Does every Poisson bracket on a commutative algebra come from a second-order deformation?

Solution 1:

I don't have a counterexample, but I have some ideas for where to look and how to construct one:

Some notation: Given a deformation (first order, formal, something in between) with a star product $a\star b=\sum \epsilon^i f_i(a,b)$, the commutator bracket gives rise to $[a,b]_{\star}=\sum \epsilon^i [a,b]_i$ where $[a,b]_i=f_i(a,b)-f_i(b,a)$. We let $J_{\star}(a,b,c)$ and $J_{\star}(a,b,c)$ be the resulting Jacobi identity expressions. Given a power series $g=\sum a_i \epsilon^i$, we let $[\epsilon^i]g=a_i$.

We have that $J_1(a,b,c)$ is contained inside $[\epsilon^2]J_{\star}(a,b,c)$ (and in no other terms), and we only get $[\epsilon^2]J_{\star}(a,b,c)=J_1(a,b,c)$ if additional terms vanish, which happens when $A$ is commutative, or if $f_2$ satisfies a particular compatibility condition, which I see no a priori reason to happen.

Additionally, we see that, if we have a first order deformation that does not lift to a second order deformation, we have no $\epsilon^2$ term to contain $J_1(a,b,c)$.

As such, you should be able to produce a counterexample with an arbitrary non-commutative algebra (almost anything should do, so I suggest looking at $T(\mathbb C^2)$), and you should be able to produce a counterexample for a commutative ring if you can find a first order deformation which does not lift to a second order one.

By the classical HKR theorem, if $A$ is a smooth $k$-algebra, we have an isomorphism of algebras $\bigwedge_A^{\bullet} \operatorname{Der}_k(A,A)\cong HH^{\bullet}(A,A)$. Moreover, this can be strengthened to an isomorphism of Gerstenhaber algebras by taking the Gertenhaber bracket on the left to be the Schouten bracket induced from the Lie algebra structure on $\operatorname{Der}_k(A,A)$. (I first saw this strengthening in a paper by Gerstenhaber and Schack, although I do not remember which one). If I'm not mistaken, we can make the isomorphism quite explicit, and so if we can find a suitable class on the left hand side, we can turn it into an actual cocycle.

So, we must ask what exactly are we looking for? If we let $\{-,-\}$ denote the Gerstenhaber bracket and $m$ the cocycle representing multiplication in $A$ (and [m] for the cohomology class, etc), we want $f_1$ such that $m+\epsilon f_1$ defines a first order deformation but there is no $f_2$ such that $m+\epsilon f_1+\epsilon^2 f_2$ defines a second order deformation. We have that $m'=\sum \epsilon^n f_n$ defines a deformation if $\{m',m'\}=0$. Note that, because all our $f_i$ are cocycles, they satisfy that $\{m,f_i\}=0$. Let us examine this more closely.

For a first order deformation $m'=m+\epsilon f_1$, our required condition (except in characteristic $2$) follows straight from $\{m,m\}=\{m,f_1\}=0$. For a second order deformation extending a first order one, $\{m+\epsilon f_1+\epsilon^2 f_2,m+\epsilon f_1+\epsilon^2 f_2\}=\epsilon^2\{f_1,f_1\},$ and so it suffices to find a cocycle which, when bracketed with itself is nonzero.

If we take $A=\mathbb{C}[x,y,z]$ (n.b., we need to take 3 variables so that $H^3(A,A)\neq 0)$, a derivation from $A$ to $A$ is determined by $d(x),d(y)$ and $d(z)$, with no compatibility conditions, and so $\operatorname{Der}_k(A,A)\cong A^{\oplus 3}$, generated as an $A$-module by $\frac{d}{dx},\frac{d}{dy},\frac{d}{dz}$. It should be straight forward to work out the Lie algebra structure on $\operatorname{Der}_k(A,A)$, and then the Gerstenhaber (Schouten) bracket. I don't know for sure that you will be able to find an element $f_1\in H^2(A,A)$ such that $\{f_1,f_1\}\neq 0$, nor do I know for sure that this will lead to a counterexample to the original problem. However, this should, at a minimum, be an instructive exercise.

Solution 2:

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.