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In the book 'Introduction to topological manifolds", I found discussions on the quotients of topological manifolds by discrete group actions.

I have some questions:

  1. Why the book focus only on quotients by discrete group actions?
  2. In 'Introduction to smooth manifolds", the quotient manifold theorem deals with Lie group actions, and as far as I know, non-zero dimensional Lie groups are not discrete. Is there a topological version of this?
  3. More specifically, I would like to know when the quotient of a topological manifold by a free action of a compact Lie group (in particular, a torus) is a topological manifold?
Link
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  • Regarding question 1, the two topics --- namely, quotients of manifolds by discrete groups, and quotients of manifolds by Lie groups --- are both quite large. Some books focus on one of those topics, some on the other. I suppose you could find a book that contains a little bit of both. – Lee Mosher May 23 '25 at 13:08
  • For your specific question, the answer is positive, you do not even need compactness: properness of the action suffices. The result is a consequence of the slice theorem due to Palais, https://math.stackexchange.com/questions/560371/orbit-space-of-a-free-proper-g-action-principal-bundle/1754706#1754706 – Moishe Kohan May 23 '25 at 16:24
  • @MoisheKohan: If I understand your answer correctly, the action is Palais-proper since a topological manifold X is metrizable and locally compact, and Lie group actions are proper. Hence we obtain a principal bundle. Does it then follow that the orbit space must be a topological manifold? – Link May 23 '25 at 18:16
  • Actually, I was too hasty. There is a counter-example already for a circle action on a 4-dimensional manifold. I will add details later. – Moishe Kohan May 23 '25 at 18:18

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To answer your specific question (number 3): No, the quotient need not be a topological manifold even if the Lie group is $S^1$. Indeed, let $B$ denote Bing's dogbone space. The product $B\times \mathbb R$ is homeomorphic to $\mathbb R^4$ but $B$ is not a topological manifold. Hence, $M=B\times S^1$ is a topological manifold as well. There is an obvious $S^1$-action on $M$ such that the quotient $M/S^1$ is $B$.

One can prove, however, that if $G$ is a Lie group, $G\times M\to M$ is a free proper topological action of $G$ on a manifold $M$, then $Q=M/G$ is a generalized manifold. In particular, if $\dim(Q)\le 2$, then $Q$ is a topological manifold.

Moishe Kohan
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  • How about closed manifoldd? – Link May 24 '25 at 03:03
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    @Link: It does not matter, Bing's construction is local. – Moishe Kohan May 24 '25 at 03:38
  • I was wondering why a $2$-dimensional generalized manifold is a manifold, but I found your other answer. Would you mind explaining why $M/G$ is a generalized manifold? Is this a consequence of Palais' theorem? – Geoffrey Sangston May 24 '25 at 11:51
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    @GeoffreySangston: Palais reduces the problem to a local question: If product of $X$ with $R^k$ is a manifold then $X$ is a generalized manifold. You first verify that $X$ is ANR using the fact that manifolds are ANR. Then check that $X$ is a homology manifold using Kunneth formula. – Moishe Kohan May 24 '25 at 16:38