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\hypersetup{pdftitle={Laurent and g-vector counterexamples for quivers with homotopies},
pdfsubject={Laurent phenomenon and g-vectors for quivers with homotopies}}
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\title{Laurent and $g$-vector counterexamples for quivers with homotopies}
\author{Qiyue Tang}
\date{29 October 2026}

\begin{document}
\maketitle
\begingroup
\renewcommand\thefootnote{}
\footnotetext{\textit{2020 Mathematics Subject Classification.} Primary 13F60; Secondary 05C20.
\newline\textit{Keywords.} Quiver mutation, homotopy, Laurent phenomenon, $g$-vector.}
\endgroup
\begin{abstract}
We give negative answers to the general Laurent-phenomenon question and
the $g$-vector sign-coherence question for quivers with homotopies posed
by Li, Liu, Mou and Pan.
\end{abstract}

\begingroup
\color{blue}
\section*{AI use and scholarly credit}
The counterexamples and their proofs were obtained in one round of
ChatGPT-6-Pro. Codex reviewed the arguments and polished the manuscript.
The results passed review by Rethlas-OSV~\cite{RethlasOSV,OSVerify}.

By posting this manuscript on this platform, I do not seek to assert
ownership of the results or claim priority. I welcome anyone interested
in preparing a finished account or checking the arguments; the eventual
scholarly credit should go to those who undertake that work.
\par
\endgroup

\section{Introduction}

Fomin and Zelevinsky introduced cluster algebras as an algebraic framework
for dual canonical bases and total positivity. Their Laurent theorem
states that every cluster variable is a Laurent polynomial in the variables
of any chosen cluster, with coefficients in the coefficient group ring
\cite[Theorem~3.1]{FZI}. Thus repeated division in the exchange relations
produces only monomial denominators in the initial variables. Their
theory of principal coefficients subsequently supplied a natural grading
and the associated $g$-vectors \cite{FZ}. The sign-coherence conjecture
\cite[Definition~6.12 and Conjecture~6.13]{FZ} says that, within one
cluster, each coordinate of these vectors has a common weak sign.
Derksen, Weyman and Zelevinsky proved this for skew-symmetric exchange
matrices using representations of quivers with potentials
\cite[Theorem~1.7]{DWZ}; Gross, Hacking, Keel and Kontsevich established
it in the skew-symmetrizable setting by scattering-diagram methods
\cite[Theorem~5.11]{GHKK}.

Extensions of cluster mutation have developed along several directions.
Huang and Li constructed infinite-rank unfoldings of acyclic
sign-skew-symmetric exchange matrices, obtaining positivity and the
constant-term property for their $F$-polynomials \cite{HL}.
Here an unfolding realizes matrix mutation by simultaneous mutations
in suitable vertex orbits of a larger quiver. Li and Pan later used
Newton polytopes to recover cluster-variable Laurent expressions from
$g$-vectors in totally sign-skew-symmetric cluster algebras with principal coefficients
\cite[Theorem~4.1]{LiPan}. In another direction, Mou constructed
scattering diagrams for Chekhov--Shapiro generalized cluster algebras
with binomially factorized exchange polynomials, identifying their
cluster variables with theta functions and obtaining positivity in the
factor coefficients \cite{Mou}. These results depend on the specified
matrix and exchange-polynomial mutation rules.

Surfaces and group actions provide a second source of extensions.
Fomin, Shapiro and Thurston described cluster complexes of marked
bordered surfaces by tagged triangulations \cite{FST}.
Paquette and Schiffler studied group actions on cluster algebras and
cluster categories, constructing generalized orbit algebras and
establishing Laurentness in their mutation-connected setting
\cite[Section~7]{PS}. Kaufman's special foldings allow folded
quivers beyond those encoded by skew-symmetrizable matrices and
include examples from punctured surfaces with self-folded triangles
\cite{Kaufman}. The stability of the vertex orbits under repeated
mutation is part of the folding condition.

Li, Liu, Mou and Pan introduced a mutation theory for loop-free quivers
with oriented $2$-cycles \cite{LLMP}. A homotopy is a normal subgroupoid
of the fundamental groupoid; it specifies which $2$-cycles are deleted.
Their mutation is involutive and extends ordinary quiver mutation
\cite[Propositions~3.18 and~3.21]{LLMP}. For marked surfaces with colored
punctures, their construction intertwines flips and homotopy mutations
\cite[Theorem~4.23]{LLMP}. We use their version~3, dated 13 October 2026.
Theorem~5.8 establishes Laurentness and positivity when the associated
cover remains weakly admissible under all orbit mutations. This includes
homotopies containing the squares of all closed walks. Problem~6.1 asks
for Laurentness for every algebra in the general theory; the authors
report computational evidence for an affirmative answer. Proposition~5.9
defines $g$-vectors through homogeneous rational functions, and the
following paragraph asks whether their sign coherence also persists.

Retaining opposite arrows changes the exchange numerators. A vertex
joined to another in both directions contributes a common variable factor
to the two exchange monomials. For comparison, the exchange polynomials
in Lam--Pylyavskyy Laurent phenomenon algebras are irreducible and are
not divisible by cluster variables \cite[Section~2.1 and Theorem~5.1]{LP}.
In the homotopy setting, both the retained multiplicities and the
subsequent homotopy-dependent deletions enter the rational recurrence.

We give negative answers to both questions. A three-vertex quiver with
six arrows produces a coefficient-free variable with an uncancelled
factor $1+x_1$ in its denominator after six mutations, where $x_1$ is
the initial variable at vertex~$1$. A four-arrow quiver produces a
mixed-sign row in its $g$-matrix after seven principal-coefficient
mutations. The quotient vertex groups are infinite cyclic, so integer
arrow labels determine every homotopy membership test. We give the
presentations, all composite arrows and deletions, and the resulting
exchange calculations. The six-arrow data also give a $g$-vector
counterexample with nonzero initial skew-symmetric difference in
Appendix~\ref{app:nonzero}. In both examples the canonical cover acquires
arrows within a deck orbit at the first mutation
(Section~\ref{sec:cover}).

\section{Homotopies and exchange relations}\label{sec:model}

Let $Q$ be a finite loop-free quiver. Regard its underlying graph as a
one-dimensional cell complex.
We use vertices $1,\ldots,n$, where $n$ is a positive integer.
Its fundamental groupoid $\pi(Q)$ has the vertices as objects and walks
modulo cancellation of an edge with its inverse as morphisms. Path
products are composed from right to left. Write $\pi_1(Q,i)$ for its
group of closed walks at vertex $i$. A directed $2$-cycle consists of two arrows $i\to j\to i$ with
$i\ne j$. A homotopy is a normal
subgroupoid containing all identity morphisms and only endomorphisms,
with a normal subgroup $H(i,i)$ at each vertex and
$pH(i,i)p^{-1}=H(j,j)$ for every walk $p:i\to j$.
It is reduced if it contains no directed $2$-cycle of $Q$.
The quotient groupoid $\pi(Q)/H$ has the same objects; walks $p,q:i\to j$
have the same image if $q^{-1}p\in H(i,i)$. Its group of endomorphisms
at $i$ is the quotient vertex group $\pi_1(Q,i)/H(i,i)$.
For closed walks based at vertex~$1$, the notation $\nc{\cdots}$ means
the smallest normal subgroupoid containing them, including their
conjugates at all vertices.

Put $p_{ij}=|\{\text{arrows }i\to j\}|$ and $P=(p_{ij})$.
Parallel arrows are counted separately even when their images modulo
$H$ agree. Set $B=P-P^{\mathsf T}$, where the superscript denotes
transpose. Mutation sequences are written in chronological order,
from left to right. A subscript $t$ records the number of mutations
already performed, and $P_t$ is the adjacency matrix after step $t$.

For the three-vertex examples, let $\mathcal Z$ be the groupoid with objects $1,2,3$ and
\[
\Hom_{\mathcal Z}(i,j)=\{(i,j;r):r\in\Z\},\qquad
(j,k;s)(i,j;r)=(i,k;r+s).
\]
An integer assignment to arrows extends uniquely to a functor
$\lambda:\pi(Q)\to\mathcal Z$. With the endpoints understood, we also
write $\lambda(p)$ for the integer component of the image of a walk $p$.
Inverses have opposite labels.
Its kernel consists of closed walks of label zero. In particular, a
zero-label arrow between distinct vertices is not an identity morphism.


\begin{lemma}\label{lem:labels}
Let $Q$ be loop-free on vertices $1,2,3$, and let $H=\ker\lambda$ be reduced. Mutation at
$k$ is computed exactly as follows.
\begin{enumerate}[label=\textup{(\arabic*)}]
\item For each two-arrow path $i\to k\to j$ with $i\ne j$, add an
arrow $i\to j$ whose label is the sum of the two labels.
\item Reverse each arrow incident with $k$ and negate its label.
\item For $i,j\ne k$, delete opposite arrows with labels $r$ and $-r$
in pairs, until no such pair remains.
\end{enumerate}
The resulting homotopy is the kernel of the label functor on the
remaining quiver. For fixed endpoints and labels, the remaining
multiplicities do not depend on the order of deletion.
\end{lemma}

\begin{proof}
Let $Q'$ be the quiver after (1)--(2). Define a functor
$\pi(Q')\to\pi(Q)/H$ by sending an unchanged arrow to its class,
a reversed arrow to the class of its inverse, and the composite arrow
for $i\xrightarrow{a}k\xrightarrow{b}j$ to the class of $ba$.
These assignments extend to walks by composition and inversion because
$\pi(Q')$ is a free groupoid. Its kernel is the premutated homotopy
of \cite[Definition~3.8]{LLMP}.

The label functor induces an injection on each morphism set of
$\pi(Q)/H$: two walks with the same endpoints and label differ by a
closed walk in $\ker\lambda=H$. Composing with this injection gives
the labels prescribed in (1)--(2), with the same kernel.
Thus a directed $2$-cycle belongs to the premutated homotopy precisely
when its labels sum to zero. Deleting all such pairs is the reduction
in \cite[Definition~3.9]{LLMP}. By Definition~3.10 the homotopy on the
remaining quiver is the kernel of the restricted functor, hence is
again the label kernel.

Opposite arrows incident
with $k$ have their old nonzero cycle label negated, and thus remain
reduced. After (3), every other opposite pair has nonzero label sum.
For fixed $i,j,r$,
only an $i\to j$ arrow of label $r$ and a $j\to i$ arrow of label $-r$
can cancel each other. Their number of cancellations is the minimum of
their two multiplicities, independently of their order.
\end{proof}

A coefficient semifield $\PP$ is a commutative multiplicative group with an auxiliary
addition $\oplus$ that is associative, commutative and distributive
over multiplication. Write $\Z\PP$ for its integral group ring.
The initial variables $x_1,\ldots,x_n$ are algebraically independent
over the fraction field of $\Z\PP$. A seed consists of a free generating
tuple $(X_1,\ldots,X_n)$ of this rational function field, a coefficient
tuple $(Y_1,\ldots,Y_n)\in\PP^n$, and a quiver with reduced homotopy.
Write $X_{i,t}$ and $Y_{i,t}$ for the variable and coefficient at vertex
$i$ after step $t$, with $X_{i,0}=x_i$. At a current seed we omit $t$;
a prime denotes the seed after one mutation. All products and sums
with a vertex index range from $1$ to $n$, and an empty product is $1$.
Mutation at $k$ leaves the other variables fixed and gives
\begin{align}
X'_k&=\frac{Y_k\prod_jX_j^{p_{jk}}+\prod_jX_j^{p_{kj}}}
{(1\oplus Y_k)X_k},\label{eq:exchange}\\
Y'_k&=Y_k^{-1},&
Y'_j&=Y_jY_k^{p_{kj}}(1\oplus Y_k)^{p_{jk}-p_{kj}}
\quad(j\ne k).\label{eq:y}
\end{align}
These are \cite[(5.1)--(5.2)]{LLMP}. For coefficient-free mutation we
use the trivial semifield $\PP=\{1\}$, so $1\oplus1=1$. For principal
coefficients we use $\Trop(y_1,\ldots,y_n)$, the free abelian
multiplicative group on $y_1,\ldots,y_n$, with
$\yt^c=\prod_{j=1}^n y_j^{c_j}$ for $c=(c_1,\ldots,c_n)^{\mathsf T}\in\Z^n$
and $\yt^a\oplus\yt^b=\yt^{\min(a,b)}$. Here $\yt=(y_1,\ldots,y_n)$
and minima are taken coordinatewise. In particular,
\begin{equation}\label{eq:tropical}
1\oplus\yt^c=\yt^{\min(c,0)}.
\end{equation}
The initial coefficients are $Y_j=y_j$, and the coefficient group ring
is $\Z[y_1^{\pm1},\ldots,y_n^{\pm1}]$. The ordinary addition in the
numerator of \eqref{eq:exchange} is distinct from $\oplus$.


