{L^\infty(D)}\}, \qquad \mathcal F_m=F_{2^{-m}}(W_{2^{-m}}). \end{equation} The subscript $h$ merely indicates the mesh on which $f$ is supported; admissible currents are continuum fields. Thus $F_h$ is the gauge $\mathfrak F_{(2,\infty)}^{D}$ of \eqref{red:eq:aniso-flow}, and $W_h$ is the square source $W_h^D$ of \eqref{red:eq:compatible-source}. The Gaussian coefficient representation is \begin{equation}\label{neg:eq:coefficients} W_h|_Q=h^{-1}(Z_Q-\overline Z),\qquad \overline Z=h^2\sum_Q Z_Q, \end{equation} with independent standard normal $Z_Q$, as in \eqref{red:eq:gaussian-coefficients}. By \cref{red:lem:uniform-gaussian-lipschitz}, the coefficient-space gauge $Z\mapsto\mathcal F_m(Z)$ is $C_{\mathrm{Lip}}$-Lipschitz with $C_{\mathrm{Lip}}$ independent of $m$. We shall prove \begin{equation}\label{neg:eq:gaussian-negative} \frac{\mathcal F_m}{\sqrt m} \xrightarrow[m\to\infty]{\mathbb P}+\infty \end{equation} for this square source, and then deduce \cref{neg:thm:main} in \cref{neg:sec:conclusion} from the eventwise comparisons of Part~\ref{umc:partII}. \section{A pathwise dual inequality} \label{neg:sec:duality} The test potentials of \cref{neg:sec:spatial} are chosen from the noise itself. The following inequality bounds their pairing with the source by the flow gauge, pathwise, so that no independence between test and noise is needed. \begin{lemma}\label{neg:lem:dual} There is an absolute constant $C_D$ such that, for every dyadic $h$, for every mean-zero Lipschitz $u:D\to\mathbb R$, and for every realized finite-dimensional source $W_h$, \begin{equation}\label{neg:eq:dual-bound} |\mathcal W(P_h^Du)| \le C_D\,F_h(W_h) \left(1+\int_D\Psi(\partial_1u,\partial_2u)\,\mathrm dx\right). \end{equation} The assertion holds pathwise even when the test $u$ is selected measurably from the same noise $\mathcal W$. \end{lemma} \begin{proof} The complementary Young bound, with an absolute $C$, is \begin{equation}\label{neg:eq:Young} |ab|\le e^{a^2}-1+C\Phi(b)\qquad (a,b\in\mathbb R). \end{equation} Indeed, if $|a|\le2\sqrt{\log(2+|b|)}$, the term $C\Phi(b)$ pays for $|ab|$. Otherwise $|b|\le e^{a^2/4}$ and $|a|\ge2\sqrt{\log2}$, so $|a|e^{a^2/4}\le C(e^{a^2}-1)$; the lower bound on $|a|$ justifies this estimate for small arguments as well. Fix $r>F_h(W_h)$ and an admissible current with $\|J_1\|_{\psi_2},\|J_2\|_\infty\le r$. The weak no-flux identity, first for smooth tests and then by Lipschitz approximation, gives \[ |\mathcal W(P_h^Du)|=\left|\int_DJ\cdot\nabla u\right| \le r\left(1+C\int_D\Phi(\partial_1u) +\int_D|\partial_2u|\right). \] Here $\int_D(e^{(J_1/r)^2}-1)\le1$ because $D$ has unit area. Taking the infimum in $r$ proves the claim. The inequality is deterministic for each current and each test, so it does not use any independence between $u$ and the noise. Since $P_h^Du$ lies in a fixed finite cell space, its pairing with the realized $W_h$ is unambiguous. \end{proof} \section{Brownian local time and arbitrarily large reward} \label{neg:sec:brownian} This section constructs the model martingale: a bounded, predictable, upper-triangular control whose terminal $\Psi$-cost is bounded by an absolute constant while its determinant reward \eqref{neg:eq:detreward} is as large as we wish. It is the only place where the endpoint structure of $\Psi$ enters. \begin{lemma} \label{neg:lem:Brownian} For every $D_0>0$ there are deterministic times $0=t_00\right\}, \qquad j=1,\dots,K, \] such that $L_t=L_j(B_{t_1}-B_{t_0},\dots,B_{t_{j-1}}-B_{t_{j-2}})$ on $[t_{j-1},t_j)$ is predictable, takes values in a fixed compact set of upper-triangular matrices with positive diagonals, and \begin{equation}\label{neg:eq:Brownian-claim} \mathbb E\,\Psi\!\left(\int_0^1L_t\,\mathrm dB_t\right) \le C_*,\qquad \mathbb E\int_0^1\sqrt{\det L_t}\,\mathrm dt\ge D_0, \end{equation} where $C_*<\infty$ is absolute, independent of $D_0$. \end{lemma} \paragraph{Heuristics.} The following formal computation explains why bounded cost and unbounded reward are compatible; it is not used in the proof. In the construction below, the control satisfies $\sigma_t\beta_t=R$ during an active stage. The reward rate $\sqrt{\det\mathcal L_t} =Y_t\sqrt R$ therefore has conditional mean $\sqrt R$, and the reward is $\sqrt R$ times the active real duration. Stage $j$ lasts $R_j/R$ in real time, so the reward is of order $\sqrt R$ as soon as $\sum_jR_j\ge R$. Each stage contributes a resource of order $q\sqrt{K_q}$, with $q=\gamma/J$ and $K_q\approx2\log(J/\gamma)$. Summing over $J$ stages gives $\sum_jR_j\approx\gamma\sqrt{2\log(J/\gamma)}$, hence a reward of order $(\log J)^{1/4}$, which is unbounded in $J$. The cost stays bounded because the two coordinates are tied together. By stage $j$, the multiplicative coordinate $Y$ has accumulated logarithmic variance $s\le2S_j$, which inflates the logarithmic weight in $\Phi$ by a factor $\sqrt{1+s}\lesssim\sqrt{S_j}$. Meanwhile the local-time stopping confines the ratio $M=X/Y$ to $|M|\le b_j=S_j^{-1/2}$. The product $|M|\sqrt{1+s}$ is therefore bounded at every stage; this is the content of \eqref{neg:eq:terminal-cost}. In this way the $\sqrt{\log}$ growth of the complementary Young function is exactly compensated by the confinement $|M|\le S_j^{-1/2}$. \begin{proof} \emph{Step 1: independent operational stages.