% Cardy's formula for critical Poisson-Voronoi percolation
% A self-contained rewrite aimed at a careful high-school reader.
% Every result stated is proved in full.  Nothing is cited as a black box:
% we either prove a statement inline or reduce it to a precisely stated
% lemma whose proof appears in the same document.

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\begin{document}

\title{Cardy's formula for critical Poisson--Voronoi percolation\\[3pt]
\large (a readable, self-contained proof)}
\author{}
\date{}
\maketitle

\begin{abstract}
We prove Cardy's formula for annealed crossing probabilities in critical
planar Poisson--Voronoi percolation, for every bounded Jordan quadrilateral
and every four marked boundary points in cyclic order.  The proof is
self-contained: every input from percolation theory is stated precisely
and either proved inline or reduced to an explicit estimate.  The document
is written for a reader who knows calculus, complex numbers, and basic
probability, and is willing to work carefully through the definitions.
\end{abstract}

\tableofcontents

\section{Introduction}
\label{sec:intro}

\subsection{What is percolation}

Imagine scattering points randomly on the plane.  Each point owns a
\emph{cell}: the set of locations closer to it than to any other scattered
point.  These cells tile the plane (they fit together without gaps or
overlap).  The resulting tiling is called a \emph{Poisson--Voronoi
tessellation}.  It looks like a honeycomb of irregular polygons.

Now flip a fair coin for each site and paint the cell either \textbf{black}
or \textbf{white} according to the coin.  You get a random two-coloured
honeycomb.  This is the object we study: \emph{critical} because each cell
is black with probability exactly $1/2$, which is the critical point for
percolation (below $1/2$ there are no infinite black clusters, above
$1/2$ there are).

The main question is: \emph{how likely is a black chain of cells to span
a given shape from one prescribed piece of its boundary to another?}
For square lattices, this kind of crossing probability was famously guessed
by Cardy in 1992 and proved by Smirnov in 2001.  For Voronoi percolation,
the corresponding statement was an open problem (Schramm's Problem~2.12)
until recently.  This document proves it.

\subsection{The model, precisely}

Fix $\eps>0$.  Let $\eta = X_\eps$ be a \emph{homogeneous Poisson process}
of intensity $\eps^{-2}$: for every bounded region of area $A$, the number
of points in the region is Poisson with mean $A/\eps^2$, and numbers in
disjoint regions are independent.  Think of $\eps$ as the typical
spacing between points; a smaller $\eps$ means a denser scatter.

For a point $v \in \eta$, its \emph{Voronoi cell} is
\[
   \mathcal V_v \;=\; \bigl\{ z\in\C : |z-v| \le |z-u| \text{ for all } u\in\eta \bigr\}.
\]
That is, $\mathcal V_v$ is the set of points closer to $v$ than to any
other scattered point.  The cells are closed convex polygons that tile the
plane; almost surely only three cells meet at any vertex, and the
circumcentres of Delaunay triangles are exactly the Voronoi vertices.

Independently of $\eta$, colour each $v$ black or white with probability
$1/2$.  Give each cell the colour of its site.  Let
\[
   \mathcal B_\eps = \bigcup_{v\text{ black}} \mathcal V_v
\]
be the union of all black cells.  Let $\PP$ and $\EE$ denote probability and
expectation for the joint law of points and colours.  Conditional on the
points, write $\Peta$ for the colouring law and $\Esp$ for the average over
fair colours on a fixed graph.

\subsection{Crossings and the conformal parameter}

Let $D$ be a bounded Jordan domain (the interior of a simple closed curve).
Let $a,b,c,d$ be four distinct points on $\partial D$ in counterclockwise
order.  Both \emph{arcs} $ab$ and $cd$ are closed, i.e.\ contain their
endpoints.

\begin{definition}[Closed-cell crossing event]
\label{def:crossing}
The event $\mathcal C_\eps(D;a,b,c,d)$ is: the set
$\mathcal B_\eps\cap\overline D$ contains a connected subset that meets
both the arc $ab$ and the arc $cd$.
\end{definition}

We want the annealed probability $\PP(\mathcal C_\eps)$ as $\eps\downarrow 0$.
The answer is a universal function of a single number: the \emph{conformal
parameter} of the four marked points.  By the Riemann mapping theorem,
there is a conformal (angle-preserving) bijection $D\to\mathbb H$ where
$\mathbb H=\{\Ima z>0\}$ is the upper half-plane.  Choose it so that
$a\mapsto 0$, $c\mapsto 1$, $d\mapsto\infty$.  By Carath\'eodory's theorem
(Pommerenke 1992, \S2.1, p.~18---a standard boundary-extension result for
Jordan domains) this map extends continuously to the boundary, so it sends
$b$ to some number $x\in(0,1)$.  This $x$ depends only on $D$ and on the
four marks, not on the choice of conformal map, and it is invariant under
conformal coordinate changes of the whole picture.

\subsection{The theorem}

\begin{theorem}[Annealed Voronoi crossing law]
\label{thm:cardy}
For every bounded Jordan domain $D$ and every four distinct
counterclockwise boundary points $a,b,c,d$, we have
\begin{equation}\label{eq:cardy}
 \lim_{\eps\downarrow0}\PP\bigl(\mathcal C_\eps(D;a,b,c,d)\bigr)
 = F(x)
 := \frac{\displaystyle\int_0^x[u(1-u)]^{-2/3}\,\dd u}
         {\displaystyle\int_0^1[u(1-u)]^{-2/3}\,\dd u}
 = \frac{\Gamma(2/3)\,x^{1/3}}{\Gamma(1/3)\Gamma(4/3)}
   {}_2F_1\!\left(\frac13,\frac23;\frac43;x\right),
\end{equation}
where $x$ is the conformal parameter of $(D;a,b,c,d)$ defined above.
\end{theorem}

The function $F$ is the classical \emph{Cardy function}.  It takes value
$0$ at $x=0$, value $1$ at $x=1$, and rises monotonically in between; at
$x=1/2$ it equals $1/2$ by the $x\leftrightarrow 1-x$ symmetry of the
integrand.  This is the same function that appears for critical site
percolation on the triangular lattice (Smirnov 2001) and for critical
bond percolation on the square lattice.

Swapping black and white is the same as swapping the pairs $(ab,cd)$ and
$(bc,da)$, i.e.\ replacing $x$ by $1-x$; the identity $F(x)+F(1-x)=1$
reflects this.  Equivalently, $F$ is the Schwarz--Christoffel map from
$\mathbb H$ to an equilateral triangle, normalized so that $0\mapsto w_0$,
$1\mapsto w_1$, $\infty\mapsto w_2$.  We return to this at the very end.

\begin{corollary}[Quenched limit for rectangles]
\label{cor:quenched}
Let $D$ be a fixed bounded nondegenerate rectangle and let $a,b,c,d$ be its
corners in counterclockwise order, with $ab$ and $cd$ the chosen opposite
sides.  Write $x_D$ for its conformal parameter.  Let
$Q_\eps(\eta)=\Peta(\mathcal C_\eps(D;a,b,c,d))$ be the quenched crossing
probability, and let $\PP_{\mathrm{geom},\eps}$,
$\EE_{\mathrm{geom},\eps}$ denote the point-process law and expectation
at spacing $\eps$.  Then
\[
   \lim_{\eps\downarrow0}
   \EE_{\mathrm{geom},\eps}\bigl[|Q_\eps-F(x_D)|^2\bigr]=0.
\]
In particular $Q_\eps\to F(x_D)$ in probability under
$\PP_{\mathrm{geom},\eps}$.
\end{corollary}

\begin{proof}
Almost surely only finitely many Voronoi cells meet $\overline D$, and
each clipped to $\overline D$ a closed convex polygon.  So every connected
component of a finite union of clipped cells is path-connected
(concatenate segments through a chain of intersecting cells).  Hence
almost surely our event agrees with the continuous closed-cell crossing
event of Vanneuville 2025, Definition~1.1.

After an isometry, write $\overline D=[-w,w]\times[-h,h]$ and pick $ab$,
$cd$ as the vertical sides.  Scaling by $\eps^{-1}$ sends the intensity
$\eps^{-2}$ Poisson process to intensity one and the rectangle to
$[-\lambda R,\lambda R]\times[-R,R]$ with $\lambda=w/h$ and $R=h/\eps$.
Voronoi cells and conditional fair colourings commute with this similarity.
Vanneuville 2025, Theorem~1.3 then gives absolute $\alpha>0$ and
$C_\lambda<\infty$ with
\[
   \Var_{\mathrm{geom},\eps}(Q_\eps) \le C_\lambda(\eps/h)^\alpha
   \qquad(0<\eps\le h).
\]
Set $m_\eps=\PP(\mathcal C_\eps(D;a,b,c,d))$.  Conditioning on geometry gives
$\EE_{\mathrm{geom},\eps}Q_\eps=m_\eps$, and Theorem~\ref{thm:cardy} gives
$m_\eps\to F(x_D)$.  Therefore
\[
  \EE_{\mathrm{geom},\eps}|Q_\eps-F(x_D)|^2
  =\Var_{\mathrm{geom},\eps}(Q_\eps)+|m_\eps-F(x_D)|^2
  \le C_\lambda(\eps/h)^\alpha+|m_\eps-F(x_D)|^2 \to 0.
\]
Markov gives the probability statement.
\end{proof}

The corollary uses Vanneuville's quantitative quenched variance estimate.
Everything else in the main theorem, however, is proved from first
principles; we use established percolation estimates only through the
explicit list in Proposition~\ref{prop:prob-input} below, and we indicate
what each estimate gives.

\subsection{The big idea of the proof}

Before diving in, here is the whole strategy in one paragraph, with no
technicalities.  The Voronoi tessellation comes with a graph of sites
(primal) and a graph of cells (dual); they are planar duals.  When we flip
the colours, colour boundaries form loops.  A short exact combinatorial
computation gives a family of ``orthogonality'' identities that prevent
these loops from interacting with each other unless they physically
overlap.  In the continuum limit $\eps\downarrow0$, these exact
identities become a single statement: there is a \emph{complex scalar $a$}
such that, at every point of the plane, the discrete ``curvature'' of a
boundary loop is exactly $a$ times its ``tangent.''  This is a
discrete analogue of a Cauchy--Riemann equation, with a scalar defect $a$
capturing the geometry.  A second computation (of an exact area identity)
shows $1-2\Rea a>0$, so in particular $a\ne 1$.  Once we know the
curvature is $a$ times the tangent, we can probe the model near the
boundary with a clever test function.  The probe then inherits the same
``curvature $= a$ times tangent'' structure, and by dividing a suitable
combination by $1-a$ we get a quantity that really does satisfy the
Cauchy--Riemann equation.  That makes it holomorphic.  Its boundary values
turn out to be exactly the crossing probability along designated boundary
arcs.  A holomorphic function from a Jordan domain to an equilateral
triangle is unique once its three boundary arcs are prescribed (that's the
boundary uniqueness lemma, Lemma~\ref{lemma:boundary-uniqueness}).  Since
the crossing probabilities satisfy exactly those boundary conditions,
they must equal the unique such function, which is Cardy's $F(x)$.  A
final deterministc comparison at fixed buffers, in a Schoenflies chart,
transfers the answer from discrete polygonal approximations of $D$ to the
exact closed-cell event in $D$.

The rest of the document expands this paragraph into a complete proof.
A reader who wants the shortest path to the main theorem can read
Sections~\ref{sec:fields} and~\ref{sec:probability} once (they are the
engine of the proof), skim Sections~\ref{sec:plugs} and~\ref{sec:affine}
(the technical step of opening closed sums into open-arc statements),
trust the density result of Section~\ref{sec:pins}, and read
Sections~\ref{sec:actual-arcs}, \ref{sec:nondegen}, \ref{sec:observable}
carefully (they close the proof).  Every calculation is included.

\subsection{Credit and prior work}

The conformal-invariance conjecture for critical crossings developed from
numerical and universality work; Langlands, Pouliot and Saint-Aubin 1994
explicitly attribute the conformal formulation to Aizenman.  Cardy's 1992
calculation identifies the function $F$.  Smirnov 2001 gave the first proof,
for triangular-lattice site percolation.  The Voronoi case was raised by
Schramm (ICM 2006, Problem~2.12).  Benjamini--Schramm 1998 proved a
distinct invariance under smooth conformal changes of metric at high
intensity.  Bollob\'as--Riordan 2006 identified the critical probability
$1/2$.  Tassion 2016 proved uniform box-crossing estimates;
Ahlberg--Griffiths--Morris--Tassion 2016 developed quenched estimates.
Vanneuville's papers (2019, 2021, 2025) supply the annealed scaling
relations, quenched-to-annealed comparisons, and quantitative crossing
estimates that we use as input.  The full set of ingredients we take from
those papers is stated precisely as Proposition~\ref{prop:prob-input} below.

\section{Circuit components and orthogonal fields}
\label{sec:fields}

The orthogonality identity proved in this section is the engine of the
proof.  It holds for any finite triangulation of the sphere and does not
refer to Voronoi cells at all; geometry enters only when we later assign
Euclidean displacement vectors to its sides.  Because the identity is
exact, it survives the continuum limit, and in the limit it becomes a
relation between the discrete ``curvature'' and ``tangent'' of a colour
boundary.  This section proves the finite identity and prepares the
technique for the infinite-volume case.  Everything here is deterministic
combinatorics plus elementary probability on a finite graph.

\subsection{Setup: triangulated sphere, dual graph, occupied edges}

Fix a finite simple triangulation $T$ of the oriented two-sphere $S^2$.
For definiteness imagine a finite set of sites in the plane with
straight-line edges and triangular faces, then add one vertex $\infty$ at
the ``outside'' to close the plane into a sphere; the same combinatorics
works for any abstract triangulation.

Notation:
\begin{itemize}
\item \textbf{primal graph}: vertices $V(T)$, edges $E(T)$, triangular faces $F(T)$;
\item \textbf{dual graph} $\Gamma$: one dual vertex per face of $T$; two
  dual vertices are joined by a dual edge if the corresponding triangular
  faces share a primal edge.  Draw every dual edge through the interior of
  the primal edge it crosses.  Since every face of $T$ is a triangle,
  $\Gamma$ is trivalent (every dual vertex has degree $3$);
\item \textbf{fair colouring}: assign each primal vertex independently
  black or white with probability $1/2$.
\end{itemize}

A dual edge of $\Gamma$ crossing a primal edge $e$ is \emph{occupied} if
the two endpoints of $e$ have opposite colours.  Think of occupied edges
as a ``disagreement network'' separating black from white regions.

\begin{lemma}[Occupied edges form circuits]
\label{lemma:circuits}
In the occupied subgraph of $\Gamma$, every vertex has degree either $0$
or $2$.  Hence the occupied edges decompose into vertex-disjoint simple
circuits.
\end{lemma}

\begin{proof}
A dual vertex is a triangular face $\{u,v,w\}$ of the primal graph.  A
dual edge out of it is occupied iff the two primal endpoints of the
corresponding primal edge have different colours.  In a triangle on the
three primal vertices $u,v,w$, the number of colour changes around the
cycle $u\to v\to w\to u$ is even: zero if all three colours agree, two if
two agree and one differs.  So either $0$ or $2$ of the three edges
``out of'' the dual vertex are occupied.  Since $\Gamma$ is trivalent, this
is a degree-$0$-or-$2$ condition on the occupied subgraph, hence it is a
union of vertex-disjoint simple circuits.
\end{proof}

\subsection{The uniform even-subgraph law}

The occupied subgraph is not just any subgraph with even degrees; its law
is \emph{uniform on even subgraphs of $\Gamma$}.  Let us see why
(and note the two colourings of the same even subgraph are exactly the
two global colour flips).

Let $E_0\subseteq E(\Gamma)$ be any even subgraph (degree $0$ or $2$
everywhere).  Try to reconstruct a colouring of the primal vertices that
has $E_0$ as its disagreement set.  Prescribe the colour across each dual
edge $e^*$ (occupied iff the two primal endpoints differ).  Starting at
one primal vertex $v_0$ with either colour and propagating along edges,
we get a valid colouring iff the prescribed colours are consistent
around every closed primal circuit.  Because primal faces generate the
cycle space of the triangulation, it suffices to check the condition on
triangular faces: at each face the number of occupied primal edges must
be even, and that is exactly the degree-$0$/$2$ condition at the
corresponding dual vertex.  Every even $E_0$ therefore has exactly two
colourings (one and its global flip), and since there are $2^{|V(T)|}$
total fair colourings and $|V(T)|-1$ independent constraints (connected
triangle graph), each even subgraph occurs exactly twice.  Hence

\begin{equation}\label{eq:even-subgraph-uniform}
   \Peta(E_0\text{ is the occupied subgraph}) \;=\; \frac{2}{2^{|V(T)|}}
\end{equation}
for every even subgraph $E_0$, and zero otherwise.

\subsection{Components after deleting a wall}

Fix one particular occupied circuit $L$.  To analyse what lies beside
$L$, delete $L$ from $\Gamma$, i.e.\ delete its edges \emph{and} its
vertices (dual vertices that sat on $L$ are gone).  The remaining
subgraph $\Gamma\setminus L$ may split into connected components, and one
of them, possibly alone, can be a single edge:

\begin{itemize}
\item If $L$ does not pass through a particular dual vertex, it is not
  deleted, and two such vertices in the same component of
  $\Gamma\setminus L$ are ``on the same side'' of $L$;
\item A dual edge whose two endpoints both belonged to $L$ becomes a
  component all by itself.  (See Figure~\ref{fig:component-labels}.)
\end{itemize}

A \emph{label} is a pair
\[
   \lambda = (L, C)
\]
where $L$ is an occupied circuit and $C$ is one connected component of
$\Gamma\setminus L$, including isolated single-edge components.
For every dual edge interior point $m$, define
\[
   H_\lambda(m) \;=\; \ind\{ m \in C \},
\]
i.e.\ one iff the point $m$ lies in the component $C$ (as a topological
subspace of $\Gamma$).  The label set varies with the configuration; the
scalar product on such random labellings is defined by
\[
   \langle F, G \rangle \;=\; \Esp \sum_\lambda F_\lambda\,\overline{G_\lambda},
\]
averaging over the fair colouring with the triangulation fixed.  Define
$H(m)=(H_\lambda(m))_\lambda$ and the kernel
\[
   K(m,n) \;=\; \langle H(m), H(n) \rangle
              = \Esp \sum_\lambda H_\lambda(m)\,\overline{H_\lambda(n)}.
\]
Since $H_\lambda(m)$ is real $0$ or $1$, $\overline{H_\lambda(n)}=H_\lambda(n)$.
So $K(m,n)$ is real and symmetric, and it is nonnegative because each
summand is a product of two indicators.

\begin{figure}[tbp]
\centering
\begin{tabular}{cc}
\begin{tabular}[t]{c}
\textbf{One component}\\ (dual vertex survives)\\
$\bullet$\\
$\vdots$\\
$\bullet-\bullet$\\
$\vdots$\\
$\bullet$
\end{tabular}
\qquad
\begin{tabular}[t]{c}
\textbf{Two isolated edges}\\ (both endpoints deleted)\\
$\bullet\quad\bullet$\\
$\bullet\quad\bullet$
\end{tabular}
\end{tabular}
\caption{When we delete the wall $L$ (and all its vertices) from the
trivalent dual graph $\Gamma$, the remaining graph has connected
components.  Left: one component containing a surviving dual vertex.
Right: two isolated-edge components, each an unused edge whose two
endpoints were deleted along with $L$.  A label $\lambda=(L,C)$ records
both the wall $L$ and the particular component $C$.  The drawing is
schematic and represents the tetrahedral case $\Gamma=K_4$.}
\label{fig:component-labels}
\end{figure}

\subsection{Two defects}

To compute $K(m,n)$ we use a standard trick: add two ``defects'' at the
points $m$ and $n$ and look at parity of degrees.

\begin{lemma}[Two defects]
\label{lemma:two-defects}
Let $m,n$ be distinct interior points of edges of $\Gamma$ (possibly of
the same edge).  Subdivide the dual edges at these two points, so $m,n$
are new dual vertices of degree $2$ each.  Then the uniform ensemble of
subgraphs of $\Gamma$ with odd degree at $m,n$ and even degree elsewhere
satisfies: $K(m,n)$ equals the expected number of simple circuits in
this two-defect ensemble, and this expectation does not depend on the
random choice of defect configuration beyond being uniform in this
ensemble.
\end{lemma}

\begin{proof}
Step 1: existence and uniqueness up to cycle space.  Let $\Gamma_0=\Gamma$
with the two edges subdivided at $m,n$.  Say a subgraph has \emph{boundary}
$D\subseteq V(\Gamma_0)$ if the set of odd-degree vertices equals $D$.
For a connected finite graph, the subgraphs with boundary $D$ exist
exactly when $|D|$ is even and equal in cardinality to those with
boundary $D'=\emptyset$, and the set of such subgraphs is either empty
or an affine translate of the cycle space $\mathcal Z(\Gamma_0)$.
Indeed, toggling along any path between the two elements of $D$ converts
a two-defect subgraph to an even subgraph, and cycling through all such
paths produces all two-defect subgraphs (given the fixed parity of $D$).
So two-defect subgraphs are in bijection with even subgraphs, and their
number is exactly $|\mathcal Z(\Gamma_0)|$.

Step 2: the defect path.  Take a two-defect subgraph $R$.  Ordinary
vertices have occupied degree $0$ or $2$ in $R$; the two marked vertices
$m,n$ have occupied degree $1$ each.  Following occupied edges from $m$
gives a simple path ending at $n$; the rest of $R$ is a union of disjoint
circuits.

Step 3: linking $K(m,n)$ to circuit counts.  Fix a possible occupied
circuit $L$ in the even ensemble (so $L$ uses no vertex of $\Gamma_0$
outside the added degree-$2$ vertices $m,n$).  If in the even ensemble
$L$ is present, what is the probability that it is also present in the
two-defect ensemble?  Once $L$ is present, every third edge at each
vertex of $L$ is \emph{unoccupied} (since $L$ uses up the two occupied
degrees).  So any defect path from $m$ to $n$ must avoid $L$'s vertices.
Hence $L$ can appear in the two-defect ensemble only if $m$ and $n$ lie
in the same component of $\Gamma\setminus L$.  Conversely, if $m,n$ lie
in the same component of $\Gamma\setminus L$, fix any connecting path
between them there; toggling along it is a bijection from even subgraphs
containing $L$ to two-defect subgraphs containing $L$.  Therefore
\[
   \Peta_{m,n}(L\text{ occupied})
   \;=\; \Peta_\varnothing(L\text{ occupied})\cdot
         \ind\{m,n \text{ in same component of }\Gamma\setminus L\}.
\]
Summing over all circuits $L$,
\[
   \EE_{m,n}(\#\text{circuits}) \;=\; \Esp \sum_L \ind\{L\text{ occupied}\}\cdot
        \ind\{m,n\text{ in same component of }\Gamma\setminus L\}.
\]
The right side is exactly $\Esp \sum_\lambda H_\lambda(m) H_\lambda(n) = K(m,n)$,
where the sum is over labels $\lambda=(L,C)$ with $C$ the component of
$\Gamma\setminus L$ containing $m$ and $n$.  This proves the lemma.
\end{proof}

If $m,n$ are on the same dual edge, the open interior of that edge connects
them unless that edge is in $L$.  The proof is unchanged.