The loop-free condition makes the numerator of \eqref{eq:exchange}
independent of $X_k$. Its two monomials have positive coefficients, so it is nonzero.
Equation~\eqref{eq:exchange} expresses $X_k$ as this same numerator
divided by $(1\oplus Y_k)X'_k$. Hence the old and new tuples generate
the same rational function field; its transcendence degree $n$ implies
that the new $n$ variables are algebraically independent. The algebra $\mathcal A(Q,H)$ is the
$\Z\PP$-subalgebra generated by all variables reached by finite
mutations from the initial seed. The Laurent property asks that all
these variables belong to $\Z\PP[x_1^{\pm1},\ldots,x_n^{\pm1}]$.

For principal coefficients, let $e_i$ be the $i$th standard basis
vector of $\Z^n$. Give the initial variables and coefficients degrees
\begin{equation}\label{eq:grading}
\deg x_i=e_i,\qquad
\deg y_j=(p_{j1}^{(0)}-p_{1j}^{(0)},\ldots,
p_{jn}^{(0)}-p_{nj}^{(0)})^{\mathsf T},
\end{equation}
where $p_{ij}^{(0)}$ is the $(i,j)$ entry of $P_0$.
Degrees add under multiplication. A polynomial is homogeneous when
all its nonzero monomials have the same degree; a quotient of nonzero
homogeneous polynomials has degree equal to the numerator degree minus
the denominator degree. This difference is independent of the
representation: equality of two fractions gives an equality of nonzero
homogeneous cross-products, whose degrees must agree. Proposition~5.9 of \cite{LLMP} gives homogeneity
of the cluster variables as rational functions. Their degrees
$g_{i,t}=\deg X_{i,t}$ are the $g$-vectors. Set $G_t=(g_{1,t}\ \cdots\ g_{n,t})$.
Coordinatewise sign coherence means that each row of $G_t$ is entirely
nonnegative or entirely nonpositive; zero entries are allowed.

\begin{theorem}\label{thm:main}
There exist loop-free quivers with reduced homotopies on three vertices,
whose quotient vertex groups are isomorphic to $\Z$, with the following
properties.
\begin{enumerate}[label=\textup{(\roman*)}]
\item A six-arrow example has a coefficient-free cluster variable outside
$\Q[x_1^{\pm1},x_2^{\pm1},x_3^{\pm1}]$ after the sequence
$(2,1,3,2,3,2)$.
\item A four-arrow example has, with principal coefficients, the cluster
\[
\left(\frac{x_3}{1+y_3},\ x_2,\ \frac{(1+y_3)x_1}{x_3}\right)
\]
after $(2,3,1,2,1,2,1)$. Its $g$-matrix, with $g$-vectors as columns, is
\[
G=\mt{0&0&1\\0&1&0\\1&0&-1}.
\]
\end{enumerate}
\end{theorem}

Part~(i) is proved in Section~\ref{sec:six} and part~(ii) in
Section~\ref{sec:four}.

\section{A non-Laurent variable over \texorpdfstring{$\Q$}{Q}}\label{sec:six}

\subsection{Initial quiver and homotopy}
Take the following six arrows.
\begin{center}
\begin{tabular}{c*{6}{c}}
\toprule
Arrow & $a$ & $b$ & $c$ & $d$ & $e$ & $f$\\
Direction & $1\to2$ & $2\to1$ & $2\to1$ & $1\to3$ & $1\to3$ & $3\to1$\\
Label & $1$ & $1$ & $0$ & $0$ & $0$ & $1$\\
\bottomrule
\end{tabular}
\end{center}
Choose the unoriented spanning tree consisting of $a,d$. Collapsing this
tree to vertex~$1$ leaves four independent circles, one for each remaining
edge. Traversing the tree to an edge and back gives the free generators
\[
u=ca,\qquad v=ba,\qquad w=d^{-1}e,\qquad z=fd
\]
of $\pi_1(Q,1)$. Define, in the entire fundamental groupoid,
\begin{equation}\label{eq:sixH}
H=\nc{vu^{-2},\ w,\ zu^{-1}}.
\end{equation}
Then
\begin{equation}\label{eq:sixgroup}
\pi_1(Q,1)/H(1,1)
=\langle u,v,w,z\mid v=u^2,\ w=1,\ z=u\rangle
\cong\langle u\rangle\cong\Z.
\end{equation}
Indeed, eliminating $v,w,z$ leaves the free group on $u$.
The labels give $(\lambda(u),\lambda(v),\lambda(w),\lambda(z))
=(1,2,0,1)$. Consequently \eqref{eq:sixgroup} identifies the quotient
map at vertex~$1$ with $\lambda$, proving equality of kernels there.
Conjugation by a connecting walk proves equality at the other vertices;
the connecting walk and its inverse have cancelling labels. Hence
\eqref{eq:sixH} is exactly $\ker\lambda$ on the whole groupoid.
Conjugating $u$ along a walk from $1$ to $i$ gives a loop of label $1$
at $i$. If $p:i\to j$ has label $r$, composing $p$ with powers of
this loop realizes every label $r+m$, $m\in\Z$, from $i$ to $j$.
Together with kernel equality, this proves $\pi(Q)/H\cong\mathcal Z$.

The directed $2$-cycles based at vertex~$1$ are $ca,ba,fd,fe$, of
labels $1,2,1,1$. Their versions at other basepoints have the same
nonzero labels. Thus $H$ is reduced. The relation $w=d^{-1}e\in H$ identifies the images of the parallel
arrows $d,e$ in the quotient groupoid. Both arrows remain in the quiver,
since deletion removes opposite arrows.

