1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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: $$ J(g\cdot p)=Ad^{*}_{g}(J(p))\subset Ad^{*}_{g}(\mathcal{O})=\mathcal{O}. $$

$\endgroup$
1
$\begingroup$

The case with non-null cohomology, when the action of the group on the moment map is not equivariant but affine, an additional term, called Souriau cocycle should be taken into account. See for example: https://www.mdpi.com/1099-4300/22/5/498

$\endgroup$

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.