A conjecture on cluster automorphisms of cluster algebras

1.   Introduction

Cluster algebras were invented by Fomin and Zelevinsky in a series of papers [9,2,10,11]. A cluster algebra is a $\mathbb Z$-subalgebra of an ambient field $\mathcal F = \mathbb Q(u_1,\cdots,u_n)$ generated by certain combinatorially defined generators (i.e., cluster variables), which are grouped into overlapping clusters. Many relations between cluster algebras and other branches of mathematics have been discovered, for example, Poisson geometry, discrete dynamical systems, higher Teichmüller spaces, representation theory of quivers and finite-dimensional algebras.

We first recall the definition of cluster automorphisms, which were introduced by Assem, Schiffler and Shamchenko in [1].

Definition 1.1 ([1]). Let $\mathcal{A} = \mathcal{A}({\bf{x}},B)$ be a cluster algebra, and $f:\mathcal{A}\rightarrow \mathcal{A}$ be an automorphism of $\mathbb{Z}$-algebras. $f$ is called a cluster automorphism of $\mathcal A$ if there exists another seed $({\bf{z}},B')$ of $\mathcal{A}$ such that

$(1)$$f({\bf x}) = {\bf z}$;

$(2)$ $f(\mu_x({\bf x})) = \mu_{f(x)}({\bf z})$ for any $x\in{\bf x}$.

Cluster automorphisms and their related groups were studied by many authors, and one can refer to [6,7,8,14,13,4,5,16] for details.

The following very insightful conjecture on cluster automorphisms is by Chang and Schiffler, which suggests that we can weaken the conditions in Definition 1.1. In particular, it suggests that the second condition in Definition 1.1 can be obtained from the first one and the assumption that $f$ is a $\mathbb{Z}$-algebra homomorphism.

Conjecture 1. [5,Conjecture 1] Let $\mathcal A$ be a cluster algebra, and $f:\mathcal A\rightarrow \mathcal A$ be a $\mathbb{Z}$-algebra homomorphism. Then $f$ is cluster automorphism if and only if there exist two clusters ${\bf{x}}$ and ${\bf{z}}$ such that $f({\bf x}) = {\bf z}$.

The following is our main result, which affirms the Conjecture 1.

Theorem 3.6 Let $\mathcal A$ be a cluster algebra, and $f:\mathcal A\rightarrow \mathcal A$ be a $\mathbb{Z}$-algebra homomorphism. Then $f$ is a cluster automorphism if and only if there exist two clusters ${\bf{x}}$ and ${\bf{z}}$ such that $f({\bf x}) = {\bf z}$.

2.   Preliminaries

In this section, we recall basic concepts and important properties of cluster algebras. In this paper, we focus on cluster algebras without coefficients (that is, with trivial coefficients). For a positive integer $n$, we will always denote by $[1, n]$ the set $\{1,2,\dots, n\}$.

Recall that $B$ is said to be skew-symmetrizable if there exists an positive diagonal integer matrix $D$ such that $BD$ is skew-symmetric.

Fix an ambient field $\mathcal{F} = \mathbb{Q}(u_1, u_2, \dots, u_n)$. A labeled seed is a pair $({\bf{x}}, B)$, where ${\bf{x}}$ is an $n$-tuple of free generators of $\mathcal{F}$, and $B$ is an $n\times n$ skew-symmetrizable integer matrix. For $k\in [1,n]$, we can define another pair $({\bf{x'}},B') = \mu_k({\bf{x}},B)$, where

$(1)$ ${\bf{x'}} = (x'_1, \dots, x'_n)$ is given by

and $x'_i = x_i$ for $i \neq k$;

$(2)$ $B' = \mu_k(B) = (b'_{ij})_{n\times n}$ is given by

where $[x]_+ = \mathrm{max}\{x,0\}$. The new pair $({\bf{x'}},B') = \mu_k({\bf{x}},B)$ is called the mutation of $({\bf{x}},B)$ at $k$. We also denote $B^\prime = \mu_k(B)$.

