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\title[Monomial immanants]{Monomial immanants of totally nonnegative matrices}
\author{Qiyue Tang}
\address{Zhili College, Tsinghua University, Beijing 100084, China}
\email{tangqy24@mails.tsinghua.edu.cn}
\date{November 9, 2026}
\subjclass[2020]{15A15, 05E05, 13F60}
\keywords{Immanant, total nonnegativity, symmetric function, cluster variety, theta function}
\hypersetup{pdftitle={Monomial immanants of totally nonnegative matrices},pdfauthor={Qiyue Tang}}
\begin{document}
\raggedbottom
\begin{abstract}
We prove that every monomial immanant is nonnegative on the totally
nonnegative real matrices, establishing a conjecture of Stembridge.
We identify the corresponding normalized symmetric function of
annular holonomy with a single cluster theta function.
\end{abstract}
\maketitle
\section{Introduction}

A real matrix is \emph{totally nonnegative} if all its minors are
nonnegative, and \emph{strictly totally positive} if all its minors
are positive. For a class function $f$ on the symmetric group $S_n$, its
immanant is the polynomial
\[
 \Imm_f(A)=\sum_{w\in S_n}f(w)\prod_{i=1}^n a_{i,w(i)},
 \qquad A=(a_{ij})_{i,j=1}^n.
\]
The determinant and permanent correspond respectively to the sign and
trivial characters. The immanants of the irreducible characters
$\chi^\lambda$, indexed by partitions $\lambda$ of $n$, interpolate
between these two examples. Stembridge proved that these irreducible
immanants are nonnegative on totally nonnegative matrices; in fact,
\[
 \Imm_{\chi^\lambda}(A)\geq \chi^\lambda(1)\det A
\]
for every such $A$ \cite[Corollaries~3.3--3.4]{Stembridge91}.
His proof combines a factorization of totally nonnegative matrices
with positivity in the group algebra of $S_n$.

The Frobenius characteristic
identifies class functions with homogeneous symmetric functions of
degree $n$:
\begin{equation}\label{eq:frobenius-characteristic}
 \ch(f)=\sum_{\rho\vdash n}\frac{f(\rho)}{z_\rho}p_\rho,
 \qquad
 z_\rho=\prod_{j\geq1}j^{m_j(\rho)}m_j(\rho)!.
\end{equation}
Here $p_j(x)=\sum_{a\geq1}x_a^j$ is the $j$th power sum,
$p_\rho=\prod_i p_{\rho_i}$, $m_j(\rho)$ is the multiplicity of
$j$ in $\rho$, and $f(\rho)$ is the common value of $f$ on that
cycle type. We write $\lambda\vdash n$ for a weakly decreasing
sequence of positive integers of sum $n$, and append zeros when
regarding it as an $n$-tuple. The characteristic of $\chi^\lambda$
is the Schur function $s_\lambda$.

For a partition $\lambda\vdash n$, let $\phi^\lambda$ be the virtual
character determined by
\[
 \ch(\phi^\lambda)=m_\lambda,
\]
where $m_\lambda$ is the sum of the distinct monomials whose exponent
sequence is a permutation of $\lambda$, padded with zeros. The
polynomial $\Imm_{\phi^\lambda}$ is the \emph{monomial immanant}
indexed by $\lambda$. The change of basis
\[
 s_\lambda=\sum_{\mu\vdash n}K_{\lambda\mu}m_\mu,
 \qquad
 \chi^\lambda=\sum_{\mu\vdash n}K_{\lambda\mu}\phi^\mu,
\]
has nonnegative integral Kostka coefficients $K_{\lambda\mu}$.
Thus nonnegativity of the monomial immanants implies nonnegativity
of every irreducible immanant. In \emph{Some conjectures for
immanants}, Stembridge formulated this stronger assertion
\cite[Conjecture~2.1]{Stembridge}. We prove it for every partition
and every matrix size.

\begin{theorem}\label{thm:main}
Let $n\geq1$ and $\lambda\vdash n$. For every totally nonnegative
real $n\times n$ matrix $A$,
\[
 \Imm_{\phi^\lambda}(A)\geq0.
\]
\end{theorem}

Stembridge's terminology in \cite{Stembridge91,Stembridge} calls a
matrix totally positive when all its minors are nonnegative; this
is the class called totally nonnegative here. His formulation also
singles out the monomial characters among all class functions.
Indeed, let $J_\mu$ be the block diagonal matrix whose blocks are
all-one square matrices of orders $\mu_1,\mu_2,\ldots$. Stembridge
proved \cite[Proposition~2.3]{Stembridge} that
\begin{equation}\label{eq:block-test}
 \Imm_{\phi^\lambda}(J_\mu)
   =\left(\prod_i\mu_i!\right)\delta_{\lambda\mu}.
\end{equation}
Here $\delta_{\lambda\mu}$ is the Kronecker delta. To see the
normalization, only permutations in the Young subgroup
$S_{\mu_1}\times S_{\mu_2}\times\cdots$ contribute. Their sum is
the order of this subgroup times the scalar product of
$\phi^\lambda$ with its induced trivial character. Under the
Frobenius characteristic the latter character is
$h_\mu=\prod_i h_{\mu_i}$, where
$h_j(x)=\sum_{a_1\leq\cdots\leq a_j}x_{a_1}\cdots x_{a_j}$.
The Hall scalar product is determined by
$\langle p_\rho,p_\sigma\rangle=z_\rho\delta_{\rho\sigma}$;
the bases $m_\lambda$ and $h_\mu$ are dual for this product.
Since $J_\mu$ is totally nonnegative, these tests force every class
function whose immanant is nonnegative on all totally nonnegative
matrices to have nonnegative coefficients in the basis
$\{\phi^\lambda\}_{\lambda\vdash n}$. The theorem proves that this
necessary condition is sufficient.

The conjecture belongs to a group of positivity questions involving
Jacobi--Trudi matrices and Hecke algebra characters. For a partition
$\beta$ with at most $n$ parts, its Jacobi--Trudi matrix is
\[
 H_\beta=(h_{\beta_i-i+j})_{i,j=1}^n,
 \qquad h_0=1,\qquad h_j=0\quad(j<0).
\]
Goulden and Jackson conjectured that irreducible immanants of these
matrices have nonnegative coefficients in the underlying variables
\cite{GouldenJackson}. Greene proved this conjecture by studying
Young's seminormal representations \cite{Greene}. Haiman then proved
the stronger Schur positivity of
$\Imm_{\chi^\lambda}(H_\beta)-\chi^\lambda(1)\det H_\beta$
\cite[Theorem~1.4]{Haiman}; Schur positivity means nonnegative
coefficients when the symmetric function is expanded in the Schur
basis.

Haiman also introduced Hecke algebra deformations of the monomial
characters and conjectured positivity of their values on normalized
Kazhdan--Lusztig basis elements \cite[Conjecture~2.1]{Haiman}.
At $q=1$, this gives a nonnegative expansion of each monomial
immanant in the Kazhdan--Lusztig immanant basis. The latter immanants
are nonnegative on totally nonnegative matrices
\cite[Proposition~2.2]{RS}, so this specialization implies
Stembridge's conjecture. The graded conjecture additionally asks
for positivity in the Hecke parameter $q$. These formulations connect
the matrix inequality with canonical bases and give stronger
coefficient assertions for particular shapes.

Several families were known before the general result. Stembridge
proved the conjecture for rectangular partitions and for
$(2,1^{n-2})$ \cite[Theorems~2.7--2.8]{Stembridge}.
Clearman, Shelton, and Skandera gave nonnegative expansions in
Temperley--Lieb immanants and path-tableau interpretations for
partitions with $\lambda_1\leq2$
\cite[Theorems~5.6--5.7]{CSS}. More recently, Dong, Wu, and Gao
proved the inequality for every hook $(a,1^k)$, $a+k=n$, under the
hypothesis that all minors of order at most $k+1$ are nonnegative
\cite[Theorem~1.2]{DongWuGao}. McNulty proved Kazhdan--Lusztig
positivity at $q=1$ for all hooks \cite{McNultyHook}, and separately
verified Haiman's graded conjecture for $n\leq10$ by exact
computation \cite{McNultyFinite}.

We prove the theorem by studying the normalized spectral function
$m_\lambda(\eig g)/\det g$ on $\PGL_n$. Pulling this function back by
the holonomy of a marked annulus places it in a cluster Poisson
algebra. A finite sequence of mutations reduces its matrix
evaluation to a tridiagonal matrix. Its characteristic coefficients
are the independent-set polynomials of a path with alternating
anchor and bridge vertices.

The identification of monomial symmetric functions evaluated on
triangular interval graphs with single theta sections was proved
in \cite[\S5.4, Theorem~5.8]{OpenAI}. We use the version of its
rotations with one bridge on each level
\cite[\S\S5.1--5.2]{OpenAI} and establish the comparison with the
full annular theta basis. A finite Newton-polytope lemma shows that every extremal
exponent of a finite family belongs to a unique theta function.
It follows that each summand of the annular spectral function
descends to the path lattice. Rotating the bridges through $n+1$
periods forces every label to be a dominant sum of anchor vectors.
A finite bound on bend increments then determines the coefficient
of each such label and identifies the spectral function with a
single theta function. Theta positivity in the original annular
chart and a simultaneous constant term give the immanant inequality.

We use the theta-function construction and positivity theorems of
Gross--Hacking--Keel--Kontsevich \cite{GHKK} and Davison--Mandel
\cite{DM}, together with the surface results of Goncharov--Shen
\cite{GS_DT,GSAtlas} and Shen \cite{Shen}. The finite form of
valuative independence needed here is proved in
Section~\ref{sec:theta}. The annular charts, all-rank word reduction,
lattice comparisons, and coefficient extraction are given
explicitly. We use $\N=\Z_{\geq0}$. For a Laurent polynomial $f$,
$[X^a]f$ denotes the coefficient of the monomial $X^a$; the same
convention applies to any explicitly named variables.

\section{The Frobenius coefficient identity}\label{sec:frobenius}

If $g$ is a complex $n\times n$ matrix, a symmetric polynomial in
$\eig g$ means its expression in the coefficients of the
characteristic polynomial of $g$. This convention also applies when
$g$ is not diagonalizable. For an invertible matrix put
\begin{equation}\label{eq:spectral-F}
 F_\lambda(g)=\frac{m_\lambda(\eig g)}{\det g}.
\end{equation}
The numerator is polynomial in the matrix entries. The denominator
is invertible on $\GL_n$, and both numerator and denominator have
scalar degree $n$. Thus $F_\lambda$ is a regular conjugation-invariant
function on $\PGL_n$.

\begin{lemma}\label{lem:frobenius}
Let $Z=\operatorname{diag}(z_1,\ldots,z_n)$. For every $n\times n$
matrix $A$,
\begin{equation}\label{eq:frobenius-coefficient}
 [z_1\cdots z_n]m_\lambda(\eig(ZA))=\Imm_{\phi^\lambda}(A).
\end{equation}
If $A$ is invertible, then
\begin{equation}\label{eq:row-constant-term}
 \Imm_{\phi^\lambda}(A)=\det(A)\,[z_1^0\cdots z_n^0]F_\lambda(ZA).
\end{equation}
\end{lemma}

\begin{proof}
The power sum $p_j(\eig(ZA))$ is $\tr((ZA)^j)$. A term of
squarefree row degree $z_1\cdots z_n$ in a product of traces
$p_\rho(\eig(ZA))$ specifies a permutation of cycle type $\rho$.
Each cycle of length $j$ has $j$ choices of starting point in its
trace, and the $m_j(\rho)$ traces of that length can be assigned to
the cycles in $m_j(\rho)!$ ways. Therefore
\[
 [z_1\cdots z_n]p_\rho(\eig(ZA))
 =z_\rho\sum_{\substack{w\in S_n\\\operatorname{type}(w)=\rho}}
       \prod_{i=1}^n a_{i,w(i)}.
\]
Apply \eqref{eq:frobenius-characteristic} with $f=\phi^\lambda$;
the factors $z_\rho$ cancel. This proves
\eqref{eq:frobenius-coefficient}. For invertible $A$, the identity
\[
 \det(ZA)=(z_1\cdots z_n)\det(A)
\]
then gives \eqref{eq:row-constant-term} by division by the indicated
row monomial.
\end{proof}

It remains to establish positive Laurent coefficients for
$F_\lambda$ in a chart whose row weights can be read off directly.
We assume $n\geq2$ until the proof of Theorem~\ref{thm:main} is
completed.
\section{An annular chart}\label{sec:annulus}

Assume $n\geq2$, put $r=n-1$, and let $G=\PGL_n(\C)$.
Let $S$ be an oriented annulus with one marked point on each boundary
component. We use its moduli space $\cX_{G,S}$ of framed local systems:
a flag is specified at each marked point, and no boundary pinning is
specified. Write $B^+,B^-$ for the opposite standard Borel subgroups,
$H=B^+\cap B^-$, and $w_0$ for the longest Weyl group element.
Choose a representative $\dot w_0\in G$.

