Properly discontinuous action and Fundamental Domain

Question

Does properly discontinuous action by a discrete group G on R^n necessarily have compact fundamental domain?

Definitions:

Properly discontinuous: For each x in R^n there is open neighborhood U of x such that gU and U does not intersect for all g in G (except identity)

(I am reading a proof of Bieberbach's theorems. The author uses the following conditions to define the group of isometries:

A group G of rigid motions in R^n is called crystallographic if

(i) for all t> 0 only finitely many $ \alpha \in G $ have the absolute value of translation part less than or equal to t. -- this is the consequence of discreteness condition, I assume.

(ii) there is some constant d such that for each $ x \in R^n $ there is an element $ \alpha \in G $ satisfying $ |a-x| \le d $ where a is the translation part of $ \alpha $

I am assuming that this second part follows from the properly discontinuous action of G. But I am unable to see the connection. )

Answer 1

Proper discontinuity does imply the existence of a fundamental domain, at least in the case of an action by isometries. You can then take, as a fundamental domain, the Dirichlet domain defined as follows:

$$D := \{ y \in \mathbb{R}^n | \forall g \in G, d(y, x) \leq d(y, gx) \}$$

where $x$ is some fixed choice of element of $\mathbb{R}^n$.

Edit: it is important that the point $x$ is chosen so that its stabiliser under the group action is trivial. If this is not the case then $D$ will fail to be a fundamental domain.

Take, as a counter example, a group $G\le O(n)$ which fixes the origin. We then have by construction $d(y,0)=d(y,g(0))$ for all $g\in G$ and $y\in \mathbb{R}^n$. Therefore, taking $x=0$ would yield $D=\mathbb{R}^n$, but this is clearly not a fundamental domain unless $G$ is trivial.

(In the general case this statement is also true, but you have to use a different definition for proper discontinuity and possibly impose some additional conditions. The Dirichlet domain construction no longer works, so the proof is much harder.)

On the other hand, the assumption of having a compact fundamental domain is stronger than proper discontinuity. For example consider the action of $\mathbb{Z}$ on $\mathbb{R}^2$ by translations (by integer multiples of, say, the first basis vector $e_1$). This action is properly discontinuous; but its fundamental domains are vertical strips of width $1$ and infinite height, so they are not compact.

Answer 2

My answer follows on from that of @IliaSmilga, which is in general correct, but omits an important condition on the point $x$.

Proper discontinuity does imply the existence of a fundamental domain, at least in the case of an action by isometries. You can then take, as a fundamental domain, the Dirichlet domain defined as follows:

$$D := \{ y \in \mathbb{R}^n | \forall g \in G, d(y, x) \leq d(y, gx) \}$$

where $x$ is some fixed element of $\mathbb{R}^n$ (for example $0$).

The point $x$ must be chosen so that its stabiliser is trivial. If this is not the case then $D$ will fail to be a fundamental domain.

Take, as a counter example, a group $G≤O(n)$ which fixes the origin. Since every element of $G$ preserves the Euclidean inner product, we have that $d(0,y)=d(0,gy)$ for all $g\in G$ and $y\in \mathbb{R}^n$. Therefore, taking $x=0$ would yield $D=\mathbb{R}^n$, but this is clearly not a fundamental domain unless $G$ is trivial.