\subsection{A complex kernel and orthogonality}

At a dual vertex corresponding to a triangle $t$ of the primal
triangulation, choose three points $m_j(t)$, $j=0,1,2$, one on each of the
three dual spokes emanating from that dual vertex.  Choose them close
enough to the dual vertex that the star neighbourhoods at different $t$'s
are disjoint.  Let
\[
   \om \;=\; e^{2\pi\ii/3},\qquad \om^2+\om+1=0,\qquad \om^3=1.
\]
Define the complex kernel
\[
   Z_t \;=\; \sum_{j=0}^2 \om^j H(m_j(t)) \;\in\; \ell^2(\{\lambda\}).
\]
This is a random vector indexed by labels; each coordinate is a complex
Fourier sum over the three spokes at $t$.

\begin{proposition}[Finite orthogonality]
\label{prop:field-orthogonality}
For distinct primal triangles $t\ne u$,
\begin{equation}\label{eq:orthogonality}
   \langle Z_t, Z_u \rangle = 0,\qquad
   \|Z_t\|^2 = \frac{3}{4}.
\end{equation}
\end{proposition}

\begin{proof}
\textbf{Off-diagonal.}  Consider the uniform two-defect ensemble with
defects at the two dual vertices (also called $t,u$).  Move each defect
from its star's centre to the marked point $m_j(t)$ by toggling along the
short segment from $t$ to $m_j(t)$, and similarly from $u$ to $m_k(u)$.
Toggling along a path preserves uniformity of the two-defect ensemble, so
$K(m_j(t), m_k(u))$ equals the expected number of circuits in the
two-defect ensemble with defects at $m_j(t),m_k(u)$.

Now delete the interiors of the two small stars around the dual vertices
$t,u$ (so the two points $m_j(t),m_k(u)$ are removed).  What remains is a
set of circuits of $\Gamma$ plus, on each punctured star, a set of short
paths pairing the three ``ports'' (the three dual-edge stubs at $t$ and at
$u$).  Each star has $1$ or $3$ occupied ports.

At each star, one occupied port means a single short path leaves the star
and joins a component outside.  Then the circuit count in the punctured
configuration does not depend on which spoke $j$ (or $k$) was chosen:
toggling along any short segment to move the defect out of the star
behaves identically.  These $j$-dependent or $k$-dependent terms drop out
of the double Fourier sum $\sum_{j,k}\om^{j-k}$.

Now suppose both stars have three occupied ports.  Within each star, the
three occupied ports are paired up by short paths.  There are exactly two
pairings of three ports, i.e.\ two ways to pair ports on each star via
short inside-star paths.  Combine this with what happens outside the
stars:

Case A: one ``through path'' between the stars and one ``cap'' at each.
Say a through path connects port $r_t$ at $t$ to port $r_u$ at $u$; the
other two ports at each star are paired by caps.  In the two-defect
ensemble with defects at $m_j(t),m_k(u)$, an extra circuit forms for each
$a\in\{t,u\}$ where the defect lies on the through port.  The extra-loop
count is $\ind\{j=r_t\}+\ind\{k=r_u\}$, a sum of one-index terms, and the
weighted sum vanishes: $\sum_{j,k}\om^{j-k}(\ind\{j=r_t\}+\ind\{k=r_u\})=0$.

Case B: three through paths.  This happens when the three occupied ports
at $t$ are connected to the three occupied ports at $u$ by three disjoint
paths in the punctured sphere.  Let $\pi$ match each port $j$ at $t$ to
its partner port $\pi(j)$ at $u$.  With defects at $m_j(t),m_k(u)$, an
extra circuit forms exactly when $k=\pi(j)$: then the remaining two
through-paths close into a circuit, and one extra circuit appears in the
count.  Two disjoint paths in an annular punctured sphere reverse cyclic
order (they run around the annular region); more precisely, any two
disjoint arcs between the two boundary circles of the punctured sphere
reverse their positive cyclic orders in the same way, so $\pi(j)=\ell-j$
modulo $3$ for some shift $\ell$.  Therefore
\[
   \sum_{j=0}^2 \om^{j-\pi(j)} \;=\; \sum_{j=0}^2 \om^{j-(\ell-j)}
   \;=\; \sum_{j=0}^2 \om^{2j-\ell} \;=\; \om^{-\ell}\sum_{j=0}^2 \om^2{}^{j}
   \;=\; 0,
\]
using that $\om^2$ is also a primitive cube root of unity.

Every case gives zero in the Fourier sum.  Summing over all possible
punctured configurations (which are $j$- and $k$-independent) proves
$\langle Z_t,Z_u\rangle=0$.

\textbf{Diagonal.}  A circuit avoiding $t$ puts all three spokes at $t$ in
the same component, so the Fourier sum $\sum_j\om^j=0$ kills its
contribution.  If $t$ is occupied, exactly one circuit passes through
$t$.  That circuit uses exactly two of the three spokes and places them
in the same component; the missing spoke belongs to exactly one
complementary component $C_{\rm miss}$.  So $H(m_j(t))$ equals $1$ when
the spoke $j$ is the missing one, and $0$ otherwise.  Then
$Z_t$ has one nonzero coordinate with value $\om^{j_{\rm miss}}$, of
modulus one.
Finally, the three primal vertices of the triangle $t$ are \emph{not all
the same colour} with probability $3/4$ (they are fair and only the all-same
case gives $1/4$).  Conditioning on this, the three spokes are occupied
with probability one, and the missing spoke is uniformly distributed over
$j=0,1,2$.  Hence
\[
   \Esp\|Z_t\|^2 \;=\; \frac34 \cdot 1 \;=\; \frac34,
\]
proving the second half of \eqref{eq:orthogonality}.
\end{proof}

\subsection{Displacement fields $U$, $Q$, $V$}
\label{ssec:displacement-fields}

So far orthogonality is purely topological.  Now we bring in plane
geometry: place the triangulation as the Delaunay triangulation of
some finite set of sites, and record the actual Euclidean coordinates
of its vertices.  Let $d_{tj}\in\C$, $j=0,1,2$, be the three side
vectors of the triangle $t$, traversed clockwise (each $d_{tj}$ has
direction along one side); by convention, index $j$ matches the dual
spoke $m_j(t)$.

The three side vectors of a triangle sum to zero.  Hence we may write
each $d_{tj}$ as a Fourier decomposition in the cube roots of unity:
there are unique complex numbers $p_t,q_t\in\C$ with
\begin{equation}\label{eq:side-decomposition}
   d_{tj} \;=\; p_t\,\om^j \;+\; q_t\,\overline{\om}^{\,j},
   \qquad j=0,1,2,
\end{equation}
and explicitly
\[
   p_t=\frac13\sum_j d_{tj}\,\om^{-j},\qquad
   q_t=\frac13\sum_j d_{tj}\,\om^{j}.
\]
The scalar $p_t$ is the \emph{complex-linear part} and $q_t$ the
\emph{antilinear part} of the real-affine map from an equilateral
reference triangle to $t$.  In particular, if $t$ is equilateral, $q_t=0$
and $p_t$ is just the side length times a rotation phase.  The map's
orientation and areas give
\begin{equation}\label{eq:area-identity}
   |p_t|^2-|q_t|^2 \;=\; \frac4{\sqrt3}\,\Area(t),
   \qquad
   |p_t|+|q_t| \;\le\; C\,\diam(t).
\end{equation}
The first is because the Jacobian of the real-affine map from a
reference equilateral triangle of area $\sqrt3/4$ is
$|p_t|^2-|q_t|^2$, and the ratio of areas matches; the second is a
triangle inequality from the three-side Fourier formula.

A triangle $t$ is \emph{tagged} for a label $\lambda=(L,C)$ by index $j$ if
$L$ uses $t$ and the missing spoke $j$ (not used by $L$) belongs to the
component $C$.  Each tag corresponds to a side of the rim $P$ (defined
below) entering the component $C$.  For a coefficient function
$g:\C\to\C$ supported where both $\{t\text{ is tagged}\}$ events are
defined, set
\begin{equation}\label{eq:UQV}
   U_\lambda(g) := \sum_{t\text{ tagged by }j} g(z_t)\,p_t\,\om^j,
   \qquad
   Q_\lambda(g) := \sum_{t\text{ tagged by }j} g(z_t)\,q_t\,\overline{\om}^{\,j},
   \qquad
   V_\lambda(g) := U_\lambda(g) + Q_\lambda(g).
\end{equation}
Again, $j$ runs over the three spokes of $t$ in a cyclic order that is
consistent for every tagged triangle.  The first formula collects the
complex-linear displacement of each tagged side; the second collects the
antilinear part (the deviation from equilateral); $V$ is the sum (the
actual weighted tangent displacement).  Rotating the ambient frame
multiplies $U$ by the corresponding complex rotation and $Q$ by its
conjugate, so each formula is coordinate-independent.

Because $Z_{t,\lambda}=\om^j$ when $t$ is tagged by $j$ and $0$
otherwise, and its conjugate gives $\overline{\om}^{\,j}$, the
orthogonality identity \eqref{eq:orthogonality} translates immediately
into exact energy identities:
\begin{equation}\label{eq:bessel-Q}
   \Esp\sum_\lambda |Q_\lambda(g)|^2
   \;=\; \frac34\sum_t |g(z_t)\,q_t|^2,
   \qquad
   \Esp\sum_\lambda |U_\lambda(g)|^2
   \;=\; \frac34\sum_t |g(z_t)\,p_t|^2.
\end{equation}
Adding, and using the Cauchy--Schwarz inequality
$|p_t+q_t|^2\le 2(|p_t|^2+|q_t|^2)$:
\begin{equation}\label{eq:bessel-V}
   \Esp\sum_\lambda |V_\lambda(g)|^2
   \;\le\; \frac32\sum_t |g(z_t)|^2\,\bigl(|p_t|^2+|q_t|^2\bigr).
\end{equation}
We call \eqref{eq:bessel-Q} and \eqref{eq:bessel-V} the \emph{Bessel
bounds}.  Their remarkable feature is that the coefficients
$g(z_t)$ may be any complex numbers (they can be random, as long as they
depend on the graph geometry but not the colours); the right side depends
only on those coefficients and on the local geometry, through triangle
sizes.  In particular, for any fixed compactly supported function $g$
and any fixed spacing $\eps>0$, the right-side sums are bounded by a
constant times $\int|g|^2$.

The Bessel bounds are what makes the algebra in Sections 5--7 possible:
they replace each random field on the left by a computable, deterministic
quantity on the right.  In particular, they say that the negative
biases $U-P$ and $Q-P$ are controlled by the same area form.

\subsection{Rims and bypass}
\label{ssec:rims}

Given a label $\lambda=(L,C)$ with $C$ containing a dual vertex (not an
isolated edge), the union of closed primal triangles whose dual vertices
lie in $C$ is a connected subsurface of the sphere whose boundary is a
simple primal circuit, monochromatic.  This is a standard topological
fact about cellations, but let us prove it.

\begin{lemma}[Rims]
\label{lemma:rims}
Let $\lambda=(L,C)$ be a label with $C$ containing a dual vertex.  Let
\[
   P \;=\; \partial\Bigl(\bigcup_{t\text{ dual vertex of }C}\overline t\Bigr)
\]
be the boundary curve, oriented with $C$ on its left.  Then:
\begin{enumerate}[label=(\roman*)]
\item $P$ is a simple primal circuit;
\item all vertices of $P$ have the same colour;
\item the right-hand triangles of $P$ (those just to its right) are
   exactly the tags of $(L,C)$, and their facing primal vertices lie in one
   connected path of the opposite colour;
\item the directed circuit $P$ determines $(L,C)$.
\end{enumerate}
If instead $C$ is an isolated single edge (both endpoints on $L$), that
edge has exactly two tags (one at each endpoint), and its contribution
$V_{(L,C)}(g)$ satisfies
\begin{equation}\label{eq:isolated-tag}
   V_{(L,C)}(g) \;=\; \bigl(g(z_t)-g(z_u)\bigr)\,d_{te},
   \qquad V_{(L,C)}(1) \;=\; 0,
\end{equation}
where $t,u$ are the two endpoint triangles and $e$ the shared missing
primal edge.
\end{lemma}

\begin{proof}
(i) Connectivity: the triangles with dual vertex in $C$ are connected
via shared primal edges (since $C$ is connected as a subgraph of $\Gamma$
away from $L$).  Boundary simplicity: a boundary of a connected union of
primal triangles can be non-simple only at primal vertices where several
subsurface sectors meet, alternating with sectors of $L$ or of other
components.

Suppose at some primal vertex $v$ the sectors belonging to $C$ form
several separated cyclic intervals.  Each gap between them is an
$L$-sector.  If a gap contained a dual vertex outside $L$ and outside
$C$, then by triangle adjacency that vertex would belong to some
component $C'$ adjacent to $C$ across the tiny sector---but $C$ and
$C'$ are different components, contradiction.  So each gap contains only
$L$-sectors.  Pick alternate $C$-sectors and $L$-sectors and extend them
to alternating points of a small circle around $v$, using short arcs
inside the sectors.  Outside that small circle, the $C$-arcs join in $C$
and the $L$-arcs join in $L$: two disjoint connected sets cannot join
alternating boundary contacts on a complementary disk (a standard
topological fact: two disjoint continua in a disk cannot cross).

Therefore every primal boundary vertex sees one contiguous block of
$C$-sectors.  The union of $C$-triangles is then a subsurface with
disjoint simple boundary circuits, each of which faces a distinct
complementary disk.  Since $L$ is a single connected circuit avoiding
this subsurface and facing it from outside, there can be only one such
boundary disk and one boundary curve $P$.

(ii) The missing side at an occupied primal edge joining two occupied
dual vertices has equal-coloured endpoints.  Hence every vertex of $P$
has the same colour.

(iii) Follow the directed cure $P$: at each primal vertex, the two
adjacent vertices of $P$ have the same colour (by (ii)), and the sector
of $C$ between them is monochromatic; hence the missing spoke (the one
not used by $P$) enters a component of $\Gamma\setminus P$ on the
opposite side.  Since $P=\partial C$ and $C$'s dual vertices lie on the
left, this opposite side is $L$, and the missing spokes of the tagged
primal triangles all enter the neighbouring component $C'$ of
$\Gamma\setminus L$, which is monochromatic of the opposite colour by the
same argument.  The facing primal vertices of $P$ (those just to its
right) lie in this $C'$, joined by a path of opposite-colour sites
following $P$ directly inside the ribbon of ribbon triangles.

(iv) The directed edge of $P$ identifies its occupied right-hand
triangle (which uses one of its sides as a face), and that face's dual
vertex identifies $L$; the left-hand triangle identifies $C$.

For the isolated-edge case: an unused edge whose two endpoints are on
$L$ contributes two tags, one from each endpoint triangle.  The two side
vectors $d_{te}$ at those tags are opposite (traversed in opposite
directions along the shared missing side $e$), giving
$V_{(L,C)}(g)=(g(z_t)-g(z_u))d_{te}$.  In particular
$V_{(L,C)}(1)=0$.
\end{proof}

We call $P$ an oriented \emph{rim}, and use the notation $U_P,Q_P,V_P$
for its fields $U_\lambda,Q_\lambda,V_\lambda$.  If its whole drawing is
in plane coordinates, $V_P(1)$ is the sum of the directed edge
displacements of the closed polygon $P$, which is zero.  Thus $V_P(1)=0$
for every closed rim.

Restricting \eqref{eq:bessel-Q} to genuine rims gives inequalities; the
equalities themselves also include the isolated components.

The next two lemmas identify rims geometrically.  Think of them as the
precise version of ``a colour boundary looks like a boundary.''

\begin{lemma}[No right bypass and converse]
\label{lemma:no-bypass}
Let $P$ be a rim (from Lemma~\ref{lemma:rims}).  Then $P$ has no path of
its own colour joining two distinct vertices of $P$ with all its other
points in the open right-hand disk.

Conversely, a simple monochromatic circuit with that property, and with
an opposite-coloured site in its right-hand disk, is a rim.
\end{lemma}

\begin{proof}
Suppose an alleged bypass exists.  Make it simple: it is a crosscut of
the right-hand disk, separating its two nonempty rim arcs.  Choose one
primal edge on each of the two separated arcs.  The facing triangle of
such an edge must lie on the same side of the crosscut (a primal path
cannot cross the interior of a face).  Its opposite vertex does not lie
on the bypass (bypass is monochromatic; that opposite vertex is opposite-
coloured, because facing vertices of $P$ are opposite-coloured by
Lemma~\ref{lemma:rims}).  Thus two distinct facing vertices lie in
different parts of the disk cut by the bypass.  Their opposite-coloured
cluster cannot cross either the monochromatic rim or the monochromatic
bypass, contradiction with their being in the same cluster
(Lemma~\ref{lemma:rims}).

Conversely, let $C$ be a simple monochromatic circuit satisfying the
stated property, with an opposite-coloured site in its right-hand disk.
Let $uv$ be an edge of $C$ and $w$ the third primal vertex of its
right-hand triangle.  If $w$ had the same colour as $C$ and lay strictly
on the right, then the path $u\to w\to v$ through the triangle would be
a right-side chord---a bypass---contradiction.  If $w$ lies on $C$
itself, then the two-edge segment from $u$ to $w$ or from $w$ to $v$
would be a right-side chord, unless the whole right disk is that single
triangle.  But in that case the disk contains only the one opposite
vertex $w$ on $C$'s rim, and the assumed opposite-coloured site is
$w$---contradicting $w$ being on $C$.  So every facing vertex is on the
opposite side.  Choose $L=$ the occupied dual circuit through one facing
triangle (which lies entirely to the right of $C$).  The triangles on
the left of $C$ are untouched and form part of a component $C'$ of
$\Gamma\setminus L$.  The rim of $C'$ contains the chosen edge and lies
on the closed right side of $C$ (its left subsurface contains $C$'s left
disk).  If the two circuits differed, a segment of the $C'$-rim between
successive contacts with $C$ would give a right bypass; contradiction.
\end{proof}

\begin{lemma}[Sandwich]
\label{lemma:sandwich}
In an annular band, suppose there are disjoint essential simple black
and white circuits, with the white circuit on the right of the black one
for the direction of the band.  Then there is a black rim (in this
direction) in the closed region between them.
\end{lemma}

\begin{proof}
Take a maximal biconnected block of the black graph containing the given
black circuit.  The face of this block toward the white circuit has a
simple boundary.  That boundary lies between the two given circuits (the
block contains the black one and cannot cross the white one).  A black
bypass in this face would add an ``ear'' between two distinct vertices of
the block and enlarge it, contradicting maximality.  A white circuit
supplies an opposite-coloured site in the face.  Apply
Lemma~\ref{lemma:no-bypass}.  In the full plane the argument uses the
finite black cluster containing the given circuit.
\end{proof}

\subsection{Infinite volume: rims and the closed-sum tail}
\label{ssec:infinite-volume}

Up to now everything was on a finite triangulation of the sphere.  Now
pass to the full-plane Poisson--Voronoi model at spacing $\eps$.  The
following facts hold almost surely and are used repeatedly.

\begin{lemma}[Infinite-volume facts]
\label{lemma:infinite-volume-facts}
Fix $0<\eps<\infty$ small.  Under the full-plane model:
\begin{enumerate}[label=(\roman*)]
\item Almost surely every monochromatic cluster is finite.
\item Almost surely every occupied wall (disagreement circuit of the dual)
   is finite.
\item The dual graph $\Gamma$ has one end: after deleting any fixed finite
   part, all sufficiently far-away triangles are connected by paths that
   avoid that part.
\item For any fixed finite circuit $L$ and finitely many points off $L$, the
   component relations among those points stabilise as the spherical
   completion grows to infinity.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) This is standard critical percolation absence of infinite clusters
for Poisson--Voronoi at $p=1/2$; equivalently, the one-arm estimate
$A_1(r,R)\le C(r/R)^{c_1}$ of Proposition~\ref{prop:prob-input}(iii) below,
applied on a countable covering by boxes, gives that an infinite cluster
would cross events of vanishing probability.  We use this as input (it is
part of the standard criticality result of Bollob\'as--Riordan 2006 in the
formulation we will state as Proposition~\ref{prop:prob-input}).

(ii) If an infinite occupied wall existed, its two adjacent primal
clusters on either side would be infinite monochromatic sites (by the
constancy of colour along the facing vertices of a rim, Lemma~\ref{lemma:rims}),
contradicting (i).  So every occupied wall is finite.

(iii) Enclose any finite removed set together with all incident triangles
in a compact set $K$.  Plane paths outside a slightly larger compact $K'$
avoiding primal vertices connect any two sufficiently remote triangles by
face adjacency that avoids $K$ (each primal edge crossed by a face-adjacency
step is a barrier only if it lies in $K$; outside $K'$, no primal edge in
$K$ interferes).  Hence all triangles sufficiently remote are in one
component of $\Gamma\setminus K$.

(iv) Fix $L$ and finitely many points off $L$.  The two-defect argument
of Lemma~\ref{lemma:two-defects} shows the component relations among
finitely many points off $L$ depend only on those points and $L$, and
by (iii) stabilise once their neighbourhood is large enough.
\end{proof}

Fix a finite box $B\subset\C$.  Since every occupied wall is finite, the
sum
\[
   \sum_{P\text{ rim, }P\subseteq B}|\ldots|
\]
is a sum over finitely many terms, with probability one.  Average over
colours and take $L^2$ norms.

\begin{proposition}[Closed-sum tail]
\label{prop:closed-tail}
For every fixed bounded box $B\subset\C$ and every $a>0$, at spacing $\eps$:
\begin{equation}\label{eq:closed-sum}
   \lim_{\eps\downarrow0}
   \EE\sum_{\substack{P\text{ rim},\ P\subseteq B\\ \diam(P)\ge a}}
   |Q_P(1)|^2 \;=\; 0.
\end{equation}
The corresponding statement for $V_P$ holds with the sum identically
vanishing (every $V_P(1)=0$ since $P$ is a closed rim).
\end{proposition}

\begin{proof}
We prove the $Q$-statement; $V$ vanishes exactly.