\subsection{Labelled mutations}
Use the sequence $(2,1,3,2,3,2)$. In Table~\ref{tab:sixlabels},
$L_{ij}^{(t)}$ is the multiset of labels on arrows $i\to j$ after
step~$t$. The column $k$ specifies the vertex mutated at step $t$;
the initial row has $t=0$. A repeated entry denotes a separate arrow,
and $\varnothing$ denotes no arrows.

\begin{table}[!ht]
\centering\small
\caption{The six-arrow example: all retained arrow labels.}\label{tab:sixlabels}
\begin{tabular}{cc*{6}{c}}
\toprule
$t$ & $k$ & $L_{12}^{(t)}$ & $L_{21}^{(t)}$ & $L_{13}^{(t)}$
& $L_{31}^{(t)}$ & $L_{23}^{(t)}$ & $L_{32}^{(t)}$\\
\midrule
0 & -- & $1$ & $1,0$ & $0,0$ & $1$ & $\varnothing$ & $\varnothing$\\
1 & 2 & $-1,0$ & $-1$ & $0,0$ & $1$ & $\varnothing$ & $\varnothing$\\
2 & 1 & $1$ & $1,0$ & $-1$ & $0,0$ & $-1$ & $0$\\
3 & 3 & $\varnothing$ & $0,-1$ & $0,0$ & $1$ & $0$ & $1$\\
4 & 2 & $0,1$ & $\varnothing$ & $0$ & $1,1$ & $-1$ & $0$\\
5 & 3 & $1$ & $\varnothing$ & $-1,-1$ & $0$ & $0$ & $1$\\
6 & 2 & $\varnothing$ & $-1$ & $-1,-1,1$ & $0$ & $-1$ & $0$\\
\bottomrule
\end{tabular}
\end{table}

Table~\ref{tab:sixdelete} lists every composite
arrow and every deletion. Here $ij:r$ means an arrow $i\to j$ with
label $r$. Two labels separated by a comma on one arrow type specify
two arrows. A pair in the last column specifies the two arrows removed.
All remaining changes are the reversals prescribed in Lemma~\ref{lem:labels}.

\begin{table}[!ht]
\centering\small
\caption{Composite arrows and deletions in the six mutations.}\label{tab:sixdelete}
\begin{tabular}{c@{\quad}l@{\qquad}l}
\toprule
Step & Composite arrows added & Pairs deleted\\
\midrule
1 & None & None\\
2 & $23:-1,-1;\quad32:0,1$ & $(23:-1,\ 32:1)$\\
3 & $12:-1;\quad21:-1,-1$ & $(12:1,\ 21:-1)$; $(12:-1,\ 21:1)$\\
4 & $31:1,0$ & $(13:0,\ 31:0)$\\
5 & $12:0;\quad21:0,0$ & $(12:0,\ 21:0)$, twice\\
6 & $13:1$ & None\\
\bottomrule
\end{tabular}
\end{table}

At step~1 all two-arrow paths through~$2$ have equal endpoints and
are omitted. At step~5 one deleted $12:0$ arrow was already present
and one was newly added. Each deleted pair has label sum zero, and
each retained opposite pair has nonzero label sum.

Counting Table~\ref{tab:sixlabels} gives the six matrices used before
the exchanges:
\begin{equation}\label{eq:sixP}
\begin{aligned}
P_0&=\mt{0&1&2\\2&0&0\\1&0&0},&
P_1&=\mt{0&2&2\\1&0&0\\1&0&0},&
P_2&=\mt{0&1&1\\2&0&1\\2&1&0},\\[4pt]
P_3&=\mt{0&0&2\\2&0&1\\1&1&0},&
P_4&=\mt{0&2&1\\0&0&1\\2&1&0},&
P_5&=\mt{0&1&2\\0&0&1\\1&1&0}.
\end{aligned}
\end{equation}

\subsection{Six exchanges and the uncancelled factor}
\begin{proof}[Proof of Theorem~\ref{thm:main}(i)]
Work over the trivial semifield with initial cluster $(x_1,x_2,x_3)$. Equations~\eqref{eq:exchange} and
\eqref{eq:sixP} give the six new cluster variables $U,V,W,Z,T,R$, in order,
\begin{equation}\label{eq:sixexchanges}
\begin{aligned}
U&=\frac{x_1(1+x_1)}{x_2},&
V&=\frac{Ux_3(1+Ux_3)}{x_1},&
W&=\frac{UV(1+V)}{x_3},\\[3pt]
Z&=\frac{W(1+V^2)}{U},&
T&=\frac{VZ(1+V)}{W},&
R&=\frac{T(1+V)}{Z}.
\end{aligned}
\end{equation}
The clusters are successively $(x_1,U,x_3)$, $(V,U,x_3)$, $(V,U,W)$,
$(V,Z,W)$, $(V,Z,T)$ and $(V,R,T)$. In particular, the fourth and
sixth exchange numerators are $W+V^2W$ and $VT+T$, respectively.
Cancelling the intermediate variables in the last three equations yields
\begin{align*}
R&=\frac{V(1+V)^2}{W}
=\frac{x_3(1+V)}{U}
=\frac{x_3}{U}+\frac{x_3^2}{x_1}+\frac{Ux_3^3}{x_1}\\
&=\frac{x_2x_3}{x_1(1+x_1)}+\frac{x_3^2}{x_1}+\frac{(1+x_1)x_3^3}{x_2}.
\end{align*}
Thus the sixth mutation produces
\begin{equation}\label{eq:strongR}
\boxed{R=
\frac{x_3\bigl[x_2^2+(1+x_1)x_2x_3+x_1(1+x_1)^2x_3^2\bigr]}{x_1(1+x_1)x_2}.}
\end{equation}

In the Laurent ring $\Q[x_1^{\pm1},x_2^{\pm1},x_3^{\pm1}]$ the element
$1+x_1$ is prime and not a unit: quotienting by it amounts to setting
$x_1=-1$, leaving $\Q[x_2^{\pm1},x_3^{\pm1}]$. The numerator of
\eqref{eq:strongR} maps to $x_3x_2^2\ne0$ in this quotient. Hence
the factor $1+x_1$ does not cancel and $R$ is not in the Laurent ring.
\end{proof}
\FloatBarrier

\section{A four-arrow counterexample to sign coherence}
\label{sec:four}

\subsection{A homotopy with one relation}
Now take
\[
a:1\to2,\qquad b:2\to1,\qquad c:1\to3,\qquad d:3\to1,
\]
with labels $1,0,0,-2$, respectively. Collapsing the spanning tree
consisting of $a,c$ leaves two independent circles, represented by
the closed walks $U=ba$ and $V=dc$ at vertex~$1$. These are free
generators of the vertex group; the letters $U,V$ in this section
denote these walks.
Define
\begin{equation}\label{eq:fourH}
H=\nc{U^2V}.
\end{equation}
Then
\[
\pi_1(Q,1)/H(1,1)
=\langle U,V\mid U^2V=1\rangle
\cong\langle U\rangle\cong\Z,
\qquad U\mapsto1,\quad V\mapsto-2.
\]
Eliminating $V=U^{-2}$ proves that the label functor has exactly
the kernel \eqref{eq:fourH}. As in the six-arrow example, conjugation
extends the kernel equality to every vertex, and $\lambda(U)=1$
makes the quotient groupoid isomorphic to $\mathcal Z$. Both directed
$2$-cycles have nonzero labels, $1$ and $-2$, so $H$ is reduced.