It can be seen that $({\bf{x'}},B')$ is also a labeled seed and $\mu_k$ is an involution.

Let $({\bf{x}},B)$ be a labeled seed. ${\bf{x}}$ is called a labeled cluster, elements in ${\bf{x}}$ are called cluster variables, and $B$ is called an exchange matrix. The unlabeled seeds are obtained by identifying labeled seeds that differ from each other by simultaneous permutations of the components in ${\bf{x}}$, and of the rows and columns of $B$. We will refer to unlabeled seeds and unlabeled clusters simply as seeds and clusters respectively, when there is no risk of confusion.

Lemma 2.1 ([2]). Let $B$ be an $n\times n$ skew-symmetrizable matrix. Then $\mu_k(B) = (J_k+E_k)B(J_k+F_k)$, where

(1) $J_k$ denotes the diagonal $n\times n$ matrix whose diagonal entries are all $1'$s, except for $-1$ in the $k$-th position;

(2) $E_k$ is the $n\times n$ matrix whose only nonzero entries are $e_{ik} = [-b_{ik}]_+$;

(3) $F_k$ is the $n\times n$ matrix whose only nonzero entries are $f_{kj} = [b_{kj}]_+$.

Definition 2.2 ([9,11]). $(1)$ Two labeled seeds $({\bf{x}},B)$ and $({\bf{x}}',B')$ are said to be mutation equivalent if $({\bf{x}}',B')$ can be obtained from $({\bf{x}},B)$ by a sequence of mutations;

$(2)$ Let $\mathrm{T}_n$ be an $n$-regular tree and valencies emitting from each vertex are labelled by $1,2,\dots,n$. A cluster pattern is an $n$-regular tree $\mathrm{T}_n$ such that for each vertex $t\in \mathrm{T}_n$, there is a labeled seed $\Sigma_t = ({\bf{x}}_t,B_t)$ and for each edge labelled by $k$, two labeled seeds in the endpoints are obtained from each other by seed mutation at $k$. We always write

The cluster algebra $\mathcal{A} = \mathcal{A}({\bf{x}}_{t_0},B_{t_0})$ associated with the initial seed $({\bf{x}}_{t_0},B_{t_0})$ is a $\mathbb{Z}$-subalgebra of $\mathcal{F}$ generated by cluster variables appeared in $\mathrm{T}_n({\bf{x}}_{t_0},B_{t_0})$, where $\mathrm{T}_n({\bf{x}}_{t_0},B_{t_0})$ is the cluster pattern with $({\bf{x}}_{t_0},B_{t_0})$ lying in the vertex $t_0 \in \mathrm{T}_n$.

Theorem 2.3 (Laurent phenomenon and positivity [11,15,12]). Let $\mathcal{A} = \mathcal{A}({\bf{x}}_{t_0}, B_{t_0})$ be a cluster algebra. Then each cluster variable $x_{i,t}$ is contained in

3.   The proof of main result

In this section, we will give our main result, which affirms the Conjecture 1.

Lemma 3.1. Let $0\neq B$ be a skew-symmetrizable integer matrix. If $B^\prime$ is obtained from $B$ by a sequence of mutations and $B = aB^\prime$ for some $a\in\mathbb Z$, then $a = \pm 1$ and $B = \pm B^\prime$.

Proof.. Since $B^\prime$ is obtained from $B$ by a sequence of mutations, there exist integer matrices $E$ and $F$ such that $B^\prime = EBF$, by Lemma 2.1. If $B = aB^\prime$, then we get $B^\prime = aE(B^\prime)F$. Also, we can have

where $s\geq 0$. By $B\neq 0$, we know $a\neq 0$. Thus $\frac{1}{a^s}B^\prime = E^s(B^\prime)F^s$ holds for any $s\geq 0$.