\subsection{The surface chart and its character lattice}

Cutting the annulus along an arc between its marked points and
transporting the two flags to a common fiber gives the description
\[
 (g,F_0,F_1)/G,
\]
where $g$ is the holonomy around the oriented core. On the locus where
$F_0,F_1$ are opposite, normalize them to the standard pair. Their
stabilizer is $H$, which acts on $g$ by conjugation. The two boundary
genericity conditions say that a standard flag and its image under
$g$ are opposite, for each of the two standard flags. Thus the resulting
open part of the moduli space is
\begin{equation}\label{eq:annular-quotient}
 G^{w_0,w_0}/\Ad H,
 \qquad
 G^{w_0,w_0}=B^+\dot w_0B^+\cap B^-\dot w_0B^-.
\end{equation}
The double-word coordinates below parametrize a dense seed torus
in this quotient.

We recall how the surface cluster coordinates restrict to this chart.
For a root $i\in\{1,\ldots,r\}$ put
\[
 E_i=I+e_{i,i+1},\qquad F_i=I+e_{i+1,i},\qquad
 H_i(t)=\operatorname{diag}(
 \underbrace{t,\ldots,t}_{i},1,\ldots,1).
\]
The matrix units are denoted by $e_{ij}$; the displayed matrices
represent elements of $G$. A signed word uses $E_i$ for the letter $i$
and $F_i$ for the letter $-i$. Its coordinates are attached to intervals
between successive occurrences of each root, with an additional
interval at each end. Evaluation inserts $H_i$ with the corresponding
coordinate in each root-$i$ interval. Since $H_i$ commutes with root
factors of every other root, its position within that interval is
immaterial. These are the double-word coordinates of
\cite{FG,GSAtlas}.

The compatibility with surface gluing includes an orientation
convention which we shall need. If $p,q$ are the boundary-oriented
pinnings of the cut rectangle and $q^*$ is the opposite pinning
obtained by reversing the boundary orientation, its transport is defined by
$q^*=gp$ in \cite[Theorem~5.3, (198)]{GSAtlas}. Consequently, for
$g=UhL$ with $U$ upper unipotent, $h\in H$, and $L$ lower unipotent, the four
flags in cyclic order are
\[
 (B^+,B^-,gB^-,gB^+),\qquad gB^-=UB^-.
\]
The two triangles cut out by the diagonal have the one-sign word
coordinates. Amalgamating their diagonal coordinates multiplies
matching coordinates and adds the exchange forms, giving the
double-word product $g$; see \cite[Theorems~5.3--5.4 and
\S13.1]{GSAtlas}. Gluing the remaining sides identifies their
pinnings up to $H$, giving precisely \eqref{eq:annular-quotient}.
In particular, the holonomy in this convention is $g$ itself.
The root label at the right endpoint of a word is $i^*=n-i$;
the boundary gluing reverses labels by the same involution
\cite[\S\S13.1.2--13.1.3, (503)--(506)]{GSAtlas}. It therefore joins the two
endpoints of the same word-root $i$.

After this gluing, the two extremal root-$i$ coordinates are replaced
by their product; all internal coordinates remain. These are
coordinates on an integral torus, as follows directly from the seam
action. Its weights on the two extremal coordinates are
$\alpha_i,-\alpha_i$, where
$\alpha_i(\operatorname{diag}(h_1,\ldots,h_n))=h_i/h_{i+1}$.
The simple roots form a $\Z$-basis of
$X^*(H)$ because $G$ is adjoint. An invariant character has equal
exponents at the two extremities of each root. Hence the endpoint
products, together with the internal coordinates, form a basis of
the invariant character lattice. In the resulting cyclic word there
is one coordinate for each interval between successive occurrences
of a root, including the interval across the cut. All these
coordinates are mutable in the unpinned surface seed: the frozen
coordinates on the two original boundary edges are deleted
\cite[\S13, opening paragraphs]{GSAtlas}. Each of the two triangles
has $n(n-1)/2-r$ interior coordinates, and the diagonal and the
glued seam each have $r$ coordinates. Thus the full seed has
$n(n-1)=r(r+1)$ mutable coordinates.

\subsection{Positive matrix parameters}

For prescribed reduced expressions of $w_0$ in the upper and lower
blocks, every strictly totally positive real matrix has a factorization
\begin{equation}\label{eq:positive-factorization}
 A=U(\mathbf u)\operatorname{diag}(t_1,\ldots,t_n)
       L(\boldsymbol\ell),
 \qquad u_a,\ell_b,t_j>0,
\end{equation}
where the elementary factors in $U,L$ occur in the prescribed orders
\cite[Theorem~23]{FZ}. The cyclic coordinates of this factorization
are obtained from the identities
\[
 I+u e_{i,i+1}=H_i(u)E_iH_i(u^{-1}),\qquad
 I+\ell e_{i+1,i}=H_i(\ell^{-1})F_iH_i(\ell),
\]
and $\operatorname{diag}(t_1,\ldots,t_n)$ represents
$\prod_iH_i(t_i/t_{i+1})$ in $G$.
Collecting the Cartan factors of root $i$ in
\eqref{eq:positive-factorization} gives, in cyclic order,
\begin{equation}\label{eq:source-gaps}
 \frac{u_{\mathrm{next}}}{u_{\mathrm{previous}}},\qquad
 \frac{t_i/t_{i+1}}{u_{\mathrm{last}}\ell_{\mathrm{first}}},\qquad
 \frac{\ell_{\mathrm{previous}}}{\ell_{\mathrm{next}}},\qquad
 o_i=\ell_{\mathrm{last}}u_{\mathrm{first}}.
\end{equation}
The first and third expressions are taken over successive occurrences
in the corresponding sign block. The initial factor
$H_i(u_{\mathrm{first}})$ and the final factor
$H_i(\ell_{\mathrm{last}})$ combine across the cyclic cut to give
$o_i$. Thus every source coordinate is positive. We call $o_i$ the
\emph{outer coordinate} of root $i$; the remaining coordinates are
internal.
\section{Reduction to a path}\label{sec:carrier}

We next make the holonomy depend on $2r$ coordinates of a mutated
annular seed. The other coordinates will remain mutable coordinates
of the full seed. The reduction is a version of the cyclic word
construction in \cite[\S\S5.4--5.6]{SS}; we give the word order and
the local identities that produce the path needed here.

\subsection{Local word transformations}

For a seed basis $(e_v)$ write $\mathsf B_{uv}=\Omega(e_u,e_v)$.
Our convention for $X$-mutation at $k$ is
\begin{equation}\label{eq:x-mutation}
 X'_k=X_k^{-1},\qquad
 X'_v=X_v\bigl(1+X_k^{-\sgn\mathsf B_{vk}}\bigr)^{-\mathsf B_{vk}}
 \quad(v\ne k).
\end{equation}
The exchange matrix is mutated by changing the signs in row and
column $k$ and, for $u,v\ne k$, by
\[
 \mathsf B'_{uv}=\mathsf B_{uv}
 +[\mathsf B_{uk}]_+[\mathsf B_{kv}]_+
 -[-\mathsf B_{uk}]_+[-\mathsf B_{kv}]_+,
 \qquad [a]_+=\max(a,0).
\]

The word seed is obtained by adding elementary tile forms. For a
positive root-$i$ tile, let $L,R$ be its two root-$i$ endpoints, and
let $j$ denote the interval of a neighboring root through the tile.
Its nonzero entries, up to skew-symmetry, are
\begin{equation}\label{eq:tile-form}
 \mathsf B_{L,R}=-1,\qquad
 \mathsf B_{L,j}=\frac12,\qquad
 \mathsf B_{R,j}=-\frac12.
\end{equation}
A negative tile
has the opposite form. This is the elementary word form of
\cite[(416)]{GSAtlas}. The same rule defines its amalgamation on
a cyclic word. For two cyclic intervals of neighboring roots, the
number of interval endpoints contributing to their pairing is even.
Their half-integral tile contributions therefore sum to an integer.

We shall use commutations, braids, mixed swaps, and merges. Opposite
signs of distinct roots commute, as do equal signs of nonadjacent
roots. These are just commutations of the corresponding elementary
matrices. The remaining transformations are single mutations, as
the following identities show.

First, with $c=x/(1+x)$, a merge is given by
\begin{equation}\label{eq:word-merge}
 \begin{split}
 E_iH_i(x)E_i&=H_i(1+x)E_iH_i(c),\\
 F_iH_i(x)F_i&=H_i(c)F_iH_i(1+x).
 \end{split}
\end{equation}
The middle column has entries $(-1,+1)$ at the two exterior
root-$i$ intervals in the positive case, and the opposite entries
in the negative case. Formula \eqref{eq:x-mutation} gives the two
Cartan factors on the right of \eqref{eq:word-merge}. The mutated
middle coordinate is $x^{-1}$ and does not occur in the shorter
product.

For a braid with $|i-j|=1$, set
\[
 a'=a(1+x),\quad d'=\frac{dx}{1+x},\quad
 b'=\frac{bx}{1+x},\quad c'=c(1+x).
\]
Then
\begin{equation}\label{eq:word-braid}
 \begin{split}
 &H_i(a)H_j(b)E_iH_i(x)E_jH_j(c)E_iH_i(d)\\
 &\hspace{8mm}=
 H_i(a')H_j(b')E_jH_j(x^{-1})E_iH_i(d')E_jH_j(c').
 \end{split}
\end{equation}
The mutating column on $(i_L,i_R,j_L,j_R)$ is
$(-1,+1,+1,-1)$. Replacing the $E$'s by $F$'s negates this column;
the same identity holds with
\[
 a'=\frac{ax}{1+x},\quad d'=d(1+x),\quad
 b'=b(1+x),\quad c'=\frac{cx}{1+x}.
\]
To check both identities, factor out $H_i(a)H_j(b)$ on the left
and $H_i(d)H_j(c)$ on the right. The two positive products have
the common $3\times3$ blocks
\[
 \begin{pmatrix}x&x+1&1\\0&1&1\\0&0&1\end{pmatrix}
 \quad(j=i+1),\qquad
 \begin{pmatrix}x&x&x\\0&x&x+1\\0&0&1\end{pmatrix}
 \quad(j=i-1).
\]
The negative products give the transposes of these blocks.
Above the affected block the common diagonal factor is $x$,
and below it the common factor is $1$.

