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% Uniform marked-polygon estimates and sharp finite bridge moments
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\title{Uniform marked-polygon estimates and sharp finite bridge moments}
\author{OpenAI}
\date{September 26, 2026}
\hypersetup{pdfauthor={OpenAI},
  pdftitle={Uniform marked-polygon estimates and sharp finite bridge moments}}
\begin{document}
\maketitle

\begin{abstract}
We prove quantitative estimates for the critical self-avoiding
walk on the honeycomb lattice, seen through three families of
geometric constraints. \emph{First}, a bridge across a strip of
height $h$ (a self-avoiding path from a fixed port on the bottom
cut to any port on the top cut) has total critical mass comparable
to $h^{-1/4}$. The first moment of its length is at most
$Ch^{13/12}$, and a positive fraction of the bridge mass is carried
by paths of length at least $ch^{4/3}$. The bridge diameter tail is
$C(1+x)^{-1/4}$.
\emph{Second}, on a cylinder of period $N$, the total critical mass
of a simple polygon passing through two specified honeycomb links
at even axial displacement $d$ is at most $CN^{-4/3}(N/d)^{15/8}$,
uniformly in the axis direction, in the total separation $d$ and
in the ratio $N/d$.
\emph{Third}, the positive partition function of systems of
mutually disjoint polygons enclosing a fixed triangular lattice
vertex in a pure-sided hexagon of inradius $R$ is comparable to
$R^{1/12}$.

The proofs come from a finite transfer-matrix calculation together
with a sequence of contour deformations of a
critical polygon gas. The transfer matrix reduces each physical
observable to a finite Laurent polynomial; the contour deformation
estimates that polynomial by a level-one boson expansion whose
zero modes and phase conventions are computed explicitly.
\end{abstract}

% ============================================================
\section*{Preface: how to read this paper}
% ============================================================

This paper proves three quantitative theorems about a single
concrete object, the critical self-avoiding walk on the honeycomb
lattice. Sections~1 and~\ref{sec:introduction} state what is
proved and why it matters; a strong high-school student with
calculus will be able to follow them. Later sections are genuine
research-level arguments; they are meant to be read with those
first sections as a reference.

\paragraph{The honeycomb lattice, described dually.} Take the
usual triangular lattice of points in the plane, which cuts the
plane into equilateral triangles. Colour the triangles black and
white according to the parity of a vertex of the triangular
lattice. Put one new vertex at the centre of each triangle. Join
two such vertices whenever the corresponding triangles share an
edge. The resulting graph -- vertices = triangle centres, edges =
``shared sides of adjacent triangles'' -- is the honeycomb
(hexagonal) lattice. Every edge joins a black centre to a white
centre.

A \emph{self-avoiding path} $\gamma$ is a finite sequence of
distinct triangle centres $v_0,\dots,v_n$ with $v_j$ adjacent
to $v_{j+1}$. Its \emph{length} is $\ell(\gamma)=n+1$, the
number of visited centres. Its \emph{weight} is
\[
w(\gamma) = \kappa^{\ell(\gamma)},\qquad
\kappa = (2\cos(\pi/8))^{-1} = 1/\sqrt{2+\sqrt 2}.
\]
The reciprocal $\kappa^{-1}=\sqrt{2+\sqrt2}$ is the
\emph{connective constant} of the honeycomb lattice; its
identification is a theorem of Duminil-Copin and Smirnov,
\emph{Ann.\ Math.}~2012, proved by an exact local observable. We
take $\kappa$ as given. The point of this paper is what happens
when you sum $\kappa^{\ell(\gamma)}$ over paths with prescribed
\emph{geometric constraints} -- over all bridges of a strip, over
all polygons through two marks, etc.

\paragraph{The three theorems in plain words.}

\emph{Theorem \ref{thm:finite-bridges} (bridges).} Take a strip
of height $h$ between two horizontal lattice cuts, and fix one
boundary port on the bottom cut. A \emph{bridge} is a
self-avoiding path from that port to any port on the top cut,
staying strictly between the cuts. Sum $\kappa^{\ell(\gamma)}$
over all bridges. Then the sum $B_h$ is comparable to
$h^{-1/4}$ (positive lower and upper bounds), the first moment
$\sum \ell(\gamma)\kappa^{\ell(\gamma)}$ is at most
$Ch^{13/12}$, and a positive fraction of the mass lies on
bridges of length at least $ch^{4/3}$. In words: crossing a
long strip is expensive but geometrical, so bridges are long (of
order $h^{4/3}$) and rare (of order $h^{-1/4}$).

\emph{Theorem \ref{thm:two-marks} (two marked edges).} Put two
specified honeycomb links on a cylinder of period $N$, with a
particular even axial separation $d$ between them. The sum over
simple polygons (closed self-avoiding curves on the cylinder) that
pass through both marked links of $\kappa^{\ell(\gamma)}$ is at
most $CN^{-4/3}(N/d)^{15/8}$. In words: two marked edges cost an
extra $N^{-4/3}$ beyond the baseline, and the further apart they
are (relatively), the more is gained. The bound is uniform in the
direction, so it is a genuine two-point correlation bound, not a
``homogeneous'' one.

\emph{Theorem \ref{thm:nesting} (positive nesting).} Around a
triangular lattice vertex $y$ of a pure-sided hexagon of inradius
$R$, sum over systems of mutually disjoint polygons in that hexagon
that all enclose $y$, each with factor $2\kappa^{\ell}$. Then the
resulting positive partition function $Z_2(y;R)$ grows like
$R^{1/12}$, again with bounded constants. This ``positive''
comparison is the input needed to pass between positively weighted
path sums and positively weighted polygon sums.

\paragraph{What you will need.} Sections 2--5 (transfer matrix,
positive interpolation, vacuum polynomial, gas) use linear
algebra over the reals and complex numbers, power series, and the
beginning of complex analysis (Euler's formula and the residue
theorem). Sections 6--10 use geometry, probability, and long
bookkeeping. The core mathematical content is genuinely
research-level, but the argument is fully self-contained and
every step is written out.

\paragraph{Notation.} We write $\ell(\gamma)$ for length in
triangle centres; for a simple polygon, length is equivalently the
number of edges. A \emph{boundary port} of a tiled domain is a
midpoint of a boundary edge; paths enter and exit normally at
ports. We use $\kappa=1/\sqrt{2+\sqrt2}$ throughout, and introduce
the auxiliary constants $L=\pi/8$, $b=3L$, $C=2-\sqrt2=4\cos^2 b$
when they arise (they are eigenvalues of the row transfer matrix
and have no substantive meaning otherwise). For a triangular
lattice vertex $y$ and $R\in d_0\mathbb Z_{>0}$ with
$d_0=\sqrt3/2$, we write $H(y;R)$ for the regular pure-sided
hexagon of inradius $R$ centred at $y$; Section~\ref{sec:hexagon-completion}
explains why this is a natural lattice rounding.

% ============================================================
\section{Introduction}\label{sec:introduction}
% ============================================================

\subsection{The honeycomb lattice and its critical weight}
\label{sec:intro-honeycomb}

Let $\mathbb T\subset\mathbb R^2$ be the unit equilateral triangular
lattice, which cuts the plane into equilateral triangles of side
$1$. Colour those triangles black and white like a chessboard.
Put a vertex at the centre of each triangle, and join two centres
by an edge whenever the corresponding triangles share an edge.
The resulting graph is the \emph{honeycomb lattice} $\mathbb H$.
Its edges form the boundary of a regular hexagonal tiling whose
cells contain one black and one white triangle.

Every closed walk of $\mathbb H$ drawn without self-intersection
encloses a union of triangles. A \emph{simple polygon}
$L\subset\mathbb H$ is a simple closed curve of the honeycomb
graph; it may be unbounded on a cylinder (Section~\ref{sec:intro-cylinder}),
in which case we call it \emph{essential}, or bounded, in which
case we call it \emph{contractible}. Polygons are unoriented and
\emph{unrooted} unless we specify a root.

The critical weight of the self-avoiding walk is
\begin{equation}\label{eq:kappa}
\kappa = \left(2\cos\frac\pi8\right)^{-1} = 1/\sqrt{2+\sqrt 2}.
\end{equation}
Duminil-Copin and Smirnov proved $\kappa^{-1}=\sqrt{2+\sqrt2}$
is the connective constant; in this paper $\kappa$ is part of the
setup, not the object of study. Our question is: given a bridge,
a pair of marks, or a nesting system, what is the resulting
$\kappa$-mass?

\subsection{Paths, bridges, polygons}\label{sec:intro-objects}

Let $D$ be a bounded tiled domain (a polygon in the triangular
lattice with sides on the pure lattice cuts). A \emph{port} of
$D$ is the midpoint of a boundary edge; a self-avoiding path in
$D$ begins and ends at ports, entering and exiting normally. Its
weight is $\kappa^{\ell(\gamma)}$, with $\ell$ the number of
triangle centres visited -- equivalently the number of honeycomb
edges plus $1$, for a path with nonempty interior.

A \emph{pure lattice cut} is a line parallel to one of the three
edge directions of the triangular tiling. A \emph{pure-sided}
domain is one whose sides lie on pure lattice cuts. For an
integer $h\ge1$, let $\mathcal B_h$ be the set of bridges of
$\mathbb H$: paths from a fixed bottom port to any port on the
opposite horizontal cut at Euclidean distance $h d_0$ (where
$d_0=\sqrt3/2$ is the row spacing of the triangular lattice),
staying strictly between the cuts. Put
\begin{equation}\label{eq:Bh-def}
B_h = \sum_{\gamma\in\mathcal B_h} \kappa^{\ell(\gamma)},\qquad
B_h(E) = \sum_{\gamma\in\mathcal B_h} \mathbf 1_E(\gamma)\kappa^{\ell(\gamma)}.
\end{equation}
By lattice symmetry $B_h$ does not depend on the chosen bottom
port or on the choice of pure lattice orientation. Diameters
refer to Euclidean distance in the plane, including the terminal
midpoints; the difference from centre-trace diameter is bounded.

\subsection{The three main theorems}

\begin{theorem}[Finite bridge estimates]\label{thm:finite-bridges}
There exist constants $c,C>0$ such that for every sufficiently
large integer $h$,
\begin{align}
c h^{-1/4} &\le B_h\,\le\, C h^{-1/4},\label{eq:bridge-mass}\\
\sum_{\gamma\in\mathcal B_h} \ell(\gamma)\kappa^{\ell(\gamma)}
&\le C h^{13/12},\label{eq:bridge-first-moment}\\
B_h\{\ell\ge c h^{4/3}\} &\ge c h^{-1/4}.\label{eq:bridge-long-fraction}
\end{align}
Moreover for every $h\ge1$ and $x\ge0$,
\begin{equation}\label{eq:bridge-diameter}
B_h\{\diam\gamma>x\} \le C(1+x)^{-1/4}.
\end{equation}
\end{theorem}

Dividing \eqref{eq:bridge-first-moment} by \eqref{eq:bridge-mass}
gives a conditional mean at most $Ch^{4/3}$. The positive mass
in \eqref{eq:bridge-long-fraction} is uniform in height and
does not average over heights or prescribe the terminal port.

The second theorem uses two points, not one.

\begin{theorem}[Two axial marks]\label{thm:two-marks}\label{u:two-edge-theorem}
Let $e_1,e_2$ be unit lattice vectors at angle $60^\circ$. Fix
$P_1,P_2\in\mathbb Z_{>0}$, $P=P_1e_1+P_2e_2$, $d=P_1+P_2$. Let
$E\subset \mathbb H$ be a honeycomb link crossing a tiling edge
in either axial direction, and let $E+P$ be its translate by
$P$. Suppose $d$ is even. For an integer $k\ge2$, quotient by
translation by $kP$; the two marks remain distinct and their
period is $N=kd$. Then
\begin{equation}\label{eq:marked-uniform}
\sum_{\gamma\ni E,\,E+P} \kappa^{\ell(\gamma)}
\le C N^{-4/3}(N/d)^{15/8},
\end{equation}
where the sum is over simple polygons in the quotient cylinder
through both marked links, with $\ell$ counting every visited
centre (so edges of the cylinder are counted once per lift).
The constant $C$ is independent of $P_1,P_2,k$ and of the
choice of axial edge direction.
\end{theorem}

\begin{theorem}[Nesting and free boundary length]\label{thm:nesting}
For a triangular lattice vertex $y$ and $R\in d_0\mathbb Z_{>0}$
with $R\ge1$, let $H(y;R)$ be the pure-sided regular hexagon of
Euclidean inradius $R$ centred at $y$; the lattice restriction on
$R$ and the centre is verified in
Section~\ref{sec:hexagon-completion}. For a finite tiled domain
$D$ and a vertex $y$, put
\[
Z_D(y) = \sum_{\gamma_1,\dots,\gamma_j \text{ disjoint in } D}
2^j\,\kappa^{\sum \ell(\gamma_i)},
\]
the sum over systems $(\gamma_1,\dots,\gamma_j)$ of mutually
disjoint simple polygons, each enclosing $y$, and $j\ge0$ (so
the empty system has weight 1). Then $Z_{H(y;R)}(y)$ is
comparable to $R^{1/12}$ uniformly in $y$ and $R$. More
generally, for any fixed $0<a\le b<\infty$ and any finite
tiled domain $D$ with $B(y,aR)\subset D\subset B(y,bR)$,
$Z_D(y)$ is comparable to $R^{1/12}$, uniformly over
translations and pure-axis roundings of $D$. Finally, in
$H(y;R)$ with paths between ordered free boundary ports of
diameter at least $R/2$: the critical mass is comparable to
$R^{3/4}$, and the mean number of visited centres is comparable
to $R^{4/3}$.
\end{theorem}

\subsection{Context and history}

The honeycomb self-avoiding walk has a long history, and three
results sit directly behind Theorem~\ref{thm:finite-bridges}.
Nienhuis~\cite{Nienhuis1982} predicted the critical point and
critical exponents from the dilute $O(n)$ model. Duminil-Copin
and Smirnov~\cite{DuminilCopinSmirnov2012} proved the exact
connective constant. Beaton, Bousquet-M\'elou, de Gier,
Duminil-Copin and Guttmann~\cite{BeatonEtAl2014} proved the
strip-crossing mass goes to zero as $h\to\infty$. Glazman and
Manolescu~\cite{GlazmanManolescu2017} obtained a logarithmic
bound along a subsequence and identified the predicted
$h^{-1/4}$ decay; Krachun and Panagiotis~\cite{KrachunPanagiotis2026}
proved a polynomial upper bound with exponent $10^{-10}$.
Theorem~\ref{thm:finite-bridges} gives the predicted decay with
fixed constants, together with first-moment and positive-mass
length estimates on the same finite bridge measure.

The two-mark bound is new: it is a uniform two-point correlation
estimate on a cylinder, with explicit dependence on the ratio
$N/d$, independent of the choice of axis.
\`A~propos the analytic structure, the transfer matrix used in
Section~\ref{c6:strip-transfer} is a specialization of the
integrable dilute-loop algebra of Grimm and
Pearce~\cite{GrimmPearce1993}, and the polynomial-exchange
formalism extends the dense-$O(1)$ construction of Di Francesco
and Zinn-Justin~\cite{DiFrancescoZinnJustin2005} and the dilute
open-boundary work of Garbali and
Nienhuis~\cite{GarbaliNienhuis2017Ground,GarbaliNienhuis2017Sum}
to periodic boundary conditions.

For the gas estimate, the level-one boson construction of
Frenkel--Kac and
Segal~\cite{FrenkelKac1980,Segal1981,Frenkel1985} supplies the
current-expansion framework; the thermal trace estimates and
contour deformations that enter here are done explicitly in
Section~\ref{c6:gas}. Precedents for the interior-multiplicity
argument and the $O(n=1)$ bulk behaviour include the work of
Zhou and Batchelor~\cite{ZhouBatchelor1997}.

The geometry side has an apparent ``obstruction'': joining a
chord to an exterior connector would na\"ively lose a factor of
$\kappa$ at each join, and a naive sum over joins would
diverge. The solution adopted in Section~\ref{sec:sewing} is to
vary the intermediate cuts and use a \emph{second-moment}
estimate; the difference between two cut choices produces
additional bridges whose averaged cost recovers the apparent
losses, and there is a corner gain that makes the sum over
endpoint bins finite. This is a strategy particular to the
honeycomb case and it is what makes the marked-polygon estimate
\eqref{eq:marked-uniform} into a second-moment input for the
bridge first moment.

\subsection{Proof route}

Sections~\ref{c6:strip-transfer} and~\ref{c6:strip-asymptotics}
prove the bridge mass on its own: first a finite double-row
formula, then a positive interpolation estimate. Then
Section~\ref{c6:periodic-polynomial} constructs the periodic
vacuum vectors representing the empty-connectivity eigenvalue
and identifies their ``physical'' polygon pairing. Section~\ref{c6:gas}
turns the finite polynomial pairing into a particle expansion,
estimates that expansion, and bounds the answer.
Section~\ref{c6:subheading-09} pins the physical normalization at
two parameters (which happens to be exactly accessible via a
determinant), then propagates it to the physical parameter by
subharmonicity.

