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Poisson maps

Recall that if $(M_1, \Pi_1)$ , $(M_2, \Pi_2)$ are Poisson manifolds then $\varphi\: M_1\to M_2$ is a Poisson map if $\varphi^*\:\Coo(M_2)\to \Coo(M_1)$ verifies

$\begin{displaymath} \varphi^*\poiss{f}{g}_{M_2}=\poiss{\varphi^* f}{\varphi^* g}_{M_1} \end{displaymath}$

for all $f, g\in\Coo(M_2)$ . Here $\varphi^*\: f\mapsto f\circ \varphi$ .

Recall also from differential geometry that having a map $\varphi\: M\to N$ , you can pull-back forms, but in general you cannot push-forward vector fields.

Let $X\in\gX(M)$ and $Y\in\gX(N)$ . Then the two vector fields and are said to be $\varphi$ -related if

$\begin{displaymath} \varphi_{*,x}(X_x)=Y_{\varphi(x)} \end{displaymath}$

for all $x\in M$ , $\varphi_{*,x}\: T_x M\to T_{\varphi(x)}N$ . This relation does not define a map. In fact you can have more than one vector field on

related to a fixed vector field on

. (Think of $\varphi\:\bR^2\to\bR^2$ , $\varphi(x,y)=(x,0)$ . Then saying that

is $\varphi$ -related to

says something only about values of

on the line

.) You can also have none. (In the example as before if $X_{(x,0)}$ and $X_{(x,1)}$ have different projections on $\im\varphi_{*,x}$ ).

If $\varphi$ is a diffeomorphism then $\varphi_*\:\gX(M)\to\gX(N)$ . Remark that you can define $\varphi$ -relation on multivectors simply by considering $\varphi_{*, x}^{\land}$ .
$\begin{prop} Let $(M_1, \Pi_1)$, $(M_2, \Pi_2)$\ be Poisson manifolds and $\var... ... $\al,\bt\in T^*_{\varphi(x)}M_2$\ and all $x\in M_1$. \end{enumerate}\end{prop}$

$\begin{proof} \mbox{} $(3)\Longleftrightarrow (4)$\ by definitions. \par \noinde... ...playmath} \poiss{f}{g} = X_f g = \bracket{X_f}{dg}. \end{displaymath}\end{proof}$

$\begin{remark} From property (3): \begin{displaymath} \rho_{\Pi_1}(x)\geq \rho_{... ...son maps between symplectic manifolds are submersions. \end{itemize}\end{remark}$
This shows that being a Poisson map between symplectic manifolds is very different from being a symplectic map (which means $\varphi\: M_1\to M_2$ , $\varphi^* \om_2=\om_1$ ). This difference is made explicit by the following two examples.
$\begin{example} \begin{align*} i & \: \bR^2\to \bR^4, \\ (q_1, p_1) & \mapsto (... ...\underbrace{\poiss{q_2\circ i}{p_2\circ i}}_{=0}. \end{displaymath}\end{example}$

$\begin{example} \begin{align*} \psi & \: \bR^4\to \bR^2, \\ (q_1, p_1, q_2, p_2... ...\land dp_1 \neq dq_1\land dp_1 + dq_2\land dp_2 . \end{displaymath}\end{example}$
This difference between morphisms in the Poisson and symplectic categories implies, obviously, that related concepts such as subobjects and quotients have different behaviours. We will see later an example of this issue when referring to submanifolds.
$\begin{prop} Let $(M_i, \Pi_i)$, $i=1,2,3$\ be Poisson manifolds. Let $\varphi\... ...and a diffeomorphism, then $\varphi^{-1}$\ is Poisson. \end{enumerate}\end{prop}$

$\begin{proof} \mbox{} \begin{enumerate} \item Obvious. \item Take $y\in M_2$, $... ..., y}. \end{align*}\item Follows immediately from (2). \end{enumerate}\end{proof}$

$\begin{examples} \mbox{} \begin{enumerate} \item Let $\phi\: \gh \to \gerg$\ be ... ...) \end{displaymath}is a Lie algebra homomorphism. \end{enumerate}\end{examples}$

$\begin{defn} Let $(M, \Pi)$\ be a Poisson manifold. A \textbf{Poisson vector fie... ...induces for all $t\in\bR$\ a local Poisson morphism $\vf_t\:M\to M$. \end{defn}$

$\begin{prop} Let $(M, \Pi)$\ be a Poisson manifold, and $X\in\gX(M)$. The follow... ...Xf, g\}+\{f, Xg\} \end{displaymath}\item $L_X\Pi = 0$. \end{enumerate}\end{prop}$

$\begin{proof} We have \begin{align*} L_X(\Pi(df, dg)) &= (L_X\Pi)(df, dg)+\Pi(L_... ... hand side is 0, and 2) $\Longleftrightarrow$\ right hand side is 0. \end{proof}$

$\begin{remark} \mbox{} \begin{enumerate} \item From $L_{[X, Y]}=[L_X, L_Y]$\ it ... ...\mathrm{std})$\ we have $\Ham(M)= \Poiss(M)=\gX(M)$. \end{enumerate}\end{remark}$

$\begin{defn} Let $d_{\Pi}\: \gX^k(M)\to \gX^{k+1}(M)$, $\Pi\mapsto [\Pi, P]$. Then $d_{\Pi}$\ is called the \textbf{Lichnerowicz coboundary}. \end{defn}$

$\begin{remark} If $\Pi$\ is Poisson, then $d_{\Pi}^2=0$. In fact $[\Pi, [\Pi, P... ...[[\Pi, \Pi], P]$\ from the Jacobi identity of the Schouten bracket. \end{remark}$

$\begin{defn} The cohomology of the complex $(\gX^{\bullet}(M), d_{\Pi})$ is call... ...owicz) cohomology} of $(M, \Pi)$\ and is denoted by $\rH^k_{\Pi}(M)$. \end{defn}$
We have $\rH^0_{\Pi}(M)=\Cas(M)$ , $[\Pi, f]=X_f$ , and $\rH^1_{\Pi}(M)=\Poiss(M)/ \Ham(M)$ .

Next: Poisson submanifolds Up: Poisson maps Previous: Poisson maps Contents

Pawel Witkowski 2006-06-26