Finally, define $H_0(t)=I$ and $H_n(t)=tI$. The two mixed identities,
valid as identities of $\GL_n$ matrices, are
\begin{equation}\label{eq:word-mixed}
 \begin{split}
 E_iH_i(x)F_i
 &=H_{i-1}(c)H_i(1+x)F_iH_i(x^{-1})
     E_iH_i(1+x)H_{i+1}(c),\\
 F_iH_i(x)E_i
 &=H_{i-1}(1+x)H_i(c)E_iH_i(x^{-1})
     F_iH_i(c)H_{i+1}(1+x),
 \qquad c=\frac{x}{1+x}.
 \end{split}
\end{equation}
Their affected $2\times2$ blocks are respectively
\[
 \begin{pmatrix}x+1&1\\1&1\end{pmatrix},\qquad
 \begin{pmatrix}x&x\\x&x+1\end{pmatrix}.
\]
Above the block both sides have diagonal factor $x$, since
$c^2(1+x)^2/x=x$, and below it both have factor $1$.
For $E_iF_i\longmapsto F_iE_i$, the middle column is $(-1,-1)$
on the exterior root-$i$ intervals and $+1$ on each neighboring-root
interval. The reverse move negates this column. Thus
\eqref{eq:word-mixed} gives precisely \eqref{eq:x-mutation}.
When $i=r$, the factor $H_n$ is scalar and is omitted in $G$.
The function $F_\lambda$ is invariant under this scalar factor.

The exchange forms can be checked on the same local intervals.
Write $b_u=\mathsf B_{uk}$ for the displayed mutating column.
Before cyclic gluing its nonzero entries are $1$ or $-1$, and
for $u,v\ne k$ the change is
\[
 \mathsf B'_{uv}-\mathsf B_{uv}
 =\begin{cases}
  1,&b_u=1,\ b_v=-1,\\
  -1,&b_u=-1,\ b_v=1,\\
  0,&\text{otherwise}.
 \end{cases}
\]
For a positive merge this changes the pairing of the two exterior
root-$i$ intervals by $-1$, giving a single positive tile.
For a positive braid, in the order $(i_L,i_R,j_L,j_R)$, the
changed upper-triangular entries are $-1,-1,+1,+1$ on
$(i_L,i_R),(i_L,j_L),(i_R,j_R),(j_L,j_R)$, respectively.
For a mixed swap the change is $-1$ from each exterior root-$i$
interval to each neighboring-root interval. Negating a local
column negates these changes. Together with the sign reversal of
the mutating row and column, these are the sums of the new tiles
in \eqref{eq:tile-form}. Entries outside the column's support are
unchanged. At a merge the shorter word uses the principal submatrix
on the remaining intervals.

\subsection{Cyclic gluing and retained coordinates}

The local transformations descend to cyclic words by the following
elementary amalgamation observation. Replace coordinates in an
identified fiber by their product and exchange entries by their
sums. Mutation at a singleton fiber commutes with this operation
provided that the entries from each other fiber to the mutating
vertex have one weak sign. For the coordinates, multiply the factors
in \eqref{eq:x-mutation}. For the exchange matrix, use
$[\sum b_j]_+=\sum[b_j]_+$ for numbers of one weak sign in the
displayed mutation formula.

In all the words below, each root occurs with each sign. For a
same-sign braid or merge, an occurrence of the opposite sign lies
outside the local patch. It separates the two exterior intervals
of that root. In a braid this holds for both roots, so exterior
entries of opposite signs are not identified. In a mixed swap,
the two exterior root-$i$ intervals can coincide, but their entries
have the same sign; the resulting entry is $2$ or $-2$. The mutating
interval is a singleton. The amalgamation observation therefore
applies to every local transformation. Changing the cut of a cyclic
word replaces a product $AB$ by $BA=A^{-1}(AB)A$ and preserves its
holonomy class.

To keep track of the full seed, call an interval \emph{active} if
it occurs in the current word evaluation. Initially all intervals
are active. At a merge, perform the mutation in the full seed and
then mark its middle interval inactive. Its coordinate still
belongs to the full mutable seed. Inductively, the active principal
submatrix is the current word matrix and the holonomy class depends
only on active coordinates. Indeed, a later mutation at an active
vertex updates active coordinates using only that coordinate and
the active principal submatrix; matrix mutation restricts to the
same submatrix. The inactive coordinates and all mixed entries
are simultaneously updated by the full mutation. This proves the
induction statement at every stage.

\subsection{The word reduction}

For integers $a,b$ write $[a\downarrow b]=(a,a-1,\ldots,b)$,
with the empty word understood when $a<b$, and put
\[
 S_b=[r\downarrow b],\qquad W_r=S_1S_2\cdots S_r,
 \qquad C_r=(1,\ldots,r),\qquad \rho(i)=r+1-i.
\]
Both $W_r$ and $\rho(W_r)$ are reduced words for $w_0$.
We choose the source word
\begin{equation}\label{eq:source-word}
 \mathsf D_0=(\rho(W_r),-W_r),
\end{equation}
where the minus sign changes each letter without reversing its
order. Formula \eqref{eq:positive-factorization} applies to this
prescribed source.

\begin{lemma}\label{lem:carrier}
There is a finite sequence of mutations of the full annular seed
\eqref{eq:source-word} for which the active cyclic word is
\begin{equation}\label{eq:terminal-word}
 E_rE_{r-1}\cdots E_1F_1F_2\cdots F_r.
\end{equation}
The holonomy conjugacy class is given by its cyclic evaluation.
The full seed has $r(r+1)$ mutable coordinates, of which $2r$
are active at the endpoint.
\end{lemma}

\begin{proof}
First consider $(\mathbf b,W_r)$, where $\mathbf b$ is any
all-negative word containing every root. We reduce its positive
block to $C_r$, restoring the order of $\mathbf b$ after each step.
The intermediate positive words are
\[
 T_{j,k}=(1,\ldots,j)[r-k\downarrow j+1]S_{j+2}\cdots S_r,
 \qquad
 0\leq j\leq r-1,\quad 0\leq k\leq r-j-1.
\]
Here $T_{0,0}=W_r$,
$T_{j,r-j-1}=T_{j+1,0}$ for $j<r-1$, and
$T_{r-1,0}=C_r$. Suppose $k\leq r-j-2$ and set $a=r-k$.
Move the first displayed letter $a$ to the left through
$1,\ldots,j$, which is possible because $a\geq j+2$.
Move it through $\mathbf b$ by opposite-sign commutations and
mixed equal-root swaps. Rotate this initial letter through the cut
to the end of the cyclic word, and commute it left through
$S_{a+2},\ldots,S_r$. It remains to perform the absorption
\begin{equation}\label{eq:word-absorption}
 S_aS_{a+1}a\longrightarrow S_aS_{a+1}.
\end{equation}

This absorption follows by induction on $r-a$. For $a=r$ it is
the merge $rr\longrightarrow r$. For $a<r$, commutations and a
single braid give
\[
 \begin{aligned}
 S_aS_{a+1}a
 &=S_{a+1}aS_{a+2}(a+1)a\\
 &\longleftrightarrow S_{a+1}S_{a+2}a(a+1)a\\
 &\longleftrightarrow S_{a+1}S_{a+2}(a+1)a(a+1).
 \end{aligned}
\]
Apply the induction hypothesis to the initial
$S_{a+1}S_{a+2}(a+1)$, then commute $a$ back through $S_{a+2}$.
The result is $S_aS_{a+1}$. The whole step has therefore changed
$(\mathbf b,T_{j,k})$ into $(\mathbf b,T_{j,k+1})$.
The displayed finite set of indices proves termination at
$(\mathbf b,C_r)$.

Apply the Dynkin-reflected version of this procedure to
$(-W_r,\rho(W_r))$, obtained from \eqref{eq:source-word} by a
change of cut. This gives $(-W_r,(r,\ldots,1))$. Change the cut
again and use the sign-reversed versions of the same local
identities to reduce the negative block. The result is
$(r,\ldots,1,-1,\ldots,-r)$, as required.

For clarity, the procedure also determines the mutation vertices.
Attach to each signed letter the label of the interval of its
root immediately following it. A commutation interchanges the
two letter-label pairs. The other update rules are
\[
 \begin{array}{rcll}
 (a,u),(-a,v)&\longmapsto&(-a,u),(a,v),&\text{mixed swap},\\
 (a,u),(b,v),(a,w)&\longmapsto&(b,u),(a,w),(b,v),&\text{braid},\\
 (a,u),(a,v)&\longmapsto&(a,v),&\text{merge}.
 \end{array}
\]
In each row the mutated interval is $u$; in the last row it is
retained as an inactive mutable coordinate. These rules are exactly
the coordinate placements in \eqref{eq:word-merge}--\eqref{eq:word-mixed}.
Each root retains at least one occurrence of each sign throughout,
so the cyclic amalgamation conditions continue to hold.
The preceding full-seed induction lifts every step to an
annular mutation. Finally, the initial word has $r(r+1)$ letters
and the terminal word has $2r$; a cyclic word has one interval
per letter. All initial coordinates remain in the full seed.
\end{proof}

\subsection{The terminal determinant and its lattice}

Denote the active interval following $E_i$ by $x_i$ and the one
following $F_i$ by $y_i$. Thus the terminal product is
\[
 E_rH_r(x_r)\cdots E_1H_1(x_1)
 F_1H_1(y_1)\cdots F_rH_r(y_r).
\]
Set
\begin{equation}\label{eq:terminal-weights}
 d_n=1,\qquad d_i=x_iy_id_{i+1},\qquad b_i=y_id_{i+1}\quad(i<n).
\end{equation}
Moving the Cartan factors to the middle gives the matrix
\begin{equation}\label{eq:terminal-form}
 M=\left(I+\sum_{i=1}^{r}e_{i,i+1}\right)
       \operatorname{diag}(d_1,\ldots,d_n)
       \left(I+\sum_{i=1}^{r}y_i e_{i+1,i}\right).
\end{equation}
The descending upper order and ascending lower order make the
outer factors bidiagonal. In particular, $M$ is tridiagonal, with
diagonal entries $d_i+b_i$ for $i<n$ and $d_n$ for $i=n$, and
\[
 M_{i,i+1}M_{i+1,i}=b_id_{i+1}.
\]

For a weighted graph $P$, let $\mathcal E_k(P)$ be the sum of the
weight products of its independent sets of size $k$.

\begin{lemma}\label{lem:path}
Let $P$ be the path with successive weights
$d_1,b_1,d_2,b_2,\ldots,b_r,d_n$. Then
\begin{equation}\label{eq:path-determinant}
 \det(I+tM)=\sum_{k=0}^{n}\mathcal E_k(P)t^k,
 \qquad \det M=\prod_{i=1}^{n}d_i.
\end{equation}
In the active order $(x_1,y_1,\ldots,x_r,y_r)$ the exchange form is
\begin{equation}\label{eq:active-band}
 \mathsf B_{j,j+1}=2,\qquad
 \mathsf B_{j,j+2}=-1,\qquad
 \mathsf B_{j,k}=0\quad(k>j+2).
\end{equation}
Its determinant is $n^2$.
\end{lemma}

\begin{proof}
Let $I_i(t)$ be the independent-set generating polynomial of the
tail path starting at $d_i$, and let $J_i(t)$ be that of the tail
starting at $b_i$, with $J_n=I_{n+1}=1$. Splitting at the first
vertex gives
\[
 I_i=J_i+td_iI_{i+1},\qquad
 J_i=I_{i+1}+tb_iJ_{i+1}.
\]
Eliminating $J_{i+1}=I_{i+1}-td_{i+1}I_{i+2}$ yields
\[
 I_i=(1+t(d_i+b_i))I_{i+1}-t^2b_id_{i+1}I_{i+2},
 \qquad I_n=1+td_n,\quad I_{n+1}=1.
\]
The trailing principal minors of $I+tM$ satisfy the same recurrence
and initial conditions. This proves the first equality in
\eqref{eq:path-determinant}. The unique independent set of size
$n$ consists of the $n$ vertices $d_i$, giving the second equality.

