\section{The tight-grading coefficient formula}\label{sec:coefficient}

We derive \eqref{eq:tight} in the strict determinant-one case used in the proof. The argument follows the triangular comparison of~\cite[Section~5.5]{BLM}. The scattering input is the existence of the consistent completion and the chamber invariance of its formal theta functions, in the form of~\cite[Theorem~2.29]{BLM}, originating in~\cite[Theorem~3.5]{GHKK}. We state the latter input below and prove the required coefficient comparison.

Fix $k,d,e,D_1,D_2$ as in \eqref{eq:ambient}. In this appendix the letters $p,q$ range over the entire rectangle
\begin{equation}\label{eq:rectangle}
 0\leq p<D_1,\qquad 0\leq q<D_2.
\end{equation}
All sums are weighted by \eqref{eq:weight}. Write $\mathcal C(p,q)$ for the sum over compatible gradings on $C(D_1,D_2)$ of vertical total $p$ and horizontal total $q$. Write $A_+(p,q)$ for the sum restricted to gradings whose selected vertical edges lie in horizontal shadows, and $A_-(p,q)$ for the sum restricted to gradings whose selected horizontal edges lie in vertical shadows. These sums equal $1$ at $(0,0)$; they are zero when either index is negative.

\subsection{Two free-edge convolutions}

\begin{lemma}\label{lem:convolution}
For every pair in \eqref{eq:rectangle},
\begin{align}
 \mathcal C(p,q)&=\sum_{s=0}^{p}[X^s]P_1(X)^{D_2-q}\,A_+(p-s,q),\label{eq:convplus}\\
 \mathcal C(p,q)&=\sum_{s=0}^{q}[Y^s]P_2(Y)^{D_1-p}\,A_-(p,q-s).\label{eq:convminus}
\end{align}
For $(p,q)\ne(0,0)$ the two sums satisfy
\begin{equation}\label{eq:shadowsupport}
 A_+(p,q)=0\quad\hbox{if }qD_1\leq pD_2,
 \qquad
 A_-(p,q)=0\quad\hbox{if }qD_1\geq pD_2.
\end{equation}
\end{lemma}
\begin{proof}
Fix the horizontal grading of total $q$. Its horizontal shadows contain exactly $q$ vertical edges by the first-return mass calculation in Section~\ref{sec:tight}. The other $D_2-q$ vertical edges may be graded freely. Indeed, if a selected vertical edge lies outside the shadow of a selected horizontal edge, the latter's first return is a strict horizontal witness. The only pairs that need a vertical witness have their vertical edge inside that horizontal shadow; every suffix under consideration then lies in the same shadow interval. Changing the grades outside all horizontal shadows therefore cannot affect these pairs. Setting all those grades to zero gives a grading counted by $A_+$, and any assignment to them reverses this operation. Their generating polynomial for the total vertical grade is $P_1(X)^{D_2-q}$. Extracting total $p$ proves \eqref{eq:convplus}. Reading backwards gives the analogous statement for the $D_1-p$ horizontal edges outside vertical shadows, proving \eqref{eq:convminus}.

For a grading counted by $A_+$ with nonzero totals, use its maximal blue intervals. Write their edge counts as $(m_i+1,n_i)$ and their initial ranks as $r_i$. The rank of the last vertical start is $r_i+D_1+D_2-(n_iD_1-m_iD_2)$, so $n_iD_1-m_iD_2>r_i\geq0$. Blue mass gives $\sum_i n_i=q$, and protected matching gives $\sum_i m_i\geq p$. Consequently
\[
 qD_1-pD_2
 =\sum_i(n_iD_1-m_iD_2)+(\sum_i m_i-p)D_2>0.
\]
If there is no selected horizontal edge, the shadow condition forces $p=q=0$, already excluded. The vertical-shadow argument in the proof of Lemma~\ref{lem:orientation} gives $pD_2-qD_1>0$ for any nonzero grading counted by $A_-$. It uses only the strict bounds in \eqref{eq:rectangle} and the rank inequalities, and proves the other support assertion.
\end{proof}

\subsection{Broken lines and chamber invariance}

Use the two-dimensional scattering diagram with the full horizontal axis carrying $P_1(X)$, the full vertical axis carrying $P_2(Y)$, and outgoing rays $\mathbb R_{\leq0}(r,s)$ for positive primitive pairs $(r,s)$ carrying $G_{r,s}(X^rY^s)$. Products around closed paths are the identity; grouping equal rays gives the ordered factorization \eqref{eq:factorization}. This is the consistent completion of the two initial walls.

