has full rank. We now claim that $\{x_i,y_i\}(p)\ne 0 $. Indeed, suppose that $\{x_i,y_i\}(p)=(\mu_1 m^+\mu_2 m^-)(p)=0$. Since $p_i=0$ implies that $(m^++m^-)(p)=0$, we would conclude as in the proof of Theorem \[th:acyclic\] that $m^+(p)=m^-(p)=0$, but that is a contradiction to our assumption that $p_j\ne 0$ for all $j<i$. Denote by $u\in  \Lambda^2 T(Y)$ the Poisson bivector. We obtain that $u_p(\frac{\delta}{\delta x_i},\cdot)\ne 0$, hence the symplectic leaf containing $p$ is not tangent to the hypersurface $x_i=0$ at $p$. It must contain a point in an analytic neighborhood of $p$ at which $x_i(p)\ne 0$ and $x_j(p)\ne 0$ for all $j<i$ by our assumption. We obtain the desired contradiction and, hence, every symplectic leaf has dimension $n$. But the manifold $X$ is connected and, therefore, cannot be a union of disjoint open submanifolds of equal dimension, hence $X$ contains only one symplectic leaf and the theorem is proved.

This result can be easily generalized to acyclic cluster algebras with invertible coefficients (using Remark \[re:Class Poisson\]). This would imply that in the set-up of locally acyclic cluster algebras it should be easy to show that the spectrum of a locally acyclic cluster algebra is a holomorphic symplectic manifold.

Ideals in Acyclic Quantum Cluster Algebras {#se:ideals 