For the exchange form, the positive and negative tiles of root $i$
each contribute $1$ to $\mathsf B(x_i,y_i)$. For consecutive roots,
adding their endpoint contributions from \eqref{eq:tile-form} gives
\[
 \begin{array}{c|rrrr}
 (u,v)&(x_i,x_{i+1})&(y_i,y_{i+1})&(y_i,x_{i+1})&(x_i,y_{i+1})\\ \hline
 \mathsf B(u,v)&-1&-1&2&0.
 \end{array}
\]
Nonadjacent roots have zero pairing. This is
\eqref{eq:active-band}. Let $p_r$ be its Pfaffian. For $r\geq2$, the first row
has entries $2,-1$ in columns $2,3$. Deleting rows and columns
$1,2$ leaves the same band matrix of size $2r-2$, contributing
$2p_{r-1}$. After deleting $1,3$, the first remaining vertex $2$
has its sole nonzero entry $\mathsf B_{2,4}=-1$, so that minor
has Pfaffian $-p_{r-2}$. The Pfaffian expansion therefore gives
\[
 p_r=2p_{r-1}-p_{r-2},\qquad p_0=1,\quad p_1=2.
\]
Thus $p_r=r+1=n$ and $\det\mathsf B=p_r^2=n^2$.
\end{proof}

The ratios of successive path weights are
$x_i=d_i/b_i$ and $y_i=b_i/d_{i+1}$. We reverse the path, so that
its forward ratios, in order, are
$(y_r,x_r,\ldots,y_1,x_1)$. This is a permutation of the actual
active coordinates.

Introduce the free vertex lattice
\[
 L=\Z D_1\oplus\Z B_1\oplus\Z D_2\oplus\cdots
       \oplus\Z B_r\oplus\Z D_n.
\]
We call $D_i$ the anchor vectors and $B_i$ the bridge vectors;
their vertex weights are the characters $X^{D_i}$ and $X^{B_i}$.
They homogenize the reversed
weights $d_{n+1-i},b_{n-i}$: specializing $X^{D_1}=1$ recovers
\eqref{eq:terminal-weights}. Give every vertex weight one, let
$L_0=\ker(\wt:L\to\Z)$, and put
\begin{equation}\label{eq:vertex-gaps}
 c_{2i-1}=B_i-D_i,\qquad c_{2i}=D_{i+1}-B_i\quad(1\leq i<n).
\end{equation}
These forward differences form a $\Z$-basis of $L_0$.
Together with $D_1$ they form a $\Z$-basis of $L$; in this
auxiliary vertex seed the gaps are mutable and $D_1$ is frozen.
More explicitly, if $V_1,\ldots,V_{2n-1}$ denote the vertices in
the displayed order, then for $\sum_ja_j=0$,
\[
 \sum_{j=1}^{2n-1}a_jV_j
 =-\sum_{j=1}^{2n-2}\left(\sum_{h=1}^{j}a_h\right)c_j.
\]
In particular, a vector of $L_0$ belongs to $\sum_j\N c_j$ precisely when
all its proper vertex prefix sums are nonpositive.

Define $\Omega$ on $L$, extending by skew-symmetry, by
$\Omega(V_{j+1},V_j)=1$ and zero on nonadjacent vertices.
Its restriction to the ordered gaps in \eqref{eq:vertex-gaps}
has first superdiagonal $-2$ and second superdiagonal $1$.
It therefore agrees with \eqref{eq:active-band} after the preceding
permutation. The character map
\begin{equation}\label{eq:active-embedding}
 \iota(y_{n-i})=X^{c_{2i-1}},\qquad
 \iota(x_{n-i})=X^{c_{2i}}
\end{equation}
identifies the actual active character lattice with $L_0$ integrally.
The vector
\begin{equation}\label{eq:vertex-radical}
 \delta=D_1+\cdots+D_n
\end{equation}
is radical: each bridge pairs with its two neighboring anchors
with opposite signs, and the anchors pair trivially with each
other. Since the restriction to $L_0$ is nondegenerate,
$\R\delta$ is the entire radical. Since
$\wt(\delta)=n$, the sublattice $L_0+\Z\delta$ has index $n$ in $L$.
We retain $L_0$ as the active integral lattice.

From now on $P$ denotes the reversed path with the formal weights
$X^{D_1},X^{B_1},\ldots,X^{B_r},X^{D_n}$.
Finally, let $Y$ denote its formal spectral alphabet, defined by
$e_k(Y)=\mathcal E_k(P)$ for $1\leq k\leq n$, where $e_k$ is the
$k$th elementary symmetric polynomial. The map
\eqref{eq:active-embedding} satisfies
\[
 \iota(d_{n+1-i})=X^{D_i-D_1},\qquad
 \iota(b_{n-i})=X^{B_i-D_1}.
\]
Thus the common factor $X^{-D_1}$ cancels between the degree-$n$
numerator and denominator of $F_\lambda$. Since
$\mathcal E_n(P)=X^\delta$, \eqref{eq:path-determinant} gives
\begin{equation}\label{eq:vertex-spectral}
 \iota(F_\lambda)=X^{-\delta}m_\lambda(Y).
\end{equation}
Both sides have vertex weight zero, so this is an identity in the
active Laurent ring.
\section{Theta functions and finite Newton polytopes}\label{sec:theta}

We use the classical theta functions of
Gross--Hacking--Keel--Kontsevich \cite{GHKK}, with the
$X$-realization and positivity results of Davison--Mandel \cite{DM}.
We first specify the realization, since its chamber convention and
its coefficient grading will both enter the proof.

\subsection{The chamber convention}

Let $M$ be the character lattice of a full unpinned seed, with
basis $e_1,\ldots,e_d$ and integral skew form $\Omega$. We write
the matrix of the form in this basis also as $\Omega$, so that
$\Omega(v,w)=v^{\mathsf T}\Omega w$. Put
\[
 C=\sum_{i=1}^d\Z_{\geq0}e_i.
\]
The positive root degree is recorded by variables
$y_1,\ldots,y_d$. The monomial belonging to a root $q\in C$ is
$x^qy^q$; in particular, the root map on the character lattice
is the identity.

It is useful to allow the form $\Omega$ to be degenerate. For
this purpose use the position lattice
\begin{equation}\label{eq:principal-data}
 \widehat M=M\oplus\Z^d,\qquad
 \widehat P(v)=(v,0),\qquad
 \widehat Q(v)=(\Omega^{\mathsf T}v,v)\in\widehat M^*,
\end{equation}
where coordinate bases identify the displayed dual lattices.
The initial wall of root $e_i$ has normal $\widehat Q(e_i)$
and factor $1+x^{e_i}y_i$. Labels and root increments lie in
$E=M\oplus0$. The second summand of $\widehat M$ is used for
wall positions; the variables $y_i$ separately record root
degrees. The initial negative chamber is nonempty, since
$(0,-\mathbf1)$ has negative pairing with each initial normal.
We always choose an endpoint $Q$ in this chamber such that
\begin{equation}\label{eq:strong-generic}
 Q\notin H
 \quad\hbox{for every proper rational linear hyperplane }
 H\subset\widehat M_{\R}.
\end{equation}
Such points are dense in every nonempty open set.

Here this negative chamber gives the usual seed expression of
an $X$-theta function. To check this integrally, consider
\begin{equation}\label{eq:principal-identification}
 T(m,a)=(\Omega^{\mathsf T}m-a,m),\qquad
 T^{-1}(u,v)=(v,\Omega^{\mathsf T}v-u).
\end{equation}
The standard principal $A$-datum has root and normal maps
\[
 P_A(v)=(\Omega^{\mathsf T}v,v),\qquad Q_A(v)=(v,0).
\]
Directly,
\[
 T\widehat P=P_A,\qquad T^*(-Q_A)=\widehat Q.
\]
Reversing a normal reverses both its crossing orientation and
the sign of its pairing in the wall action, so it leaves that
action unchanged. Thus our negative chamber corresponds to
the positive chamber for the principal $A$-datum.
Moreover,
\[
 T(\nu,0)=(\Omega^{\mathsf T}\nu,\nu)
\]
is precisely the integral embedding which defines the
$X$-theta functions in \cite[\S\S4.3--4.4 and 4.7.2]{DM}.
This integral comparison fixes the labels and the chamber
convention also for a degenerate full form.
The construction in \cite[\S4.7.2]{DM} permits generic endpoints
in the full principal position space and identifies its chamber
charts with the usual $X$-cluster charts.

We recall the properties of positive scattering used below.
Its wall supports are rational cones in central hyperplanes,
and each joint cone has a rational linear span of codimension at
least two. Modulo any
fixed positive root degree, only finitely many walls are
nontrivial. A broken line is a piecewise linear path
$\Gamma:(-\infty,0]\to\widehat M_{\R}$, ending at $Q$, whose
velocity on a segment with attached exponent $m\in E$ is
$-m$. It has finitely many nontrivial bends. At a joint-free
crossing, the next monomial is chosen from the wall action on
the preceding monomial. The theta function sums the resulting
terminal monomials with their coefficients.

The positive scattering theorem gives a representative in
which these coefficients are nonnegative integers. In the
chamber just fixed, the principal expression is pointed:
\begin{equation}\label{eq:theta-principal}
 \Theta_\nu(x,y)
   =x^\nu\sum_{q\in C}a_{\nu,q}x^qy^q,
 \qquad a_{\nu,q}\in\Z_{\geq0},\qquad a_{\nu,0}=1.
\end{equation}
Strong positivity also gives, coefficientwise in the principal
completion,
\begin{equation}\label{eq:strong-domination}
 \Theta_{k\nu}\ \leq\ \Theta_\nu^k
 \qquad(k\in\Z_{>0}).
\end{equation}
Give $x^my^q$ the degree $m-q\in M$. Then $\Theta_\nu$ is
homogeneous of degree $\nu$, so a product term indexed by
$\nu+\mu+r$ carries the factor $y^r$. The positive product expansion is
\[
 \Theta_\nu\Theta_\mu
 =\Theta_{\nu+\mu}
  +\sum_{0\ne r\in C}\alpha_r y^r\Theta_{\nu+\mu+r},
 \qquad \alpha_r\in\N.
\]
Iterating its leading term proves \eqref{eq:strong-domination}.
We use the classical
specializations of \cite[Theorems 3.9--3.10]{DM}; their
compatibility with the $X$-construction follows from
\cite[Lemma 4.6 and Proposition 4.8]{DM}.

\subsection{Finite principal expressions}

Suppose that the actual $X$-theta function $\vartheta_\nu$ is
a Laurent polynomial in the chosen seed. Then
$\Theta_\nu$ in \eqref{eq:theta-principal} is finite as well.
To see this, remove the coefficient monomial $y^q$ on the
displayed support graph. Distinct $q$ have distinct character
exponents $\nu+q$, so this is an injective coefficientwise map
to the completion in the cone $C$. By the preceding principal
comparison, its image is the formal seed expansion of
$\vartheta_\nu$. Laurentness of the latter therefore forces
all but finitely many $a_{\nu,q}$ to vanish. After this
observation one can write
\[
 \vartheta_\nu(x)=\Theta_\nu(x,1)
\]
as an ordinary substitution in a finite polynomial.

For a nonzero Laurent polynomial $f$ write $\Newt(f)$ for the
convex hull of its character exponents, and put
\[
 v_w(f)=\min_{a\in\Supp(f)}\langle w,a\rangle
 \qquad(w\in M_{\R}^*).
\]
The following finite form of valuative independence will be
needed. It is a special case of \cite[Theorem 3.10]{CMMM}.
We give a proof from positive scattering and
\eqref{eq:strong-domination}.

\begin{lemma}\label{lem:newton}
Let $\nu_s$, $s\in S$, be a nonempty finite set of distinct
theta labels whose seed expressions $\vartheta_{\nu_s}$ are
Laurent polynomials. If $c_s\in\C^\times$, then
$\sum_{s\in S}c_s\vartheta_{\nu_s}\ne0$ and
\begin{equation}\label{eq:newton}
 \Newt\left(\sum_{s\in S}c_s\vartheta_{\nu_s}\right)
   =\conv\bigcup_{s\in S}\Newt(\vartheta_{\nu_s}).
\end{equation}
Equivalently, for every $w\in M_{\R}^*$,
\begin{equation}\label{eq:finite-valuation}
 v_w\left(\sum_{s\in S}c_s\vartheta_{\nu_s}\right)
   =\min_{s\in S}v_w(\vartheta_{\nu_s}).
\end{equation}
\end{lemma}

\begin{proof}
All principal expressions in the statement are finite by the
preceding argument. We work at the single endpoint $Q$ chosen
in \eqref{eq:strong-generic}. We extend each covector on $M_\R$
to $\widehat M_\R$ by zero on the auxiliary summand.