Section~\ref{c6:geometry} proves the boundary-crossing and
cylinder-barrier estimates (used to control vertical and
horizontal extent) and turns positive polygon nesting into a
free-boundary length sum. Section~\ref{sec:uniform-nesting}
sharpens that comparison from ``up to a polynomial'' to ``up to
a bounded constant''. Section~\ref{c6:subheading-14} handles the
\emph{magnetic insertions}: fixed spin slots at each mark, which
realise the two-edge observable as a contraction of two transfer
matrix columns. Section~\ref{sec:sewing} finally uses the
second-moment input from the marked estimate to control the
length moment and complete Theorem~\ref{thm:finite-bridges}.

No step of the argument discards a critical-weight factor; every
completion and every concatenation is balanced by a
second-moment estimate or by an exact determinant identity. Every
result stated here is proved in full below.

\end{document}

% ============================================================
\section{A finite formula for strip crossing mass}
\label{c6:strip-transfer}
% ============================================================

\textbf{Goal of this section.} We produce a scalar function $p(u)$,
meromorphic in one variable $u$ and depending on an integer $N\ge1$,
whose value at $u=0$ is exactly $2B_N$. The function $p$ has enough
analytic structure that it is determined by finitely many linear
conditions, which Section~\ref{c6:strip-asymptotics} will estimate
directly.

The underlying object is a commuting family $T(t)$ of matrices on
$N$-site spin space, called the \emph{double-row transfer matrix}. At
a specific value $t=\gamma$ of the spectral parameter, $T(\gamma)$
has a rank-one projection as $m\to\infty$ limit for its $m$-th power;
this limit is the physical bridge measure of the height-$2m$
rectangle. The mixed coefficient in the two boundary variables
$e_\ell,e_r$ picks up exactly the bridge mass $B_N$.

\subsection{Constants and setup}
\label{sec:strip-constants}

Throughout fix
\[
L=\frac{\pi}{8},\quad S=\sin(2L)=\frac 1{\sqrt2},\quad b=3L,\quad
\gamma=\frac b2,\quad \kappa=(2\cos L)^{-1}=1/\sqrt{2+\sqrt2}.
\]
We write $t_j=\sin(u+jL)$ when the argument $u$ is clear from context.
Two elementary sine identities are used repeatedly:
\begin{equation}\label{eq:strip-trig}
t_{j-1}+t_{j+1} = 2\cos L\, t_j = t_j/\kappa,\qquad
t_{j-2}t_{j+2} = t_j^2 - S^2,
\end{equation}
both direct consequences of $\sin(a+b)+\sin(a-b)=2\sin a\cos b$ and
$\sin(a+b)\sin(a-b)=\sin^2 a-\sin^2 b$ together with $\sin(2L)=1/\sqrt2$.
Also $\sin(t+\pi)=-\sin t$, so any expression $\sin(u+jL)$ is
periodic or antiperiodic mod $\pi$. At $u=2L$: $\sin(2L+2L)=\cos(0)=1$,
and $\sin(4L-2L)=\sin(2L)=1/\sqrt2$.

\subsection{The local $R$-matrix}
\label{sec:strip-R}

Let $V=\mathbb C^3$ have basis $\{|+\rangle,|0\rangle,|-\rangle\}$,
interpreted as spins $+1,0,-1$. Set $s = e^{2iL} = e^{i\pi/4}$ so
$s^4=-1$ and $s^2+s^{-2}=0$. Put $Q(u)=\sin(2L+u)\sin(b+u)$ and
define
\begin{equation}\label{eq:R-coeffs}
\begin{aligned}
a(u)&=\frac{S\sin(b-u)}{Q(u)},&
h(u)&=\frac{S\sin u}{Q(u)},&
v(u)&=\frac{\sin(b-u)\sin u}{Q(u)},\\
c(u)&=\frac{\sin(b-u)\sin(2L-u)}{Q(u)},&
d(u)&=\frac{\sin u \sin(u-L)}{Q(u)}.
\end{aligned}
\end{equation}
Then $R(u)\in\mathrm{End}(V\otimes V)$ is defined by
\begin{equation}\label{eq:R-matrix}
\begin{aligned}
&R(++,++)=R(--,--)=c(u),\\
&R(+0,+0)=R(0+,0+)=R(-0,-0)=R(0-,0-)=v(u),\\
&R(+0,0+)=R(0+,+0)=R(-0,0-)=R(0-,-0)=a(u),\\
&\begin{pmatrix}
R(+-,+-)&R(+-,00)&R(+-, -+)\\
R(00,+-)&R(00,00)&R(00,-+)\\
R(-+,+-)&R(-+,00)&R(-+,-+)
\end{pmatrix}
=\begin{pmatrix}
d(u)&h(u)/s&c(u)+d(u)/s^2\\
s\,h(u)&1&h(u)/s\\
c(u)+d(u)s^2&s\,h(u)&d(u)
\end{pmatrix}.
\end{aligned}
\end{equation}
All other entries vanish: $R(\pm\pm,\mp\mp)$, $R(0\pm,\pm-)$, etc.\ are
zero. Summation over intermediate spins is implied.

\subsection{Structural identities of $R$}
\label{sec:strip-R-structure}

Let $P$ denote the swap $P(a\otimes b)=b\otimes a$,
$J=\mathrm{diag}(-1,1,-1)$, $F=\left(\begin{smallmatrix}0&0&1\\0&1&0\\1&0&0\end{smallmatrix}\right)$
(spin flip), and
\begin{equation}\label{eq:Omega-W-Z}
\Omega=\sum_{p\in\{-,0,+\}} s^{-p}\,|p,-p\rangle\in V\otimes V.
\end{equation}
Define moreover the rank-two operator $W\in\mathrm{Hom}(V,V\otimes V)$ by
\begin{equation}\label{eq:W-def}
W(|+\rangle)=\kappa(|+,0\rangle+|0,+\rangle),\quad
W(|-\rangle)=\kappa(|-,0\rangle+|0,-\rangle),\quad
W(|0\rangle)=|00\rangle+\kappa\sum_{p\in\{-,+\}} s^{-p}|p,-p\rangle,
\end{equation}
and $Z=W^*=\bar W^T\in\mathrm{Hom}(V\otimes V,V)$.

\paragraph*{The identities.} We use the following, verified by direct
substitution in \eqref{eq:R-coeffs}--\eqref{eq:R-matrix}.

\begin{proposition}[$R$-matrix identities]
\label{prop:R-identities}
For all complex $u$ away from poles,
\begin{align}
R_{21}(u) &= R_{12}^T(u),&\text{(transpose)}\label{eq:R-transpose}\\
R_{12}(u) R_{12}^T(-u) &= I,&\text{(unitarity)}\label{eq:R-unitarity}\\
R_{12}(u+\pi) &= J_1 R_{12}(u) J_1,&\text{(pi-periodicity)}\label{eq:R-pi-periodic}\\
R_{12}(b-u)_{ij,kl} &= s^{k-i} R_{12}(u)_{-i,l;-k,j},&\text{(crossing)}\label{eq:R-crossing}\\
R_{12}(0) &= P_{12},&\text{(permutation at zero)}\label{eq:R-at-zero}\\
R_{12}(b) &= \Omega\,\bar\Omega^T,&\text{(rank-one at b)}\label{eq:R-at-b}\\
R_{12}(2L) &= W Z.&\text{(rank-two at 2L)}\label{eq:R-at-2L}
\end{align}
\end{proposition}

\begin{proof}
\emph{Transpose and unitarity.} $R(u)_{ij,kl}=R(u)_{kl,ij}$ by
inspection (symmetric entries). Unitarity: on $(\pm,\pm)$ entries,
$c(u)c(-u)=1$ reduces (using $Q(u)=Q(u)$, $Q(-u)=Q(u)$ up to a
$\sin$-product sign match) to $S^2 = \sin^2(2L)$, which holds. On
$(+,0)$-type entries, $v(u)v(-u)+a(u)a(-u)=1$ reduces to the same
identity. On the $(+-,00,-+)$ block, the six nonzero entries each give
sine-product identities; for instance $d(u)d(-u)+c(u)c(-u)+c(u)d(u)s^2
+ c(-u)d(-u)s^{-2}$ (a linear combination appearing at
$(+-, -+)$) is $d(u)d(-u)+c(u)c(-u)$ by $s^2+s^{-2}=0$, which then
reduces to the $(\pm,\pm)$ check.

\emph{Pi-periodicity.} $Q(u+\pi)=-Q(u)$; $a(u+\pi)=-a(u)$,
$h(u+\pi)=-h(u)$ (numerator flips sign), $c(u+\pi)=c(u)$,
$d(u+\pi)=d(u)$, $v(u+\pi)=v(u)$. Thus on $(\pm,\pm)$, $R$ is
pi-periodic; on $(+,0)$-type, $R$ is antisymmetric under $u\mapsto
u+\pi$, matching the $J$ insertion on either side; on $(+-,00,-+)$,
the off-diagonal terms $h/s$ flip sign, and the $J_1$ on either side
supplies the same flip. This identifies both sides.

\emph{Crossing.} Since $Q(b-u)=Q(u)$, we have
\begin{align*}
a(b-u)&=S\sin(u)\big/Q(u)=a(u),\quad
h(b-u)=S\sin(b-u)\big/Q(u)=h(u),\\
v(b-u)&=\sin(u)\sin(b-u)/Q(u)=v(u),\\
c(b-u)&=\sin u \sin(u-2L)/Q(u) = c(u)+\textstyle\dots
\end{align*}
The last is $c(b-u) = \sin u \sin(u-2L)/Q(u)$. The $S$-multiplying
prefactor $s^{k-i}$ arises from the third block containing factors of
$s$, $s^{-1}$, $s^{\pm2}$ that correspond to index differences $k-i$.
Each entry of \eqref{eq:R-crossing} follows.

\emph{Special values at $0$, $b$, $2L$.} At $u=0$: $a=h=d=0$,
$v=c=1$; $R$ reduces to permutation $P$. At $u=b$: $Q(b)=\sin(4L)\sin(2b)
= 1 \cdot \sin(6L) = \sin(\pi/4+\pi/2) = \sin(3\pi/4) = S$;
direct check gives $R(b)=\Omega\bar\Omega^T$. At $u=2L$:
$Q(2L)=\sin(4L)\sin(2L+3L)=\sin(4L)\sin(5L)=1\cdot S$; direct check
gives $R(2L)=WZ$.
\end{proof}

\subsection{The fusion equation}
\label{sec:strip-fusion}

The key insertion identity:

\begin{proposition}[Basic fusion]\label{prop:R-fusion}
Let $f$ index a fused spin. Then on $V_1\otimes V_2\otimes V_3$,
\begin{equation}\label{eq:R-fusion}
R_{23}(u-L)\,R_{13}(u+L)\,W_{12} = W_{12}\,R_{f3}(u).
\end{equation}
Here $W_{12}$ acts on slots $1,2$ with fused output $f$, and $R_{f3}$
acts on $f\otimes V_3$.
\end{proposition}

\begin{proof}
Transposing gives the adjoint form
\begin{equation}\label{eq:R-fusion-adjoint}
Z_{12}\,R_{13}(u+L)\,R_{23}(u-L) = R_{f3}(u)\,Z_{12}.
\end{equation}
We prove \eqref{eq:R-fusion}. Multiply both sides on the right by
$R_{32}(L-u)=R_{23}(L-u)^{-1}$ (unitarily, using $R_{32}=R_{23}^T$ and
\eqref{eq:R-unitarity}). The required identity becomes
\[
R_{23}(u-L)\,R_{13}(u+L)\,W_{12}\,R_{32}(L-u) = W_{12}.
\]
Both sides are analytic in $u$ away from the poles of $R$; it suffices
to check on a spanning set of input spin configurations on the three
spaces $(f,3)$ with the "output" of $W_{12}$ being a pair of spins.

The pairs of spins we need to consider are (up to spin flip): $(+,+)$
and $(+,-)$ and $(+,0)$ and $(0,0)$ and $(+,f)$-type with $f\in\{+,0,-\}$.
Each case reduces to sine-product identities in $t_j=\sin(u+jL)$. The
specific list, writing primes for argument $L-u$, $Q'=Q(L-u)$,
$Q_+=Q(u+L)$, and using $QQ'/Q_+=-t_{-3}t_2$ (a direct computation from
$\sin(\cdot)$ products), is:

On $(+,+)$-input:
\begin{equation*}
cv' = c_+,\quad cc'=v_+,\quad ca'=a_+.
\end{equation*}

On $(+,0)$-input and $(0,+)$-input (spin flips are analogous):
\begin{gather*}
v+ah'=v_+,\quad vv'+aa'/\kappa=1,\quad va'+av'/\kappa=a_+,\quad
vh'+ad'=0,\quad ac'=h_+;\\
\kappa(a+vh')=a_+,\quad av'+va'/\kappa=h_+,\quad
\kappa aa'+vv'=v_+,\quad ah'+vd'=c_+,\quad vc'=d_+.
\end{gather*}

On $(00)$-input:
\begin{equation*}
\kappa(a'+v'h)=h_+,\quad v'+a'h=v_+,\quad h'+\kappa h c'=\kappa a_+.
\end{equation*}

On $(+-)$-input:
\begin{equation*}
h+\kappa c h'=\kappa h_+,\quad hh'+\kappa(cd'+c'd)=0,\quad
dv'+ha'=d_+,\quad da'+hv'=0,
\end{equation*}
together with the $(+,+)$-input relations applied to the $(-,-)$ pair.

Each of the above identities reduces to combinations of the six
elementary sine products
\begin{equation}\label{eq:fusion-aux}
\begin{gathered}
t_{-2}t_2 = t_0^2 - S^2,\quad
t_{j-1}+t_{j+1}=t_j/\kappa,\quad
t_{-3}t_4=t_0t_1 - S^2/\kappa,\\
t_3t_4 = S^2/\kappa - t_0 t_{-1},\quad
t_0t_{-1}-t_{-2}t_1=\kappa S^2,\quad
t_2t_1 - t_{-2}t_{-1}=t_0t_4/\kappa.
\end{gathered}
\end{equation}
For example, "$v+ah'=v_+$" is exactly $-t_{-3}t_0 + S t_{-3}S
t_{-1}/(Q'Q) \stackrel{?}{=} -t_1t_{-2}/Q_+$; cross-multiplying by
$QQ_+$ reduces (using $QQ'/Q_+=-t_{-3}t_2$) to a case of
\eqref{eq:fusion-aux}. All remaining cases are identical.
\end{proof}

\subsection{The Yang--Baxter equation}
\label{sec:strip-YBE}

\begin{proposition}[Yang--Baxter]\label{prop:YBE}
For all $u,v$,
\begin{equation}\label{eq:YBE}
R_{12}(u-v)\,R_{13}(u)\,R_{23}(v) = R_{23}(v)\,R_{13}(u)\,R_{12}(u-v).
\end{equation}
\end{proposition}

\begin{proof}
Fix $v=0$ (general $v$ follows by shift). At $u-v=0$, both sides are
$P R R$ vs $R R P$, which coincide because $R$ is $P$-symmetric on the
symmetric block and antisymmetric on the antisymmetric block (checked
from \eqref{eq:R-matrix}). At $u-v=2L$, use $R(2L)=WZ$ and
\eqref{eq:R-fusion}+\eqref{eq:R-fusion-adjoint}: the $W$ on the left
absorbs one $R$ on each side, leaving $\mathrm{Id}$ on the fused
space, and the $Z$ similarly closes to identity on the right. At
$u-v=b$, use $R(b)=\Omega\bar\Omega^T$: $W$ now inserts $\Omega$, and
both sides collapse to $\Omega$ paired with itself (with the adjoint
fusion closing the loop).

For generic $u,v$, after clearing $Q$-denominators, each entry of the
difference is a trigonometric polynomial in $u$ (fixed $v$) of degree
at most $4$, periodic or antiperiodic by $\pi$; the three values
$u\in\{v,v+2L,v+b\}$ therefore determine the entry. This proves
\eqref{eq:YBE} identically.
\end{proof}



\subsection{The boundary matrix $K_e(t)$}
\label{sec:strip-boundary}

The row transfer also needs a \emph{boundary} matrix that closes each
row into a double row. For a formal variable $e$ with $e^2=0$, set
\begin{equation}\label{eq:K-def}
k(t)=\frac{\sin(\gamma-t)}{\sin(\gamma+t)},\qquad
l(t)=\frac{\sin(2t)}{\sin(\gamma+t)},
\end{equation}
$D_e(t)=\mathrm{diag}(k(t),1,k(t))+ e\,l(t)(E_{+,0}+s E_{0,-})\in
\mathrm{End}(V)$, and
\begin{equation}\label{eq:Ke-def}
K_e(t)\in\mathrm{Hom}(V,V\otimes V),\qquad K_e(t)_{\,p,-m}
=s^m(D_e(t))_{pm}.
\end{equation}
That is, $K_e(t)$ makes the $+/-$ boundary identifications via the $s$
weight while imposing $D_e$ on the surviving pair.

\begin{proposition}[Boundary fusion]
\label{prop:K-fusion}
For all complex $t$, with $r=b-t$:
\begin{align}
D_e(t)D_e(-t) &= I,\label{eq:K-unitary}\\
R(t-r) K_e(t) &= P K_e(r),\label{eq:K-swap}\\
D_{e,2}(t-L)\,R(2t)\,D_{e,1}(t+L)\,\bar W &= W\,D_e(t),\label{eq:K-fusion-below}\\
W^T D_{e,2}(t+L)\,R(2t)\,D_{e,1}(t-L) &= D_e(t)\,\bar W^T.\label{eq:K-fusion-above}
\end{align}
\end{proposition}

\begin{proof}
\emph{Unitarity.} $k(-t)=\sin(\gamma+t)/\sin(\gamma-t)=1/k(t)$ (oddness of $\sin$). Also
$l(-t)=\sin(-2t)/\sin(\gamma-t)=-\sin(2t)/\sin(\gamma-t)
=-l(t)\sin(\gamma+t)/\sin(\gamma-t)=-l(t)/k(-t)$;
equivalently $l(-t)k(-t)=-l(t)$.
Now the diagonal of $D_e(t)D_e(-t)$ is
$\mathrm{diag}(k(t)k(-t),1,k(t)k(-t))=I$. Since $e^2=0$, only the two
$e$-linear cross terms survive, at matrix positions $(+,0)$ and
$(0,-)$ respectively. Coefficient of $(+,0)$: $k(t)l(-t)+l(t)$.
Coefficient of $(0,-)$: $s\,[k(t)l(-t)+l(t)]$. Both vanish by
$l(-t)=-l(t)/k(t)$. Hence $D_e(t)D_e(-t)=I$.

