# Properly discontinuous actions and discrete groups in complete Riemannian manifolds.

I was reading the article "The Geometries of 3-manifolds" by Peter Scott and in the end of page 406 he states the following:

If $$G$$ acts properly discontinuously on a space $$X$$, then $$G$$ is a discrete subset of the space of all continuous functions $$X \to X$$ with the compact-open topology. The converse is false, in general, but is true if $$X$$ is a complete Riemannian manifold and $$G$$ is a group of isometries of $$X$$.

How do I prove the last statement?

If $$X$$ is a complete Riemannian manifold and $$G$$ is a group of isometries of $$X$$ acting properly discontinuously on $$X$$, then $$G$$ is discrete.

Observation: We say a group $$G$$ of homeomorphisms of $$X$$ acts properly discontinuously on $$X$$ if for every compact $$K \subset X$$ the set $$\{g \in G: gK \cap K \neq \emptyset\}$$ is finite.

• Do you mean, if $G$ is a group of isometries that is discrete subset in the compact-open topology on $\operatorname{Hom}(X)$, with $X$ a complete Riemannian manifold, then $G$ acts properly discontinuously on $X$? To me, that is the last statement in the quote. Commented May 3, 2019 at 2:37
• If $X$ is a complete Riemannian manifold, then the Hopf-Rinow theorem says that $X$ is a complete metric space, and so all the closed and bounded subsets of $X$ are compact. Can't you use that somehow? Commented May 3, 2019 at 2:57
• @RyleeLyman I mean exactly that.
– Hugo
Commented May 3, 2019 at 3:45
• A form of this question was asked many times, e.g. math.stackexchange.com/questions/1493767/…. I am sure it was also answered at MSE. There are many ways to prove this result, for instance, using Arzela-Ascoli theorem. Commented May 3, 2019 at 18:10
• @MoisheKohan: Why your linked post is a duplicate of this post? it is earlier than this one!! Commented Apr 22, 2020 at 18:38

## 1 Answer

Suppose that $$X$$ is a (complete) metric space which satisfies the Heine-Borel property (every closed and bounded subset is compact). For instance, you can take $$X$$ to be a complete connected (finite dimensional) Riemannian manifold equipped with Riemannian distance function.

Then Arzela-Ascoli theorem implies that for every sequence of isometries $$f_i: X\to X$$ such that there exists $$p\in X$$ and $$R$$ for which $$d(p, f_i(p))\le R$$ for all $$i$$, there exists a subsequence $$(f_{i_j})$$ which converges to an isometry uniformly on compacts in $$X$$.

Given this, let us prove

Lemma. Suppose that $$\Gamma$$ is a discrete subgroup of $$Isom(X)$$ (the isometry group of $$X$$) equipped with the topology of uniform convergence on compacts. Then $$\Gamma$$ acts properly discontinuously on $$X$$.

Proof. Suppose not. Then there exists a compact $$K\subset X$$ and an infinite sequence of distinct elements $$\gamma_i\in\Gamma$$ such that $$\gamma_i K\cap K\ne \emptyset$$. Taking $$p\in K$$ and $$R=2diam(K)$$, we conclude that for each $$\gamma_i$$, $$d(\gamma_i(p), p)\le R$$. Therefore, by the above observation, $$(\gamma_i)$$ contains a convergent subsequence $$(\gamma_{i_j})$$. Taking the sequence of products $$\alpha_j:= \gamma_{i_j}^{-1} \gamma_{i_{j+1}},$$ we conclude that $$\alpha_j\to id$$ uniformly on compacts. (I am using here the property that $$Isom(X)$$ with topology of uniform convergence on compacts is a topological group.) Hence, $$\Gamma$$ is not a discrete subgroup of $$Isom(X)$$. A contradiction. qed

• Cool! Thank you!
– Hugo
Commented May 3, 2019 at 23:35
• How does one see that the Arzela-Ascoli Theorem imply the existence of this subsequence?
– Dai
Commented Mar 30, 2021 at 14:08
• @Dai: Isometries form an equicontinuous family. Then you verify that the limit of a convergent sequence of isometries is an isometry. Commented Mar 30, 2021 at 14:13
• Still confused how does the property $d(p,f_i(p))<R$ play a role here?
– Dai
Commented Mar 30, 2021 at 14:17
• @Dai: Yes: AA theorem in its usual form requires compactness. What you have is only compactness of images of balls. For each $r\in {\mathbb N}$ you consider $B_r=\bar{B}(p,r)$ and note that $f_i(B_r)\subset \bar{B}(p, r+R)$ and the latter is compact. Now, you use AA theorem plus a diagonal subsequence argument to construct a convergent subsequence. Commented Mar 30, 2021 at 14:21