\section{The affine theta algebra and its symplectic form}\label{sec:geometry}
We first work over an algebraically closed field $\kk$ of characteristic zero. Positivity and boundedness give finite multiplication on all theta labels. Write $C=\Can(\Aprin)$ and form the graded algebra
\[
 S_\Xi=\bigoplus_{d\ge0}\ \bigoplus_{m\in d\Xi\cap\LL}\kk\tht_mT^d,
\]
where $T$ is an independent degree-one variable and every $\tht_mT^d$ has degree $d$.

For a normal variety $Y$, $K_Y$ denotes a canonical Weil divisor. The notation $K_Y+D\sim0$ means that this divisor is linearly equivalent to zero. Equivalently, its associated rank-one reflexive sheaf is trivial. We use the openness theorem \cite[Lemma~8.42]{GHKK}: in a proper flat family over a regular characteristic-zero base, with flat boundary, the simultaneous conditions of normality and Cohen--Macaulayness, reduced divisorial boundary, log canonicity, trivial $\omega(D)$, and $H^1(\mathcal O)=0$ persist near a fiber satisfying them. Here log canonicity has its usual discrepancy meaning: on a resolution every discrepancy is at least $-1$. The toric central fiber below satisfies all five conditions.

\begin{lemma}\label{lem:compactification}
The algebra $C$ is a finitely generated normal domain. There is a normal projective Cohen--Macaulay variety $Y$ with reduced boundary divisor $D$ such that
\[
 Y\setminus D=\Spec C,\qquad (Y,D)\text{ is log canonical},\qquad K_Y+D\sim0.
\]
\end{lemma}
\begin{proof}
The lattice semigroup of pairs $(m,d)$ with $m\in d\Xi$ is finitely generated by Gordan's lemma. For each Hilbert basis element take the corresponding $\tht_mT^d$. Products have leading term prescribed by addition, with coefficient one; all other labels differ by $Pn\ne0$ and have strictly larger $\ell$-value. Each degree slice is finite. Decompose $(m,d)=\sum_j(m_j,d_j)$ into Hilbert basis elements. The product of their theta monomials equals $\tht_mT^d$ plus a finite linear combination of $\tht_{m+Pn}T^d$ with $n\ne0$. Each extra label belongs to $d\Xi$ by positivity and satisfies $\ell(m+Pn)>\ell(m)$. Starting from the largest $\ell$-value in the finite set $d\Xi\cap\LL$, subtract these already constructed terms. This produces $\tht_mT^d$ in the generated algebra and terminates for every label in that slice. Thus $S_\Xi$ is finitely generated. Evaluation at $T=1$ maps $S_\Xi$ onto $C$, since absorption places each label in some degree. Its kernel is $(T-1)$: for a finite expression in the kernel, raise each of its homogeneous terms to a common degree by multiplication by powers of $T$. The resulting expression is zero by theta independence, and the change of each term is divisible by $T-1$. Thus $C=S_\Xi/(T-1)$. Theta independence and Laurent finiteness embed these algebras in the initial Laurent ring and its polynomial extension.

For each output $r=p+q+Pn$, introduce independent mutable parameters $t_i$ and replace the multiplication coefficient by $\alpha(p,q,r)t^n$. In a triple product, both bracketings leading from inputs $p,q,s$ to output $r$ have total charge $N$ determined by $PN=r-p-q-s$. Their parameter monomials are both $t^N$. Associativity of the original structure constants therefore gives associativity of the new product coefficient by coefficient. Denote the resulting graded algebra by $\mathscr S$. The same triangular induction proves finite generation over $\kk[t_i:i\in J]$, while every homogeneous component has the indicated finite free theta basis.

