MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a really basic question about cluster algebras and cluster varieties. According to the definition of Fomin-Zelevinsky a cluster algebra is generated by a bunch of polynomial rings inside the ring of Laurent polynomials. It means that the corresponding variety is covered by a bunch of affine spaces (all which have some $\mathbb G_m^n$ in common). Let's call this variety $X$.

Question: Is it true that all the above maps from $\mathbb A^n\to X$ are open embeddings (a priori it is only a birational map)? A related question: is it true that $X$ is always a smooth variety?

Edit: Of course, the question is wrongly asked as the maps are from $X$ to $\mathbb A^n$.

share|cite|improve this question
Which maps $\mathbb{A}^n\to X$? There's a map from $\mathbb{A}^n$ minus its coordinate axes (induced by the embedding in Laurent polynomials in each cluster) which is an open embedding, and a map $X\to \mathbb{A}^n$ given by each cluster. The latter are very, very much not embeddings. – Ben Webster Feb 27 at 16:36
Yes, you are right, I got confused. But David below gave exactly the answer I wanted. – Alexander Braverman Feb 27 at 20:11
up vote 9 down vote accepted

A bunch of points: $\def\Spec{\mathrm{Spec}\ }$

• Let $A$ be a cluster algebra over a field $k$, let $(x_1, \ldots, x_n)$ be a cluster and let $L$ be the Laurent polynomial ring $k[x_1^{\pm}, \ldots, x_n^{\pm}]$. I imaging your intended question is whether the map $\Spec L \to \Spec A$ is an open immersion. (You ask about a map $\mathbb{A}^n \to X$, but there isn't a natural such map; $\Spec L$ is a torus, not affine space.)

The answer is yes. The question is equivalent to asking whether $L$ is a localization of $A$. I claim that $L = A[x_1^{-1}, \ldots, x_n^{-1}]$. Proof: On the one hand, $A \subset L$ (by the Laurent phenomenon) and $x_1^{-1}$, ..., $x_n^{-1} \in L$ (obviously), so $A[x_1^{-1}, \ldots, x_n^{-1}] \subseteq L$. On the other hand, $L$ is generated by the $x_j$ and their reciprocals, and these are all in $A[x_1^{-1}, \ldots, x_n^{-1}]$, so $L \subseteq A[x_1^{-1}, \ldots, x_n^{-1}]$. $\square$.

• $\Spec A$ is not generally the union of cluster tori. This is true even in the simplest case: The extended exchange matrix $\left( \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right)$ gives the cluster algebra $$\mathbb{C}[x,x', y^{\pm 1}]/x x'-y-1.$$ The two tori are $xy \neq 0$ and $x' y \neq 0$. The point $(x,x',y) = (0,0,-1)$ is in neither torus.

• Some sources define "the cluster variety" as the (quasi-affine) union of cluster tori. If you take that as the definition, it is obviously smooth.

• But I imagine what you care about is whether $\Spec A$ is smooth (or possibly $\Spec U$, where $U$ is the upper cluster algebra. No, that doesn't have to be smooth. (The singular points are not in any of the cluster tori, of course.) I think the simplest example is the Markov clsuter algebra, with $B$-matrix $\left( \begin{smallmatrix} 0 & 2 & -2 \\ -2 & 0 & 2 \\ 2 & -2 & 0 \end{smallmatrix} \right)$. This ring is not finitely generated, and there is a maximal ideal generated by all cluster variables where the Zariski tangent space is infinite dimensional. If you like at the upper cluster algebra instead, it is $k[\lambda, x_1, x_2, x_3]/x_1 x_2 x_3 \lambda - x_1^2- x_2^2 - x_3^3$, which is singular along the line $x_1 = x_2 = x_3 = 0$.

For another example, which doesn't have the $A$ versus $U$ issue, look at the $A_3$ cluster algebra with no frozen variables. From Corollary 1.17 in Cluster Algebras III, this is generated by $(x_1, x_2, x_3, x'_1, x'_2, x'_3)$ module the relations $$x_1 x'_1 = x_2+1,\ x_2 x'_2 = x_1 + x_3,\ x_3 x'_3 = x_2+1.$$

Look at the point $(x_1, x_2, x_3 , x'_1, x'_2, x'_3) = (0, -1, 0, 0,0,0)$. The Jacobian matrix of these $3$ equations with respect to the $6$ variables is $$\begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{pmatrix}$$ which has rank two, so this is a singular point.

• We do have Theorem 7.7 of Greg Muller's Locally Acyclic Cluster Algebras -- if the cluster algebra is locally acyclic, the $B$-matrix has full rank and our ground field has characteristic zero, then $\Spec A$ is smooth.

See Muller 1 and Benito-Muller-Rajchgot-Smith 2 for more on singularities of cluster varieties.

share|cite|improve this answer
As I recall Gekhtman-Shapiro-Vainshtein distinguish the quasi-affine "cluster manifold" (the union of tori) from the "cluster variety" (that $Spec$, i.e. the affinization). – Allen Knutson Feb 27 at 23:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.