# Free and proper action of a closed subgroup of a Lie group

I'm taking a course on Riemannian geometry and in my homework set I'm asked to prove that the (left) action of a closed subgroup $H$ of a Lie group $G$ on $G$ is free and proper.

To prove that it is free is easy, but how do I prove properness? By definition, one should check that the preimage by the map $\Phi: G \times M \rightarrow M \times M$ which takes $(g,x)\mapsto (g\cdot x,x)$ of any compact set in $M \times M$ is a compact set in $G \times M$, $M$ being a smooth manifold.

In the Google preview of the book "Lie Groups" by Duistermaat and Kolk, page 56, it is said that the left action of $G$ on $G$ is free and proper because the map $(g,x)\mapsto (g\cdot x,x)$ is bijective: $G \times G \rightarrow G \times G$, and the inverse $(g,x)\mapsto (g\cdot x^{-1},x)$ is continuous, thus mapping compact sets to compact sets. I understand this reasoning, but if I try to do the same for a (closed) subgroup $H$, where does the closed assumption come into play? This is what I don't get.

I've also checked this question. It is actually the same problem I'm trying to solve. Although I understand the scheme of the proof in the answer given there, I'm not sure how that reconciles with the definition of properness I gave above, again, that the preimage by $\Phi$ of any compact set in $G \times G$ should be a compact set in $H \times G$.

EDIT: I think I found a very easy solution for the problem, working with yet another, equivalent, definition of a proper action, that is, an action $A: H \times G \rightarrow G$ is proper iff for any sequence $\{h_n\}$ in $H$ and any convergent sequence $\{g_n\}$ in $G$, such that $\{h_n\cdot g_n\}$ converges, the sequence $\{h_n\}$ admits a convergent subsequence which converges in $H$.

So, we can write $h_n=(h_n\cdot g_n)\cdot g_n^{-1}, \forall n$ because we are in a group. But the sequence $\{h_n\cdot g_n\}$ converges by hypothesis, and the sequence $\{g_n^{-1}\}$ also converges because $\{g_n\}$ converges. So the product sequence, $h_n$, is a convergent sequence and every subsequence of it converges to the same limit. With $H$ being closed, this limit is in $H$, and so we conclude that $H$ acts properly on $G$.

Anyway, I still don't understand well why the three definitions, the one I gave initially, the one in which the answer to the linked question is based and this last one related to sequences, are equivalent. But the proof seems to come out very easily this way.

• Since $G$ is locally compact and Hausdorff you can probably use the fact that the properness of $\Phi$ is equivalent to proving $\Phi$ is closed and inverse of a singleton is compact. – R_D Oct 30 '15 at 4:14
• Thank you. Yes, that appears to be yet another equivalent definition of a proper action. I've found a simple solution based on another definition (see my edit). I still have to think about the equivalence of the various definitions though. – Strider Oct 30 '15 at 6:26

The map $\Phi : H \times G \to G \times G$ defined by $\Phi(h,g) = (hg,g)$ is the composition of the inclusion $H\times G \to G \times G$ with the map $\Psi:G \times G \to G \times G$ given by $\Psi(g',g)= (g'g,g)$.
As you remarked, $\Psi$ is proper since it is a homeomorphism. Further, since $H$ is closed in $G$, then $H\times G$ is closed in $G\times G$, and in particular the inclusion $H\times G \to G \times G$ is proper (the intersection of a compact subset with a closed subset in any topological space is compact).
In other words, $\Phi$ is the composition of two proper maps; thus it is proper as well.