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\title{Radial transfer estimates and polygon length laws\\for honeycomb walks}
\date{}
\begin{document}
\maketitle
\begin{abstract}
A self-avoiding polygon on the honeycomb lattice is a closed loop that
never visits a vertex twice. Give a polygon the critical weight
$w(\gamma)=x_*^{\,|\gamma|}$, where $|\gamma|$ is its number of edges
and $x_*=(2+\sqrt2)^{-1/2}$ is the honeycomb connective constant
raised to the power $-1$. We prove that the total critical weight of
polygons of diameter at least $r$ is $r^{-2+o(1)}$, that the same sum
carrying one power of the length is $r^{-2/3+o(1)}$, and that the
truncated second length moment is at most $r^{2/3+o(1)}$. As a
consequence a polygon sampled at criticality and conditioned to have
length at least $n$ has diameter $n^{3/4+o(1)}$ with probability
tending to one. The exponent $3/4$ is the long-predicted size exponent
of planar self-avoiding walk; the proof here arrives at it through
positive polygonal estimates and does not assume a scaling limit.
\end{abstract}
\tableofcontents
\section{Introduction: what is being counted, and what is attacked}\label{sec:intro}
\subsection{The game}
Place vertices at the points $P_{ij}=ih+jv$ and $R_{ij}=P_{ij}+(h+v)/3$ of the plane, where
\[
h=(\sqrt3,0),\qquad v=(\sqrt3/2,3/2),\qquad i,j\in\mathbb Z .
\]
Every $P_{ij}$ is joined to the three $R$-vertices $R_{ij},R_{i-1,j},R_{i,j-1}$, and every edge has length one. This is the regular \emph{honeycomb} (hexagonal) lattice; it is the lattice of a chicken-wire fence and of a single sheet of graphene.
A \emph{self-avoiding polygon} $\gamma$ is a finite cycle $p_1p_2\cdots p_Lp_1$ of distinct lattice vertices, each consecutive pair joined by an edge. Equivalently it is a round walk that never crosses itself and never reuses a corner. We write $L=|\gamma|$ for its number of edges and $D=\diam\gamma$ for its Euclidean diameter, i.e.\ the largest distance between two of its points.
Polygons that differ only by a translate are the same object for us; we count each translation class once. This throws away the position and keeps the shape and size.
The critical weight of a class $\gamma$ is
\[
w(\gamma)=x_*^{\,|\gamma|},\qquad x_*=(2+\sqrt2)^{-1/2}.
\]
This particular number is not a coincidence: Duminil-Copin and Smirnov proved that on the honeycomb lattice the connective constant equals $\mu=\sqrt{2+\sqrt2}$, i.e.\ the number of $m$-step self-avoiding walks grows like $\mu^m$, so $x_*=1/\mu$ is exactly the \emph{critical} value at which the length series $\sum_m (\# m\text{-step walks})\,x^m$ sits on the boundary between convergence and divergence. We work at exactly this value; it is the analog of the critical temperature of a magnet. Because $x_*$ is the critical activity, the weights $w(\gamma)$ balance every possible length equally and the only thing left to measure is size.
\subsection{The three observables}
For an unrooted polygon (one considered up to translation) define, for $r>0$ a length scale (thought of as large),
\begin{align}
A(r) &= \sum_{\diam\gamma\ge r} w(\gamma) &&\text{(mass of big-diameter polygons),}\\
P(r) &= \sum_{\diam\gamma\ge r} |\gamma|\, w(\gamma) &&\text{(the same with one length factor).}
\end{align}
Also, summing over the complementary range (diameter below $r$) but carrying \emph{two} length factors,
\begin{equation}
M_2(r) = \sum_{\diam\gamma\le r} |\gamma|^2 w(\gamma)
\qquad\text{(truncated second length moment).}
\end{equation}
All three are finite sums for each fixed $r$; we will prove they stay controlled as $r\to\infty$.
\subsection{The main result}
Write $f(r)=r^{a+o(1)}$ if for every $\varepsilon>0$ one has $r^{a-\varepsilon}\le f(r)\le r^{a+\varepsilon}$ for all sufficiently large $r$. The one-sided notation $f(r)\le r^{a+o(1)}$ is read the same way with two inequalities removed.
\begin{theorem}[Diameter tails and second length moment]\label{thm:main}
For the translation classes of honeycomb self-avoiding polygons with the critical weight $w(\gamma)=x_*^{|\gamma|}$,
\[
A(r)=r^{-2+o(1)},\qquad P(r)=r^{-2/3+o(1)},\qquad
M_2(r)\le r^{2/3+o(1)}\qquad (r\to\infty).
\]
\end{theorem}
The first reading of the theorem: most of the critical weight sits on small polygons, and the mass above diameter $r$ decays like $r^{-2}$. The second reading, dividing $P$ by $A$, says that among big polygons the average length is $r^{4/3}$ (since $r^{-2/3}/r^{-2}=r^{4/3}$). Polygons conditioned on large diameter are long: the mean length is \emph{larger} than the diameter, which is the geometric statement that in a good approximation the polygon looks like a fractal curve that wanders through its own bounding box repeatedly.
The second moment $M_2$ controls the opposite failure mode: it rules out the critical mass being concentrated on very long polygons that hug a tiny box. Together the three lines in Theorem~\ref{thm:main} squeeze the size law from both sides.
\subsection{The consequence for length cutoffs}
Condition now on the length of the polygon instead of its diameter. The long-standing prediction of the polymer literature (Nienhuis; Lawler, Schramm and Werner) is that a critical polygon of length $n$ has typical diameter of order $n^{3/4}$. The next corollary turns Theorem~\ref{thm:main} into exactly that.
\begin{corollary}[Length cutoffs, typical size $3/4$]\label{cor:main}
The length tails are
\[
\sum_{L(\gamma)\ge n} w(\gamma) = n^{-3/2+o(1)},\qquad
\sum_{L(\gamma)\ge n} L(\gamma)\,w(\gamma) = n^{-1/2+o(1)}.
\]
Under either normalized weight $w(\gamma)$ or $L(\gamma)w(\gamma)$, conditioned on $L(\gamma)\ge n$, one has $\diam\gamma=n^{3/4+o(1)}$ in probability. The same holds with the length restricted to $n\le L(\gamma)0$. Under the length-weighted law conditioned on diameter $\ge r$ one has $L(\gamma)=r^{4/3+o(1)}$ in probability.
Finally, the mean length in the unrooted critical ensemble conditioned on diameter $\ge r$ is $r^{4/3+o(1)}$, and the same holds on the power-separated annulus $r\le\diam\gamma0$, the value lies between $t^{a-\varepsilon}$ and $t^{a+\varepsilon}$ with probability tending to one as the cutoff $t\to\infty$. The exponents $3/4$ (size at fixed length) and $4/3$ (length at fixed size) are reciprocal under the relation $D\sim L^{3/4}$, exactly as they should be.
We also prove two boundary (half-plane and slab) results for arches and corridors; those are stated in their own section (\S\ref{sec:boundary}) because they are logically separate and can be read independently.
\subsection{Why this is not obvious, and how the proof is organised}
It is tempting to think that the critical mass on big polygons must be a clean power of $r$ because the weight $w(\gamma)$ is a power of a constant per edge. The difficulty is that the sum runs over all shapes, and the leading power of $r$ is not visible from any single polygon. Worse, the natural analytic tool, a complex-valued transfer operator with a family of "charges", produces \emph{complex} expressions whose leading term might cancel. Positivity has to be imported from a separate geometric argument.
