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For a Lie algebra, $\mathfrak{g}$, one has an equivalence of categories between Mod($\mathfrak{g}$) and Mod($U(\mathfrak{g})$), where $U(\mathfrak{g})$ is the universal enveloping algebra of $\mathfrak{g}$.

Let $P=(A,\{,\})$ be a Poisson algebra over $A=k[x_1,...,x_n]$.

An $A$-module $P$ is a Poisson module if there is a bilinear product $\{,\}_M:A \times M \rightarrow M$ such that the following hold for all $a,b \in A$ and all $m \in M$:

  1. $\{\{a,b\},m\}_M = \{a,\{b,m\}_M\}_M - \{b,\{a,m\}_M\}_M$;
  2. $\{a,bm\}_M = \{a,b\}_M + b\{a,m\}_M$;
  3. $\{ab,m\}_M = a\{b,m\}_M + b\{a,m\}_M$.

One can construct the enveloping algebra of $P$ in much the same way as one constructs $U(\mathfrak{g})$. In particular,

$U(P) = k\langle x_1,...,x_n \mid x_ix_j-x_jx_i - \{x_i,x_j\} \text{ for all } 1 \leq i,j \leq n\rangle$, $k$ a field.

Does the same equivalence exist between $U(P)$-modules and Poisson $P$-modules?

Here is the example I have in mind. Let $A=k[x,y,z]$ and define a Poisson bracket on $A$ by $\{x,y\}=z^2$, $\{y,z\}=x^2$, $\{z,x\}=y^2$. Then the universal enveloping algebra for $P$ should be

$k\langle x,y,z \mid xy-yx=z^2, yz-zy=x^2, zx-xz=y^2 \rangle$.

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Your construction of $U(P)$ makes no mention of the multiplication on $P$, so it is just the universal enveloping algebra of the underlying Lie algebra of $P$. Whatever a Poisson $P$-module is (I am not familiar with this notion) it presumably involves the multiplication on $P$, so there's no reason to expect this. –  Qiaochu Yuan Dec 20 '12 at 20:38
I've added the definition of a Poisson module in case that makes a difference. –  J. Gaddis Dec 20 '12 at 20:50
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2 Answers

up vote 1 down vote accepted

The answer to your question is negative. For any Poisson algebra $P$, there does exist an associative algebra, which is denoted by $P^e$, such that the module category over $P^e$ is equivalent to the category of Poisson modules over $P$. See http://arxiv.org/pdf/1102.0366v1.pdf for the details. However, $P^e$ is not constructed the way you mentioned. In fact, one should think $P^e$ as the analog of $A\otimes A^{op}$ of an associative algebra $A$. So in some sense, the size of $P^e$ is "doubled" from $P$. As for your example, $P^e$ is a non-commutative algebra of six generators of GK-dimension 6.

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You're right. I came across this definition a little while after posting this question. Though, I prefer the presentation in Sei-Qwon Oh's original paper, "Poisson enveloping algebras" and subsequent work with Eun-Hee Cho. –  J. Gaddis Oct 20 '13 at 14:54
You are right. In Oh's original paper the construction is very explicit. While the one given in the paper I mentioned is actually quite easy to work with as well. –  Guangbin Zhuang Nov 2 '13 at 16:56
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(Edited:) I don't have an answer to the question as stated (although I still expect it to be no), but I just want to point out that this construction is not all that closely analogous to the construction of the universal enveloping algebra. The universal enveloping algebra $U(\mathfrak{g})$ can be canonically defined without making any choices (e.g. a choice of basis of $\mathfrak{g}$) whereas in this definition you are singling out a particular choice of generators.

The analogous construction would be to take the left adjoint of the forgetful functor from algebras to Poisson algebras, but 1) this does not agree with your construction, 2) it gives an equivalence for a different notion of Poisson module, and 3) it does not seem to be an interesting notion of Poisson module.

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Maybe I'm doing a poor job of writing my definition. I would think in this case that $U(P)=K[x]$. I'll try another edit. –  J. Gaddis Dec 20 '12 at 22:18
@linearfish: then I don't understand your definition. What are the $x_i$? –  Qiaochu Yuan Dec 20 '12 at 22:26
They are the basis elements for the Poisson algebra. Sorry, that probably wasn't clear in my post. –  J. Gaddis Dec 20 '12 at 22:27
@linearfish: what basis elements? In your example the $x_i$ are generators. How do you know that the resulting algebra is independent of a choice of generators, and what do you do if $P$ isn't a polynomial algebra? –  Qiaochu Yuan Dec 20 '12 at 22:28
@linearfish: that is highly nonstandard. The standard term for that is "finitely generated." In my experience, in the phrase "finite-dimensional algebra" the term "finite-dimensional" means "as a module" (a quick Google search will verify this). –  Qiaochu Yuan Dec 20 '12 at 22:32
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