Let $A$ be a Poisson $\mathbb{C}$-algebra: $A$ has a the structure of a complex commutative algebra and at the same time it carries the structure of a Lie algebra, with Lie bracket $\{\cdot, \cdot\}$. The latter should be a biderivation of the commutative multiplication, as well as the Lie multiplication.

Just as the definition of an $A$-module may be seen as an abstraction of left multiplication of $A$ on itself, we can define a Poisson $A$-module to be a vector space $V$ together with two multiplicative rules $A \times V\rightarrow V$ satisfying compatibility conditions. For more detail see one of the original papers by Farkas.

My question is the following:

Is every simple Poisson module finitely generated as an $A$-module?

I expect that the answer is `no' although I have been unable to find a counterexample or indeed any reference which considers this question. I am especially interested in the case where $A$ is an affine algebra, ie. finitely generated and reduced. Proposition 1.1 in the paper cited above states that when $A$ is symplectic every Poisson module is induced by a $\mathcal{D}$-module, so this may be a fertile source of counterexamples: is every simple $\mathcal{D}(A)$-module finitely generated over $A$?

Thanks in advance!

Edit: Following Ben's response I am now specifically looking for examples where $A$ is a reduced Poisson algebras, ie. no nilpotent elements.


1 Answer 1


This is definitely not true; in fact, it's easy to construct counterexamples with $A$ finite dimensional.

For any Lie algebra $\mathfrak{g}$, you can define a Poisson algebra structure on $\mathbb{C}\oplus \mathfrak{g}$ such that $(a_1+X_1)(a_2+X_2)=a_1a_2+a_1X_2+a_2X_1$ for $a_i\in \mathbb{C}$ and $X_i\in \mathfrak{g}$ (you can more fancily define this as $\mathrm{Sym}(\mathfrak{g})/ \mathfrak{g}^2$); the Poisson bracket is defined by $\{a_1+X_1,a_2+X_2\} =[X_1,X_2]$. The simple Poisson modules over this algebra are precisely the simple modules over the Lie algebra (with the "commutative" action of $\mathfrak{g}$ trivial, and the "bracket" action being the representation). It's well known that finite dimensional Lie algebras have lots of simple infinite dimensional modules; for example, consider the Verma module for a generic highest weight over $\mathfrak{sl}_2$. These aren't finitely generated over $A$, since a finitely generated $A$ module is finite dimensional.

EDIT: I hadn't noticed the comment about reducedness in the question, but you can think of these same examples as modules over $\mathrm{Sym}(\mathfrak{g})$; more generally, as argued in this paper of David Jordan, these sort of quotients show up as $A/I^2$ whenever you have a maximal ideal $I$ which is Poisson (and every finite dimensional simple is pulled back from this construction).

  • $\begingroup$ Thanks for pointing that out. As I say I'm especially interested in the case where $A$ is reduced as a commutative algebra. I guess I should add that as a hypothesis to discount examples such as these. $\endgroup$ Feb 6, 2017 at 20:30
  • $\begingroup$ @LewisTopley Adding "reduced" doesn't help. These examples are quotients of $\mathrm{Sym}(\mathfrak{g})$. $\endgroup$
    – Ben Webster
    Feb 6, 2017 at 20:32
  • $\begingroup$ Thanks a lot, I see that these modules aren't finitely generated over $S(\mathfrak{g})$ $\endgroup$ Feb 6, 2017 at 20:40

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