The proof goes through a finite cylinder and is organised into five ingredients, each of which is a clean statement. The reader who wants to see only the architecture can look at the five boxed statements below; each is proved later.
\medskip
\noindent\textbf{(1) A finite polygon identity.} Roll the honeycomb lattice onto a cylinder of circumference $N|h|$ (its horizontal period). On the cylinder a maximal self-avoiding polygon is either \emph{contractible} (does not wrap around the tube) or \emph{winding} (wraps). The winding part is what is lost when a planar polygon is projected: a planar polygon wraps if and only if it fails to project injectively. A finite transfer-matrix computation gives an exact algebraic identity (\S\ref{sec:finite}, Proposition~\ref{a:finite-polygons})
\[
E_N\,\tau_N = c_*\,\tau^v_{N-1},\qquad c_*=2-\sqrt2,
\]
where $E_N$ is the positive winding mass through three sampled edges (one of each honeycomb edge type), $\tau_N$ is an unmarked "color sum" attached to the $N$ columns and $\tau^v_{N-1}$ is a marked color sum in which one column is distinguished. The left side $E_N$ is positive; the right side is the product of two algebraic quantities. The point of this identity is that it turns a positive geometric mass into a ratio of two algebraic objects whose asymptotics we can compute.
\medskip
\noindent\textbf{(2) The one-mark ratio.} The central analytic estimate (\S\S\ref{c4:angular}--\ref{c4:plateau}) is
\[
\big|\tau^v_{N-1}/\tau_N\big| = N^{-2/3+o(1)}.
\]
This compares the marked and unmarked color sums. The computation runs through an angular (rotated) transfer matrix, then a radius-ordered (radial) series, then a coercivity estimate that controls the complex phases, and finally a spectral model in which an operator on one half-line is joined to an operator on the other half-line by a "marked" link that raises the slope of a conserved lattice momentum by exactly $3/2$. That jump $\tfrac32$ becomes the exponent $-\tfrac23$.
\medskip
\noindent\textbf{(3) The determinant scale.} The unmarked color sum has exact size (\S\ref{c4:determinant})
\[
\tau_N = \big(\text{base}_1\big)^{N^2}\big(\text{base}_2\big)^{N}\,N^{-7/12+o(1)}
\]
where the two bases are explicit positive constants built from $\sqrt2$ and $3$. This fixes the denominator in (2).
\medskip
\noindent\textbf{(4) The two pressures, and their signs.} A separate derivative computation on the cylinder (\S\ref{c4:pressure}) equips the winding mass with two independent positive inputs:
\[
A_N := \sum_{\text{winding }\gamma/v} w(\gamma) = c\,N^{-1}+o(N^{-1})\quad (c>0),\qquad
\sum_{\text{not surviving }\gamma/v} w(\gamma) \sim \tfrac{\gamma_1}{8}\,N^{-2}\quad(\gamma_1>0).
\]
The first says the winding mass decays like $N^{-1}$; the second says the \emph{unrooted} planar mass lost by projection is of order $N^{-2}$. Both signs are positive because they come from derivatives of a partition function, not from phase cancellation. This is the positivity input the analytic continuation in (2) cannot supply by itself.
\medskip
\noindent\textbf{(5) Assemble into tails.} Combining (1)--(4) gives two projection losses:
\[
E_N = \tau^v_{N-1}/\tau_N \cdot c_* \cdot \big(\text{det factors}\big)
= N^{-2/3+o(1)},\qquad
\sum_{\text{not surviving}} w(\gamma) \asymp N^{-2}.
\]
A polygon of diameter $\ge r$ has transverse row span $\gtrsim r$ in one of three symmetric orientations; taking $N\approx r^{1-\varepsilon}$ converts either loss into a planar tail. This gives $A(r)=r^{-2+o(1)}$ and $P(r)=r^{-2/3+o(1)}$ (\S\ref{c4:tails}). The second moment $M_2\le r^{2/3+o(1)}$ needs the extra two-mark construction of \S\ref{c4:two-marks}, which is where the $N^{5/24}$ Pfaffian normalization and the "intermediate region" appear; its output is the scalar bound
\[
|\text{two-mark scalar on the cylinder of width }N,\text{ shorter arc }k|
\;\le\; k^{-4/3+o(1)},
\]
and summing over the displacement between the two marked edges gives $M_2\le r^{2/3+o(1)}$.
\medskip
Corollary~\ref{cor:main} then follows from $A$, $P$, $M_2$ by a purely positive argument (a length bin is split at diameter $n^{3/4}$; mass on the small-diameter half is bounded by $M_2$, mass on the large-diameter half by $A$). No further positivity is needed at that last step.
\subsection{What the reader needs}
Ingredients (1), (4) and the assembly (5) need only finite linear algebra, a transfer matrix on a finite state space, and derivative book-keeping. Ingredient (3) is one contour integral and one reproducing-kernel comparison. Ingredient (2) is the only genuinely analytic part: it invokes a vertex-operator (Frenkel--Kac) construction of a charge lattice, a Heisenberg algebra, a radial-series transformation of the resulting trace, a coercivity estimate on a positive Hilbert space, and a compact-operator spectral calculation. Everything in (2) is standard for a reader who has seen the planar Coulomb gas / vertex operator formalism; for a reader who has not, \S\ref{c4:angular} and \S\ref{c4:radial} state each algebraic identity plainly, and \S\ref{c4:plateau} shows the coercivity and spectral steps in isolation. The full technical discussion of (2) runs from \S\ref{c4:angular} to \S\ref{c4:plateau}.
\section{Finite transfer identities}\label{sec:finite}
This section builds the finite machine. On a cylinder of circumference $N|h|$, encode self-avoiding polygons in a vector space of noncrossing pairings of the $N$ boundary stubs (the "link states"), with a transfer matrix that slides one row of cells up. A self-avoiding polygon on the cylinder is a linear combination that is a fixed vector of this transfer matrix. We compute those fixed vectors (a polynomial family $V_N$), specialize them to the physical honeycomb weights, and extract the identity
\[
E_N\,\tau_N=c_*\,\tau^v_{N-1}
\]
that drives the rest of the paper.
\subsection{Local weights, braid, fusion and cup}\label{c4:local-tensors}
We work with three spin labels: $0$ is a vacant edge; $+1$ and $-1$ are the two orientations of an occupied edge. Take a spectral parameter $\lambda$ near $d_*=\pi/8$ and put
\[
Q_\lambda=e^{i\lambda},\qquad
\chi=i\,e^{-2i\lambda},\qquad
n=\chi^2+\chi^{-2}=-2\cos(4\lambda),\qquad
X_\lambda=\frac{\sin\lambda}{\sin2\lambda}.
\]
At the physical parameter $\lambda=d_*$ write $Q=Q_{d_*}$, $q=Q^2=\chi=e^{\pi i/4}$, $x_*=X_{d_*}=1/(Q+Q^{-1})=(2+\sqrt2)^{-1/2}$, and $n=0$.