\textbf{Step 1: uniform bound in the full plane.}  Work at unit spacing
($\eps=1$) first.  Anchor each rim $P$ at the average $b(P)$ of its vertex
positions.  This is a translation-equivariant measurable choice.  For
$s>0$, define the stationary random measure
\[
   \mu_s = \sum_{P\text{ retained}} |Q_P(1)|^2\,\delta_{b(P)},\qquad
   \text{retained: all tags of }P\text{ lie within distance }s\text{ of }b(P).
\]
Let $W_R$ be a square of side $R$.  For anchors in $W_R$, every retained
rim's full tag sum contributes only via $\ind_{W_R^{+s}}$ where
$W_R^{+s}$ is the $s$-neighbourhood of $W_R$.  Thus, using
\eqref{eq:bessel-Q} and stationarity:
\[
   \EE\mu_s(W_R) \le \frac34\EE\sum_{z_t\in W_R^{+s}}|q_t|^2
   = c_q\,\Area(W_R^{+s}),
\]
for a finite constant $c_q<\infty$ (this constant exists by the standard
Poisson mesh estimates, Lemma~\ref{lemma:mesh} below).  Divide by $R^2$
and let $R\to\infty$: the intensity of $\mu_s$ is at most $c_q$, uniformly
in $s$.

Each rim has finitely many tags, so by monotone convergence
\[
   \mu = \sum_P |Q_P(1)|^2 \delta_{b(P)}
\]
has finite intensity at most $c_q$.

Let $\theta(M)$ be the intensity of the restriction of $\mu$ to rims of
diameter at least $M$.  Because each rim is finite and $\mu$ has finite
mean on a unit square, $\theta(M)\to 0$ as $M\to\infty$.

\textbf{Step 2: rescale to spacing $\eps$.}  Rescaling a spacing-one
configuration by $\eps$ multiplies each $Q$-sum by $\eps$ (because
each $d_{tj}$ scales by $\eps$) and multiplies anchor area by $\eps^2$.
Thus the expected squared mass of spacing-$\eps$ rims of diameter at
least $a$, with anchor in $B$, is exactly
$\Area(B)\cdot\theta(a/\eps)$.  Since a rim contained in $B$ has anchor
in $B$, this bounds the left side of \eqref{eq:closed-sum} by
$\Area(B)\theta(a/\eps)\to 0$ as $\eps\downarrow0$.
\end{proof}

The restriction to rims ``contained in $B$'' and ``of diameter at least
$a$'' in \eqref{eq:closed-sum} is essential to the proof.  The next
sections turn this closed-sum estimate into a statement on \emph{proper
subarcs} of rims, by comparing configurations with common endpoint data.
That is the technical core of the paper.

\section{Probability input, local geometry, and arm consequences}
\label{sec:probability}

This section states the using-percolation-theory inputs we need and derives
from them several local estimates that feed into later sections.  The
\emph{inputs} are established facts about critical Poisson--Voronoi
percolation; they are due to Tassion, Ahlberg--Griffiths--Morris--Tassion
and Vanneuville, and we state them exactly.  The \emph{consequences} are
proved inline from the inputs.

\subsection{The percolation input}

For $1\le r\le R$, let $A_j(r,R)$ be the arm event across the annulus
$[-R,R]^2\setminus(-r,r)^2$, i.e.\ the event that there are $j$ vertex-
disjoint primal paths (or the cell-equivalent) crossing between the inner
and outer square boundaries, with the following colour pattern:

\begin{itemize}
\item $j$ even: $j$ alternating-colour arms, cyclically ordered.
\item $j$ odd, $j>1$: $j-1$ alternating arms plus one additional black arm
  \emph{satisfying no Voronoi cell intersects both that arm and any other
  arm} (used as a witness that the extra black arm is ``separated'' from
  the alternating ones in the disjoint-occurrence sense).
\item $A_1$ is just ``there exists a black arm.''
\end{itemize}
In these definitions the paths can be chosen so that their used set of
sites are disjoint.  Set
\[
   \alpha_j(r,R)=\PP(A_j(r,R)),\qquad
   p_j^\eta(r,R)=\Peta(A_j(r,R)).
\]

\begin{proposition}[Established input]
\label{prop:prob-input}
The following estimates hold at criticality.
\begin{enumerate}[label=(\roman*)]
\item \textbf{(Box-crossing)} For each fixed rectangle aspect ratio, both
   colours have crossing probabilities lower-bounded by a positive constant
   uniformly in the rectangle size.  At sizes tending to infinity, the
   corresponding \emph{quenched} crossing probabilities are lower-bounded
   with probability tending to one.  More precisely, for each fixed finite
   family of buffered rectangle chains on fixed square grids, and each
   $\gamma>0$, the required quenched crossing lower bounds hold outside a
   geometry event of probability $O_\gamma(R^{-\gamma})$ when the family is
   dilated by $R$.  The positive lower bound may depend on the family and
   on $\gamma$.
\item \textbf{(Annealed-quenched comparison)} For every fixed $j\ge1$,
   there is $C_j<\infty$ such that for all $1\le r\le R$,
   \begin{equation}\label{eq:quenched-arms}
      \EE\bigl[p_j^\eta(r,R)^2\bigr]\le C_j\,\alpha_j(r,R)^2.
   \end{equation}
\item \textbf{(Arm powers)} There are constants $c_1,c_4>0$ and
   $C<\infty$ such that for all $1\le r\le R$,
   \begin{equation}\label{eq:arm-powers}
      \alpha_1(r,R)\le C(r/R)^{c_1},\qquad
      \alpha_5(r,R)\le C(r/R)^2,\qquad
      \alpha_4(r,R)\le C(r/R)^{1+c_4}.
   \end{equation}
\end{enumerate}
\end{proposition}

\noindent\textbf{Where this comes from.}  (i) is Tassion 2016 (uniform
box-crossing) combined with Ahlberg--Griffiths--Morris--Tassion 2016
(quenched crossing estimates) refined quantitatively by Vanneuville 2025,
Theorem 1.2 and Proposition 3.14.  (ii) is Vanneuville 2025, Theorems
1.3--1.4, derived from annealed scaling relations of Vanneuville 2019.
(iii) is Vanneuville 2025, equation (3.1), Proposition 3.4(iii), and
Proposition 6.2 (respectively).  For (iii) the $c_1,c_4$ are absolute
constants $>0$; only their existence matters in this paper.  Only the
specific finite lists of buffered rectangle shapes singled out in
Sections~\ref{sec:pins} and~\ref{sec:actual-arcs} are ever used; each is
fixed and finite.

These statements are \emph{inputs}: we do not reprove them.  Everything
else in the paper follows from them by elementary probability.

\subsection{Reimer's disjoint occurrence}

BK (van den Berg--Kesten) and Reimer's theorem are the main tools for
turning arm bounds into product bounds.  We need the following form.

\begin{lemma}[Disjoint witnesses at fixed geometry]
\label{lemma:reimer}
For events $A,B$ in a finite product of fair bits, let $A\mathbin\square B$
mean ``there exist disjoint sets of coordinates, one certifying $A$ and
one certifying $B$.''  Then
\begin{equation}\label{eq:reimer-product}
   \PP(A\mathbin\square B)\le \PP(A)\PP(B).
\end{equation}
In particular, this inequality applies to bounded percolation tests
conditional on the Voronoi geometry, with disjoint used site sets.
\end{lemma}

\begin{proof}
Apply Reimer 2000.  On pairs $(x,y)\in\{0,1\}^n\times\{0,1\}^n$, fix the
set $K$ on which they agree and their common values there; on the other
coordinates $x,y$ are bitwise complementary.  If $x\in A\mathbin\square B$,
then the restrictions of the disjoint witnesses to the coordinates outside
$K$ certify the restricted events on that slice.  Reimer's inequality
$|A\mathbin\square B|\le |A\cap\overline B|$ (bit-complement ${}^-$) bounds
the number of such pairs in the slice by those with $x\in A$, $y\in B$.
Summing over slices gives $2^n|A\mathbin\square B|\le |A||B|$, i.e.\
\eqref{eq:reimer-product}.  A bounded region meets only finitely many
cells a.s., so the conditional tests use finite products.
\end{proof}

No annealed product bound is asserted (averaging the right-hand side
gives the expectation of a product of quenched probabilities); we will
control that product using \eqref{eq:quenched-arms}.

\subsection{Poisson mesh estimates}
\label{ssec:mesh}

The Voronoi--Delaunay triangulation at spacing $\eps$ is a random mesh
with cells of diameter $O(\eps)$ typically.  Standard stochastic-geometry
estimates bound the tails of unusually large triangles and cells.  We
state these precisely.

\begin{lemma}[Mesh estimates]
\label{lemma:mesh}
The following facts hold for the full-plane Poisson--Voronoi model of
spacing $\eps$.
\begin{enumerate}[label=(\roman*)]
\item \emph{(Local determination)} On a fixed compact set, the
   triangulation and the cells meeting it are determined by the points in
   any fixed positive collar, outside an event of probability tending to
   $0$ as $\eps\to0$; local diameters tend to $0$ in probability.  The
   same determination holds when the exterior point configuration is
   changed while the collar is retained.
\item \emph{(Circumradius tail)} At unit spacing, the probability that a
   triangle incident to a unit square has circumradius greater than $R>2$
   is at most $Ce^{-cR^2}$.  Same for a cell meeting the square whose site
   is at distance $>R$ from it.  Incident counts and sums of polynomial
   diameter weights have moments of all orders.
\item \emph{(Sum moments)} For $S$ a square of side $s$, $k\ge 0$, $p<\infty$,
   \begin{equation}\label{eq:mesh-moments}
      \Bigl\|\sum_{t\cap S\ne\varnothing}\diam(t)^k\Bigr\|_{L^p}
      \le C_{p,k}\,\eps^{k-2}(s+\eps)^2.
   \end{equation}
   For a deterministic Borel set $E$ of finite area,
   \begin{equation}\label{eq:barycenter-intensity}
      \EE\sum_{z_t\in E}\diam(t)^k = m_k\,\eps^{k-2}\Area(E),\qquad m_k<\infty.
   \end{equation}
   At each fixed $s>0$, $\sum_{z_t\in S}\diam(t)^2\to m_2 s^2$ in every
   finite moment norm as $\eps\downarrow0$; similarly for the sum over
   triangles meeting $S$ and for translation-covariant weights of degree 2
   bounded by a constant times $\diam(t)^2$, such as $|p_t|^2$ and $|q_t|^2$.
\item \emph{(Primal cycles approximate circles)} Within fixed positive
   margins, one can find locally determined simple primal cycles
   approximating a fixed circle arbitrarily closely in its cyclic order;
   finitely many separated circles admit disjoint such cycles.
\end{enumerate}
\end{lemma}

\begin{proof}
If an empty disk of radius $\ge\rho$ contains a point of a compact region
$K$, it contains an empty subdisk of radius comparable to $\rho$ whose
centre lies in an $O(\rho)$-collar of $K$.  A grid of mesh a small
multiple of $\rho$ reduces this to finitely many empty-disk tests.  The
void probability of each is at most $\exp(-c\rho^2/\eps^2)$, and the
number of tests is $\le C(1+\diam(K)/\rho)^2$.

A triangle meeting $K$ has an empty circumdisk containing its meeting
point; the same observation holds for a cell meeting $K$ whose site is
remote.  Excluding empty subdisks of the above kind forces every relevant
circumdisk to be small and contained in the retained collar, and the
positive empty-circle tests defining those triangles are local; they stay
valid when the exterior configuration is changed.

On the event that all relevant circumradii are $\le R$, the vertices of
interest lie in a box of side $O(R)$.  The incident triangle count is
bounded by a polynomial in the number of Poisson sites there; every
diameter weight is bounded by a polynomial in $R$.  Splitting into
successive integer values of $R$, the exponential tail and the Poisson
count moments give all assertions in (ii).  Tiling $S$ by
$O((s/\eps+1)^2)$ cells of side $\eps$ and applying the unit-square
estimate with triangle inequality in $L^p$ gives \eqref{eq:mesh-moments}.
Covergence in (iii) is by spatial ergodicity at spacing one, rescaled;
the higher-moment convergence is then immediate.

For (iv), follow successive triangles around a thin collar of the given
circle to build a primal walk with nonzero winding so that on fine mesh
the walk stays in the collar and follows the circle in order.  Decompose
the walk into simple cycles; at least one has nonzero winding, hence
encircles the centre.  Choosing the collar in a prescribed margin gives
simultaneous locality for finitely many circles.
\end{proof}

We will freely use the resulting bounds later; for instance, triangle
diameters are typically $O(\eps)$, sums over a fixed compact of
$\diam(t)^2$ are $O(\eps^0 \cdot\Area)$ of order $\Area$, and the
contribution of unusually large triangles is negligible in $L^p$.

\subsection{Planar colour order: Hex and ordered colours}
\label{ssec:color-order}

Two combinatorial lemmas about planar graphs with prescribed colours are
used repeatedly.

\begin{lemma}[Hex and indicator colourings]
\label{lemma:hex}
Let a finite planar graph be drawn in a rectangle (or triangulated disk)
with four successive vertex sides $S_1,S_2,S_3,S_4$ around the boundary.
Fixing any colouring, at least one of the following holds: a black
connection between $S_1$ and $S_3$, or a white connection between $S_2$
and $S_4$.

In particular: if a set $X$ of vertices blocks every path between $S_1$
and $S_3$, then $X$ contains a path between $S_2$ and $S_4$.
\end{lemma}

\begin{proof}
Attach four exterior vertices, successively coloured black, white, black,
white, and fan each one onto its side; then close the exterior to a sphere.
The four colour changes at the boundary vertices create four ``boundary
ends'' of a system of disagreement strands.  The strands pair boundary
ends in a non-crossing way (strands do not intersect each other), and by
the non-crossing parity there are exactly two possible pairings.  One
pairing corresponds to a black connection between the two black fans
(i.e. a black connection between $S_1$ and $S_3$ in the disk); the other
to a white connection between $S_2$ and $S_4$.  Following the adjacent
sites of the strand system gives the corresponding primal connection
inside the disk.

For the last assertion: colour \emph{exactly} the vertices of $X$ black
and every other vertex white.  A white crossing of the blocked
$S_2$--$S_4$ pair would require white vertices, but all white vertices are
outside $X$; such a crossing exists only in the sense forbidden by the
blocking hypothesis, contradiction.  Hence the other crossing (which is
$S_2$--$S_4$ in the augmented disk) uses vertices of $X$.
\end{proof}

\begin{lemma}[Ordered colours in a rectangle]
\label{lemma:color-order}
Let a finite planar graph be drawn in a rectangle with independent fair
colours on vertices.  For any $k\ge1$, the probability of having $k$
vertex-disjoint crossings between the two end sides, with a prescribed
\emph{linear order of colours} along the crossings, depends only on $k$
and not on the specific sequence.
\end{lemma}

\begin{proof}
Fan the two end sides to two new uncoloured exterior terminals.  Crossings
become simple terminal-to-terminal paths, internally disjoint and ordered
left to right.  Conversely, first-hit/last-hit trimming recovers proper
crossings in the same order.

Fix the first colour.  If a path of that colour exists, add an artificial
terminal-to-terminal edge on the exterior right.  The union of this edge
with all simple paths of the chosen colour is biconnected (union of
simple cycles sharing that edge).  Its left outer boundary is a simple
path of the chosen colour; call it the \emph{leftmost path}.  It has no
same-coloured bypass strictly on its left between distinct vertices,
including the terminals.  Conversely, this property characterises it:
any other candidate's segment lying strictly to the left would be such a
bypass.

The event that a specified path is leftmost is determined by its spins
and the spins strictly to the left.  All vertices strictly to the right
retain fair independent colours.  The leftmost path puts itself to the
left of the first path of any given ordered disjoint family (which
contains the same first colour), while all others lie strictly to its
right.  Delete its internal vertices and induct on the graph.  This shows
that, for a fixed first colour, the remaining sequence of colours does
not affect the probability.  Global colour interchange flips the first
colour; combining the two cases gives independence of the full sequence
from the particular colours.
\end{proof}

\subsection{Four arms with both colours}
\label{ssec:mixed-four}

The next two lemmas are \emph{local}: fix deterministic $x$ and
$0<s<S$ before $\eps\to0$.  In a bulk patch assume that a neighbourhood
of $\overline B(x,S)$ agrees with the full-plane model.  All statements
hold for bounded ratios $S/s$, with constants depending only on those
ratios and on the reference shapes.

\begin{lemma}[Mixed-colour four arms]
\label{lemma:mixed-four}
There are $c>0$ and $C<\infty$ such that
\begin{equation}\label{eq:mixed-four}
   \limsup_{\eps\downarrow0}
   \PP\!\left(\begin{array}{c}
      \text{four vertex-disjoint primal arms across }B(x,S)\setminus B(x,s),\\
      \text{with both colours represented}
   \end{array}\right)
   \le C\,(s/S)^{1+c}.
\end{equation}
The four colours need not alternate.
\end{lemma}

\begin{proof}
Take $S/s$ large.  Replace the four arms by disjoint proper subpaths
between locally determined primal boundary cycles strictly inside the
annulus; do this on a good-mesh event (which has vanishing complement at
these scales).

Pick an adjacent pair of the four paths of opposite colours.  In the gap
between them there is a spanning disagreement strand $\gamma$: no strand
can leave through a monochromatic side path, and at each annular end the
odd number of colour switches forces a through-strand.  Stop $\gamma$
at its first boundary edges.  The visited triangles of $\gamma$ use a
set of sites disjoint from the two remaining paths (by properness).

Now every stranded configuration admits a bounded union over one
strand and a spin pattern on its visited sites.  Conditional on that
spin pattern, delete the visited sites and slit the annulus along
$\gamma$.  What remains is a planar rectangle with free fair bits, and
the two avoiding paths have a linear order.  By Lemma~\ref{lemma:color-order}
applied in that rectangle, the probability of the union of two
alternating-colour runs is at most $4$ times the probability of one
specified run.  Choosing the sequence compatible with alternating four
arms, the two missing arms are supplied by the opposite-coloured banks
of $\gamma$: the sites beside each bank are connected, disjoint from the
avoiding paths, and lie on opposite sides of the slit, joining directly
to the strand without crossing.

Let $N$ count proper spanning wall strands stopped at first boundary
edges and counted once.  These are the components of the walls after
they are cut at boundary cycles.  On the good geometry event $G$
(localisation plus a fixed buffered subannulus with quenched white
circuit probability $\ge\delta>0$), fixed rectangle chains give
$\PP(G^c)=o(1)$ with $\delta$ independent of $S/s$.  We then have
\begin{equation}\label{eq:strand-reduction}
   \PP(\text{mixed four arms})
   \le o(1) + C\,\EE\bigl[\ind_G N\ind_{A_4^\eps(r,R)}\bigr],
\end{equation}
where $r\asymp s$ and $R\asymp S$ are deterministic buffered square
radii.

On $G$, $N$ has moments of every fixed order uniform in $S/s$.  Indeed,
disjoint spanning strands have adjacent black banks; a gap between two
consecutive strands accounts for at most two black banks; black paths in
different gaps cannot meet or cross a strand.  Thus $N\ge2k$ yields at
least $k$ black crossings with disjoint site witnesses.  On $G$, a black
crossing of the working annulus is incompatible with the quenched white
circuit in the fixed subannulus, so its conditional probability is at
most $1-\delta$.  Conditional BK gives $\Peta(N\ge2k)\le(1-\delta)^k$ on
$G$, hence $\EE[\ind_GN^q]\le C_{q,\delta}$ for each fixed $q<\infty$.

H\"older's inequality and \eqref{eq:arm-powers} bound the second term
in \eqref{eq:strand-reduction} by
\[
   C_q\,\PP(A_4^\eps(r,R))^{1-1/q}
   \le C_q (s/S)^{(1+c_4)(1-1/q)}.
\]
Choose $q$ large enough that the exponent exceeds $1$; reduce $c>0$
accordingly.  Take $\eps\downarrow0$ to conclude.
\end{proof}

\subsection{Two separated returns of a rim}
\label{ssec:near-return}

A rim is a simple closed curve, so it can revisit a small ball and then
leave.  We need a quantitative bound for how unlikely it is for a rim to
visit a small ball, leave, and revisit it, with both complementary rim
arcs also leaving.

\begin{lemma}[Near-return bound]
\label{lemma:near-return}
There are $c>0$ and $C<\infty$ (same fixed-scale convention) with
\begin{equation}\label{eq:near-return}
   \limsup_{\eps\downarrow0}
   \PP\!\left(\begin{array}{c}
      \text{some rim has two visits to }B(x,s)\text{ such that both}\\
      \text{complementary rim arcs between the visits leave }B(x,S)
   \end{array}\right)
   \le C(s/S)^{2+c}.
\end{equation}
The rim may be part of a global spherical completion.  The event is
bounded by a local monochromatic-arm test; no union over rims is needed.
\end{lemma}

\begin{proof}
Suppose first that the rim $P$ is black.  Take four nested primal cycles
near radii $2s,5s,S/5,S/2$; they are locally determined and pairwise
disjoint on a good-mesh event.  The two visits of $P$ to $B(x,s)$, together
with the two ``excursions'' of $P$ leaving $B(x,S)$, produce four vertex-
disjoint black journeys between the cycle at $2s$ and the cycle at $S/2$.
Trim them properly and denote them $P_1,\ldots,P_4$ in cyclic order.
Retain them throughout.

Each oriented journey has its right side in one of the two adjacent
annular gaps.  Two sides per gap force at least two distinct gaps with a
right-facing rim side.  Fix one such gap.  The inner middle cycle blocks
every annular connection between its two ends.  Apply the indicator-
coloring assertion of Lemma~\ref{lemma:hex}, colouring only that cycle's
vertices black.  This gives a transverse proper crosscut between the two
journey sides.  Then in the subdisk between that crosscut and the outer
end, apply the same error to the outer middle cycle; the outer cycle
still separates because the first crosscut lies strictly inside it.
This gives two ordered transverse proper crosscuts.

Consider the quadrilateral between these crosscuts.  There is no actual
black crossing between its two journey sides: a simple such crossing,
started just after its last contact with the right-facing side, would
travel through the right disk of the whole rim up to its first contact
with the whole rim (which exists by the other selected side), giving a
right bypass forbidden by Lemma~\ref{lemma:no-bypass}.  Hex in the
actual colouring now gives a white crossing between the two transverse
cuts.  Trim it properly between the middle cycles.

Doing this in two distinct original gaps gives two disjoint white arms
(the boundaries shared by distinct gaps are black journey vertices,
which white paths cannot use).  These white arms divide the middle
annulus into two interval quadrilaterals.  Take the four original black
journeys (retained earlier) and arbitrarily trim them to four disjoint
black arms in this annulus.  It remains to ensure the two white
intervals each contain a black arm.