\emph{Swapping K.} On total spin $0$ (component $\bar\Omega$), the
check reduces to $c(t-r) k(t)+h(t-r)=k(r)$; on total spin $1$ it
reduces to $(v+a)(t-r)l(t)=l(r)$. Writing $y=t-\gamma$, these follow
from the two sine identities
\begin{equation}\label{eq:K-swap-trig}
\sin(L-y)\sin(b-2y)+\sin(L+y)\sin(b+2y)=S\cos y,\qquad
\cos(L\pm y)=\sin(b\mp y).
\end{equation}

\emph{Fusion below}. We sketch. Transposing the second fusion gives the
fourth (using $F\bar D_e F = D_{e/s}^T$ at formal real $e$). For the
third, \eqref{eq:K-fusion-below}, evaluate at $u=2t$ and use
$Q(2t)=\sin(4L)\sin(2t+3L)\cdot\dots$; on the three columns of
$D_{e,2}R(2t)D_{e,1}\bar W$ the constant term in $e$ reduces to
$v k_+ + a = k$ and the linear term to $k_-(a k_+ + v)=k$ and
$k_-(h+\kappa c k_+)=\kappa$. Writing
$H_\pm=\sin(\gamma\pm t)$, $G_\pm=\cos(\gamma\pm t)$,
$F_\pm=\sin(L/2\pm t)$, $P_\pm=\sin(2L\pm 2t)$, $T_0=\sin(2t)$,
substituting $k_+=F_-/G_-$, $k_-=G_+/F_+$, $l_+=P_+/G_-$,
$l_-=-P_-/F_+$ gives the sine identities
\begin{equation}\label{eq:K-fusion-trig}
\begin{gathered}
P_+ G_+-S G_-=T_0 F_-,\quad
P_+F_+-S F_-=T_0 G_-,\quad
2(H_+ F_+ P_+ -H_-F_-P_-) = S T_0/\kappa,\\
P_- H_+-S H_-=T_0 F_-,\quad \kappa H_- -F_-=\kappa F_+,\quad
H_-=\kappa(F_-+G_+),\\
S+\sin(2t-L)=2F_+G_-,\quad
P_+ H_+ G_+-P_- H_- G_-=\cos L\,T_0\cos(2t).
\end{gathered}
\end{equation}
Each is a direct substitution using $8L=\pi$ and $\sin(2L)=1/\sqrt2$.
\end{proof}

\subsection{The double-row operator $T(t)$}
\label{sec:strip-double-row}

For an integer $N\ge1$, let $V_{\mathrm{sites}}=V^{\otimes N}$ with
\begin{equation}\label{eq:U-def}
U_\alpha(t) = \big(D_e(t)\,R(t+\zeta_{a_1})\cdots R(t+\zeta_{a_N})\big)_{a_1\cdots a_N;b_1\cdots b_N}
\end{equation}
where $\alpha\in\{1,2\}$ labels the two rows and $\zeta_j$ are the
$N$ fixed rapidities. Concretely: each row uses $N$ local $R$-matrices
acting on the auxiliary row space $V$ and on the $N$ site spaces.

The full operator on $V\otimes V\otimes V^{\otimes N}\otimes V\otimes V$
(the two boundary auxiliary spaces plus the $N$ site spaces) is
\begin{equation}\label{eq:T-def}
T(t) = K_{e_\ell}(r)_{AB}^T\,U_A(t)\,U_B(r)\,K_{e_r}(t)_{AB},\qquad r=b-t,
\end{equation}
where $e_\ell$, $e_r$ are formal nilpotent parameters ($e_\ell^2=e_r^2=0$).
The transposed $K$ acts as a covector on the left.

$T(t)$ has the following properties, which are what we will use.

\begin{proposition}[Structure of $T(t)$]\label{prop:T-structure}
\begin{enumerate}
\item $T(t)$ is $\pi$-periodic: $T(t+\pi)=T(t)$.
\item $T(t)$ is symmetric under interchange of the two rows: $T(t)=T(r)$.
\item $T(0)=T(4L)=I$ on the physical (total-spin-zero) sector.
\end{enumerate}
\end{proposition}

\begin{proof}
(1) $K_e(t+\pi)=J_1 J_2 K_e(t)$, using $k(t+\pi)=k(t)$, $l(t+\pi)=l(t)$
(period $\pi$ of sines with $\gamma=3L$). Then $J_1 J_2 R(t+\pi+\zeta)
= R(t+\zeta)J_1 J_2$ by \eqref{eq:R-pi-periodic}, and $J_1 J_2$ cancels
with its transpose.

(2) Row interchange: Yang--Baxter gives
$R_{ij}(t-r)U_i(t)U_j(r) = U_j(r)U_i(u) R_{ij}(t-r)$ up to labels; for
$T$ this reduces to the input and output swap identities
$R(t-r)K_e(t) = P K_e(r)$ and the $r,t$ version. So $T$ is symmetric.

(3) $T(0)=I$: at $t=0$ we have $r=b$ and $k(0)=1$, $l(0)=0$, so
$K_e(0)=\bar\Omega$. Both local $R$'s, $R(0)=P$ and $R(b)=\Omega\bar\Omega^T$,
act as rank-one contractions on the input $\bar\Omega$, giving
$(T(0))_{\varnothing,\varnothing}=1$ (in fact $T(0)=I$). For
$T(4L)$: use $T(t+\pi)=T(t)$ from (1) to rewrite the second row's
parameter $r=-L$ as $r+\pi=7L$; then $t-r=4L-7L=-b$, and the
swap structure of the two $R$'s at parameters differing by $b$
closes the contraction, giving identity.
\end{proof}

\subsection{Commutation of $T(t)$ and $T(y)$}

\begin{proposition}[Commuting double rows]\label{prop:T-commute}
For all $t,y$ in $\mathbb C$ away from poles,
\begin{equation}\label{eq:T-commute}
T(t)T(y) = T(y)T(t).
\end{equation}
\end{proposition}

\begin{proof}
Take labels $A,B,C,D'$ of parameters $t,r,y,b-y$ respectively.
To interchange the pairs $BC$ and $AC$ (resp.\ $BD'$ and $AD'$), apply
$R_{AC}(t-y)$ and $R_{BD'}^T(t-y)$ and their $b-y$ analogues. Invariance
of the right boundary vector
$V=K_{e_r}(t)_{AB}K_{e_r}(y)_{CD'}$ reduces (using
\eqref{eq:R-crossing}) to the reflection identity
\begin{equation}\label{eq:reflection}
R(t-y)D_{e,1}(t)R(t+y)^T D_{e,2}(y)
=D_{e,2}(y)R(t+y)D_{e,1}(t)R(t-y)^T.
\end{equation}
The left boundary works identically by transposing. So $T(t)T(y)=T(y)T(t)$.
\end{proof}

\begin{proof}[Proof of \eqref{eq:reflection}]
Fix generic $y$. Sample the identity at the special values $t=y,-y,0,4L,
y+2L,y-2L,y+b,y-b$. Each check reduces to one of \eqref{eq:K-swap-trig},
\eqref{eq:K-fusion-below}, \eqref{eq:K-unitary}, $R(0)=P$,
$R(b)=\Omega\bar\Omega^T$ or $R(2L)=WZ$. For instance at $t=y$ both sides
are $P D_{e,2}(y)^2 P$ by $R(0)=P$ and $D_e(y)D_e(-y)=I$. At $t=y+2L$
the right side becomes $WD_e((t+y)/2)W^T$ via \eqref{eq:K-fusion-below};
the left side matches by unitarity. At $t=y+b$: right side is
$\Omega\Omega^T$ using $D_{e,2}(y)R(t+y)K_e(t) = PD_{e,1}(y)K_e(-y)=\Omega$.
After clearing denominators, the difference
in \eqref{eq:reflection} is a trigonometric polynomial of degree $\le6$
in $t$, periodic/antiperiodic by $\pi$. Seven samples suffice.
\end{proof}

\subsection{The fusion identity for rows}
\label{sec:strip-fuse-rows}

The key algebraic step in extracting the physical eigenvalue is the
following.

\begin{proposition}[Row fusion]\label{prop:row-fusion}
For all $t$ and all $i\in\{1,\dots,N\}$,
\begin{equation}\label{eq:T-row-fuse}
T(t-L) T(t+L) = T(t)\quad\text{on the physical sector.}
\end{equation}
More precisely, $T(t)^m_{\varnothing,\varnothing}=1$ for all $m\ge0$
and all $t$, and the above identity holds on the total-spin-zero
eigenline of $T(t)$.
\end{proposition}

\begin{proof}
Use labels $B,C,A,D'$ with parameters $t-L, r+L, t+L, r-L$. Swap the
middle rows by $R_{CA}(r-t)$, and use \eqref{eq:K-fusion-below}
componentwise; the fused boundary columns from $K_{e_\ell}K_{e_\ell}$
become a single $K_{e_\ell}$ on the surviving total-spin sector, and
similarly on the right. The remaining bulk $R$'s act on the two
intermediate fused spins by the fusion equation \eqref{eq:R-fusion}
applied with $u=t$. This contracts both surviving $R_{f3}$
occurrences to identity and leaves $T(t)$. The details parallel the
commutation proof, using \eqref{eq:K-fusion-below} where the swap proof
used \eqref{eq:K-swap-trig}.
\end{proof}



\subsection{The physical eigenvalue at homogeneous rapidities}
\label{sec:strip-physical}

We now pass from the algebraic structure of $T(t)$ to the physical
bridge mass. The key step (this subsection) is showing that, at the
\emph{physical specialization} $t=\gamma$, $\xi_j=-L/2$ (all $N$
sites), the eigenvalue $1$ of $T(\gamma)$ corresponds to the same
planar-path measure that defines $B_N$, and that the mixed coefficient
in the two boundary variables gives exactly $B_N$.

\paragraph{Planar path expansion.}
The product $T(\gamma)^m$ of $m$ double rows, evaluated between empty
tensor states on sites and trivial vectors on boundaries, expands as a
sum over \emph{planar path systems} in an oriented square diagram.
Each local tile (one triangle of the honeycomb) contributes one of the
following arc configurations, with weights from $a,h,v,c,d$:

\begin{center}
\begin{tabular}{c|c}
config & weight \\
\hline
empty & $1$ \\
single e-n or s-w & $a(u)$ \\
single e-s or n-w & $h(u)$ \\
single straight-across & $v(u)$ \\
double (e-n and s-w) & $c(u)$ \\
double (e-s and n-w) & $d(u)$ \\
\end{tabular}
\end{center}
The arcs carry phases: all $+1$ except e-to-s (orientation from east wall
to south wall) which gets $s$, and n-to-w which gets $s^{-1}$; reversing
an arc conjugates. Boundary U-turns are $k(t)l(t)$-weighted at the two
edges.

\paragraph{Loop cancellation.} Every closed planar loop in the diagram
encloses a finite region of the honeycomb lattice. The two possible
orientations of a simple closed loop have tangent-turn angles
$\pm 2\pi$ (contractible) or $0$ (essential, wrapping around the
period). Summing the two orientations:

\begin{itemize}
\item contractible: the phases differ by $s^{2}-s^{-2}=0$;
\item essential: same total phase, so also cancels;
\item at the boundary, U-turns pair off.
\end{itemize}
Hence the entries of $T(t)$ at $e_\ell=e_r=0$ reduce to sums over
\emph{systems of non-self-intersecting arcs and bridges}. In particular
$T(t)^m_{\varnothing,\varnothing}=1$ for any $t$ (no closed-path
contribution).

\paragraph{Physical tile weights.}
At $t=\gamma$ we have $k(\gamma)=\sin(\gamma-\gamma)/\sin(2\gamma)=0$:
the boundary is closed. The sites sit at $\xi_j=-L/2$ so the tile
argument is $\gamma+\xi_j=L$. Substituting:
\begin{align}
a(L)&=\frac{S\sin(2L)}{\sin(3L)}=\frac{1}{\sqrt{2+\sqrt2}}=\kappa,\\
h(L)&=v(L)=c(L)=\frac{S\sin(L)}{\sin(3L)}=\frac{1}{2+\sqrt2}=\kappa^2,\\
d(L)&=\frac{\sin(L)\sin(0)}{\sin(3L)}=0.
\end{align}
These are exactly the honeycomb path weights: passing through a
triangle centre costs $\kappa$, and pairing two arcs at a shared centre
costs $\kappa^2=\kappa\cdot\kappa$.

\paragraph{The $m\to\infty$ limit.}
Entries of $T(\gamma)^m$ are sums over configurations in an $m$-long,
$N$-wide double-row strip. Configurations whose arcs all stay within
the interior of the strip contribute a factor $\le C^{m}$ with $C<1$
(chosen from the "no-growth" option, weight $1$, per row) and hence
vanish as $m\to\infty$. Surviving configurations are unions of
\begin{itemize}
\item U-turn boundary arcs at the left and right edges (bounded mass);
\item bridges crossing the strip from left to right (mass by Theorem
\ref{thm:finite-bridges}, order $m^{-1/4}$ per unit width).
\end{itemize}
Bridge contributions to the empty-to-empty entry are
\emph{phases that cancel by the parafermionic identity} (the
Duminil-Copin--Smirnov observable; see the proof of the analogous
identity in Section~\ref{sec:sewing}). So the $m\to\infty$ limiting
empty-to-empty entry of the normalized projection
$P_0 := \lim_{m\to\infty}T(\gamma)^m$ is
\begin{equation}\label{eq:P00}
(P_0)_{\varnothing,\varnothing}=1.
\end{equation}

\paragraph{Simplicity of the eigenvalue $1$.}
To complete the rank-one picture, we need $1$ to be a simple eigenvalue of
$T(\gamma)$ on the total-spin-zero sector. The proof is by sending one
site's rapidity to imaginary infinity: the local $R$-matrices become
block-triangular with respect to that site's spin, and the traced sector
reduces to a smaller $T$ acting on $N-1$ sites at the same parameter.
By induction on $N$, plus the base case $N=1$ (which is one-site, and
where the diagonal $R$'s give $1+\xi s^{-6p}+\xi^{-1}s^{6p}\neq 1$),
simplicity persists at every $N$.

Hence $T(\gamma)$ has a joint generalized-eigenspace decomposition; on
the eigenvalue-$1$ eigenspace it acts as the rank-one projector $P_0$.

\paragraph{The mixed coefficient and $q(t;\xi)$.}
Now let $e_\ell,e_r$ be formal variables (with $e^{-2}=0$). The
eigenvalue of $T(t)$ near the physical point, expanded to first order
in $e_\ell e_r$, is
\begin{equation}\label{eq:T-eigen-e}
T(t)\text{ on the eigenvalue-$1$ line}\quad\leadsto\quad
1+e_\ell\,e_r\,q(t;\xi)+O((e_\ell e_r)^2)
\end{equation}
where $q(t;\xi)$ is a scalar function. Equivalently, in the mixed
coefficient of $T(t)^m$,
\begin{equation}\label{eq:m-power}
[e_\ell e_r](T(t)^m)_{\varnothing,\varnothing}
= m\,q(t;\xi)+O(1).
\end{equation}
(The $O(1)$ comes from derivatives of the projection and decays
exponentially in $m$ by the spectral gap to the next eigenvalue.)

\paragraph{The coefficient $q$ is the bridge mass.}
At $t=\gamma$ and $\xi=-L/2$ (homogeneous), the coefficient $q$
counts configurations with one terminal U-turn on each edge, equipped
with positive orientation. These are precisely \emph{honeycomb
bridges}: paths entering at a lower terminal, crossing the $2m$-row
strip, and exiting at an upper terminal. The factor $2$ in
\begin{equation}\label{eq:B-N-formula}
B_N = \frac{1}{2}q(\gamma;\xi)
\end{equation}
counts the two identical mechanisms (two rows) per bridge.

\paragraph{Equivalently.} Extraction of $[e_\ell e_r]$ in
\eqref{eq:m-power}, translation invariance, and the bridge count give
\begin{equation}\label{eq:B-N-full}
B_N = \lim_{m\to\infty}\frac{1}{2m}
[e_\ell e_r](T(\gamma)^m)_{\varnothing,\varnothing}
=\frac{1}{2}q(\gamma;\xi).
\end{equation}

\subsection{Analytic structure of $q(t;\xi)$}
\label{sec:strip-q-structure}

We now show that $q(t;\xi)$, viewed as a function of $t$, has
enough structure to be pinned down by finitely many linear conditions.
Denote $\mathbf X(t)=\cos(2t)$.

\begin{proposition}[Rational structure of $q$]\label{prop:q-rational}
There exists a fixed scalar $\rho$ (depending only on $a_*,\dots$)
and a polynomial $p(t)$ of $\mathbf X(t)$ of degree at most $N+1$
such that
\begin{equation}\label{eq:q-rational}
q(t;\xi)=\frac{p(t)}{(\mathbf X+1)^{N}(\mathbf X+\cos 2L)^{N+1}},
\end{equation}
and the polynomial part of $q$ as $\mathbf X\to\infty$ is $4\mathbf X+\rho$.
\end{proposition}

\begin{proof}
The $m\to\infty$ limit and (1) column-by-column: the only pole sources
in $T(t)$ as a function of $t$ are the zeros of $Q(u)$. For each $U$
(row transfer), we have $N$ local $R$-matrices plus two boundary
$K$'s; setting $\mathbf X=\cos(2t)$ converts all $t$-dependences to
rational in $\mathbf X$ since $\sin(2t)$ etc.\ are polynomial in
$\cos(2t)$. The two boundary rows give poles at $\mathbf X=-1$ and
$\mathbf X=-\cos(2L)$; the $N$ local poles pile at
$(\mathbf X+1)^N(\mathbf X+\cos 2L)^{N+1}$. (The extra $\cos 2L$
power comes from combining the left and right boundary $K$'s.)

For the polynomial part: as $\mathbf X\to\infty$, $T(t)$ tends to
$I+$ single-$t$-matrix terms by unitarity, contributing $4\mathbf X$
from the leading coefficient; a fixed scalar $\rho$ traces the linear
correction.
\end{proof}

\begin{proposition}[Zeros of $q$]\label{prop:q-zeros}
The following hold:
\begin{equation}\label{eq:q-zeros-1}
q(3L/2;\xi)=q(5L/2;\xi)=0.
\end{equation}
Moreover, as an identity in the argument $u=t-\gamma$,
\begin{equation}\label{eq:q-zeros-2}
q(\gamma+u-L;\xi)+q(\gamma+u+L;\xi)-q(\gamma+u;\xi)
\text{ has a zero of multiplicity}
\begin{cases}
\ge 2N & \text{at } u=0,\\
\ge N & \text{at } u=L.
\end{cases}
\end{equation}
\end{proposition}

\begin{proof}
First, $q(0)=q(4L)=0$: at $t=0$ and $t=4L$, $T(t)=I$ on the
physical sector (Proposition~\ref{prop:T-structure}(3)), so the
eigenvalue $1$ is exact and the $e_\ell e_r$ coefficient vanishes.
Using $T(t)=T(b-t)$ from Proposition~\ref{prop:T-structure}(2), we
have $q(b)=q(0)=0$ as well. Passing to $p(u)=q(\gamma+u)$ and
using $b=3L$, $\gamma=3L/2$:
\[
p(3L/2)=q(3L/2+\gamma)=q(3L)=q(b)=0,\qquad
p(5L/2)=q(5L/2+\gamma)=q(5L/2+3L/2)=q(4L)=0.
\]
This proves \eqref{eq:q-zeros-1}.

For \eqref{eq:q-zeros-2}: consider
$T(t-L)T(t+L)-T(t)$ acting on the eigenspace. Using fusion and the
matrix identities, the difference vanishes to order $2N$ at $t=\gamma$
(since $2N$ local $R$-contractions align) and to order $N$ at $t=b/2$
(from the boundary $K$'s). Taking mixed coefficients gives
\eqref{eq:q-zeros-2}.
\end{proof}

\subsection{The scalar crossing function}
\label{sec:strip-p}

At homogeneous $\xi_j$ (i.e., at the physical rapidities with correction
$-L/2$ already absorbed), define
\begin{equation}\label{eq:p-def}
p(u)=q(\gamma+u;\xi),\qquad \mathbf X(u)=\cos(2(\gamma+u)).
\end{equation}
Then $p$ is even: $p(u)=p(-u)$ (from $T(t)=T(r)$, $r=b-t$,
so $p(u)=q(\gamma+u)=q(b-\gamma-u)=q(\gamma-u)=p(-u)$). And $p$
is $\pi$-periodic since $T$ is. So $p$ is a function of $\mathbf X$, and
\eqref{eq:q-rational} with a change of variable gives
\begin{equation}\label{eq:p-rational}
p(u)=\frac{P(\mathbf X(u))}{(\mathbf X+1)^N(\mathbf X+\cos 2L)^{N+1}}.
\end{equation}

The conditions \eqref{eq:q-zeros-1}, \eqref{eq:q-zeros-2} translate to:
\begin{itemize}
\item $P(\mathbf X(3L/2))=P(\mathbf X(5L/2))=0$, from
$p(3L/2)=p(5L/2)=0$ via \eqref{eq:q-zeros-1}.
\item The combination $p(u-L)+p(u+L)-p(u)$ vanishes to order $2N$ at
$u=0$ and order $N$ at $u=L$.
\end{itemize}

We will use the following functional-analytic representation in the
next section. For $u$ in the horizontal strip $|\Re u|\le 5L/2$,
\begin{equation}\label{eq:p-functional}
p(u)=\mathcal L\!\left(\frac{C_u}{D}\right),
\end{equation}
where
\[
C_u(x)=\cosh(u\sqrt x),\quad D(x)=2C_L(x)-1,
\]
$\mathcal L$ is a fixed linear functional, and $x=z^2$.

The value $2B_N=2\cdot\frac{1}{2}q(\gamma)=q(\gamma)=p(0)$ becomes
\begin{equation}\label{eq:p-at-zero}
2B_N=p(0)=\mathcal L(1/D).
\end{equation}

\subsection{The finite functional}
\label{sec:strip-finite-functional}

We now pin down $\mathcal L$. Set $\mu_0(x)=C_{L/2}(x)$,
$\mu_1(x)=C_{3L/2}(x)$, and
$M=\mathrm{span}\{1,x,\dots,x^{N-1};\partial_m^j C_m(x)|_{m=L}\}_{j=0}^{N-1}$.
Then:

\begin{proposition}[Finite functional]\label{prop:finite-functional}
There is a unique linear functional $\mathcal L$ on the span of
$M$, $C_{L/2}$, $C_{3L/2}/D$ such that:
\begin{align}
\mathcal L(f) &= 0\ \text{ for every }f\in M,\label{eq:L-M}\\
\mathcal L(C_{L/2}) &= 0,\label{eq:L-C-L-2}\\
\mathcal L(C_{3L/2}-C_{L/2}/D) &= 0.\label{eq:L-C-3L-2}
\end{align}
For this functional, $2B_N=\mathcal L(1/D)$ for every $N$.
\end{proposition}

\begin{proof}[Outline]
Apply partial fractions to $p(u)$ in $\mathbf X$. Standard identities
\begin{equation}\label{eq:cheb-integral}
\int_0^\infty\frac{\cosh(uz)\sinh(mz)}{\sinh(\pi z/2)}\,dz
=\frac{1}{2}\left[\tan(m+u)+\tan(m-u)\right]=\frac{S\sin(2m)}{\mathbf X+\cos 2m}
\end{equation}
(at $z$ real, $x=z^2$) give the interpolation basis $\{1,x,\dots,x^{N-1},
\partial_m^j C_m(x)|_{m=L}\}_j$. Conditions \eqref{eq:q-zeros-2}
of Section~\ref{sec:strip-q-structure} translate to
\eqref{eq:L-M}; \eqref{eq:q-zeros-1} translate to
\eqref{eq:L-C-L-2} and \eqref{eq:L-C-3L-2}. The remaining free
coefficient of $1/D$ is $\mathcal L(1/D)=p(0)=2B_N$.

The interpolation basis $\{1,x,\dots,x^{N-1}\}\cup
\{\partial_m^j C_m(x)|_{m=L}\}_{j=0}^{N-1}$ is $2N$-dimensional,
matching the space of rational functions with pole structure
$(\mathbf X+1)^{-N}(\mathbf X+\cos 2L)^{-(N+1)}$ and polynomial
degree $\le 2N+1$ subject to the two homogeneous constraints
\eqref{eq:q-zeros-1} and \eqref{eq:q-zeros-2}. The identity
\eqref{eq:cheb-integral} is a standard residue computation:
contour shift to $z=2ik$, $k\ge1$, picks up the simple residues of
$\sinh(\pi z/2)^{-1}$ and produces a Fourier series in $u$ that
sums to the displayed $\tan$-combination.
\end{proof}



\section{The positive interpolation estimate}\label{c6:strip-asymptotics}

Section~\ref{c6:strip-transfer} reduced the strip bridge mass to a
single linear functional $\mathcal L$ on a $(2N+2)$-dimensional space.
Here we evaluate $\mathcal L(1/D)$ up to an absolute constant, obtaining
$B_N\asymp N^{-1/4}$. The engine is a determinant-density representation
and an asymptotic estimate of characteristic products.

\subsection{Spaces, measure, determinant density}

Retain from Proposition~\ref{prop:finite-functional}:
\begin{itemize}
\item measure $d\mu(x)=zD(x)\,dz/\sinh(4Lz)$, $x=z^2$;
\item $M_N=\mathrm{span}\{1,x,\dots,x^{N-1};\partial_m^jC_m(x)|_{m=L}\}_{j=0}^{N-1}$;
\item $\mathcal L(M_N)=\mathcal L(C_{L/2})=
\mathcal L(C_{3L/2}-C_{L/2}/D)=0$;
\item $2B_N=\mathcal L(1/D)$.
\end{itemize}

Set
\begin{align}
c_1&=C_{L/2},\quad c_3=C_{3L/2},\quad
h_0=1/D,\quad h_1=c_1/D,\quad g(x)=\sinh(L\sqrt x)/\sqrt x,\\
F_0&=M_N+\mathbb Cc_1,\quad F=F_0+\mathbb Cc_3,\quad G=M_N+\mathbb Cg.
\end{align}
Note $g(x)=\int_0^L C_t(x)\,dt$.

Let $d=2N+1=\dim F_0$. For ordered $X=(x_1<\dots<x_d)$ in $(0,\infty)^d$,
let $\det F_0(X)$, $\det G(X)$ denote evaluation determinants (rows =
basis functions, columns = points, jet values at coincident points).
Put
\begin{equation}\label{eq:nu-density}
\nu = \frac{\det F_0(X)\det G(X)\prod_{i=1}^d d\mu(x_i)}
{\int\det F_0\det G\prod d\mu}.
\end{equation}
By Lemma~\ref{lem:jet-positivity} below this is a probability density.

\begin{lemma}[Jet positivity]\label{lem:jet-positivity}
Enumerate "labels" $\beta\in\{0,L^2\}$ each with multiplicity $N$, plus
one $g$-row inserted between the two label blocks. For any ordered
$X_1<\dots<X_d$ in $(0,\infty)$, every jet-determinant formed from
complete jets of $C_{\sqrt\beta}(x)$ at the labels is positive; at
confluent points it is nonnegative. In particular $M_N$, $F_0$, $F$, $G$
are Chebyshev on $[0,\infty)$, counting multiplicities.
\end{lemma}

\begin{proof}
$C_{\sqrt\beta}(x)=\sum_{j\ge0}\beta^jx^j/(2j)!$ has nonnegative
Taylor coefficients, strictly positive for consecutive low degrees.
Ordered power determinants on positive nodes are positive by Rolle and
coalescing; confluent cases are nonnegative limits. Expanding jet rows
in these powers and using Cauchy--Binet (with nonnegative minors by
$\beta\ge0$) gives the ordered jet determinants; the $g$ row is a
positive integral of $C_t$ rows, $0<t<L$.
\end{proof}

\begin{proposition}[Determinant density]\label{prop:det-density}
For $H$ with $\int|Hf|d\mu<\infty$ ($f\in M_N+\mathbb Cg$):
\begin{equation}\label{eq:L-in-terms-of-nu}
\frac{\mathcal L(H)}{4a_*}=\nu[e_{-4}H]+s_*\,\nu[e_0H],
\end{equation}
with $e_{-a}H=H(-a)-I_{F_0,X}H(-a)$ the residual of $F_0$-interpolation
at nodes $X$, and $s_*=\rho/(4a_*)$ the scalar in $\mathcal L$.
\end{proposition}

\begin{proof}
Apply Andr\'eief's identity
$\int\det F_0\det G\prod d\mu=\det(\int F_0G^Td\mu)$ row by row: the
integral $\int Hf\,d\mu$ against the density reduces to a projective
$\nu$-average; point evaluations at $-4$ and $0$ enter as residual
evaluation vectors because $H$ is fixed on the discrete $\mathcal L$
component. See Tracy--Widom \cite{TracyWidom1998} for the identity
itself.
\end{proof}

We will use the Schur complement identity on the appended rows $c_3,
h_0$ (eliminating $c_3$ from the $F_0$ determinant):
\begin{equation}\label{eq:schur-identity}
\nu[e_0c_3]\,\nu[e_{-4}h_0]-\nu[e_{-4}c_3]\,\nu[e_0h_0]
=\nu\!\left[(e_0c_3)E_{-4}h_0\right],
\end{equation}
where $E_y$ is the residual with row $c_3$ appended before the
$h_0$-residual.

Also define the tilted measure
\begin{equation}\label{eq:nu-tilt}
d\nu_3(X)\propto -e_0c_3(X)\,d\nu(X).
\end{equation}
Since $-e_0c_3>0$ by jet positivity ($d$ is odd), $\nu_3$ is a
probability density.

\subsection{Characteristic products}

\begin{lemma}[Characteristic products]\label{lem:char-products}
Fix $0<y<z$ and $r(x)=(x+y^2)/(x+z^2)$. For either $\eta=\nu$ or
$\eta=\nu_3$ and any $N$-subset of the coordinates,
\begin{equation}\label{eq:char-products}
\eta\!\left[\prod_{i=1}^Nr(x_i)\right]\asymp N^{b(y-z)/\pi}.
\end{equation}
In the growing regime $y^2=2z^2=2N^{3/2}$ (where $1\le r\le2$)
the same expectation is at most $\exp(o(N))$.
\end{lemma}

\begin{proof}[Sketch]
\emph{Interlacing.} Let $m=\dim V=\dim B$, and consider the roots of
$\det\int V B^T(x-\lambda)d\mu'$ for either $(F_0,G_t)$ vs $(M,M)$,
or $(A,G_t)$ vs $(U,U)$ (here $A=\{f\in F:f(0)=0\}$, $U=\{f\in M:f(0)=0\}$).
The roots of the smaller pair interlace those of the larger. This is
the Chebyshev sign-change argument: any nonzero real analytic function
orthogonal to a dimension-$m$ Chebyshev space flips sign at least $m$
times; consecutive interlacing subspaces cannot share a positive root.
Deform all $\beta$-labels to 0 continuously; interlacing is preserved.

\emph{Reduction to polynomial.} In the symmetric case ($M=M$ or $U=U$),
the expectation is the determinant $\det(r|_H)$ of the compression of
$r$ to $H$ in $L^2(\mu)$. By $L^2(\mu)$ approximation, $H$ lies between
the polynomial spaces $\mathcal P_{<N}$ and $\mathcal P_{<CN}$ for
a large absolute $C$, with operator-norm error $e^{-N}$. Hence the
determinant changes by $1+o(1)$.

\emph{Polynomial determinants.} For $\mu_c$ the probability measure
$dp/\cosh(cp)$ on $p=\sqrt x$, put $\alpha=\pi/(2c)$ and
\begin{equation}\label{eq:poly-gen}
A_m(x)=\cos(\alpha m)C_m(x),\quad
\int A_mA_n\,d\mu_c=\frac{1}{1-w_mw_n},\quad w_m=\tan(\alpha m).
\end{equation}
(The identity follows from
$\int C_mC_nd\mu_c=\frac12(\sec\alpha(m+n)+\sec\alpha(m-n))$, itself a
direct Fourier transform.) This gives explicit orthogonal polynomials
$P_j$ and permits precise evaluation:
$K_{n,\mu_c}(-v^2)\asymp n^{2cv/\pi}$ for fixed $v>0$, where
$K_{n,W'}(v)=\sup_{\deg h\le n}|h(v)|^2/\int|h|^2dW'$ is the
normalization. Then
\begin{equation}\label{eq:poly-det}
R_{W,n}(r^2)=K_{n,W'}(-y^2)/K_{n,W'}(-z^2)
\asymp N^{2b(y-z)/\pi}.
\end{equation}
This is the Gram cofactor identity on degrees $\le n$ for the
hyperplanes spanned by $(x+y^2)x^j$ and $(x+z^2)x^j$, $0\le j<n$.

\emph{Cauchy--Schwarz.} Since $r^2$ is the product of two compressed
determinants with $r$ and $r'=r\mu/\mu$, the square-root asymptotic
\eqref{eq:char-products} follows from \eqref{eq:poly-det} plus Jensen
on $\log r$.
\end{proof}

The exponentially small probability of $\{x_{N-1}<N^{3/2}\}$ follows
by applying the growing regime: there the product is at least
$(3/2)^{N-1}$, forcing its expectation under a probability to be tiny.

\subsection{Interpolation residuals}

\begin{lemma}[Residual bound]\label{lem:residual-bd}
Let $\lambda_*=\pi/(3L)=8/3$. For $A\in\{e,E\}$ and $a\in\{0,4\}$ and
$j\in\{0,1\}$, and for $v_i$ the interpolation nodes in increasing order:
\begin{equation}\label{eq:residual-bd}
0\le A_{-a}h_j\le C\prod_{i=1}^N\frac{v_i+a}{v_i+\lambda_*^2}
\end{equation}
uniformly in $X$. For $h_0$ the analogous lower bound with
$\prod_{i=1}^d$ also holds.
\end{lemma}

\begin{proof}[Sketch]
\emph{Translate.} Put $w=x+a$, so evaluation is at $0$ with strictly
positive ordered nodes.

\emph{Monodromy.} In this coordinate, each residual against the jet
space is a \emph{convex} combination of monomial residuals with powers
$0,\ldots,N-1$ and at most one additional power. This uses Cauchy--Binet
in the evaluation and error determinants, where the auxiliary minors
$D_J(0)$ are positive with derivative ratio bounded by
$\theta_N=\frac{\sum\beta_i}{2(2N+1)}$; for large $N$ the minors remain
positive after translating by $a$.

\emph{Laplace structure.} Write $h_j(w-a)=h_j(-a)\mathbb E e^{-wZ_j}$,
where $Z_j\ge0$ is a random variable. Explicitly, the product formula
for $D(x)=2C_L(x)-1$ (via $\cosh$-product) gives
$D(x)\propto(x+\lambda_*^2)\prod_{k\ge2}(x+s_k)$ with
$s_k=((2\pi k+\pi/3)/L)^2$; similarly $c_1$ and $(c_3-c_1)$ have
product form. Hence
$\mathbb P(Z_j\ge t)\asymp e^{-\lambda_*^2t}$ (for $h_1$, one pairs
each numerator factor with a distinct smaller $s_k$, splitting off
the smallest rate by a change-of-measure uniform bound).

\emph{Monotone interpolation.} Monomial residuals are monotone in
vertex powers: the $e^{tw}w^j$ coefficient matrix is totally
nonnegative (exponential of the upper shift times $t$), so the
old interpolant's coefficient at each node has sign $(-1)^i$; a
single-power increase changes the residual by a sign-consistent
amount. Reduce to the pure-exponential case:
\[
\mathbb E[e^{-(\lambda_*^2-a)Y}]=\prod_{i}\frac{v_i+a}{v_i+\lambda_*^2}
\]
since $Y$ is a sum of exponential variables with rates $v_i+a$.
\end{proof}

Under the measures of \eqref{eq:nu-density}, the omitted $O(N)$
factors in \eqref{eq:residual-bd} above $N^{3/2}$ give $e^{o(N^{-1/2})}$;
and the small-node probability is exponentially small. Thus
\begin{equation}\label{eq:P-7-estimates}
\nu[-e_0c_3]\gtrsim N^{-1/2},\quad
|\nu[e_0h_i]|\le N^{-1+o(1)},\quad
|\nu[e_{-4}h_i]|=O(1),
\end{equation}
and the tilted target
\begin{equation}\label{eq:P-7-target}
\nu_3[E_{-4}h_0]\asymp N^{-1/4}.
\end{equation}

\subsection{Increasing-weight comparison}

The final size estimate on $\nu_3[-e_0c_3]$ uses monotone coupling.

\begin{lemma}[Increasing-weight comparison]\label{lem:weight-comp}
Let $\mu_b\propto dx/\cosh(b\sqrt x)$ and $\nu_b$ be the determinant
density built from $\mu_b$ and the same Jacobi jet structure. The
ratio of base measures $\mu_b/\mu$ (up to constants) is the increasing
function
\begin{equation}\label{eq:weight-ratio}
\phi(x)=\frac{z(\cosh(4z)+\cosh(2z)-\cosh(3z))}{\sinh(4z)},\quad z=L\sqrt x,
\end{equation}
with increasing series-coefficient ratios
$\frac{2j+1}{4}(1+(1/4)^j-(9/16)^j)/1$. Therefore, for any bounded
coordinatewise increasing function $f$ on $(0,\infty)^d$:
$\nu_b[f]\le C\nu[f]$.
\end{lemma}

\begin{proof}[Sketch]
The MTP$_2$ property for determinant-product densities gives positive
association in the strict order. The one-coordinate heat-bath ratio
is monotone in the other coordinates, using Plücker's two-by-two
identity: fixing columns, the determinant with choice $t'$ over
choice $s$ at one coordinate is monotone in others. Interpolating to
$\mu$ via increasing density ratio $\phi$ then gives the comparison.
See Bogso \cite[Theorem 2.27]{Bogso2014} for the MTP$_2$ formulation.
\end{proof}

Under $\mu_b$, the interpolant of $A_m$ at 0 is a rational function
in $w=w_m$ whose denominator has roots $w$-labels of $G_t$;
the residual at $w=1$ is
\begin{equation}\label{eq:residual-explicit}
\frac{1}{1+s}\prod_i\frac{1+sb_i'}{1-b_i'}\prod_i\frac{1-a_i'}{s+a_i'},
\end{equation}
where $a_i',b_i'\in[0,1/3]$ are the $w$-labels of $F_0$ and $G_t$.
This is uniformly $\gtrsim d^{-1/2}$ on $1<s<1+1/d$, giving
$\nu_b[-e_0c_3]\gtrsim d^{-1/2}$.

\subsection{Assembly}

Combine \eqref{eq:schur-identity}, \eqref{eq:P-7-estimates}, and
Lemma~\ref{lem:weight-comp}. Write
$A_y=\nu[e_yc_3]$, $H_{j,y}=\nu[e_yh_j]$
($y\in\{0,-4\}$). The constraint $\mathcal L(c_3-h_1)=0$ determines
$s_*=-\frac{A_{-4}-H_{1,-4}}{A_0-H_{1,0}}$; the denominator is
nonzero for large $N$ since $|A_0|\gtrsim N^{-1/2}$ and
$|H_{1,0}|\le N^{-1+o(1)}$.

The exact form of $\mathcal L(h_0)/(4a_*)$ is then
\begin{equation}\label{eq:L-exact}
\frac{\mathcal L(h_0)}{4a_*}
=\frac{A_0\,\nu_3[E_{-4}h_0]
+H_{1,-4}H_{0,0}-H_{1,0}H_{0,-4}}
{A_0-H_{1,0}}.
\end{equation}
The numerator has leading term $A_0\nu_3[E_{-4}h_0]\asymp N^{-1/4}\cdot
N^{-1/2}=N^{-3/4}$; the remaining quotient contributes
$O(N^{-1/2+o(1)})$, negligible. So
$\mathcal L(h_0)/(4a_*)\asymp N^{-1/4}$, hence
$B_N\asymp N^{-1/4}$.

With this, Theorem~\ref{thm:finite-bridges} items \eqref{eq:bridge-mass}
are proved. The remaining items (first-moment and long-bridge) are
deferred to Section~\ref{sec:sewing}, where the two-mark estimate is
applied.



\section{The periodic vacuum and its physical pairing}\label{c6:periodic-polynomial}

We need a finite formula for the polygonal partition on a periodic
honeycomb cylinder. The formula expresses that partition as a ratio
$G_N(v)/(2f_N(v)^2)$ of two explicit trigonometric polynomials,
evaluated at homogeneous spectral variables $v_1=\cdots=v_N=0$. The
interesting content (that these two-dimensional sums count the same
polygons) is proved through a family of polynomial vacuum vectors.

\subsection{The cylinder and the finite formula}

Take the lozenge-cylinder of period $N$. Put corners $A,B$ of tiles
on a common horizontal line, $A$ just before column $1$ and $B$ just
after column $m$, with $m+n=N$. Let $\mathcal Z_N(y)$ sum systems of
finitely many disjoint simple polygons in the cylinder, each with
weight $\kappa^{\rm length}$ and a factor $y+y^{-1}$. Every polygon
in a system separates $A$ from $B$; the empty system is included.

Introduce spectral variables $v_1,\dots,v_N$, one per column. Put
$C=2-\sqrt2=4\cos^2 b$ and
\begin{equation}\label{eq:R_s_R_c_D}
R_s(d)=\frac{-1}{\sin(d+L)\sin(d-L)},\quad R_c(d)=\cos^2 d,\quad
D(d)=\sin(d+2L)\sin(d-2L)\sin(d+b)\sin(d-b).
\end{equation}
Sum over site labels $\alpha_j\in\{+1,0,-1\}$, with total sum in the
two groups $\{1,\dots,m\}$ and $\{m+1,\dots,N\}$ equal to a common $h$.
Weight every configuration by $y^h\prod_jc_{\alpha_j}$ with
$c_\pm=\sqrt C$, $c_0=2$, and by the two-body factors
\begin{equation}\label{eq:P-kernel}
P_{\alpha\alpha'}(v_j-v_i)=\begin{cases}
1,&(\alpha,\alpha')\in\{(+,+),(-,-)\},\\
H(v_j-v_i),&(\alpha,\alpha')=(0,0),\\
H(v_j-v_i-L),&(\alpha,\alpha')\in\{(+,0),(0,-)\},\\
H(v_j-v_i-L)H(v_j-v_i-2L),&(\alpha,\alpha')=(+,-),
\end{cases}
\end{equation}
with $P_{\alpha\alpha'}(d)=P_{\alpha'\alpha}(-d)$, and with $H=R_s$
for a same-group pair, $H=R_c$ across groups. Same-group pairs are
additionally multiplied by $D(v_j-v_i)$. The resulting sum is
\begin{equation}\label{eq:G-def}
G_N(v) = \sum_{\{\alpha_j\},\ \sum_{1}^m\alpha_j=\sum_{m+1}^N\alpha_j}
y^{\sum\alpha_j}\prod_jc_{\alpha_j}\prod_{i<j}P_{\alpha_i\alpha_j}(v_j-v_i).
\end{equation}

The denominator is a Pfaffian. Put
\begin{equation}\label{eq:B-h-def}
B(u)=\frac{\sin(u+b)\sin(u-b)}{\sin u},\qquad h(u)=\frac{\cos u}{B(u)}.
\end{equation}
Let $\mathsf h_N$ be the skew matrix with upper entries
$(\mathsf h_N)_{jk}=h(v_k-v_j)$, $j<k$; for odd $N$ append a last
column of ones and a row of minus ones with bottom-right entry $0$.
Set $a_{2l}=2^l/\sqrt 2$, $a_{2l+1}=(-2)^l$, and
\begin{equation}\label{eq:f_N}
f_N(v)=a_N\prod_{j<k}B(v_k-v_j)\,\mathrm{Pf}\,\mathsf h_N.
\end{equation}
Explicitly, $f_0=1/\sqrt2$ and $f_1=1$, and for even size
\begin{equation}\label{eq:f-even}
2f_{2m'}^2=4^{m'}\prod_{i<j}B(v_j-v_i)^2\,\det[h(v_j-v_i)]_{i,j=1}^{2m'}.
\end{equation}
The apparent singularities in~\eqref{eq:f_N} are removable
(Lemma~\ref{lem:poly-vacuum} below); $f_N$ is a trigonometric polynomial.

\begin{proposition}[Homogeneous physical pairing]\label{prop:physical-pairing}
For $y>0$,
\begin{equation}\label{eq:physical-pairing}
\mathcal Z_N(y)=\left.\frac{G_N(v)}{2f_N(v)^2}\right|_{v_1=\cdots=v_N=0}.
\end{equation}
The denominator is nonzero there, and the identity holds also as a
Laurent-polynomial identity in $y$.
\end{proposition}

The proof is by polynomial interpolation: the right-hand side is
identified with the physical polygon partition through a finite
family of fusion conditions.

\subsection{Periodic eigenlines}

Let $K(\xi)_{pp}=\xi^p$ denote the spin twist ($p\in\{+1,0,-1\}$)
and $T_\xi(t)$ the single-row transfer matrix
\begin{equation}\label{eq:T-xi-def}
T_\xi(t)=\mathrm{Tr}_a\,K_a(\xi)R_{a1}(t+v_1)\cdots R_{aN}(t+v_N),
\qquad\xi=i\ \text{or }-i,
\end{equation}
acting on total-spin-zero site states. The empty site vector is
denoted $\varnothing$; site order is site 1 first even though operator
action is from the right.

\begin{lemma}[Polynomial vacuum]\label{lem:poly-vacuum}
For $N\ge1$ and generic $(v_j)$, $T_i(t)$ has a simple common eigenline
with eigenvalue $1$. Its normalized eigenvector $w^i$ has
$w^i_\varnothing=1$, and $\Psi_N:=f_Nw^i$ (with $\Psi_0=1/\sqrt 2$)
is a trigonometric polynomial of degree $\le N-1$ in each $v_j$,
with the following properties:
\begin{enumerate}
\item\emph{Parity.} On a component with site spin $p_j$, $\Psi_N$ is
$(-1)^{N-1+p_j}$-parity in $v_j\mapsto v_j+\pi$.
\item\emph{Fusion.} With remaining list $U$,
\begin{equation}\label{eq:psi-fusion-2L}
\Psi_N(u,u+2L,U)=\prod_{z\in U}\cos(z-u-L)\cdot W_{12}\Psi_{N-1}(u+L,U),
\end{equation}
\begin{equation}\label{eq:psi-fusion-b}
\Psi_N(u,u+b,U)=2\cos b\prod_{z\in U}\cos(z-u-L)\cos(z-u-2L)\cdot\Omega_{12}\Psi_{N-2}(U).
\end{equation}
\item\emph{Infinity.} As $|v_N|\to i\infty$,
$\Psi_N(v)/\prod_{j<N}\cos(v_N-v_j)\to\sqrt2\Psi_{N-1}(v_1,\dots,v_{N-1})\otimes|0\rangle_N$.
\item\emph{Wheel zero.} $\Psi_N$ vanishes at the wheel
$(v_1,v_2,v_3)=(u,u+2L,u+5L)$ (order $\ge1$; order $\ge2$ for twist $-i$).
\end{enumerate}
\end{lemma}

\begin{proof}[Proof sketch]
\emph{Eigenvalue and simplicity.} Let $T_j=T_\xi(-v_p+jL)$ at any site $p$.
Yang--Baxter and spin conservation give
\begin{equation}\label{eq:fusion-T}
T_0T_2=T_1,\quad T_0T_3=I,\quad T_1T_3=T_2.
\end{equation}
(Check by applying $R_{2p}(0)R_{1p}(2L)=I$ on the image of $W$.)
Any finite product at common twist has $\varnothing,\varnothing$ entry
$1$ (loop cancellation, as in Section~\ref{sec:strip-physical}).
Judicious imaginary-limit sampling reduces the simplicity on total spin
zero to a base case $N=1$.

\emph{Vector normalization.} We set $w^\xi=P^\xi|\varnothing\rangle$,
$\ell^\xi=\langle\varnothing|P^\xi$, both rational with empty entry $1$,
and related by crossing $\ell^\xi(v)=w^{\xi^{-1}}(-v)^T$. Exchange and
fusion of $w$ follow as in Section~\ref{sec:strip-physical}: at
$v_q-v_p\in\{0,2L,b\}$, the local $R_{pq}$ and its fusions act as stated.

\emph{Degree bound.} For each row $T_\xi(t)$ on the vacuum line,
Fourier-Laurent expansion gives
\[
T_\xi(t)=I+\sum_k[A_k\cot(t+v_k+2L)+B'_k\cot(t+v_k+3L)],
\qquad\sum_k(A_k+B'_k)=0.
\]
Applying the fusion equations to $T_j=T_\xi(-v_p+jL)$ gives commuting-operator
annihilator rows $\mathcal A_{p,0}=c_0-c_1+c_2+X_0c_2$ and
$\mathcal A_{p,1}=c_1-c_2+c_3+X_1c_3$. Collecting into operator
matrix $\mathcal A$, one has
$\det\mathcal A=(\det H_{\mathcal A})P^\xi$, where $H_{\mathcal A}$
is the cotangent kernel $r(u)=\cot(u+3L)-\cot(u+4L)+\cot(u+5L)$.
The scalar Pfaffian $J_N=\mathrm{Pf}(H_{\mathcal A})\prod_{j<k}p(v_k-v_j)$,
\begin{equation}\label{eq:p-pfaffian}
p(u)=\frac{\sin(u+2L)\sin(u-2L)\cos u\ [\sin(u+b)\sin(u-b)]^2}
{\sin^2u\,\sin(u+L)\sin(u-L)},
\end{equation}
is a polynomial of trigonometric degree $\le 3(N-1)$ per variable.
Next, the two-site computation
\begin{equation}\label{eq:J-2-sites}
J_2=\cos u\,[S t_*-(t_*^2/2)\cos 2u],\qquad t_*=-\frac{1}{1+S},
\end{equation}
identifies three simple roots modulo $\pi$: one real at $u\equiv\pi/2$,
and two complex from $\cos 2u=-1-\sqrt2$. This gives generic simplicity
at $v_j-v_k\in\{L,2L\}$. Combining with
fusion via $W$ and $\Omega$, the total pole degree of $P^\xi$ is
bounded by $3(N-1)$, hence $d_N^i+d_N^{-i}\le 3(N-1)$.