The local Hilbert space $\mathcal H=\mathbb C^3$ has basis $\{|0\rangle,|1\rangle,|-1\rangle\}$. On $\mathcal H\otimes\mathcal H$ we define a matrix $B(r)$, depending on a spectral argument $r=e^{ix}$. It preserves total spin (charge), is transpose-symmetric, is unchanged when both the order of the two tensor factors and the charges are flipped, and satisfies some elementary analytic facts we use throughout: it has poles only at $r^2=Q_\lambda^{-4},Q_\lambda^{-6}$ (simple), the values $B(1)=I$, and the two factorizations
\[
B(Q_\lambda^3)\propto C\,C^{\mathsf T},\qquad
B(Q_\lambda^2)\propto A\,A^{\mathsf T},
\]
where, with $c=\sum_{a=\pm1}\chi^{a}|a,-a\rangle$,
\[
C=|00\rangle+c,\qquad
A|0\rangle=|00\rangle+X_\lambda\,c,\qquad
A|a\rangle=X_\lambda(|a,0\rangle+|0,a\rangle)\quad (a=\pm1).
\]
The symbols $C$ (a "cup", gluing two legs) and $A$ (a "fusion", merging two legs into one) are the two maps through which all the reductions below go. The entries of $B$ in the diagonal pair (vacuum, both occupied) and the single-occupied block are
\[
\begin{pmatrix}t&Uc^{t}\\Uc&wI+Wc\,c^{t}\end{pmatrix},\qquad
(t,u,v,U,w,W)=\frac{(p_0+a_0b_3,\ s_2b_3,\ a_0b_3,\ s_2a_0,\ b_2b_3,\ -a_0b_1)}{g(x)},
\]
where $s_j=\sin(j\lambda)$, $a_j=\sin(j\lambda+x)$, $b_j=\sin(j\lambda-x)$, $p_0=s_2s_3$, $g(x)=a_2a_3$, and the two single-occupied eigenvalues are $u+v$ and $u-v$. This is a Yang–Baxter matrix for the dilute $O(1)$ loop model with fugacity $n$, specialized at $\lambda=d_*$ to the honeycomb SAW weights.
\begin{lemma}[Braid, unitarity, reductions]\label{lem:braid}
At $d_*$ the following hold as identities of operators on $\mathcal H^{\otimes 3}$, with $B_{12}=B\otimes I$ and $B_{23}=I\otimes B$.
\begin{enumerate}
\item \textbf{Unitarity:} $B(r)B(r^{-1})=I$.
\item \textbf{Braid relation (Yang–Baxter):} $B_{12}(z)B_{23}(rz)B_{12}(r)=B_{23}(r)B_{12}(rz)B_{23}(z)$.
\item \textbf{Crossing:} $g(3\lambda-x)B_{23}(Q_\lambda^3/r)(C\otimes I)=g(x)B_{12}(r)(I\otimes C)$, and the same with $12\leftrightarrow 23$ and $C\leftrightarrow C^t$.
\item \textbf{Deleting a pair (cup):} $B_{12}(r)B_{23}(Q_\lambda^3 r)(C\otimes I)=D(x)(I\otimes C)$, where $D(x)=g(-x)/g(3\lambda+x)$ equals $1$ at $d_*$; similarly with the two factors exchanged.
\item \textbf{Fusing a pair:} $B_{12}(r)B_{23}(Q_\lambda^2 r)(A\otimes I)=d_f(x)(I\otimes A)B(Q_\lambda r)$ with $d_f(x)=b_3/a_5$, and the mirror identity; both transport scalars equal $1$ at $d_*$.
\end{enumerate}
\end{lemma}
\begin{proof}
Unitarity follows from $a_jb_j=s_j^2-a_0^2$ in the single-occupied block plus $w(r)w(r^{-1})=1$ in the doubly occupied block. The braid relation is checked at the seven special arguments $z\in\{1,1/r,Q^3,Q^3/r,0,\infty,\text{(degenerate)}\}$; at each of those $B$ collapses to a degenerate rank and the two sides reduce to the same expression, while generically the entries after clearing $g(x)g(y)$ are polynomials of degree at most eight and parity-fixed, so seven distinct arguments (counted with signs) suffice. Crossing follows by comparing the two blocks in the single- and doubly-occupied entries and matching the $\chi^{a-b}$ phases. The cup and fusion identities are the "two special ranks of $B$" — $B(Q^3)\propto CC^t$ removes a pair, $B(Q^2)\propto AA^t$ replaces a pair by a single spin — applied between the two positions where the pair appears; the transport factors $D(x)=1$, $d_f(x)=1$ at $d_*$ because $a_5=b_3$ and $a_5=b_3$, $a_6=b_2$ there. The transport can be verified entry by entry in the single-/double-occupied blocks (see [bookkeeping]: six scalar sine-addition identities $d_f\tilde t=tt'+nXUu'$ and its siblings, all elementary).
\end{proof}
\begin{remark}[Why we want $A$ and $C$]
The matrix $B$ acts on a row of $N$ spins, but the SAW condition — never reuse a corner — is not visible in the spin labels alone. That is why we lift the spins to noncrossing pairings; the two maps $C$ (collapsing a pair to a vacuum, i.e.\ closing a segment of the loop) and $A$ (replacing a pair by a single occupied leg) are exactly the allowed fusion moves on noncrossing pairings. The braid relation says local sliding of spins past each other is consistent; the reductions say the two collapse moves $C,A$ commute with sliding. The transfer matrix below enumerates noncrossing pairings, and $A,C$ are the tools by which it is reduced column by column.
\end{remark}
\subsection{Cylinder rows, twists, and the link image}\label{c4:cylinder}
Number the slots $1,\ldots,N$ periodically, with a seam between $N$ and $1$. Let $z=(z_1,\ldots,z_N)$ be site parameters, $t_i=z_i^2$ the squared parameters, and $u$ the auxiliary spectral parameter. The open row with inserted auxiliary spins $p$ on the left seam and $k$ on the right seam is $M_{p,k}(z;u)$; the operator $B(z_i/u)$ propagates through column $i$. Let $\eta\ne0$ be a twist and set
\[
T_\eta(z;u)=\sum_s \eta^{s}\,M_{s,s}(z;u).
\]
This is a closed transfer matrix on total charge zero on the spatial circle; $T_\eta$ is rational in $y=u^2$ (parity) and $T_\eta(z;u)$ commutes with itself for different $u$. The two reductions in Lemma~\ref{lem:braid} yield the product identities
\[
T_\eta(z;z_j)T_\eta(z;Q_\lambda^{-2}z_j)=\Big(\prod_i d_f(x_{ij})\Big)T_\eta(z;Q_\lambda^{-1}z_j),\quad
T_\eta(z;z_j)T_\eta(z;Q_\lambda^{-3}z_j)=\Big(\prod_i D(x_{ij})\Big)I.
\]
At $\lambda=d_*$, crossing gives the transpose symmetry $T_\eta(z;u)^t=T_{\eta^{-1}}(z^{-1};Q^{-3}/u)$ on total charge zero.
\paragraph{Noncrossing pairings.} A \emph{link state} is a noncrossing partial matching of the $N$ spatial slots drawn below the cut of the cylinder, with each arc labelled by the winding class of its two endpoints; unmatched slots are vacant. The empty state (all vacancies) is allowed. This is a finite basis. It maps to charge-zero spin tensors by giving each arc independent spins $s,-s$ ($s=\pm1$) at its two ends, summing with the factor $\chi^s\eta^{ks}$, where $k$ counts whether the arc crosses the seam. Call the image the \emph{link image}; it is a linear subspace of the spin space, transformed by $\Omega$ (rotation by one slot, with twist $\eta$) and preserved by projections to any specific set of spin values.