Use the sequence $(2,3,1,2,1,2,1)$. Its complete retained arrow data
are in Table~\ref{tab:fourlabels}.

\begin{table}[!ht]
\centering\small
\caption{The four-arrow example: all retained arrow labels.}\label{tab:fourlabels}
\begin{tabular}{cc*{6}{c}}
\toprule
$t$ & $k$ & $L_{12}^{(t)}$ & $L_{21}^{(t)}$ & $L_{13}^{(t)}$
& $L_{31}^{(t)}$ & $L_{23}^{(t)}$ & $L_{32}^{(t)}$\\
\midrule
0 & -- & $1$ & $0$ & $0$ & $-2$ & $\varnothing$ & $\varnothing$\\
1 & 2 & $0$ & $-1$ & $0$ & $-2$ & $\varnothing$ & $\varnothing$\\
2 & 3 & $0$ & $-1$ & $2$ & $0$ & $\varnothing$ & $\varnothing$\\
3 & 1 & $1$ & $0$ & $0$ & $-2$ & $1$ & $0$\\
4 & 2 & $0$ & $-1$ & $\varnothing$ & $\varnothing$ & $0$ & $-1$\\
5 & 1 & $1$ & $0$ & $\varnothing$ & $\varnothing$ & $0$ & $-1$\\
6 & 2 & $0$ & $-1$ & $\varnothing$ & $\varnothing$ & $1$ & $0$\\
7 & 1 & $1$ & $0$ & $\varnothing$ & $\varnothing$ & $1$ & $0$\\
\bottomrule
\end{tabular}
\end{table}

The first two steps only reverse arrows. Step~3 adds $23:1$ and
$32:0$, which do not cancel. Step~4 adds $13:2$ and $31:0$; these
cancel the old arrows $31:-2$ and $13:0$, respectively. Step~5 only
reverses arrows. Step~6 adds $13:1$ and $31:-1$, which cancel each
other. Step~7 again only reverses arrows. This lists every composite
and deletion, and every retained opposite pair has nonzero label sum.

These deletions can also be checked as words in the initial groupoid.
At step~4 the new $1\to3$ and $3\to1$ arrows have images $d^{-1}$
and $c^{-1}$, respectively, so their deletions with the old images
$d$ and $c$ already hold by free cancellation. At step~6 the two
new arrows have images $cU$ and $Ud$. Their directed cycle at
vertex~$1$ has image $(Ud)(cU)=UVU=U^{-1}(U^2V)U$, a conjugate of
the defining relator. This justifies the deletion directly by
\eqref{eq:fourH}.

Counting the arrows gives
\begin{equation}\label{eq:symmetricP}
\begin{gathered}
P_0=P_1=P_2=\mt{0&1&1\\1&0&0\\1&0&0},\qquad
P_3=\mt{0&1&1\\1&0&1\\1&1&0},\\
P_4=P_5=P_6=P_7=\mt{0&1&0\\1&0&1\\0&1&0}.
\end{gathered}
\end{equation}
Every $P_t$ in this sequence is symmetric.

\subsection{Principal coefficients and degrees}
\begin{proof}[Proof of Theorem~\ref{thm:main}(ii)]
Symmetry reduces \eqref{eq:exchange}--\eqref{eq:y} to
\begin{equation}\label{eq:symmetricexchange}
\begin{aligned}
Y'_k&=Y_k^{-1},& Y'_j&=Y_jY_k^{p_{kj}}\quad(j\ne k),\\
X'_k&=\frac{1+Y_k}{1\oplus Y_k}
\frac{\prod_jX_j^{p_{kj}}}{X_k}.
\end{aligned}
\end{equation}
The coefficient tuples obtained from the initial tuple
$(y_1,y_2,y_3)$ are listed in Table~\ref{tab:Y}.

\begin{table}[!ht]
\centering\small
\caption{Coefficient tuples after each mutation.}\label{tab:Y}
\begin{tabular}{ccl}
\toprule
$t$ & $k$ & $(Y_{1,t},Y_{2,t},Y_{3,t})$\\
\midrule
0 & -- & $(y_1,y_2,y_3)$\\
1 & 2 & $(y_1y_2,y_2^{-1},y_3)$\\
2 & 3 & $(y_1y_2y_3,y_2^{-1},y_3^{-1})$\\
3 & 1 & $((y_1y_2y_3)^{-1},y_1y_3,y_1y_2)$\\
4 & 2 & $(y_2^{-1},(y_1y_3)^{-1},y_1^2y_2y_3)$\\
5 & 1 & $(y_2,(y_1y_2y_3)^{-1},y_1^2y_2y_3)$\\
6 & 2 & $((y_1y_3)^{-1},y_1y_2y_3,y_1)$\\
7 & 1 & $(y_1y_3,y_2,y_1)$\\
\bottomrule
\end{tabular}
\end{table}

The factors $(1+Y_k)/(1\oplus Y_k)$ at the seven exchanges are,
in order,
\[
1+y_2,\quad 1+y_3,\quad 1+y_1y_2y_3,\quad
1+y_1y_3,\quad 1+y_2,\quad 1+y_1y_2y_3,\quad 1+y_1y_3.
\]
At the first four exchanges the current coefficient has nonnegative
exponents, so its tropical sum with $1$ is $1$. At the last three it
is the reciprocal of a monomial with nonnegative exponents. For
$c\in\Z_{\geq0}^n$, the identity
\[
\frac{1+\yt^{-c}}{1\oplus\yt^{-c}}
=\frac{1+\yt^{-c}}{\yt^{-c}}=1+\yt^c
\]
gives the remaining factors.
Using the adjacency matrices in \eqref{eq:symmetricP}, the seven
exchanges are
\begin{align}
X_{2,1}&=\frac{(1+y_2)x_1}{x_2},&
X_{3,2}&=\frac{(1+y_3)x_1}{x_3},\label{eq:firsttwo}\\
X_{1,3}&=\frac{(1+y_1y_2y_3)X_{2,1}X_{3,2}}{x_1}
\notag\\
&=\frac{(1+y_2)(1+y_3)(1+y_1y_2y_3)x_1}{x_2x_3},\label{eq:third}\\
X_{2,4}&=\frac{(1+y_1y_3)X_{1,3}X_{3,2}}{X_{2,1}}
\notag\\
&=\frac{(1+y_3)^2(1+y_1y_2y_3)(1+y_1y_3)x_1}{x_3^2},\label{eq:fourth}\\
X_{1,5}&=\frac{(1+y_2)X_{2,4}}{X_{1,3}}
=\frac{(1+y_3)(1+y_1y_3)x_2}{x_3},\label{eq:fifth}\\
X_{2,6}&=\frac{(1+y_1y_2y_3)X_{1,5}X_{3,2}}{X_{2,4}}=x_2,\label{eq:sixth}\\
X_{1,7}&=\frac{(1+y_1y_3)X_{2,6}}{X_{1,5}}
=\frac{x_3}{1+y_3}.\label{eq:seventh}
\end{align}
The variables at the other vertices remain unchanged at each step.
The resulting cluster is
\begin{equation}\label{eq:finalfour}
\left(\frac{x_3}{1+y_3},\ x_2,\ \frac{(1+y_3)x_1}{x_3}\right).
\end{equation}

