$G$ a (finite dimensional) Lie group. $M$ smooth manifold. $A:M\times G\to M$ a Lie group action (left or right). $M/G$ the quotient space with quotient topology.

A sufficient condition for $M/G$ to possess a unique smooth manifold structure such that $\pi:M\to M/G$ is a smooth surjective submersion is when the following holds true for the action:

  1. smooth

  2. free

  3. proper

  4. effective (faithful)

How can these conditions be weakened? Effectiveness is not needed. Freeness is not needed, but the stabilizer groups should be all isomorphic, so that there is just one orbit type and the stratification by orbit types gives just one stratum. But properness and smoothness are hard to relax, right? Smoothness is of course necessary, but properness doesn't seem to be.

Consider the action of $\mathbb R$ on $S^1$ by $(\theta,t)\mapsto (t+\theta) \,\text{mod} \,2\pi$. This is effective, smooth, transitive, not free and not proper, but the quotient is just a point, so a perfectly valid manifold. The stabiliser/isotropy group of the orbit is $\mathbb Z$, that's why it's not free and since $\mathbb Z$ is non-compact the transitive action is not proper. So is there a weaker condition than properness? Or is my argument not valid?


1 Answer 1


I can provide provide only a partial answer:

Let $A$ be a (right) Lie group action which is smooth and free, and there is some topology and smooth structure on $M / G$ making the quotient $\pi: M \rightarrow M / G$ into a smooth surjective submersion, then $A$ is proper.

To see this, one has to know two things:

1) Let $P$ and $N$ be any smooth manifolds. Let $\pi: P \rightarrow N$ be a smooth surjective submersion. Assume that there is a smooth (right) action $A: P \times G \rightarrow P$ which is free, and its orbits are precisely the fibers of $\pi$.

Then there is a unique (up to an equivalence) local trivialization making $\pi: P \rightarrow M$ into a principal $G$-bundle.

Hint of the proof: Find local sections of $\pi$ and use them to define local trivialization maps.

2) Let $\pi: P \rightarrow N$ be a principal $G$-bundle with a right action $A: P \times G \rightarrow P$. Then $A$ is proper.

Hint of the proof: Group action $A$ locally looks as a right multiplication in the second component of the Cartesian product $U \times G$. This action is easily seen to be proper.

As the quotient map $\pi: M \rightarrow M/G$ satisfies the assumptions of 1), we can then use 2) to show that the action $A$ has to be proper.


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