Equivalent definitions of properly discontinuous action

Question

Let $G$ be a subgroup of isometries of a metric space $X$. I want to prove that the following statements are equivalent:

For a non-empty compact subset $K$ of $X$, $g(K) \cap K \neq \phi$ for only finitely many $g\in G$.
Each orbit $Gx$ of $X$ is locally finite i. e. any compact subset $K$ of $X$ intersects only finitely many elements of $Gx$.

I am following the book $\textbf{Fuchsian Groups}$ by $\textbf{Svetlana Katok}$. The second statement is definition of properly discontinuous action according to this book. While in Wikipedia a properly discontinuous action is defined as the first statement. I want to show that both the definitions are equivalent but couldn't get any lead. How to prove this?

Does this answer your question? Properly discontinuous action: equivalent definitions — C.F.G, May 15 '20 at 08:41
@C.F.G: Nope, this is not a duplicate: In the linked question there is no assumption that the action is isometric, which makes all the difference. Without this assumption, the two definitions are indeed not equivalent, but, in the context of the current question, the two definitions are equivalent. — Moishe Kohan, May 15 '20 at 16:46
BTW, I think, you are misstating Katok's definition, see here for a detailed discussion. — Moishe Kohan, May 15 '20 at 16:49
@MoisheKohan Definition 2 is Katok's definition from the book mentioned in the question on page 27. The link you have given also mentions the same definition in Edit 3 section. — Priyanka, May 15 '20 at 17:21
Yes and this is not quite the same that you wrote. The correct definition is that for each compact $K\subset X$ there are only finitely many $g\in G$ such that $g(x)\in K$. It is not the same as $Gx\cap K$ is finite. — Moishe Kohan, May 15 '20 at 18:43
@MoisheKohan I understood it very well. This is a nice proof. Thank you very much. — Priyanka, May 19 '20 at 21:30
@MoisheKohan Why is $g(x) \in K$ for only finitely many $g\in G$ not same as $Gx \cap K$ is finite? $Gx$ is orbit of $x$ and contains elements of the form $g(x)$ where $g\in G$. — Priyanka, May 19 '20 at 21:34

score 3 · Answer 1 · answered May 15 '20 at 20:02

First of all, the correct interpretation of the definition given in Katok's book is:

Definition. Given an isometric $G$-action on a metric space $X$, the orbit $Gx$ is said to be locally finite if for every compact $K\subset X$, the subset $$ \{g\in G: gx\in K\}\subset G $$ is finite. In other words, if we equip $G$ with the discrete topology, then the orbit map $o_x: G\to X, o_x(g)=gx$, is proper.

Example. Take any infinite group $G$ acting on itself via left multiplication. Consider the set $X:= G\sqcup \{p\}$; we extend the action of $G$ to $X$ by requiring that every $g\in G$ fixes $p$. Equip $X$ with the discrete metric ($d(x,y)=1$ iff $x\ne y$). This metric is obviously $G$-invariant. Then for every compact (equivalently, finite) subset $K\subset X$ and every $x\in X$, the intersection $$ Gx\cap K $$ is a finite subset of $X$. However, the action of $G$ on $X$ is not proper in the usual sense (since $p$ has infinite stabilizer). Thus, assuming local finiteness as stated in your question is not enough for proper discontinuity of an isometric action.

Theorem. Let $X$ be a metric space and $G$ is a group acting isometrically on $X$. Then the following are equivalent:

$G$ acts on $X$ properly discontinuously. (I.e. if we equip $G$ with the discrete topology, the action $G\times X\to X$ is proper.)
$G$ acts on $X$ such that all $G$-orbits are locally finite in the sense of Definition.

Proof. (a) Suppose 1 holds, but 2 fails, i.e. there is a compact $K$ and a point $x\in X$ such that $$ S=\{g\in G: gx\in K\} $$
is infinite. Then the subset $T\subset G$ consisting of products $gh^{-1}$, $g, h\in S$, isinfinite. Moreover, for each $t\in T$, $tK\cap K\ne \emptyset$. This contradicts the assumption 1. (Note that this part did not require an isometric action, just a continuous $G$-action.)

(b) Suppose that 2 holds but 1 fails. Consider $K\subset X$ such that $$ \{g\in G: gK\cap K\ne \emptyset\} $$ is infinite. Thus, there exists a sequence $x_n\in K$ and an infinite sequence of pairwise distinct elements $g_n\in G$ such that $g_n(x_n)=y_n\in K$. By compactness of $K$, after passing to a subsequence, we can assume that $x_n$ converges to some $x\in K$ and $y_n$ converges to some $y\in K$. Since the $G$-action is isometric, $g_n(x)$ also converges to $y$ (just use the triangle inequality and $G$-invariance of the metric). Now, form the compact subset $K\subset X$ equal to $$ \{y\}\cup \{y_n: n\in {\mathbb N}\}. $$ (I will leave it to you to verify compactness: Prove that every sequence in $K$ contains a convergent subsequence.) We then have $g_n(x)\in K$ for all $n\in {\mathbb N}$, contradiction the assumption 2. qed

Equivalent definitions of properly discontinuous action

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