# On quotient manifold theorem

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?

• Could you state or link the answer you found, and describe what about that answer you find unsatisfactory? Dec 26, 2020 at 16:48

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!

• 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 $G$! May 8, 2021 at 11:08
• 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! Jul 2, 2022 at 9:12
• 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 $\chi(\pi(p))=\chi(z)=p$. Jul 2, 2022 at 13:49