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A moment map $\mu$ is defined when one has a Hamiltonian $G$-action on a symplectic manifold $M$, for some Lie group $G$. My question is, what are the geometric interpretations of the kernel and critical points of the moment map?

For the kernel $\mu^{-1}(0)$, I've read here https://en.m.wikipedia.org/wiki/Moment_map that it is invariant under $G$. If one assumes 0 is a regular point and that $G$ acts freely on the kernel, then $\mu^{-1}(0)/G$ is the symplectic quotient $M//G$. But how do we know that the kernel is a manifold?

I've also read that the critical points of the moment map (which are the points where all its derivatives vanish) correspond to fixed points of the group action when $G$ is a product of circle actions. Is this true for nonabelian $G$?

Meer Ashwinkumar
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  • I'm not familiar with this subject, but is $0$ a regular value of $\mu$? If so, then it follows that $\mu^{-1}(0)$ is a dimension $0$ submanifold of $G$. – Alex G. Sep 19 '15 at 12:27
  • Yes, it is a regular point, how do we show that it is a dimension 0 submanifold? – Meer Ashwinkumar Sep 19 '15 at 12:32
  • I looked at the definition of $\mu$ again and I made a mistake in my previous comment. $\mu^{-1}(0)$ will have codimension $\dim G$ in $M$ by the following theorem from elementary differential topology: If $f: M\to N$ is a smooth map between manifolds of dimensions $m$ and $n$, respectively, and $y\in N$ is a regular value of $f$, then $f^{-1}(y)$ is a submanifold of $M$ of dimension $m-n$. – Alex G. Sep 19 '15 at 12:36
  • Thanks for the explanation. What if 0 is not a regular point, can we still say anything about the dimension and other properties of the kernel? – Meer Ashwinkumar Sep 19 '15 at 12:44
  • To be very careful, the condition is that $0$ is a regular value, which means that every point in $\mu^{-1}(0)$ is a regular point. If this is not the case then there is no guarantee that $\mu^{-1}(0)$ is a submanifold, so you would only be able to view it as a $G$-set (assuming that it is preserved by the action of $G$). – Alex G. Sep 19 '15 at 12:48
  • Thanks. This may seem random, but do you know on any relation between the kernel and Lagrangian submanifolds of $M$? – Meer Ashwinkumar Sep 19 '15 at 12:54

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