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The symplectic foliation

$\begin{defn} Let $S$\ be a distribution on a manifold $M$. An \textbf{integral} ... ...mersed submanifold $N$\ of $M$\ such that $(N, i_N)$\ is an integral. \end{defn}$

$\begin{remark} Integral manifolds are immersed submanifolds but not necessarily ... ...$). In particular integral submanifolds are not necessarily closed. \end{remark}$

$\begin{defn} A distribution is \textbf{fully integrable} if for every $x\in M$\ ... ...gral $(N, h)$\ of $S$\ such that $x\in h(N)$ (maximal at each point). \end{defn}$
Frobenius theorem: A constant rank differentiable distribution is fully integrable if and only if it is involutive, i.e. for all

-sections of

, $[X, Y]\in S$ .

For a regular Poisson manifold $(M, \Pi)$ of rank , the characteristic distribution is of constant rank, and also involutive, due to $[X_f, X_g]=X_{\poiss{f}{g}}$ . Therefore it is fully integrable. Each regular Poisson manifold thus, comes equipped with a regular foliation.

Furthermore on the tangent space to the leaf $(\im\char93 _{\Pi})_x$ it is always possible to define a natural antisymmetric nondegenerate bilinear product.

Computations similar to those of proposition , allow to prove that if is the maximal integral containing , then there is a symplectic 2-form $\om_N$ on such that $\om_N=h^*\om_{\Pi}$ , where $(\om_{\Pi})_x\in\La^2(\im \char93 _{\Pi})^*_x$ is determined by the above bilinear product.

We will call this foliation the symplectic foliation of .
$\begin{example} $M=\bR^{2n+p}$, $\Pi=\sum_{i=1}^n\del_{q_i}\land\del_{p_i}$. The... ....} \end{displaymath}On each leaf $\om=\sum_{i=1}^ndq_i\land dp_i$. \end{example}$
Remark: Not every foliation is a symplectic foliation. In fact, first of all leaves need to carry a symplectic structure, a condition which already puts some topological restriction (e.g. you cannot have symplectic structure on $\bS^{2n}$ if ). Furthermore more delicate obstructions depend on how the symplectic forms vary from leaf to leaf [].

Now we want to generalize this statement to non regular Poisson manifolds.
$\begin{thm}[Weinstein's splitting theorem] Let $(M, \Pi)$\ be a Poisson manifold... ...ends only on coordinates $y_1, \hdots, y_k$ and $\varphi_{ij}(x_0)=0$. \end{thm}$

$\begin{proof} Induction on $n$, $\rho_{\Pi}(x_0)=2n$. \noindent If $n=0$\ there ... ...right summand which is a Poisson bivector on $M$\ of rank $2(n-1)$. \end{proof}$
In the symplectic case this theorem recovers a well-known result:
$\begin{cor}[Darboux theorem] Let $(M, \om)$\ be a symplectic manifold and $x_0\i... ...laymath} \om\vert _U = \sum_{i=1}^n dq_i\land dp_i. \end{displaymath} \end{cor}$
In analogy with this last statement coordinates generated by the splitting theorem are also called Darboux coordinates centered at .
$\begin{example} Let $\gerg^*$\ be the dual of a Lie algebra $\gerg$\ and $\Pi=\s... ...re is no symplectic term and all functions $\vf_{ij}$\ are linear. \end{example}$

$\begin{defn} Let $(M, \Pi)$\ be a Poisson manifold. A \textbf{hamiltonian path} ... ...a piecewise Hamiltonian curve $\ga$\ on $M$\ connecting $x$\ to $y$. \end{defn}$

$\begin{lem} Hamiltonian relation is an equivalence relation. \end{lem}$

$\begin{proof}(exercise) Reflexive - trivial; symmetric - change backwards the ti... ...nsitive - concatenation of Hamiltonian paths is an Hamiltonian path. \end{proof}$

$\begin{defn} Connected components of equivalence classes of this relation are called \textbf{symplectic leaves} of $(M, \Pi)$. \end{defn}$

$\begin{prop} Each symplectic leaf is a maximal integrable submanifold of $(M, \Pi)$. \end{prop}$

$\begin{proof}(sketchy) Let $F$\ be a leaf, $x\in F$, $T_x F\subset \im\char93 _{... ...en by $y_1=\hdots=y_p=0$\ therefore $F$\ is an immersed submanifold. \end{proof}$

$\begin{prop} Let $(M, \Pi)$\ be a Poisson manifold. On each symplectic leaf $F$\... ...structure such that the inclusion map $i\: F\to M$\ is a Poisson map. \end{prop}$

$\begin{proof} Let $x\in F$\ and $F_x$\ has splitting coordinate neighbourhood $(... ...isson, it is enough to check it on brackets of coordinate functions. \end{proof}$

$\begin{examples} \mbox{} \begin{enumerate} \item Each symplectic manifold is a s... ...m, called the Kirillov-Kostant-Souriau form (KKS). \end{enumerate}\end{examples}$

$\begin{prop} Let $(M, \Pi)$\ be a Poisson manifold. Casimir functions are consta... ...ves are contained in connected components of level sets of Casimirs). \end{prop}$

$\begin{proof} Let $F$\ be a leaf, $f$\ a Casimir function. We want to prove that... ...laymath} X_g f=\{g,f\}=0\quad\forall\; g\in\Coo(M). \end{displaymath}\end{proof}$

$\begin{prop} Let $(M, \Pi)$\ be a Poisson manifold, $\dim M=m$. Let $x_0\in M$\ ... ...\Pi}(x_0)=m$. Then the symplectic leaf through $x_0$\ is open in $M$. \end{prop}$

$\begin{proof} Around $x_0$\ rank is constant and equal $\dim M$. Therefore $(\im... ...path on $M$ exiting $x_0$\ is locally Hamiltonian, hence the thesis. \end{proof}$

$\begin{prop} Let $(M, \Pi)$\ be a Poisson manifold. Let $f_1, \hdots, f_p$ be Ca... ...\ and is smooth, then its connected components are symplectic leaves. \end{prop}$
Remark: But $\Ga_{1,c}\cap\hdots\cap\Ga_{p, c}$ may have $\dim\neq\rank \Pi_0$ .

Let us consider a general polynomial Poisson bracket on $\mathbb R^n$ , i.e. a Poisson bracket such that $\poiss{x_i}{x_j}=P_{ij}(x_1,\ldots,x_n)$ . A function $f\in \Coo(\bR^n)$ is a Casimir function if and only if $\poiss{x_i}{f}=0$ for any $i=1,\dots, n$ . This can be rewritten as $\sum_j P_{ij}\del_{x_j}f=0$ . Therefore has to be a smooth solution of a system of linear first order PDE's. If, as in this case, we are considering a linear Poisson structure the system has constant coefficients (which are the structural constant of the Lie algebra) and its solutions can be explicitly determined.
$\begin{example} Consider $\gsu(2)^*\cong \bR^3$. Its Lie-Poisson bivector is \be... ...nique $\SU(2)$\ invariant volume form on the sphere of radius $r$. \end{example}$

$\begin{example}(Natsume-Olsen Poisson sphere) \begin{displaymath} \bS^2\subset \... ... \zeta=\bar{\zeta}=0,\quad z=1 & \text{ north pole.} \end{align*}\end{example}$

Next: Schouten-Nijenhuis bracket Up: Poisson Geometry Previous: The sharp map Contents

Pawel Witkowski 2006-06-26