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!

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Qiaochu Yuan