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Poisson cohomology

Let us recall the definition of Poisson cohomology. Let $(M, \Pi)$ be a Poisson manifold. Consider the cochain complex $(\gX^k(M), d_{\Pi})$ , where

$\begin{displaymath} d_{\Pi}\: \gX^k(M)\to\gX^{k+1}(M),\quad P\mapsto [\Pi, P], \end{displaymath}$

(1)

where

is the Schouten bracket. Then $d_{\Pi}^2=0$ as a consequence of the graded Jacobi identity together with $[\Pi, \Pi]=0$ . Remark that the Poisson tensor itself always defines a

-cocycle and, thus, a Poisson cohomology class. When $[\Pi]=0$ the Poisson manifold is said to be exact. We would like now to give a different, more explicit expression for this coboundary operator.
$\begin{prop} In the above hypothesis, for all $P\in\gX^k(M)$\ and for all $\al_i... ...s, \Hat{\al_i}, \hdots, \Hat{\al_j}, \hdots, \al_k). \end{displaymath}\end{prop}$

$\begin{proof} % latex2html id marker 4093Let us first remark that the formula ... ...d is decomposable) tha claim holds true for all $k$--vector fields. \end{proof}$

$\begin{remark} From this expicit expression it would be tempting to say that the... ...is is one of the reasons why such cohomology is seldom considered. \end{remark}$

$\begin{remark} Let $f$\ be a Casimir function for $(M, \Pi)$. Let $P\in\gX^k(M)$... ...ucture of $\rH^0_{\Pi}(M)=\Cas(M)$-module on each $\rH_{\Pi}^k(M)$. \end{remark}$

$\begin{prop} The external product of multivector fields induces an associative a... ...displaymath}this product will be called the \textbf{Poisson product}. \end{prop}$

$\begin{proof} \begin{displaymath}[\Pi, P\land Q]=[\Pi, P]\land Q + (-1)^{p-1}P\l... ...erties are a trivial consequence of analogous properties of $\land$. \end{proof}$

$\begin{remark} In a similar way it is easy to verify that also the Schouten brac... ...property is connected to the Jacobi identity for $[[\Pi, \Pi], Q]$. \end{remark}$

$\begin{remark} $\rH^k_{\Pi}$\ is not functorial. In fact given a Poisson map $\v... ...ked already, only the weaker notion of $\vf_*$-relatedness survive. \end{remark}$

$\begin{thm} Let $(M, \Pi)$\ be a Poisson manifold. The sharp map intertwines the... ...rmore, if $M$\ is symplectic then $\char93 _{\Pi}$\ is an isomorphism. \end{thm}$

$\begin{proof} Here $\char93 _{\Pi}$\ is extended to $k$-forms as \begin{displaym... ...ble at the chain level and therefore it remains such on cohomology. \end{proof}$

$\begin{prop} Let $\gerg$\ be a Lie algebra, and $\gerg^*$\ the dual vector space... ...here on the left $\rH^k_L$\ is the Lie algebra cohomology of $\gerg$. \end{prop}$

$\begin{remark} To complete the list of basic examples consider that if $(M, 0)$ ... ...(M)$--modules is not always satisfied by Poisson cohomology groups. \end{remark}$

$\begin{thm}[Mayer-Vietoris sequence for Poisson cohomology] Let $(M, \Pi)$\ be a... ...\xrightarrow{\del} \rH^{k+1}_{\Pi}(U\cup V)\to \hdots \end{displaymath}\end{thm}$

$\begin{proof} Here one basically recalls how the proof of Mayer-Vietoris theorem... ...$. The usual arguments, based on the snake lemma, prove the theorem. \end{proof}$

Subsections

Pawel Witkowski 2006-06-26