} Each successful stage gains about $q\sqrt{\log(1/q)}$, at the price of a failure probability $O(q)$. With $J$ stages and $q=\gamma/J$, the total resource $\sum_jR_j$ defined below grows like $\sqrt{\log J}$, while the probability of some failure stays small. Fix a small absolute $\gamma>0$ and an integer $J\gg1$. Put \[ \begin{gathered} q=\frac\gamma J,\qquad K_q=\log(1+q^{-2}),\qquad S_j=2^j,\qquad b_j=S_j^{-1/2},\qquad \ell_j=q b_j,\\ w_j(v)=\frac{S_j}{K_q(v+\ell_j^2)},\qquad 0\le v\le b_j^2. \end{gathered} \] In stage $j$ let $U$ be a fresh standard real Brownian motion started at $0$, and stop it at \[ T_j=\tau_{\ell_j}\wedge\tau_{\{\pm b_j\}}\wedge b_j^2, \] where $\tau_{\ell}$ is the first time that Tanaka-normalized local time at $0$ reaches $\ell$. A stage succeeds when $\tau_{\ell_j}$ is the first stopping event; then $U_{T_j}=0$. Put \[ R_j=\int_0^{T_j}\sqrt{w_j(v)}\,\mathrm dv. \] Tanaka's formula at the bounded stop yields $\mathbb E|U_{T_j}|=\mathbb EL^0_{T_j}\le\ell_j$, so $\mathbb P(\text{exit at }\pm b_j)\le q$. L\'evy's local-time identity and Brownian reflection imply \begin{equation}\label{neg:eq:local-tail} c\min\{1,\ell/\sqrt v\} \le\mathbb P(\tau_\ell>v) \le C\min\{1,\ell/\sqrt v\} \qquad(\ell,v>0). \end{equation} The probability of reaching the time cap without success is therefore at most $Cq$. For a sufficiently small absolute $c_0$ and $\ell_j^2\le v\le c_0b_j^2$, subtracting the $O(q)$ total failure probability from the left side of \eqref{neg:eq:local-tail} gives \begin{equation}\label{neg:eq:successful-tail} \mathbb P(T_j>v,\ \text{stage $j$ succeeds}) \ge c\frac{\ell_j}{\sqrt v}. \end{equation} Consequently, by the tail integral formula, \begin{equation}\label{neg:eq:resource-bounds} \begin{aligned} \mathbb E[R_j;\,\text{success}]&\ge \frac{c\sqrt{S_j}\,\ell_j}{\sqrt{K_q}} \int_{\ell_j^2}^{c_0b_j^2}\frac{\mathrm dv}{v} \ge c q\sqrt{K_q},\\ 0\le R_j&\le\frac{\sqrt{S_j}}{\sqrt{K_q}} \int_0^{b_j^2}\frac{\mathrm dv}{\sqrt{v+\ell_j^2}} \le\frac2{\sqrt{K_q}}. \end{aligned} \end{equation} Each operational stage succeeds with probability at least $1-Cq$. Thus the probability that all $J$ stages succeed is at least $1-CJq\ge1-C\gamma$. Conditioned on this event, the resources remain independent. Their sum $S_R$ satisfies \[ \mu_J:=\mathbb E[S_R\mid\text{all succeed}] \ge cJq\sqrt{K_q}=c\gamma\sqrt{K_q}\longrightarrow\infty, \qquad \operatorname{Var}(S_R\mid\text{all succeed}) \le\frac{2}{\sqrt{K_q}}\mu_J. \] Fix any prescribed real-time resource $R>0$. Once $\gamma$ is chosen sufficiently small and $J$ is large enough that $\mu_J\ge2R$, Chebyshev's inequality gives $\mathbb P(S_R0$, the deterministic partition and the terminal estimate are then as claimed, with an absolute $C_*$ after increasing it once. \end{proof} \section{Realization by spatially adapted Gaussian bubbles} \label{neg:sec:spatial} This section transfers the Brownian model of \cref{neg:lem:Brownian} to test potentials on $D$ (\cref{neg:lem:spatial}). The model is realized without sampling Brownian motion in physical space: the Brownian increments are replaced by bounded innovations, obtained by following the children of a uniform spatial point through a tree of triangles. The family of bubbles is chosen with the help of white noise, but the resulting spatial estimates hold for every fixed realization of the chosen signs. \paragraph{Reference geometry.} Let $T=\{(x_1,x_2):x_1,x_2\ge0,\ x_1+x_2\le1\}$. Choose four disjoint small axis-parallel dyadic squares strictly inside $T$, whose centers form a square, and put a translated tent $(r-\|x-c\|_\infty)_+$ in each, with checkerboard signs. Their sum $b$ is compactly supported in $\operatorname{int}T$, piecewise affine, and satisfies \begin{equation}\label{neg:eq:affine-moments} \int_T b(x)\,\ell(x)\,\mathrm dx=0 \quad\text{for every affine }\ell. \end{equation} The checkerboard arrangement cancels both the total mass and the first moments; this is what will give Gaussian orthogonality between ancestors and descendants. The law of $\nabla b(x)$ for uniform $x\in T$ is supported on $0,\pm e_1,\pm e_2$ and symmetric under signed coordinate permutations. Choose $c_b>0$ so that $|T|^{-1}\int_T\nabla b\otimes\nabla b=c_b^2 I$, and denote by $\nu$ the law of $c_b^{-1}\nabla b(U)$ for uniform $U\in T$. Then $\nu$ is bounded, centered, has covariance $I$, and is invariant under signed coordinate permutations. Fix a sufficiently fine finite dyadic triangulation into congruent right-isosceles triangles of leg $2^{-k_0}$ on which $b$ is affine. The integer $k_0$ and template $b$ are absolute. \paragraph{Two conditional laws.} The next two lemmas serve different purposes. In the first, we condition on earlier Gaussian observations to justify freshness of the next selected bubble. In the second, we freeze all bubble signs and condition on the spatial branch to obtain the i.i.d.