1
$\begingroup$

I worked now some time with coisotropic actions of Liegroups on manifolds. But there is one key fact, that I don't understand, although it is very central in my considerations.

Let $(M,\omega)$ be a symplectic manifold and $G$ a connected Liegroup acting on a connected manifold $M$ by symplectomorphisms and lets assume we have a momentum map $$\Phi \colon M \to \mathfrak{g}^*,$$ which is $G$-equivariant w.r.t. to the $G$-action on $M$ and the coadjoint-action on $\mathfrak{g}^*$.

Let $\mathcal{O}$ be a coadjoint $G$-orbit in $\Phi(M)$. Assuming that $\Phi$ has clean intersection with $\mathcal{O}$ (i.e. $\Phi^{-1}(\mathcal{O})$ is a submanifold of $M$ and $T_x \Phi^{-1}(\mathcal{O})=(d_x\Phi)^{-1}(T_\alpha \mathcal{O})$) I understand that the orbit $G.x$ is coisotropic for all $x \in \Phi^{-1}(\mathcal{O})$, iff $G$ acts locally transitively on $\Phi^{-1}(\mathcal{O})$.

Now Guillemin & Sternberg always work with compact Liegroups in their book "Symplectic techniques in Physics". So to understand this problem, I want to focus on $G$ compact.

They now say, that $G.x$ is coisotropic for some open and dense subset $\Sigma \subset M$, iff $G$ acts locally transitively on $\Phi^{-1}(G.\alpha)$ for generic orbits $G.\alpha$ in $\Phi(M)$.

I understand this last fact as: The set $\Theta \subset \mathfrak{g}^*$ defined as $$\Theta := \{ \alpha \in \Phi(M) \ | \ G \text{ acts locally transitively on } \Phi^{-1}(G \cdot \alpha)\}$$ is open and dense in $\Phi(M)$, w.r.t. the subspace topology in $\mathfrak{g}^*$.

But I really don't understand why the openness and denseness of $\Sigma$ (the set of coisotropic orbits) is equivalent to the openness and denseness of $\Theta$ (the set of orbits $G. \alpha$ in $\Phi(M)$, such that $G$ acts locally transitively on $\Phi^{-1}(G\alpha)$).

Could someone give a detailed explanation, why this could be true?

$\endgroup$

1 Answer 1

3
$\begingroup$

One of the defining properties of the moment map is $$ \omega(\xi,\eta x)=\langle d\Phi_x(\xi),\eta)\text{ for all } x\in M, \xi\in T_xM,\eta\in\mathfrak g. $$ This implies readily $$ \mathfrak gx=(\ker d\Phi_x)^\perp. $$ That the orbit $Gx$ is coisotropic means $(\mathfrak g x)^\perp\subseteq\mathfrak gx$ and is therefore equivalent to $\ker d\Phi_x\subseteq\mathfrak gx$. The set of (non-critical) points $x$ with $\ker d\Phi_x=T_x\Phi^{-1}(\alpha)$, $\alpha:=\Phi(x)$, is open and dense in $M$. In these points, coisotropy of $\mathfrak gx$ is equivalent to $T_x\Phi^{-1}(\alpha)\subseteq\mathfrak gx$ or, because of $\mathfrak g_\alpha x\subseteq T_x\Phi^{-1}(\alpha)$, to $$ T_x\Phi^{-1}(\alpha)=\mathfrak g_\alpha x. $$ This is precisely the property that $G_\alpha$ acts locally transitively on $\Phi^{-1}(\alpha)$. This, in turn, means that $G$ acts locally transitively on $\Phi^{-1}(G\alpha)$.

$\endgroup$
2
  • $\begingroup$ So we don't require, that the set $\Theta$ defined in my question, has to be an open and dense subset in $\Phi(M)$? It means only, that the set $N = \{ x \in M \ | \ \Phi(x) \in \Theta\}$ has to be open and dense in $M$, because that's "exactly" the set of points, such that the orbits through this points are coisotropic (modulo some critical points)? $\endgroup$
    – Olorin
    Jun 5, 2016 at 14:19
  • $\begingroup$ That's correct. By the way, if $G$ is connected and compact and $M$ is compact then $\Theta=M$. This follows from Kirwan's connectedness theorem. Thus, $\Theta$ may be bigger than $\Sigma$. $\endgroup$ Jun 5, 2016 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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