Hamiltonian actions and contractible loops - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T02:33:03Z http://mathoverflow.net/feeds/question/117420 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117420/hamiltonian-actions-and-contractible-loops Hamiltonian actions and contractible loops Guangbo Xu 2012-12-28T17:00:33Z 2012-12-30T11:36:39Z <p>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 $\mu: M \to {\mathfrak g}^*$.</p> <p>We assume that the moment map is proper in case $M$ is noncompact. </p> <p>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.</p> <p>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.</p> <p>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.</p> <p>Is there any proof or counter-example? Or should we add some conditions to guarantee this?</p> http://mathoverflow.net/questions/117420/hamiltonian-actions-and-contractible-loops/117482#117482 Answer by Dmitri for Hamiltonian actions and contractible loops Dmitri 2012-12-29T06:56:00Z 2012-12-30T11:36:39Z <p>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:</p> <p>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.</p> <p><em>One more counterexamle</em>. 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. </p> <p>So we see that in the case the Lie group is not simply-connected it always admits a "bad" action. </p>