# Universal enveloping algebra of a Poisson algebra

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. – linearfish Dec 20 '12 at 20:50

(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.
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. – linearfish 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
@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