Let $G$ be a compact Lie group which acts by symplectomorphisms on a symplectic manifold $(M,\omega)$. Futhermore let $\mu \colon M \to \mathfrak g$ be a moment map for this action. Denote by $\Omega = \mu^{-1}(0)$ and $M_{\rm red}$ the reduced space $\Omega/G$ (we do not assume that $0$ is a regular value). It is known that $G$ has a principal orbit $M_{(H)}$ in $M$ which is open and dense. Let $(H)$ be its orbit type for $H$ a closed subgroup of $G$.

On the other side following Theorem 5.9 (together with Remark 5.10) of Sjamaar and Lerman every connected component of $M_{\rm red}$ has a unique open and dense piece $(M_{red})_{(\tilde H)} = (M_{(\tilde H)} \cap \Omega)/G$ for $\tilde H$ another closed subgroup of $G$.

Question: Are $(H)$ and $(\tilde H)$ of the same orbit type? If not is there an example such that $(H)\neq (\tilde H)$?

  • $\begingroup$ Already the action of $G=U(1)$ on $M=\mathbb{C}$ with moment map $\mu(z)=\vert z\vert^2$ seems to be a counterexample. Then $H=1$ while $\tilde H=G$. Am I missing something? $\endgroup$ – Friedrich Knop Apr 20 '16 at 8:43
  • $\begingroup$ $(\tilde{H})$ should be the smallest orbit type for $M$ whose orbit type subset $M_{(\tilde{H})}$ has non-empty intersection with $\mu^{-1}(0)$. $\endgroup$ – Tobias Diez Apr 20 '16 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.