The span $\mathscr I$ of $\tht_mT^d$ with $g(m)<d$ is an ideal, since $g(r)\le g(p)+g(q)$. Its quotient is degreewise free as well. Therefore
\[
 (\mathscr Y,\mathscr B)=
 \bigl(\Proj\mathscr S,\Proj(\mathscr S/\mathscr I)\bigr)\longrightarrow\mathbb A^J
\]
is a proper flat family with flat boundary. For a homogeneous element $u$, the localization $\mathscr S[u^{-1}]$ is a filtered direct limit of the flat base module $\mathscr S$, with transition maps multiplication by $u$. It is flat. Its degree-zero part is a direct summand and hence flat, giving the claim on each chart $D_+(u)$. The same argument applies to $\mathscr S/\mathscr I$. At $t=0$ the product is addition of lattice labels; the saturated cone semigroup gives a normal projective toric variety. A lattice point in the cone is interior exactly when it lies on no facet. Consequently the ideal spanned by interior monomials is the intersection of the monomial prime ideals corresponding to the facets. It is radical, and its Proj is the reduced union of the torus-invariant prime divisors. The toric variety is projective because $\Xi$ is bounded; its cone semigroup is saturated because $k(m,d)$ lying in the cone for an integer $k>0$ forces $(m,d)$ to lie there. Normal toric varieties are Cohen--Macaulay, and the toric boundary cancels the canonical divisor. On a toric log resolution every exceptional toric divisor has discrepancy $-1$, so the pair is log canonical. Projective toric varieties have $H^1(\mathcal O)=0$.

These are the five central-fiber hypotheses of \cite[Lemma~8.42]{GHKK}, and the base is regular. Hence all five properties hold on an open neighborhood of zero. The torus weights
\[
 \operatorname{wt}(\tht_m)=m,\quad \operatorname{wt}(t_i)=-Pe_i,
 \quad\operatorname{wt}(T)=0
\]
give a contracting one-parameter action on the base using the cocharacter $-\ell$. Explicitly, $-\ell$ assigns weight $1$ to every $t_i$, so the orbit of a base point $a$ extends to the map $\mathbb A^1\to\mathbb A^J$, $\lambda\mapsto\lambda a$, and sends $0$ to the toric fiber. The inverse image of the good open set contains $0$ and therefore a nonzero $\lambda$. The torus action identifies the fibers at $a$ and $\lambda a$. Hence the properties hold at $t=(1,\ldots,1)$.

Let $Y=\Proj S_\Xi$ and let $D$ be this fiber of the boundary. The interior ideal has radical equal to the radical of $(T)$. Indeed, for $g(m)<d$, sufficiently large powers of $\tht_mT^d$ have all outputs in $(kd-1)\Xi$ and are divisible by $T$. Conversely $T=\tht_0T$ is in the interior ideal. Thus $Y\setminus D=D_+(T)=\Spec C$. Its normality follows from that of $Y$.
\end{proof}

\begin{lemma}\label{lem:poisson}
Define the bracket of characters by
\[
 \{z^u,z^v\}=-u^{\mathsf T}\Lambda v\,z^{u+v}.
\]
It restricts to $C$. The inverse of its bivector is a regular symplectic form on the smooth locus of $\Spec C$.
\end{lemma}
\begin{proof}
For a wall of charge $n$, write its primitive normal as $\nu=cRn$, $c>0$. Then
\[
 \{z^{Pn},z^u\}=\pair{Rn}{u}z^{Pn+u}.
\]
The wall map is the exponential of a Hamiltonian derivation: for a binomial wall $(1+z^{Pn})^a$, take the Hamiltonian
\[
 ca\sum_{j\ge1}\frac{(-1)^{j+1}}{j^2}z^{jPn}.
\]
Its bracket with $z^u$ is $a\nu(u)z^u\log(1+z^{Pn})$. Since this derivation fixes $z^{Pn}$, exponentiation gives $z^u(1+z^{Pn})^{a\nu(u)}$. Reversing the crossing negates the Hamiltonian. Thus every wall map preserves the bracket. The auxiliary charge variables make these operations continuous: Euler derivatives preserve each charge and a Hamiltonian term raises it by a positive amount. Let $\beta(p,q,r)$ denote the formal theta coefficient of $\{\tht_p,\tht_q\}$. The positive-charge triangular change from theta functions to characters and its degreewise inverse show that its possible outputs lie in $p+q+P\NN^J$. For a fixed output $r=p+q+Pn_0$, only charges at most $n_0$ componentwise contribute. Choose a finite truncation containing all those charges. At an endpoint $Q$ sufficiently close to $r$, each relevant normal $\nu$ with $\nu(r)\ne0$ has $\nu(Q)\nu(r)>0$. A backward final ray $Q+\RR_{\ge0}r$ meets no such hyperplane; it is parallel to the others. Hence a line with final exponent $r$ has no last nontrivial bend in this truncation. Its input must also be $r$. Consequently the coefficient of $\tht_r$ equals the coefficient of $z^r$ for any formal theta expansion under consideration, and linear coefficient extraction gives
\[
 \begin{split}
 \alpha(p,q,r)&=\sum_{e_\gamma+e_\eta=r}c_\gamma c_\eta,\\
 \beta(p,q,r)&=\sum_{e_\gamma+e_\eta=r}c_\gamma c_\eta
                     (-e_\gamma^{\mathsf T}\Lambda e_\eta).
 \end{split}
\]
Here $e_\gamma,e_\eta$ are the final exponents and $c_\gamma,c_\eta>0$ their coefficients; both sums use the same finite set of pairs. Thus $\alpha(p,q,r)=0$ implies $\beta(p,q,r)=0$. In particular the bracket has finite support and preserves $C$. The formal identities for Jacobi and Leibniz hold degreewise and descend through the injective Laurent realization.