\smallskip
\noindent\emph{Rational broken lines.}
At a joint-free crossing, combine the wall factors having the
same supporting hyperplane. Choose its primitive normal
$\ell$ so that $\ell(m)>0$ for the incoming exponent $m$.
If an individual normal has absolute multiplicity $b$ relative
to $\ell$, include its factor to the power $b$. The resulting
positive factor $f$, with constant term one, has crossing rule
\[
 x^m\longmapsto x^m f^{\ell(m)}.
\]
Every exponent in $f$ lies in $\ker\ell$. Coefficient degrees
are understood throughout. A \emph{rational bend} is a choice
\begin{equation}\label{eq:rational-bend}
 m'=m+\ell(m)r,\qquad
 r\in\conv_{\Q}\Supp_x(f),
\end{equation}
where $\Supp_x(f)$ is the set of character exponents occurring in
$f$, and $\conv_{\Q}$ is the set of their finite rational convex
combinations. Thus $\ell(r)=0$, and the bend is
the shear
\[
 T_r=I+r\otimes\ell,\qquad
 T_r^{-1}=I-r\otimes\ell.
\]
It fixes its supporting hyperplane pointwise. An ordinary
bend is a rational bend, since an exponent of
$f^{\ell(m)}$ is a sum of $\ell(m)$ exponents of $f$.

Consider a rational broken line with initial exponent $\nu$
and finitely many nontrivial bends. At each bend choose a
finite expression
\[
 r_i=\sum_j b_{ij}q_{ij},\qquad
 b_{ij}\in\Q_{\geq0},\quad \sum_jb_{ij}=1,\quad
 q_{ij}\in\Supp_x(f_i).
\]
Choose a positive integer $k$ clearing all denominators of
the slopes and of the numbers $\ell_i(m_{i-1})b_{ij}$.
The positive expansion of
$f_i^{k\ell_i(m_{i-1})}$ then contains the required increment
$k\ell_i(m_{i-1})r_i$. Multiplying all slopes by $k$ and
rescaling time turns the path into an ordinary broken line
with initial exponent $k\nu$. If $m$ was its original
terminal exponent, it follows that
\begin{equation}\label{eq:rational-newton}
 km\in\Supp(\vartheta_{k\nu})
       \subseteq\Supp(\vartheta_\nu^k),
 \qquad m\in\Newt(\vartheta_\nu).
\end{equation}
Here \eqref{eq:strong-domination} first makes
$\Theta_{k\nu}$ finite, because $\Theta_\nu^k$ is finite.
Consequently all the supports used in
\eqref{eq:rational-newton} are finite. Positivity shows that
the support of $\vartheta_\nu^k$ consists of sums of $k$
members of $\Supp(\vartheta_\nu)$, which proves the last
inclusion.

If $f_i$ is an infinite product of parallel wall factors,
each monomial used in the displayed convex combinations
requires only finitely many factors, by local finiteness in
each root degree. The remaining factors contribute their
constant term. Thus the same denominator-clearing argument
applies. In particular, an ordinary line whose terminal
exponent minimizes a covector on $\vartheta_\nu$ also
minimizes it among all rational lines with the same initial
label and endpoint.

\smallskip
\noindent\emph{Deformation with a fixed endpoint.}
Let a minimizing line have successive exponents
$m_0,\ldots,m_\ell$ and normalized bends $r_i$, with
\[
 T_i=I+r_i\otimes\ell_i,\qquad
 m_i=T_i m_{i-1},\qquad
 \ell_i(m_{i-1})>0.
\]
When testing a crossing at which the line is straight, include
it in this finite list with $r_i=0$. Put
\[
 A_i=T_\ell\cdots T_{i+1},\qquad A_\ell=I.
\]
Suppose that at crossing $j$ an admissible $r'_j$ has a smaller
pairing than $r_j$ with the transported covector $A_j^*w$.
Replace $r_j$ by
$r_j(s)=(1-s)r_j+sr'_j$, where $s>0$ is rational, and leave
all other normalized bends unchanged. Starting with the same
$m_0$, compute the new exponents forward. Starting with the
same endpoint, compute the bend positions backward:
\begin{equation}\label{eq:backward-positions}
 \begin{split}
 z_{\ell+1}(s)&=Q,\\
 \tau_i(s)&=-\frac{\ell_i(z_{i+1}(s))}
                    {\ell_i(m_i(s))},\\
 z_i(s)&=z_{i+1}(s)+\tau_i(s)m_i(s)
 \qquad(i=\ell,\ldots,1).
 \end{split}
\end{equation}
At $s=0$ the denominators are nonzero, the segment lengths
$\tau_i$ are positive, and the specified wall factors contain
the bend points in the relative interiors of their supports.
For all sufficiently small $s$ these finitely many conditions
persist. Choose beforehand the finitely many factors
witnessing the convex combinations for the $r_i$ and $r'_j$.
Their support conditions persist as well. Extra factors and
all unprescribed crossings are traversed using constant terms.
The initial geometric ray may move; its exponent remains
$m_0$.

It remains to check admissibility for the entire scattering
diagram, which can have infinitely many joints. For rational
$s$, the shears
\[
 A_i(s)=T_\ell(s)\cdots T_{i+1}(s)
\]
are rational invertible maps. They unfold the spatial
segments to one straight line. More precisely, with time
zero at the endpoint, on the $i$th segment
\begin{equation}\label{eq:unfolded-line}
 A_i(s)\Gamma_s(t)=Q-tm_\ell(s).
\end{equation}
Indeed, all transformed velocities equal $-m_\ell(s)$, and
the transformed segments join because $T_i(s)$ fixes the
$i$th bend hyperplane pointwise.

If a segment met a joint cone $J$, then
\[
 Q\in A_i(s)\spanR(J)+\R m_\ell(s).
\]
The right side is a proper rational linear subspace of
$\widehat M_{\R}$: $\spanR(J)$ has codimension at least two,
and only one direction has been added. This contradicts
\eqref{eq:strong-generic}. If a segment lay in a wall hyperplane
$H$, its direction would satisfy $m_i(s)\in H$, and hence
$m_\ell(s)=A_i(s)m_i(s)\in A_i(s)H$. Equation
\eqref{eq:unfolded-line} would then imply $Q\in A_i(s)H$,
again a proper rational hyperplane. Thus all crossings
are transverse and the perturbed path avoids every joint of
the full diagram for every sufficiently small rational $s$.

The later shears and the incoming exponent at crossing $j$
were fixed. Hence the terminal exponent changes by
\[
 m_\ell(s)-m_\ell
   =s\,\ell_j(m_{j-1})A_j(r'_j-r_j).
\]
Its pairing with $w$ is strictly negative, contradicting
\eqref{eq:rational-newton}. We have proved that a minimizing
line makes a minimizing normalized choice at every crossing,
including its straight crossings. Call a line with this
property \emph{taut}.

\smallskip
\noindent\emph{Backward uniqueness.}
Choose $w$ so that its restriction to $E$ is nonzero on
every nonzero rational vector of $E$. Every rational
invertible shear preserves this property on $E$. A local
minimum of a transported covector on
$\conv_{\Q}\Supp_x(f)$, when attained, is therefore attained
at a unique point $r_*$. For a taut line, the outgoing
exponent $m$ and the outgoing transported covector determine
the preceding exponent:
\[
 m_{\mathrm{in}}=m-\ell(m)r_*.
\]
The normal component is unchanged by the bend, so its
positive value $\ell(m)$ is known. The shear
$I+r_*\otimes\ell$, and hence the preceding transported
covector, are known as well.

Consequently two taut lines having the same endpoint and
terminal exponent must have the same initial label. For a
precise backward comparison, take the finite union of their
actual nontrivial bends. Starting from their common terminal
segment, take the latest remaining bend. At this crossing
both lines are taut, even if one of them was originally
straight there. The preceding formula forces the same
incoming segment for both. Repetition over the finite union
proves the assertion. If the
common terminal exponent is zero, invertibility of the bend
shears forces the initial label to be zero; the constant
theta function is handled by the same argument.

\smallskip
\noindent\emph{The vertices cannot cancel.}
Let
\[
 \mathcal P=\conv\bigcup_{s\in S}\Newt(\vartheta_{\nu_s}).
\]
Fix a vertex $v$ of this finite polytope. There is a nonempty
open cone of covectors exposing $v$ as the unique minimum on
$\mathcal P$. Choose $w$ in that cone satisfying the
genericity used in the preceding paragraph. If $v$ occurred
in two theta functions, choose an ordinary broken line
contributing it in each. Both lines minimize $w$, and hence
are taut. Backward uniqueness would make their initial
labels equal. Thus $v$ belongs to exactly one theta function
in the given family. Its coefficient in the proposed sum is
a nonzero scalar $c_s$ times a positive integer, and cannot
vanish. Every vertex of $\mathcal P$ survives, while no
support exponent of the sum lies outside $\mathcal P$. This
proves \eqref{eq:newton}, including nonvanishing of the sum.
Taking minima of linear forms gives
\eqref{eq:finite-valuation}.
\end{proof}
\section{A finite expansion on the active lattice}\label{sec:descent}

Let $\cX$ be the framed $\PGL_n$-space of the annulus from
Section~\ref{sec:annulus}. All coordinates of a full unpinned
seed remain present, including the inactive mutable
coordinates produced by Lemma~\ref{lem:carrier}. We denote by
$\Ocl(\cX)$ the intersection of the Laurent rings of all
mutation-equivalent seeds.

\subsection{The full annular expansion}

\begin{lemma}\label{lem:finite-expansion}
For $\lambda\vdash n$, the annular holonomy function
\[
 F_\lambda(g)=\frac{m_\lambda(\eig g)}{\det g}
\]
belongs to $\Ocl(\cX)$. In every full seed it has an expansion
\begin{equation}\label{eq:full-expansion}
 F_\lambda=\sum_{s\in S}c_s\vartheta^{\mathrm{full}}_{\nu_s},
 \qquad |S|<\infty,
\end{equation}
where the labels $\nu_s$ are distinct, $c_s\ne0$, and each
$\vartheta^{\mathrm{full}}_{\nu_s}$ is a regular theta
function.
\end{lemma}

\begin{proof}
By \eqref{eq:spectral-F}, $F_\lambda$ is a regular class
function on $\PGL_n$. Its pullback by holonomy to the space
with boundary pinnings is therefore regular.

Shen's regularity theorem
\cite[Theorem 1.1 and Definition 2.1]{Shen} identifies the
regular functions on the pinned space with the intersection
of the Laurent rings over all its mutation-equivalent seeds.
Forgetting the boundary pinnings deletes their frozen
coordinates and gives precisely the framed cluster
$X$-structure \cite[Theorem 8.3, \S8.2.3, \S20.1.3, (704), (708)]{GSAtlas}.
Every mutation word of an
unpinned seed lifts to the same word in the pinned seed:
the updates of the mutable coordinates and of their
principal exchange submatrix do not use the deleted frozen
coordinates.

Write $x$ for the unpinned coordinates and $z$ for those
boundary frozen coordinates in a lifted chart. The pullback
of $F_\lambda$ is Laurent in $x,z$ and belongs to $\C(x)$.
Algebraic independence gives
\begin{equation}\label{eq:laurent-intersection}
 \C[x^{\pm1},z^{\pm1}]\cap\C(x)=\C[x^{\pm1}].
\end{equation}
This proves Laurentness in every unpinned seed, including
seeds obtained by arbitrary mutation words.

