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Coinduced Poisson structures

Let $\vf\: M_1\to M_2$ be a surjective map. Then if we want it to be Poisson, then $\Pi_2$ is uniquely determined by $\Pi_1$ .
$\begin{defn} A surjective mapping from a Poisson manifold can be Poisson for at ... ...ucture on $M_2$\ is \textbf{coinduced} via $\vf$\ from that on $M_1$. \end{defn}$

$\begin{prop} Let $(M_1, \Pi_1)$\ be a Poisson manifold. If $\vf\:(M_1, \Pi_1)\to... ...ymath}is constant along the fibers of $\vf$\ for all $f,g\in\Coo(M)$. \end{prop}$

$\begin{proof} If $\poiss{\vf^* f}{\vf^*g}_{M_1}$\ is constant then define \begin... ...depend on $\phi$, hence the left hand side is constant along fibers. \end{proof}$

$\begin{example} Consider $\bS^2\xrightarrow{\vf}\bRP^2$\ the projection being gi... ... surjective map $\vf\:\bS^2\to\bRP^2$ you have the same condition. \end{example}$

$\begin{prop} Let $(M_1, \Pi_1)$\ be a Poisson manifold. Let $\vf\: M_1\to M_2$ b... ...iltonian vector fields, then $M_2$\ has coinduced Poisson structure. \end{prop}$

$\begin{proof} Take $f,g\in\Coo(M_2)$. We want to prove that $\poiss{\vf^* f}{\vf... ...\la}(\poiss{\vf^* f}{\vf^*g}_{M_1})=0 \end{displaymath}hence thesis. \end{proof}$

Pawel Witkowski 2006-06-26