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Poisson Morita equivalence

Take $(M, \Pi)$ to be Poisson, $(S, \om)$ symplectic,
$\begin{align*} \char93 _{\Pi} &\: T^*M\to TM \\ \char93 _{\om^{-1}} &\: T^*S\to TS \\ \flat_{\om} &\: TS\to T^*S \end{align*}$
Say we have $\rho\: S\to M$ surjective submersion,

$\begin{displaymath} \rho_{*,p}\: T_pS\to T_{\rho(p)}M. \end{displaymath}$

For $p\in S$ , $x=\rho(p)$ , $\rho^{-1}(x)$ is a closed submanifold. We have
$\begin{align*} \ker(\rho_{*,p}) &= \{ v\in T_p S : \rho_{*,p}(v)=0\} = T_p \rho^... ...\rho^{-1}(x))\\ &= \{X^{\om}_{\rho^*(f)} : f\in \rho^*(\Coo(M))\}. \end{align*}$

$\begin{defn} Two Poisson manifolds $(M_1, \Pi_1)$\ and $(M_2, \Pi_2)$\ form a \t... ...e pair is called \textbf{full} if $\rho_1$, $\rho_2$\ are surjective. \end{defn}$

$\begin{remark} We have seen that in some sense a symplectic realization of $M_1$... ...ed module over $M_1$''. The dual pair is thus a notion of bimodule. \end{remark}$
Our task now is to unravel this definition.
$\begin{prop} % latex2html id marker 5647Let $(S, \om)$\ with $\rho_i\: (S, \om... ...o symplectic orthogonality of tangent spaces if fibers are connected. \end{prop}$

$\begin{proof} \begin{displaymath} \ker((\rho_1)_{*,p}) = T_p\rho_1^{-1}(x) = (T... ...noindent The argument can be reversed provided fibers are connected. \end{proof}$

$\begin{example} Let $S$\ be a symplectic manifold, $J\: S\to \gerg^*$\ constant ... ...tions only after restriction to an open subset $U$\ of $\gerg^*$. \end{example}$

$\begin{prop} Let $(M_i, \Pi_i)$, $i=1,2$, be Poisson manifolds. Let $(S, \om)$ b... ...ic leaves of $M_1$\ and $M_2$, inducing homeomorphism on leaf spaces. \end{prop}$

$\begin{proof} The basic idea is the following. Take a leaf $F_1$\ in $M_1$. Cons... ...h}It remains to show that it is homeomorphism of topological spaces. \end{proof}$

$\begin{lem} Let $f\: S^{(n)}\to M^{(m)}$\ be a submersion, $n\geq m$, $F$ is $(m... ...k)$-dimensional submanifold of $S$\ given by $\phi_1=\hdots=\phi_k=0$. \end{lem}$

$\begin{defn}[\cite{xum}] Two Poisson manifolds are called \textbf{Poisson Morita... ... of $\rho_1, \rho_2$\ are connected, simply connected. \end{enumerate}\end{defn}$

$\begin{displaymath} \xymatrix{ & S \ar[dl]_{\rho_1} \ar[dr]^{\rho_2} & \\ M_1 && M_2 } \end{displaymath}$

$\begin{remark} Despite its name Poisson Morita equivalence is not an equivalence... ...class of objects on which a relation indeed defines an equivalence: \end{remark}$

$\begin{defn} Poisson manifolds Poisson Morita equivalent to themeselves are called \textbf{integrable}. \end{defn}$
Reason for the name is that the associated Lie algebroid can be integrated to a Lie grupoid.
$\begin{prop} Let $(M_1, \om_1)$\ and $(M_2, \om_2)$\ be symplectic manifolds. Th... ...nnected symplectic manifold is Poisson-Morita equivalent to a point. \end{prop}$

$\begin{proof} Let \begin{displaymath} \xymatrix{ & S \ar[dl]_{\rho_1} \ar[dr]^{\... ...ar[dl]_{\rho_1} \ar[dr]^{\rho_2} & \\ M_1 && M_2 } \end{displaymath}\end{proof}$

$\begin{example} Let $(S, \om)$\ be a connected, simply connected symplectic mani... ...he second component composed with the cotangent bundle projection. \end{example}$

$\begin{prop}[Lu-Ginzburg] Poisson-Morita equivalent manifolds have isomorphic fi... ...ology $\rH^1_{\Pi}(-)$, but can have non-isomorphic $\rH^k_{\Pi}(-)$. \end{prop}$

$\begin{remark} With some more work one can prove that the induced map between set of leaves is in fact a heomeomorphism of topological spaces. \end{remark}$

$\begin{remark} The first Poisson cohomology and modular class are Poisson-Morita invariants. \end{remark}$

Pawel Witkowski 2006-06-26