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

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?

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My guess: you are remembering Gerstenhaber's deformation theory for associative algebras. His paper on the subject (the first one) is extremely readable. – Mariano Suárez-Alvarez Aug 3 '11 at 4:44
@Mariano: thanks for reminding me that I should reread that paper. I remember getting a little lost in the second section and, frightened and confused, forgetting a little of what I had learned from the first... – Qiaochu Yuan Aug 3 '11 at 5:05
These notes seem relevant. They don't directly answer your question, but it gives indications for some of them. – Aaron Aug 3 '11 at 6:13
@Qiaochu: A bit of an aside: would you care to write a tag wiki summary for deformation theory? Since you created the tag I assume you know what it means. – Willie Wong Aug 3 '11 at 19:47

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.

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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.

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Under the HKR isomorphism, we see that elements of $HH^2(A,A)$ are regular bivectors, and the condition that we can lift a first order deformation to a second order deformation is exactly the condition that the bivector is a Poisson bivector. There is clearly some a deep geometric connection here. – Aaron Aug 5 '11 at 8:06