Assume by contradiction that $a\neq \pm1$, then when $s$ is large enough, $\frac{1}{a^s}B^\prime$ will not be an integer matrix. But $E^s(B^\prime)F^s$ is always an integer matrix. This is a contradiction. So we must have $a = \pm 1$ and thus $B = \pm B^\prime$.

A square matrix $A$ is decomposable if there exists a permutation matrix $P$ such that $PAP^{\rm T}$ is a block-diagonal matrix, and indecomposable otherwise.

Lemma 3.2. Let $0\neq B$ be an indecomposable skew-symmetrizable matrix. If $B^\prime$ is obtained from $B$ by a sequence of mutations and $B = B^\prime A$ for some integer diagonal matrix $A = diag(a_1,\cdots,a_n)$, then $A = \pm I_n$ and $B = \pm B^\prime$.

Proof. If there exists $i_0$ such that $a_{i_0} = 0$, then the $i_0$-th column vector of $B$ is zero, by $B = B^\prime A$. This contradicts that $B$ is indecomposable and $B\neq 0$. So each $a_{i_0}$ is nonzero for $i_0 = 1,\cdots,n$.

Let $D = diag(d_1,\cdots,d_n)$ be a skew-symmetrizer of $B$. By $B = B^\prime A$ and $AD = DA$, we know that

By the definition of mutation, we know that $D$ is also a skew-symmetrizer of $\mu_k(B)$, $k = 1,\cdots,n$. Since $B^\prime$ is obtained from $B$ by a sequence of mutations, we get that $D$ is a skew-symmetrizer of $B^\prime$. Namely, we have that both $B^\prime D$ and $BD = (B^\prime D)A$ are skew-symmetric. Since $0\neq B$ is indecomposable, we must have $a_1 = \cdots = a_n$. So $A = aI_n$ for some $a\in\mathbb Z$, and $B = aB^\prime$. Then by Lemma 3.1, we can get $A = \pm I_n$ and $B = \pm B^\prime$.

Lemma 3.3. Let $B = diag(B_1,\cdots,B_s)$, where each $B_i$ is a nonzero indecomposable skew-symmetrizable matrix of size $n_i\times n_i$. If $B^\prime$ is obtained from $B$ by a sequence of mutations and $B = B^\prime A$ for some integer diagonal matrix $A = diag(a_{1},\cdots,a_{n})$, then $a_j = \pm 1$ for $j = 1,\cdots,n$.

Proof. By the definition of mutation, we know that $B^\prime$ has the form of $B^\prime = diag(B_1^\prime,\cdots,B_s^\prime)$, where each $B_i^\prime$ is obtained from $B_i$ by a sequence of mutations. We can write $A$ as a block-diagonal matrix $A = diag(A_1,\cdots,A_s)$, where $A_i$ is a $n_i\times n_i$ integer diagonal matrix. By $B = B^\prime A$, we know that $B_i = B_i^\prime A_i$. Then by Lemma 3.2, we have $A_i = \pm I_{n_i}$ and $B_i = \pm B_i^\prime$ for $i = 1,\cdots,s$. In particular, we get $a_j = \pm 1$ for $j = 1,\cdots,n$.

Lemma 3.4. Let $\mathcal A = \mathcal A({\bf{x}}, B)$ be a cluster algebra, and $f : \mathcal{A} \rightarrow \mathcal{A}$ be a $\mathbb Z$-homomorphism of $\mathcal{A}$. If there exists another seed $({\bf{z}},B')$ of $\mathcal{A}$ such that such that $f({\bf x}) = {\bf z}$, then $f(\mu_x({\bf x})) = \mu_{f(x)}({\bf z})$ for any $x\in{\bf x}$.

Proof. After permutating the rows and columns of $B$, it can be written as a block-diagonal matrix as follows.

where $B_1$ is an $n_1\times n_1$ zero matrix and $B_j$ is nonzero indecomposable skew-symmetrizable matrix of size $n_j\times n_j$ for $j = 2,\cdots,s$.