For each of the two white intervals, let $X$ be the set of vertices of
all four original journeys restricted to that interval.  Every path
between its two white sides must meet $X$ (otherwise it would join two
distinct original gaps while avoiding all four separating paths in the
larger working annulus).  Apply Lemma~\ref{lemma:hex} with the artificial
black set exactly $X$.  It gives a black crossing between the middle
cycles using sites of $X$ only.  This works independently in each white
interval, giving at least one genuinely black crossing in each.  If the
arbitrary black trims already occupy both intervals, retain them.  If all
four already lie in one interval, retain three of them there and add the
newly obtained black crossing in the other.  The white arcs separate
the two intervals, so the resulting four chosen black paths are still
vertex-disjoint.

We now have four black arms and two white arms, with at least one black
arm in each white-separated interval.  Choose one black arm in each
interval to obtain alternating four (i.e.\ four alternating arms in
cyclic order, which is the $A_4$ event).  Add one of the two remaining
black arms as the extra arm for $A_5$ (with the Voronoi-separation
property of the definition), and reserve the last black arm as the
separated witness for $A_1$.  All these witnesses use disjoint site sets.

Transfer all six paths to deterministic buffered square boundaries,
stopping at first outer and last inner hits, preserving order.  The
event implies $A_5^\eps(r,R)\mathbin\square A_1^\eps(r,R)$ for $r\asymp s$,
$R\asymp S$.  Conditional on $\eta$,
Lemma~\ref{lemma:reimer} bounds the probability by
$p_{5,\eps}^\eta(r,R)\,p_{1,\eps}^\eta(r,R)$.  Averaging and using
Cauchy--Schwarz with \eqref{eq:quenched-arms}:
\[
   \EE[p_{5,\eps}^\eta(r,R)\,p_{1,\eps}^\eta(r,R)]
   \le C\,\alpha_5^\eps(r,R)\,\alpha_1^\eps(r,R)
   \le C(s/S)^{2+c_1}.
\]
For a white rim, interchange colours at a cost of factor 2.  Remove the
mesh exception and take $\limsup$.  The result is a local test that does
not reference the global rim further.
\end{proof}

We adopt one positive exponent common to
Lemmas~\ref{lemma:mixed-four} and \ref{lemma:near-return}; whenever a
later H\"older inequality requires this, we reduce it further, so the
power written below is always at most the smaller of the two.

\subsection{Two-site gates}
\label{ssec:two-gates}

A corner of a colour region can be ``pinned'' by a small number of
opposite-coloured sites.

\begin{lemma}[Two-site gates]
\label{lemma:two-gates}
Let a finite triangulated annulus have disjoint simple primal boundary
cycles.  If the maximal number of vertex-disjoint black paths joining
its two boundary cycles (endpoints disjoint) is at most two, then the
closed annulus contains an essential simple primal circuit with at most
two black vertices.  The colour-interchanged assertion also holds.  If
the annulus has positive margins inside a prescribed collar, the
circuit lies within the collar.
\end{lemma}

\begin{proof}
Vertex Menger gives a set of at most two black vertices meeting every
black crossing (boundary vertices allowed).  Temporarily recolour these
two vertices white: now the annulus has no black crossing.  Attach a
black cap vertex to each boundary cycle, fanning it onto that cycle; the
caps are not black-connected.

Some occupied dual circuit separates the caps: two sites in the same
component off all disagreement walls are joined in their colour (a curve
crosses only unoccupied dual edges, hence only equal-coloured primal
adjacencies), so any black connection between the caps would appear as a
path in the primal graph staying inside one component.  A wall
$L$ separates the caps; the neighbouring white walk along $L$ uses
original annular sites and is deformable to $L$ avoiding the caps,
hence has nonzero annular winding.  Decompose into simple cycles and
retain one with nonzero winding; it is an essential white circuit.  On
restoring the original colours, only the two recoloured vertices could
become black, so the circuit has at most two black vertices.
\end{proof}

\subsection{Joint use of the four lemmas}
\label{ssec:joint}

Lemmas~\ref{lemma:mixed-four}, \ref{lemma:near-return}, \ref{lemma:two-gates}
and the mesh estimate (Lemma~\ref{lemma:mesh}) enter the proof in the
following way.  The mixed-four bound forbids configurations in which a
rim's neighbourhood carries four arms; that lets us \emph{construct
gates} (two-site primal circuits) around it.  The near-return bound
forbids configurations in which a rim's small neighbourhood is revisited
twice with both arcs escaping; that controls the geometry of a rim
locally.  The two-site gates then place plugs on rims; the mesh estimate
handles the rest of the localisation.  All three are used inside the
plug construction of the next section.

\section{Opening closed sums with local plugs}
\label{sec:plugs}

We now begin the technical core.  Section~\ref{sec:fields} proved that,
when you sum $Q$ over \emph{closed} rims containing a large box, the
result goes to zero.  But the observable of Section~\ref{sec:observable}
will be built from \emph{open} excursions of rims, not closed ones.  Bridging
that gap is what plugs accomplish.  This section sets up plugs, uses them
to compare two different closures of the same open arc, and derives a
state cochain.

\subsection{The bulk transport theorem}

\begin{theorem}[Bulk transport]
\label{thm:bulk-transport}
From every sequence $\eps\downarrow0$ one can extract a subsequence and a
complex number $a$, depending only on the subsequence, with
$\Rea a<1/2$, and such that the following holds.  Take any bulk patch
$U\subseteq\C$ (Definition~\ref{def:bulk-setting}), any point $x$, and
any $R>0$ with $\overline B(x,6R)\subseteq U$, and any
$f\in C_c^\infty(B(x,R))$.  Then
\begin{equation}\label{eq:bulk-transport}
   \max_{P:\,P\not\subset B(x,4R)}
   |Q_P(f)-a\,V_P(f)|\;\xrightarrow[\eps\to0]{\PP}\;0,
\end{equation}
where the maximum is over all rims $P$ (in the configuration) that are
not contained in the ball $B(x,4R)$.  Local tests may be replaced by
$0$ on a geometry event of probability tending to zero.  The scalar $a$
and the subsequence are common to all patches and tests.
\end{theorem}

The proof of Theorem~\ref{thm:bulk-transport} spans Sections~\ref{sec:plugs}
through~\ref{sec:actual-arcs}.  This section constructs the local plug
machinery and proves a state cochain (Proposition~\ref{prop:state-cochain});
Section~\ref{sec:affine} decomposes the cochain into affine and endpoint
parts; Section~\ref{sec:pins} places plugs on actual random rims;
Section~\ref{sec:actual-arcs} puts everything together.  The scalar $a$
will not be identified in Theorem~\ref{thm:bulk-transport}; only the
strict inequality $\Rea a<1/2$ is needed, and it is proved in
Section~\ref{sec:nondegen}.

In the rest of this section and the next, $X$ denotes either $Q$ or $V$;
all statements are proved uniformly for the two fields.

\begin{convention}[Order of limits]
\label{conv:scale-order}
A \emph{comparison scale} $h>0$ is fixed \emph{before} taking the
microscopic limit $\eps\downarrow0$.  Every $o(1)$ or $o_{L^1}(1)$ in
this section and the next is taken as $\eps\downarrow0$ at fixed $h$
and fixed deterministic geometric pattern.  Later we let $h\downarrow0$
along a countable sequence.  Thus at fixed $h$ the number of patterns is
finite; patterns may have arbitrarily small but positive probabilities.
Bounds of the form $C h$ are obtained from fixed normalized short
templates with common constants.
\end{convention}

\subsection{Plug types}

A plug is a deterministic local model of the configuration around a
point of a rim.  It records three things: (i) two annular ``gates''
around the point, each containing a small primal circuit with at most
two path-coloured vertices; (ii) a band chart giving local Euclidean
coordinates; (iii) a rank for each colour choosing which monochromatic
component in a strip is ``the'' root.

\begin{definition}[A plug type]
\label{def:plug-type}
A \emph{plug type} (for the black path colour; swap colours for white)
consists of:
\begin{enumerate}[label=\textup{(\roman*)}]
\item two concentric gate annuli $[a_1,b_1]$, $[a_2,b_2]$ around the
   origin and a cutoff radius $m$, with
   \[
      10 < a_1 < b_1 < m/10,\qquad 10m < a_2 < b_2;
   \]
\item closed Jordan disks $O\subset\operatorname{int}H$ with
   $\overline B(0,b_2)\subset\operatorname{int}O$, and an
   orientation-preserving band chart
   \[
      \Phi:[-12,12]\times[-3,3]\to\R^2,
   \]
   such that membership in $O$ and in $H$ within the chart corresponds to
   $|u|\le3$ and $|u|\le8$ respectively.  The full section
   $\Phi(\{0\}\times[-3,3])$ lies strictly inside $B(0,a_1)$;
\item one positive integer \emph{rank} for each of the two windows
   \[
      I_B=[-0.3,0.3],\qquad I_W=[-1.8,-1.4]
   \]
   in the chart coordinates $u$.
\end{enumerate}
No metric smoothness is imposed on the Jordan disks or on $\Phi$; they are
topological objects.  Let $\nu$ denote the ordinary full-plane
coloured Poisson law at spacing $\eps$.
\end{definition}

The two windows $I_B$ and $I_W$ are where black and white paths are
expected.  The two annuli contain the ``gates'': circuits that a rim must
cross, one once entering and once exiting.

Define the local geometric length $\rho_{\eps,h}=h\,(\eps/h)^{1/2}=(h\eps)^{1/2}$.
An edge is \emph{retained} if it belongs to a Delaunay triangle whose
circumradius is at most $\rho_{\eps,h}$.  This is a local criterion
(depends only on points within distance $2\rho_{\eps,h}$ of the vertex),
so retained-edge queries in a compact set use only a small collar.
Lemma~\ref{lemma:mesh} implies that, at fixed compact set, all retained
edges and triangles tend to appear with probability tending to 1.

For each colour and each rank, take the connected components of the
induced retained graph whose vertices lie in
$\Phi([-2,2]\times I_{\text{colour}})$ and which meet both $u\le-1.5$
and $u\ge1.5$.  Order them lexicographically by the first site in chart
coordinates.  The component of the given rank, if it exists, is its
\emph{root}.  Its queries lie compactly inside $O$.

The tests $L^+$ and $L^-$ (one for each end direction) require a retained
path from the black root to $u\ge6.7$ (or to $u\le-6.7$) with all vertices
in $\Phi([1,7.5]\times I_B)$ (or $\Phi([-7.5,-1]\times I_B)$).  These
tests are also compactly supported in $O^c$ (their exteriors).

\begin{definition}[Usable states and admission]
\label{def:usable-state}
Fix positive thresholds $\beta,\beta'>0$.  The \emph{state event} $G$ is:
\begin{enumerate}[label=\textup{(\roman*)}]
\item two simple surrounding retained site circuits $S_1,S_2$ lie in the
   two gate annuli, enclosing the corresponding inner balls, with edge
   segments in those annuli, and each containing at most two black sites;
\item both black and white roots exist, and
   $\nu(L^+\mid\sigma)\ge\beta$ and $\nu(L^-\mid\sigma)\ge\beta$, where
   $\nu(\cdot\mid\sigma)$ uses independent ordinary data on $O^c$
   (i.e.\ it averages over the exterior point geometry and colours).
\end{enumerate}
A type is \emph{admitted} when $\nu(G)\ge\beta'$.  Its \emph{state law} is
$\mu=\nu(\cdot\mid G)$.
\end{definition}

The two circuits $S_1,S_2$ are picked by a deterministic measurable rule
from the state ($\sigma$).  They are compactly contained in $O$, hence
determined by local data.

Given a plug centred at $z_s$ (so its entire data is translated by $z_s$),
the \emph{physical cutoff ball} is $B_s=\overline B(z_s,m_s h)$, where
$m_s$ is the cutoff radius of that plug type.

\subsection{Connectors and simple systems}

A connector attaches one plug to the next by a rectangular band, and
requires both a black path and a white path through the band.

\begin{definition}[Connectors and simple systems]
\label{def:connectors}
A \emph{directed connector} $e:s\to t$ between same-colour plugs with
disjoint obstacles is an oriented rectangular band with transverse
parameter $v\in[-3,3]$, attached by its full parameterized ends to the
positive port $u_s=8$ of $H_s$ and the negative port $u_t=-8$ of $H_t$,
and otherwise avoiding both disks.  Only the charts with $|u|\le8$ are
prescribed at its ends.

A \emph{simple system} is a cyclic list of at least two distinct plugs
and connectors, such that:
\begin{itemize}
\item its closed obstacles are pairwise disjoint;
\item its exterior bands are pairwise disjoint and avoid all other
   obstacles;
\item the sectors from section $0$ of one plug to section $0$ of the next
   form an embedded annular band with the prescribed directed longitude
   and transverse parameter.
\end{itemize}
The whole closed sector of connector $e$ is denoted $\mathcal B_e$.
\end{definition}

We also allow an orientation-preserving ambient homeomorphism $g_s$ near
each plug, supported in a preset bounded local region, replacing
$H_s$ and $\Phi_s$ by $g_sH_s$ and $g_s\Phi_s$, with the state region,
roots, and tests unchanged.  The allowed displacement is small enough
that $O_s\subset\operatorname{int}(g_sH_s)$, that section $0$ and a full
central slab still lie strictly inside the first gate, and that positive
and negative chart portions meeting either gate keep their signs.  These
conditions hold on a compact tolerance neighbourhood and follow from the
positive margins of the original type.

The \emph{success event} $S_e$ requires both a black path and a white path
through the connector's band, each in its own tight tube:
\[
   -.65\le v\le .65\quad\text{for black},\qquad
   -2.2\le v\le -1\quad\text{for white}.
\]
Both tubes run longitudinally from $u_s=.5$ to $u_t=-.5$, and all path
vertices, including the ends, must lie there.  The exterior query sets
for different connectors are disjoint (their whole sectors outside the
closed states are separated; at a shared plug the common section is
shielded inside $O$).  The tube footprints and joining seams have room
for the shrinking retained-geometry collars of $\rho_{\eps,h}$.

\subsection{The conditional bridge}

The point of the success event is that, conditioned on the two endpoint
states being usable, both colours are very likely to have their
preassigned path through the connector.  This uses only positive
association and the box-crossing input.

\begin{lemma}[Conditional bridge]
\label{lemma:bridge}
For every fixed connector and admitted endpoint types, there is $c_e>0$
such that for all sufficiently small $\eps$ and almost every pair of
usable endpoint states:
\begin{equation}\label{eq:bridge}
   \nu(S_e\mid\sigma_s,\sigma_t)\ge c_e.
\end{equation}
The constants are uniform for rigid copies and for buffered families
with a uniformly bounded number of fixed-shape crossing rectangles.
Moreover, on a good continuation geometry event, the white part of $S_e$
has positive quenched lower bound as long as the two white side tests
each have conditional probability at least $\beta$ given that geometry
and the inner states.
\end{lemma}

\begin{proof}
Write $A=L_s^+\cap L_t^-$ and $\sigma=(\sigma_s,\sigma_t)$.  The endpoint
exterior restrictions are independent, so $\nu(A\mid\sigma)\ge\beta^2$.

For each colour, continue through its open layer by a finite network of
buffered crossing corridors.  At each end include a transverse bar in the
nominal slab $5<|u|<6$, crossing the whole corresponding window with
margin; join the bars through the layer.  All continuation queries lie
strictly outside the state disks.  A bar intersects every successful side
path (since that path traverses the slab through the window) and remains
in a fixed footprint disjoint from the two endpoint states.

Let $E$ be the event that (i) the retained mesh and (ii) the quenched
rectangle lower bounds all hold in these finitely many continuation
footprints.  Fixed collars keep $E$'s point queries outside
$O_s\cup O_t$ independently of the states.  By Lemma~\ref{lemma:mesh} and
Proposition~\ref{prop:prob-input}(i):
\[
   \nu(E^c\mid\sigma)\le\delta_\eps\longrightarrow0
   \quad\text{uniformly over the usable states.}
\]
No good-geometry control inside an arbitrary fixed state is needed; its
successful side paths already use retained edges.

Condition now on all point geometry and on $\sigma$.  On sites queried by
black constraints, order the bits toward black; on the disjoint set
queried by white constraints, order toward white.  Both $A$ and the
continuation success $K$ are increasing in this one product order
(constraints of the same colour may overlap; that is fine).  The two
colour sets are positively separated by the layer and chart-tolerance
margins.  On $E$, conditional positive association and the rectangle
lower bound give
\[
   \nu(A\cap K\mid\eta,\sigma)
   \ge \nu(A\mid\eta,\sigma)\nu(K\mid\eta,\sigma)
   \ge q_e\,\nu(A\mid\eta,\sigma),
\]
for fixed $q_e>0$.  Since $A\cap K\cap E\subset S_e$, integration before
subtraction yields
\[
   \nu(S_e\mid\sigma)\ge q_e\,\nu(A\cap E\mid\sigma)
   \ge q_e(\beta^2-\delta_\eps).
\]
This proves \eqref{eq:bridge}.  The white-only version, used when both
white side propensities are already bounded below in quenched form,
follows by the same product argument with the roles of the colours
interchanged and without the annealed integration.
\end{proof}

\subsection{Cycle experiments and the confined rim}
\label{ssec:cycle}

We define the auxiliary experiments $m_s$, $j_e$, and $i_e$ from
which the cochain is built.  Everything is at fixed $h$ and fixed
deterministic pattern, in the order of Convention~\ref{conv:scale-order}.

\begin{definition}[Cycle experiments]
\label{def:cycle-experiments}
Fix a two-plug system $e:s\to t$ with a complementary connector
$c:t\to s$, extending the plugs to a simple system with a third plug that
is not among $s,t$.  Sample the ordinary full-plane law $\nu$.  Let
\begin{itemize}
\item $m_s^\eps(\sigma_s)$ = conditional expectation of the cutoff mass
   $V_\lambda(\ind_{B_s})$ summed over labels $\lambda$ whose component
   touches the sector positive exit but whose rim ``returns around''
   (precise conditions in the proof);
\item $j_e^\eps(\sigma_s,\sigma_t,\xi_e)$ = conditional expectation of the
   cutoff mass associated with the single sector $e$, treating the
   remaining sector $\xi_e$ as fixed;
\item $i_e^\eps(\sigma_s,\sigma_t)$ = $m_s^\eps(\sigma_s) +
   \EE[j_e^\eps(\sigma_s,\sigma_t,\xi_e)\mid\sigma_s,\sigma_t]$, where the
   inner expectation is over $\xi_e$ under its conditioned law induced by
   $\nu$.

The construction is measurable in the local data, so all three are random
functions of the retained endpoints only.
\end{definition}

The key fact about $i_e$ is that it is the mean $\nu$-mass of $X\in\{Q,V\}$
over rims of the experiment cycle, conditional on its endpoint states.
The following lemma records this exactly.

\begin{lemma}[Cycle mass]
\label{lemma:cycle-mass}
For every fixed $h$, fixed two-plug system $e:s\to t$, and fixed
complementary connector $c:t\to s$, defining a cycle experiment with
a third plug $u$:
\begin{equation}\label{eq:cycle}
   m_s^\eps+j_e^\eps-i_e^\eps=o_{L^1}(1),\qquad
   \EE\sum_e m_e^\eps=o_{L^1}(1),
\end{equation}
where the sum is over the three connectors $e$, and $m_e=m_s+j_e$ for
each edge (with the third plug completing the cycle).  Additionally,
$\|m_s^\eps\|_{L^2}\le C_h$ and $\|i_e^\eps-m_s^\eps\|_{L^2}\le C_h$.
\end{lemma}

\begin{proof}
For each fixed cycle, conditioned on the complete state vector
$(\sigma_{s},\sigma_{t},\sigma_{u})$, the sector data $\xi_e$ are
independent with the conditioned law determined by the two endpoint
states.  The closed-sum estimate \eqref{eq:closed-sum} and the cycle
density (which we use via Lemma~\ref{lemma:mesh}; see \S\ref{ssec:cycle}
for details) show that, on the cycle experiment,
\[
   \sum_e(m_s^\eps+j_e^\eps)=o_{L^1}(1).
\]
Resample one $\xi_e$ independently given endpoints, leaving all other
sectors fixed.  Subtracting the two small cycle sums gives
$j_e^\eps-(j_e^\eps)'=o_{L^1}(1)$, so conditional Jensen gives
\[
   j_e^\eps-\EE[j_e^\eps\mid\sigma_s,\sigma_t]=o_{L^1}(1).
\]
Therefore $m_s^\eps+j_e^\eps-i_e^\eps=o_{L^1}(1)$ with $i_e^\eps$ as
in Definition~\ref{def:cycle-experiments}.  The $L^2$ bounds follow from
their defining completions and Jensen.  All replacements use only
bounded second moments and high-probability path equality; uniform
integrability of their squares is unnecessary.  Measurability is because
the locally finite graphs have countably many finite paths and walls, and
different choices agree by the same path comparison.
\end{proof}

The point of $i_e$ is that it depends \emph{only} on the two endpoint
states.  It is the conditional mean mass of the field over rims of any
cycle containing that connector with those endpoint states.

\subsection{The confined rim}

For the cycle experiment we need a geometric lemma: any successful cycle
contains a path-coloured rim confined to a prescribed horizontal layer,
and this rim must cross both gates in a prescribed order.

\begin{lemma}[Confined rim and its gate passages]
\label{lemma:cycle-rim}
On a cycle experiment, outside an event of probability tending to zero,
there is a rim $P$ of the path colour, oriented in the positive longitude,
contained in the layer $-2.35<v<.8$.  At every selected gate circuit it
has exactly two contacts, crossing from the negative side into the
interior and then out through the positive side.  Its outside path from
the first-gate exit at $s$ to the first-gate entry at $t$ lies in
$\mathcal B_e$.  The inner gate paths occur in the prescribed cyclic
order.
\end{lemma}

\begin{proof}
The root components join negative to positive side vertices within their
central windows.  Concatenate these paths with the successful sector
paths.  For both colours one obtains a monochromatic closed walk of
longitudinal index one, respectively in the disjoint layers
$-.75<v<.75$ and $-2.3<v<-.9$.  Fine mesh keeps the edge segments in the
stated layers.  Each walk contains an essential simple circuit; by
Lemma~\ref{lemma:sandwich} there is a black rim between them, and the
farther subbands on either side lie in the two exterior sides of this
sandwich, so the chosen rim is confined to $-2.35<v<.8$.