\ innovation law and Jensen's stopping inequality. \begin{lemma} \label{neg:lem:fresh} Suppose that a finite sequence of nonzero Hilbert-space vectors is chosen predictably from an isonormal Gaussian process: at step $k$, the vector $b_k$ is measurable with respect to previously revealed pairings $\mathcal W(b_1),\dots,\mathcal W(b_{k-1})$, takes one of finitely many values on each finite history, and is orthogonal to every previous vector on that history. Let $\mathcal G_{k-1}$ denote that history. Then, conditionally on $\mathcal G_{k-1}$, $\mathcal W(b_k)$ is centered Gaussian with variance $\|b_k\|_2^2$. For every finite-valued $\mathcal G_{k-1}$-measurable Hilbert-space vector $v_k$, \begin{equation}\label{neg:eq:conditional-sign} \mathbb E\left[\operatorname{sgn}\mathcal W(b_k) \;\mathcal W(v_k)\mid\mathcal G_{k-1}\right] =\sqrt{\frac2\pi}\, \frac{\langle v_k,b_k\rangle}{\|b_k\|_2}. \end{equation} \end{lemma} Here $v_k$ may itself depend on the previous observations and need not be orthogonal to previous directions. The finite-valued assumption is the one used below, and it makes all random isonormal pairings unambiguous. \begin{proof} We proceed inductively through the finite decision tree. On any branch determined by previously observed pairings, the selected directions are fixed vectors, and the event selecting that branch is measurable with respect to their Gaussian coordinates. Conditional on the actual observed values, the restriction of white noise to the orthogonal complement of their span remains a fresh isonormal Gaussian process. The next direction belongs to this complement, so its normalized pairing is a fresh standard Gaussian. Decompose $v_k=\langle v_k,b_k\rangle b_k/\|b_k\|_2^2+r_k$. Although $\mathcal W(r_k)$ may have a past-measurable component, its remaining centered component is conditionally independent of $\mathcal W(b_k)$, and the past-measurable component has zero pairing with its centered sign. Since $\mathbb E|G|=\sqrt{2/\pi}$ for $G\sim N(0,1)$, the formula follows. An induction over the finite histories justifies the conditioning. \end{proof} \begin{lemma} \label{neg:lem:branch} Fix a bounded continuous simple control $L$ as in \cref{neg:lem:Brownian} and a deterministic assignment $Q\mapsto s(Q)\in\{\pm1\}$ of signs to all potential parents. Let $U$ be uniform on $D$, independent of an auxiliary sequence of independent $\nu$-distributed vectors, and let $(\mathcal S_k)_{k=0}^n$ be a filtration with the following properties. \begin{enumerate}[label=(\alph*),leftmargin=2.5em] \item \emph{Complete congruent subdivisions.} Before microstep $k$, the active triangle containing $U$ is $\mathcal S_{k-1}$-measurable and has the form $Q=x_Q+d_QAO_QT$, where $A$ is the current chart, $O_Q$ is a signed coordinate permutation, and $x_Q,d_Q,A,O_Q$ are $\mathcal S_{k-1}$-measurable. Its children are the images of the reference triangulation under the same affine map. \item \emph{Signs fixed before revelation.} The increment recorded at microstep $k$ is \[ \xi_k=s(Q)\,O_Q\,c_b^{-1}\nabla b(\widehat U), \qquad \widehat U=d_Q^{-1}O_Q^TA^{-1}(U-x_Q)\in T, \] and $\mathcal S_k$ is generated by $\mathcal S_{k-1}$ and the child of $Q$ containing $U$. \item \emph{Predictable retiling.} At the end of a control block, each active triangle is retiled into complete triangles of the form in (a) for the new chart, plus a discarded remainder. Which retained triangle contains $U$, or whether $U$ lies in the discarded remainder, is $\mathcal S$-measurable before the next microstep. After $U$ is discarded, the increments $\xi_k$ are the subsequent auxiliary $\nu$-vectors, which are added to $\mathcal S_k$. \item \emph{No further information.} Apart from (a)--(c), $\mathcal S_k$ carries no information on the position of $U$ inside its current triangle. \end{enumerate} The bubble construction of \cref{neg:lem:spatial}, with all signs frozen after the white noise is fixed, satisfies (a)--(d). Then the ideal increments $\xi_1,\dots,\xi_n$ are independent with law $\nu$. The actual gradient path is the ideal vector martingale stopped at the first discarded region. In particular, for the convex $\Psi$ in \eqref{neg:eq:PhiPsi}, \begin{equation}\label{neg:eq:stopped-Jensen} \mathbb E_{U\sim\mathrm{Unif}(D)} \Psi(G_{\tau}(U)) \le\mathbb E_{U,\mathrm{aux}}\Psi(G_n), \end{equation} where $G_k$ is the corresponding ideal gradient martingale, $\tau$ is the geometric stopping index, and the expectation on the right also includes the independent innovations appended after loss. \end{lemma} Once all signs are frozen, \cref{neg:lem:branch} is a statement about the uniform point $U$ and the auxiliary innovations only; the white noise plays no role in it. \begin{proof} Fix all bubble signs. With $U$ uniform on $D$, the initial current triangle is one of two congruent halves of $D$. Conditional on its identity, $U$ is uniform in that triangle. At a microstep, the complete affine child triangulation partitions the parent into congruent equal-area triangles. Thus the normalized reference increment has law $\nu$, irrespective of the parent, its signed permutation $O$, and