The annulus with one marked point on each boundary is an
admissible surface in \cite[Definition 1.1]{GS_DT}.
The cluster Donaldson--Thomas transformation of
\cite[Theorem 1.3]{GS_DT} supplies both signs of the initial
basis rays in the associated $\mathcal A$ cluster complex.
Their positive
span is the whole space, so this complex is contained in no
proper linear half-space. The full theta-basis criterion
\cite[Definition 8.27 and Proposition 8.28]{GHKK} therefore
applies to the actual unpinned $X$-seed. It identifies its
upper algebra with the vector space having the regular
theta functions as a basis. This is an ordinary vector-space
basis, and hence gives the finite expansion
\eqref{eq:full-expansion}; see also
\cite[Theorem~1.4(i)]{GS_DT}. The construction uses the character
lattice of this seed throughout.
\end{proof}

\subsection{Eliminating the retained directions}

At the terminal full seed, write
\begin{equation}\label{eq:block-form}
 M=M_I\oplus M_J,\qquad
 \Omega=\begin{pmatrix}B&K\\-K^{\mathsf T}&\Omega_J\end{pmatrix}.
\end{equation}
Here $I$ denotes the active gaps and $J$ the retained
inactive vertices, and $\Omega_J$ is the form on $M_J$.
The matrix $B$ is the active form
computed in \eqref{eq:active-band}; in particular,
$\det B=n^2\ne0$. The following descent also applies in any
full seed satisfying the same hypotheses.

\begin{lemma}\label{lem:descent}
Suppose a nonzero finite linear combination of distinct
regular full theta functions depends only on the active
variables $x_I$, and the active form $B$ is nonsingular.
Then each summand is a Laurent polynomial in $x_I$ alone.
Its label lies in $M_I$, and its seed expression equals the
theta function of the active datum with that same integral
label.
\end{lemma}

\begin{proof}
Let
\[
 F=\sum_{s\in S}c_s\vartheta^{\mathrm{full}}_{\nu_s},
 \qquad c_s\ne0.
\]
The ring intersection \eqref{eq:laurent-intersection}, now
with $x_I,x_J$, makes $F$ a Laurent polynomial in $x_I$.
For every $j\in J$ both $v_{e_j^*}(F)$ and
$v_{-e_j^*}(F)$ equal zero. Lemma~\ref{lem:newton} therefore
makes the corresponding two valuations of each summand
nonnegative. Every monomial in every summand has its
$j$th exponent both nonnegative and nonpositive, and hence
equal to zero.

Each principal refinement is finite and has the form
\eqref{eq:theta-principal}. Its leading coefficient is one,
so $(\nu_s)_J=0$. For every nonzero coefficient in that
refinement the character exponent is $\nu_s+q$; its retained
part is zero as well. Thus $q_J=0$ for every term. These
conclusions hold for each theta function separately.

Now take the continuous quotient of the root-degree
completion obtained by setting
\[
 y_j=0\qquad(j\in J).
\]
The monoid $\Z_{\geq0}^I\times0$ is a saturated face of the
full free positive monoid. Degrees add under Lie brackets
and under consecutive bends, so a term with positive
retained degree cannot contribute to degree zero in the
retained variables. The quotient diagram is therefore the
consistent completion with only the active initial roots.
Their root and normal maps, before the auxiliary enlargement,
are
\begin{equation}\label{eq:full-active-normals}
 P_0(v)=\binom{v}{0},\qquad
 Q_0(v)=\binom{B^{\mathsf T}v}{K^{\mathsf T}v}.
\end{equation}
In particular, the mixed block $K$ remains in the normals.

Theta functions of this quotient are the retained-degree-zero
parts of the full theta functions. This can also be seen
directly from broken lines: a line of total retained degree
zero has only active nontrivial bends, and deleted walls are
crossed using their constant terms. Conversely, a line in
the quotient can be regarded in the full diagram in this
way. At the fixed endpoint \eqref{eq:strong-generic}, the
unfolding argument \eqref{eq:unfolded-line} excludes the
additional joints of the full diagram as well. Since each
summand was already proved to have only retained degree
zero, it equals its quotient theta function.

It remains to identify the latter with the active theta
function. The comparison takes place in real position
spaces. Without auxiliary directions it is
\[
 \pi(u,z)=u+B^{-1}Kz,
 \qquad u\in(M_I)_{\R},\quad z\in(M_J)_{\R}.
\]
In the enlargement \eqref{eq:principal-data}, write the
extra position as $a=(a_I,a_J)$ and use
\begin{equation}\label{eq:active-position-projection}
 \pi(u,z,a)=u+B^{-1}(Kz+a_I).
\end{equation}
For every active root $v$,
\[
 \big\langle\widehat Q((v,0)),(u,z,a)\big\rangle
   =v^{\mathsf T}(Bu+Kz+a_I)
   =v^{\mathsf T}B\pi(u,z,a).
\]
Moreover, $\pi$ fixes the active root directions. The
pullback of the active scattering diagram consequently has
exactly the initial normals
\eqref{eq:full-active-normals}, with the auxiliary components
included. Its wall Lie brackets are the same, since the
pairing of two active roots is $v^{\mathsf T}Bw$. Pullback
preserves consistency and outgoingness; uniqueness of
consistent completion identifies it with the quotient
diagram.

A broken line with active initial exponent has only active
exponents, so its retained and auxiliary positions $z_0,a_0$
are constant. Applying $\pi$ sends it to an active broken
line. The inverse lift is
\[
 \gamma(t)\longmapsto
 \bigl(\gamma(t)-B^{-1}(Kz_0+(a_0)_I),\,z_0,\,a_0\bigr).
\]
The normal identity preserves each crossing and its
coefficient, and identifies the negative chambers. A
strongly generic endpoint projects to one which is strongly
generic in the active position space, because $\pi$ is a
rational surjection. This proves the claimed equality of
theta expressions.

Only real positions have been changed. Roots, labels and
character exponents remain in the original integral active
lattice. Since the retained components of all labels are
zero, their distinctness is preserved by the descent.
\end{proof}

Apply Lemma~\ref{lem:descent} to
\eqref{eq:full-expansion}. Lemma~\ref{lem:carrier} makes
$F_\lambda$ an active rational function, and
Lemma~\ref{lem:finite-expansion} gives the required full
regularity. Thus every summand in the terminal chart is
an active theta Laurent polynomial.

\subsection{The vertex lattice}

We now use the integral identification
\[
 \iota:M_I\xrightarrow{\ \sim\ }L_0
\]
of \eqref{eq:active-embedding}. Recall that
$L$ has basis
$D_1,B_1,\ldots,B_{n-1},D_n$, that every vertex has weight
one, and that
\[
 L_0=\ker(\wt),\qquad
 \delta=\sum_{i=1}^nD_i,\qquad \wt(\delta)=n.
\]
The forward gaps $c_1,\ldots,c_{2n-2}$ form an integral
basis of $L_0$. Together with the frozen vector $D_1$ they
form an integral basis of $L$. We use the vertex seed with
these mutable roots and with frozen vector $D_1$.

By \eqref{eq:vertex-radical}, $\delta$ is in the radical
of the vertex form. The active and vertex scattering
diagrams compare by the real projection
\begin{equation}\label{eq:radical-position-projection}
 \rho:L_{\R}\longrightarrow(L_0)_{\R},\qquad
 \rho(z)=z-\frac{\wt(z)}n\delta.
\end{equation}
Indeed, $\rho$ fixes all root directions $c_j$, and for
every such root
\[
 \Omega(c_j,\rho(z))=\Omega(c_j,z).
\]
The initial walls, their graded wall groups, and their
outgoing completions therefore agree under pullback. For
a weight-zero label, all exponents of a broken line have
weight zero. The entire path has constant position weight,
so \eqref{eq:radical-position-projection} is just a
translation of it in the radical direction. Translation
back by the prescribed multiple of $\delta$ gives the
inverse comparison and preserves all coefficients. Hence
the active theta function of a label $\nu\in L_0$ equals
the vertex theta function $\vartheta^L_\nu$.

The projection acts on real positions. The character lattice
remains $L_0$, and
$L_0+\Z\delta$ has index $n$ in $L$, since weight identifies
$L/(L_0+\Z\delta)$ with $\Z/n\Z$.

Radical translation gives the individual identity
\begin{equation}\label{eq:radical-shift}
 X^\delta\vartheta^L_\nu=\vartheta^L_{\nu+\delta}.
\end{equation}
A broken line $\Gamma(t)$ for $\nu$
is changed to $\Gamma(t)-t\delta$ for $\nu+\delta$.
Wall supports are invariant under translation by $\delta$,
and all their normal pairings with $\delta$ vanish.
Consequently the crossings, their times, and their
coefficients are unchanged. The terminal exponent increases
by $\delta$, proving \eqref{eq:radical-shift}; this is also
\cite[Proposition 3.18]{DM}. All theta expressions are
homogeneous for the weight grading, since every root has
weight zero.

Let $Y$ denote the spectral variables of the path, so
$e_k(Y)=\mathcal E_k(P)$ as in Section~\ref{sec:carrier}.
The homogenized identity
\eqref{eq:vertex-spectral} is
\[
 \iota(F_\lambda)=X^{-\delta}m_\lambda(Y).
\]
It incorporates the normalization $d_n=1$ of the terminal
matrix. Multiplying the descended finite expansion by
$X^\delta$ and using \eqref{eq:radical-shift} gives
\begin{equation}\label{eq:finite-vertex}
 m_\lambda(Y)=\sum_{s\in S}c_s\vartheta^L_{m_s},
 \qquad
 m_s=\iota(\nu_s)+\delta,\qquad \wt(m_s)=n.
\end{equation}
Here $\nu_s$ is identified with its active component, the
labels remain distinct, and each summand is a finite
Laurent polynomial in the present chart. The shift is made
only in the vertex lattice, where $\delta$ is radical.
The next section identifies all coefficients in this
finite expansion.
\section{Rotations and the spectral theta function}\label{sec:rotations}

We now identify the summands in the finite expansion
\eqref{eq:finite-vertex}. Throughout this section $n\geq2$ and
$\lambda\vdash n$. We use the vertex lattice and its form from
Section~\ref{sec:carrier}. For a graph $\Gamma$ on a vertex basis of $L$, write
\[
 \mathcal E_k(\Gamma)=\sum_{\substack{S\text{ independent in }\Gamma\\ |S|=k}}
 X^{\sum_{v\in S}v}.
\]
The notation $f(Y_\Gamma)$ means substitution of $\mathcal E_k(\Gamma)$ for
$e_k$ in a symmetric polynomial $f$ in $n$ variables. The rotations
below have one bridge on each level. They specialize the
interval-graph rotations and monomial-section identity of
\cite[\S\S5.1--5.2, Theorem~5.8]{OpenAI}. We give the finite
coefficient argument needed for the annular comparison.

\subsection{The local rotation}

Place $D_i$ at time $i$ and $B_i$ at time $i+\eta_i$, where the
phases $\eta_1,\ldots,\eta_{n-1}$ are distinct elements of $(0,1)$.
Join two vertices when the difference of their times has absolute
value less than one. The skew form is specified by
\begin{align}
 \Omega(D_i,D_j)&=0,
 &\Omega(B_i,D_j)&=
 \begin{cases}
  1,&j=i,\\[-2pt]
  -1,&j=i+1,\\[-2pt]
  0,&\text{otherwise},
 \end{cases}\notag\\
 \Omega(B_{i+1},B_i)&=\mathbf1_{\eta_i>\eta_{i+1}},
 &\Omega(B_j,B_i)&=0\quad(j>i+1).
 \label{eq:phase-form}
\end{align}
Here $\mathbf1_{\mathcal C}$ is $1$ when the condition $\mathcal C$
holds and $0$ otherwise. The reverse pairings follow by skew-symmetry. Initially take
$\eta_i=\theta_i$, where $0<\theta_1<\cdots<\theta_{n-1}<1$; the
graph is then the path of Lemma~\ref{lem:path}.

