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Suppose a Lie group acts smoothly (but not necessarily properly or freely) on a smooth manifold $M$. Show that each orbit is an immersed submanifold of $M$; which is embedded if the action is proper.

This is problem 15 of chapter 21 of Lee's smooth manifolds book I was searching to solve the above problem and I'm now a little confused writing an easy argument for this. I found the following argument but this is not easy enough for me to understand:

  1. For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$.
  2. The quotient space $G/G_p$ has a unique smooth manifold structure, and $G$ acts transitively on this smooth manifold by left multiplication.
  3. The orbit map $\mathscr O_p\colon G\to M$ given by $\mathscr O_p(g) = g\centerdot p$ descends smoothly to the quotient $G/G_p$ to give a smooth map $F_p\colon G/G_p\to M$ that is a bijection from $G/G_p$ onto the orbit $G\centerdot p$.
  4. The map $F_p$ is equivariant with respect to the left actions of $G$ on $G/G_p$ and $M$, and therefore it has constant rank.
  5. Because it is a constant-rank injection, $F_p$ is a smooth immersion, and therefore its image $G\centerdot p$ is a smoothly immersed submanifold of $M$.

Could anyone help me prove this?

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So I am going to do this step by step. So I belive it's clear that $G_p$ is a closed subgroup of $G$, and we can consider the proper free action $G\times G_p\rightarrow G_p$ by left translation , and since $G_p$ is a closed subgroup it also has a manifold structure and we can consider the space $G/G_p$. Now we get a bijection from $G/G_p$ to the orbit $g.p$ of $p$ since we are doing the quotient by the points that would fix the orbit. Then you have that $F_p$ is an equivariant map because $g.F_p(hG_p)=g(hp)=(gh)p=(gh)G_p p=F_p(ghG_p)$. Using the fact that the action of $G$ in $G_p$ is transitive and the equivariant rank theorem you have that $F_p$ must have constant rank and then using the fact that locally for suitable coordinates $F_p$ looks like $(x_1,...,x_r)\rightarrow (x_1,...,x_k,0,...,0)$ where $k$ is the rank of $F_p$, but since $F_p$ is injective this forces $k=r$ and so you get that it's an immersion, and so since the image of $F_p$ is the orbit of $p$ and that $F$ is an injective immersion you get the desired result.

I don't know what much could be added , is there anything that needs clarification ?

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  • $\begingroup$ Thank you what about the time the action is proper? $\endgroup$ Jan 18 '21 at 5:42
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    $\begingroup$ If the action is proper will you have that it's an embedded submanifold , since we can consider again the injective immersion from $G/G_p\rightarrow M$, and if we prove that this is a proper map we have that the orbit is actually embedded. But I belive this just comes from using the commutative diagram and the fact that $\phi$ is proper we have that $\hat \phi^{-1}_p(K)=\pi\circ (\phi^{-1}_p)(K)$ @Reza. $\endgroup$
    – Something
    Jan 18 '21 at 8:27
  • $\begingroup$ Also there's another thing one might add, if the action is not proper you have something stronger than just an immersed submanifold you will have that the orbit at that point will be an initial submanifold , if you don't know the definition of this there's not point in going into too much detail but I just thought I should add this property. $\endgroup$
    – Something
    Jan 18 '21 at 8:40

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