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 $\mathbb A^n$. 



Cluster algebras and cluster varieties
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](http://arxiv.org/abs/math/0305434), 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](http://arxiv.org/abs/1111.4468) -- 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](http://arxiv.org/abs/1111.4468) and Benito-Muller-Rajchgot-Smith [2](http://arxiv.org/abs/1404.4399) for more on singularities of cluster varieties.