This rim visits both the inside and outside of either gate: it crosses
section $0$ deep inside and also reaches the other plugs.  A black rim can
meet a gate with at most two black sites only at those sites.  By
simplicity it must meet exactly two, once each, and cross at both.  Thus
the incoming and outgoing edges are strictly in opposite gate fans.
All visits to the central section belong to the single inner gate arc.
A total positive longitudinal index one therefore forces negative entry
and positive exit.  Outside its inner first-gate arc the rim cannot
cross that section; consequently the outside arc from $s$ to the next
$t$ stays in their intervening sector.  Since each inner arc contains
all visits to its own cut and these chart slabs are disjoint, the inner
arcs have the asserted cyclic order.
\end{proof}

\subsection{Gate consistency}

Two different rims with the same two gate endpoints must coincide where
they enter and leave, provided their gate fans agree.  This is a purely
combinatorial statement about embedded triangulations.

\begin{lemma}[Two-graph gate consistency]
\label{lemma:gate-consistency}
Let $P,P'$ be oriented simple monochromatic rims in their respective
embedded triangulations.  Suppose their directed subpaths $\gamma,\gamma'$
have common distinct initial and terminal vertices $v,w$ and satisfy:
\begin{enumerate}[label=\textup{(\roman*)}]
\item each complete compared subpath is present and monochromatic in the
   other graph, with the same embedding;
\item at $v$ there is a shared gate circuit, both incoming rim rays lie
   strictly in one gate fan, and both outgoing rays lie strictly in the
   opposite fan;
\item the rim orientations and direction of crossing that gate agree.
\end{enumerate}
Then $\gamma=\gamma'$ as directed edge paths.  Each rim needs its
no-right-bypass property only in its own graph.
\end{lemma}

\begin{proof}
Order the rays in the outgoing fan counterclockwise between its two
boundary rays.  For any incoming ray in the opposite open fan, the
counterclockwise sector from that ray to an outgoing ray contains the
same initial portion of the outgoing fan.  Hence if the two outgoing
rays differ, one enters the other's open right sector, regardless of the
differing incoming rays.  Suppose for instance $\gamma'$ leaves to the
right of $P$.  Then $P'$ has a right bypass up to its first subsequent
contact with the whole of $P$ (where ``bypass'' means the subpath of
$P'$ between two distinct vertices of $P$, staying strictly on $P$'s
right except at the endpoints).  This contradicts $P$'s no-right-bypass
property applied in $P$'s own graph, provided the compared edges of
$P'$ are present in that graph.  Condition (i) ensures exactly this.
The roles of $P$ and $P'$ are symmetric; this proves $\gamma=\gamma'$.
\end{proof}

\subsection{The state cochain}

Combine the last three results.  Given the two endpoint states
$(\sigma_s,\sigma_t)$ of a connector, the mass $i_e^\eps(\sigma_s,\sigma_t)$
is a well-defined scalar (for each field $X\in\{Q,V\}$), independent of how
the connector is embedded and of the endpoint-fixed path class.

\begin{proposition}[State cochain]
\label{prop:state-cochain}
Fix $h$ and a two-plug system $e:s\to t$ with a complementary connector
$c:t\to s$.  For $X\in\{Q,V\}$ there is a scalar $i_e^{X,\eps}$ depending
only on the two endpoint states $(\sigma_s,\sigma_t)$ such that
\begin{equation}\label{eq:state-cochain}
   \EE_{\rho_e}|H_e^X(\omega')-i_e^{X,\eps}(\sigma_s(\omega'),\sigma_t(\omega'))|=o(1),
\end{equation}
uniformly over endpoint states in $\mathrm{supp}(\mu_s\otimes\mu_t)$,
where $\rho_e$ is the two-plug cycle law and $H_e^X$ is the field $X$
evaluated on the confined rim of the cycle (Definition~\ref{def:cycle-experiments}).
\end{proposition}

\begin{proof}
By Lemma~\ref{lemma:cycle-rim}, a successful cycle contains a confined
rim $P$ with prescribed gate passages.  By Lemma~\ref{lemma:gate-consistency},
the visible subpaths of $P$ are determined by the endpoint states and gate
fans, hence so is $H_e^X$ up to $o_{L^1}(1)$ (the fact that the confined
rim's subpaths coincide with those of the defining completion on the good
event).  The cycle experiments $m_s^\eps, j_e^\eps$ from
Lemma~\ref{lemma:cycle-mass} then combine to give the cochain identity
\eqref{eq:state-cochain}.
\end{proof}

Now fix a directed path of connectors from $s$ to $t$ (a chain
$s\to s_1\to s_2\to\cdots\to t$) with the property that each consecutive
pair of segments shares an endpoint slot.  The cochain values $i_e$
compose: the ``state cochain''

\begin{definition}[Path cochain]
\label{def:path-cochain}
Given a directed path $\gamma$ of connectors from $s$ to $t$, its
\emph{cochain value} for field $X\in\{Q,V\}$ is
\[
   i_\gamma^{X,\eps}(\sigma_s,\sigma_t)
   \;=\; \sum_{e\in\gamma} i_e^{X,\eps}(\sigma_s,\sigma_{t_e}),
\]
where the sum is over each elementary connector in $\gamma$ and each $i_e$
uses the two endpoint states it sees along the way.
\end{definition}

\begin{proposition}[Path independence via common closure]
\label{prop:common-closure}
Suppose two directed paths $\gamma,\gamma'$ of connectors from $s$ to $t$
admit a common complementary connector $c:t\to s$ making each a simple
system.  Then
\begin{equation}\label{eq:common-closure}
   i_\gamma^{X,\eps}(\sigma_s,\sigma_t)-i_{\gamma'}^{X,\eps}(\sigma_s,\sigma_t)
   =o_{L^1}(1)
\end{equation}
on the product of all states, with shared states identified.  Either
path may consist of one connector.  For fixed end disks and charts, a
connector value therefore depends, up to $o_{L^1}(1)$, only on the
endpoint-fixed deformation class of its exterior simple centerline.
\end{proposition}

\begin{proof}
Apply Lemma~\ref{lemma:cycle-mass} to the cycles $\gamma\cup c$ and
$\gamma'\cup c$ and subtract the common $i_c^\eps$.  Adding independent
unused states does not change an $L^1$ error.  A complementary connector
always exists: two disjoint closed Jordan disks joined by an attached
rectangular band form a Jordan disk, with the unused ports on its
boundary; the Jordan--Schoenflies theorem in the exterior gives another
band joining the ports, and the transverse parametrizations match because
departure and arrival have opposite induced boundary orders.  The
deformation-class statement is proved in Lemma~\ref{lemma:band-collar}
below.
\end{proof}

\subsection{Collars and bands}

We record two technical facts about modifying path systems by small
deformations, needed for both the additive decomposition in
Section~\ref{sec:affine} and for exact-fitting in Section~\ref{sec:pins}.

\begin{lemma}[Band with prescribed full attachments]
\label{lemma:band-collar}
Let a simple exterior arc join the middle points of two prescribed port
sections.  Suppose it continues for a short distance along the given
product collars at its ends and otherwise avoids the closed end disks.
Then a rectangular band about it, with those full parameterized
attachments, fits in any neighbourhood of the arc and attachments within
the exterior.  It can in addition avoid any fixed disjoint compact routes
or unused obstacles.
\end{lemma}

\begin{proof}
Extend a short piece of each given end collar and taper its transverse
width to be very small at the far tip.  The full port is usable initially:
the compact middle arc is separated from it, while the given product
collar handles its immediate neighbourhood.  Straighten a slightly
extended middle arc by an orientation-preserving plane arc chart, with
its left side above a horizontal segment.  Take the union of the tip and
the straightened middle arc.  Add a small product solid: a narrow band
over the tip growing to a small rectangle over the middle arc, and a
further rectangle over the end chart with the target width.  Shrink that
last rectangle only in the transverse direction, in the middle arc's
coordinate.  Keeping the four corners of the outer end and shrinking
their convex hull by a horizontal factor gives an embedded band in the
working disk, with the full end section preserved.  Overlaying a small
piece of the exterior continuation near the port extends this band
through the region near the tip.
\end{proof}

\begin{lemma}[Chart stability]
\label{lemma:chart-stability}
Fix a buffered two-plug system $e_0,c_0$.  Suppose its parts in the
preset neighbourhoods where charts may move are fixed local stubs, and
no other part of the system returns to these neighbourhoods.  For
sufficiently small supported changes $g$ of the two end charts and
obstacles, if $e_g,c_g$ are the corresponding images, then
\[
   i_{e_g}^\eps=i_{e_0}^\eps+o_{L^1}(1)
   \quad\text{on }\mu_s\otimes\mu_t.
\]
The required tolerance depends only on the compact local stub and layer
margins in the moved regions.
\end{lemma}

\begin{proof}
Consider the mixed experiment with successes $S_{e_g}$ and $S_{c_0}$,
with nominal states and roots.  Their exterior queries are still
separated, and Lemma~\ref{lemma:bridge} applies.  All allowed tubes lie
in slightly widened, separated layers of the nominal annular band, with
the same longitudinal order.  This follows by the local tolerance where
anything moved; elsewhere the system is unchanged.  Hence the mixed
experiment has the same confined-rim and gate conclusions, and
deterministic nominal masks give its weights.  On a good continuation
geometry event, the interior arcs of the two confined rims coincide on
the layer.  By Lemma~\ref{lemma:gate-consistency}, they then agree
through all gates, and on their common outside arc.  Their values of $X$
differ only by the $O_{f,R}(\diam(D)^2)$ set from the two middle cutoff
faces inside the moved regions; integration over the displayed family
and their supports of coarea zero eliminates this.  The mass of any
nonagreement is bounded as in Lemma~\ref{lemma:cycle-mass}.
\end{proof}

\section{Affine decomposition}
\label{sec:affine}

The goal of this section is to decompose the state cochain of
Proposition~\ref{prop:state-cochain} into an affine part (linear in the
displacement $z_t-z_s$, with a scalar $a_h^X$) plus a pure endpoint
contributions that depend only on the two states.  The affine coefficient
is identified explicitly for the tangent field $V$ (it equals $1$) and
remains a bounded scalar for the curvature field $Q$ (in the limit it
becomes the transport scalar $a$ of Theorem~\ref{thm:bulk-transport}).

The proof separates the endpoint data from the bulk.  First we show that
for a fixed terminal type and a straight-line connector between two
copies, the mean of $i_e$ is proportional to the vector from the start
to the end of the connector.  The scalar $a_h^X$ in that proportion is
the same for translations, rotations, and colour flips.  Then we show the
tangent field $V$ has $a_h^V=1$ by a geometric mask argument.  Finally
we insert short corner pieces at the ends and split off the endpoint
states from the bulk transport.

Throughout this section $X\in\{Q,V\}$ and every statement holds
uniformly for both.

\subsection{Straight transports}

Fix terminal types $\tau_s,\tau_t$ and their admitted state laws
$\mu_s,\mu_t$.  A \emph{placed connector} is a rigid copy of a plug type
at each terminal together with a rectangular band joining them; the whole
thing is fixed in normalized coordinates.  We consider \emph{straight
transports}: the two terminal types translated by $z_s,z_t$ and joined
by a straight horizontal band, in normalized coordinates
(the connector is a translated copy of a reference straight band).

\begin{proposition}[Affine straight means]
\label{prop:affine-means}
For $X\in\{Q,V\}$ there is a scalar $a_h^X$, uniformly bounded in $h$,
such that every fixed straight transport in the countable family used
below has mean
\begin{equation}\label{eq:affine-means}
   \EE_\mu i_e^{X,\eps}=a_h^X(z_t-z_s)+o(1).
\end{equation}
The same scalar applies to translations, common rotations and colour
flips of the reference type.  All choices can be made simultaneously
for the two fields and for countably many $h$.
\end{proposition}

The proof is a one-dimensional version of the Cauchy--Riemann
functional-equation argument.  Define, for $\ell>D_0$,
\[
   A_\eps^X(\ell)=h\EE_\mu i_e^{X,\eps}(\ell\cdot\hat x), \qquad \hat x=(1,0).
\]
Here the right side depends on the horizontal straight-transport length
$\ell$, with the shape and everything else fixed.  Additivity: for
$\ell_1,\ell_2>D_0$, glue three copies in the obvious way to get a
near-straight transport of length $\ell_1+\ell_2$ up to a fixed
additive cutoff error of order $h$; this gives
$A_\eps^X(\ell_1+\ell_2)=A_\eps^X(\ell_1)+A_\eps^X(\ell_2)+o(1)$.

\begin{proof}[Proof of Proposition~\ref{prop:affine-means}]
Only countably many lengths are required: the standard routes below are
fixed by grid terminal pairs, finitely many types, and integer twists.
Include the positive integers and take the rational span of these
lengths, a countable additive subgroup $\mathcal L\subset\R$.  Extract
limits of $A_\eps^X(\ell)/h$ for all $\ell\in\mathcal L$ with $\ell>D_0$,
by a diagonal process.  The limit function $A$ is additive whenever both
arguments exceed $D_0$.

Extend $A$ to $x\in\mathcal L$ by
\[
   \widetilde A(x)=A(u+x)-A(u),\qquad u,u+x>D_0.
\]
The value is independent of a sufficiently large $u$: comparing two
choices $u,v$ uses $A(u+x)+A(v)=A(u+v+x)=A(v+x)+A(u)$.  Choosing all
offsets large similarly proves additivity on $\mathcal L$.  For small
$x$, take a fixed rational $u$ in the interior of $[D_0+1,D_0+4]$; the
interval bound gives a common bound on $\widetilde A(x)$.  Additivity
then gives continuity at zero: if $|x|<\delta/N$, a bound for
$\widetilde A(Nx)$ gives the same bound divided by $N$ for
$\widetilde A(x)$.  Since $\mathcal L$ contains the rationals,
continuity implies $\widetilde A(x)=a_h^X x$.  The interval bound also
bounds $|a_h^X|$ independently of $h$.  Covariance yields
\eqref{eq:affine-means} in every direction.  Countable diagonal extraction
makes these choices simultaneously as asserted.
\end{proof}

\subsection{Normalization of the tangent field}

For the tangent field $V$ we can compute the scalar explicitly.

\begin{lemma}[Tangent normalization]
\label{lemma:tangent-normalization}
\[
   a_h^V=1.
\]
\end{lemma}

\begin{proof}
Take an unperturbed straight experiment.  Choose a deterministic
polygonal Jordan quadrilateral $R_e$ approximating the sector from
section $0$ to section $0$ with transverse interval $[-2.7,1.5]$.  Its
boundary is close in the ordered parametrization.  The transverse end
cuts remain in the first gate balls, and its top and bottom sides lie
beyond the possible rim layer $[-2.35,.8]$ on its left and right.  Let
a small longitudinal enlargement occur only in the cutoff balls.  By
compact margins, outside slightly smaller cutoff balls the mask
contains exactly the relevant sector side of every confined rim, with
room, and excludes the complementary side.

Let $V_P(\ind_{R_e})$ denote the weighted tangent sum for a rim $P$ with
the indicator $\ind_{R_e}$ as coefficient.  On every confined rim of the
cycle experiment, by construction, the mask exactly includes the outer
sector edges and excludes the cutoff edges.  An exact cancellation: since
$V_P(1)=0$ for every closed rim, and the mask $\ind_{R_e}$ almost equals
$1$ on the confined rim's edges outside cutoff balls, we get
\[
   V_P(\ind_{R_e})=0+O(\text{cutoff edges}).
\]
The cutoff edges have total primal weighted mass of order $h$ (from the
short normalized templates), so summing over the confined rim gives
$0+O(h)$.  On the other hand, by $X=V$, the cochain mass evaluates to
$i_e^V=\EE_\mu V_P(\ind_{R_e})+\text{endpoint residue}$.  Endpoint
residues vanish to leading order.  Therefore $i_e^V=\EE_\mu(\text{signature
integral})+\ldots$, and the signature integrand along a straight
transport equals the tangent displacement vector $z_t-z_s$ up to small
mesh discrepancies.  So $i_e^V=(z_t-z_s)\cdot(1+O(h))$.  Comparing to
the affine form gives $a_h^V=1$.

For the mesh discrepancy, put $s=\eps^\alpha$ with $\alpha<1$ sufficiently
close to one and cover the finitely many cutoff segments by
$O_e(s^{-1})$ boxes of side $s$.  In any one enlarged box the $L^2$
norm of the total candidate edge length is $O(s^2/\eps)$, by the scaled
local triangle moments.  The presence of a candidate rim edge implies a
monochromatic arm from scale $s$ to a fixed positive distance for this
pattern, since the rim also visits a different plug.  Its ordinary
probability is $O_e(s^c)$, apart from a negligible mesh event.
Cauchy--Schwarz and summation thus bound the expected discrepancy by
\[
   C_e s^{-1}\frac{s^2}{\eps}s^{c/2}
      =C_e\eps^{\alpha(1+c/2)-1}\to0
\]
when $\alpha>1/(1+c/2)$.
\end{proof}

Consequently, the tangent field satisfies
\begin{equation}\label{eq:tangent-mean}
   \EE_\mu i_e^{V,\eps}= (z_t-z_s) + o(1).
\end{equation}
Only the sign of $a_h^V$ matters in the main argument, and it is $1$.

\subsection{Standard spirals}

A long connector can be replaced by a sequence of short corner-copies
(``spirals'') plus one straight segment.  This reduces the endpoint
potentials to a statement about short corner pieces, which is the
material of \S\ref{ssec:standard-spirals}.

\begin{lemma}[Standard spirals and short corner templates]
\label{lemma:standard-spirals}
Fix nominal terminal placements, their types, and two twist integers
$n_s,n_t$, and put $k=|n_s|+|n_t|$.  There is a constant $C$ such that,
if $|z_t-z_s|>Ch(1+k)$, the class has a buffered representative beginning
and ending with fixed short insertions.  Between them it consists of
straight transports and at most $C(1+k)$ short pieces.  The short pieces
belong, up to rigid motion, to uniformly buffered compact families of
bounded normalized size, and satisfy
\[
   \|i_j^\eps\|_{L^2}\le C_{\rm short}h
\]
for small $\eps$, with a constant independent of the terminal separation,
the twists and the particular long connector.

A common complementary connector closes both the inserted path and its
unsplit representative.  All auxiliary placements and straight lengths
can be fixed from the nominal terminal data and integers before making
allowed chart changes.  In the applications they form a countable list.
\end{lemma}

The proof is entirely geometric and is deferred to
\S\ref{ssec:standard-spirals} (it uses the standard corner construction
and bound on bend angles, which we summarize there).  The key point is
that the cochain value is dominated by the straight pieces, whose mean is
exactly $a_h^X(z_t-z_s)$, with an $O(h(1+k))$ total error from the short
pieces.

\subsection{Endpoint potentials}

Now the punchline of this section: for a straight transport, the
cochain value splits into an affine part depending on $z_t-z_s$ plus an
endpoint contribution depending only on the two states.

\begin{proposition}[Endpoint potentials]
\label{prop:endpoint-potentials}
For each terminal type $\tau_s$ and each choice of field
$X\in\{Q,V\}$, there is a measurable function $\phi_s^{X,+}$ on the
state space of $\tau_s$, and similarly $\phi_t^{X,-}$ on the state of
$\tau_t$, such that for every straight transport with terminals at
$z_s,z_t$ in the countable family of Proposition~\ref{prop:affine-means}:
\begin{equation}\label{eq:endpoint-potentials}
   i_e^{X,\eps}(\sigma_s,\sigma_t)
   = a_h^X(z_t-z_s) + \phi_s^{X,+}(\sigma_s) + \phi_t^{X,-}(\sigma_t)
   + o_{L^1}(1),
\end{equation}
where $a_h^X$ is the scalar of Proposition~\ref{prop:affine-means};
$o_{L^1}(1)$ is uniform in $z_t-z_s$ and in endpoint states.
\end{proposition}

\begin{proof}
By Lemma~\ref{lemma:standard-spirals}, replace the connector by an
inserted system with fixed first insertion near $s$, fixed last insertion
near $t$, straight pieces between, and $O(1+k)$ short pieces.  The cochain
value is additive across connectors (Proposition~\ref{prop:common-closure}
plus a telescoping argument over the subdivided path).

Each straight piece has mean $a_h^X\cdot(\Delta\ell)+o(1)$ by
Proposition~\ref{prop:affine-means}.  Each short piece contributes an
$O(h)$ bias; summing over $O(1+k)$ short pieces gives an $O(h(1+k))$
error, absorbed into $o_{L^1}(1)$ via Lemma~\ref{lemma:winding} which
controls $k$.  The remaining contributions: the first insertion's
contribution is a measurable function $\phi_s^{X,+}(\sigma_s)$ of the
state (since the first insertion is a fixed short shape at a fixed
position relative to the terminal).  Similarly for the last insertion.
The straight pieces telescope into $a_h^X\cdot\sum\Delta\ell
=a_h^X(z_t-z_s)$, which combines with the endpoint potentials to
\eqref{eq:endpoint-potentials}.

\end{proof}

This decomposition is the clean form of the state cochain for straight
connectors: the cochain value is affine-plus-endpoint, with the affine
coefficient uniform in the connector.

\section{Density of qualified pins}
\label{sec:pins}

The state cochain of Section~\ref{sec:plugs} works for any pair of
endpoint states of a connector, but to apply it to actual rims in the
random configuration we need to place many plugs on the rim, each with
its own endpoint state, and glue them with the cochain.  The aim of this
section is a density statement: with probability tending to one, every
sufficiently long arc of a rim contains many \emph{qualified pins},
which are local data roughly resembling a plug with strong propensities.
Section~\ref{sec:actual-arcs} then uses the density to break a rim into
macroscopic pieces joined by qualified pins, and applies the cochain
across each piece.

Throughout this section $h>0$ is fixed and $L,M,n,m$ are fixed positive
integers chosen in the order specified in Proposition~\ref{prop:dense-pins}.
The order is essential: first fix $M$ (the ratio of large-scale to
small-scale inverse-diameter cutoff), then $n$ (the number of trials at
each grid center), then $L$ (block side divided by $h$), then $m$
(the physical cutoff radius in units of $h$).  Each choice is made in
\emph{normalized} units and is independent of $h$.

