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.

Please add tags as you see fit.

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.

Please add tags as you see fit.