Converse To Quotient Manifold Theorem [Exercise in Lee Smooth Manifolds]

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?

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.

• Thanks! This does it I think. To show that $G$ acts properly on a principal bundle with structure group $G$, I think you can just use an equivalent definition of properness [Lee-prop 9.13] which is that $G$ acts properly on $M$ iff for each $p_i \in M$ and $g_i\in G$ with $p_i\to p\in M$ and $g_i p_i \to q \in M$, a subsequence of $g_i$ converges in $G$. Using this it is basically trivial to show that $G$ acts properly, just take a local trivialization including both limit points, and then it remains to check that if $r_k,g_k \in G$, $r_k \to r$, $r_kg_k\to q$ then $g_k$ converges. Very nice!
– Otis
Aug 25, 2010 at 0:59

The following is an addendum to Sam's answer, meant to fill-in missing details.

Part 1. Suppose that $$G\times M\to M$$ is such that the quotient $$B=M/B$$ has structure of a smooth manifold such that the quotient map $$q: M\to B$$ is a submersion. Then $$q: M\to B$$ is a topological principal $$G$$-fiber bundle.

Proof. Pick a $$G$$-orbit $$Gx\subset M$$. Our goal is to find a $$G$$-invariant neighborhood $$W$$ of $$Gx$$ in $$M$$ such that $$W$$ is $$G$$-equivariantly homeomorphic to a product $$U\times G$$, $$U$$ is an open subset of $$B$$, and $$G$$ acts trivially on the first factor and by left multiplication on the second factor. Equivariance of a homeomorphism $$f: U\times G\to U$$ means: $$f(g(u,h))=g f(u,h)$$ for all $$u\in U, g\in G, h\in G$$.

Since $$q$$ is a submersion, there exists, $$V$$, a neighborhood of $$x$$ in $$M$$ and a diffeomorphism $$g: V\to U\times {\mathbb R}^n$$, where $$U=q(V)$$ and $$q|_V= p_U\circ g$$, where $$p_U: U\times {\mathbb R}^n\to U$$ is the projection to the first factor, $$g(x)=(q(x), 0)$$. Here $$n+dim(B)=dim(M)$$.

Set $$U':= g^{-1}(U\times \{0\})$$. Such a subset $$U'$$ of $$V$$ is called a slice through $$x$$ for the $$G$$-action on $$M$$. By the construction, every $$G$$-orbit intersects $$U'$$ in at most one point and $$U'$$ is a smooth submanifold of dimension $$d=dim(B)$$ in $$M$$. Now, consider the orbit map $$f: G\times U'\to M, \quad f(g,y)= gy.$$ This map is smooth and 1-1 (since each $$G$$-orbit intersects $$U'$$ at most once). Moreover, $$dim(G\times U')=dim(M)$$. Hence, by the invariance of domain theorem, $$f$$ is a homeomorphism to its image, which is an open subset $$W\subset M$$. Clearly, $$W$$ contains $$Gx$$. Moreover, by the construction, $$f$$ is $$G$$-equivariant.

Remark. i. With more work, one can prove that $$f$$ is also a diffeomorphism, but I do not need this.

ii. The map $$f$$ as above plays the role of "spreading $$V$$ out'' in Sam's answer.

This concludes the proof of Part 1.

Part 2. If $$M\to B$$ is a principal $$G$$-fiber bundle (where $$G$$ is a Lie group and $$M, B$$ are topological manifolds), then the $$G$$-action on $$M$$ is proper.

This properness property holds even in much greater generality, see Lemma B in my answer here.