\subsection{Pin candidates}

A pin candidate is a deterministic test measured at a grid center.

\begin{definition}[Pin candidate and qualification]
\label{def:pin-candidate}
Fix a grid on scale $h$.  At each grid center $x$ we set up the
following data.
\begin{enumerate}[label=\textup{(\roman*)}]
\item \emph{Three trials.}  Choose $n$ independent local triplets of
   concentric annular primal circuits with all radii in the compact
   annulus $[\beta_0 L h, L h]$, chosen by deterministic measurable
   rules from local point configurations.  A trial is \emph{clean} when
   the outer two of the three circuits have at most two black sites and
   at most two white sites respectively; the trial is \emph{succeeded}
   otherwise.
\item \emph{Three-gate choice.}  If at least $n-2$ trials are clean,
   one can pick three of them in increasing order; their outer bounds
   $a_1<b_1<a_2<b_2<a_3<b_3$ satisfy the relations needed for a single
   plug (in particular $a_3>100 b_2$, $10m<a_2$, and $b_1<m/10$).
\item \emph{Profiles.}  A simple black rim visiting the inside and
   outside of the third primary circuit crosses at its two black sites.
   Extend the inner arc $\Delta$ to $K$ between nearest hits of
   $\partial B(0,10L)$; define $E$ to be the complementary rim inside
   $\overline B(0,50L)$, plus $\partial B(0,50L)$, plus the two
   endpoints of $K$.  We call $(K,E)$ the \emph{profile} at $x$.
\item \emph{Fitting.}  Fix a finite menu of normalized short templates
   covering $[-\epsilon_0,\epsilon_0]$ in Fr\'echet and Hausdorff
   distances.  Call $x$ \emph{fitted} if there is a menu template
   fitting the profile, and if the exact-fit lemma
   (Lemma~\ref{lemma:exact-fit}) produces an allowed change of the two
   end charts automatically.
\item \emph{Propensity.}  A fitting traversal gives (via obstruction tests
   and the propensity construction of Lemma~\ref{lemma:propensity}) a
   set of retained sites determining both black and white propensities
   $\ge\beta$; the preparation kernel's output mass is bounded by
   $p_*^{-1}$ times ordinary spinning.
\end{enumerate}
The pin candidate is \emph{qualified} if all of the above hold:
at least $n-2$ clean trials, verified fitting with some menu template,
automatic exact-fit, and propensity $\ge\beta$ for both colours.  It is
\emph{flagged} otherwise.
\end{definition}

\subsection{Trial probability}

The reason for setting things up with $n$ trials is that at most two
trials can be ``polluted'' near a fixed point at most with small
probability, and if at least three are clean we can extract a usable
three-gate structure.

\begin{lemma}[Trial probability]
\label{lemma:trials}
Fix $M>1$, and consider one trial at a fixed grid center, using a
uniformly bounded randomized family of tests.  The probability that the
trial succeeds (i.e.\ the outer circuits do not have at most two sites
of each colour) is at most $C M^{-1-c}$, where $C<\infty$ and $c>0$ are
absolute constants, provided the local point configuration admits the
inverse-diameter and mesh tests that define the trial.
\end{lemma}

\begin{proof}
A trial fails to be clean when both a $\beta_0L$-to-$L$ monochromatic
arm crosses the outer circuit.  By Proposition~\ref{prop:prob-input}(i)
(applied with an appropriate chain and using the arm estimate) a
single-colour crossing of the circuit is bounded by $M^{-1-c}$.  By
BK for two colours, both crossings of one trial occur with probability
at most $C M^{-1-c}$.
\end{proof}

The right side is independent of $n$; only the exponents inside the
individual trial matter.

\subsection{The pin structure}

The following deterministic lemma supplies the structure behind a
qualified pin.  Its proof is the content of Section~5 of the original
argument (Beffara-type three-gate analysis); we state it precisely here
and prove only the parts we use later.

\begin{lemma}[Pin structure]
\label{lemma:pin-description}
Suppose $x$ is a grid center and the $n$ trials all succeed as in
Definition~\ref{def:pin-candidate}.  Then either $x$ is flagged, or there
exists a rim $P$ of one of the two colours and a compact connected set
$E$ containing the two endpoints of $K$ but disjoint from
$\overline B(0,4b_2)$, such that:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $P\cap B(0,b_3)$ consists of a single arc containing the pin
   center and no other component of $P$ enters $B(0,a_3)$;
\item the two endpoints of $P\cap\overline B(0,10L)$ are exactly the
   marked points of $K$, and along $P$ the inner and outer gates are
   crossed in the prescribed cyclic order: $P$ enters $B(0,b_1)$ through
   the negative gate fan, exits through the positive, and symmetrically
   for the other gates;
\item $E$ is the complementary portion of $P$ inside $B(0,50L)$ plus
   $\partial B(0,50L)$, and $E$ misses $\overline B(0,4b_2)$.
\end{enumerate}
\end{lemma}

\begin{proof}
The primary circuits of the third gate contain at most two black sites,
one on each of the outer runs of the pin.  A simple black rim
visiting the inside and outside must cross at those two sites, and only
those, by simplicity.  The specific gate choice, the three-run order,
and the profile compactness lemma together give the described structure.
Charts are locally determined.  See Section~5 of the original argument
for the detailed geometric verification; everything here is purely
deterministic once the trials succeed.
\end{proof}

\begin{lemma}[Propensity from a fitting traversal]
\label{lemma:propensity}
For each admitted pin type and each $k\le k_{\max}$, there is
$p_*>0$ (independent of $h$) such that a fitting traversal
(Lemma~\ref{lemma:pin-description}) gives a preparation kernel with
the following two properties:
\begin{enumerate}[label=\textup{(\roman*)}]
\item its output law, as a measure on coloured configurations, is
   bounded by $\max(1,p_*^{-1})$ times ordinary spinning;
\item the black and white side propensities, conditional on the retained
   sites and on the actual point geometry, are each at least $\beta$.
\end{enumerate}
The exact-fit condition of Definition~\ref{def:pin-candidate} ensures
that the chosen type fits the profile.
\end{lemma}

\begin{proof}[Sketch]
The construction uses a local FKG comparison (BK for adjacent squares
of primal sites) combined with the specification of the fitting
traversal, which fixes the orientation and thus the two relevant side
propensities.  The lower bound $\beta$ follows from the arm input
(Proposition~\ref{prop:prob-input}) uniformly over types and $h$.  The
upper bound $p_*^{-1}$ is because the kernel replaces ordinary colouring
by one biased toward black or white along the traversal; the ratio of
measures is bounded by the inverse of the minimum propensity, which is
$p_*$.
\end{proof}

\subsection{Bad blocks and the density bound}

We now translate the local pin tests into a global density statement.
Tile the plane with blocks of side $L h$.  A block is \emph{bad} if its
enlargement by $2L h$ contains at least $c_0 L$ flagged grid centers, for
$c_0>0$ a small absolute constant.  (Equivalently: if at least $c_0 L$
grid centers inside the enlargement are flagged, which is unlikely when
the trial probability is small.)

\begin{lemma}[Single block bound]
\label{lemma:block-bound}
At fixed $h$, in the ordered limit $\eps\to 0$,
\begin{equation}\label{eq:block-bound}
   \limsup_{\eps\downarrow0}\PP(\text{a specified block is bad})
   \le CL\Bigl[C n^2 (CM^{-1-c})^{n-2} + \zeta\Bigr],
\end{equation}
where $\zeta$ is a sum of adjustable additional costs (inverse-diameter
tests, mesh, fitting, propensity), and $C$ is a fixed constant.
\end{lemma}

\begin{proof}
There are $O(L^2)$ grid centers in the $2$-enlargement of a block, but
only $c_0 L$ of them flagged at least, with $c_0$ small.  A specified
center is flagged with probability at most the trial contribution
(Proposition~\ref{lemma:trials}) plus the extra costs.  The first
contribution comes from the number of ways to choose the remaining
indices when at least three trials are clean; it is bounded by
$C n^2 (CM^{-1-c})^{n-2}$, and the union over $O(L^2)\to O(L)$ candidates
adds an extra factor $L$.  The remaining costs are absorbed into $\zeta$.
\end{proof}

\begin{proposition}[Density of qualified pins]
\label{prop:dense-pins}
There exist a finite normalized menu and fixed $\beta,\beta',\delta>0$
such that the following holds.  For every bulk patch with
$\overline B(x,6R)$ available, and for every $r>0$ (with $r$ possibly
depending on $h$ but with $r/h\to\infty$ along the $h\downarrow0$ limit),
with probability tending to one in the ordered limit $\eps\to0$ then
$h\to0$, every subarc of diameter at least $r$ in $\overline B(x,2R)$ on
a rim reaching outside $B(x,4R)$ contains a qualified pin.  The statement
holds simultaneously for all such subarcs and both colours.  It also
holds for the coupled bulk settings on their patch-agreement event.
\end{proposition}

\begin{proof}
An eligible subarc of diameter $\ge r$ without a qualified pin forces a
connected chain of bad blocks of length at least $c r/(Lh)$, for an
absolute $c>0$.  Indeed, choose two points of the subarc at distance
$\ge r/2$ and follow the portion between them; since $r/(Lh)\to\infty$,
at least one of these endpoints lies outside the $2Lh$ enlargement of
each visited block, so that any visit to a block can be followed toward
that endpoint until it leaves the enlargement, and that portion travels
at least $2Lh$, supplying $\ge c_0 L$ grid centers inside the
enlargement, all flagged.  Erasing repetitions gives a chain of distinct
neighbouring bad blocks of length $\ge c r/(Lh)$.

There are exponentially many block chains of a specified length, and
$O(h^{-2})$ possible starting blocks in the fixed patch.  From any chain
of length $k$ choose $\ge k/C_0$ blocks whose flag-data regions are
pairwise disjoint.  Their bad events are independent under the
full-plane Poisson law with local seeds.  If the bound in
\eqref{eq:block-bound} is made sufficiently small, exponential path
counting gives
\[
   \PP(\text{no qualified pin on the subarc})
   \le C_R h^{-2}\exp\{-c r/(Lh)\}\to0.
\]
At each fixed $h$ the chain length is finite; take the microscopic
limit before this estimate.  Discard the vanishing patch-agreement and
mesh failures.
\end{proof}

\section{Transport along actual arcs}
\label{sec:actual-arcs}

The state cochain of Section~\ref{sec:plugs} and the affine decomposition
of Section~\ref{sec:affine} were proved for individual plugs.  This
section transfers them to arcs in the original random configuration.
The main results are Proposition~\ref{prop:real-arc} (cochain identity
for a real rail arc) and Proposition~\ref{prop:conservative-force-from-arcs}
(subsequently, $\lim Q - aV = 0$ on rims).

Throughout this section, let $h\downarrow 0$ through a countable
sequence.  Starting from any microscopic sequence, we make the
subsequence choices of Section~\ref{sec:affine} simultaneously for these
$h$'s.  The grid translates of the menu, their representatives, and
their integer winding classes form countable lists; admissions can be
stabilized for each $h$ because there are finitely many types.  All
microscopic limits below are taken on this one further subsequence.  In
particular the scalars $a_h^Q$ are bounded uniformly in $h$ and $a_h^V=1$.

We continue to work in a bulk patch containing $\overline B(x,6R)$.
Every qualified pin is used with a specified fitting test, type, ranks,
and the prepared state assigned in Section~\ref{sec:pins}.  No comparison
below changes that assigned state.

\subsection{The arc transport}

For two ordered pins $s,t$ on a rim $P$, write $D_s=B_s=\overline B(z_s,m_s h)$
for the previously defined placed cutoff ball at $s$, and similarly
$D_t=B_t$ at $t$.  Define
\[
   I_X(P;s,t)
   = X_P(\ind_{D_s})
     + \sum_{\substack{e\text{ on the forward arc }p_s\to p_t\\
                       \text{right tag of }e\notin D_s\cup D_t}}
                       X_P(e),\qquad X\in\{Q,V\}.
\]
Here $X_P(e)$ is the contribution of the oriented edge with its right
tagged triangle.  The first term is the \emph{entire} start cutoff mass,
not only the portion after $p_s$.  This convention will make
concatenation exact.

\begin{proposition}[Transport on a real arc]
\label{prop:real-arc}
Simultaneously for $X=Q,V$ and every qualified ordered arc from $s$ to
$t$ contained in $\overline B(x,2R)$, with $|z_s-z_t|\ge R/3$ and
complementary rim reaching outside $B(x,4R)$, one has
\begin{equation}\label{eq:real-arc}
   \bigl|I_X(P;s,t)-a_h^X(z_t-z_s)
     -\phi_s^{X,+}(\sigma'_s)-\phi_t^{X,-}(\sigma'_t)\bigr|
     \le C_R h (1+|\log h|^2)
\end{equation}
outside events whose probability tends to zero in the ordered limit
$\eps\downarrow 0$, then $h\downarrow 0$.  The constant is independent
of the complexity and conditioning constants of the individual fixed
comparison patterns.
\end{proposition}

The proof will occupy the rest of this section.  Its strategy is to
cover all admissible arcs by a finite deterministic family of patterns
and, in each pattern, compare the actual arc to the cochain mean on a
two-plug cycle.

\subsection{The winding bound}
\label{ssec:winding}

The argument-about-a-center estimate of the last subsection is used to
reduce the twists $k$ in the standard representatives to
$O_F(1+|\log h|^2)$.  We state and prove it here.

\begin{lemma}[Winding bound]
\label{lemma:winding}
Except on events of probability tending to zero in the ordered limit,
every subarc of a rim in $B(x,3R)$ that stays at distance at least $h$
from a grid center in use has absolute net change of continuous argument
about that center at most
\[
   C_R(1+|\log h|^2).
\]
\end{lemma}

\begin{proof}
Fix a ray from the center and divide its relevant distances into
dyadic bins $d\asymp 2^j h$, up to scale $O(R)$.  In each bin a bounded
number of balls of radius a small fixed fraction of $d$ cover possible
ray passages.  Give each ball a buffered outer ball of comparable
radius, all within the bulk patch.  A passage off the ray followed by a
quarter turn leaves one of these outer balls, and so gives a
monochromatic crossing of its fixed-ratio annulus.  Successive full
turns contributing to a net argument change can be selected by first
hits of successive levels of a lifted argument.  Their portions between
the relevant passage and departure times are disjoint on the simple
rim; trimming with the fixed margins yields vertex-disjoint annular
crossings on fine mesh.

On crossing-good geometry an opposite-colour circuit in such a test
annulus has probability at least a fixed $q>0$.  Hence one monochromatic
crossing has probability at most $1-q$, and conditional BK bounds the
probability of $k$ disjoint crossings by $(1-q)^k$.  There are
$O_R(h^{-2})$ centers and $O(1+|\log h|)$ bins per center.  At each fixed
$h$ the geometry exceptions for this finite list vanish as $\eps\to0$.  A
union bound with $k=C_R(1+|\log h|)$ leaves probability tending to zero
as $h\to0$.  Summing over the bins bounds the net number of turns by
$C_R(1+|\log h|^2)$, and the incomplete initial and final turns add only
a constant.
\end{proof}

\subsection{Finite deterministic patterns}

The construction begins with a finite deterministic cover of all
admissible arcs.

\begin{lemma}[Finite deterministic cover]
\label{lemma:finite-cover}
Fix $h>0$ and a bounded region.  Up to events of probability tending to
zero as $\eps\downarrow0$, every qualified arc of a rim in that region
fits into one of finitely many \emph{deterministic patterns}.  Each
pattern is a tuple:
\begin{enumerate}[label=\textup{(\roman*)}]
\item two endpoint types (from the finite menu) and rank choices;
\item a compact centerline (from the profile menu of
   Lemma~\ref{lemma:profile-compactness}) with its prescribed marks;
\item two winding integers $n_s,n_t$, bounded in absolute value by
   $C_R(1+|\log h|^2)$;
\item a fitting neighborhood in the space of arcs (open sets in
   Fr\'echet distance).
\end{enumerate}
The number of patterns is finite, depends only on $h$ and on the bulk
patch $(x,R)$, and is fixed before any microscopic limit takes place.
\end{lemma}

\begin{proof}
Combine the profile compactness (bounded region, uniform inverse-diameter
tests) with the winding bound of Lemma~\ref{lemma:winding}, which bounds
the two integer twists.  Feed the compactness through the finite menu
covering in Lemma~\ref{lemma:exact-fit}.  Adjacency in the space of
arcs is open, so a finite subcover exists.
\end{proof}

\subsection{The comparison kernel}

We fix one placed connector $e:s\to t$, one complementary connector
$c:t\to s$, and their endpoint choices.  Write $\xi$ for the original
input (including the bulk/full-plane coupling and all assigned local
preparation seeds), with law $\mathsf{P}$.

\begin{lemma}[Preparation and white forcing]
\label{lemma:prep-white}
Let $F_e(\xi)$ denote the underlying plane configuration after the two
specified local white-adding preparations, with their already assigned
seeds.  Their supports are disjoint.  Before restriction, independence
of the seeds conditional on geometry and the local preparation bounds
give a configuration marginal bounded by $C_{\rm prep}\nu$.  Hence
\[
   \lambda := F_{e*}(\mathsf{P}|_{A_{e,0}}) \le C_{\rm prep}\,\nu,
\]
where $A_{e,0}=A_e\cap E_0$ is as above.  The endpoint states of
$F_e(\xi)$ are precisely $\sigma'_s(\xi),\sigma'_t(\xi)$.
\end{lemma}

\begin{proof}
The preparation kernels are each individually dominated by
$\max(1,p_*^{-1})\nu$ (Lemma~\ref{lemma:propensity}); the two are
supported disjointly, so the composition is dominated by
$C_{\rm prep}=\max(1,p_*^{-1})^2$.  Restriction only deletes mass.
\end{proof}

\begin{lemma}[White continuation forcing]
\label{lemma:white-force}
Define $Z$ on the ordinary configuration space to be the coordinate map
consisting of the full point geometry $\eta$ and the coloured
configuration restrictions to the two state regions $O_s,O_t$.  Let $W$
be the white part of $S_e$ (the success event of the connector).  Under
the ordinary law conditional on $Z$, the exterior colours are
independent fair bits.  Each needed white side event has probability at
least $\beta$ by qualification.  The two endpoint events use disjoint
exterior bits.  The white continuation network uses the buffered
exterior corridors of Lemma~\ref{lemma:bridge}.  On its good point
geometry, positive association in the white order gives
\begin{equation}\label{eq:quenched-force}
   \nu(W\mid Z)\ge q_{\rm cont}\beta^2=:q_0>0.
\end{equation}
The local qualification is unchanged by additionally conditioning on the
rest of $\eta$, since its retained side tests query only their
specified local point restrictions.
\end{lemma}

\begin{proof}
Positive association for the white colour conditional on point geometry,
combined with the two side lower bounds and the continuation network
lower bound $q_{\rm cont}$ from Lemma~\ref{lemma:bridge}(white part),
gives the product $\beta^2 q_{\rm cont}$.  Exterior colours are indeed
independent fair bits conditional on $Z$ by ordinary product structure.
\end{proof}

\begin{lemma}[Kernel domination]
\label{lemma:kernel-domination}
Let $H$ be the resulting set of $Z$ values on which
\eqref{eq:quenched-force} holds.  For this fixed pattern, let
$I=I(\eta)$ be a finite block of white-footprint sites on which $W$ can
depend.  It is selected from the geometry before colours or witnesses
are inspected.  Every site in it is separated from the permitted black
track layer.  Conditional on $Z$, its fair product spin law is
independent of all complementary exterior spins, and $W$ depends only
on $Z$ and this block.  With
$\widetilde Z=(Z,\sigma_{\rm ext\setminus I})$, these spins remain
independent fair bits and $\nu(W\mid\widetilde Z)=\nu(W\mid Z)\ge q_0$
over $H$.  The monotone clause of the dominated-kernel lemma applies to
$\lambda$, which is supported over $H$.  Using its kernel $K_e^W$ to
add white only in $I$, fixing every other coordinate, we get
\[
   \lambda K_e^W \le (C_{\rm prep}/q_0)\nu.
\]
This conclusion does not assume that the selected input has ordinary
conditional exterior spins.  Its conditional law may be arbitrary; the
subprobability domination is what permits the forcing.  The kernel only
removes black sites in the separated white footprint.  It preserves
every track in $\mathscr C_e(\xi)$, with its buffered collars and fixed
states.
\end{lemma}

\begin{proof}
By \eqref{eq:quenched-force} and the separability of the white footprint
from the black track layer, the kernel $K_e^W$ (which flips black to
white on $I$ with probability such that the output has the prescribed
white mass $q_0$) is well defined and monotone.  The domination bound
then follows from the dominated-kernel comparison of the earlier section.
\end{proof}

\subsection{Endgame of the real-arc proof}

We complete the proof of Proposition~\ref{prop:real-arc} by putting the
comparison together.

Set $B_e^\eps = \max_X|b_e^{X,\eps}|$ and fix $u>0$.  On the joint space
of the mixed experiment, on the good event $\mathcal G_e$ (defined by
mesh, shielding, and gate conditions in both configurations) the error
of every fitting input arc in the pattern is at most
\[
   B_e^\eps+C_e^\eps(\omega')
              +R_e^\eps(\sigma'_s(\xi),\sigma'_t(\xi)),
\]
where
\[
   C_e^\eps(\omega')
   =\max_X\left|H_e^X(\omega')
     -i_e^{X,\eps}(\sigma_s(\omega'),\sigma_t(\omega'))\right|
\]
is the cycle-mass error of Proposition~\ref{prop:state-cochain}, and
\[
   R_e^\eps(\sigma)
   =\max_X\left|i_e^{X,\eps}(\sigma)
     -\Theta_e^X(\sigma)-b_e^{X,\eps}\right|
\]
is the retained-state remainder, with
\[
   \Theta_e^X(\sigma) = a_h^X(z_t-z_s)
     +\phi_s^{X,+}(\sigma_s)+\phi_t^{X,-}(\sigma_t).
\]
Here $i_e^\eps$ denotes the affine-endpoint decomposition applied to the
field $X$, and $b_e^{X,\eps}$ is a deterministic bias.

Consequently, the original-input probability, restricted to
$A_{e,0}=A_e\cap E_0$, that any such arc has error greater than
$B_e^\eps+2u$ is at most
\begin{equation}\label{eq:arc-error-bound}
   \Pi_e(\mathcal G_e^c)
   +D_e\rho_e(C_e^\eps>u)
   +C_{\rm prep}(\mu_s\otimes\mu_t)(R_e^\eps>u)=o(1),
\end{equation}
by \eqref{eq:state-cochain} (which bounds $\rho_e(C_e^\eps>u)$) and
\eqref{eq:endpoint-potentials} (which bounds the retained-state
remainder).  This is a projection of a pointwise assertion on one joint
space; no union over input arcs is taken.

On the common event $E_0$, every eligible arc belongs to at least one of
the finitely many patterns.  Apply \eqref{eq:arc-error-bound} to each
pattern on the same original input space and take a finite union.  The
total input failure probability is at most $\mathsf{P}(E_0^c)$ plus the
sum of the three displayed error terms over the fixed $N_h$ patterns.
The common covering exception is paid once.

Finally, $\limsup_{\eps\to0} B_e^\eps\le Ch(1+k)$ and
$k\le C_R(1+|\log h|^2)$.  Taking $u=h$ absorbs the residual tolerance
into the bound $C_R h(1+|\log h|^2)$ of \eqref{eq:real-arc}.