Suppose that $\eta_i$ is the smallest phase. Put
\begin{equation}\label{eq:rotation-roots}
 p_i=B_i-D_i,\qquad q_i=D_{i+1}-B_i,\qquad
 \delta_i=D_{i+1}-D_i=p_i+q_i.
\end{equation}
Replace this phase by a phase larger than all the others and replace
the corresponding vertex vector by
\begin{equation}\label{eq:bridge-rotation}
 B_i'=B_i+\delta_i.
\end{equation}
All other vertex vectors are unchanged. Only the relative order of
the phases matters.

\begin{lemma}\label{lem:rotation}
The operation \eqref{eq:bridge-rotation} is mutation at the actual
gap $p_i$, followed by an interchange of the two gap indices on
level $i$. If $\Gamma'$ is the resulting graph, then
\begin{equation}\label{eq:rotation-polynomial}
 \mathcal E_k(\Gamma)=\mathcal A_{p_i}\bigl(\mathcal E_k(\Gamma')\bigr),
 \qquad
 \mathcal A_p(X^a)=X^a(1+X^p)^{\Omega(p,a)}.
\end{equation}
Consequently the same identity holds with $\mathcal E_k$ replaced by
$f(Y_\Gamma)$ for any symmetric polynomial $f$. Individual theta
labels transform by
\begin{equation}\label{eq:rotation-label}
 T_p(m)=m+[-\Omega(p,m)]_+p.
\end{equation}
\end{lemma}

\begin{proof}
The smallest-phase condition and \eqref{eq:phase-form} give
\[
 \Omega(p_i,D_i)=1,\qquad
 \Omega(p_i,D_{i+1})=-1,\qquad
 \Omega(p_i,B_i)=1,
\]
and zero pairing with every other vertex. In the integral vertex
seed with mutable basis $(p_1,q_1,\ldots,p_{n-1},q_{n-1})$ and
frozen vector $D_1$, the nonzero entries in the column of $p_i$ are
\begin{equation}\label{eq:rotation-column}
 \Omega(q_i,p_i)=2,\quad
 \Omega(p_{i+1},p_i)=-1,\quad
 \Omega(q_{i-1},p_i)=-1,\quad
 \Omega(D_1,p_i)=-\mathbf1_{i=1},
\end{equation}
with nonexistent indices omitted. The basis mutation
$e\mapsto e+[\Omega(e,p_i)]_+p_i$ for $e\ne p_i$, and
$p_i\mapsto-p_i$, therefore changes the two gaps on this level to
$(-p_i,q_i+2p_i)$ and fixes the frozen vector. Interchanging these
two indices gives
\[
 p_i'=B_i'-D_i=2p_i+q_i,\qquad
 q_i'=D_{i+1}-B_i'=-p_i.
\]
The other gaps are unchanged. The only changed bridge pairings are
$\Omega(B_i,B_{i-1})=1$ becoming $0$, and
$\Omega(B_{i+1},B_i)=0$ becoming $1$. These are precisely the
changes obtained by moving the smallest phase to the largest.
Thus \eqref{eq:phase-form} holds in the new vertex basis. By the
full-seed retention in Lemma~\ref{lem:carrier}, this mutation is
also a mutation of the full annular seed; all retained coordinates
and mixed exchange entries undergo that full mutation.

To prove the polynomial identity, fix an independent set $S$ outside
$\{D_i,B_i,D_{i+1}\}$. Set
\[
 a=\mathbf1_{B_{i-1}\notin S},\qquad
 b=\mathbf1_{B_{i+1}\notin S},
\]
taking the corresponding value to be $1$ at a boundary. Use
$x=X^{D_i}$, $y=X^{B_i}$, $z=X^{D_{i+1}}$, and
$y'=X^{B_i'}=yz/x$. If $u$ records cardinality, the possible
extensions of $S$ in the old and new triples contribute, respectively,
\begin{align*}
 &1+u\{a(x+y)+bz\}+u^2abxz,\\
 &1+u\{ax+b(z+y')\}+u^2abxz.
\end{align*}
Indeed, the old bridge has the same outside neighbors as $D_i$,
and the new bridge has the same outside neighbors as $D_{i+1}$.
Both anchors can be chosen together. The automorphism
$\mathcal A_{p_i}$ fixes all outside monomials, and
\[
 \mathcal A_{p_i}(x)=x+y,\qquad
 \mathcal A_{p_i}(z+y')=z,\qquad
 \mathcal A_{p_i}(xz)=xz.
\]
It takes the second extension polynomial to the first. Multiplying
by $X^{\sum_{v\in S}v}$ and summing over $S$ proves
\eqref{eq:rotation-polynomial}. Since the elementary symmetric
functions generate the symmetric polynomial ring, it proves the
assertion for $f$ as well. Formula~\eqref{eq:rotation-label} is the
individual $X$-theta mutation formula in the negative-chamber
normalization of Section~\ref{sec:theta}; see
\cite[Appendix~A.3]{DM}.
\end{proof}

We apply these rotations in the repeated order
$1,2,\ldots,n-1$. This is the successive smallest-phase order.
One such sequence is called a period. After a period the phase
order is the original one, so the original numerical phases may
be used again. After $k$ periods the vertex vectors are
\begin{equation}\label{eq:period-vertices}
 D_i,\qquad B_i^{(k)}=B_i+k\delta_i.
\end{equation}
Every intermediate graph contains the anchor--bridge path and
has the anchors as its unique independent set of cardinality $n$.
In particular,
\begin{equation}\label{eq:rotated-determinant}
 \mathcal E_n(\Gamma)=X^\delta,\qquad \delta=D_1+\cdots+D_n,
\end{equation}
throughout the sequence.

\subsection{The labels are pure and dominant}

Each summand in \eqref{eq:finite-vertex} comes from a regular theta
function of the full annular variety. After an active mutation its
rational expression still involves only the active coordinates:
their transformation is autonomous. Its full expression is
Laurent by regularity, so the Laurent-ring intersection used in
Lemma~\ref{lem:descent} makes it a finite active Laurent polynomial
in the new chart. An active basis change has an integral matrix
$U$ with $\det U=\pm1$, and hence
$\det(U^{\mathsf T}BU)=\det B=n^2$.
Apply that lemma anew in this full seed. It
identifies the summand with the active theta having its transported
label and retains the full mixed normals in making the comparison.

Index the charts by the number $t\geq0$ of individual rotations,
with $\Gamma_0=P$. Write $f_s^{(t)}$ for these full theta expressions in a rotated
chart, and $\iota_t$ for its integral active-to-vertex embedding.
The individual central shift gives finite Laurent polynomials
\[
 \Psi_s^{(t)}=X^\delta\iota_t(f_s^{(t)}),
 \qquad
 P_t=m_\lambda(Y_{\Gamma_t})=\sum_s c_s\Psi_s^{(t)}.
\]
The second equality follows from Lemma~\ref{lem:rotation} and
\eqref{eq:rotated-determinant}. For a real covector on $L$, extend
its pullback along $\iota_t$ to the full seed lattice. Applying
Lemma~\ref{lem:newton} there and then making the common affine
shift by $\delta$ gives
\begin{equation}\label{eq:rotated-newton}
 \Newt(P_t)=\conv\bigcup_s\Newt\bigl(\Psi_s^{(t)}\bigr).
\end{equation}
Each $\mathcal E_k(\Gamma_t)$ is an ordinary homogeneous vertex polynomial
of degree $k$. Hence $P_t$ is an ordinary homogeneous vertex
polynomial of degree $n$. Its support is contained in the
simplex
\[
 \Delta_n=\left\{\sum_v a_vv:\ a_v\geq0,
                  \ \sum_v a_v=n\right\}.
\]
Equation~\eqref{eq:rotated-newton} puts the support of each
$\Psi_s^{(t)}$ in this simplex. In particular, its pointed label
has nonnegative integral vertex coordinates in every rotated chart.

\begin{lemma}\label{lem:dominant}
Every label in \eqref{eq:finite-vertex} is of the form
\begin{equation}\label{eq:pure-dominant}
 m=\sum_{i=1}^n\mu_iD_i,
 \qquad \mu_1\geq\cdots\geq\mu_n\geq0,
 \qquad \sum_i\mu_i=n.
\end{equation}
Such a label is fixed by every subsequent rotation.
\end{lemma}

\begin{proof}
Write a current label as
$m=\sum_j\nu_jD_j+\sum_j\kappa_jB_j$. At a level-$i$ rotation put
$\epsilon=[-\Omega(p_i,m)]_+$. Reexpressing
\eqref{eq:rotation-label} in the basis with
$B_i'=B_i+D_{i+1}-D_i$ gives
\begin{equation}\label{eq:label-coordinates}
 \kappa_i'=\kappa_i+\epsilon,\qquad
 \nu_i'=\nu_i+\kappa_i,\qquad
 \nu_{i+1}'=\nu_{i+1}-\kappa_i'.
\end{equation}
All other coordinates stay fixed. Thus each bridge coefficient is
nondecreasing. Define
\[
 \tau_i=n-\sum_{j\leq i}(\nu_j+\kappa_j).
\]
The simplex inclusion gives $0\leq\tau_i\leq n$. An event at
another level leaves $\tau_i$ unchanged, while an event at level
$i$ decreases it by $\kappa_i'$. After $n+1$ periods,
\[
 0\leq\tau_i^{\mathrm{end}}
   =\tau_i^{\mathrm{start}}-
       \sum_{a=1}^{n+1}\kappa_i^{(a)}
   \leq n-(n+1)\kappa_i^{(1)},
\]
where $\kappa_i^{(a)}$ is the coefficient after the $a$th event
at that level. Nonnegativity and integrality imply
$\kappa_i^{(1)}=0$. Its initial value and its first increment
are therefore both zero. This holds at every level. Consequently
all events in the first period fixed the original label, which
was already pure. Their inequalities $\epsilon=0$ say
$\mu_i-\mu_{i+1}\geq0$, proving \eqref{eq:pure-dominant}.
Finally,
\[
 \Omega(p_i+a\delta_i,m)=\mu_i-\mu_{i+1}\geq0
 \qquad(a\geq0),
\]
so \eqref{eq:rotation-label} fixes $m$ at every later event.
\end{proof}

\subsection{A finite coefficient extraction}

We next recover the scalar coefficient of each pure label. The
argument concerns only finitely many root degrees. In the original
vertex seed put
\[
 c_{2i-1}=p_i,\qquad c_{2i}=q_i,\qquad
 C^+=\sum_{j=1}^{2n-2}\N c_j.
\]
If $m$ is as in
\eqref{eq:pure-dominant}, then
\begin{equation}\label{eq:target-box}
 h=m-nD_1=\sum_{j=1}^{2n-2}h_jc_j
   =\sum_{i=1}^{n-1}
       \left(\sum_{a>i}\mu_a\right)(c_{2i-1}+c_{2i}),
 \qquad 0\leq h_j\leq n.
\end{equation}
Every pure nonnegative label $\nu$ of weight $n$ also belongs to
$nD_1+C^+$. A broken line from $\nu$ with final exponent $m$
has total increment $m-\nu\in C^+$. Each of its nonzero bend
increments therefore belongs to the finite set
\begin{equation}\label{eq:relevant-roots}
 R_m=\left\{\sum_{j=1}^{2n-2}a_jc_j:\ a_j\in\N,\ a_j\leq h_j,
                                      \ \sum_j a_j>0\right\}.
\end{equation}

For wall positions it is convenient to use covectors on $L$.
Set $V(u)=\Omega(\,\cdot\,,u)$. A wall of degree $r$ is contained
in $r^\perp$, and a broken-line segment carrying exponent $m$
has covector velocity $-V(m)$. Since the radical is $\R\delta$
and the form on $(L_0)_\R$ is nondegenerate, $V$ maps $L_\R$
onto $\delta^\perp$ and restricts to an isomorphism
$(L_0)_\R\to\delta^\perp$. Wall data are constant along the
radical in the exponent-position realization. For $Q=V(u)$, we
write $\vartheta^L_{\nu,Q}$ for the section defined using any lift
$u$. Two lifts differ by a real multiple of $\delta$; translating
each broken line by that difference preserves all coefficients.