the already fixed sign. Conditional on any retained new-chart triangle after a retiling, $U$ is again uniform in that triangle. The portion discarded at a retiling is identified \emph{before} the next increment; thereafter fresh independent artificial $\nu$-increments supply the ideal continuation. By (b)--(d), the filtration records the active parent, the selected child and its innovation, each retiling/survival decision, and the appended innovations after loss; it does not disclose the exact location of $U$ inside the current triangle. Hence, conditional on $\mathcal S_{k-1}$, the point $\widehat U$ is uniform in $T$ on survival, and by the invariance of $\nu$ under signed coordinate permutations the next ideal increment always has law $\nu$. Iterated conditioning proves the claimed i.i.d.\ law and the martingale property of $G_k$. Discarding is a stopping time for this filtration, and conditional Jensen for $G_n$ at the stopped $\sigma$-algebra yields \eqref{neg:eq:stopped-Jensen}. The geometric decisions may depend on previous innovations through the chosen charts; predictability is sufficient. \end{proof} \begin{lemma} \label{neg:lem:spatial} Fix a simple control $L$ supplied by \cref{neg:lem:Brownian} and put $a_t=\sqrt{\det L_t}$. There are mean-zero random piecewise affine Lipschitz potentials $u_m$ on $D$, built from $\mathcal W$ and $h=2^{-m}$, such that \begin{align} \int_D\Psi(\partial_1u_m,\partial_2u_m)\,\mathrm dx &\le\mathbb E\Psi\!\left(\int_0^1L_t\,\mathrm dB_t\right) +o_L(1)\quad\text{almost surely}, \label{neg:eq:spatial-cost}\\ \liminf_{m\to\infty}\frac{\mathbb E\,\mathcal W(P_h^Du_m)}{\sqrt m} &\ge c_*\,\mathbb E\int_0^1a_t\,\mathrm dt, \label{neg:eq:spatial-reward} \end{align} where $c_*>0$ is absolute. The error and the minimum admissible $m$ may depend on the fixed control, but $c_*$ does not. \end{lemma} \begin{proof} The potential $u_m$ is a sum of signed bubbles on an adaptive tree of triangles. Steps~1 and~2 build the tree and prove the cost bound \eqref{neg:eq:spatial-cost} for every choice of signs; Steps~3 and~4 average over the Gaussian signs and prove the reward bound \eqref{neg:eq:spatial-reward}. \emph{Step 1: charts and depths.} Let $0=t_0<\cdots0$ and choose once and for all the simple control of \cref{neg:lem:Brownian}. By \cref{neg:lem:spatial,neg:lem:dual}, for all sufficiently large $m$, \[ |\mathcal W(P_h^Du_m)|\le C_D\mathcal F_m(C_*+2) \quad\text{almost surely}, \] while \[ \liminf_{m\to\infty} \frac{\mathbb E\mathcal W(P_h^Du_m)}{\sqrt m} \ge c_* D_0. \] Consequently \begin{equation}\label{neg:eq:expectation-divergence} \liminf_{m\to\infty}\frac{\mathbb E\mathcal F_m}{\sqrt m} \ge\frac{c_*D_0}{C_D(C_*+2)}. \end{equation} The denominator and $c_*$ are independent of $D_0$. We first send $m\to\infty$ for this fixed control, and then let $D_0$ range over arbitrarily large fixed values. This proves $\mathbb E\mathcal F_m/\sqrt m\to\infty$. The mesh-independent Lipschitz estimate \cref{red:lem:uniform-gaussian-lipschitz} and the Gaussian concentration inequality imply, for every $t>0$, \[ \mathbb P\{\mathcal F_m\le\mathbb E\mathcal F_m-t\} \le\exp\left(-\frac{t^2}{2C_{\rm Lip}^2}\right). \] For any fixed $A$, eventually $\mathbb E\mathcal F_m\ge2A\sqrt m$. Substituting $t=A\sqrt m$ gives \eqref{neg:eq:Gaussian-divergence}. \end{proof} The equivalence of \cref{red:thm:main} alone would only show that no fixed constant bounds $M_N/s_N$ with probability tending to one. To obtain divergence in probability, we transfer the estimates at the level of probabilities. \begin{proof}[Proof of \cref{neg:thm:main}] Choose the dyadic mesh $h_N$ of \eqref{red:eq:h-choice}, with $A_0s_N\le h_N<2A_0s_N$ and $m(N)=-\log_2h_N\asymp\log N$. On the high-probability histogram ellipticity event of \cref{red:lem:hist-ellipticity}, the atomic-to-histogram comparison \cref{red:prop:atomic-hist} and the endpoint transport--flow theorem \cref{red:thm:transport-flow-endpoint} give, with constants independent of $N$, \begin{equation}\label{neg:eq:eventwise-matching} F_{h_N}\big(P_{h_N}^D(\mu_N-\nu_N)\big) \le C\big(M_N+h_N\big). \end{equation} In the coupling of \cref{red:thm:KMT} there is an exact divergence corrector $Q_{h_N}$ with \begin{equation}\label{neg:eq:eventwise-KMT} P_{h_N}^D(\mu_N-\nu_N) =\sqrt{\frac2N}\,W_{h_N}+\operatorname{div}Q_{h_N}, \qquad\max\{\|Q_{h_N,1}\|_{\psi_2},\|Q_{h_N,2}\|_\infty\} \le C s_N \end{equation} on an event of probability $1-o(1)$ (the finite horizontal coordinate has an even smaller $o_{\mathbb P}(s_N)$ correction). The triangle inequality for the minimum-flow gauge (\cref{red:lem:flow-lipschitz}) then yields, pathwise on this event, \begin{equation}\label{neg:eq:transfer} \sqrt{\frac2N}\,F_{h_N}(W_{h_N}) \le C\big(M_N+h_N+s_N\big). \end{equation} For each fixed $A<\infty$, the event $\{M_N\le As_N\}$ therefore implies $\{\mathcal F_{m(N)}\le C_A\sqrt{m(N)}\}$ on the good coupling event. By \cref{neg:thm:Gaussian}, \[ \mathbb P(M_N\le As_N) \le\mathbb P\{\mathcal F_{m(N)}\le C_A\sqrt{m(N)}\}+o(1) \longrightarrow0. \] This proves \eqref{neg:eq:negative-result}. For