TITLE: Which commutative algebras admit a nonzero Poisson bracket?
QUESTION [7 upvotes]: Let $A$ be a commutative algebra, not necessarily unital, over a field $k$ (of characteristic not equal to $2$, or even equal to $0$, if it helps). A second-order formal deformation of $A$ is a $k[h]/h^3$-bilinear associative product $\star$ on $A[h]/h^3$ such that quotienting by $h$, we obtain the original product on $A$. Writing such a product as
$$a \star b = ab + h m_1(a, b) + h^2 m_2(a, b), a, b \in A$$
it's not hard to verify that $\{ a, b \} = m_1(a, b) - m_1(b, a)$ is a Poisson bracket on $A$, that is, a Lie bracket satisfying the Leibniz rule $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}$. Given a nonzero Poisson bracket on $A$, it is interesting to ask whether we can find a formal deformation (replace $k[h]/h^3$ with $k[[h]]$) which gives rise to it as above ("deformation quantization"). 
But of course we can't ask this question until we have a nonzero Poisson bracket in the first place. So:

Which commutative algebras admit a nonzero Poisson bracket? 

If there is no reasonable description in general feel free to restrict to the finitely-generated case or smooth functions on manifolds etc.
What I know: any polynomial algebra in $2$ or more variables admits a nonzero Poisson bracket (take the symmetric algebra on a nonabelian Lie algebra). Any nonzero Poisson bracket gives a nonzero element of the alternating part of the second Hochschild cohomology $H^2(A, A)$, so if this group is trivial then no such brackets exist. I doubt this implication can be reversed in general, but I don't know a counterexample. If you do, I have a math.SE question you should answer!

REPLY [4 votes]: Smooth functions on a manifold always admit lots of Poisson structures. Indeed, one can construct Poisson structures on $\mathbb R^n$ with support in $(-1,1)^n$ and any prescribed value $\pi^{ij}e_i \wedge e_j$ of the Poisson-Bivector at the origin. This is done in two steps: 
At first one constructs $n$ commuting vector fields $X_i$ with support in $(-1,1)^n$.
Then on just defines the $\pi:=\pi^{ij} X_i\wedge X_j$.