Put $\ell=n+1$ and retain the initial phases
$0<\theta_1<\cdots<\theta_{n-1}<1$. Define a covector by
\begin{equation}\label{eq:far-endpoint}
 Q_\ell(D_i)=\frac{n+1}{2}-i,\qquad Q_\ell(p_i)=\ell-\theta_i.
\end{equation}
Then $Q_\ell(\delta)=0$ and $Q_\ell(\delta_i)=-1$.
The final gaps from \eqref{eq:period-vertices} satisfy
\begin{equation}\label{eq:final-negative}
 Q_\ell(p_i+\ell\delta_i)=-\theta_i<0,\qquad
 Q_\ell(q_i-\ell\delta_i)=-1+\theta_i<0.
\end{equation}
Thus this covector lies in the final negative chamber. Every
earlier cut has the form $p_i+a\delta_i$, $0\leq a<\ell$, and
\begin{equation}\label{eq:identity-ray}
 \begin{split}
 Q_\ell(p_i+a\delta_i)&=\ell-\theta_i-a>0,\\
 \Omega(p_i+a\delta_i,m)&=\mu_i-\mu_{i+1}\geq0.
 \end{split}
\end{equation}
Consequently the whole ray $Q_\ell+sV(m)$, $s\geq0$, lies on the
identity side of every geometric mutation. Indeed, the geometric
$\mathcal X$-map on positions is
$u\mapsto u+[\Omega(u,p)]_+p$
\cite[Appendix~A]{DM}. Since $\Omega(u,p)=-Q(p)$, it is the
identity when $Q(p)>0$, and the same applies to the whole lifted
ray $u+s m$.

\begin{lemma}\label{lem:delta}
There is a generic endpoint $Q$ in the final negative chamber,
arbitrarily close to $Q_\ell$, for which
\begin{equation}\label{eq:coefficient-delta}
 [X^m]\vartheta^L_{\nu,Q}=\delta_{\nu,m}
\end{equation}
for every pair of pure dominant labels $m,\nu$ of weight $n$.
\end{lemma}

\begin{proof}
We first work with the displayed covector $Q_\ell$. Suppose that a
wall with a relevant nonzero degree $r\in R_m$ could meet the
ray $Q_\ell+sV(m)$. In the mutation of scattering diagrams,
\cite[Proposition~A.1 and Lemma~A.2]{DM} split each non-cut wall
into its two half-space pieces, transport each piece and its wall
function by the corresponding linear map, and place each
transported wall function in the new positive-root group.
We apply this construction to the descended active diagram and
pull it back along \eqref{eq:radical-position-projection}.
Choose the representatives recursively by this half-space
construction along the rotation sequence, as in
\cite[Construction~1.30]{GHKK}. Thus every transported wall
function belongs to the positive-root group of the current seed.
By \eqref{eq:identity-ray}, every local map near the hypothetical
intersection is the identity, and no mutation cut meets the ray.
The actual wall piece and its nonzero degree $r$ are therefore
unchanged there through the sequence. The same degree must belong
to the final positive gap cone
\[
 C_\ell^+=\sum_{i=1}^{n-1}
       \bigl(\N(p_i+\ell\delta_i)+\N(q_i-\ell\delta_i)\bigr).
\]
This statement applies to each transported wall piece. A bend
increment that is a positive multiple of its wall direction also
belongs to $C_\ell^+$; we use that actual increment as $r$.

Write $r=\sum_j a_jc_j\in R_m$ and put
$\kappa_i(r)=a_{2i-1}-a_{2i}$. In the original ordered vertex
basis the prefix sum of the coefficients of $r$ through $B_i$
is $-a_{2i}$. Expressing $r$ in the final vertex basis
\eqref{eq:period-vertices} changes this prefix sum to
\[
 -a_{2i}+\ell\kappa_i(r).
\]
Every proper vertex prefix of a vector in a forward-gap cone is
nonpositive: if the vector is
$\sum_j b_j(v_{j+1}-v_j)$ with $b_j\geq0$, its prefix through
$v_j$ is $-b_j$. Thus $r\in C_\ell^+$ gives
\begin{equation}\label{eq:cone-prefix}
 -a_{2i}+\ell\kappa_i(r)\leq0.
\end{equation}
Since $a_{2i}\leq n<\ell$ and $\kappa_i(r)$ is integral, it follows
that $\kappa_i(r)\leq0$ for every $i$. We obtain
\begin{align}
 Q_\ell(r)&=\sum_i(\ell-\theta_i)\kappa_i(r)-\sum_i a_{2i}<0,
 \label{eq:root-endpoint-negative}\\
 \Omega(r,m)&=\sum_i\kappa_i(r)(\mu_i-\mu_{i+1})\leq0.
 \label{eq:root-slope-negative}
\end{align}
The first inequality is strict: either some $\kappa_i(r)<0$,
or all are zero and the nonzero vector $r$ has some $a_{2i}>0$.
These two inequalities imply
$(Q_\ell+sV(m))(r)<0$ for every $s\geq0$. The supporting hyperplane
$r^\perp$ cannot meet the ray, a contradiction.

We can make the endpoint generic while preserving this argument.
There are finitely many earlier cuts, finitely many final gaps,
and finitely many nonzero lattice vectors $r=\sum_j a_jc_j$ with
$0\leq a_j\leq n$. Preserve the
strict inequalities in \eqref{eq:final-negative}, the first line
of \eqref{eq:identity-ray}, and
\eqref{eq:root-endpoint-negative} for those vectors belonging to
$C_\ell^+$. They hold on a neighborhood of $Q_\ell$ in
$\delta^\perp$. Choose in its inverse image under $V$ an
exponent-position endpoint outside every proper rational linear
hyperplane, and let $Q$ be its image. Such endpoints exist by
surjectivity of $V$. This perturbation changes only the real endpoint;
all labels remain in $L$. Inequality
\eqref{eq:root-slope-negative} is independent of the endpoint.
The same exclusion of relevant walls consequently holds for
$Q+sV(m)$, simultaneously for every pure dominant target of
weight $n$.

Any broken line with final exponent $m$ has its backward terminal
segment on this ray. A last nontrivial bend would have an increment
in $R_m$ and would lie on one of the excluded walls. There is
therefore no nontrivial bend. The straight line has coefficient
one and initial label $m$, which proves
\eqref{eq:coefficient-delta}. If $m$ is radical, the projected
ray is a point and the same inequalities still exclude every
possible last bend. Equivalently, a bend of degree $r$ into $m$
would have input pairing
$\Omega(r,m-r)=\Omega(r,m)=0$, so its wall fixes that monomial.
\end{proof}

\subsection{Identification of the spectral function}

\begin{lemma}\label{lem:spectral-theta}
In the terminal full annular seed, the function
$F_\lambda$ of \eqref{eq:spectral-F} is the regular theta function
\begin{equation}\label{eq:spectral-theta}
 F_\lambda=
 \vartheta^{\mathrm{full}}_{\mu_\lambda},\qquad
 \mu_\lambda=\sum_{i=1}^n(\lambda_i-1)D_i.
\end{equation}
Here $\lambda$ is padded by zeros, and
$\mu_\lambda\in L_0$ is interpreted through the actual active
gap lattice, with zero component on every retained vertex.
\end{lemma}

\begin{proof}
By Lemma~\ref{lem:dominant}, the finite vertex expansion consists
of pure dominant labels of weight $n$. Denote the scalar
coefficient of a label $m$ by $c_m$, setting it to zero if the
label is absent. Choose the endpoint $Q$ of Lemma~\ref{lem:delta}.
Write $\vartheta^{L,0}_{\nu,Q}$ and
$\vartheta^{L,\mathrm{fin}}_{\nu,Q}$ for the sections in the original
and final scattering diagrams, respectively. The geometric
mutation maps fix a lift of $Q$ and are the identity near it by
\eqref{eq:identity-ray}, so their induced linear maps on attached monomials are also
the identity. The label maps fix every pure dominant label by
Lemma~\ref{lem:dominant}. Individual theta mutation compatibility
\cite[Proposition~A.3, equation~(108), and Appendix~A.3]{DM}
therefore gives
\[
 \vartheta^{L,0}_{\nu,Q}=\vartheta^{L,\mathrm{fin}}_{\nu,Q}.
\]
The right side is the final negative-chamber expression.
There are finitely many summands. Ordinary chart
compatibility and Lemma~\ref{lem:rotation} therefore identify
their sum with the finite polynomial
$m_\lambda(Y_{\Gamma_{\mathrm{fin}}})$ in the final vertex variables. It follows
from \eqref{eq:coefficient-delta} that
\[
 c_m=[X^m]m_\lambda(Y_{\Gamma_{\mathrm{fin}}}).
\]

The final vertices form an integral basis of $L$, since
\eqref{eq:period-vertices} adds only integral combinations of
anchors to bridges. A monomial with a positive bridge coordinate
cannot equal the pure-anchor monomial $X^m$. We may consequently
set all final bridge variables to zero when computing this
coefficient. The remaining graph consists of the independent
anchors, so, for $m=\sum_i\mu_iD_i$,
\begin{equation}\label{eq:scalar-coefficient}
 c_m=[z_1^{\mu_1}\cdots z_n^{\mu_n}]
               m_\lambda(z_1,\ldots,z_n).
\end{equation}
Among weakly decreasing nonnegative tuples $\mu$ of sum $n$,
this coefficient is one precisely when $\mu=\lambda$, and is
zero otherwise. Thus the finite vertex expansion is the single
theta function $\vartheta^L_{\sum_i\lambda_iD_i}$.
Undoing the shift by the radical vector $\delta$ and using the
individual descent of Lemma~\ref{lem:descent} gives
\eqref{eq:spectral-theta}. Since it is one of the original full
theta summands, it is regular on the full annular cluster variety.
\end{proof}
\section{Proof of the immanant inequality}\label{sec:finish}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
For $n=1$ the assertion is $a_{11}\geq0$. Suppose $n\geq2$ and first
assume that every minor of $A$ is strictly positive. The prescribed
double reduced word in Section~\ref{sec:carrier} gives a factorization
\[
 A=U(\mathbf u)\operatorname{diag}(t_1,\ldots,t_n)
       L(\boldsymbol\ell)
\]
with every parameter positive \cite{FZ,FZBruhat}. Consequently all the
annular source coordinates in \eqref{eq:source-gaps} are positive.
By Lemma~\ref{lem:spectral-theta} and theta positivity,
$F_\lambda$ is a finite Laurent polynomial with nonnegative integral
coefficients in these coordinates.

For $Z=\operatorname{diag}(z_1,\ldots,z_n)$ we have
\[
 ZA=(ZUZ^{-1})(Z\operatorname{diag}(t_1,\ldots,t_n))L.
\]
Every upper parameter of root $i$ is multiplied by $z_i/z_{i+1}$,
each $t_i$ by $z_i$, and every lower parameter is unchanged.
Substituting in \eqref{eq:source-gaps} shows that all internal gaps
are unchanged, whereas the outer gap satisfies
\begin{equation}\label{eq:outer-row-weight}
 o_i\longmapsto (z_i/z_{i+1})o_i\qquad(1\leq i<n).
\end{equation}
The characters $z_i/z_{i+1}$, $1\leq i<n$, are multiplicatively
independent. Hence a source Laurent monomial has zero $z$-weight if and
only if its exponent on every $o_i$ is zero. Combining
\eqref{eq:row-constant-term} and \eqref{eq:outer-row-weight}, we obtain
\begin{equation}\label{eq:immanant-ct}
 \Imm_{\phi^\lambda}(A)
   =\det(A)\,\CT_{o_1,\ldots,o_{n-1}}F_\lambda.
\end{equation}
The constant term on the right is a Laurent polynomial with
nonnegative coefficients in the internal source coordinates. Its
value is nonnegative because those coordinates are positive; also
$\det(A)>0$. This proves the desired inequality for strictly
totally positive $A$.

The strictly totally positive real matrices are dense in the
totally nonnegative matrices, including the singular ones
\cite[Theorem~11 and the paragraph following it]{FZ}. The immanant
is a polynomial in the entries of $A$. Passing to the limit proves
the inequality for every totally nonnegative $A$.
\end{proof}
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\end{document}