Declare $T$ central. The support inclusion gives $\{(S_\Xi)_d,(S_\Xi)_e\}\subset(S_\Xi)_{d+e}$ and preserves the interior ideal. The bracket extends by the quotient rule on homogeneous charts to $Y$, with $D$ a Poisson divisor. At the generic point of each boundary component, the terms of the bivector containing a normal direction are divisible by its local equation. Consequently its top wedge defines a section of the rank-one reflexive sheaf $\mathcal O_Y(-K_Y-D)$. By Lemma~\ref{lem:compactification} this sheaf is trivial, and its global sections are constants. On the initial torus the top wedge is nonzero because $\Lambda$ is invertible. It is therefore nowhere zero on the smooth part of $Y\setminus D$. The inverse form is regular there and is closed, as it is the log-symplectic form on a dense torus. Its volume is a nonzero constant multiple of the initial logarithmic volume form.
\end{proof}

\section{Divisors omitted by the initial tori}\label{sec:divisors}
Let $x_i=z^{(f_i,0)}$ and $y_i=z^{(0,e_i)}$ be the initial characters. Initial and adjacent cluster variables belong to $C$. Laurent finiteness gives
\[
 C[(\prod_{i\in J}x_i)^{-1}]
   =\kk[x_i^{\pm1},y_i^{\pm1}:i\in I].
\]
For $i\in J$, let $x_i'$ be the adjacent variable obtained by mutation at $i$. Its exchange relation is $x_ix_i'=F_i$, where
\[
 F_i=y_i\prod_{j:B_{ji}>0}x_j^{B_{ji}}
           +\prod_{j:B_{ji}<0}x_j^{-B_{ji}}.
\]
Both products range over $j\in I$; an empty product is $1$. Invert the other mutable variables. The intersection of the initial and $i$th adjacent Laurent rings is the rank-one algebra
\[
 R_i=\kk[\text{other initial characters}^{\pm1},x_i,x_i']/(x_ix_i'-F_i).
\]
To see the equality, the coefficient of each $x_i^{-j}$ must be divisible by $F_i^j$, which replaces that term by a multiple of $(x_i')^j$. This intersection contains the corresponding localization of $C$ and is generated by elements of it. Principal coefficients make $F_i$ linear in the independent $y_i$, with unit leading coefficient. Hence $R_i$ is smooth and $x_i=x_i'=0$ has codimension two. A divisor missed by the initial and all adjacent tori must therefore give positive order to at least two mutable $x_i$.

