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

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). 


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. $V$. 