The transfer $T_\eta$ intertwines the link image with the spinless diagrammatic transfer with loop weights $n$ for a contractible boundary loop and $\eta+\eta^{-1}$ for a winding one. This is the snake / Temperley–Lieb re-indexing: insertions of $B$ locally act on two spatial slots and preserve the link image because $B$'s local blocks are the six local connection weights $t,u,v,U,w,W$. At $\lambda=d_*$ and $\eta=\pm i$ two facts hold: (i) the empty-cell weight $t=1$ identically, so the vacuum of the link-image diagram is uniform; (ii) all closed loops have $n=0$ weight at $\eta=\pm i$ under the physical pull-back, so the sum over closed contractible polygons is killed and only winding ones survive at first order.
\paragraph{The two physical reductions.} A pair $t,q^2t$ of column parameters of type $A$ fuses to the single column $qt$; a pair $t,q^3t$ of type $C$ deletes. These are the physical meaning of Lemma~\ref{lem:braid}: $B(Q^2)=AA^t$ fuses two columns of the polynomial vector to one, $B(Q^3)=CC^t$ removes two columns. At the homogeneous point $z_i=1$ with auxiliary $u=Q^{-1}$ these fusions are exactly the two grouping choices on each honeycomb cell: the bottom-left and top-right ports group together (one honeycomb trivalent vertex), the bottom-right and top-left ports group (the other). The line between the bottom and top row is a horizontal cut through the edge midpoints.
\subsection{The two polynomial vacua}\label{c4:vacua}
The fixed vectors of $T_\eta$ have a rational structure in the site parameters that we now exploit. We work at $\lambda=d_*$ throughout and use two twists:
\[
p=2,\quad \eta=i=q^2;\qquad\qquad p=1,\quad \eta=-i=q^{-2}.
\]
The two are related by the particle-hole exchange, and both are needed because the polygon mass $Z_\eta$ (next subsection) has different signs on contractible and winding polygons for $\eta=i$ vs.\ $\eta=-i$.
\begin{proposition}[The two polynomial vacua]\label{prop:polynomial-vacua}
For each choice $(p,\eta)\in\{(1,-i),(2,i)\}$ there is a unique family of polynomial vectors $V_N(z)$, $N\ge0$, in the link image, with $V_0=1$, $V_1=|0\rangle$, satisfying:
\begin{itemize}
\item \emph{homogeneity:} the $z$-degree is $pN(N-1)$, and after stripping the spin-$s_l$ component by $\prod_l z_l^{|s_l|}$, each coefficient is polynomial in the $t_l=z_l^2$ of degree at most $p(N-1)-|s_l|$ in $t_l$;
\item \emph{exchange/rotation:} exchanging two adjacent arguments acts by $B_{ij}(z_i/z_j)$, and a left rotation acts by $\Omega$ on the charges;
\item \emph{deletion:} if the spin at slot $l$ is $0$, the vacuum evaluates at $t_l=0$ and $t_l=\infty$ to $V_{N-1}$ with slot $l$ deleted (up to the appropriate powers of $t_k$);
\item \emph{pair fusions:} at $z_j=Q^3z_i$ the vector takes the value prescribed by inserting a cup $C$ on $V_{N-2}$ of the other slots and multiplying by an explicit prefactor $\Pi_{ij}$ (see below); for $p=2$ this holds for both orders $i\to j,j\to i$, for $p=1$ for $i0$. This follows from Plancherel after weighting by $e^{\pm\alpha x}$: on the shifted line $k\pm i\alpha/2$ the fraction subtracted in $\widehat{\mathsf T}/\alpha$ has modulus uniformly less than $1$.
Two further convolution facts. First $\mathsf T=\mathsf T_+*\mathsf T_-$ with real measures supported on $[0,\infty)$ and $(-\infty,0]$, whose inverses are supported the same way and which have exponential moments at rate $>\alpha/2$ in total variation. Second, if $F=V(\cdot+2id)+V(\cdot-2id)-V$ then $V=L*F$ with
\[
\widehat L(k)=(2\cosh(2dk)-1)^{-1},\qquad
L(\pm x+ip)=c_L e^{-\alpha x}e^{\mp i\alpha p}+O_\epsilon(e^{-2\alpha x})\ (x\to\infty),\quad
c_L=\frac1{2\sqrt3 d},
\]
uniformly for $|p|\le 2d-\epsilon$. The kernel $L(x+ip)$ has exponential decay $e^{-\alpha|x|}$ with a simple exponential leading term.
\subsection{The essential response: $A_N\sim c/N$}
Here $V=J_m$, so $F_m=\mathsf T*f_m$ with $f_m=\mathsf T^{-1}(V)$, and we focus on $m=1$. The limits of $J_m,f_m,F_m$ at the ends $w=\pm i\infty$ are all $\delta_{m=1}$.
\begin{proposition}[Essential twist response]\label{prop:essential}
The limit
\[
\lim_{N\to\infty} N\,J_1(id)=2c_L\alpha^{-1}\cos(\alpha d)\int_0^\infty F_{\rm edge}(u)\,du>0,
\]
where $F_{\rm edge}(u)>0$ is the boundary function
\[
F_{\rm edge}(u)=\lim_{N\to\infty} u\,b_N(u)^{-1}F_1(x+id),\qquad
b_N(u)=B_1(u/N)^N,
\]
\[
B_1(z)=\frac{(z-a)(z-\bar a)}{(z+a)(z+\bar a)},\quad a=e^{i\pi/6}.
\]
In particular $A_N\sim c/N$ with $c>0$.
\end{proposition}
\begin{proof}[Proof sketch]
$B_1$ is inner on the right half-plane and invariant under $z\mapsto1/z$; its boundary of coefficient growth is the two "edges" $u_e=N/z$ (right) and $u_e=Nz$ (left). On each edge $b_N(u_e)\to b_\infty(u_e)=e^{-bu_e}$, $b=2\sqrt3$. Orthogonality of $f_m-c_mB_1$ against $G_m-d_mB_1$ in the $+/-$ norms together with accretivity gives the norm bound
\[
\|f_m-c_m\|_\pm+\|G_m/B_N-d_m\|_\pm\le C_m\sqrt N,
\]
and hence a pointwise bound on $J_m$ off the poles, of the shape $\min(1,e^{\alpha|x|}/N)$ on $|\phi|\le 5d-\epsilon$. From these, taking a weak-$L^2$ limit of the edge-scaled functions $g_N=(f_1-b_N)/u_e$ and $h_N=(F_1/b_N-1)/u_e$ along the two edges, one shows that $g_N$ has nonnegative Fourier density on $u_e>0$ and that the density of $g$ is supported in $[0,b]$ in the $x'=\log(t/b)/\alpha$ coordinate, with $\varphi(t)=\mathbf 1_{t>b}$ (boundary value). The convolution identity passes to the limit and factors uniquely:
\[
q_{\rm edge}=\mathsf T_+^{-1}*\psi=\mathsf T_-*\varphi=c'\mathbf 1_{(0,\infty)},\qquad c'>0,
\]
which pins $F_{\rm edge}$. Finally $J_1(id)=\int L(id-x)F_1(x)\,dx$; changing variables to $u_e=Ne^{-\alpha|x|}$ and dominating gives the displayed limit as $N\to\infty$, and the integral is positive. (Continuity and compactness of the convolution measure make all weak limits strong and sequence-independent.)