The initial adjacency matrix is symmetric, so \eqref{eq:grading} gives
$\deg y_j=0$. The final $g$-vectors, in the order of
\eqref{eq:finalfour}, are $e_3,e_2,e_1-e_3$. Thus
\[
G_7=\mt{0&0&1\\0&1&0\\1&0&-1},
\]
whose third row has both a positive and a negative entry.
\end{proof}

The first variable of \eqref{eq:finalfour} also fails the Laurent
property over the principal coefficient group ring: $1+y_3$ is a
nonunit prime in $\Z[y_1^{\pm1},y_2^{\pm1},y_3^{\pm1}]
[x_1^{\pm1},x_2^{\pm1},x_3^{\pm1}]$, since setting $y_3=-1$
identifies its quotient with the nonzero integral domain
$\Z[y_1^{\pm1},y_2^{\pm1},x_1^{\pm1},x_2^{\pm1},x_3^{\pm1}]$.
It therefore cannot divide the unit $x_3$.
Over the trivial semifield the final cluster is
$(x_3/2,x_2,2x_1/x_3)$, which is Laurent over $\Q$ but not over $\Z$.

\FloatBarrier

\section{The recurrence for symmetric adjacency matrices}\label{sec:mechanism}

For any $n$ and $v=(v_1,\ldots,v_n)^{\mathsf T}\in\Z^n$, write
$\xt^v=\prod_{i=1}^n x_i^{v_i}$, where $\xt=(x_1,\ldots,x_n)$.
The symmetric case separates these variable exponents from the
coefficient factors.

\begin{proposition}\label{prop:symmetric}
For a quiver on $n$ vertices, if the initial adjacency matrix is symmetric,
every mutated adjacency
matrix is symmetric. With principal coefficients every cluster variable
has the form
\[
X_{i,t}=R_{i,t}(\yt)\xt^{g_{i,t}},\qquad
R_{i,t}\in\Q(\yt)\setminus\{0\},\qquad g_{i,t}\in\Z^n.
\]
At the current seed, write $R_i,g_i$ for its coefficient factors and
exponent vectors. Write its coefficient at $k$ as $Y_k=\yt^{c_k}$,
where $c_k\in\Z^n$, and put
\[
h_k=\frac{1+\yt^{c_k}}{\yt^{\min(c_k,0)}}.
\]
Then mutation at $k$ satisfies
\begin{align}
g'_k&=-g_k+\sum_jp_{kj}g_j,\label{eq:grecurrence}\\
R'_k&=h_k\,\frac{\prod_jR_j^{p_{kj}}}{R_k},\label{eq:Rrecurrence}
\end{align}
while $g'_j=g_j$ and $R'_j=R_j$ for $j\ne k$.
\end{proposition}

\begin{proof}
For distinct $i,j\ne k$, premutation adds $p_{ik}p_{kj}$ arrows $i\to j$
and $p_{jk}p_{ki}$ arrows $j\to i$. These numbers agree if $P$ is
symmetric. Reversal of arrows incident with $k$ preserves symmetry,
and each deletion subtracts one arrow in each direction. The diagonal
entries remain zero because premutation omits loops. This proves
the first assertion inductively. Equation~\eqref{eq:symmetricexchange}
then holds throughout. Initially $R_i=1$, $g_i=e_i$, and $\deg y_j=0$. Substituting
$X_j=R_j\xt^{g_j}$ into \eqref{eq:symmetricexchange} gives
\[
X'_k=\frac{1+\yt^{c_k}}{\yt^{\min(c_k,0)}}
\frac{\prod_jR_j^{p_{kj}}}{R_k}
\xt^{-g_k+\sum_jp_{kj}g_j}.
\]
The first two factors are nonzero rational functions of $\yt$ and
have degree zero. The last factor has the displayed exponent vector,
proving \eqref{eq:grecurrence}--\eqref{eq:Rrecurrence} by induction.
\end{proof}

For an irreducible coefficient factor $\phi$ in the Laurent polynomial
ring $\Q[y_1^{\pm1},\ldots,y_n^{\pm1}]$, define $\nu_\phi(f)$ for
a nonzero rational function $f$ as the multiplicity of $\phi$ in its
numerator minus its multiplicity in its denominator. Unique
factorization makes this independent of the fraction representing $f$
and gives $\nu_\phi(fg)=\nu_\phi(f)+\nu_\phi(g)$.
Applying this identity to \eqref{eq:Rrecurrence} gives
\begin{equation}\label{eq:valuation}
\nu_\phi(R'_k)=\nu_\phi(h_k)-\nu_\phi(R_k)
+\sum_jp_{kj}\nu_\phi(R_j).
\end{equation}
The normalizing factor is the polynomial
\[
h_k=\yt^{c_k-\min(c_k,0)}+\yt^{-\min(c_k,0)},
\]
since both exponent columns are nonnegative.
In the four-arrow example, take $\phi=1+y_3$. Immediately before
the last exchange, the valuations of $(R_1,R_2,R_3)$ are $(1,0,1)$.
The last exchange has $h_1=1+y_1y_3$ and $p_{12}=1$, $p_{13}=0$.
Since $1+y_1y_3$ becomes $1-y_1\ne0$ at $y_3=-1$, its
$\phi$-valuation is zero. Thus \eqref{eq:valuation} gives
$\nu_\phi(R'_1)=0-1+0=-1$.

