A symplectic manifold gives rise to a Poisson algebra. If the symplectic form is exact, how is this revealed in the algebra?

$\begingroup$ Side remarks: on a compact manifold it cannot be exact, since w^n is volume nondegenerate form and if it is exact volume would be zero. On the other hand any symplectic manifold M multiplied by R^2 will give an example of such situation d(pdq w)=(dpdq) w. $\endgroup$– Alexander ChervovOct 8, 2012 at 18:14
2 Answers
It seems to me that there is a (quasi)isomorphism between the de Rham algebra and the dg algebra of polyvector fields equipped with the differential $[\pi,]$ (where $\pi$ is the Poisson structure corresponding to the symplectic form).
Through this isomorphism the equation $d\omega=0$ is sent to $[\pi,\pi]=0$, and the equation $\omega=d\lambda$ is sent to $\pi=[\pi,V]$, where $V$ is a vector field.
On the level of the Poisson algebra of functions it tells you that for any two functions $f,g$, we have (up to a sign) $$ \{f,g\}=V(\{f,g\})\{V(f),g\}+\{f,V(g)\} $$
Algebraically you can say that there is a derivation $V$ for the product such that the Poisson bracket is its own derived bracket w.r.t. $V$.

$\begingroup$ Do you have a reference or simple argument for the quasiisomorphism? $\endgroup$ Oct 8, 2012 at 12:19

1$\begingroup$ The fact that Poisson and De Rham cohomology are isomorphic on symplectic manifolds is not difficlt. The Poisson bivector defines a "sharp map" from kforms to kmultivector fields, just by extending the map between 1forms and 1vector fields given by contractipn with a 2bivector. This map intertwines coboundaries and intertwines cup products. It is an isomorphism already at the chain level, being invertible. Every book on Poisson manifold has this remarked; maybe the easiest source (depends on tastes) is Vaisman "lectures on the geometry of Poisson manifolds" birkhauser. $\endgroup$ Oct 8, 2012 at 17:27

$\begingroup$ @igor khavkine: there is an isomorphism $TX\to T^*X$ that sends a vector $v$ to the oneform $\omega(v,)$. It extends to an algebra morphism from polyvectors to forms (wedge product on both sides). You can chrck in local coordinates that $[\pi,]$ is sent to $d$. $\endgroup$– DamienCOct 8, 2012 at 17:28

1$\begingroup$ So, the quasiisomorphism is actually an isomorphism of dg algebras. $\endgroup$– DamienCOct 8, 2012 at 17:31
The closest to algebraic I know of is the following:
Let L be the lie algebra of infinitesimal symplectic transformations $L_X\omega=0$, here $\omega$ is the symplectic form, and let $L_1$ be the Lie algebra of infinitesimal conforma symplectic transformations $L_X\omega+k_X\omega=0$ for a constant $k_X$. It is easily seen that $L$ is an ideal in $L_1$. Then Lichnerowicz proved the following:
If $\omega$ is exact $L=[L_1,L_1]$; if $\omega$ is not exact $L=L_1$.
All such statements are quite easy to prove; i am not sure but I think they are contained in "les varietes des Poisson et leurs algebres de Lie associes" in Journ. diff Geom. 1977. I cannot be at present more precise with the reference, sorry.
*ADDED *
itmaybe obvious to anyone but the link between my answer and Damien answer is the following: take $\pi=\omega^{1}$ and contract with $df\wedge dg$. On one hand bo answer are a rephrasing of the fact that $[\omega]=0$, Damien in terms of Poisson cohomology (his condition states that the Poisson bivector is a coboundary).