# Why is the definition of a proper group action the way it is?

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.

• 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]). Commented Sep 18, 2017 at 19:34
• 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 $K, K'$. Commented Sep 18, 2017 at 19:41
• @DanRust do you mean $\{g : K \cap gK \neq \emptyset \}$ is compact?
– D_S
Commented Sep 22, 2017 at 19:10
• @D_S yes sorry, I did mean that (after taking closure)! Commented Sep 22, 2017 at 19:15
• 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 $$p: G \times X \rightarrow X \times X, (g,x) \mapsto (x,g.x)$$
– D_S
Commented Sep 22, 2017 at 19:19

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.