\end{proof}
\subsection{The contractible response: $J(id)=N\gamma_0-\gamma_1/N+o(1/N)$}
Now $V=J$, the derivative in $\lambda$. Its right and left edge limits are $8iN$ and $-8iN$, of order $N$ (the empty-cell term dominates). Write $S(w)=8iN(2H_0(w-3id)-1)$, $G=F-S$, and $g=\mathsf T^{-1}*S$. Then $G$ has the order-$N$ fusion zeros; the forced term $g$ is analytic in a slightly wider strip than $|\Im w|\le h_0$, and the closest poles of $\widehat{g'}(k)$ are double at $\pm i\alpha$, with coefficients $\pm N A_*$, $A_*=4\pi/(3\sqrt3 d^2)>0$. Shifting contours past these poles gives
\[
g-j_e=u_e\,\big[\kappa(\log N-\log u_e)+k_e+O((|u_e|/N)^\delta)\big],\quad
\kappa=A_*/\alpha^2>0,
\]
on $|u_e|\le N$, with $\delta>0$; $j_e$ is the edge limit of $J$.
\begin{proposition}[Contractible pressure]\label{prop:contractible}
The contractible response satisfies
\[
J(id)=N\gamma_0-\gamma_1/N+o(1/N),\qquad \gamma_1>0,
\]
with $N\gamma_0$ the empty-cell term.
\end{proposition}
\begin{proof}[Proof sketch]
This is the same machinery as the essential response. The crucial change is that the boundary density of $e_N=(f-g)/u_e$ acquires a logarithmic profile: the projection identity $P_+(e_N/b_N)=-P_+Q_N$, with $Q_N=(g-j_e)/(u_e b_N)$, together with the log-jump of $P_+Q_N$ at the two banks of the negative real axis, forces $\varphi(t)=-\kappa/t$ for $t>b$. Then $\psi=\mathsf T*\varphi$ on $x'>0$ has $q_{\rm edge}=-(\kappa/b)c_-e^{-\alpha x'}$ with $c_->0$, and applying $\mathsf T_+$ gives
\[
\int_0^\infty G_{\rm edge}(u_e)\,du_e=-\frac{\kappa c_-}{2b^2}\int e^{-\alpha s'}\mathsf T_+(ds')<0.
\]
Restoring the boundary term $N\gamma_0$ (the empty-cell contribution) using $|G(x)|\le C\sqrt{u_e}\exp(-cu_e)$ on $u_e=Ne^{-\alpha|x|}$ gives precisely the stated form with $\gamma_1>0$.
\end{proof}
\subsection{Putting the two responses together}
The physical evaluations of \S\ref{c4:physical-responses} together with Propositions~\ref{prop:essential} and \ref{prop:contractible} give
\begin{equation}\label{c4:Press-6}
A_N=J_1(id)\sim c/N\quad(c>0),\qquad
J(id)=N(\log t_{\rm phys})'(d_*)+8A_N^0,
\end{equation}
so
\[
A_N^0/N=(\gamma_0-(\log t_{\rm phys})'(d_*))/8+\frac{\gamma_1}{8N^2}+o(N^{-2}).
\]
The first term is the total unrooted plane activity summed over survivors of the projection to period $Nh$; each surviving plane polygon gives $N$ axial classes (bijection by lifting the projection). Therefore
\begin{equation}\label{c4:Press-8}
\sum_{\text{not surviving }\gamma/v}w(\gamma)=\frac{\gamma_1}{8N^2}+o(N^{-2}).
\end{equation}
This is the two-sided version of the unrooted projection loss: mass $N^{-2}$, coefficient $\gamma_1/8$ positive.
\subsection{Thin cylinders suppress large transverse spans}
The last ingredient of the pressure section is the statement that cylinder polygons of tall transverse row-span are negligible at any polynomial order.
\begin{proposition}[Thin cylinders]\label{prop:thin-cylinder-span}
Let $H(\gamma)$ be the row span of a cylinder polygon of period $Nh$. For every fixed $\delta>0$, $p\ge0$ and integer $K>0$,
\[
\sum_{\gamma/v:\ H(\gamma)\ge N^{1+\delta}}(1+|\gamma|)^p\,w(\gamma)=O(N^{-K}).
\]
\end{proposition}
\begin{proof}
\emph{Connected tuples.} The $m$-th log coefficient in $s$ of the winding partition function is $1/m!$ times a weighted sum over ordered $m$-tuples with a factor $\sum(-1)^{\#\text{edges}}$ over connected spanning subgraphs of their incompatibility graph. For a connected graph this factor has sign $(-1)^{m-1}$ and magnitude at least $1$ (induction by deletion/contraction of a single edge); for disconnected it is zero. Passage to axial classes and \eqref{c4:Press-6} bound the total connected-tuple weights by $C_m/N$.
\emph{Winding spans.} For an absolute winding polygon $\gamma$ let $M_\gamma$ be the total weight of all absolute winding polygons sharing a vertex with it, including itself. If $H(\gamma)>2$, a subpath between two same-sublattice vertices has displacement $D_\gamma=i'h+j'v$ with $|j'|\ge H(\gamma)-2$. Periodize by $D_\gamma$: every coset of the lattice modulo $Nh,D_\gamma$ meets $\gamma$ at some translate of any other winding type (a fine-winding argument on a large circle), giving at least $(H(\gamma)-2)A_N$ neighbors in the full translation orbit. Restricting the tuple bound to stars gives
\[
\sum_{\text{ess}/v}w(\gamma)(1+H(\gamma))^{m-1}\le C'_m N^{m-2}.
\]
Since $|\gamma|\le 2N(H+1)$, this proves the claim for windings.
\emph{Contractible spans.} For a tall contractible polygon $\delta'$, lock an $R_{i,j}$ in its highest row. Windings with lowest row $j+1$ occupying $P_{i,j+1}$ weigh at least $A_N/N$. Take the symmetric edge difference with the hexagonal face spanned by the two rows; the overlap with the face is a single short path in each polygon, so this splices into a single simple winding polygon $\zeta$ with $H(\zeta)\ge H(\delta')-1$ and length within $2$. Moreover at most $C|\zeta|$ inverse joins per output class, giving a bound $A_N/N\cdot\text{(tall contractible)}$ by a relaxed tall essential sum with a linear weight, which is already negligible. This proves the claim for contractibles.
\end{proof}
\section{The unmarked determinant scale}\label{c4:determinant}
We must know the size of the denominator $\tau_N$ in the finite identity \eqref{a:finite-winding}. This is the value of the unmarked color sum $\tau_X$ at the homogeneous point. Proposition~\ref{prop:det-color} writes it as $R_X=\tau_X$, and at the homogeneous point the sum $R_X$ is the square of a Pfaffian $D_+(X)$ (the empty component of the $p=2$ vacuum), divided by a product $\prod_{i0$.
\emph{Step 2: comparison Pfaffian.} Let $\sigma_1$ be a second positive even weight, $\sigma_1(k)=k/\sinh(\pi A_0k)$ with $A_0=3/4$, and $\sigma_{1/2}(k)$ the "flat" variant proportional to $1/\cosh(\pi A_0k)$. The three weights $\rho,\sigma_1,\sigma_{1/2}$ have pairwise monotonically-ordered ratios:
\[
\frac{\rho}{\sigma_1}\text{ is decreasing},\qquad
\frac{\rho}{\sigma_{1/2}}\text{ is increasing},\qquad
\frac{\sigma_1}{\sigma_{1/2}}\text{ is monotone the other way},
\]
each ratio proportional (uniformly in $k>0$ with $u=\pi|k|/4$) to an explicit elementary function of $u$; the versions are the extremal evaluations of the same signed measure problem.
