Properly discontinuous group action

A group $𝐺$ acting on a topological space $𝑋$ is called properly discontinuous¹ iff every $𝑥 \in 𝑋$ has a neighbourhood $𝑈$ such that for every $𝛾 1, 𝛾 2 \in 𝐺$ with $𝛾 1 \neq 𝛾 2$ , $𝛾 1 𝑈 \cap 𝛾 2 𝑈 = \emptyset$ . #m/def/topology

Properties

A properly discontinuous group action is necessarily free.
If $𝐺$ is also topological group and acts continuously, then the orbit map $𝐺 \to 𝐺 𝑥$ is a homeomorphism of discrete topological spaces.

Proof of properties 1–2

Let $𝐺$ be a topological group acting on $𝑋$ continuously and properly discontinuously.

Assume that $𝐺$ does not act freely, i.e. there exist $𝛾 1, 𝛾 2 \in 𝐺$ with $𝛾 1 \neq 𝛾 2$ such that $𝛾 1 𝑥 = 𝛾 2 𝑥$ for some $𝑥$ . Then for any neighbourhood $𝑈$ of $𝑥$ , $𝛾 1 𝑥 = 𝛾 2 𝑥 \in 𝛾 1 (𝑈) \cap 𝛾 2 (𝑈)$ , violating proper discontinuity. Thus $𝐺$ acts freely.

Now consider the orbit of a point $𝑥$ with its subspace topology and the corresponding orbit map $𝑜 𝑥 : 𝐺 \to 𝐺 𝑥 : 𝛾 \mapsto 𝛾 𝑥$ .

Assume there exists $𝛾 1 𝑥 \in 𝐺 𝑥$ with ${𝛾 1 𝑥}$ not open in $𝐺 𝑥$ . Let $𝑈$ be an open neighbourhood of $𝑥$ in $𝑋$ . Since $𝛾 1$ is a homeomorphism, $𝛾 1 𝑈$ is open in $𝑋$ , and thus $𝛾 1 𝑈 \cap 𝐺 𝑥$ is open in $𝐺 𝑥$ , so at least one distinct point $𝛾 2 𝑥$ is contained in $𝛾 1 𝑈$ . Then $𝛾 2 𝑥 \in 𝛾 1 𝑈 \cap 𝛾 2 𝑈$ , violating proper discontinuity. Therefore $𝐺 𝑥$ must be discrete.

Now clearly the orbit map $𝑜 𝑥$ is continuous and bijective (injectivity by freeness, surjectivity by construction). Thus every singleton ${𝛾 1}$ in $𝐺$ is the preïmage of a singleton ${𝛾 1 𝑥}$ in $𝐺 𝑥$ and is therefore open. Therefore $𝐺$ is discrete, and $𝑜 𝑥$ is a homeomorphism, since the inverse $𝑜 - 1 𝑥$ is continuous as a map between discrete spaces.

Orbit space of a properly discontinuous group action $𝑋 / 𝐺$ covers $𝑋$ .

#state/tidy | #lang/en | #SemBr

German eigentlich diskontinuierlich ↩

Properly discontinuous group action

Properties

Footnotes