Without loss of generality, we assume that $f(x_i) = z_i$ for $1\leq i \leq n$.

Let $x_k^\prime$ and $z_k^\prime$ be the new obtained variables in $\mu_k({\bf{x}}, B)$ and $\mu_k({\bf{z}}, B')$. So we have

Thus

Note that the above expression is the expansion of $f(x_k^\prime)$ with respect to the cluster $\mu_k({\bf z})$. By

we can get

Since both $\prod\limits_{i = 1}^nz_i^{[b_{ik}]_+}+\prod\limits_{i = 1}^nz_i^{[-b_{ik}]_+}$ and $\prod\limits_{i = 1}^nz_i^{[b_{ik}^\prime]_+}+\prod\limits_{i = 1}^nz_i^{[-b_{ik}^\prime]_+}$ is not divided by any $z_i$, we actually have

So for each $k$, there exists an integer $a_k\in \mathbb{Z}$ such that $(b_{1k}, b_{2k}, \dots , b_{nk})^{\rm T} = a_k(b'_{1k}, b'_{2k}, \dots , b'_{nk})^{\rm T}$. Namely, we have $B = B^\prime A$, where $A = diag(a_1,\cdots,a_n)$. Note that $B$ has the form of

where $B_1$ is an $n_1\times n_1$ zero matrix and $B_j$ is a nonzero indecomposable skew- symmetrizable matrix of size $n_j\times n_j$ for $j = 2,\cdots,s$. Applying Lemma 3.3 to the skew-symmetrizable matrix $diag(B_2,\cdots,B_s)$, we can get $a_{j} = \pm 1$ for $n_1+1,\cdots,n$. Since the first $n_1$ column vectors of both $B$ and $B^\prime$ are zero vectors, we can just take $a_1 = \cdots = a_{n_1} = 1$. So for each $k$, we have $a_k = \pm1$ and

Hence,

Thus we get

So $f(\mu_x({\bf x})) = \mu_{f(x)}({\bf z})$ for any $x\in{\bf x}$.

Lemma 3.5. Let $\mathcal A = \mathcal{A}({\bf x}, B)\subseteq\mathcal F$ be a cluster algebra, and $f$ be an automorphism of the ambient field $\mathcal F$. If there exists another seed $({\bf{z}},B')$ of $\mathcal{A}$ such that $f({\bf x}) = {\bf z}$ and $f(\mu_x({\bf x})) = \mu_{f(x)}({\bf z})$ for any $x\in{\bf x}$. Then

(i) $f$ is an automorphism of $\mathcal{A}$;

(ii) $f$ is a cluster automorphism of $\mathcal{A}$.

Proof. (ⅰ) Since $f$ is an automorphism of the ambient field $\mathcal F$, we know that $f$ is injective.

Since $f$ commutes with mutations, we know that $f$ restricts to a surjection on $\mathcal{X}$, where $\mathcal{X}$ is the set of cluster variables of $\mathcal A$. Because $\mathcal{A}$ is generated by $\mathcal{X}$, we get that $f$ restricts to an epimorphism of $\mathcal{A}$.

Hence, $f$ is an automorphism of $\mathcal A$.

(ⅱ) follows from (ⅰ) and the definition of cluster automorphisms.

Theorem 3.6. Let $\mathcal A$ be a cluster algebra, and $f:\mathcal A\rightarrow \mathcal A$ be a $\mathbb{Z}$-algebra homomorphism. Then $f$ is a cluster automorphism if and only if there exist two clusters ${\bf{x}}$ and ${\bf{z}}$ such that $f({\bf x}) = {\bf z}$.

Proof. "Only if part": It follows from the definition of cluster automorphism.

"If part": It follows from Lemma 3.4 and Lemma 3.5.

Acknowledgments

This project is supported by the National Natural Science Foundation of China (No.11671350 and No.11571173) and the Zhejiang Provincial Natural Science Foundation of China (No.LY19A010023).