Let $G$ be a topological group acting continuously on a topological space $X$. This means that $G \times X \rightarrow X$ is a continuous function.

A continuous map $Y \rightarrow Z$ is said to be proper if the preimage of a compact set is compact.

If $G$ acts continuously on $X$, then the action is said to be proper if the map

$$G \times X \rightarrow X \times X, (g,x) \mapsto (x, g\cdot x)$$

is proper.

I would have expected the definition of a proper group action to be that $G \times X \rightarrow X$ is proper, not $G \times X \rightarrow X \times X$. Why is the definition the way it is? What is the most natural way to think about this?

There are two situations in which I have encountered this notion which I want to understand better:

1 . An analytic Lie group $H$ over a local field of characteristic zero is acting properly and freely on an analytic manifold $X$, and for each $x \in X$, the map $h \mapsto h.x$ is an immersion $H \rightarrow X$. Then some result in Bourbaki, Differential and Analytic Manifolds says that the quotient space $H \setminus X$ has the natural structure of an analytic manifold.

2 . There is an arrangement $\mathscr H$ of hyperplanes in a real affine space $E$, and $W$, the group of affine transformations in $E$ generated by the reflections about the hyperplanes, is assumed to stabilize the set of hyperplanes $\mathscr H$ and act properly on $E$. This is the situation for many results described in Chapter V of Bourbaki, Lie Groups and Lie Algebras.

  • 3
    $\begingroup$ It might help to know that if $X$ is Hausdorff, then the definition of a proper action is equivalent to the property that: for all compact $K \subset X$, the set $\{g \mid K \cap gK = \emptyset\}$ has compact closure in $G$. It's a kind of separation property for a topological group action. It means that 'not many' elements of the group translate $K$ in such a way to intersect itself (where here, 'not many' means only a compact set of elements [after taking closure]). $\endgroup$
    – Dan Rust
    Commented Sep 18, 2017 at 19:34
  • 1
    $\begingroup$ Also note that map $f: G × X → X$ is proper iff $f^{-1}[K]$ is compact for every compact $K$, while the mentioned map $G × X → X × X$ is proper iff $f^{-1}[K] ∩ (G × K')$ is compact for every compacta $K, K'$. $\endgroup$
    – user87690
    Commented Sep 18, 2017 at 19:41
  • $\begingroup$ @DanRust do you mean $\{g : K \cap gK \neq \emptyset \}$ is compact? $\endgroup$
    – D_S
    Commented Sep 22, 2017 at 19:10
  • $\begingroup$ @D_S yes sorry, I did mean that (after taking closure)! $\endgroup$
    – Dan Rust
    Commented Sep 22, 2017 at 19:15
  • $\begingroup$ But if $G$ acts properly on $X$, then $\{g \in G : K \cap gK \neq \emptyset \}$ is already compact, since it is the image of the compact set $\{ (g, x) : x \in K, g.x \in K \} = p^{-1}(K \times K)$, where $$p: G \times X \rightarrow X \times X, (g,x) \mapsto (x,g.x)$$ $\endgroup$
    – D_S
    Commented Sep 22, 2017 at 19:19

1 Answer 1


Observe that for a point $x_0\in X$, the pre-image of the compact subset $\{x_0\}\subset X$ under the action map $G\times X\to X$ is $\{(g,g^{-1}\cdot x_0):g\in G\}\subset G\times X$. If the action map is proper, then this is a compact subset of $G\times X$. Since the restriction of the first projection to this subset defines a continuous surjection onto $G$, the group $G$ itself has to be compact. Thus requiring the action map to be proper brings you back to the case of compact groups.


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