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?


1 Answer 1


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 $\Phi^{-1}(G\alpha)$.

  • $\begingroup$ 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

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