Problem Let $(M, \omega)$ be a symplectic manifold. Let $G$ be a connected, compact Lie group acting on $M$. Let $J: M \rightarrow \mathfrak{g}^{*}$ be the moment map. Let $\eta$ be a regular value of $J$ and let $\mathcal{O}$ be the orbit of $\eta$ under the coadjoint action, i.e. $\mathcal{O} = \left\{Ad_{g^{-1}}^{*} \eta \mid g \in G \right\}$.
Prove that $i: J^{-1} (\mathcal{O}) \rightarrow M$ is a coisotropic submanifold.
Attempt: Let $p \in J^{-1} (\mathcal{O})$. Then by definition, I have to show that $T_p (J^{-1} (\mathcal{O}))^{\omega} \subset T_p (J^{-1} (\mathcal{O}))$.
I know the following, that $T_q (J^{-1} (\eta))^{\omega} = T_q (G \cdot q)$ where $G \cdot q = \left\{ \Phi(g,q) \mid g \in G \right\}$ is the orbit and $q \in J^{-1} (\eta)$. Also, by standard differential geometry, since $\eta$ is a regular value, we have $T_q J^{-1} (\eta) = \text{ker} (T_q J)$.
Also, there is a result that $$ T_q (G_{\eta} \cdot q) = T_q (G \cdot q) \cap T_q(J^{-1} (\eta)). $$ It does not follow that $J^{-1} (\mathcal{O})$ is a coistropic submanifold of $M$, if I would prove that for every $\zeta \in \mathcal{O}$, the inverse image $J^{-1} (\zeta)$ is a coisotropic submanifold (which might be false, not sure)?
So how do I figure out what $T_p (J^{-1} (\mathcal{O}))^{\omega}$ is?
Best Answer
1) You should first argue why $J^{-1}(\mathcal{O})\subset M$ is a submanifold in the first place. To do so, it is enough to remark that $J:M\rightarrow\mathfrak{g}^{*}$ is transverse to $\mathcal{O}\subset\mathfrak{g}^{*}$, i.e. for all $q\in J^{-1}(\mathcal{O})$ we have $$ d_q J(T_q M)+T_{J(q)}\mathcal{O}=\mathfrak{g}^{*}. $$ This is the case because $J(q)\in\mathcal{O}$ is also a regular value, so that $d_q J(T_q M)=\mathfrak{g}^{*}$.
2) Now let $p\in J^{-1}(\mathcal{O})$ and assume that $J(p)=\zeta\in\mathcal{O}$, i.e. $p\in J^{-1}(\zeta)$. Since $T_{p}J^{-1}(\zeta)\subset T_{p}J^{-1}(\mathcal{O})$, we have $$ (T_{p}J^{-1}(\mathcal{O}))^{\omega}\subset (T_{p}J^{-1}(\zeta))^{\omega}=T_{p}(G\cdot p). $$ So in order to conclude, it is enough to show that $G\cdot p\subset J^{-1}(\mathcal{O})$. This inclusion holds because $J$ is equivariant: $$ J(g\cdot p)=Ad^{*}_{g}(J(p))\subset Ad^{*}_{g}(\mathcal{O})=\mathcal{O}. $$