Hamiltonian actions and contractible loops

2012-12-28T17:00:33Z

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?

Answer by Dmitri for Hamiltonian actions and contractible loops

2012-12-29T06:56:00Z

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.

Hamiltonian actions and contractible loops - MathOverflow

Hamiltonian actions and contractible loops

Answer by Dmitri for Hamiltonian actions and contractible loops