A broken line consists of a piecewise linear path $\gamma:(-\infty,0]\to\mathbb R^2$ and a monomial $aX^{m_1}Y^{m_2}$ attached to each linear segment. Here $m=(m_1,m_2)\in\ZZ^2$ is its exponent and $a$ is its coefficient. Its velocity on that segment is $-m$. We start with exponent $(-D_1,-D_2)$ and coefficient $1$. At a wall with function $f$, choose a primitive integral normal $n$ with $\langle n,m\rangle>0$ for the incoming exponent. The outgoing monomial is one term of
\[
 aX^{m_1}Y^{m_2}f^{\langle n,m\rangle}.
\]
This is the bending rule; the constant term permits crossing without a bend. A broken line avoids the origin and ends at a point off the walls. Its weight is the coefficient of its final monomial. All increments of exponents are nonnegative in each coordinate.

The formal theta function at an endpoint is the sum of these final monomials. The chamber-invariance theorem asserts that this sum is the same for endpoints in a common chamber of a consistent scattering diagram whose coordinates are linearly independent over $\QQ$. The theorem is formal: it holds coefficient by coefficient in the completed monomial ring. Polynomial identity and the homogeneous specialization described after \eqref{eq:tight} then permit arbitrary values of the initial coefficients.

Choose an irrational real number $\varepsilon>0$ with
\begin{equation}\label{eq:epsilon}
 \varepsilon<\min\left\{\frac ed,\frac1{D_1},\frac1{D_2}\right\}.
\end{equation}
These bounds ensure the estimates below throughout the full rectangle \eqref{eq:rectangle}. The endpoints $Q_+=(1,\varepsilon)$ and $Q_-=(\varepsilon,1)$ lie in the first-quadrant chamber. Let $\Theta(p,q)$ be the coefficient of $X^{-D_1+p}Y^{-D_2+q}$ in their common theta function.

Every contributing segment has positive velocity in both coordinates: its exponent increments are bounded by the final pair $(p,q)$ in \eqref{eq:rectangle}. Along the path the quantity
\[
 L=\gamma_2m_1-\gamma_1m_2
\]
is constant. It is constant on segments because $\gamma'=-m$, and at a bend because the exponent increment is parallel to the ray containing $\gamma$. Its final values are
\[
 L=(D_2-q)-\varepsilon(D_1-p)>0\quad\text{at }Q_+,
 \qquad
 L=\varepsilon(D_2-q)-(D_1-p)<0\quad\text{at }Q_-.
\]
At a positive horizontal crossing $L$ is positive, and at a positive vertical crossing it is negative. Thus a path to $Q_+$ last enters the first quadrant through the positive horizontal axis, and one to $Q_-$ through the positive vertical axis.

Let $B_+^s(p,q)$ be the weighted sum of paths to $Q_+$ for which this last crossing chooses the term of degree $s$ in $X$ from the horizontal-wall factor. Define $B_-^s(p,q)$ using the term of degree $s$ in $Y$ at the last positive vertical crossing. Write $B_+(p,q)=B_+^0(p,q)$ and $B_-(p,q)=B_-^0(p,q)$, and set these quantities to zero at negative indices.

\begin{lemma}\label{lem:blconvolution}
The broken-line sums satisfy
\begin{align}
 \Theta(p,q)&=\sum_{s=0}^{p}[X^s]P_1(X)^{D_2-q}\,B_+(p-s,q),\label{eq:blplus}\\
 \Theta(p,q)&=\sum_{s=0}^{q}[Y^s]P_2(Y)^{D_1-p}\,B_-(p,q-s),\label{eq:blminus}
\end{align}
and the support conditions \eqref{eq:shadowsupport} hold with $A_\pm$ replaced by $B_\pm$.
\end{lemma}
\begin{proof}
For $Q_+$ the wall power at the last crossing is $D_2-q$, since its primitive inward normal is $(0,-1)$. Consider a path counted by $B_+(p-s,q)$. Its positive horizontal crossing is at
\[
 \left(1-\frac{(D_1-p+s)\varepsilon}{D_2-q},0\right).
\]
Scale its entire prefix before that crossing about the origin by the positive factor
\begin{equation}\label{eq:rescale}
 \frac{1-(D_1-p)\varepsilon/(D_2-q)}
      {1-(D_1-p+s)\varepsilon/(D_2-q)}.
\end{equation}
Both numerator and denominator are positive by \eqref{eq:epsilon}. Radial scaling preserves all wall incidences, segment directions, and monomial choices; rescaling segment times preserves their prescribed velocities. At the new crossing choose the degree-$s$ term of $P_1^{D_2-q}$ and append the final segment of velocity $(D_1-p,D_2-q)$ to $Q_+$. The construction is invertible: remove that choice and scale the prefix by the reciprocal factor. Its weight is multiplied by $[X^s]P_1^{D_2-q}$. Summing over $s$ proves \eqref{eq:blplus}. Interchanging coordinates proves \eqref{eq:blminus} with $P_2^{D_1-p}$.

