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Let's say a Lie group $G$ acts on $M$. Then the orbit is diffeomorphic to the quotient of the group with the stabilizer ${\rm Orb}(x) \cong G/{\rm Stab}(x)$.

What are the exact requirement for the statement. I.e. has the group act freely or transitive...?

Shaun
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NicAG
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  • If you assume properness of the action then this theorem holds. Without properness, it (in general) fails. – Moishe Kohan Jul 01 '21 at 12:33
  • Thanks. Is there no way around, i.e. if $M$ is a manifold and everything is smooth? – NicAG Jul 01 '21 at 12:39
  • Just think about the action of ${\mathbb Z}$ on the circle generated by a rotation of infinite order. Or, if you want an example of a connected Lie group, take an isometric effective action of ${\mathbb R}$ on the torus $T^2$. – Moishe Kohan Jul 01 '21 at 13:23
  • I think the only key requirement is just Orb($x$) is an embedded submanifold of M, and each of the other conditions is just to deduce it. – Tian LAN Oct 14 '23 at 21:26

1 Answers1

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There is no "golden standard" here. The most common assumptions under which this theorem holds are:

  1. The orbit $Gx\subset M$ is open.

or

  1. The orbit $Gx\subset M$ is closed.

or

  1. The $G$-action $G\times M\to M$ is proper (which implies 2).

See my answers here, here and here.

Most of the work goes into proving that the natural map $G/G_x\to Gx$ is a homeomorphism. Once this is done, the diffeomorphism part follows from the Constant Rank Theorem (either the immersion or the submersion case).

Moishe Kohan
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  • Thanks a lot. In the case of a principal bundle, I know that the orbit is diffeomorphic to the fiber, i.e. the group. But which of these assumptions above holds in that case. I do not assume a proper group action in my definition. But that the base and total space are manifolds and that the group is a Lie group and the action is free and transitive. – NicAG Jul 01 '21 at 19:39
  • In the case of a principal fiber bundle, the action is proper. Transitive case is covered by the item (1) in my list (and (2), of course). – Moishe Kohan Jul 01 '21 at 19:47
  • But why is it proper. I do not assume it from the beginning. I just assume that $P,M$ is a manifold, $G$ a Lie group and that the action is fiber preserving, free and transitive and that the local trivialization is $G$-equivariant. – NicAG Jul 01 '21 at 20:08
  • @NicAG It is a nice exercise in understanding the definitions, one can find a detailed proof somewhere at MSE. Try to prove it for one hour, if you can't prove it in one hour, I will add a proof or a link – Moishe Kohan Jul 01 '21 at 20:29
  • https://math.stackexchange.com/questions/3233/converse-to-quotient-manifold-theorem-exercise-in-lee-smooth-manifolds – Moishe Kohan Jul 01 '21 at 21:17
  • Ah ok. Thanks. That is what I was looking for – NicAG Jul 02 '21 at 00:07