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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
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 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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