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?

  • $\begingroup$ My guess: you are remembering Gerstenhaber's deformation theory for associative algebras. His paper on the subject (the first one) is extremely readable. $\endgroup$ – Mariano Suárez-Álvarez Aug 3 '11 at 4:44
  • $\begingroup$ @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... $\endgroup$ – Qiaochu Yuan Aug 3 '11 at 5:05
  • $\begingroup$ These notes seem relevant. They don't directly answer your question, but it gives indications for some of them. $\endgroup$ – Aaron Aug 3 '11 at 6:13
  • $\begingroup$ @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. $\endgroup$ – 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.

| cite | improve this answer | |

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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. $\endgroup$ – Aaron Aug 5 '11 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.