I would like help with the following problem (chapter 9, #4) from Lee's Smooth Manifolds [its not homework, I'm reading it and I got stuck on this one]
If a Lie group $G$ acts smoothly and freely on a smooth manifold $M$ and the orbit space $M/G$ has a smooth manifold structure such that the quotient map $\pi: M\to M/G$ is a smooth submersion, then $G$ acts properly.
Its kind of a converse to the standard theorem about quotienting a manifold by a group action.
Any hints/help?
Best Answer
I'm pretty sure this is true. Here's my suggested approach. There's a step I don't know how to do, so hopefully someone can fill in the gap, or you can figure it out as an exercise. (And please post a comment if you do, as I'm curious!)
First, show that $M$ is a principal fiber bundle with structure group $G$. That is, locally on $M/G$, $M$ looks like $(M/G)\times G$. To do this, use that $M\to M/G$ is a submersion to find neighborhoods $U$ on $M/G$ and $V$ on $M$ where $M\to M/G$ looks like the projection onto several coordinates. The coordinates you "forget" in this projection give a chart on some neighborhood of $G$. (This is why the fibers of submersions are smooth manifolds; in this case, the fiber is $G$ since the action is free.) Now use translation by the group law to "spread $V$ out" so that the entire preimage of $U$ in $M$ is diffeomorphic to $U\times G$, with the restriction of $M\to M/G$ to this open being the projection onto the first factor. This ensures that $M$ is a principal $G$-bundle.
This reduces the problem to showing that $G$ acts properly and freely on any principal fiber bundle with structure group $G$ (freeness being clear). To show that $G\times M\to M\times M$ is proper, it suffices to show that it is closed and that the preimage of a point in compact. The latter condition is obvious, by working locally in a trivializing neighborhood of that point. I don't immediately see how to get the closedness of the map, but I've convinced myself it's true in some examples.