How can I tell whether a Poisson structure is symplectic "algebraically"?

Question

My primary motivation for asking this question comes from the discussion taking place in the comments to What is a symplectic form intuitively?.

Let $M$ be a smooth finite-dimensional manifold, and $A = \cal C^\infty(M)$ its algebra of smooth functions. A derivation on $A$ is a linear map $\{\}: A \to A$ such that $\{fg\} = f\{g\} + \{f\}g$ (multiplication in $A$). Recall that all derivations factor through the de Rham differential, and so: Theorem: Derivations are the same as vector fields.

A biderivation is a linear map $\{,\}: A\otimes A \to A$ such that $\{f,-\}$ and $\{-,f\}$ are derivations for each $f\in A$. By the same argument as above, biderivations are the same as sections of the tensor square bundle ${\rm T}^{\otimes 2}M$. Antisymmetric biderivations are the same as sections of the exterior square bundle ${\rm T}^{\wedge 2}M$. A Poisson structure is an antisymmetric biderivation such that $\{,\}$ satisfies the Jacobi identity.

Recall that sections of ${\rm T}^{\otimes 2}M$ are the same as vector-bundle maps ${\rm T}^*M \to {\rm T}M$. A symplectic structure on $M$ is a Poisson structure such that the corresponding bundle map is an isomorphism. Then its inverse map makes sense as an antisymmetric section of ${\rm T^*}^{\otimes 2}M$, i.e. a differential 2-form, and the Jacobi identity translates into this 2-form being closed. So this definition agrees with the one you may be used to of "closed nondegenerate 2-form".

Question: Is there a "purely algebraic" way to test whether a Poisson structure is symplectic? I.e. a way that refers only to the algebra $A$ and not the manifold $M$?

For example, it is necessary but not sufficient that $\{f,-\} = 0$ implies that $f$ be locally constant, where I guess "locally constant" means "in the kernel of every derivation". The easiest way that I know to see that it is necessary is to use Darboux theorem to make $f$ locally a coordinate wherever its derivative doesn't vanish; it is not sufficient because, for example, the rank of the Poisson structure can drop at points.

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Answer 1

In the purely algebraic setting, Daniel Farkas proved in his beautiful paper [Farkas, Daniel R. Characterizations of Poisson algebras. Comm. Algebra 23 (1995), no. 12, 4669--4686. MR1352562] that a Poisson-simple linear Poisson algebra over an algebraically closed field is a regular symplectic domain, a partial converse of the much easier fact that a commutative regular affine domain which is symplectic is Poisson-simple. There are examples of non symplectic Poisson-simple polynomial algebras, though.

Answer 2

The moral answer is "Yes. A Poisson structure is symplectic if and only if the algebra has no interesting Poisson ideals."

The idea is this: a Poisson ideal is one which is closed under the operation of Poisson bracket with every function (not to be confused with a coisotropic ideal, which one closed under taking the bracket of two elements in the ideal). Obviously, the vanishing set of a Poisson ideal is a Poisson submanifold, so if you're symplectic, you'd better not have any whose vanishing set is a submanifold (not empty or the whole space). And if you have a Poisson submanifold, than the ideal vanishing on the submanifold is certainly Poisson.

So basically up to nebulous concerns about how the Nullstellensatz works for manifolds, that's the right answer.