TITLE: Does every Poisson bracket on a commutative algebra come from a second-order deformation?
QUESTION [12 upvotes]: Let $A$ be a commutative algebra over a field $k$ (of characteristic not equal to $2$ to be safe). Recall that $f : A \otimes A \to A$ is a Hochschild $2$-cocycle if it satisfies
$$f(ab, c) + f(a, b) c = f(a, bc) + a f(b, c)$$
and recall that $\{ -, - \} : A \otimes A \to A$ is a Poisson bracket if it is a Lie bracket such that $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}.$ Until recently I was pretty sure that if $f$ is a $2$-cocycle, then $f(a, b) - f(b, a)$ is a Poisson bracket (which only depends on the image of $f$ in $H^2(A, A)$). I can prove that $f(a, b) - f(b, a)$ is alternating and satisfies the Leibniz rule, but I can't prove the Jacobi identity and now I'm no longer sure it holds in general, although I don't know how to construct a counterexample.

What's a counterexample to this statement?

The correct statement seems to be that $f(a, b) - f(b, a)$ satisfies the Jacobi identity if the first-order deformation of $A$
$$a \star b = ab + \epsilon f(a, b)$$
defined by $f$ extends to a second-order deformation. Is this condition necessary? That is,

Does every Poisson bracket on $A$ extend to a second-order deformation?

I recall that there is a nice way of checking whether $f$ extends in this way involving $H^3(A, A)$, but I don't remember where I saw it or what it was. (Edit: it can be found in Gerstenhaber's original paper on the subject. I think the answer to the question ought to be "no," but I don't know an explicit counterexample.)
Also, is there a nice name for an alternating bilinear map on an algebra that satisfies the Leibniz rule but not the Jacobi identity? An alternating biderivation? A quasi-Poisson bracket?

REPLY [2 votes]: Here is a counterexample.  Take $A$ to be the unital $k$-algebra generated by $x_1,x_2,x_3$ with trivial multiplication.
Any map such that $f(1,1)=0 = f(1,x_i) = f(x_i,1)$ and $f(x_i,x_j)=\sum_k a_{ijk}x_k$ is a cocycle.  So pick $f(x_1,x_2)=x_1, f(x_1,x_3)=x_2$ and the rest zero.  Then $[x_1,[x_2,x_3]]=0$, $[x_3, [x_1,x_2]]=-x_2$, $[x_2, [x_3,x_1]]=0$ and the Jacobi identity fails.
Algebras with trivial multiplication are homologically badly behaved in some senses.  For example, Mathieu's examples of Poisson algebras that cannot be quantized arise this way: see Keller's "Notes for an introduction to Kontsevich's quantization theorem" 1.4.