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I want to solve the following exercise from Lee's book on smooth manifolds. 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.

This question was asked before somewhere but the answer is not satisfactory. Can someone give some hint on how to solve this problem?

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    $\begingroup$ Could you state or link the answer you found, and describe what about that answer you find unsatisfactory? $\endgroup$
    – Kajelad
    Dec 26, 2020 at 16:48

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We show this by using the characterization of proper actions via sequences:

Let $(g_{i})\subseteq G$ and $(p_{i})\subseteq M$ be sequences such that $p_{i}\to p$ and $g_{i}\cdot p_{i}\to q$. We need to find a convergent subsequence of $(g_{i})$.

Observe that $\pi\colon M\to M/G$ is a smooth submersion. This means that there is an open set $V\subseteq M/G$ and a smooth section of $\pi$ defined on $V$ (that is, a map $\chi\colon V\to M$ such that $\pi\circ \chi=\operatorname{Id}_{V}$ and $p\in \chi(V)$). Define a map $\Phi\colon G\times V\to \pi^{-1}(V)$ by letting $\Phi(g,z)=g\cdot\chi(z)$. This map is bijective and $G$-equivariant, so it has constant rank along the orbits $G\times \{s\}$ for all $s\in V$. Therefore, if we can prove that the rank of $\Phi$ is globally constant, we get that $\Phi$ is a diffeomorphism. This is because, by taking a point of the form $(e,s)$, we get that $\Phi_{*(e,s)}(T_{(e,s)}(G\times V))=T_{\chi(s)}(G\cdot \chi(s)))+\chi_{*s}(T_{s}(M/G))$. Since $\chi$ is an immersion and those two summands have trivial intersection, we get that $\Phi$ has constant rank in $\{e\}\times V$, so $\Phi$ has constant rank everywhere and is therefore a diffeomorphism.

Now, since $p_{i}\to p$, we can suppose that $\pi(p_{i})\in W$ for all $i$, where $W\subseteq V$ is an open subset containing $p$ and having compact closure in $V$. Therefore, we can write $\Phi^{-1}(p_{i})=(h_{i},s_{i})$ for $h_{i}\in G$, $s_{i}\in V$. Notice that $\pi(p_{i})=\pi(\Phi(h_{i},s_{i}))=\pi(h_{i}\cdot \chi(s_{i}))=\pi(\chi(s_{i}))=s_{i}$, so that $p_{i}=h_{i}\cdot \chi(\pi(p_{i}))$. Also, $p=\chi(\pi(p))=e\cdot \chi(\pi(p))$. Since $p_{i}\to p$, $h_{i}\to e$, because $\Phi$ is a diffeomorphism.

Furthermore, $g_{i}\cdot p_{i}\to q$, so that $\pi(p_{i})\to \pi(q)\in \overline{W}\subseteq V$. We can therefore write $q=h\cdot \chi(\pi(q))$ for some $h\in G$. Also, $g_{i}\cdot p_{i}=(g_{i}h_{i})\cdot \chi(\pi(p_{i}))\to q=h\cdot \chi(\pi(q))$. Again, since $\Phi$ is a diffeomorphism, this implies that $g_{i}h_{i}\to h$, so $g_{i}=g_{i}h_{i}h_{i}^{-1}\to h$. We get that $(g_{i})$ is a convergent subsequence.

Hope this helps!

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  • $\begingroup$ As a curiosity: you can do this procedure backwards to prove the Quotient Manifold Theorem without the need for heavy machinery (i.e. the Frobenius Theorem). The construction of $\Phi$ gives a hint for how to construct "adapted charts" for the action because a small enough open set in $M/G$ can be identified with a submanifold in $M$ transverse to all orbits passing through it. Also, $\Phi$ is a trivialization, so we implicitly proved that $M\to M/G$ is a fiber bundle with base $M/G$ and fiber $G$! $\endgroup$ May 8, 2021 at 11:08
  • $\begingroup$ Why do you say that $\chi(\pi(p)) = p$? For all we know we only have $\pi(\chi(x)) = x$. Not the other way round! $\endgroup$ Jul 2, 2022 at 9:12
  • $\begingroup$ By construction, $p\in \chi(V)$, so we may find some $z\in V$ such that $p=\chi(z)$. But $\pi(p)=\pi(\chi(z))=z$, so $\chi(\pi(p))=\chi(z)=p$. $\endgroup$ Jul 2, 2022 at 13:49

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