the one-cloud statement, fix any deterministic reference $z^{(N)}\in D^N$ and let $p_N(A;z^{(N)})$ be the probability of matching one uniform sample to it with both coordinate costs at most $As_N$. For two independent samples, the corresponding events are independent and each has probability $p_N(A;z^{(N)})$. Composing the two permutations via their common reference, and using the triangle inequality for the discrete Luxemburg norm and for the supremum, produces a two-sample matching of cost at most $2As_N$. Thus \[ p_N(A;z^{(N)})^2\le\mathbb P(M_N\le2As_N)\longrightarrow0. \] The right side is independent of the reference; taking the supremum over $z^{(N)}\in D^N$ proves \eqref{neg:eq:reference-negative}. Finally, the symmetry of the sample law under exchange of axes proves the other hard endpoint. \end{proof} \clearpage \phantomsection \addcontentsline{toc}{part}{Appendix} \appendix \section{Geometric estimates for the adaptive construction}\label{app:geometry} This appendix collects the deterministic geometric estimates behind the adaptive construction of Part~\ref{umc:partI}. Throughout, $K>1$ is the distortion bound between consecutive metrics, $c_0=c_0(K)$ the inball ratio, $s$ the terminal scale, and $\theta,D_0,\vartheta$ the Whitney parameters of Section~\ref{sec:geom}. Appendix~\ref{app:birthchart} compares a cell's sizes in its birth metric and in later metrics; these contraction bounds feed the ancestor-decay factor $\vartheta\sqrt K$ used in Sections~\ref{sec:bootstrap} and~\ref{sec:phase}. Appendix~\ref{app:kernelchain} proves the bounds \eqref{eq:singular} and \eqref{eq:support} on the ray kernel, which are the input of Lemma~\ref{lem:pk}. Appendix~\ref{app:allbranches} gives the finite Whitney partition with boundary receivers used in Lemma~\ref{lem:whitney}, fixes the order of the constants, and records the indexing of actual refinements along a revealed ancestral chain. \subsection{Chart contraction and terminal cells}\label{app:birthchart} We use the metric convention \[ d_G(x,z)=|G^{-1/2}(x-z)|,\qquad \operatorname{diam}_G E=\sup_{x,z\in E}d_G(x,z). \] \begin{lemma}\label{lem:birthchart} Fix $K>1$. At an active generation let $Q$ be a square of side $\ell$ in its birth chart $G$, where $\det G=1$. Let $G^+$ be the predictably selected next metric, with \[ K^{-1}G\preceq G^+\preceq KG,\qquad \det G^+=1. \] Let $D=\operatorname{diam}_{G^+}Q$. Suppose that $D>D_0s$ and that $Q$ is split by the balanced Whitney construction of Lemma~\ref{lem:whitney}. Write $C_A$ for a uniform bound such that every terminal union created by this split has $G^+$-diameter at most $C_As$. Then the following claims hold simultaneously: \begin{align} \sqrt{2/K}\,\ell&\le D\le \sqrt{2K}\,\ell,\label{eq:appD}\\ \ell(S)&\le \theta\sqrt{2K}\,\ell \quad\text{for every nonterminal child }S,\label{eq:appside}\\ \operatorname{diam}_{G^+}S &\le (C_A\sqrt{2K}/D_0)\,\ell \quad\text{for every terminal union }S.\label{eq:appterm} \end{align} In particular, for any prescribed $\vartheta>0$ one can choose $\theta>0$ sufficiently small, and subsequently $D_0$ sufficiently large, so that (i) every new nonterminal side obeys $\ell(S)\le\vartheta\ell$ and (ii) every newly terminal union obeys $\operatorname{diam}_{G^+}S\le\vartheta\ell$. The choices depend only on the predetermined shape and clipping parameters, not on $s$, the depth, the parent location, or the eccentricity of $G$ in fixed physical coordinates. A parent declared terminal without splitting has $\operatorname{diam}_G Q\le\sqrt K D_0s$. \end{lemma} \begin{proof} The transformation from birth coordinates to new coordinates is \[ A=(G^+)^{-1/2}G^{1/2}. \] Since $K^{-1}G\preceq G^+\preceq KG$, its singular values lie in $[K^{-1/2},K^{1/2}]$. The diameter of a Euclidean square of side $\ell$ is $\sqrt2\ell$; this proves \eqref{eq:appD}. The Whitney upper cap in the new chart is $\ell(S)\le\theta D$, which combined with \eqref{eq:appD} proves \eqref{eq:appside}. For a terminal child $\operatorname{diam}_{G^+}S\le C_As$. Since the parent was split, $D>D_0s$, and by $D\le\sqrt{2K}\ell$ we have $s<\sqrt{2K}\ell/D_0$. This proves \eqref{eq:appterm}. Take $\theta<\vartheta/\sqrt{2K}$ and $D_0>C_A\sqrt{2K}/\vartheta$, together with the other finite lower bounds of Lemma~\ref{lem:exactcover}. The constant $C_A$ of the receiver construction is determined before $D_0$ is chosen and is independent of $D_0$. Finally, a no-split parent with $D\le D_0s$ satisfies \[\operatorname{diam}_G Q=\sqrt2\ell\le\sqrt K D\le\sqrt K D_0s.