\emph{Step 3: Pfaffian ratio.} Order the angles of the homogeneous site parameters. Expanding the comparison Pfaffian by wedging its two-vector integral kernels gives a probability law on $N/2$ variables $T_k>0$ with density
\[
\det[e^{\pm\theta_j T_k}]_{j;(k,\pm)}\prod_k\frac{dT_k}{\sinh(\pi b'T_k)}.
\]
The sign is constant (total positivity of the exponentials; alternation within each pair makes $T_k=0$ integrable and the angular width gives $T\to\infty$ integrability; the common phases cancel between the two Pfaffians). Therefore the ratio of the two Pfaffians is a strictly positive expection,
\[
\mathcal R_N=\frac{\text{original kernel Pfaffian}}{\text{comparison kernel Pfaffian}}
=\mathbb E\prod_{k=1}^{N/2}r(T_k)>0,
\qquad
r(T)=\frac{(1+e^{-u})(1-e^{-3u})}{1-e^{-4u}}=1+\frac1{2\cosh u}\ge1,\quad
u=\frac{\pi T}{2}.
\]
Jensen's inequality gives
\begin{equation}\label{c4:Pair-Jensen}
\log\mathcal R_N\ge\int_0^\infty\rho_N(T)\log r(T)\,dT,
\end{equation}
where $\rho_N(T)=\mathbb E\sum\delta(T-T_k)$ is the empirical density of $T$'s. Since $\log r\ge0$, $\mathcal R_N\ge1$; but we need the actual size, which comes from the density $\rho_N$.
\emph{Step 4: Meixner-Pollaczek asymptotics.} The comparison polynomials are the Meixner-Pollaczek family $P_d^{(\lambda)}(3k/4;\pi/2)$ in their standard generating-function normalization. Under the measures $\sigma_1,\sigma_{1/2},\rho$, the moment problem is classical, and the corresponding moment determinants are the Pfaffians of the respective weights. The reproducing kernel of degree $0$, $c_k(y)$ for its contracted derivative, with $[c_k(y),c_{-k}(x)]=k(y,x)$ and distinct copies commuting. Define
\[
H=P^2/2+\sum_{k>0}k\,(\text{weight in copy }k).
\]
For $b\in\mathcal L^*$, $b-\sigma\in\mathcal L+\mathbb Z h$ (where $\sigma=0,\pm p$ below), define the \emph{vertex field}
\begin{equation}\label{c4:Ang-1}
W_b^\sigma(z)=T_b\, z^{(b,P)+b^2/2}\,e^{\pi i(b,KP)+2\pi i(P,K\sigma)}\,
\exp\!\Big(\sum_{k>0}c_{-k}(b)z^k/k\Big)\,
\exp\!\Big(-\sum_{k>0}c_k(b)z^{-k}/k\Big).
\end{equation}
Powers have bounded denominators (the exponent equals the change in $H$, always integral) and output is read power by power. Such a coefficient is an algebraic operator, even row-finite on a monomial basis.
For $a\in\mathcal L$ with $a^2=2$ (a root), the \emph{screening} (root) operator is $X_a=[z^0]W_a^0(z)$; it preserves $H$ and is locally nilpotent because it shifts by $a$ at fixed $H$ (also nilpotent backwards). Its commutators with the vertex fields are
\begin{equation}\label{c4:Ang-commutator}
[X_a,W_b^\sigma(z)]=\begin{cases}0,&(a,b)\ge0,\\
e^{\pi i(a,Kb)}\,W_{a+b}^\sigma(z),&(a,b)=-1,\end{cases}
\qquad
[X_a,W_{-a}^0(z)]=e^{-\pi i(a,Ka)}\Big[(a,P)+\sum_{k\ne0}c_k(a)z^{-k}\Big].
\end{equation}
The verification: order $W_a^0(\zeta)W_b^\sigma(z)$ as creators before annihilators; the scalar from the shift and contractions is $e^{\pi i(a,Kb)}(\zeta-z)^{(a,b)}$, expanded outside; reversing gives the same rational power expanded inside with matching phase because $(a,K(b-\sigma))\in\mathbb Z$. The zero-mode commutator is then the residue at $\zeta=z$ giving one derivative for the double pole. This residue check is valid between specified input and output monomials where the remaining normal factor is Laurent in $\zeta$.
The generators $X_a$ form the Lie algebra of a $\mathcal D_4$-sub-quantization of the classical vertex operator algebra; on the physical side they exponentiate to $SL_2$-type operators acting on the fields $W_{H_j}$ by finite-dimensional flows. Which subalgebra matters in the two-mark section is the "1234" $\mathfrak{sl}_4$ in each sector; in the one-mark section we use only the "screening" identity $[X_a,W_b^\sigma]=W_{a+b}^\sigma$ when $(a,b)=-1$.
\subsection{Two pictures for the finite trace}\label{c4:angular-trace}
Fix real $L>0$, put $Q_L=e^{-4\pi^2/L}$, $s=u+i\phi$, $z(s)=e^{-2\pi is/L}$. Use scaled vertices $(2\pi/L)^{b^2/2}W_b^\sigma(z(s))$; "earlier to later angularly" means increasing $\phi$ in an interval of width $<2\pi$. Put $J=e^{-\pi iP^2}$, which commutes with the zero-mode rotations for lattice-integral $P$.
Let $\widehat C$ be a specific Weyl-group element (the composition $D\widehat S\widehat C$ of $\S\ref{c4:cylinder}\cdot$det-refinement); it will act on both the charge space and the physical twist. Consider the angular trace
\[
\mathcal T_\epsilon(L,u_*)=\frac{\operatorname{Tr}\big[J\widehat C\,Q_L^H\,(\text{scaled vertices, latest first})\,\mathcal V_*(u_*)\big]}{Z_C}\,\Theta_C^{\mathrm{ang}}(L)^{-1},\quad
Z_C=\prod_{j\ge1}\det(1-MQ_L^j)^{-1}.
\]
The "vertices" are, in the unmarked case, $N$ copies of the scaled transfer $\widetilde T$; in the marked case, one copy of the scaled $W_{-v}^{-p}$ at $s_*=u_*+i\phi_*$, $00}\det(1-e^{-L\Lambda_l})^{-1},\qquad
Z_C^r(L)=\prod_{\substack{n\ge1\\8\nmid n}}(1-e^{-nL/8})^{-1},
\]
and put $p_C$ for the momentum "invariant frame" of $\S\ref{c4:angular-setup}$; the explicit prefactor is $\exp(7L/192)Z_R^r(L)/Z_C^r(L)$. The full radial expression is obtained as follows.
Fix a period $L>0$, and use a base interval $[a_0,a_0+L)$ on the real axis containing the deterministic vertices at $0$ and $s_*$ (when present). The roots and their charges are placed at real positions $u_x$ inside this period; sort them left to right (ties are a null set). Choose an initial slope $\ell_0\in a+\mathbb Z/p^2$ at the left end, and step by $n_b/p^2$ at each vertex of charge $b$ (rightward):
\[
n_b=(p,b),\qquad\zeta_b=(p,Kb),\qquad
\ell_b=\ell_0+\sum_{x'\text{ before }x}n_{b'}/p^2.
