Let $(M, \omega)$ be a symplectic manifold and let $G$ be a compact, connected Lie group acting on it. Let $J: M \rightarrow \mathfrak{g}^{*}$ be the moment map. Assume that $\eta$ is a regular value of $J$, and let $\mathcal{O}$ be the orbit of $\eta$ under the coadjoint action.
Problem: Prove that if $\zeta \in \mathcal{O} \subset \mathfrak{g}^{*}$, then $\zeta$ is a regular value.
Attempt: I know that $\zeta$ is a regular value iff $\forall p \in J^{-1} (\zeta)$, we have $\mathfrak{g}_p = 0$ where $\mathfrak{g}_p$ is the isotropy subalgebra. This is equivalent to saying $\mathfrak{g}_p^{0} = \mathfrak{g}^{*}$, where $\mathfrak{g}_p^{0}$ denotes the annihilator. I don't understand how to show that $\mathfrak{g}_p = 0$?
Best Answer
This is a consequence of the map $J:M\rightarrow\mathfrak{g}^{*}$ being equivariant. I will denote by $\phi_{g}:M\rightarrow M$ and $Ad_{g}^{*}:\mathfrak{g}^{*}\rightarrow\mathfrak{g}^{*}$ the $G$-actions on $M$ and $\mathfrak{g}^{*}$, respectively. Equivariance of $J:M\rightarrow\mathfrak{g}^{*}$ means that $$ Ad_{g}^{*}\circ J = J\circ\phi_{g}\hspace{1cm}\forall g\in G. $$ Choose $p\in J^{-1}(\zeta)$. Since $\zeta$ lies in the orbit of $\eta$, we there exists $g\in G$ such that $Ad_{g}^{*}(\zeta)=\eta$. Taking differentials in the above equality, we get a commutative diagram
$\require{AMScd}$ \begin{CD} T_{p}M @>d_{p}J>> \mathfrak{g}^{*}\\ @V d_{p}\phi_{g} V V @VV Ad_{g}^{*} V\\ T_{\phi_{g}(p)}M @>>d_{\phi_{g}(p)}J> \mathfrak{g}^{*} \end{CD}
Note that $J(\phi_{g}(p))=Ad_{g}^{*}(J(p))=Ad_{g}^{*}(\zeta)=\eta$. Since $\eta$ is a regular value, the bottom map in the diagram is surjective. Since the vertical maps are isomorphisms, also the top map is surjective.