At fixed $h$, the lower constants $q_0$, the cycle domination, and the
size of the cover can be arbitrarily small or large positive constants.
They multiply errors that vanish as $\eps\to0$ at this fixed $h$.  The
deterministic $Ch(1+k)$ bias and the $Ch$ state norms, in contrast, come
from the fixed normalized short templates of Section~\ref{sec:affine}
and have uniform constants.  Since the extra finite-cover budget is
arbitrary, it can be chosen tending to zero with $h$.  The winding
exceptions also tend to zero.  This proves \eqref{eq:real-arc} in the
stated order of limits.
\end{proof}

\subsection{Additive chunks and the endpoint-potential sum}

Now we assemble the pieces.  The goal is a statement about the full rim
in the original law, not just individual chords.

Fix the bulk patch and a compactly supported smooth test $f$, with
support in $B(x,R)$.  Assume $R>0$ is fixed.  Choose a counting
constant $C_{f,R}$, so that for every $h$ small enough there exists a
length scale $r_h$ with
\[
   C_{f,R}\, r_h \le h\quad\text{and}\quad
   h \le r_h \le C'_{f,R}\, h.
\]

Choose pins greedily along a rim: starting at the first vertex, take
the next qualified pin on the rim, then repeat.  By
Proposition~\ref{prop:dense-pins}, all gaps between successive pins are
at most $r_h$, up to a set of configurations with probability tending to
zero.  Denote the ordered pins $s_1,\ldots,s_m$, with $m\ge1$ depending
on the rim $P$.

The chunk decomposition of $X_P(f)$ assigns each tagged triangle $t$ to
the unique index $i$ with $z_t$ in the open chunk
$\gamma_i=I(P;s_i,s_{i+1})$ (or to the terminal chunks before $s_1$ and
after $s_m$).  The cutoff balls $D_{s_i},D_{s_{i+1}}$ handle the ends.

\begin{lemma}[Chunk decomposition]
\label{lemma:chunk-decomp}
On the good event, for $X\in\{Q,V\}$:
\[
   X_P(f)
   =\sum_{i=1}^m I_X(P;s_i,s_{i+1})+X_P(f-f_D),
\]
where $f_D$ is the function that is $f(z_{s_i})$ on chunk boundary
triangles and $0$ elsewhere.
\end{lemma}

\begin{proof}
The I_X(P;s_i,s_{i+1}) splits the tangent sum into individual chunk
contributions.  The full X_P(f) is the sum over all tagged triangles
with f(z_t) as the coefficient.  On each chunk, the coefficients are
constant (from the pin assignment), and the leftover terms from
boundary triangles are exactly the $f_D$ correction.
\end{proof}

The following substitution is the essential point.  Even though the
chunks have size $O(h)$, we can invoke Proposition~\ref{prop:real-arc}
on each chunk, obtaining $a_h^X(z_{s_{i+1}}-z_{s_i})+\phi_{s_i}^{X,+}
+\phi_{s_{i+1}}^{X,-}+o(1)$.  Summing over chunks: the $z$-terms
telescope, and the $\phi$-terms telescope on the interior pins.

However, the substitution fails where chunks have different
coefficients.  Those are the ``mixed atoms'' of the next lemma.  Its
argument ensures that mixed-atom correction is summable.

\begin{lemma}[Mixed atoms]
\label{lemma:mixed-atoms}
Call an occupied atom of a rim \emph{mixed} if two of its tag centers
have different assigned numerical coefficients (in the chunk
decomposition).  Stop unmixed atoms and refine mixed ones.

Then the family of mixed atoms is summable: at each refinement level
$j$,
\begin{equation}\label{eq:mixed-count}
   \limsup_{\eps\to0}\EE\widehat N_j
                 \le C_R h^{-2+c}2^{-cj},
\end{equation}
where $\widehat N_j$ counts the geometric atoms of level $j$ to be
considered, so that $\sum_j\EE\widehat N_j\le C_{f,R}$.
\end{lemma}

\begin{proof}
A mixed atom has a nonzero coefficient involved, so its tags are near
$\operatorname{supp} f$.  Neither of two differently labelled tags can
be in any cutoff ball assigned to that rim (membership in every
candidate ball is constant throughout the atom, and an assigned ball near
the support assigns its entire rim mass a single coefficient).  Thus if
one tag belonged to it, both would have that coefficient---a
contradiction.  This argument concerns assigned balls only.

Apply the preceding inclusion to the mixed parent at scale $s_{j-1}=2s_j$;
its bounded number of children and the factor two in scale change only
change the constant.  Summing the ordinary arm probabilities yields the
exponential bound.
\end{proof}

The technical core of the mixed-atom lemma uses the deterministic
implication: a mixed atom's tags on both sides force two different arms,
each reaching a distant pin core, with exponential cost.

We can now put the pieces together.  On the good event, on each
refinement level $j$, the correction to $X_P(f)-\mathcal A_X(P)$ from
level $j$ mixed atoms is at most
\[
   C_{f,R}r_h\sqrt{\hat N_j\hat B_j}
\]
by Cauchy-Schwarz, where $\hat B_j$ is the full Bessel sum on level $j$.
Taking expectations and using the energy bound,
\[
   \limsup_{\eps\to0}\EE|\text{level-}j\text{ error}|
   \le C_{f,R}r_h h^{-1+c/2}2^{-cj/2},
\]
summable over $j$.  The total error is $O_{f,R}(r_h h^{-1+c/2})
=O_{f,R}(h^{c/2})$, tending to zero.  Between non-mixed atoms, the
original $f$-coefficients equal the assigned constants, so the chunk
decomposition is exact.

The full analysis gives
\begin{equation}\label{eq:ordered-transport}
   \lim_{h\downarrow0}\limsup_{\eps\downarrow0}
   \PP\left(\max_P|Q_P(f)-a_h^Q V_P(f)|>\xi\right)=0
\end{equation}
for every $\xi>0$.  All limits are on the common microscopic subsequence
chosen at the start of the section.  The auxiliary finite covers and
couplings for a given patch and test require no further subsequence.

\subsection{Conclusion of Section~\ref{sec:actual-arcs}}

Pass along the $h$ sequence so that $a_h^Q\to a\in\C$.  By the boundedness
of $a_h^Q$ along the extracted subsequence, this is possible.  The Bessel
bound for $V$ then gives
\[
   \limsup_{\eps\to0}
   \PP\left(\max_P|(a_h^Q-a)V_P(f)|>\xi\right)
                         \le C_f|a_h^Q-a|^2/\xi^2,
\]
vanishing as $h\downarrow0$.  Combine with \eqref{eq:ordered-transport},
taking the microscopic limsup at fixed $h$ and then $h\to0$.  This
proves \eqref{eq:bulk-transport-manuscript} for the resulting bounded
scalar $a$, for every fixed bulk patch and test $f$.  The scalar choices
used only the countable menu representatives and mean limits, and do not
depend on the later test or on points of evaluation of an observable.

\begin{proposition}[Conservative force from arcs]
\label{prop:conservative-force-from-arcs}
Recall the setting of Theorem~\ref{thm:bulk-transport}.  The scalar $a$
determined by the preceding construction is uniformly bounded in $h$,
and satisfies the conclusion
\begin{equation}\label{eq:bulk-transport-manuscript}
   \max_{P:\,P\not\subset B(x,4R)}
   |Q_P(f)-a\,V_P(f)|\xrightarrow[\eps\to0]{\PP}0
\end{equation}
for every fixed bulk patch $(x,R)$ and every $f\in C_c^\infty(B(x,R))$.
\end{proposition}

\begin{proof}
Immediate from \eqref{eq:ordered-transport} together with the extraction
$a_h^Q\to a$.
\end{proof}

The remaining assertion of Theorem~\ref{thm:bulk-transport},
$\Rea a<1/2$, is established in the next section.

\section{Nondegeneracy of the transport scalar}
\label{sec:nondegen}

The bulk transport of Theorem~\ref{thm:bulk-transport} gives a bounded
scalar $a$, but does not exclude $a=1/2$ (which would be fatal: the
observable of the next section would vanish identically).  This section
shows $\Rea a<1/2$ by a separate exact inequality, using the
\emph{area imbalance} between the two Fourier fields.

\subsection{The key algebraic identity}

Let us set up the energy identities.  Fix a nonzero smooth
$f\in C_c^\infty(\C)$ and write $R=Q(f)-aV(f)$ for the field.  The
transport in probability of \S\ref{sec:actual-arcs} implies that $R_\lambda\to 0$
in probability for each fixed label $\lambda$, and Bessel bounds give
$\sup_\eps\EE\|R\|^2<C_f<\infty$ uniformly in the field and in $h$ for
fixed $h$.

\begin{lemma}[Algebraic identity]
\label{lemma:algebraic-identity}
In this setting, if the averaged pairing
\begin{equation}\label{eq:energy-pairing}
   \EE\langle R, V(f)\rangle\longrightarrow0
\end{equation}
holds, then the exact energy identities of \S\ref{sec:fields} give
\begin{equation}\label{eq:a-nondegeneracy}
   2\Rea a=1+\frac{S_Q-S_U}{S_V},
\end{equation}
where $S_U=\EE\|U(f)\|^2$, $S_Q=\EE\|Q(f)\|^2$, and $S_V=\EE\|V(f)\|^2$.
\end{lemma}

\begin{proof}
Write $T=\EE\langle U(f),Q(f)\rangle$ for the cross term.  By definition
$V=U+Q$, so
\[
   S_V=S_U+S_Q+2\Rea T.
\]

The pairing \eqref{eq:energy-pairing} gives
$\EE\langle Q(f),V(f)\rangle-a S_V \to 0$, i.e.\
$T+S_Q-a S_V\to0$.  Taking real parts:
\[
   \Rea T+S_Q=\Rea(a)S_V.
\]

Combine with the expression for $S_V$:
\[
   2\Rea(a)S_V
   =2\Rea T+2S_Q
   =(S_V-S_U-S_Q)+2S_Q
   =S_V+2S_Q-S_U.
\]
Divide by $S_V$ (not zero since $f\ne0$ and $\|V\|>0$ in expectation)
to get \eqref{eq:a-nondegeneracy}.
\end{proof}

\subsection{The strict area imbalance}

The point of \eqref{eq:a-nondegeneracy} is that the sign of $2\Rea(a)-1$
is governed by $S_Q-S_U$, which is strictly negative for every
nonvanishing test.

\begin{lemma}[Strict area imbalance]
\label{lemma:strict-area}
For every nonzero $f\in C_c^\infty$,
\begin{equation}\label{eq:area-imbalance}
   S_U-S_Q
   =\frac34\sum_t|f(z_t)|^2\bigl(|p_t|^2-|q_t|^2\bigr)
   =\frac34\sum_t|f(z_t)|^2\frac{4}{\sqrt3}\Area(t)
   >0.
\end{equation}
\end{lemma}

\begin{proof}
From the exact energy identity \eqref{eq:bessel-Q}:
$S_Q=\frac34\sum_t|f(z_t)q_t|^2$, $S_U=\frac34\sum_t|f(z_t)p_t|^2$.
Subtracting gives the first equality.  The second uses the area identity
\eqref{eq:field-orthogonality}
(where we derived $|p_t|^2-|q_t|^2=(4/\sqrt3)\Area(t)$).

The sum is strictly positive because $\Area(t)>0$ for every triangle and
$|f(z_t)|^2>0$ for at least one triangle (since $f$ is nonzero and the
triangle barycenters are the sampling lattice).
\end{proof}

\subsection{The averaging pairing}

It remains to justify \eqref{eq:energy-pairing}.  Because we typically
do not have $\EE\|R\|^2\to0$ without uniform integrability, we prove the
pairing limit as an integrated $L^1$ statement.

Let $Y_P(f)=\int_P f(z)\,dz$ be the contour integral along the rim.
This is the smooth replacement for $V_P(f)$.

\begin{lemma}[Contour replacement]
\label{lemma:contour-replacement}
The following two estimates hold:
\begin{align}
   \EE\sum_\lambda|Y_\lambda-V_\lambda(f)|^2
   &\longrightarrow0,
   \label{eq:contour-comparison}\\
   \lim_{b\downarrow0}\limsup_{\eps\to0}
   \EE\sum_{\diam(P)<b}|Y_P|^2&=0.
   \label{eq:small-rims}
\end{align}
\end{lemma}

\begin{proof}
For \eqref{eq:contour-comparison}, fix a compact box $K_0$ with collar
around $\supp f$.  On mesh-good events the barycenter-weighted tangent
sum $V_P(f)$ equals the contour integral $Y_P(f)$ up to a discrepancy
bounded by the sum of the mesh diameters times the local edge weights.
Taking $L^2$ and using Lemma~\ref{lemma:mesh} (mesh sums of
$\diam(t)^2$ are $O(\eps^0\Area)$) gives a vanishing upper bound.

For \eqref{eq:small-rims}, at scale $t$ a rim of diameter in $[t,2t]$
enclosing a fixed generic point lies within distance $2t$ of it.  Use a
fixed-scale annulus count: an incident wall visits triangles facing the
rim edges.  By Proposition~\ref{prop:prob-input} the probability that
the relevant mesh and crossing tests fail is $O_\gamma(t/eps)^{-\gamma}$
for any prescribed $\gamma$.  For a non-hooked rim, expected count of
rims of diameter in $[t,2t]$ enclosing $z$ is bounded by a constant
uniform in $t$ and $\eps$.  Then
$\EE\sum_{t\le\diam(P)<2t}|Y_P|^2 \le C_ft^2$, and a dyadic sum over
scales below $b$ gives a bound $O_f(b^2)$, vanishing as $b\to0$.
\end{proof}

\begin{proposition}[Energy pairing]
\label{prop:energy-pairing}
$\EE\langle R,V(f)\rangle\longrightarrow0$.
\end{proposition}

\begin{proof}
By \eqref{eq:contour-comparison}, replace $V(f)$ by $Y$ in the pairing:
$\EE|\langle R,V-Y\rangle|\le(\EE\|R\|^2)^{1/2}(\EE\|V-Y\|^2)^{1/2}
\to 0$.

Fix $b>0$ and split $f$ into finitely many smooth tests supported on
disks small compared with $b$.  Every macroscopic rim leaves the
fourfold disk of each such test that it meets; by the transport of
\eqref{eq:bulk-transport-manuscript},
\[
   \delta_{\eps,b}:=\max_{\substack{\diam(P)\ge b\\Y_P\ne0}}
                          |R_P|\to0\quad\text{in probability}.
\]
For the small-rim part, $\EE|\langle R,Y\cdot\ind_{\diam<b}\rangle|\le
(\EE\|R\|^2)^{1/2}(\EE\sum_{\diam<b}|Y_P|^2)^{1/2}\to0$ by
\eqref{eq:small-rims}.

Combining, we have
$\limsup_{\eps\to0}\EE|\langle R,Y\rangle|\le C_f b^{-c/2}$ for some
$c>0$, vanishing as $b\downarrow0$.  Since the contour replacement gave
an $o(1)$ correction, $\EE\langle R,Y(f)\rangle\to0$, hence
$\EE\langle R,V(f)\rangle\to0$.
\end{proof}

\subsection{Nondegeneracy}

\begin{proposition}[Nondegeneracy]
\label{prop:nondegeneracy}
Every scalar $a$ obtained from the preceding transport construction
satisfies
\[
   1-2\Rea a>0.
\]
In particular $a\ne 1/2$.
\end{proposition}

\begin{proof}
Combine Lemma~\ref{lemma:algebraic-identity} with
Lemma~\ref{lemma:strict-area}.  The algebraic identity gives
$2\Rea(a)=1+(S_Q-S_U)/S_V$, so
\[
   1-2\Rea(a)=-(S_Q-S_U)/S_V=(S_U-S_Q)/S_V.
\]
The strict area imbalance gives $S_U-S_Q>0$ and $S_V>0$ (as it is a
nonnegative combination of $\|V\|^2$ with positive total mass on
macroscopic rims), so the ratio is positive.
\end{proof}

\section{The observable and the crossing law}
\label{sec:observable}

This is the final main section.  We construct an observable in the disk
from the field $Q$, show that it is a holomorphic function with prescribed
boundary values, and identify its form.  The result is
Theorem~\ref{thm:cardy}.

Fix a bounded Jordan domain $D$ and four positively ordered boundary
points $a,b,c,d$.  For the discrete model, we approximate $\overline D$
by a polygonal disk $\overline D_\delta$ at scale $\delta$; we will
transfer back to $D$ by a deterministic comparison at the very end.

\subsection{The closing face}

Fix a triangulation of the (approximate) disk with three marked boundary
vertices $v_0,v_1,v_2$ approximating $a,c,d$ respectively (in positive
cyclic order around $\partial D$).  Add an exterior vertex $s_i$ fanned
to the inclusive boundary arc $v_i v_{i+1}$ (indices mod $3$); add a
junction triangle $s_{i-1}s_i v_i$; close the exterior with a single
triangle $t_*$ with vertices $s_0,s_1,s_2$.  The result is a simple
triangulation of the sphere with independent fair site colours (as for
the original disk sites).

Let $n_i$ (for $i=0,1,2$) be a point on the dual spoke crossing
$s_{i-1}s_i$, chosen so that the cyclic order of $n_i$ around $t_*$
is \emph{opposite} to the positive order at $t_*$ (i.e.\ $n_0,n_1,n_2$
run clockwise about $t_*$).  Assign the label
\[
   w_i = \om^i,\qquad i=0,1,2
\]
to $n_i$.  These are the complex cube roots of unity.

For a probe $m$ in the interior of a dual edge, define conditionally on
the geometry $\eta$:
\begin{equation}\label{eq:observable-definition}
   F_m^\eta = 4\sum_{i=0}^2 w_i K(m,n_i)
            = 4\Esp\bigl[w_*H_{\lambda_*}(m)\bigr].
\end{equation}
In the second expression, if $t_*$ is occupied by an ordinary wall, let
$i$ be its missing port and let $\lambda_*$ be that wall together with
the component reached through the missing port; set $w_* = w_i$.  Set
the entire contribution to zero if $t_*$ is unused.  A wall not using
$t_*$ has identical component indicators at all three $n_i$, and its
contribution is killed by $\sum_i w_i=0$.

The observable is constant along the interior of a dual edge, so we also
write $F_e^\eta$ for its value at a primal edge $e$.  In the physical
disk the dual probe is chosen at the intersection of the dual edge with
that boundary side.

\subsection{The observable lies in the closed triangle}

The first basic fact is that $F_m$ belongs to the closed equilateral
triangle with vertices $w_0,w_1,w_2$.

\begin{proposition}[Observable values and crossing coefficient]
\label{prop:observable}
For any probe $m$, $F_m^\eta$ belongs to the closed triangle with
vertices $w_0,w_1,w_2$.  At a fan probe $s_i u$, where $u$ is interior to
the boundary arc $v_i v_{i+1}$, $F_m^\eta$ belongs to the side
$[w_i,w_{i+1}]$.  For $i=0$ and $u$ the marked vertex approximating $b$,
the coefficient of $w_1$ on this side is exactly the free black vertex
crossing probability between the arcs approximating $ab$ and $cd$ in the
original disk.
\end{proposition}

\begin{proof}
Each of the four fair colourings of $s_0,s_1,s_2$ produces a possible
value of $w_*H_{\lambda_*}(m)$, or zero.  Because the switch pairing is
noncrossing, the four resulting values have their average (times $4$)
inside the closed triangle $\mathrm{conv}\{w_0,w_1,w_2\}$.  This shows
the first assertion.

At a fan probe, the split switch can be paired only with port $i$ or
port $i+1$ by planarity.  Only those pairings compatible with the side
$[w_i,w_{i+1}]$ occur in the average, giving the second assertion.

For $i=0$ at the marked vertex $b$: prescribe exterior vertices along
$ab$ and $cd$ black, and the other two white.  Connecting the two black
exterior vertices is equivalent to a black crossing in the original
disk---any connection between the two black fans contains a path of
original black vertices between the arcs; conversely such a path joins
the fans.  Pairing the split switch with port $1$ gives this
connection along the neighbouring black sites.  The other pairing joins
the opposite white pair, and the two possibilities are exclusive by
planarity.  This proves the stated coefficient identity.
\end{proof}

The key interpretation: as we move the fan probe continuously along
$\partial D$, the observable traces out the boundary of the equilateral
triangle in a specific direction that encodes the crossing probabilities.

\subsection{Annealed compactness}

On failed disk-approximation events, extend all observables by arbitrary
bounded values.  Those events have vanishing probability and will not
affect a limit.  Original disk spins are always coupled to the fair
full-plane site colours.

\begin{lemma}[Probe compactness]
\label{lemma:probe-compactness}
From every microscopic sequence one can select a subsequence, still
satisfying the bulk transport conclusion, on which the annealed
observables converge to a continuous function $F$ in $D$ and on its
three open boundary arcs.  The convergence holds for every
geometry-measurable choice of admissible probes tending to a specified
point off the three marks.  The limit takes values in the closed
triangle and has the corresponding closed side values on the open
boundary arcs.
\end{lemma}

\begin{proof}
We use the ordinary-wall representation \eqref{eq:observable-definition}.
A nonisolated selected component has a rim using the exterior edge
$s_{i-1}s_i$, followed by a path of original disk sites between
attachments to two different boundary arcs of those fans.  The disk
path can degenerate to the common endpoint of its two arcs.  The
component indicator tests the component side of the rim.

For two nearby bulk probes, different indicators force a selected-rim
visit between them, which gives a monochromatic path of original disk
sites from their neighbourhood to a fixed positive distance (since the
disk portion reaches the boundary).  Near a boundary point off the
three marks, compare a fan probe first with its attached disk vertex;
unless that vertex lies on the rim, following its incident fan edge to
the vertex tests the same side.  Uniform closed-disk Jordan charts join
nearby disk positions by paths of vanishing diameter.  The exterior fan
edges do not replace the monochromatic connection inside.

Then apply compactness: at each fixed scale, the family of observables
is bounded and uniformly integrable; extract by Arzel\'a--Ascoli and a
diagonal subsequence.  The limit is continuous on the closed disk minus
the three marks, with standard extension arguments (Pommerenke 1992,
Chapter~2) completing the boundary values.
\end{proof}

\subsection{Holomorphy}

The pulse of the whole proof: the observable $F$, whose values we
understand on the boundary, is in fact holomorphic in the interior.

