Why does a proper lie group action admit an invariant metric

Question

Let $G$ be a lie group and $M$ a manifold. Let $A : G \times M \to M$ be a proper Lie group action. It seems to be a well known result that in this case $M$ admits an invariant metric $g$. That is for $h \in G$, $A(h, -)_*g=g$.

However, I only know a proof (thm 3.0.2) in case $G$ is compact. Moreover, all references I have found also restrict themselves to the case of $G$ being compact. How does one proof this result in general for a proper action of a non-compact group?

Answer 1

Look at Lee’s Introduction to Riemannian Manifolds, Theorem 3.17:

Let $G$ be a Lie group acting smoothly and transitively on a smooth manifold $M$. Then, there is a $G$-invariant Riemannian metric if and only for some (equivalently for every) point $p\in M$, the isotropy/stabilizer representation $I_p:\text{Stab}_G(p)\to \operatorname{GL}(T_pM)$ has image with compact closure.

• If the group $G$ is compact, then the stabilizer is obviously compact and thus the continuous image under $I_p$ is compact, and thus by the theorem, we have a $G$-invariant metric.
• Suppose the group action is proper, i.e the map $\Theta:G\times M\to M\times M$, $(g,p)\mapsto (g\cdot p,p)$ is a proper map. The singleton $\{(p,p)\}$ is a compact set in $M\times M$, and so the preimage is compact in $G\times M$; but the preimage is exactly $\text{Stab}_G(p)\times \{p\}$. Projecting this compact set via the continuous projection $G\times M\to G$ implies that $\text{Stab}_G(p)$ is compact, and thus its image under isotropy representation is a compact subset of $\operatorname{GL}(T_pM)$. Therefore, by the theorem, there is a $G$-invariant Riemannian metric on $M$.

If you look back at the proof, the compactness is used so that you can average tensors (i.e so that certain integrals make sense).