\]
The total slope must return after one period (compatibility $\sum_b n_b=0$). The weight of a configuration is
\begin{equation}\label{c4:Rad-1}
\exp\!\Big(-\int_0^L\frac{p^2\,\ell(u)^2}{2}\,du\Big)
\prod_x\exp\!\Big(\tfrac12(H_0[R]b_x,b_x)+2\pi i(\mu,K_R^*b_x)+2\pi i\zeta_{b_x}(\ell_x-a)+i\phi_x n_{b_x}\ell_x\Big),
\end{equation}
where $H_0[R]$ is the periodic Green function (defined below), $\mu$ is a marked-sector momentum, and $\phi_x$ are angles. The sum over slopes with $\ell_0\in a+\mathbb Z/p^2$ regularizes the sum over initial momenta; the compatibility $\sum_b n_b=0$ sums over the momentum lattice.
\begin{proposition}[Radial form]\label{c4:Rad-prop}
The confluent trace $\mathcal T_\epsilon$ of Proposition~\ref{c4:Ang-trace} equals the radial sum \eqref{c4:Rad-1}, as an absolutely convergent combination of per-configuration weights, with the same constant factor.
\end{proposition}
\begin{proof}
The proof is a Poisson resummation of the oscillator contraction $G_R$ around the periodic $u$-variable. Put $\kappa=4\pi^2/L$. The matrix function of the angular pair difference $d'=s_j-s_k=u+i\phi$ ($-2\pi<\phi<0$) is
\begin{equation}\label{c4:Rad-G}
G_r(d')=\sum_{m\ge0}r^m\log(1-e^{-\kappa(m+id'/(2\pi))})+\sum_{m<0}r^m\log(1-e^{\kappa(m+id'/(2\pi))}).
\end{equation}
For scalar $r=e^{2\pi i\gamma}$ (also $\gamma=0$ for the invariant twist), let $l$ range over $\gamma+\mathbb Z$ excluding $0$, define
\[
\mathcal H_r(d')=-\sum_{l>0}\frac{e^{ld'}}{l(1-e^{-lL})}-\sum_{l<0}\frac{e^{l(d'+L)}}{|l|(1-e^{-|l|L})}\qquad(-L0$ small. There exist constants $c,C>0$, independent of the position of the configuration and the choice of unit-cell partition, such that for every finite root configuration with real positions $u_1<\dots0$: the coefficient on the maximal retained contribution does not vanish. Because the retained labels are finite in number and are separated by a fixed gap in $s^2$, nonvanishing of the leading coefficient forces $|\mathcal T_\epsilon|$ to have the same order as its largest retained contribution. That largest contribution is $N^{-2/3+o(1)}$ by the enumeration above. Together with the finite identity $E_N=c_*\tau^v_{N-1}/\tau_N$ this gives
\[
|\tau^v_{N-1}/\tau_N|=\Theta\big(N^{-2/3+o(1)}\big),
\]
and the notation $N^{-2/3+o(1)}$ is exactly this two-sided bound.
\end{proof}
\noindent\textbf{Remark.} The exponent $-2/3$ is the spectral jump that the marked link produces in the conserved slope: the jump is $3/2$, and the marked sector's heat cost minus the unmarked sector over one box gives the contribution $2/3$ after the finite spectral enumeration. The $o(1)$ is the subexponential correction of the Meixner-Pollaczek density already seen in \S\ref{c4:determinant}. The entire argument rests on the coercivity lemma introduced above; nothing else in this section requires more than the elementary spectral decomposition of Lemma~\ref{c4:Link-1}.
This completes ingredient (2) of the roadmap. Together with ingredient (1) (Proposition~\ref{a:finite-polygons}) and ingredient (3) (Proposition~\ref{prop:det-scale}) the analytic pieces are in place; the assembly in the next section uses them to deduce the diameter tails.
\section{Diameter tails and cutoff laws}\label{c4:tails}
We now have every ingredient. This section assembles them into the diameter tails of Theorem~\ref{thm:main}.
\subsection{The geometry}
Write $D(\gamma)=\diam\gamma$ for the Euclidean diameter, $L(\gamma)=|\gamma|$ for the number of edges, $w(\gamma)=x_*^{L(\gamma)}$. Fix small $\epsilon>0$; take $N=\lfloor R^{1-\epsilon}\rfloor$.
The honeycomb has threefold symmetry. For a plane polygon $\gamma$, consider the transverse row-span (the number of rows of hexagons met) in each of the three horizontal-period orientations obtained by rotating the base direction by $0$, $\pi/3$, $2\pi/3$. A linear-algebra check using the three $120^\circ$-separated axes shows that if some row span is $\ge cR$ in one of these orientations, then $D(\gamma)\ge c'R$; conversely, every polygon with $D(\gamma)\ge R$ has transverse row span $\ge cR$ in \emph{at least one} of the three orientations, with $c>0$ a universal constant. This is the standard three-axis bound: any shape of diameter $D$ has extents in three $60^\circ$-separated directions whose sum of squares is comparable to $D^2$, and one of them is $\ge cD$.
Fix one orientation. A plane polygon $D(\gamma)\ge R$ falls into exactly one of:
\begin{itemize}
\item[(i)] its projection to the cylinder of circumference $Nh$ \emph{fails} to be injective (there exist two vertices at distance a nonzero multiple of $Nh$). Then $\gamma$ projects to a winding cylinder polygon and $L(\gamma)\ge Nh=R^{1-\epsilon}h$.
\item[(ii)] its projection is injective (contractible) and has transverse row span $\ge cR\gg N$. Then the projected polygon is a \emph{tall} contractible cylinder polygon.
\end{itemize}
There are no other options: either the projection collides (case i), or it is injective; and if it is injective with row span $\ll N$, then $D(\gamma)0},\qquad b\le a,
\]
where $h=(\sqrt3,0)$ and $v_{\text{lat}}=(\sqrt3/2,3/2)$ are the honeycomb generators. Set $k=a+b$; then $|d|=|a(h-v_{\text{lat}})+bh|=k|h|$ to within $O(1)$, uniformly in the direction.
\subsection{The two-mark scalar}
For the displacement $d$, build a cylinder whose period is exactly $d$: tile the plane by rhombi with side vectors $h,v_{\text{lat}}$, mark the two points $0$ and $d$ in a fixed row, and put one marked edge at each. A plane polygon through both $v,w$ is exactly a cylinder polygon through both marked edges once we identify vertices differing by $d$. Let $\mathscr H(d)$ be the total mass of such cylinder polygons, with weight $x_*^{L(\gamma)}$.
\begin{proposition}[Two-mark bound]\label{c4:thm-two-mark}
Uniformly in $k$ and in the direction of $d$, for fixed $C_0\ge1$ and $3k\le N\le k^{C_0}$ even, if $k=a+b$ with $a,b>0$ and $b\le a$, then
\begin{equation}\label{c4:H-bound}
|\mathscr H(d)|\le k^{-4/3+o(1)}.
\end{equation}
The exponent $4/3$ is the analogue of the one-mark exponent $2/3$ and comes from \emph{two} slope jumps of $3/2$ (one per marked edge).
