0
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.