\] \end{proof} \begin{lemma} \label{lem:ancestorindexed} Suppose the sequence of active ancestors of a cell $Q_k$ is $Q_j\supset Q_{j+1}\supset\cdots\supset Q_k$, with each $Q_p$ a nonterminal square of side $\ell_p$ in $G_p$ for $j\le p\le k$, and \[ \ell_{p+1}\le\vartheta\ell_p,\qquad K^{-1}G_p\preceq G_{p+1}\preceq KG_p. \] Then \begin{equation}\label{eq:appancestor} \frac{\operatorname{diam}_{G_j}Q_k}{\ell_j} \le\sqrt2(\vartheta\sqrt K)^{k-j}. \end{equation} If instead $Q_k$ is a terminal union born from the last active parent $Q_{k-1}$, the same inequality holds with $\sqrt2$ replaced by an absolute fixed constant. \end{lemma} A terminal leaf is never refined again, so the lemma concerns only actual refinements. \begin{proof} By multiplication of the consecutive Loewner comparisons, $G_j^{-1}\preceq K^{k-j}G_k^{-1}$, whence $\operatorname{diam}_{G_j}Q_k\le K^{(k-j)/2}\operatorname{diam}_{G_k}Q_k$. For a square $\operatorname{diam}_{G_k}Q_k=\sqrt2\ell_k$ and $\ell_k\le\vartheta^{k-j}\ell_j$. For a terminal union use the terminal version of Lemma~\ref{lem:birthchart} at the last step, giving the same product with a fixed prefactor (Figure~\ref{fig:metric-refinement}). Once terminal the cell is frozen. \end{proof} \begin{figure}[H] \centering \begin{tikzpicture}[>=Latex,font=\footnotesize,line join=round] % square in birth chart \draw[line width=.8pt] (0.1,.55) rectangle (2.2,2.65); \node[align=center] at (1.15,3.05) {birth chart $G_j$}; \node at (1.15,1.6) {$Q_j$}; \draw[<->,thin] (.1,.28)--(2.2,.28); \node[anchor=north] at (1.15,.23) {$\ell_j$}; \draw[->,line width=.8pt] (2.45,1.6)--(4.10,1.6); \node[align=center] at (3.26,2.12) {predictable\\metric update}; \node[align=center] at (3.26,1.18) {$G_j\longmapsto G_j^+$}; % same parent in updated chart, sheared box \draw[line width=.8pt] (4.42,.56)--(6.83,.56)--(7.35,2.64)--(4.94,2.64)--cycle; \draw[fill=black!16,line width=.75pt] (5.1,1.0) rectangle (5.79,1.69); \draw[fill=black!30,line width=.75pt] (6.14,.83)--(6.77,.83)--(6.67,1.36)--(6.14,1.36)--cycle; \node[align=center] at (5.85,3.05) {updated chart $G_j^+$}; \node[align=center] at (5.85,.18) {$D=\diam_{G_j^+}Q_j$}; % branches \draw[->,line width=.75pt] (7.60,1.95)--(9.02,2.50); \draw[->,line width=.75pt] (7.60,1.26)--(9.02,.69); \draw[fill=black!12,line width=.8pt] (9.25,2.19) rectangle (10.22,3.16); \node[anchor=west] at (10.35,2.7) {active $Q_{j+1}$}; \node[anchor=west] at (10.35,2.28) {$\ell_{j+1}\le\vartheta\ell_j$}; \draw[fill=black!30,line width=.8pt] (9.26,.38)--(9.76,.42)--(9.97,.77)--(9.39,.96)--cycle; \node[anchor=west] at (10.35,.73) {terminal $S$}; \node[anchor=west] at (10.35,.32) {final reveal only}; \end{tikzpicture} \caption{One actual refinement step. The parent is a square in its birth metric $G_j$ but is generally oblique in the updated chart $G_j^+$. A newly active child receives birth metric $G_{j+1}=G_j^+$; a terminal union receives its last increment and is not refined further. Comparing diameters in earlier birth metrics also incurs the factor $\sqrt K$ per actual chart transition.} \label{fig:metric-refinement} \end{figure} \paragraph{Finite depth and area floor.} All the square sides and diameter contractions in Lemmas~\ref{lem:birthchart} and~\ref{lem:ancestorindexed} are measured in the specified birth charts. A split-created terminal union enters only through the estimate for its final transition; it is not a square in a subsequent chart and is never split again. The factor $\sqrt K$ in \eqref{eq:appancestor} comes in addition to the factor $\sqrt{2K}$ in the newborn-side bound \eqref{eq:appside}, and each is counted exactly once. Every active square is born with side at least $2s$, by construction; its physical area is the square of that side because $\det G=1$. Since $\ell_p\le\vartheta^p\ell_0$ along a chain of active squares, the number of strict refinements is at most \[ J_0\le 2+\frac{\log(\ell_0/(2s))}{|\log\vartheta|} =O(\log(1/s))=O(m). \] Every terminal union contains an entire uncovered $s$-square; therefore all cells, at every revealed stage, have area at least $s^2$. \subsection{Ray kernel estimates}\label{app:kernelchain} We work on a convex parent in the intrinsic chart of Section~\ref{sec:geom} and, as in the proof of Lemma~\ref{lem:pk}, drop the hats: $Q$, $x,y$, $K_Q$ and $\omega_Q$ stand for $\widehat Q$, $z,w$, $\widehat K_Q$ and $\widehat\omega_Q$. Thus $D=\diam Q$, $B(z_0,c_0D)\subset Q$, and $\omega_Q(z)=D^{-2}\omega((z-z_0)/D)$ is the rescaled bump of \eqref{eq:ray}, with $\omega\ge0$ smooth, of integral one and supported in $B(0,c_0/4)$. Then $\int_Q\omega_Q=1$ and $\|\nabla^j\omega_Q\|_\infty\le C_jD^{-2-j}$ for $j=0,1$. All constants below may depend on $c_0$ and the fixed bump, but not on $D/s$, the chart eccentricity, or the random realization. \begin{lemma}\label{lem:app-ray} For $x,y\in Q$, $x\ne y$, the vector kernel \[ K_Q(x,y)=(x-y)\int_1^\infty t\omega_Q(y+t(x-y))\,dt \] satisfies \[ |K_Q(x,y)|\le C|x-y|^{-1},\qquad |\nabla_xK_Q(x,y)|+|\nabla_yK_Q(x,y)| \le C|x-y|^{-2}. \] It vanishes when $|x-y|>C\operatorname{dist}(x,\partial Q)$. \end{lemma} \begin{proof} Write $r=|x-y|$. On the support of the integrand, $z=y+t(x-y)\in\operatorname{supp}\omega_Q\subset Q$, so $tr=|z-y|\le D$. The integral therefore ends by $D/r$. The bounds on the bump give \[ |K_Q(x,y)|\le CrD^{-2}\int_1^{D/r}t\,dt\le C/r. \] Differentiating the prefactor contributes at most $CD^{-2}\int_1^{D/r}t\,dt\le C/r^2$. Differentiating the bump in $x$ or $y$ contributes at most $CrD^{-3}\int_1^{D/r}t^2\,dt\le C/r^2$. These are the asserted gradient bounds. For the support assertion, choose a supporting line $P$ for the convex domain which realizes $\operatorname{dist}(x,\partial Q)$, with nonnegative inward signed distance $\delta_P$. A point $z$ in the bump support satisfies $\delta_P(z)\ge cD$ for fixed $c>0$, whereas $\delta_P(y)\ge0$. As $x=(1-t^{-1})y+t^{-1}z$, one gets $\operatorname{dist}(x,\partial Q)=\delta_P(x)\ge cD/t$. Combining with $r\le D/t$ proves the support assertion. \end{proof} In the notation of Section~\ref{sec:geom}, Lemma~\ref{lem:app-ray} is exactly \eqref{eq:singular} and \eqref{eq:support}. \subsection{Finite Whitney geometry and actual-reveal indexing} \label{app:allbranches} We record the finite partition of a clipped parallelogram and the indexing of actual refinements. The stopped scalar process itself is treated in Proposition~\ref{prop:stopped}. All assertions about disjointness of cells, pointwise values, and Gaussian evaluations are understood up to Lebesgue-null cell boundaries, with a fixed half-open dyadic convention. Continuous versions of the local Gaussian fields are used on the closures of individual children. The constants below never depend on the physical eccentricity of the old metric. \begin{lemma} \label{lem:exactcover} Let $Q\subset\mathbb R^2$ be convex with $B(z_0,c_0D)\subset Q$, $D=\diam Q$, and let $\mathcal D_s$ be a translated, fixed, nested dyadic lattice of $s$-squares. Fix $A>8/c_0$ and $\gamma>(A+\sqrt2)/2+2$. For any $\theta\in(0,1/8)$ there is a constant $C_A=C(c_0,A,\gamma,\theta)$ such that, if $D>D_0s$ with $D_0>\max\{2A,4/\theta\}$, the following algorithm gives a finite partition, modulo null boundaries. \begin{enumerate}[label=(\roman*),leftmargin=2.5em] \item Select the inclusion-maximal dyadic squares $B\subset Q$ whose side $2^js$ satisfies $2s\le 2^js\le\theta D$ and $\dist(B,\partial Q)\ge\gamma 2^js$. Keep them as individual children. \item Among the remaining $s$-squares, retain as \emph{receivers} those which are wholly contained in $Q$. All such squares lie within $C_A s$ of $\partial Q$. \item For every remaining $s$-square $R$ meeting both the interior and the complement of $Q$, take the measurable clipped piece $R\cap Q$ and attach it to one receiver within $C_A s$, using deterministic lexicographic tie breaking. \item Each receiver, together with all pieces assigned to it, is a single terminal child. A receiver with no assigned pieces is also a terminal child. \end{enumerate} The selected large squares and resulting terminal children are pairwise disjoint modulo null sets and cover $Q$ modulo null sets. Every terminal child $S$ satisfies \[ s^2\le |S|\le C_A s^2,\qquad \diam S\le C_A s,\qquad \sup_{z\in S}\dist(z,\partial Q)\le C_A s. \] If $B$ is a selected large square, its side $\ell_B$ satisfies $\ell_B\le\theta D$ and $\gamma\ell_B\le\dist(z,\partial Q)\le C_A\ell_B$ for $z\in B$. In particular the partition has the pointwise Whitney comparisons needed for the projected kernel in Lemma~\ref{lem:pk}. \end{lemma} \begin{proof} Any $s$-square meeting $Q$ is either contained in a selected large square, is a full uncovered square, or has nonempty intersection with $Q$ and with its complement. Half-open boundaries make these three assignments disjoint modulo null sets. Inclusion-maximal squares in the same dyadic lattice are disjoint or nested, so distinct selected squares are disjoint. The dyadic hierarchy is finite because its sides are between $s$ and $\theta D$. If a full uncovered $s$-square $R$ had an admissible $2s$ ancestor, that ancestor would lie in some selected square; this contradicts uncoveredness. Since $2s<\theta D$, the $2s$ ancestor fails either containment in $Q$ or the boundary-distance inequality. In either case its points, and therefore all of $R$, are within $(2\gamma+3\sqrt2)s$ of $\partial Q$. Thus there are no uncovered squares far from the boundary, regardless of the orientation or translation of the lattice. For a clipped $R$, take $z\in R\cap Q$, so $\dist(z,\partial Q)\le\sqrt2s$. Set $z_t=(1-t)z+tz_0$, $t=As/D<1/2$. By the inball and convexity, and then by the Lipschitz property of boundary distance, \[ Ac_0s\le\dist(z_t,\partial Q) \le(A+\sqrt2)s,\qquad |z_t-z|\le As. \] Since $Ac_0>8$, the unique half-open $s$-square $R'$ with $z_t\in R'$ is contained in $Q$. Were $R'$ covered by a selected dyadic square of side $L\ge2s$, then $\dist(z_t,\partial Q)\ge\gamma L\ge2\gamma s$, contrary to $2\gamma>A+\sqrt2$. Thus $R'$ is uncovered and is an eligible receiver. Its distance from $R$ is at most $(A+2\sqrt2)s$. This construction works at an acute corner even when $R$ meets two boundary sides; no side is chosen, only the inball center. Every piece assigned to a given receiver originates in one of the at most $C(A)$ lattice squares in a fixed-radius $O_A(s)$ neighborhood. The receiver supplies area at least $s^2$, and the union of at most $C(A)$ clipped pieces has area at most $C_A s^2$ and diameter at most $C_A s$. The boundary-distance assertion follows either from the uncovered-full-square estimate or the clipped-square estimate. Finally, for a selected $B$ of side $\ell$, its parent $B^\uparrow$ of side $2\ell$ is not admissible. If it fails containment, the distance to $\partial Q$ is at most $2\sqrt2\ell$ on $B$; if it fails clearance, the upper bound is $(2\gamma+2\sqrt2)\ell$; if it fails the cap, then $D<2\ell/\theta$ and every point of $Q$ is within $D$ of the boundary. The lower bound on $B$ is its admissibility. These are the stated uniform comparisons. The proof is independent of the relative translation of the lattice and of all angles except through the fixed inball ratio $c_0$. \end{proof} \paragraph{Choice of constants.