\emph{Wheel zeros.} Consider $\Psi$ at $(v_1,v_2,v_3)=(u,u+2L,u+5L)$
in the order $1,2,3$ with remaining list $Y$. Perturbing
$v_2=u+2L+z$, $v_3=u+5L+z'$ and using~\eqref{eq:psi-fusion-2L}:
at $z=0$, $\Psi\in\mathrm{im}(W_{12})$; on $z'=z$ and $z'=0$ we get
specific columns $b_0,c_0$ proportional to $\Omega_{23}$ and to
$K_3 J^{\rm spin}_1 \Omega_{31}$ (composed with site-3 factors). The
column is in $\mathrm{im}(W_{12})$ only at specific spin coefficient
identities; matching at spin $\pm1$ then at spin $0$ forces both
$\gamma,\alpha=0$ (for $\xi=-i$), giving vanishing of order $\ge2$.

\emph{The Pfaffian normalization.} The candidate $f_N$ of
\eqref{eq:f_N} also satisfies the fusions~\eqref{eq:psi-fusion-2L},
\eqref{eq:psi-fusion-b} (using $B(d)B(d-b)=-\cos(d-L)\cos(d-2L)$ and
induction on $N$) and has the required degree, parity and symmetry.
The base cases $f_0,f_1,f_2$ match by direct computation. A polynomial
of degree $\le N-1$ with fixed parity and $2(N-1)$ prescribed values
modulo $\pi$ is unique, giving $f_N=\Psi_N^\varnothing$.

The remaining claims on $\Psi_N$ (degree control, parity) follow from
the fusion and normalized-vector structure.
\end{proof}

The equality of the exact degree to $3(N-1)$ is then obtained from
divergence at $v_2-v_1\in\{b\}$ in the double fusion~\eqref{eq:psi-fusion-b}:
the wheel factor there forces $d_N^i\ge N-1$ and $d_N^{-i}\ge 2(N-1)$.
Matching \eqref{eq:f_N} gives the exact scalar fusion identities
\begin{align}
\Psi_N(u,u+2L,U) &= c_1\prod_{z\in U}\cos(z-u-L)\cdot W_{12}\Psi_{N-1}(u+L,U),\label{eq:exact-fusion-2L}\\
\Psi_N(u,u+b,U) &= c_2\prod_{z\in U}\cos(z-u-L)\cos(z-u-2L)\cdot\Omega_{12}\Psi_{N-2}(U),\label{eq:exact-fusion-b}
\end{align}
with $c_1=1$ (after normalization) and $c_2=2\cos b$
(cross-checked at $N=2,3$ and by induction).

\subsection{The physical contraction}

\begin{proof}[Proof of Proposition~\ref{prop:physical-pairing}]
Define
$Z(v)=\ell^{-i}(v)(\prod_{j\le m}K_j(\delta))w^i(v)$,
with $\delta=i y$. Then $U=2f_N^2Z$ is a trigonometric polynomial
of degrees $\le 2(N-1)$ in each variable, $\pi$-periodic by spin-sign
covariance, and separately symmetric in the two groups (spin
conservation plus exchange).

Both $U$ and $G_N$ (from~\eqref{eq:G-def}) share:
\begin{itemize}
\item separate symmetry in the two groups,
\item $\pi$-periodicity in each $v_j$,
\item exact fusion to the smaller list at fused rapidities
$(v_{j+1}-v_j)\in\{0,\pm2L,\pm b\}$ (with normalizations from $f_N$),
\item growth behavior at imaginary infinity by~\eqref{eq:psi-fusion-b},
\item initial conditions $G_{N=0}=1$, $G_{N=1}=1$ (identical to $U$).
\end{itemize}
By induction on $N$, $G_N=U$. Indeed, the difference $G_N-U$ has zero
degree after dividing by the fusion-denominator product and vanishes
at each step. The base cases $N=0,1$ match.

Evaluating at $v_1=\cdots=v_N=0$ with the sandwiched $K_j(\delta)$'s:
\begin{enumerate}
\item $\mathcal Z_N$ at homogeneous $v$ has a natural expansion as an
oriented sum over polygon systems; the contractible and essential
sectors get phases $i$ and $\pm i$ (reversing conjugates), and the
`left-side switch' factor $(-1)^{1_{A\in D'}}$ plus
$\delta^{1_{A\in D'}-1_{B\in D'}}$ from the sandwich. Reversing
orientation inverts the factor, so a polygon contributes $0$ unless
$A,B$ are separated, giving weight $y+y^{-1}$ exactly.
\item The tile weights in $G_N$ without phases are honeycomb path
weights.
\item The empty system gives $1$, matching $(\varnothing,\varnothing)$
entry.
\end{enumerate}
For positive $y$, exhaustion proves $\mathcal Z_N=G_N/(2f_N^2)|_{v=0}$,
and the equality extends to a Laurent polynomial identity.
\end{proof}



\section{The two-line gas expansion}\label{c6:gas}

Section~\ref{c6:periodic-polynomial} expressed the physical cylinder
partition as a finite trigonometric ratio. We now expand that ratio as
a grand-canonical sum over \emph{particles}, and control the sum by a
coercivity bound on the kernel. The output is a rigorous upper bound
on $\log|P_m|$ (Proposition~\ref{prop:gas-upper}) which
Section~\ref{c6:subheading-09} saturates.

\subsection{Normalization and the target}

Put
\begin{equation}\label{eq:P-m-normalized}
P_m(e^r)=\frac{G(e^r)}{C^mD(0)^{m(m-1)}}.
\end{equation}
With $L=\pi/8$, $b=3L$, $C=2-\sqrt2$, $\widehat\rho(p)=1/(2\cosh Lp-1)$,
and $V(s)$ the real log of $R_s(is)R_c(is)$ for
$R_s(z)=-1/[\sin(z-L)\sin(z+L)]$, $R_c(z)=\cos^2z$, put
$I=\int_{\mathbb R}V(s)\rho(s)\,ds$.

\begin{proposition}[Homogeneous gas upper bound]\label{prop:gas-upper}
As $m\to\infty$, locally uniformly in $|\Im r|<.54\pi$:
\begin{equation}\label{eq:gas-upper}
\log|P_m(e^r)|\le m^2I+\left(\frac38+\frac{2}{3\pi^2}\Re r^2\right)\log m+o(\log m).
\end{equation}
\end{proposition}

\subsection{Residue particles}

Each lowering of a site label in the finite double-pole formula
produces one or two residue particles. A particle $d$ carries a color
sign $q_d\in\{\pm1\}$, a species, a contour, and a real position $s_d$.

The elementary mode potential is $e^{mV(w)}$, and the pair weights are
\begin{equation}\label{eq:pair-weight}
t_{gh}(w)=\frac{R_{gh}(iw)}{R_{gh}(iw+L)R_{gh}(iw-L)},
\end{equation}
with measure $\tau dz/(2\pi i)$ and $\tau=2\sin(2L)/\sqrt C$.

After periodization (replacing each $\sinh(w+ijL)$ by
$\frac12S_T(w+ijL)$ with $S_T(w)=2\sinh w\prod_{n\ge1}(1-e^{-2nT+2w})
(1-e^{-2nT-2w})$, and multiplying the integrand by
$\sum_le^{-2Tl^2+4l\sum q_jw_j}$) we obtain a $T$-periodic
$P_{m,T}(e^r)$, agreeing with $P_m(e^r)$ at $e^{-2T}=0$. We work on
$P_{m,T}$ and return by maximum modulus in the nome $e^{-T}$.

\subsection{Species and offsets}

After contour manipulation and fusion, five species remain in each
color. Their weights $w_i$ (exponent of $y$), offsets $\delta_i$
(multiples of $L$), and chemical activities $\sigma_i$ are:
\begin{center}
\begin{tabular}{c|ccccc}
$i$ & $0$ & $1$ & $2$ & $3$ & $4$ \\
\hline
label & $S$ & $D$ & $I_+'$ & $I_-'$ & $F$ \\
$w_i$ & $1$ & $1$ & $-1$ & $1$ & $2$ \\
$12\delta_i$ & $6$ & $1$ & $5$ & $-3$ & $-1$ \\
$\sigma_i$ & $0$ & $1$ & $1$ & $-1$ & $0$
\end{tabular}
\end{center}
The neutral-winding constraint on total particle counts is
\begin{equation}\label{eq:winding-neutrality}
\sum_{g\in\{c,-\}}\sum_{i=1}^5w_i\nu_{g,i}=0.
\end{equation}

\subsection{The interaction kernel}

The particle kernel for two particles at real frequency $p$ is
\begin{equation}\label{eq:Z-def}
Z^{(c)}_{ij}(p)=\frac{2\pi}{p}e^{(\delta_i-\delta_j)Lp}F^{(c)}_{ij}(e^{Lp}),
\end{equation}
with $F^{(c)}_{ij}(z)$ expressed through an auxiliary function $H(a;\sigma)$
that sums residues on the unit circle:
\begin{equation}\label{eq:H-aux}
H(a;\sigma)(z)=\sum_{j=0}^7h_j\frac{z^{[a+j]_8}}{z^8-1}+\frac{1-\sigma}{2}h_{[-a]_8},
\quad h=(2,-1,1,-2c,2c,-2c,1,-1).
\end{equation}
Let $H_*=H(0;0)$, $A_0=\sum_{a\in(2)}H(a;1)$, $A_1=\sum_{a\in(2,3)}H(a;-1)$,
$A_2=1$, $A_3=-1$, $A_4=A_1+A_3$. For $0\le i,j<4$, set
\begin{equation}\label{eq:F-ij}
F^{(c)}_{ij}(z)=P_{ij}(z)+\frac{A_i(z)A_j(z^{-1})}{H_*(z)}
+(-1+2c)\operatorname{sgn}(i-j){\mathbf 1}_{\{i,j\}=\{1,3\}},
\end{equation}
$F_{i4}=F_{i1}+F_{i3}+{\mathbf 1}_{i=3}-{\mathbf 1}_{i=1}$,
$F_{4i}(z)=-F_{i4}(z^{-1})$, $F_{44}=F_{11}+F_{13}+F_{31}+F_{33}$,
where $P_{ij}$ collects lower-derivative residue rules
$P_{ij}(z)=\sum_{a\in b_i,k\in b_j}H(a-k;\operatorname{sgn}(j-i))$ for $i,j<2$
and $\operatorname{sgn}(j-i)$ for $i,j\ge2$, and $0$ otherwise.

On the two $R$-lines (before fusion), the matrix symbol has a clean form.
Let $K=K_{RR}$ and $\widehat\rho(p)=1/(2\cosh Lp-1)$; then
\begin{equation}\label{eq:K-R-est}
K=(1-e(p))K_0,\quad
e(p)=\frac{\sinh(3Lp)-\sinh(2Lp)}{\sinh(4Lp)}\in(0,1),
\end{equation}
\begin{equation}\label{eq:K-0}
K_0=\frac{2\pi}{p}\begin{pmatrix}\coth(bp)&-\operatorname{csch}(bp)\\
-\operatorname{csch}(bp)&\coth(bp)\end{pmatrix},\quad V_R=K{\mathbf 1}\widehat\rho.
\end{equation}
(Here $V_R$ denotes the potential contribution.)

\subsection{Coercivity}

The key analytic input is a positive-definite bound on $Z$.

\begin{proposition}[Coercivity]\label{prop:coercivity}
Put $h_*=3\pi b/4$ and $\alpha=2\pi^2/h_*$, and let $W=(w,w)^t$ be the
weight column. Then for every real $p\ne0$ there is $\eta>0$ with
\begin{equation}\label{eq:Z-coercive}
Z_{\rm s}(p)\succcurlyeq\frac{\eta\alpha}{p^2}WW^t+\epsilon_0(1+p^2)^{-1}{\bf I},
\end{equation}
for some $\epsilon_0>0$ independent of $T$ and $m$, where $Z_s=(Z+Z^t)/2$.
\end{proposition}

\begin{proof}[Sketch]
On each $R$-line, $K=(1-e(p))K_0$ with $e(p)\in(0,1)$ and $K_0$
positive definite for $p\ne0$. The leading behavior at small $p$ is
$K_0\sim2\pi/(p)(\coth bp{\bf I}-\operatorname{csch}bp\,P)$ where
$P$ is the color-swap permutation matrix; the eigen-decomposition is
common and gives the $WW^t/p^2$ leading term. Away from $p=0$, the
residual is bounded below by $\epsilon_0(1+p^2)^{-1}{\bf I}$ after
partial-fraction expansion of the explicit $F$-rules.
\end{proof}

\begin{equation}\label{eq:Z-small-p}
Z(p)=\alpha WW^t/p^2+M_*/p+O(1)\quad(p\to0),
\quad M_*\text{ real skew}.
\end{equation}
\begin{equation}\label{eq:Z-large-p}
Z(p)=\frac{2\pi}{|p|}{\mathbf I}+2ff^t+O(e^{-\epsilon|p|})\quad(|p|\to\infty),\quad f={\mathbf e}_4.
\end{equation}

\subsection{The grand sum}

\begin{theorem}[Grand sum]\label{thm:grand-sum}
For every integer $m\ge1$ and every sufficiently large real $T$,
$P_{m,T}(e^r)$ equals
\begin{equation}\label{eq:grand-sum}
\sum_{\{\nu_{g,i}:\,(\ref{eq:winding-neutrality})\}}
\frac{1}{\prod_{g,i}\nu_{g,i}!}
\,e^{c\cdot 2\pi iTr\,\sigma^t\nu/(4h_*)-S_{\rm bulk}(T)-S_{\rm gas}[\nu;r]},
\end{equation}
with:
\begin{itemize}
\item $S_{\rm bulk}(T)=\frac{1}{2T}\sum_{p\ne0}\widehat J^tK(p)\widehat J\,\widehat\rho(p)^2
-\frac{1}{2}(T/3-2\log2)(M_A-M_B)^2$,
a real quadratic form;
\item $S_{\rm gas}[\nu;r]=-\frac{1}{2T}\sum_{p\ne0}\sum_{d,e}Z_{g_de_d}(p)e^{-ip(s_d-s_e)}-\frac{1}{T}\sum_{\text{linear}}\dots$,
the pairwise and single-particle interactions against real positions
$s_d\in(-T/2,T/2)$;
\item $\varepsilon_\nu\in\{\pm1\}$ is a fixed phase depending only on the multiplicity pattern;
\item $c$ is a fixed scalar.
\end{itemize}
The sum and integral converge absolutely, and the sum localizes on
\begin{equation}\label{eq:gas-localization}
\max_d|\Re s_d|\lesssim\log m+\text{const}.
\end{equation}
\end{theorem}

\subsection{Convergence and localization}

\begin{proof}[Sketch]
Proposition~\ref{prop:coercivity} bounds the off-diagonal $Z_s$ terms
from below by a positive quadratic form. After regularizing the diagonal,
the sum over $\nu_{g,i}$ is bounded by a geometric series in
$\sum_{g,i}\nu_{g,i}$ weighted by a Gaussian-decaying kernel in the
positions $s_d$ and a bounded activity from the chemical potential. For any configuration with a particle at $|\Re s_d|>\log m+\text{const}$,
the one-particle suppression from $S_{\rm bulk}$ dominates the local
gain, and the sum over such configurations is at most $e^{-\epsilon m}$.
Consequently the whole sum is dominated by the localized region
\eqref{eq:gas-localization}.
\end{proof}



\section{Saturation at the two physical parameters}\label{c6:saturation}

We have the upper bound $\log|P_m(e^r)|\le m^2I+(3/8+2\Re(r^2)/(3\pi^2))
\log m+o(\log m)$ for $|\Im r|<.54\pi$ (Proposition~\ref{prop:gas-upper}),
plus an exact evaluation of $P_m(i)$ via a determinant at the "empty
polygon" value. We now show both bounds are sharp and take their ratio.

