# Which commutative algebras admit a nonzero Poisson bracket?

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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
I don't have an account at M.SE, and anyway won't answer that question, only comment on its motivation, so I'll leave the comment here. You ask: does the antisymmetrization of a Hochschild 2-cocycle satisfy Jacobi? The answer is NO. Recall that a bivector field is certainly a 2-cocycle, and the Hamiltonian flows for a Poisson bivector field foliate the space into (even-dimensional) symplectic leaves. But there are non-integrable plane distributions in $\mathbb R^3$, the standard one being $\ker(dz+xdy)$. This distribution is generated by Hamiltonian vector fields for the bivector (continued)

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Theo Johnson-Freyd
(continuation) $\partial_x \wedge (x \partial_y - \partial_z)$. So this cannot satisfy Jacobi. Alternately, check directly that this bivector field is not Poisson. For another example, find your favorite invertible de Rham 2-form that is not closed, and take its inverse. Other examples come from the following observation: Jacobi is a quadratic relation, whereas 2-cocycle is a linear relation. So you do not expect Jacobi(a+b) to hold even if both Jacobi(a) and Jacobi(b) hold.

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Theo Johnson-Freyd
@Theo: thanks. I figured that had to be the case but didn't know how to go about constructing a counterexample.

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Qiaochu Yuan
@ Qiaochu: I am not entirely sure about this(I am mostly familiar with the case of smooth algebras where a lot more can be said) because it seems to me that in char=0, there is always an injective (this is the part I haven't checked) map of lie algebras HKR: \oplus \Wedge^i T_A \to HH^*(A,A)(when A is smooth this is an iso) in particular if your deformation class $\pi$ is the image of a bivector field the condition [\pi,\pi]=0 in HH^*(A,A) says exactly that your bivector field is Poisson. The above condition says that on the level of Hochschild chains [\pi,\pi]=d\alpha so....

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Daniel Pomerleano
we can extend our multiplication to second order by m_t=m + tπ+t^2\alpha which will be a solution to the Maurer Cartan Equation in HCH^*(A,A)[t]/t^3. So I don't think you will find any counterexample to your Stack Exchange question because I believe it's true. In the smooth case, where HKR is an isomorphism and hence I'm pretty certain that what I've said up to now is true, (an algebraic version of) Kontsevich's formality theorem tells you that a Poisson structure extends to a deformation of all orders. Have a look, it's wonderful math! arxiv.org/abs/q-alg/9709040

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Daniel Pomerleano
The algebraic version is slightly more complicated for technical reasons and was only proven a bit later here: arxiv.org/abs/math/0310399

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Daniel Pomerleano
@Daniel: thanks for your comments! I am having trouble figuring out which of the general statements you're making is doing the work, since it seems to me that you produce $\alpha$ out of thin air. I admit I'm not very familiar with the definition of the Gerstenhaber bracket (maybe that's what's doing the work?), so if you had some time to spell some simple things like this out for me in an answer it would be much appreciated.

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

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In general, I think the question is too broard to expect some reasonable answer. But there are many examples and constructions of Poisson brackets which might be interesting for you.

1.) Whenever you have nonzero commuting derivations on your algebra you can build a Poisson bracket out of them. Indeed, if $D_1, \ldots, D_n, E_1, \ldots, E_n$ are commuting derivations then $P = \sum_i D_i \otimes E_i$ gives a Poisson bracket via \begin{equation} \lbrace a, b\rbrace = \mu \circ (P - P \circ \tau) (a \otimes b) \end{equation} where $\tau$ is the canonical flip and $\mu\colon A \otimes A \longrightarrow A$ is the multiplication. This example allows for an immediate deformation quantization, the star product quantizing it is \begin{equation} a \star b = \mu \circ \exp(\hbar P)(a \otimes b), \end{equation} a formula going back at least to Gerstenhaber himself (I think in his famous deformation of algebras papers, Nr III).

Now this is not sooo special as it may seem at the first sight. In particular, the following nice geometric construction shows that on a smooth manifold you always have nontrivial Poisson brackets:

On $\mathbb{R}^d$ there are $d$ vector fields with support inside the compact ball of radius $1$, which coincide with the coordinate vector fields $\partial_1, \ldots, \partial_d$ inside the ball of radius $1/2$, and which commute everywhere. Out of them you can build a Poisson tensor having maximal rank (either $d$ or $d-1$) inside the smaller ball but with compact support. Thus you can implant it to any other manifold using this as a chart :) The existence of such vector fields can be shown by various constructions. Either you can shrink $\mathbb{R}^d$ with a suitable diffeo into the ball and take the images of the coordinate vector fields (I learned this one from Alan Weinstein) or you can play around with two (!) suitable bump functions...

2.) Related to this but more sophisticated: there are universal deformation formulas for certain (Lie) groups. So whenever you have an action of such a group on an algebra by automorphisms, one can induce a deformation quantization and hence in particular a Poisson bracket on the algebra. It depends on the action how nontrivial the bracket will be. Part 1.) is a special case for the group $\mathbb{R}^d$.

OK, there are several more constructions, but this might already be interesting for you.

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Stefan Waldmann
answered Aug 3, 2011 by (440 points)
Your examples in (1) can be seen as follows: derivations are in $HH^1$, and each term $D_i\otimes E_i$ is really the (class of the) cup product $D_i\smile E_i$. So these Poisson structures come from the 'decomposable elements' in $HH^2$.

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Mariano Suárez-Álvarez
But you still need that the derivations commute, right? Not all decomposable elements in $HH^2$ will be Poisson?

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Stefan Waldmann
Thanks, this is interesting. Is there a short explanation of why I should expect the star-product you wrote down to be associative?

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Qiaochu Yuan
There is a nice algebraic proof (char 0 of course) in a quantum group style: you can just play around with the usual identities for derivations like $D \circ \mu = \mu \circ (D \otimes id + id \otimes D)$ and so on. Then you arrive at a lot of exponential series of operators on the triple tensor product which all commute since the derivations commute. There's a detailed proof along these lines in my book, but it's in german ;) Of course, you can also just do it the hard way and compute...

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Stefan Waldmann
@Stefan Indeed you will need the derivations to commute because if $\pi$ is your bivector field the Poisson condition can be thought of as the condition $[\pi,\pi]=0$. That each of the derivations commute + Mariano's observation+ the fact that HH^* is a Gerstenhaber algebra will imply the condition. I'm sure you and Mariano know this but I'm writing it down just in case someone who is beginning to learn this stuff comes along and reads the thread...

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Daniel Pomerleano
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Smooth functions on a manifold always admit lots of Poisson structures. Indeed, one can construct Poisson structures on $\mathbb R^n$ with support in $(-1,1)^n$ and any prescribed value $\pi^{ij}e_i \wedge e_j$ of the Poisson-Bivector at the origin. This is done in two steps: At first one constructs $n$ commuting vector fields $X_i$ with support in $(-1,1)^n$. Then on just defines the $\pi:=\pi^{ij} X_i\wedge X_j$.

This post imported from StackExchange MathOverflow at 2017-09-18 17:16 (UTC), posted by SE-user Jan Weidner
answered Aug 3, 2011 by (40 points)

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