For a path counted by $B_+$, all nonconstant bends occur before the last positive horizontal crossing. They are on the negative vertical axis or in the third quadrant. Write a bend point as $-\lambda(r,s)$ with $\lambda>0$, $r\geq0$, $s>0$, and the incoming velocity as $(v_x,v_y)$. Choosing exponent increment $j(r,s)$, where $j>0$, changes the velocity to $(v_x-jr,v_y-js)$. Since
\[
 L=\lambda(sv_x-rv_y)>0,
\]
the change in velocity slope is
\[
 \frac{v_y-js}{v_x-jr}-\frac{v_y}{v_x}
 =\frac{j(rv_y-sv_x)}{v_x(v_x-jr)}<0.
\]
The denominators are positive by \eqref{eq:rectangle}. At nonzero total increment this gives
\[
 \frac{D_2-q}{D_1-p}<\frac{D_2}{D_1},
 \quad\text{equivalently}\quad qD_1>pD_2.
\]
If there are no bends, the total increment is $(0,0)$ and the unique straight path has weight $1$. At $Q_-$ the sign of $L$ is reversed, so velocity slopes increase and $qD_1<pD_2$. These are the required support statements and initial values.
\end{proof}

\begin{lemma}\label{lem:triangular}
For all pairs in \eqref{eq:rectangle}, $A_+(p,q)=B_+(p,q)$ and $A_-(p,q)=B_-(p,q)$.
\end{lemma}
\begin{proof}
Equating the two convolutions for $\mathcal C$ gives
\begin{align*}
 A_+(p,q)-A_-(p,q)
 &=\sum_{s=1}^{q}[Y^s]P_2^{D_1-p}A_-(p,q-s)\\
 &\quad-\sum_{s=1}^{p}[X^s]P_1^{D_2-q}A_+(p-s,q).
\end{align*}
The same equation holds for $B_\pm$ by chamber invariance and Lemma~\ref{lem:blconvolution}. The initial values are $1$ at $(0,0)$. Every term on the right has smaller sum of indices. At a nonzero pair, the support conditions determine at least one of the two left-hand terms to be zero, so the equation determines the other. Induction on $p+q$ proves equality. In fact equality $qD_1=pD_2$ occurs in the strict rectangle only at the origin, since $D_1,D_2$ are coprime.
\end{proof}

\subsection{The determinant-one coefficient}

\begin{proposition}\label{prop:coefficient}
With \eqref{eq:ambient}, the identity \eqref{eq:tight} holds.
\end{proposition}
\begin{proof}
By Lemma~\ref{lem:orientation}, the tight gradings of totals $(kd,ke)$ are precisely those counted by $A_+(kd,ke)$. By Lemma~\ref{lem:triangular}, this sum is $B_+(kd,ke)$. We show that a path counted by the latter has exactly one nonconstant bend and that it reads the coefficient $[z^k]G_{d,e}(z)$.

Its exponent increments sum to $(kd,ke)$, and its last positive horizontal crossing has zero increment. While it lies in the third quadrant, the polar angle of its position increases strictly: its derivative is $L/|\gamma|^2>0$. Thus the ratios $r/s$ of the negative ray directions at successive bends strictly decrease. The negative horizontal axis cannot carry a bend because there $L<0$. At the first bend, positivity of $L$ gives $r/s<D_1/D_2$, and therefore the same strict bound holds at every bend, including any negative vertical bend of ratio zero.

If there were more than one distinct direction, the weighted mean of their ratios would be $d/e$. Collect the increments with ratio strictly greater than $d/e$ and let their sum be $(A,B)$. This group is nonempty, $0<B\leq ke<D_2$, and
\[
 \frac de<\frac AB<\frac{D_1}{D_2}.
\]
Since $eD_1-dD_2=1$, it follows that
\[
 0<eA-dB<\frac{B}{D_2}<1,
\]
contradicting integrality. All increments therefore have direction $(d,e)$. Strict angular monotonicity allows only one encounter with that ray, so there is exactly one nonconstant bend.

There is a unique geometric path with that bend and those initial and final velocities. Its bend point and the time from the bend to $Q_+$ are respectively
\[
 -\bigl((D_2-ke)-(D_1-kd)\varepsilon\bigr)(d,e),
 \qquad e-d\varepsilon.
\]
The coefficient in parentheses and the time are positive by \eqref{eq:epsilon}. Adding this time times the final velocity $(D_1-kd,D_2-ke)$ to the bend point gives $(1,\varepsilon)$, since $eD_1-dD_2=1$. Extending backwards with velocity $(D_1,D_2)$ gives the required initial segment; all other crossings choose constant terms.

The primitive normal at this bend is $(-e,d)$. Its pairing with the incoming exponent $(-D_1,-D_2)$ is $1$. Thus the wall is used to the first power, and the term of increment $(kd,ke)$ has coefficient $[z^k]G_{d,e}(z)$. Consequently $B_+(kd,ke)=c_k$, proving \eqref{eq:tight}.
\end{proof}