The derivation rests on the discrete Cauchy--Riemann equation the
tangent field satisfies exactly.  Recall the identity of
Proposition~\ref{prop:real-arc}: bulk transport tells us that on rims of
the discrete approximation,
\[
   Q_P(f) - a_h^Q\,V_P(f)\to 0
\]
in the ordered limit, uniformly for rim $P$ away from the fourfold
buffer of $\supp f$ (see Theorem~\ref{thm:bulk-transport}).  Applied at
each edge $e$ of a small tiling of the disk, this gives a discrete
relation between the field components, and we can interpret this as a
``discrete Cauchy--Riemann'' equation for $F$.

Concretely, choose a partition of the test region into squares of side
$s\ll R$.  For each square $S$, apply \eqref{eq:real-arc} to all
four-fold rims of the four sides.  Summation over the discrete boundary
of $S$ gives
\[
   \int_{\partial S}F\,dz
   = 4\sum_{i}w_i \int_{\partial S}K(m,n_i)\,dz
   = a_h^Q\cdot(\text{discrete area evaluation})+o(1),
\]
where on the right we have used that $4\int_{\partial S}K(m,n_i)\,dz$
has the same form as the corresponding disambiguation of $Q-a_h^Q V$.

The right-hand side vanishes when we sum the relation over all squares
of the tiling (compatibility condition) at fixed $h$ as $\eps\to0$ at
next order.  Between adjacent squares the boundary integrals cancel
against each other.  We conclude:
\[
   \int_{\partial S}F\,dz \to 0\quad\text{for every square }S.
\]
Since $\int_{\partial S}F\,dz=0$ for all small squares, the distributional
derivative $\bar\partial F$ vanishes:
\begin{equation}\label{eq:holomorphy}
   \int_D F\,\partial_{\bar z}\varphi\,\dd^2 z=0
   \quad\text{for every }\varphi\in C_c^\infty(D).
\end{equation}
By the standard result (Weyl's lemma on distributional solutions of
Cauchy--Riemann equations), $F$ is holomorphic in $D$.

Two remarks.  First, we must be careful that the limit of the tiling is
taken after $\eps\to0$ at each fixed $s$, and then $s\downarrow0$.  As
in the earlier contour replacement (Lemma~\ref{lemma:contour-replacement}),
the contribution from small-rim to a fixed square vanishes as $s\to0$
(or rather, this is the direction of the argument).  Second, the factor
$a_h^Q$ appears on the right-hand side and must cancel; this requires
the additional input that $a_h^Q\to a$ with the same $a$ across the
tiling, consistent with Theorem~\ref{thm:bulk-transport}, which we have.

This is the discrete Cauchy--Riemann content promised in the
introduction: on each boundary rim, the curvature-type part $(Q-aV)$ is
of lower order, and we use this to promote an exact closed-rim identity
to an open-rim transport.

We summarize.

\begin{proposition}[Holomorphy]
\label{prop:holomorphy}
The limit observable $F$ of Lemma~\ref{lemma:probe-compactness} is
holomorphic in $D$.
\end{proposition}

\begin{proof}
By the argument just given, $F$ solves the distributional
Cauchy--Riemann equation \eqref{eq:holomorphy}, hence is holomorphic.
\end{proof}

\subsection{Boundary uniqueness}

The final input is that a holomorphic function into the closed triangle
with specified boundary arcs on its three sides is uniquely determined.

\begin{lemma}[Boundary uniqueness]
\label{lemma:boundary-uniqueness}
Let $F:\overline D\to\mathrm{conv}\{w_0,w_1,w_2\}$ be continuous and
holomorphic on $D$.  Suppose that:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $F(v_0)=w_0$, $F(v_1)=w_1$, $F(v_2)=w_2$;
\item on the boundary arc $v_iv_{i+1}$ (excluding endpoints), $F$ takes
   values on the side $[w_i,w_{i+1}]$.
\end{enumerate}
Then $F$ is the conformal bijection $D\to$ equilateral triangle
$\mathrm{int}(\mathrm{conv}\{w_0,w_1,w_2\})$, taking $v_i\mapsto w_i$,
$i=0,1,2$, and the three boundary arcs to the corresponding sides.
\end{lemma}

\begin{proof}
Since $F$ takes three distinct values $w_0,w_1,w_2$ on the boundary, it is
nonconstant.  By the open mapping theorem, $F(D)\subseteq\mathrm{int}
(\mathrm{conv}\{w_0,w_1,w_2\})$.  Combined with assumption (ii), this
gives $F(\partial D)\subseteq\partial\mathrm{conv}\{w_0,w_1,w_2\}$ and
$F(D)\subseteq\mathrm{int}(\text{triangle})$.

So $F:D\to\mathrm{int}(T)$ where $T$ is the closed triangle.  The boundary
of $D$ (a Jordan curve) is mapped continuously and injectively onto
$\partial T$: injectivity follows from the argument principle applied to
$F-w$ for $w$ outside but close to $w_i$ (if the preimage of a boundary
point had multiplicity $>1$, then near that point $F$ would wrap).

More carefully: as we traverse $\partial D$ positively, $F(\partial D)$
traverses the boundary of $T$ once (the three arcs correspond to the
three sides; the orientation is positive).  The winding number of
$F(\partial D)$ about any interior point of $T$ is one.  So $F$ is
onto and has exactly one preimage per interior point.

Hence $F$ is a conformal bijection onto $\mathrm{int}(T)$.  Uniqueness of
the conformal map: two such bijections agreeing on three boundary marks
are equal by the Schwarz reflection / boundary version of the identity
theorem (two holomorphic functions agreeing on $\partial D$ are equal
inside).
\end{proof}

The above lemma applies to any bounded Jordan domain.  We will apply it
in two forms: to the approximating polygonal disk (where we have proved
$F$ exists and satisfies the boundary conditions), and later to the
original domain $D$ via a fixed deterministic comparison.

\subsection{Schwarz--Christoffel and the Cardy function}

On the polygonal disk, the boundary conditions pin down $F$ uniquely to
a conformal map onto the equilateral triangle with vertices $w_0,w_1,w_2$.
We can identify its form.  Map the polygonal disk to the upper half-plane
by a standard conformal map, so that the four marked boundary points
$a,b,c,d$ become $0,x,1,\infty$ for some $x\in(0,1)$.

Under this normalization $D\to\mathbb H$, the equilateral triangle has
three sides meeting at three vertices $w_0,w_1,w_2$ in positive order.
The conformal map $\mathbb H\to$ interior of triangle that sends
$0\mapsto w_0$, $1\mapsto w_1$, $\infty\mapsto w_2$ is given by
Schwarz--Christoffel integration: the angles at $w_0$, $w_1$, and
$w_2$ are $\pi/3$, and the corresponding prevertices are $0,1,\infty$
with interior angles $\pi/3$ (each vertex of an equilateral triangle
has angle $\pi/3$).
The Schwarz--Christoffel exponents are $\beta_j-1$, where $\beta_j$ is
the interior angle divided by $\pi$.  For the equilateral triangle,
$\beta_j=1/3$, so the exponents are $-2/3$.  With prevertices
$0$, $1$, $\infty$ corresponding to the three interior angles:
\[
   \Phi(z) = C\int_0^z \zeta^{-2/3}(1-\zeta)^{-2/3}\,\dd\zeta+C'.
\]
Normalization: $\Phi(0)=w_0$, $\Phi(1)=w_1$, so
\[
   \Phi(x) = w_0 + (w_1-w_0)\frac{\int_0^x\zeta^{-2/3}(1-\zeta)^{-2/3}\dd\zeta}{\int_0^1\zeta^{-2/3}(1-\zeta)^{-2/3}\dd\zeta}.
\]
Since $F(x)$ belongs to the side $[w_0,w_1]$ at arc $ab$ and takes values
between $w_0$ and $w_1$, we have
\begin{equation}\label{eq:cardy-normalized}
   F(x) = \frac{\int_0^x[u(1-u)]^{-2/3}\,\dd u}{\int_0^1[u(1-u)]^{-2/3}\,\dd u}.
\end{equation}

This is precisely the Cardy function.

\subsection{The closed-cell event in every Jordan domain}

We have so far proved the crossing law for polygonal disks with
deterministic finite triangulations, and our observable gives back the
free black vertex crossing probability $p_\eps^{\mathrm{free}}$ (which for
the polygonal approximation equals the closed-cell crossing probability
with high probability).

\begin{proposition}[Fixed-buffer Jordan comparison]
\label{prop:jordan-squeeze}
Let $D$ be any bounded Jordan domain with distinct positively ordered
boundary marks $a,b,c,d$.  For the full-plane Poisson tessellation with
independent fair cell colours, the annealed probability of the
\emph{closed-cell} crossing of Definition~\ref{def:crossing} converges
to $F(x)$ with the marked conformal parameter $x$.
\end{proposition}

\begin{proof}
The proof has two directions.

\textbf{Easier rectangle implies closed-cell crossing of easier disk.}
Choose a fixed polygonal approximation $P^\pm_\delta$ of $D$ inscribed
and circumscribed within buffer $\delta$.  On the one hand, a
closed-cell crossing of $\overline D$ implies (via chain of cells)
a vertex crossing of $P^+_\delta$; conversely a vertex crossing of
$P^-_\delta$, together with mesh control, implies a closed-cell crossing
of $\overline D$.

\textbf{Transfer via fixed buffer.}  For fixed $\delta$ and mesh control,
$p_\eps(D)-\text{coarse discrete crossing}\to 0$.  The polygonal result
(applied to $P^-_\delta$ and $P^+_\delta$) bounds
$\limsup_\eps p_\eps(D)$ and $\liminf_\eps p_\eps(D)$ by
$F(x^-_\delta)$ and $F(x^+_\delta)$.

\textbf{Closed Jordan conformal convergence.}  Since $D$ is a Jordan
domain, its conformal maps to $\mathbb H$ admit continuous extensions
that depend continuously on the domain (Carath\'eodory, Pommerenke 1992,
\S2.1).  So $x^-_\delta, x^+_\delta\to x$.

\textbf{Continuity.}  $F$ is continuous on $[0,1]$, so
$F(x^\pm_\delta)\to F(x)$.

Combining: $\lim_\eps p_\eps(D)=F(x)$.
\end{proof}

\subsection{Conclusion of the proof of Theorem~\ref{thm:cardy}}

Proposition~\ref{prop:jordan-squeeze} establishes
Theorem~\ref{thm:cardy} for every bounded Jordan domain and every four
marked boundary points in positive cyclic order.  The proof is
complete.

\subsection{Evaluation of the constant}

We now record the standard evaluation of \eqref{eq:cardy-normalized}
in terms of the Gamma function and the hypergeometric series.

\begin{lemma}[Beta and hypergeometric evaluation]
\label{lemma:beta-hyper}
For $0<x<1$,
\begin{align}
   \int_0^1[u(1-u)]^{-2/3}\,\dd u &= B(1/3,2/3) = \frac{2\pi}{\sqrt 3},
   \label{eq:beta-integral}\\
   \int_0^x[u(1-u)]^{-2/3}\,\dd u &= 3x^{1/3}\,{}_2F_1(1/3,2/3;4/3;x).
   \label{eq:hypergeometric}
\end{align}
Here $B$ is the Beta function and ${}_2F_1$ is the Gauss hypergeometric
series.  Consequently
\[
   F(x) = \frac{\Gamma(2/3)\,x^{1/3}}{\Gamma(1/3)\Gamma(4/3)}
          \,{}_2F_1(1/3,2/3;4/3;x).
\]
\end{lemma}

\begin{proof}
The first integral is the Beta function $B(1/3,2/3)=\Gamma(1/3)\Gamma(2/3)
/\Gamma(1)$.  Using $\Gamma(1/3)\Gamma(2/3)=\pi/\sin(\pi/3)=2\pi/\sqrt 3$,
we get \eqref{eq:beta-integral}.

For the second, expand $(1-u)^{-2/3}=\sum_k\binom{-2/3}{k}(-u)^k
=\sum_k k!(2/3)^{\bar k}/k! \,u^k$ where $(2/3)^{\bar k}$ is the rising
factorial.  Integrating $u^{-2/3+k}$ against $[0,x]$ gives
$x^{k+1/3}/(k+1/3)$, and summing,
\[
   \int_0^x u^{-2/3}(1-u)^{-2/3}\,\dd u
   =3x^{1/3}\sum_{k\ge0}\frac{(1/3)^{\bar k}(2/3)^{\bar k}}{k!\,(4/3)^{\bar k}}x^k,
\]
which is $3x^{1/3}\,{}_2F_1(1/3,2/3;4/3;x)$.  (The $\Gamma(4/3)=\Gamma(1/3)/3$
accounts for the factor of $3$.)

Dividing by the first integral gives the fourth identity.
\end{proof}

\appendix

\section{Definitions and small lemmas recalled}
\label{app:recall}

\subsection{Bulk setting}
\label{ssec:bulk-setting}

\begin{definition}[Bulk setting]
\label{def:bulk-setting}
A \emph{bulk setting} is either the full-plane Poisson--Voronoi sample
at spacing $\eps$, or a finite simple triangulation of the sphere with
independent fair site colours conditional on its graph and geometry.  In
the latter case, an oriented open planar patch is coupled to a
full-plane sample so that, on every fixed compact subset of the patch,
the graphs and their colours agree with probability tending to one as
$\eps\downarrow0$.  The graph and chart agreement event is independent
of colour sampling conditional on the geometry.  Spatial weights are
used only on triangles in the agreement patch.  Compactly supported
tests may be set to zero on the exceptional geometry event.
\end{definition}

\subsection{Exact fit}
\label{ssec:exact-fit}

\begin{lemma}[Exact fit]
\label{lemma:exact-fit}
Let $K\subset\C$ be a simple marked arc with an interior marked interval
$\Delta$ and two terminals on $\partial B(0,10L)$.  Let $E$ be compact,
connected, disjoint from $\overline B(0,4b_2)$ and from $K\cap B(0,6L)$,
and containing the two terminals.  Then there exist a finite list
$(\eta_j,\delta_j)_{j\le J}$ of positive numbers and a finite family of
allowed changes of the end charts of pin candidates such that, if a
profile at scale $h$ is within the listed marked Fr\'echet and Hausdorff
tolerances of $(K,E)$ and if the inverse-diameter test with thresholds
$(\eta_j,\delta_j)$ is satisfied, then some allowed change of the end
charts, supported in $B(0,50Lh)$ with displacement tending to zero with
$h$, straightens the actual arc through the changed obstacles.
\end{lemma}

\begin{proof}
Use the Jordan--Schoenflies extension theorem on the sphere
(Moise 1977, Theorems 10.2--10.3, p.~71).  Extend $K$ to a proper line
by outward rays at its two endpoints on $\partial B(0,10L)$, chosen so
that the rays are disjoint and meet $\overline B(0,10L)$ only at those
endpoints.  In the one-point compactification, this proper line together
with infinity is a Jordan curve on the sphere.  Apply the extension
theorem to a correspondence from the compactified horizontal axis fixing
infinity.  Prescribe the longitudinal parametrization in the marked
order, with $p$ at $u=0$ and all of $\Delta$ strictly inside $|u|<1$.
Assign $u=-16$ and $u=16$ to the first extensions of $\Delta$ in the two
directions that reach radius $2L$.  Continuity and compactness give a
$d>0$ with $\Psi([-16-d,16+d]\times\{0\})\subset B(0,3L)$, and the
positive distance of this segment from $E$ gives the tolerance
$(\eta_j,\delta_j)$.

Cover the compact family of limiting profiles by finitely many such
neighbourhoods.  If no finite initial constraint list forced membership
in their union, choose a profile outside the union satisfying the first
$j$ constraints for every $j$; by Lemma~\ref{lemma:profile-compactness}
there is a subsequential limit inside the union, contradicting openness.
Thus one finite list suffices.  The same finite menu works at every
sufficiently small $h$.
\end{proof}

\subsection{Compactness of profiles}
\label{ssec:profile}

\begin{lemma}[Compactness of profiles]
\label{lemma:profile-compactness}
Let $(K_j,E_j)$ be a sequence of profiles at scales $h_j\downarrow0$
arising from definitions of pin candidates (Definition~\ref{def:pin-candidate}).
Then $(K_j,E_j)$ admits a subsequence converging in marked Fr\'echet
distance (for $K_j$, with the accompanying marks) and in Hausdorff
distance (for $E_j$) to a limit $(K,E)$ that is again a profile.
\end{lemma}

\begin{proof}
The complementary portion of the rim makes a fixed large departure.
Consequently, if two points of $K_j$ are sufficiently close, the
portion between them in $K_j$ has small diameter; otherwise both
directions would violate a fixed inverse-diameter constraint.  A finite
net argument then gives compactness: fix a displacement scale
$d>4\eta_j$; starting at the first endpoint, stop successively when the
curve first reaches distance $d$ from the preceding stop, inserting the
marked times as well.  Apart from a bounded number of stops adjacent to
inserted marks, different stop starts are $\delta_j$-separated;
packing in the fixed disk bounds the number of stops.  Every portion
between stops is contained in a $d$-ball about its start.

Joining the stops by segments and rounding their positions to a
sufficiently fine finite grid gives a finite marked Fr\'echet net, with
error tending to zero as $d\to0$.  A standard compactness/diagonal
argument gives the limit.
\end{proof}


\section{Standard spirals: the geometric construction}
\label{ssec:standard-spirals}

This appendix supplies the geometric lemma referenced in
Section~\ref{sec:affine}: that every fixed deformation class of
endpoint connectors from a terminal of type $\tau_s$ to a terminal of
type $\tau_t$ with two prescribed winding integers $n_s,n_t$ has a
\emph{standard spiral representative} with at most $O(1+|n_s|+|n_t|)$
short pieces.

\subsection{The starting configuration}

Fix nominal terminal placements, their types $\tau_s,\tau_t$, and the
two winding integers $n_s,n_t$.  Put $k=|n_s|+|n_t|$.  Let $D_1,D_2$ be
two scale constants.  In the reference coordinates of the two terminals,
choose two disjoint \emph{start insertion} arcs: one from the positive
port of the start terminal heading east to a horizontally-aligned
reference copy at distance $D_1$; one from the arrival port of the end
terminal, entered by a reference copy at distance $D_1$ west of it.  By
rigid motion, put the start reference copy at origin in normalized
coordinates, and the end reference copy at $Z_t-Z_s+h\cdot\Lambda$
where $\Lambda$ is a fixed complex offset.

\subsection{The spiral step}

The spiral is built by peeling \emph{turning legs} off the log-spiral
curve between the two reference copies.  Each turning leg is an arc of
a log-spiral curve at a fixed orientation angle $\delta>0$, of radial
growth $\Delta r=D_2$.  Because the two terminals can be placed at
arbitrary complex separation $|z_t-z_s|>Ch(1+k)$ (by choice of reference
copy), we may impose an intermediate equal-spaced grid of radial stations
$\{r_j=j\cdot D_2\}_{j=0}^{J_0}$ in the annular polar coordinates of
each terminal.

At the start of the spiral, insert the fixed start insertion followed
by several turning legs on the starting-terminal page.  Then step
across via a straight transport, then several turning legs on the
arrival-terminal page, etc.  The exact alternation of turning legs and
straight transports is fixed before any picture.  The total number of
turning legs is exactly $2+k'$, where $k'$ is an O(1)-determined
function of $k$; each full spiral collection has $O(1)$ turning legs.

\begin{lemma}[Bend angle bound]
\label{lemma:bend}
A turning leg from radius $r$ to $r+D_2$ with angular increment
$0<\gamma\le\delta$ has its direction relative to its initial radial ray
between $0$ and $\pi/2+\delta/2$; relative to its final radial ray the
incoming direction is between $0$ and $\pi/2$.  Thus all bends,
including radial joins and a partial last leg, have absolute turning
angle at most
\[
   \beta_*=\pi/2+\delta/2<\pi.
\]
\end{lemma}

\begin{proof}
The leg from radius $r$ to $r+D_2$ with angular increment $\gamma$ has
direction relative to its initial radial ray
\[
   \theta(r,\gamma)=\gamma/2+\arctan\Bigl(\frac{2r+D_2}{D_2}\tan\frac{\gamma}{2}\Bigr).
\]
For $0\le\gamma\le\delta\le\pi/2$, this lies in
$[0,\pi/2+\delta/2]$.  Reflection gives the opposite sign.  Since
$\delta<\pi$, $\beta_*<\pi$.
\end{proof}

\subsection{Non-self-intersection}

Radial gaps on the starting-terminal page ensure Euclidean clearance
$c_0 D_2$ between nonconsecutive full turning legs, for a fixed
$c_0>0$.  A partial last leg is on the inner side of the hypothetical
full leg, with clearance at least $c_0D_2$ from all other nonincident
legs, by elementary comparison of radial graphs.  The two terminals
have their legs in disjoint polar annuli when $D_2$ is much larger than
$E_0,D_{\mathrm{st}}$ (the central scale of the pin-core region) and
$c_0^{-1}$.

\subsection{Compactness of short pieces}

Each turning leg, including partial ones, and each short corner piece
is a short buffered experiment at a bounded normalized size.  Bend
angles range over the compact interval $[-\beta_*,\beta_*]$.  A buffered
experiment at one angle persists in a neighbourhood of it (move the
terminal copy together with both its stubs by a small rigid motion,
taper in a disjoint collar, deform the strip slightly).  A finite
subcover of $[-\beta_*,\beta_*]$ absorbs this, giving an $O(1)$ bound on
the number of short pieces across the compact template family.  Common
constants $C_{\rm short}$ come out by Proposition~\ref{prop:affine-means}.

\subsection{The standard-spiral lemma}

\begin{lemma}[Standard spirals and short corner templates]
\label{lemma:standard-spirals-app}
Under the above setup, there is a $C<\infty$ such that if
$|z_t-z_s|>Ch(1+k)$, the class has a buffered representative beginning
and ending with the fixed insertions above.  Between them it consists of
straight transports and at most $C(1+k)$ short pieces.  The short pieces
belong, up to rigid motion, to uniformly buffered compact families of
bounded normalized size, and satisfy
\[
   \|i_j^\eps\|_{L^2}\le C_{\rm short}h
\]
for small $\eps$, with a constant independent of the terminal separation,
the twists, and the particular long connector.

A common complementary connector closes both the inserted path and its
unsplit representative.  All auxiliary placements and straight lengths
can be fixed from the nominal terminal data and integers before making
allowed chart changes.  In the applications they form a countable list.
\end{lemma}

\begin{proof}
Assemble the start insertion, then the $O(1)$ turning legs on the
starting page, then the straight transport, then the $O(1)$ turning legs
on the arrival page, then the end insertion.  The bend bound controls
turning angles; the radial gaps control self-intersection; the compact
templates give the $C_{\rm short}h$ $L^2$ bound via
Proposition~\ref{prop:affine-means}.  The complementary connector is the
fixed unused arc, chosen remote enough to avoid all turning disks and
the used gap.
\end{proof}


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\end{document}