Over the trivial semifield, the same argument gives
$X_{i,t}=2^{r_{i,t}}\xt^{g_{i,t}}$ with integer exponents $r_{i,t}$,
and
\begin{equation}\label{eq:two}
r'_k=1-r_k+\sum_jp_{kj}r_j.
\end{equation}
For the seven-step example the exponent triples are
\[
(0,0,0),\ (0,1,0),\ (0,1,1),\ (3,1,1),\ (3,4,1),\
(2,4,1),\ (2,0,1),\ (-1,0,1).
\]



\section{The associated coverings}\label{sec:cover}

A quiver covering is a surjective map on vertices and arrows inducing
bijections on the incoming and outgoing arrows at each vertex.
A deck transformation is an automorphism of the covering quiver that
commutes with the covering map. A connected covering is normal when
its deck transformations act transitively on each vertex fiber.
It is weakly admissible when its covering quiver
has no oriented $2$-cycles and no arrow joins two vertices in the same
deck-transformation orbit. An orbit mutation applies ordinary quiver
mutation once at every vertex in one such orbit. Global weak
admissibility requires weak admissibility after every finite sequence
of orbit mutations; see \cite[Section~2.3]{LLMP}. The mutations within
one orbit commute because its vertices are pairwise nonadjacent.
For the locally finite covers below, each arrow is affected by only
finitely many of these mutations.

For either label assignment, the connected normal cover corresponding
to $H=\ker\lambda$ has vertices $(i,r)$ with $r\in\Z$. An arrow
$i\to j$ of label $\ell$ lifts to
\[
(i,r)\longrightarrow(j,r+\ell)\qquad(r\in\Z).
\]
The covering map forgets the second coordinate. Every base arrow has
one lift with any prescribed initial or terminal vertex, giving the
required local bijections. A walk of label $m$ lifts from $(i,r)$ to
$(j,r+m)$, so a closed walk lifts to a closed walk exactly when its
label is zero. Its subgroupoid of closed lifts is therefore $H$.
The realization of every integer label between any two base vertices
proves connectedness. Translations $(i,r)\mapsto(i,r+s)$, $s\in\Z$,
are deck transformations and act transitively on each fiber, proving
normality.
Initially it is weakly admissible: base arrows have distinct endpoints,
and a lifted directed $2$-cycle would project to a directed $2$-cycle
of total label zero, which the reduced condition excludes.

In the four-arrow example, the first orbit mutation is at the orbit
above vertex~$2$. For every $r$, the path
\[
(1,r)\xrightarrow{a}(2,r+1)\xrightarrow{b}(1,r+1)
\]
creates an arrow $(1,r)\to(1,r+1)$ in the ordinary Fomin--Zelevinsky
premutation of the cover. There was no arrow within this orbit, and
all new within-orbit arrows point from level $r$ to level $r+1$.
Consequently no opposite arrow can cancel it. The mutated cover
has an arrow between two vertices in one deck orbit and is no longer
weakly admissible.

The same obstruction occurs in the six-arrow example. The paths
$ca$ and $ba$ through the first mutated orbit create arrows
$(1,r)\to(1,r+1)$ and $(1,r)\to(1,r+2)$, respectively, with no
opposite arrows. Thus its canonical covering also loses weak
admissibility immediately. In the base quiver these paths have equal
endpoints and are omitted by the homotopy premutation rule, allowing
the subsequent base mutations to continue.



\appendix
\section{Row sign incoherence with nonzero initial \texorpdfstring{$B$}{B}}
\label{app:nonzero}

\begin{proposition}\label{prop:nonzero}
For the six-arrow initial quiver and homotopy of Section~\ref{sec:six},
principal-coefficient mutation along $(2,1,3,2,3,2,3,1)$ gives
\[
G_8=\mt{-1&1&0\\0&-1&0\\5&3&1}.
\]
\end{proposition}

\begin{proof}
Here
\[
B_0=P_0-P_0^{\mathsf T}
=\mt{0&-1&1\\1&0&0\\-1&0&0}\ne0,
\]
and the columns of $-B_0$ are the degrees of the initial coefficients.


\subsection{The last two quiver mutations}
The additional label rows are given in Table~\ref{tab:extension}.
They continue Table~\ref{tab:sixlabels}.

\begin{table}[H]
\centering\small
\setlength{\tabcolsep}{5pt}
\caption{The last two label rows of the eight-step sequence.}\label{tab:extension}
\begin{tabular}{cc*{6}{c}}
\toprule
$t$ & $k$ & $L_{12}^{(t)}$ & $L_{21}^{(t)}$ & $L_{13}^{(t)}$
& $L_{31}^{(t)}$ & $L_{23}^{(t)}$ & $L_{32}^{(t)}$\\
\midrule
7 & 3 & $-1,-1$ & $-1$ & $0$ & $1,1,-1$ & $0$ & $1$\\
8 & 1 & $1$ & $1,1$ & $-1,-1,1$ & $0$ & $\varnothing$ & $0,0,0,-2,-2$\\
\bottomrule
\end{tabular}
\end{table}

Step~7 adds $12:-1,-1,1$ and $21:-1$, and deletes one pair
$(12:1,21:-1)$. Step~8 adds $32:0,0,0,0,-2,-2$ and $23:-1$;
it deletes $(23:0,32:0)$ and $(23:-1,32:1)$. All other changes are
reversals, and no further zero-sum pair remains. The matrices before
these two exchanges are therefore
\[
P_6=\mt{0&0&3\\1&0&1\\1&1&0},\qquad
P_7=\mt{0&2&1\\1&0&1\\3&1&0}.
\]

\subsection{Coefficient exponents and degrees}
For $j=1,2,3$, let $c_j\in\Z^3$ be the exponent column in
$Y_j=\yt^{c_j}$, and let $g_j=\deg X_j$ at the current seed.
Directly from \eqref{eq:y}, mutation at $k$ gives
\begin{equation}\label{eq:Crecur}
c'_k=-c_k,\qquad
c'_j=c_j+p_{kj}c_k+(p_{jk}-p_{kj})\min(c_k,0)\quad(j\ne k).
\end{equation}
The two numerator terms in \eqref{eq:exchange} have the same degree
precisely when
\begin{equation}\label{eq:homcheck}
-B_0c_k+\sum_i p_{ik}g_i=\sum_i p_{ki}g_i.
\end{equation}
Under this equality their common degree is the right-hand side, while
the normalizing monomial has degree $-B_0\min(c_k,0)$. Consequently
\begin{equation}\label{eq:Ggeneral}
g'_k=-g_k+\sum_i p_{ki}g_i+B_0\min(c_k,0),
\qquad g'_j=g_j\quad(j\ne k).
\end{equation}
Starting with $c_j=g_j=e_j$ for $j=1,2,3$, the integer computations are recorded in
Tables~\ref{tab:C8} and~\ref{tab:G8}. Every tuple in these tables
denotes a column vector. Table~\ref{tab:degrees} gives the two numerator
degrees and the normalizer degree at every exchange, using the seed
immediately before that exchange. The sum of the third and fourth
columns equals the fifth column, verifying \eqref{eq:homcheck}.
Subtracting the old $g_k$ and the last column from the fifth column
gives the new degree in Table~\ref{tab:G8}.