\subsection{The saturation statement}

\begin{auxproposition}[Saturation at $y=1$ and $y=i$]\label{prop:saturation}
\begin{equation}\label{eq:saturation-aux}
\log P_m(1)=Im^2+\frac38\log m+o(\log m),\qquad
\log P_m(i)=Im^2+\frac5{24}\log m+o(\log m).
\end{equation}
Equivalently, the nesting ratio satisfies
\begin{equation}\label{eq:nesting-ratio}
\mathcal Z_{m,m}(1)=\frac{P_m(1)}{P_m(i)}=m^{1/6+o(1)}.
\end{equation}
\end{auxproposition}

\subsection{The reference determinant at $y=i$}

At $y=i$ the finite polynomial $G$ becomes $2f_N^2$: only the empty
polygon system contributes. In its determinantal expression replace
the kernel $h(d)=(\cot(d+b)+\cot(d-b))/2$ by
$h_0(d)=-\lambda\tan(\lambda d)/2$, $\lambda=\pi/(2b)=4/3$. The
normalized homogeneous value of this modified partition evaluates
exactly, by Schur's Pfaffian identity, to
\begin{equation}\label{eq:reference-det}
P_0^{(\lambda)}=\big[\lambda^4\sin^8(b)/D(0)\big]^{m^2}=e^{Im^2}.
\end{equation}
Indeed, Schur's Pfaffian of $G_{jk}=(X_j-X_k)/(X_j+X_k)$ with
$X_j=e^{2i\lambda v_j}$ equals $\prod_{i<j}(X_i-X_j)^2/
\prod(X_i+X_j)^2$ (clearing denominators, both sides alternate with
the same Vandermonde degree; the coefficient is fixed by a base case),
and $B(u)\tan(\lambda u)\to-\lambda\sin^2b$ at coincidence while
$D(0)=C\sin^4b$.

We are reduced to bounding the ratio $P_m(i)/P_0^{(\lambda)}$.

\subsection{Bounding the ratio at $y=i$}

Write $P_m(i)/P_0^{(\lambda)}$ as the product over $m$ "duplicate
pairs" $v_p,v_p+L$ of the linear-perturbation ratios
\begin{equation}\label{eq:duplicate-ratio}
\prod_{k=1}^m\frac{\det(1+\Delta_k)}{\det(1)},
\end{equation}
where $\Delta_k$ is the rank-one modification between the $h$ and $h_0$
kernels restricted to the $k$-th pair.

\emph{Weyl chamber decomposition.} Decompose the integration domain by
the partition of $\{1,\dots,m\}$ into the subsets $\Omega^A$, where $A$
is the unique subset containing the up-to-relabeling largest $|z_j|$.
On each $\Omega^A$, moving the rapidities with $j\in A$ to imaginary
infinity is pole-free within the chamber, and the two kernels (for $h$
and $h_0$) differ by a bounded factor with a Sobolev/$L_p$ estimate
from the linear-differential operator computation.

For each $k$, one gets uniformly
\begin{equation}\label{eq:weyl-bound}
\big|\det(1+\Delta_k)\big|\le e^{-\epsilon\,\mathrm{dist}(A_k,
\partial\mathrm{chamber})}.
\end{equation}
Summing chambers and applying maximum modulus:

\begin{equation}\label{eq:log-modulus-total}
\log\big|P_m(i)/P_0^{(\lambda)}\big|\le\frac5{24}\log m+O(1).
\end{equation}

Matching lower bound: restrict to configurations with all duplicates
within bounded distance of center, on which $\det(1+\Delta_k)=1-o(1)$.
This gives $|P_m(i)/P_0^{(\lambda)}|\ge m^{5/24-o(1)}$, hence
\begin{equation}\label{eq:P-m-i-final}
\log P_m(i)=Im^2+\frac5{24}\log m+o(\log m).
\end{equation}

\subsection{Propagation to $y=1$}

Set $r=\log y$ and put
\begin{equation}\label{eq:f-m}
f_m(r)=\frac{\log|P_m(e^r)|-Im^2}{\log m}
-\left(\frac38+\frac{2\Re(r^2)}{3\pi^2}\right).
\end{equation}
From Proposition~\ref{prop:gas-upper}, $f_m\le o(1)$ locally
uniformly in $|\Im r|<.54\pi$. From \eqref{eq:P-m-i-final} at
$r=i\pi/2$ (i.e. $y=i$), $f_m(i\pi/2)\to0$.

The function $f_m$ is subharmonic. Apply the subharmonic area
mean inequality on a finite chain of disks starting at $r=i\pi/2$ and
reaching any compact subset of $|\Im r|<.54\pi$: the two ends of the
chain give $f_m\to0$ in local $L^1$ throughout the strip. In
particular at $r=0$:
\begin{equation}\label{eq:saturation-at-1}
\log P_m(1)=Im^2+\frac38\log m+o(\log m).
\end{equation}

Combining \eqref{eq:P-m-i-final} and \eqref{eq:saturation-at-1}:
\begin{equation}\label{eq:nesting-ratio-concl}
\mathcal Z_{m,m}(1)=P_m(1)/P_m(i)=m^{1/6+o(1)},
\end{equation}
as claimed.
\end{proof}



\section{Crossings, cylinder barriers, and positive nesting}\label{c6:geometry}

Section~\ref{c6:strip-asymptotics} gave the strip-mass estimate
$B_k\asymp k^{-1/4}$. Here we extract (a) escape and confinement
estimates, (b) essential-polygon formation on a cylinder, (c) positive
polygon nesting, and (d) a boundary-occupancy identity. These are the
geometric inputs for Section~\ref{sec:sewing}.

All constants are uniform in $N$, $h$, and $H$.

\subsection{The critical weight at the port convention}

Adapting the Hammersley--Welsh argument to boundary-port convention
confirms that $\kappa$ is the critical self-avoiding-walk weight at
fixed starting triangle. The route uses strip bridges: for $x<\kappa$
the mass of weak bridges of height $k$ is at most $Ce^{-ck}$, and
decomposition over weak half-space walks gives the susceptibility
bounds. Constants depend only on $x$.

\subsection{Boundary crossings}

\begin{proposition}[Boundary crossings]\label{prop:crossings}
Fix a convex pure-sided tiled domain $D$ and a port $p$ on its
boundary (including strips and half-planes by exhaustion). Then:
\begin{enumerate}
\item The critical mass of completed paths of diameter $\ge r$ is at
most $C(1+r)^{-1/4}$.
\item The half-plane mass $K_l$ from a base port to another at lattice
distance $l$ is decreasing in $l$ and satisfies $K_l\asymp l^{-5/4}$.
\end{enumerate}
\end{proposition}

\begin{proof}[Sketch]
\emph{Contour comparison.} The signed exit identity (Section~\ref{c6:strip-flux}
with $\beta=3/8$) gives
\begin{equation}\label{eq:positive-exit}
\sum_{\gamma:p\to\partial D}\kappa^{\ell(\gamma)}
\cos(\beta W(\gamma))=1,\quad c\le\cos(\beta W(\gamma))\le1.
\end{equation}
Restricting to a convex subdomain and using $c=\cos(\beta\pi)>0$ gives
the first assertion.

\emph{Triangle escape.} For a centered triangle $T(p,n_j,t)$ of width
$t$, the exit mass $S(t)$ to each sloping side is comparable to
$(1+t)^{-1/4}$. The upper bound comes from the strip embedding; the
lower bound from the base-return identity
$2\cos(\beta\pi/3)S=1-cA_T$ and the diameter bound.

\emph{Monotonicity.} A non-crossing identity for $K$ and an intermediate
port between sources $a$ and $Q=P+1$ gives
\begin{equation}\label{eq:K-monotone}
(K_{P-a}-K_{Q-a})\sin(\beta\pi)=D_1\sin(2\beta\pi)\ge0,
\end{equation}
using the flux rule with an artificial arc below the base line. Combined
with the diameter bound over a window of length $\asymp l$: $K_l\lesssim l^{-5/4}$.

\emph{Deep exits.} For an exit from $T(a,n_0,t)$ through the side with
outward normal $n_1$, at depth $q=n_0\cdot(p_1-a)$,
\begin{equation}\label{eq:deep-exit}
\mathop{\mathrm{average}}_{t\in[\alpha R,AR]}\sum_{q\ge\delta R}k(t,q)
\ge c_{\alpha,A,\delta}R^{-1/4},
\end{equation}
for $\delta$ small, by translating along the base to align the exit
ports and cutting strictly before all starts.

\emph{Three exits and matching lower bound.} Iterating the deep-exit
construction at $p_1,p_2$ builds a three-exit path. Averaging over $t$
and $R$ gives mass $\gtrsim R^{-3/4}$ of paths with $q\ge\delta R$ and
$X\ge\eta R$, whence a second-moment bound
\begin{equation}\label{eq:three-exit-second-moment}
\mu[Y^2]\le CR^{-5/4},
\end{equation}
using the difference bridges with $B_{|t-t'|/d_0}$. Cauchy--Schwarz gives
$\mu[Y>0]\gtrsim R^{-1/4}$ and the matching lower bound for $K_l$ by
monotonicity.
\end{proof}

\subsection{Narrow-channel bridges}

\begin{lemma}[Narrow-channel bridges]\label{lem:narrow-channel-bridges}
For every sufficiently large absolute constant $A$, there are
constants $C,R_0>0$ such that, for every strip of height $H$, every
starting port on one cap, and every integer $M\ge1$ with $H/M\ge R_0$,
the critical mass of bridges whose transverse displacement from the
starting port stays at most $4AH/M$ throughout is at least
\begin{equation}\label{eq:channel-bridge-mass}
e^{-CM}(H/M)^{-1/4}.
\end{equation}
\end{lemma}

\begin{proof}[Sketch]
Put $s=H/M$ and choose $M-1$ internal levels independently in windows
of length $\asymp s$ around equally spaced cuts. Each piece has height
between $s/2$ and $2s$ and transverse excursion $\le As$, with
endpoint shift directed towards the original axis. Reflection retains
at least half the mass after the excursion restriction; the strip
bound on each piece gives $cs^{-1/4}$; Cauchy--Schwarz over the lists
of internal levels gives \eqref{eq:channel-bridge-mass}.
\end{proof}

\subsection{Essential polygons in a slab}

\begin{proposition}[Slab essential polygons]\label{prop:slab-essential-polygons}
There are absolute constants $c,C>0$ such that, for every sufficiently
large integer $N$ and every horizontal slab $S$ of height $CN$ on the
cylinder of period $N$:
\begin{equation}\label{eq:slab-essential-mass}
\sum_{\gamma\subset S,\ \gamma\text{ essential simple}}
\kappa^{\ell(\gamma)}\ge c.
\end{equation}
\end{proposition}

\begin{proof}[Sketch]
Build confined arches via collar decompositions at two ports
$p,q$ separated by $D\in[N/2-\epsilon N,N/2+\epsilon N]$. Each collar
uses \eqref{eq:deep-exit}, \eqref{eq:channel-bridge-mass}, and one
$S(t)$ factor, so a collar pair has mass $\gtrsim N^{-3/2}$, and the
connector adds another $N^{-5/4}$ factor by the $K_l$ estimate.
Averaging over the collar-cut lists and Cauchy--Schwarz gives a
slab-mass bound independent of the period. Joining upper and lower
confined arches and averaging over their shared base level gives
\eqref{eq:slab-essential-mass}.
\end{proof}

\subsection{Positivity of the negative polymer polynomial}

Let $D$ be the finite cylinder between horizontal row-boundary caps,
with height $H=h d_0$ and quotient by $N$ steps. Let $\mathcal P$ be
its set of simple polygons. For $0\le\beta'\le3/8$, put
$\alpha=\pi\beta'/3$, $x=(2\cos\alpha)^{-1}$, and
\begin{equation}\label{eq:omega-weight}
\omega(L)=x^{|L|}\cdot\begin{cases}
-2\cos(4\alpha),&L\text{ contractible},\\
-2\cos(2\alpha),&L\text{ essential}.
\end{cases}
\end{equation}
For $\mathcal A\subseteq\mathcal P$:
\begin{equation}\label{eq:Q-def}
Q(\mathcal A)=\sum_{\substack{\mathcal F\subseteq\mathcal A\\
\text{polygons in }\mathcal F\text{ disjoint}}}
\prod_{L\in\mathcal F}\omega(L).
\end{equation}

\begin{auxproposition}[Positivity]\label{prop:Q-positive}
For every $N\ge1$, $h\ge1$, $0\le\beta'\le3/8$, $\mathcal A\subseteq\mathcal P$:
$Q(\mathcal A)>0$.
\end{auxproposition}

\begin{proof}[Sketch]
Short periods $N=1$ or $h=1$ follow from the explicit row-cycle
formulas. For $N,h\ge2$: elementary polygons (six-centre loops around
interior lattice points) form a connected overlap family meeting every
triangle centre, so the incompatibility graph is connected. The signed
exit identity gives
\begin{equation}\label{eq:Q-expression}
Q(\mathcal P)=\sum_{\gamma:p\to\partial D}x^{|\gamma|}
e^{i\beta'W(\gamma)}Q(\{L:L\cap\gamma=\varnothing\}).
\end{equation}
Continue from $\beta'=0$, $x=1/2$ where defect terms are nonnegative,
to $\beta'=3/8$. A minimal-zero subset $\mathcal A$ must be all of
$\mathcal P$ by connectedness; but a fixed bridge through every level
gives a positive residual polynomial, contradiction.
\end{proof}

\subsection{Deletion and free-energy bounds}

At the endpoint parameters replace $\omega(L)$ by the essential-only
weight $-t\kappa^{|L|}$ with $t\in[0,\sqrt2]$. Positivity holds by
multiaffinity. The deletion ratio is
\begin{equation}\label{eq:cylinder-deletion-ratio}
H_H(\gamma;t)\ge\exp\Big(t\!\sum_{\ell\ \text{essential},\ \ell\cap\gamma\ne\varnothing}\!\kappa^{|\ell|}\Big).
\end{equation}
Combining \eqref{eq:cylinder-deletion-ratio} with
\eqref{eq:slab-essential-mass} and \eqref{eq:strip-upper} gives
\begin{equation}\label{eq:cylinder-free-energy}
-\log Z_H(\sqrt2)\le CH/N\qquad(H\ge N),
\end{equation}
and hence
\begin{equation}\label{eq:cylinder-integrated-ratio}
\int_1^{\sqrt2}\sum_{\ell\text{ essential}}\kappa^{|\ell|}
H_H(\ell;t)\,dt\le CH/N.
\end{equation}

\subsection{Cylinder barriers and height tails}

\begin{auxproposition}[Barriers and tails]\label{prop:cylinder-barriers}
On the infinite cylinder of period $N$:
\begin{enumerate}
\item The individual critical weights of essential polygons meeting a
fixed horizontal level have bounded sum.
\item From a fixed bottom port in a semi-infinite cylinder, arches
reaching height $s$ have mass at most $Ce^{-cs/N}$.
\item For $s\ge1$, all polygons meeting a fixed level with height span
in $[s,2s]$ have total mass at most $\operatorname{poly}(N,s)
e^{-cs/N}$.
\end{enumerate}
\end{auxproposition}

\begin{proof}[Sketch]
Combine \eqref{eq:cylinder-deletion-ratio} with
\eqref{eq:slab-essential-mass}. A polygon of span $s$ meets order
$s/(CN)$ disjoint slabs, each forcing positive-size cost:
$H_H(\gamma;t)\gtrsim e^{c's/N}$. Sum by spans between $(jN,(j+1)N)$.
The case $N=2$ uses an explicit two-strip surgery on the two choices of
upward-centre; $N=1$ is a row-cycle count.
\end{proof}

\subsection{Local nesting and boundary occupancy}

\begin{proposition}[Local nesting]\label{prop:local-nesting}
Let $Z(R)$ be the positive partition sum of pairwise disjoint simple
polygons in the plane, factor $2$, critical weight, supported within
radius $R$ of a fixed triangular-lattice point, including the empty
system. Then
\begin{equation}\label{eq:Z-R}
Z(R)=R^{1/12+o(1)}.
\end{equation}
\end{proposition}

\begin{proof}[Sketch]
On a cylinder, disregard essential separators at bounded multiplicative
cost by \eqref{eq:cylinder-deletion-ratio}. Contractible separators
lift to plane polygons around the marked point; the barrier bounds
\eqref{eq:cylinder-barriers} discard those with excessive vertical or
horizontal span. Systems inject. Hence
$\mathcal Z\lesssim Z(CN^{1+2\xi})^2$ for any $\xi>0$, and combining
with \eqref{eq:nesting-ratio} and taking $\xi\downarrow0$ gives
\eqref{eq:Z-R}.
\end{proof}

\begin{proposition}[Boundary occupancy]\label{prop:occupancy}
Let $D$ be a finite convex tiled domain and let $e$ be an internal
honeycomb link crossing a tiling edge $[y,z]$. If $U_D(e)$ is the
critical mass of completed external paths using $e$ with ordered free
boundary ports, then
\begin{equation}\label{eq:occupancy}
U_D(e)\asymp Z_D(y)+Z_D(z)
\end{equation}
with absolute constants.
\end{proposition}

\begin{proof}[Sketch]
The cut-link identity gives a signed exit mass
$S_e=Z_e+(M_y+M_z)/\sqrt2$, where $Z_e$ is the enclosing-polygon
partition in the cut domain and $M_y,M_z$ are contributions of a
polygon that uses $e$ with $y$ or $z$ inside ($Z_D(y)=Z_e+2M_y$,
$Z_D(z)=Z_e+2M_z$). Reversing path prefixes and summing over sources
yields \eqref{eq:occupancy}.
\end{proof}



\section{Bounded-factor nesting estimates}\label{sec:uniform-nesting}

The saturation bound of Section~\ref{c6:saturation} identifies a nonzero
normalization at exponent level. We now sharpen two inputs to bounded
factors: the geometric comparison between plane and cylinder, and the
analytic estimate of the normalized edge contributions. Both are used
in the sewing argument (\S\ref{sec:sewing}).

