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Let $(M, \omega)$ be a symplectic manifold and $G$ be a compact Lie group. Suppose we have a Hamiltonian $G$-action on $M$, with moment map $\mu: M \to {\mathfrak g}^*$.

We assume that the moment map is proper in case $M$ is noncompact.

The question is: for any loop $\gamma: S^1 \to G$, and a point $x\in M$, is the loop $t\mapsto \gamma(t) x$ a contractible loop in $M$? So we assume neither $G$ or $M$ is simply-connected.

We may assume that $\gamma$ is actually a 1-parameter subgroup of $G$, generated by a vector $\xi \in {\mathfrak g}$. In the case $M$ is compact, we restrict the moment map to this subgroup, which is equivalent to a real valued function $\mu_\gamma$. Then the gradient flow of this function should push the loop to a critical point, which is a fixed point of this subgroup. Hence this shows that the loop is contractible.

Now if $M$ is noncompact, the gradient flow doesn't necessarily converge to a critical point (could escape to $\pm \infty$). Note that the real valued function $\mu_\gamma$ is not necessarily proper. So the above method fails. But I still guess that the loop should be contractible.

Is there any proof or counter-example? Or should we add some conditions to guarantee this?

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There are counter-examples, hope they answer your question completely, just take any non-simply connected $G$ and consider its action on $T^*G$. The simplest case is:

Let $M$ be the cylinder $S^1\times \mathbb R$ with the symplectic form $ds \wedge dt$. Then the Hamiltonial $H=t$ defines an $S^1$-action on the cylinder.

One more counterexamle. Consider just the action of $SO(3)$ on its cotangent space. Clearly this action is Hamiltonian. Let us take the subgroup $S^1\subset SO(3)$ that represents the non-zero element of $\pi_1(SO(3))$. Obviously all the orbits of the action of this $S^1$ on $T^*(SO(3))$ will not be contractible.

So we see that in the case the Lie group is not simply-connected it always admits a "bad" action.

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Thanks. But if we consider moment maps which are like "quadratic functions", for general compact Lie group, is there still counter-examples? Or is there some other conditions to guarantee the similar situation in the compact case (i.e., the convergence of gradient flow)? –  Guangbo Xu Dec 29 '12 at 15:35
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Dear Guangbo, unfortunately I don't undrestand what you mean by ""if we consider moment maps which are like "quadratic functions", for general compact Lie group"". What do you mean by "general compact Lie group"? What is "moment maps which are like "quadratiс functions""? Probably you should give the example you have in mind. –  Dmitri Dec 30 '12 at 11:27
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