\begin{table}[H]
\centering\small
\caption{All coefficient exponent columns for the six-arrow data.}\label{tab:C8}
\begin{tabular}{ccccc}
\toprule
$t$ & $k$ & $c_1$ & $c_2$ & $c_3$\\
\midrule
0 & -- & $(1,0,0)$ & $(0,1,0)$ & $(0,0,1)$\\
1 & 2 & $(1,2,0)$ & $(0,-1,0)$ & $(0,0,1)$\\
2 & 1 & $(-1,-2,0)$ & $(2,3,0)$ & $(2,4,1)$\\
3 & 3 & $(3,6,2)$ & $(4,7,1)$ & $(-2,-4,-1)$\\
4 & 2 & $(11,20,4)$ & $(-4,-7,-1)$ & $(2,3,0)$\\
5 & 3 & $(15,26,4)$ & $(-2,-4,-1)$ & $(-2,-3,0)$\\
6 & 2 & $(13,22,3)$ & $(2,4,1)$ & $(-4,-7,-1)$\\
7 & 3 & $(1,1,0)$ & $(-2,-3,0)$ & $(4,7,1)$\\
8 & 1 & $(-1,-1,0)$ & $(0,-1,0)$ & $(5,8,1)$\\
\bottomrule
\end{tabular}
\end{table}

\begin{table}[H]
\centering\small
\caption{All $g$-vector columns for the same principal-coefficient pattern.}\label{tab:G8}
\begin{tabular}{ccccc}
\toprule
$t$ & $k$ & $g_1$ & $g_2$ & $g_3$\\
\midrule
0 & -- & $(1,0,0)$ & $(0,1,0)$ & $(0,0,1)$\\
1 & 2 & $(1,0,0)$ & $(2,-1,0)$ & $(0,0,1)$\\
2 & 1 & $(3,-2,2)$ & $(2,-1,0)$ & $(0,0,1)$\\
3 & 3 & $(3,-2,2)$ & $(2,-1,0)$ & $(8,-5,3)$\\
4 & 2 & $(3,-2,2)$ & $(12,-8,7)$ & $(8,-5,3)$\\
5 & 3 & $(3,-2,2)$ & $(12,-8,7)$ & $(10,-7,8)$\\
6 & 2 & $(3,-2,2)$ & $(1,-1,3)$ & $(10,-7,8)$\\
7 & 3 & $(3,-2,2)$ & $(1,-1,3)$ & $(0,0,1)$\\
8 & 1 & $(-1,0,5)$ & $(1,-1,3)$ & $(0,0,1)$\\
\bottomrule
\end{tabular}
\end{table}

\begin{table}[H]
\centering\footnotesize
\setlength{\tabcolsep}{4pt}
\caption{Numerator and normalizer degrees before each exchange.}\label{tab:degrees}
\begin{tabular}{cccccc}
\toprule
Step & $k$ & $-B_0c_k$ & $\sum_i p_{ik}g_i$ & $\sum_i p_{ki}g_i$
& $-B_0\min(c_k,0)$\\
\midrule
1 & 2 & $(1,0,0)$ & $(1,0,0)$ & $(2,0,0)$ & $(0,0,0)$\\
2 & 1 & $(2,-1,1)$ & $(2,-1,1)$ & $(4,-2,2)$ & $(0,0,0)$\\
3 & 3 & $(3,-2,2)$ & $(5,-3,2)$ & $(8,-5,4)$ & $(0,0,0)$\\
4 & 2 & $(6,-4,4)$ & $(8,-5,3)$ & $(14,-9,7)$ & $(0,0,0)$\\
5 & 3 & $(3,-2,2)$ & $(15,-10,9)$ & $(18,-12,11)$ & $(0,0,0)$\\
6 & 2 & $(-3,2,-2)$ & $(13,-9,10)$ & $(10,-7,8)$ & $(-3,2,-2)$\\
7 & 3 & $(-6,4,-4)$ & $(10,-7,9)$ & $(4,-3,5)$ & $(-6,4,-4)$\\
8 & 1 & $(1,-1,1)$ & $(1,-1,6)$ & $(2,-2,7)$ & $(0,0,0)$\\
\bottomrule
\end{tabular}
\end{table}

In the seventh exchange $c_3=(-4,-7,-1)^{\mathsf T}=\min(c_3,0)$
and $-B_0\min(c_3,0)=(-6,4,-4)^{\mathsf T}$. Its new degree is
\[
-(10,-7,8)^{\mathsf T}+(3,-2,2)^{\mathsf T}+(1,-1,3)^{\mathsf T}
-(-6,4,-4)^{\mathsf T}=(0,0,1)^{\mathsf T}.
\]
For the eighth exchange $c_1=(1,1,0)^{\mathsf T}$, so $\min(c_1,0)=0$, and
\[
g'_1=-(3,-2,2)^{\mathsf T}+2(1,-1,3)^{\mathsf T}
+(0,0,1)^{\mathsf T}=(-1,0,5)^{\mathsf T}.
\]
The resulting matrix has a mixed-sign first row:
\begin{equation}\label{eq:G8}
\boxed{G_8=\mt{-1&1&0\\0&-1&0\\5&3&1}.}\qedhere
\end{equation}
\end{proof}

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\url{https://github.com/OSVerify/Rethlas-OSV}.

\bibitem{OSVerify}
Fan Nie, Haotian Ye, Shurui Liu, Zihao Wang, Stefano Ermon, and James Zou,
\emph{One-Sided Verification Under Limited Expert Review},
arXiv:2610.18089v2, 2026.
\url{https://arxiv.org/abs/2610.18089v2}.

\end{thebibliography}

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\noindent\textsc{Zhili College, Tsinghua University, Beijing 100084, China}\par
\noindent\textit{Email address:}\enspace
\mbox{\href{mailto:tangqy24@mails.tsinghua.edu.cn}{\texttt{tangqy24@mails.tsinghua.edu.cn}}}\par
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