# Why does a proper lie group action admit an invariant metric?

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?

• I would have said: pick a metric at your favorite point on $M$, and then just move it around. (We do need the action to be transitive for this.) But this doesn't require the action to be proper, so I'm missing something. Apr 24 at 17:50
• For completeness of the question, you should include references to the proofs in the compact case. Apr 24 at 17:51
• @hunter: Yes, you need transitivity for this to work. Your argument will fail if point-stabilizers are noncompact. This is where you will need properness of the action. Apr 24 at 17:54

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).

• Yes, the answer misses the point of the question. Apr 24 at 20:04
• But, in case of the theorem in Lee, we assume that the group action is transitive. Whilst the result seems to follow just from properness. And in fact, I think that because of the invariant linearisation of Lie group actions (Lie groups, Duistermaat and Kolk thm 2.4.1), we have that the condition stated in the theorem by Lee is equivalent to properness. Apr 24 at 20:08
• Lol I read the question, then read the comments, then convinced myself the question had “transitive” in it as well and the question was merely about generalizing to non-compact $G$. Apr 24 at 23:03

You can find a proof of this result in Theorem 2 on page 7 of

Koszul, J. L., Lectures on groups of transformations. Notes by R. R. Simha and R. Sridharan, Tata Institute of Fundamental Research Lectures on Mathematics and Physics. Mathematics. 32. Bombay: Tata Institute of Fundamental Research. 97 p. (1965). ZBL0195.04605.

freely available here. (For some reason, I forgot to check Koszul's notes before commenting.)