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

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

2010-03-31T19:26:44Z

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 by Ben Webster for How can I tell whether a Poisson structure is symplectic "algebraically"?

2010-04-01T01:17:52Z

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.

Answer by Mariano Suárez-Alvarez for How can I tell whether a Poisson structure is symplectic "algebraically"?

2010-04-01T03:22:09Z

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.