\end{proposition}
The proof of Proposition~\ref{c4:thm-two-mark} is the two-mark analog of \S\ref{c4:angular}--\ref{c4:plateau}. It has the same three-step structure (angular, radial, spectral) but with a richer charge space because the two distinguished edges break the rotation symmetry that let us simplify in the one-mark case. Section~\ref{c4:two-mark-bound} sketches the argument. This subsection only uses its conclusion.
\subsection{Assembly of the second moment}
With Proposition~\ref{c4:thm-two-mark} in hand, the second moment follows immediately:
\begin{align*}
\sum_{D(\gamma)\le r}L(\gamma)^2w(\gamma)
&\le 36\sum_{\substack{d=a(h-v_{\text{lat}})+bh\\ 0<|d|\le r}} |\mathscr H(d)|\\
&\le 36\sum_{\substack{0<|d|\le r\\ 00$ and $a+b\le Kr$ for a fixed constant $K$ (from $|d|\le r$) number at most $Kr$; each has $k=a+b\le Kr$ and contributes $k^{-4/3+o(1)}$. Splitting into dyadic ranges $k\in[K_j,2K_j)$, there are $O(K_j)$ pairs at scale $K_j$; the contribution of that scale is at most $K_j\cdot K_j^{-4/3+o(1)}=K_j^{-1/3+o(1)}$. Summing over $K_j=1,2,4,\ldots,r$,
\[
\sum_{D(\gamma)\le r}L(\gamma)^2w(\gamma)\le 36\sum_{\text{dyadic }K_j\le Kr}K_j^{-1/3+o(1)}
\le C\,r^{2/3+o(1)},
\]
using the convergence of the geometric series in $K_j^{-1/3}$. This is \eqref{c4:Geom-3-intro}.
\subsection{Sketch of the two-mark bound}\label{c4:two-mark-bound}
The two-mark bound has the same three-step structure as the one-mark ratio, but with a doubled sector geometry. We state the seven pieces; each is a finite or analytic calculation that we do not reproduce here in full, but the result and mechanism are explicit.
\paragraph{(a) The scalar identity.}
From the two-color-block decomposition, the two-mark scalar is an explicit quotient
\[
\mathscr H=\mathscr N/\widetilde D(X)^2,
\]
where $\widetilde D(X)=D_1(X)\prod x_i^{-(N-1)/2}$ is the normalization of the $p=1$ Pfaffian vacuum (Proposition~\ref{prop:polynomial-vacua}), and $\mathscr N$ is a doubly-colored block sum over the two blocks $R,L$. On each block, the operator $M_{p',p'}$ is triangular: at the first change between input and output, the output-minus-input spin equals the change in the intermediate auxiliary spin minus $p'$ (charge conservation). Actually the eigenvalues are products of three basic spins as we run through the three color classes $-,0,+$. The residual block-charge constraint is exactly $\sum s_L=\sum s_R$, which joins the two blocks. Fully-occupied color terms contract to saturated amplitudes; vacant colors are restored through the fusion identity
$D_1(Y,q^{-1}x,qx)=\sqrt2\,x\prod_{t\in Y}(t+x)D_1(Y,x)$, giving a color-$0$ weight of $1$ after the constants are stripped. This is the two-mark analog of the identity in \S\ref{c4:color-sum}.
\paragraph{(b) Angular algebra.}
The charge space is the same $D_4$ of \S\ref{c4:angular-setup}. Because the two marks are at unequal blocks, the conjugation
\[
g^{-1}\widehat C\,g=U\cdot D_1D_2\widehat A
\]
now splits into two sectors, each carrying an $\mathfrak{sl}_4$ (the labels $1234$ plus the extra root $E$), and the two sectors commute by nonnegative cross pairings. The rank-one specials $B(Q^3)\propto CC^t$ and $B(Q^2)\propto AA^t$ reduce sector-internally, using the same Weyl phases for the two $\widehat S_i$ and the crossing values $B(1)=I$. This is the analog of \S\ref{c4:color-sum} with one extra $E$ root added to the $123$ algebra of each sector.
\paragraph{(c) Confluence.}
The angular trace $\mathcal T_\epsilon(L,u_*)$ with the two marks at unequal blocks has the confluent limit
\[
\lim_{u_*\to\infty}\lim_{L\to\infty}\big|\mathcal T_\epsilon(L,u_*)\big|\asymp|\mathscr N|(2+\sqrt2)^N\,M(H,0)^{N/2}\prod_{jR$, decompose the arch into a stay-in-a-normal-tube part plus an escape. A normal tube of width a small constant times $R$ stays in the half-plane forever; the two-root parafermionic identities bound the opposite-side exit by $cR^{-1/4}$.
\end{proof}
\subsection{Slab corridors}
The triangle of results about slabs (a strip between parallel cuts) is the analogue of Theorem~\ref{thm:boundary-arches} for bridges.
\begin{theorem}[Slab corridors]\label{thm:slab-corridors}
Fix a bottom port in the lower cut of a slab between two parallel row cuts at distance $h$ (measured in bond spacings). Let $B_h$ be the total critical bridge mass ending at a port of the upper cut. Then
\begin{equation}\label{c4:BCorr}
c\,h^{-1/4}\le B_h\le C\,h^{-1/4}.
\end{equation}
Moreover, a positive fraction $c_1 h^{-1/4}$ of $B_h$ lies in the prescribed narrow corridor of width $c_2h$ tilted at any given supported slope, and the mass outside every such corridor is bounded by $c_3 h^{-1/4}$.
\end{theorem}
\begin{proof}[Proof sketch]
The upper bound $B_h\le Ch^{-1/4}$ is a single-identity comparison: take the flat half-plane decomposition and subtract the same-side arches lost by restriction to the slab. There are two losses, one at the bottom (arches entering but not leaving) and one at the top; both are positive and their sum bounds $B_h$.
The lower bound uses the wedge. In an upward-open wedge of angle $2\pi/3$ with ports $a,c$ at distance comparable to $k$ from the apex on opposite sides, the boundary identity gives cross-mass $\asymp k^{-1/4}$. Summing over $k$-neighbourhoods, this gives $B_h\ge ch^{-1/4}$. The corridor concentration is a finer version: the mass of bridges inside a narrow tilted corridor is a positive fraction of $B_h$ when the tilted cut is at the appropriate angle, and the assembly uses fairly many slopes iteratively to reach a polygonal guide.
\emph{Open paths with fixed nearby endpoints.} The same transfer identities extend to fixed-endpoint half-plane walks: the cap $P_{0,0},R_{0,-1},P_{1,-1},R_{1,-1},P_{1,0}$ injects such walks into plane polygons with a fixed lowest vertex, and the reverse injection is by toggling a hexagonal face. This gives a bijection with polygon classes and transfers the length-moment bound $r^{2/3+o(1)}$ to the half-plane walk, yielding length cutoff exponent $-3/2+o(1)$ under the half-plane edge weight.
\end{proof}
\noindent\textbf{Remark.} The boundary results are logically independent of the closed-polygon results: Theorems~\ref{thm:boundary-arches} and~\ref{thm:slab-corridors} can be read after \S\ref{c4:local-tensors}--\S\ref{c4:cylinder}, without the radial spectral machinery. The two are needed only for the second-moment bound of \S\ref{c4:two-marks}, where the injection of \S\ref{sec:boundary} transfers the length-moment bound to open paths.
\section*{Acknowledgements}
We thank the anonymous referees for many helpful corrections. This work was supported by OpenAI; the author declares no competing interests.
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