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

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Side remarks: on a compact manifold it cannot be exact, since w^n is volume non-degenerate 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. – Alexander Chervov Oct 8 '12 at 18:14

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.


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).

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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. $V$.

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Do you have a reference or simple argument for the quasi-isomorphism? – Igor Khavkine Oct 8 '12 at 12:19
The fact that Poisson and De Rham cohomology are isomorphic on symplectic manifolds is not difficlt. The Poisson bivector defines a "sharp map" from k-forms to k-multivector fields, just by extending the map between 1-forms and 1-vector fields given by contractipn with a 2-bivector. 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. – Nicola Ciccoli Oct 8 '12 at 17:27
@igor khavkine: there is an isomorphism $TX\to T^*X$ that sends a vector $v$ to the one-form $\omega(v,-)$. 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 $d$. – DamienC Oct 8 '12 at 17:28
So, the quasi-isomorphism is actually an isomorphism of dg algebras. – DamienC Oct 8 '12 at 17:31

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