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?


2 Answers 2


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.

  • 1
    $\begingroup$ 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! $\endgroup$
    – Otis
    Commented 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .