2
$\begingroup$

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 $M$?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '16 at 7:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.