Let $(M, \{.,.\})$ be a Poissonmanifold and $B$ the corresponding Poissontensor. Now in this context, a embedded submanifold $C \subset M$ is called coisotropic, if $B^\#(TC^\circ) \subset TC$.

For embedded submanifolds it is just the same as to say:

for $f,g \in C^\infty(M)$ with $f|_C= g|_C=0$ we have $\{f,g\}|_C=0$.

Now having a Poissonbracket, I'll restrict myself to a symplectic leaf $S \subset M$, so we get a symplectic form $\omega$ on $S$.

Now if $C$ is coisotropic as submanifold of $S$ in the symplectic case, do I know, that $C$ is also coisotropic in $M$ in the Poisson-case?

In my special case, I'm working on the dual of a liealgebra (so $\mathfrak{g}$ is the liealgebra and $\mathfrak{g}^*$ the dual) and have the Lie-Poisson-bracket, i.e. $\{f,g\}(\alpha) = <\alpha, [d_\alpha f, d_\alpha g]>$ with $\alpha \in \mathfrak{g}^*$, the canonical identification of $(\mathfrak{g}^*)^* \cong \mathfrak{g}$ and $[.,.]$ the Liebracket on $\mathfrak{g}$.

If $G$ is a Liegroup with that Liealgebra $\mathfrak{g}$, then the coadjoint-orbits on $\mathfrak{g}^*$ are symplectic manifolds. So if $S$ is such an orbit, and $C$ is a submanifold of this orbit, which is a coisotropic submanifold of $S$, does that imply, that $C$ is a coisotropic submanifold of $M$?


1 Answer 1


The answer is 'Yes', at least with some additional transversality conditions.

Corollary (1.2.6) in Weinstein's Coisotropic calculus and Poisson groupoids states that

Let $M$ be a submanifold of the Poisson manifold $P$ which has clean intersection with each symplectic leaf of $P$. Then $M$ is coisotropic if and only if its intersection with each symplectic leaf is coisotropic in $P$, or, equivalently, in the symplectic leaf.

  • $\begingroup$ And assuming that $S \subset M$ is a embedded submanifold and $C$ is an embedded submanifold of $S$, the transversality conditions should be fullfilled, don't they? $\endgroup$
    – Feanoris
    Jun 7, 2016 at 7:39

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