Let us define a properly discontinuous action of a group $G$ on a topological space $X$ as an action such that every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$. I would like to prove that this property is equivalent to, having given $G$ the discrete topology and in the $X$ locally compact Hausdorff case, the map $G \times X \rightarrow X \times X$ given by $(g, x) \mapsto (x, gx)$ being proper (i.e. closed and preimage of compact sets is compact) plus the action being free.

I have managed to prove one direction, that is, if the action is proper and free with $G$ having the discrete topology then it is properly discontinuous. I'm having trouble though with the other direction. Here is an attempt: let's denote by $\rho : G \times X \rightarrow X \times X$ the map $\rho(g, x) = (x, gx)$. Suppose $K \subset X \times X$ is compact. We wish to show $\rho^{-1}(K)$ is compact. Let $(g_i, x_i)$ be a net in $\rho^{-1}(K)$. Then $\rho(g_i, x_i) = (x_i, g_i x_i)$ admits a convergent subnet, so passing to it we may assume $x_i \rightarrow x$ and $g_i x_i \rightarrow y$. Essentially we must now find a way to prove $g_i$ converges, but I can't seem to do this. Any hints?


1 Answer 1


These properties are not equivalent. Here's a counterexample: Let $X=\mathbb R^2\smallsetminus\{(0,0)\}$, and define an action of $\mathbb Z$ on $X$ by $n\cdot (x,y) = (2^n x, 2^{-n} y)$. This is properly discontinuous by your definition, but it's not a proper action. The subset $K \times K \subseteq X\times X$ is compact, where $K = \{(x,y): \max(|x|,|y|)=1\}$, but $\rho^{-1}(K\times K)$ contains the sequence $(n, (2^{-n},1))$, which has no convergent subsequence.

I think one reason for your confusion is that different authors give different definitions of "properly discontinuous." Topologists concerned primarily with actions that determine covering maps often give the definition you gave:

(i) Every $x \in X$ has a neighborhood $U$ such that $gU \cap U \neq \emptyset$ implies $g = e$.

This is necessary and sufficient for the quotient map $X\to X/G$ to be a covering map. However, in order for the action to be proper (and thus for the quotient space to be Hausdorff), an additional condition is needed:

(ii) If $x,x'\in X$ are not in the same $G$-orbit, then there exist neighborhoods $U$ of $x$ and $U'$ of $x'$ such that $gU\cap U' = \emptyset$ for all $g\in G$.

When $X$ is a locally compact Hausdorff space and $G$ is a discrete group acting freely on $X$, the action is proper if and only if both conditions (i) and (ii) are satisfied. Differential geometers, who are typically concerned with forming quotient spaces that are manifolds, are more apt to define "properly discontinuous" to mean both (i) and (ii) are satisfied.

Because of this ambiguity (and because the term "properly discontinuous" leads to oxymoronic phrases such as "a continuous properly discontinuous action"), Allen Hatcher in his Algebraic Topology coined the term covering space action for an action satisfying condition (i). I've adopted that terminology, and I use free and proper action for an action satisfying (i) and (ii) (at least for locally compact Hausdorff spaces). I sincerely hope the term properly discontinuous will eventually die out.

You can find more about these issues in the second editions of my books Introduction to Topological Manifolds (Chapter 12) and Introduction to Smooth Manifolds (Chapter 21).

  • 15
    $\begingroup$ +1 for the answer, +10 for 'I sincerely hope the term properly continuous will die out' (I've found it as confusing as the OP) and +1000 for the books. $\endgroup$ Sep 16, 2017 at 22:19
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    $\begingroup$ @MoziburUllah: Thank you! I'll try to use my 1000 points wisely. :-) $\endgroup$
    – Jack Lee
    Sep 16, 2017 at 23:03
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    $\begingroup$ For future reference, since this is a standard example: The quotient space in this example, $Y=X/G$, where $G={\mathbb Z}$, is also non-Hausdorff. The distinct points of $Y$ which cannot be separated by disjoint neighborhoods are, for instance, the projections of $(1,0)\in X$ and $(0,1)\in X$. $\endgroup$ Nov 8, 2021 at 17:03

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