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 ${,}: Aotimes A to A$ such that ${f,-}$ and ${-,f}$ are derivations for each $fin A$. 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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