\subsection{Deletion ratio and the first-hit comparison}

Let $D$ be either a finite cylinder between flat caps, or a finite
convex tiled polygon in the plane. Write
\begin{equation}\label{eq:Q-D}
Q_D=\begin{cases}
Z_H(\sqrt2),&D\text{ cylinder},\\
1,&D\text{ plane},
\end{cases}
\end{equation}
the positive essential-polygon partition. For a set $K$ of occupied
centres:
\begin{equation}\label{eq:theta-D-K}
\Theta_D(K)=Q_{D\setminus K}/Q_D.
\end{equation}

\begin{lemma}[Deletion monotonicity]\label{lem:deletion-monotone}
$\Theta_D(K)$ is increasing in both the set of available polygons and
the deleted set $K$, for $0\le t\le\sqrt2$.
\end{lemma}

\begin{proof}
For a single deleted polygon $v$:
\begin{equation}\label{eq:single-deletion-ratio}
Q_P/Q_{P\setminus v}=1-t\kappa^{|v|}\cdot Q_{P\setminus N[v]}/Q_{P\setminus v},
\end{equation}
where $N[v]$ is the closed conflict neighbourhood. By positivity
\eqref{eq:Q-def} the ratio is $\ge1$. Induction on $|P|$ (deleting
neighbours successively in the smaller set) gives the multi-deletion
inequality; extra neighbours only increase the ratio.
\end{proof}

For a simple polygon $L\subset D$ and a boundary source $a$, let
$P(a,L)$ be the real part of the normalized signed mass of stems that
first hit $L$. Contour identity:
\begin{equation}\label{eq:first-hit-comparison}
1\ge P(a,L)\ge
\sum_{\substack{\gamma:a\to\partial D\\ \gamma\cap L\ne\varnothing}}
\kappa^{|\gamma|}\cos(\beta W(\gamma))H_D(\gamma)\ge0.
\end{equation}

\subsection{Boundary-flux primitive}

\begin{auxlemma}[Primitive of boundary flux]\label{lem:flux-primitive}
Choose a dual-edge chain $\Gamma$ from a boundary lattice vertex $y_0$
to an interior vertex $y$. Define $\operatorname{cr}_\Gamma(\gamma)$ to
be the signed number of $\Gamma$-crossings of a completed path and put
\begin{equation}\label{eq:current-I}
I=\Im\sum_{a\in\partial D}\sum_{\gamma:a\to\partial D}
\operatorname{cr}_\Gamma(\gamma)\,\kappa^{|\gamma|}e^{i\beta W(\gamma)}H_D(\gamma).
\end{equation}
Then
\begin{equation}\label{eq:flux-primitive-bound}
2\!\!\sum_{L\text{ contractible enclosing }y}\!\!\kappa^{|L|}H_D(L)\sum_aP(a,L)
\le|I|+\sqrt2\!\!\sum_{L\text{ essential}}\!\!\kappa^{|L|}H_D(L)\sum_aP(a,L).
\end{equation}
The essential sum is absent in the plane.
\end{auxlemma}

\begin{proof}[Sketch]
Evaluate the same prefix current $J$ two ways.
\emph{(i)} Telescoping along $\Gamma$ via the cut-link identity of
Prop.~\ref{prop:occupancy}: $J=-(\mathcal Y(y)/Q_D-1)/\sqrt2$.
\emph{(ii)} Continuing each prefix to its first exit gives
$J=I+2A+E$, where $A$ is the contractible enclosing sum and $|E|\le
\sqrt2\sum_{\text{essential}}\kappa^{|L|}H_D(L)\sum_aP(a,L)$ (turn
increments $\pm4\pi/3$ versus $\pm2\pi/3$). Comparing gives the
inequality.
\end{proof}

\subsection{The annular bound}

\begin{proposition}[Planar annular bound]\label{prop:annular-bound}
For every $s\ge1$ and every interior lattice point $y$:
\begin{equation}\label{eq:annular-mass}
\sum_{\substack{L\text{ contractible enclosing }y\\
\operatorname{diam}(L)\in[s,2s]}}\kappa^{|L|}\le C.
\end{equation}
\end{proposition}

\begin{proof}[Sketch]
Take a regular lattice hexagon of inradius $\asymp s$ around $y$. Only
chords that separate $y$ from $y_0$ contribute to $I$, with bounded net
crossing; summing the boundary diameter bound over the sources gives
$|I|\le Cs^{3/4}$.

A polygon enclosing $y$ with diameter in $[s,2s]$ fits inside a
doubled hexagon and has transverse span $\ge cs$ in some pure direction.
For order $s$ sources on the corresponding cap, transverse tubes of
width $\ll s$ force intersection of every confined path with $L$, so
\eqref{eq:channel-bridge-mass} and \eqref{eq:first-hit-comparison} give
$\sum_aP(a,L)\ge cs^{3/4}$. The essential sum vanishes in the plane; the
auxlemma gives the bound.
\end{proof}

\subsection{Bounded cost of large specified lifts}

\begin{auxproposition}[Bounded cost of large lifts]\label{prop:large-lifts}
Fix $c_2>0$. On a cylinder of period $N$, the total individual critical
weight of contractible polygons enclosing a fixed mark $y$ whose
specified lifts have diameter greater than $c_2N$ is at most $C_{c_2}$,
independently of $N$.
\end{auxproposition}

\begin{proof}[Sketch]
\emph{Good heights.} Put $F_H(t)=-\log Z_H(t)$. Superadditivity of
$F_H$ in $H$ plus the free-energy bound \eqref{eq:cylinder-free-energy}
make
\begin{equation}\label{eq:good-height}
a_j=\frac{NF_{H_j}(\sqrt2)}{H_j},\qquad H_j=d_0N2^j,
\end{equation}
a nondecreasing sequence in $[0,C]$. For any $u>0$, at most $4rC/u$
indices (for $k=2^r$ a fixed dyadic integer depending on $u$) violate
$F_{H_j}(\sqrt2)-kF_{H_j/k}(\sqrt2)\le uH_j/(4N)$, and then
\begin{equation}\label{eq:good-height-conclusion}
\sum_{L\text{ essential in }H_j\text{-slab}}\kappa^{|L|}H_{\text{slab}}(L)
\le e^{uH_j/N}.
\end{equation}

\emph{Good-height contractible sums.} For a good height $H$,
\begin{equation}\label{eq:good-contractible}
\sum_{\substack{L\text{ contractible enclosing }y\\
\operatorname{span}_{\text{vert}}\in[H/4,H/2]}}\kappa^{|L|}\le Ce^{-cH/N},
\end{equation}
by binning their minimum levels into windows of width $\delta N$ (bounded
number of bins by good-height hypothesis), applying
\eqref{eq:flux-primitive-bound} in an encompassing slab, and using
\eqref{eq:cap-first-hit} and the slab-bound \auxref{prop:cylinder-barriers}
to control both sides.

\emph{Tilted cylinder projection.} Dyadic bands with vertical span
$\ge cN$ project injectively onto a cylinder tilted by $60^\circ$, of
period $\asymp$ span. The tilted cylinder has its own dyadic-height
decomposition; by the above, all but boundedly many heights are good, so
\begin{equation}\label{eq:large-lifts-bound}
\sum_{\substack{L\text{ contractible enclosing }y\\
\text{lift-diam}>c_2N}}\kappa^{|L|}\le C_{c_2}.
\end{equation}
The remaining finite collection of exceptional bands is bounded by the
planar annular bound \eqref{eq:annular-mass} applied to the corresponding
diameter intervals.
\end{proof}

\subsection{Classwise convergence of the normalization}

\begin{auxproposition}[Classwise convergence]\label{prop:classwise-convergence}
For $r\in\{0,i\pi/2\}$ put $\sigma(r)=3/8+2r^2/(3\pi^2)$ and
$\mathcal A_m(r)=e^{-Im^2}m^{-\sigma(r)}P_m(e^r)$. There exist a fixed
integer $Q\ge1$, a $\delta>0$, and positive constants
$\mathfrak a_b(r)$, $0\le b<Q$, such that
\begin{equation}\label{eq:classwise-convergence}
\mathcal A_m(r)=\mathfrak a_{m\bmod Q}(r)+O(m^{-\delta}).
\end{equation}
Consequently
\begin{equation}\label{eq:uniform-cylinder-ratio}
\mathcal Z_{m,m}(1)=P_m(1)/P_m(i)\asymp m^{1/6}.
\end{equation}
\end{auxproposition}

\begin{proof}[Sketch]
Work with the periodic grand sum at period
$T=2a_0^{-1}\log m+g+i\theta$, $|\theta|\le\pi/2$, for a suitably large
fixed $g$. The stability estimate \eqref{eq:Z-coercive} and
localization \eqref{eq:gas-localization} continue to hold with the
coercivity error $O(1/g)\mathcal Q$ absorbed by $g$ large; the period
graph is fixed-mod-$g$ independent of $m$ on half intervals.

Multiplying the lifted polynomial by
$e^{-2Am^2e^{-2T}}e^{-Im^2}m^{-\sigma(r)}$ produces an
$F_m(\theta,r)$ whose circle mean is exactly $\mathcal A_m(r)$. We show
$F_m$ is Cauchy at exponential speed along each class modulo $Q$.

\emph{Scalar factor.} The quadratic bulk contribution gives
$m^{3/8}$; the logarithmic determinant gives
$\pi T/(16L)=\frac{5}{24}\log m$ plus constant; combining with the
edge integrals gives $m^{\sigma(r)}$ exactly.

\emph{Edge integrals.} Retain only particles within $\epsilon T'$ of an
edge, at most $T'$ particles, charge $j$ matching the right edge. The
omitted mass is $e^{-cT'}$, and standard small-slope deformation gives
agreement of the two retained integrands up to $e^{-cT'}$.

\emph{Discrete phases.} The $m$-dependent phase factors are roots of
unity from fixed logarithmic branch multiplies; the entire phase
depends only on $m\bmod Q$ for a suitable finite $Q$ chosen as a common
multiple of the (finitely many) root-of-unity denominators. The
retained count is at most $T'$, so $\|F_{m'}-F_m\|$ is
$\lesssim e^{-cT'}$ for $m'=m+Q$. This proves
\eqref{eq:classwise-convergence}.
\end{proof}



\section{Uniform correlations of two prescribed edges}\label{u:marked-uniform}

We prove the third main theorem: the two-edge marked-polygon bound
\eqref{eq:marked-uniform}, uniform both in the staircase direction and
in the ratio $N/d$. The argument has three parts: (a) identify the
two-edge observable as a contraction of polynomial columns, (b) in the
balanced (antipodal) case reduce it to a two-line gas and estimate,
(c) for unequal intervals control the two long columns outside a small
core.

\subsection{Setup}

Keep the triangular-lattice directions $e_1,e_2$ and identical
translated marks $E,E+P$ with $P=P_1e_1+P_2e_2$, $P_1,P_2>0$. Thus
$d=P_1+P_2$ is even, the cylinder period is $kP$ with $k\ge2$, and
$N=kd$. The polygon weight is $\kappa^{\#\text{triangle centres visited}}$.
The two restrictions (edge plus displacement) place both marks on a
monotone staircase of period $P$. The parity of $d$ makes both
intervals between the marks even in staircase steps.

\subsection{Finite marked contraction}

Fix integers $k\ge1$, $0\le l,h\le k$, and $m,n\ge0$. Take two lists
of common slots $A$ (size $m$) and $B$ (size $n$), and four extra lists
$E_b,H_b$ (sizes $l,h$) and $E_t,H_t$ (sizes $k-l,k-h$). Columns below
have total length $m+n+l+h$, above $m+n+2k-l-h$.

\begin{auxproposition}[Finite marked contraction]\label{prop:marked-contract}
For $\delta\ne0$, the contraction
\begin{equation}\label{eq:Y-def}
Y=\big\langle\Psi_b\big|\,\delta^{\sum_A p_j}\big|\Psi_t\big\rangle,
\end{equation}
of the columns (below with physical rapidities, above with negated
rapidities, spin-fixed extra lists $-,+,+,-$) satisfies
\begin{equation}\label{eq:finite-marked-contraction}
Y=Q_{lh}\,\mathcal J_{\rm ext}\!\!
\sum_{\substack{a_j\in\{-1,0,1\}\\
\sum_Aa_j-\sum_Ba_j=h-l}}
(i/\delta)^{\sum_Aa_j}\,\mathcal W(a)\,\mathcal M(a),
\end{equation}
where the right-hand data are specified below.
\end{auxproposition}

Here:
\begin{itemize}
\item $Q_{lh}=\tfrac12C^{k/2}s^{-k}(-s^{-1})^ls^{-h}$.
\item $J(u)=\cos(u+L)\cos(u+2L)$, and $\mathcal J_{\rm ext}$ is a
product of $J$-factors between extra slots in each list.
\item $\mathcal W(a)=\prod_jc_{a_j}\prod_{j<j'}P_{a_ja_{j'}}(v_{j'}-v_j)
\prod_{j<j',\,\text{same list}}D(v_{j'}-v_j)$, with $P$, $D$ from
Section~\ref{c6:periodic-polynomial}.
\item $M^1_+(u)=(1,\cos(u+L),J(u))$, $M^1_0=1$ extracted by
convention, and $M^2_a(u)=M^1_{-a}(-u)$.
\item $\mathcal M(a)=\prod_{j\in A\cup B}\prod_{e\in\text{extras}}
M^{d(j,e)}_{a_j}(x_e-v_j)$ with $d(j,e)\in\{1,2\}$ determined by list
and spin type.
\end{itemize}

\begin{proof}[Sketch]
Divide both sides by the within-extra-list $J$-products; the resulting
polynomial is separately symmetric in each list. Compare cap
specializations of adjacency: at each extra-common pair, one
specialization at difference $b$ either eliminates both slots or
converts a common slot on the opposite side into an extra of opposite
spin, preserving $k$. The scalar fusion identities
$J(u-b)=J(-u)$ and $D(u)=J(u)J(-u)$ close the recursion. Degrees
after division match on both sides with definite parities, leaving
only a constant factor, fixed by the base case $m+n=0$ (equivalently
by saturation of the two fixed-spin columns).
\end{proof}

\subsection{Physical interpretation}

\begin{auxproposition}[Physical two-edge mass]\label{prop:physical-two-edge}
For $k=2$, $l=h=1$, $\delta=i$, and physical rapidities zero:
\begin{equation}\label{eq:physical-two-edge}
\frac{Y}{f_{m+n+2}(0)^2}
\end{equation}
is the critical mass of individual simple polygons through the two
prescribed row edges on the cylinder, counted once per polygon.
\end{auxproposition}

\begin{proof}[Sketch]
At $k=2$, $l=h=1$ and $\delta=i$, the extra fixed spins at $E$, $H$
force two oriented arcs from $E$ to $H$, one departing upward and one
departing downward, uniquely determined by the polygon's local side
at the marks. Two arcs form the polygon in one orientation only.
The relative signs (from turns, seams, and the phase insertion) yield
a positive weight; the identity extends by height exhaustion.
\end{proof}

\subsection{The antipodal case}

When the two marks are antipodal (i.e.\ $P_1=P_2=m+1$ and $k=2$), the
two-line gas of Section~\ref{c6:gas} directly estimates the marked mass.

\begin{proposition}[Antipodal bound]\label{prop:antipodal-bound}
On the homogeneous cylinder of even period $2m+2$ with antipodal row
sites, the two-edge marked mass satisfies
\begin{equation}\label{eq:antipodal-bound}
K_2\le(m+1)^{-4/3+o(1)}.
\end{equation}
\end{proposition}

\begin{proof}[Sketch]
The two extra slots $(E,H)$ supply $J(v)^2$ at an unlowered common
site plus additional factors $\cos^2z/\cos^2(z+b)$ per lowering, which
amount to a shift of the magnetic potential by a bounded factor. Normalize
by $Q_{11}C^mD(0)^{m(m-1)}J(0)^{4m}$; this gives the constant
$P'_m/P_M(i)$ with $M=m+1$, exactly the ratio used in Section~\ref{c6:gas}.
The gas estimate at this shifted potential gives the claimed bound.

\emph{The magnetic test potential.}
In the residue lift, using the periodized variable $w=i\log$, the
elementary potential at $z=iw$ is $e^{MV(w)}[-\sin(z+L)/\sin(z-L)]$.
The bracket is replaced by
\begin{equation}\label{eq:U-T}
U_T(w)=-\frac{S_T(w-iL)}{S_T(w+iL)}
\left(e^{-3iL}\frac{S_T(w-T/2+iL)}{S_T(w-T/2+4iL)}\right)^2,
\end{equation}
which is $T$-periodic, tends to the original bracket, is analytic in
$e^{-T}$ near zero, and has a zero at $w=iL$ restoring pole
multiplicity $m$. The first ratio's continuous log increases by
$12iL$ over a period, compensated by the edge ratio's $-12iL$.
\end{proof}