} Fix the clipping constant $K>1$. A square in its birth metric transformed to the updated chart has singular values between $K^{-1/2}$ and $K^{1/2}$; thus it has an inball ratio at least $c_0=c_0(K)>0$. Fix $A>8/c_0$, then $\gamma>(A+\sqrt2)/2+2$. These determine the terminal union constant $C_A$, independently of $s$ and of $D_0$. Fix the contraction target $\vartheta$; then choose $\theta<\vartheta/\sqrt{2K}$ and \[ D_0>\max\{2A,4/\theta,C_A\sqrt{2K}/\vartheta\}. \] Increasing $D_0$ never changes $C_A$ or the selected-square comparisons. Consequently Lemma~\ref{lem:birthchart} applies uniformly to the exact finite partition, and by reducing $\vartheta$ we enforce $\vartheta\sqrt K<1$ and subsequently $(\vartheta\sqrt K)^{2\beta}(1-\eta)^{-2a_*}<\rho_1$. \begin{lemma} \label{lem:actualreveal} Let $Q_0\supset Q_1\supset\cdots\supset Q_b$ be the cells of the \emph{actual} revealed ancestral chain of a point, where $Q_j$ is born with metric $G_j$, and each revealed increment $V_j$ is routed in $G_j$. Suppose every strict transition from $Q_{p-1}$ to $Q_p$ satisfies the comparison and side/diameter bounds of Lemma~\ref{lem:birthchart}. If $Q_b$ is a square or a newly terminal union, then for every $1\le j\le b$, \[ \frac{\diam_{G_j}(Q_b)}{\ell_j} \le C(\vartheta\sqrt K)^{b-j}, \] where $\ell_j$ is the side of $Q_j$ in its birth metric when $Q_j$ is a square; for a newly terminal $Q_j$ its last intrinsic diameter serves as the length at $j$. If $Q_b$ is stopped directly, without a further reveal, its $G_b$-diameter is at most $\sqrt K D_0s$. At padded times $k>b$, $U_k=U_b$, $G_k=G_b$ and $T_k=T_b$ on this leaf. No new factor $(\vartheta\sqrt K)^{k-b}$ is assigned to its geometry, and no new Gaussian envelope is charged. \end{lemma} \begin{proof} For a chain of square children, the intrinsic birth-side contractions multiply to $\ell_b\le\vartheta^{b-j}\ell_j$. As in the proof of Lemma~\ref{lem:ancestorindexed}, the Loewner comparisons imply $G_j^{-1}\preceq K^{b-j}G_b^{-1}$ without any commutativity assumption; this gives the asserted ancestor diameter bound. If the final child is a newly terminal union, apply the terminal-diameter estimate only on the final transition and use the ordinary side contractions earlier. If the parent stops without splitting, its new-chart diameter is $\le D_0s$, while the old/new metric comparison gives $\diam_{G_b}Q_b\le\sqrt K D_0s$. Frozen padded times have no strict transitions. \end{proof} \paragraph{Stopping convention.} Lemma~\ref{lem:actualreveal} counts only actual refinements. Once a leaf is terminal, its field and memory are frozen; only the decreasing clock continues to act on its exponential integrand. At a failed fresh-envelope test, the numerical modular is killed. These are the conventions of the stopped-supermartingale argument in Proposition~\ref{prop:stopped}. On successful completion, the terminal partition and success event are measurable with respect to the revealed history, so the unrevealed source is the conditional Gaussian bridge of Lemma~\ref{lem:bridge}. \section*{Use of AI tools} \addcontentsline{toc}{section}{Use of AI tools} The starting point of this work was the previous project \cite{umc:AGGT26} on the finite $p$ cost case. Intense interactions with LLMs spanning several weeks led to a first proof of the Ultimate matching conjecture in the subcritical case, leaving the critical case (as well a the endpoint cases) seemingly out of reach. Somewhat unexpectedly to us, a solution for the critical finite exponent case was finally found during another round of exploration, see the 'monolithic' file \texttt{finite-exponents.pdf} in the \href{https://github.com/grottofrancesco/UltimateMatchingTheorem.git}{repository}. After this GPT-6 quickly obtained the exact asymptotic for the endpoint cases, see the 'monolithic' file \texttt{endpoints.pdf} in the same repository. The content of this file was mostly generated and proof-checked by the LLMs during interactions with us aimed at understanding the 'monolithic' versions. We have an overall understanding of the proofs but plan to provide a fully digested version shortly. \begin{thebibliography}{99} \bibitem{umc:AKT84} M.~Ajtai, J.~Koml\'os and G.~Tusn\'ady, \emph{On optimal matchings}, Combinatorica \textbf{4} (1984), 259--264. \bibitem{AGS} L.~Ambrosio, N.~Gigli and G.~Savar\'e, \emph{Gradient Flows in Metric Spaces and in the Space of Probability Measures}, Birkh\"auser, 2008. \bibitem{umc:AG19} L.~Ambrosio and F.~Glaudo, \emph{Finer estimates on the 2-dimensional matching problem}, J. \'Ec. polytech. 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