\begin{lemma}\label{lem:valuation}
Suppose such a prime divisor exists in $\Spec C$, with normalized valuation $v$. For a nonzero rational function, $v$ is its integer order of vanishing at the generic point of that divisor; $v(0)=+\infty$. Its character valuation is the homomorphism $a:\LL\to\ZZ$ defined by $a(m)=v(z^m)$. Write $a=ka_0\in\LL^*$, where $a_0$ is primitive and $k\ge1$. There are a primitive charge $n_0\in\NN^J$ supported on at least two indices, a character $H=z^b$ with $b=Pn_0$, and $c\in\kk^*$, for which the following holds. Choose a character $Z=z^u$ with $\pair{a_0}{u}=1$, and complete $H$ to coordinates $H,S_1,\ldots,S_{d-2}$ on the torus with character lattice $\ker a_0$, where $d=\operatorname{rank}\LL$. Then
\begin{equation}\label{eq:valuation}
 v\left(\sum_r Z^r f_r(H,S)\right)
   =\min_{f_r\ne0}\{kr+\operatorname{ord}_{H=c}f_r\}
\end{equation}
for finite Laurent polynomials. Here $\operatorname{ord}_{H=c}$ is the multiplicity of the factor $H-c$ over $\kk(S_1,\ldots,S_{d-2})$.
\end{lemma}
\begin{proof}
All coefficient and original frozen characters are units, so $a$ is nonnegative and supported in mutable $x$-coordinates. At the generic point, normality gives a discrete valuation ring. It lies in the smooth locus in characteristic zero. Let $\tau$ be a uniformizer. Writing each character as $z^m=\tau^{ka_0(m)}u_m$ with a unit $u_m$ gives
\[
 d\log z^m=ka_0(m)\frac{d\tau}{\tau}+d\log u_m.
\]
The logarithmic expression for the symplectic form is obtained by inverting the character bracket matrix $-\Lambda$. For a valuation-zero function, an overline denotes its image in the residue field of the valuation ring. The coefficient of $d\tau/\tau$ must vanish after taking residues, since the form is regular at this divisor. Contracting its logarithmic expression with $a_0$ gives the character direction $\QQ\Lambda^{-1}a_0$. Here $(a_0)_x$ is the column of values of $a_0$ on the characters $x_i$, and it is zero at frozen indices. Formula~\eqref{eq:lambda} identifies the ray with the image under $P$ of the mutable entries of $\Delta(a_0)_x$. Choose primitive $n_0$ on this ray and set $b=Pn_0$. Saturation of $P$ makes $b$ primitive; skewness gives $\pair{a_0}{b}=0$. The residue identity is $d\log\overline H=0$. In characteristic zero over an algebraically closed field, it gives $\overline H=c\in\kk^*$.

Primitivity of $a_0$ splits $\LL=\ZZ u\oplus\ker a_0$. Every character in $\ker a_0$ has valuation zero, whereas $v(Z)=k>0$. Thus the coordinate ring of the smooth model $X_0=\mathbb A^1_Z\times\Spec\kk[\ker a_0]$ embeds in the valuation ring. The inverse image of its maximal ideal defines the center, which lies in $(Z,H-c)$. Suppose its center has codimension $c_0$. At its generic point choose regular parameters $Z,t_2,\ldots,t_{c_0}$ and lifts of a separating transcendence basis for the center. The torus volume is a valuation-zero unit times $dZ/Z\wedge dt_2\wedge\cdots\wedge dt_{c_0}$ wedged with the remaining basis differentials. Indeed, if $t_i=\tau^{v(t_i)}u_i$, then
\[
 dt_i=\tau^{v(t_i)}\left(v(t_i)u_i\frac{d\tau}{\tau}+du_i\right).
\]
In a nonzero wedge at most one factor contributes $d\tau/\tau$, since its wedge with itself is zero. The $dZ/Z$ factor has no positive prefactor, while the other center parameters contribute $\tau^{v(t_i)}$. The remaining differentials are regular. This proves the following order bound for the initial logarithmic volume form $\Omega$:
\[
 0=\operatorname{ord}_v\Omega\ge\sum_{i=2}^{c_0}v(t_i)-1\ge c_0-2.
\]
Here the left equality is Lemma~\ref{lem:poisson}. Thus $c_0=2$, the center is $(Z,H-c)$, and $F:=H-c$ satisfies $v(F)=1$.

Put $V=Z/F^k$. The exact identity
\[
 \frac{dZ}{Z}\wedge\frac{dH}{H}
   =-\frac{dF}{c+F}\wedge\frac{dV}{V}
\]
shows that the order-zero coefficient of the volume contains
$dF\wedge d\log\overline V\wedge d\log\overline S_1\wedge\cdots\wedge d\log\overline S_{d-2}$.
It is nonzero, so $\overline V$ is transcendental over $\kk(\overline S)$. If $M$ is the minimum in \eqref{eq:valuation}, the residue after division by $F^M$ is
\[
 \sum_{kr+\operatorname{ord}_{H=c}f_r=M}
   \left.(F^{-\operatorname{ord}_{H=c}f_r}f_r)\right|_{H=c}\overline V^{\,r}.
\]
Distinct $r$ give distinct powers of a transcendental element, so this sum is nonzero. This proves the formula. For a quotient of two Laurent polynomials of the same valuation, the displayed leading expressions identify its residue with an element of $\kk(\overline S,\overline V)$. Every valuation-zero rational function is a quotient of Laurent polynomials of the same valuation. Conversely the units $S_i$ and $V$ realize these generators in the residue field. Thus the residue field is exactly $\kk(\overline S,\overline V)$.
\end{proof}