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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.

Hugo
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    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. – Robbie Lyman May 03 '19 at 02:37
  • 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? – Ivo Terek May 03 '19 at 02:57
  • @RyleeLyman I mean exactly that. – Hugo May 03 '19 at 03:45
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    A form of this question was asked many times, e.g. https://math.stackexchange.com/questions/1493767/a-problem-of-a-discrete-group-of-smooth-isometries-acting-discontinuously-on-a-s?rq=1. I am sure it was also answered at MSE. There are many ways to prove this result, for instance, using Arzela-Ascoli theorem. – Moishe Kohan May 03 '19 at 18:10
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    @MoisheKohan: Why your linked post is a duplicate of this post? it is earlier than this one!! – C.F.G Apr 22 '20 at 18:38
  • @C.F.G: The timing is irrelevant. What is relevant is that one post contains an answer to another. Arguably, I could have answered another post and voted to close this one as a duplicate. I do not remember why. – Moishe Kohan Apr 22 '20 at 18:44

1 Answers1

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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

Moishe Kohan
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  • Cool! Thank you! – Hugo May 03 '19 at 23:35
  • How does one see that the Arzela-Ascoli Theorem imply the existence of this subsequence? – Dai Mar 30 '21 at 14:08
  • @Dai: Isometries form an equicontinuous family. Then you verify that the limit of a convergent sequence of isometries is an isometry. – Moishe Kohan Mar 30 '21 at 14:13
  • Still confused how does the property $d(p,f_i(p))<R$ play a role here? – Dai Mar 30 '21 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. – Moishe Kohan Mar 30 '21 at 14:21
  • Oh I see. Thanks! – Dai Mar 30 '21 at 14:24