Next: Poisson homology Up: Poisson cohomology Previous: Modular class Contents

Computation for Poisson cohomology

Let us consider the quadratic Poisson structure on $\bR^2$

$\begin{displaymath} \Pi_0(x,y)=(x^2+y^2)\del_x\land\del_y. \end{displaymath}$

We want to prove the following
$\begin{prop}[Ginzburg] The Poisson cohomology of $(\bR^2, \Pi_0)$\ is given by \... ...0}(\bR^2) &= \bR\langle \del_x\land\del_y, \Pi_0\rangle. \end{align*}\end{prop}$

$\begin{proof} % latex2html id marker 4386To make computations easier let us id... ...oefficients and has isolated singular points (this can be weakened). \end{proof}$

$\begin{example} Let $\SU(2)$\ have Poisson structure we already mentioned. The a... ...aves - the north and south pole, and complement which is a 2-leaf. \end{example}$

$\begin{example} Take the stereographic projection from the south pole, i.e. \beg... ...1_{\Pi}(\bS^2)$ is nontrivial then $\rH^1_{\Pi}(\bS^2)\simeq \bR$. \end{example}$

Pawel Witkowski 2006-06-26