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Coisotropic submanifolds

Let $(M, \Pi)$ be a Poisson manifold,

a submanifold of

and

its conormal bundle defined as:

$\begin{displaymath} N^*C=\{\al\in T^*M : \scalar{\al}{v}=0 \quad\forall\; v\in TC\}. \end{displaymath}$

$\begin{defn} $C$\ is called \textbf{coisotropic sumbanifold} of $M$\ if \begin{displaymath} \char93 _{\Pi}(N^* C)\subseteq TC \end{displaymath}\end{defn}$

$\begin{remark} On symplectic manifolds, for a submanifold $N$\ of $M$\ you consi... ...TN\supseteq TN^{\perp\om} & \quad \textbf{coisotropic}. \end{align*}\end{remark}$

$\begin{exer} Prove that if $(M, \Pi)$\ is the Poisson manifold associated to a s... ...\subseteq TC \word{iff} (TC)^{\perp\om}\subseteq TC. \end{displaymath}\end{exer}$

$\begin{prop} The following are equivalent \begin{enumerate} \item $C$\ is coisot... ...h that $f\vert _C=0$, $X_f\vert _C$ is tangent to $C$. \end{enumerate}\end{prop}$

$\begin{proof} The point here is that if $I=\{f\in\Coo(M) \vert f\vert _C=0\}$\ t... ... we have easily (3) $\implies$\ (1) $\implies$\ (2) $\implies$\ (3). \end{proof}$

$\begin{remark} \mbox{} \begin{itemize} \item If $C$\ is Poisson submanifold then... ...$C$\ is coisotropic then $I$\ is a Poisson subalgebra. \end{itemize}\end{remark}$

$\begin{exer} Let $\gh$\ be a Lie subalgebra in $\gerg$. Prove that $\gh^{\perp}$\ is a coisotropic submanifold in $\gerg^*$. \end{exer}$

$\begin{thm} $\vf\: (M_1, \Pi_1)\to (M_2, \Pi_2)$\ is a Poisson map if and only i... ...end{displaymath}is a coisotropic submanifold of $M_1\times \Bar{M_2}$. \end{thm}$
The notation:

$\begin{displaymath} M_1\times \Bar{M_2} = (M_1\times M_2, \Pi_1\oplus (-\Pi_2)) \end{displaymath}$

with product Poisson structure.
$\begin{proof} We have \begin{displaymath} T_{(x, \vf(x))}\Ga_{\vf}=\{(v, \vf_{*,... ...playmath}which is one of the conditions equivalent to being Poisson. \end{proof}$

$\begin{defn} Let $C$\ be a coisotropic submanifold of $(M, \Pi)$ and let $I:=\{ ... ...h} N(I):=\{g\in \Coo(M) : \poiss{g}{I}\subseteq I\}. \end{displaymath}\end{defn}$

$\begin{prop} $N(I)$\ is a Poisson subalgebra of $\Coo(M)$, $I$\ is a Poisson ideal of $N(I)$\ and therefore $N(I) / I$\ is a Poisson algebra. \end{prop}$

$\begin{proof} From the Jacobi identity we get the first part: \begin{displaymath... ...ar93 _{\Pi}N^*C)\}\subseteq \text{Poisson manifold} \end{displaymath}\end{proof}$

$\begin{prop} A submanifold $C$\ is coisotropic if and only if $f\vert _C=0$ and $g\vert _C=0$\ implies $\poiss{f}{g}\vert _C=0$. \end{prop}$

$\begin{remark} Is it true that $C$\ is a coisotropic submanifold of $M$\ if and ... ...isotropy to coisotropy in the leaves (see for example \cite{vai}). \end{remark}$

Subsections

Pawel Witkowski 2006-06-26