# Proper discontinuity and existence of a fundamental domain

I am currently teaching a topics course where I talk about some discrete groups acting properly. A student asked a very basic question that stumped me: what is the precise relationship between proper discontinuity and existence of a fundamental domain? Namely, let $G$ be a discrete group acting on a topological space $X$. Recall that the action is properly discontinuous if for every compact subset $K \subset X$, the set $$\{g \in G | gK \cap K \neq \emptyset \}$$ is finite. A fundamental domain is a subset $U \subset X$ such that its images by all elements of $G$ are disjoint and cover $X$ (possibly up to some "thin" set, e.g. the boundary of $U$).

Is there any theorem saying that the action is proper if and only if there exists a "sufficiently nice" fundamental domain on $X$, possibly with some additional conditions?

This seems to be taken for granted by everyone, and I often used it as an explanation of what a "properly discontinuous action" means intuitively. But to my extreme surprise, a search for precise references brought up almost nothing. I only found one statement of this sort: Theorem 3.6 in T. Drumm and V. Charette, "Complete Lorentzian 3-manifolds" (in Contemporary Mathematics, vol. 639 "Geometry, Groups and Dynamics", pp. 43--72). This statement seems to be actually wrong, as the well-known counterexample of the cyclic group generated by $\left( \begin{smallmatrix} 2 & 0 \\ 0 & 1/2 \end{smallmatrix} \right)$ acting on a closed quadrant (or on $\mathbb{R}^2 \setminus \{0\}$) seems to contradict it (unless I missed some additional assumptions they made somewhere).

Possible restrictions needed to make it work:

• Clearly we need to impose some regularity condition on $U$, to prevent people from just picking a point in every orbit at random and pretending that the resulting mess is a fundamental domain. One that is often used is being regular in the topological sense, i.e. open with $\overset{\circ}{\overline{U}} = U$, or (which is closely related) closed with $\overline{\overset{\circ}{U}} = U$.
• We need a local finiteness condition (e.g. every compact set meets only finitely many tiles), to exclude the counterexample above.
• Points with nontrivial stabilizers are annoying, as pointed out in the comments. We must either restrict our attention to free actions, or modify the definitions somehow. (It would still be nice to be able to deal with orbifolds.)
• Maybe things are simpler for groups acting on smooth manifolds by diffeomorphisms? Or if we have some sort of measure? Or something else?

I tried to think a little about this on my own. The "if" part seems more or less doable by hand. For the "only if" part however, I don't really know how to proceed. There is indeed a well-known theorem that says that if you have a proper action, then you have a nice (Hausdorff) quotient. But then you would have to somehow prove that by "cutting open" this quotient in a suitable way, you can flatten it and lift it to a fundamental domain - which seems really hard to do in general. Case in point: there is one example of proper actions from my field (namely the Margulis spacetimes - see for example "Properly Discontinuous Groups of Affine Transformations: A Survey" by H. Abels) which was found in 1983, but the description of the fundamental domain (the "crooked planes" construction) was not done until 1992.

• I'd like the definitions to be stated explicitly in the text. (This is my constant request). – Włodzimierz Holsztyński Oct 7 '16 at 19:12
• Here's a link to Drumm-Charette's paper (a preprint version); there's no Theorem 3.6, but probably you mean Theorem 3.1.6: indico.ictp.it/event/a14294/session/33/contribution/138/…, – YCor Oct 7 '16 at 22:53
• Drumm-Charette define a fundamental domain as a closed subset $F$ that's closure of its interior, whose $G$-translates cover the space, such that the $G$-translates of the interior of $F$ are pairwise disjoint when $g$ ranges over $G$. – YCor Oct 7 '16 at 22:57
• Their Theorem 3.1.6 states that if there's a fundamental then the action is proper (or properly discontinuous, if one likes old-fashioned cumbersome terminology). Indeed this is not true, for your favorite example: the set of $(x,y)$ with $x\ge 0,y\ge 0, x+y>0$, with action of the diagonal matrix $(2,1/2)$. Then $A\cup B$ is a fundamental domain in their sense, when $A$ is the segment $([1,2],0)$ and $B$ is the rectangle $\{(x,y):x\ge 0,y\in [1,2]$. But the action is not proper. – YCor Oct 7 '16 at 23:02
• Your definition is not satisfied by all actions of finite groups (which should be proper). You cannot exclude the possibility that some element of finite order has a set of fixed points of non-empty interior. – YCor Oct 7 '16 at 23:06

I will assume that you are interested in group actions on connected manifolds: In the case of more general spaces it is not even completely clear what a fundamental domain means since an element of finite order can fix a nonempty open subset.

Definition. Let $M$ be a manifold (in any category you like, DIFF, PL or TOP). Let $G\times M\to M$ be a proper action of a discrete group. An open subset $D\subset M$ is a fundamental domain for the action if:

1. The interior of the closure of $D$ equals $D$.

2. $G$-orbit of $cl(D)$ equals $M$.

3. $g(D)\cap D=\emptyset$ for all $g\in G -\{1\}$.

4. The collection of subsets $\{g cl(D): g\in G\}$ is locally finite.

I will prove that every smooth proper effective action on a smooth connected manifold admits a fundamental domain. First, you construct a $G$-invariant Riemannian metric $g$ on $M$; let $d$ be the corresponding distance function on $M$. This is done using a partition of unity on the quotient orbifold $M/G$. Now, since $M$ is connected, $G$ is countable and the action is effective, there is a point $x\in M$ not fixed by any $g\in G$. Now, define $D$ to be the Dirichlet fundamental domain: $$D= \{y\in M: d(y,x)< d(y, gx) \forall g\in G - \{1\} \}.$$
I will leave you to verify that $D$ has the right properties (this is very standard).

Suppose now that $M$ is a connected topological n-manifold and $G\times M\to M$ be an effective proper action of a discrete group. In order to show that there is a fundamental domain you can use a theorem of Morton Brown or, rather its generaization by Berlanga. Consider $X$ which is the quotient $M/G$ minus the projection of the union of fixed point sets of nontrivial elements of $G$. Then $X$ is open and dense in $M$. By Brown's theorem in the case when $X$ is compact and Berlanga's theorem in the general case, $X$ contains a dense open subset $U$ homeomorphic to $R^n$. Now, take $D$ to be a component of the preimage of $U$ in $M$. This will satisfy conditions 1--3 but I am not at all sure about 4 even in the case when $M/G$ is compact. Proving 4 would require some serious work.

You probably do not want to unload any of these on your students. in this case, it is best to work in the simplicial category with triangulated manifolds. You can find a proof in this case in the book by Seifert and Threlfall "A textbook on topology". The construction is quite simple and easy to explain even to undergraduate students taking an introductory topology class. Namely, Let $T$ denote the quotient-triangulation of $M/G$. Let $\alpha$ denote the 1-skeleton of the dual triangulation (vertices of $\alpha$ are facets and the edges of $\alpha$ are codimension 1 faces, called panels). Let $\tau\subset \alpha$ be a maximal subtree. Now, take the union $U$ of all open facets of $T$ with all the open panels. This union is simply-connected. Now, take $D$ to be a connected component of the preimage of $U$ in $M$. This will be a fundamental domain: Part 4 works because of the combinatorial nature of the construction since the triangulation of $M$ is locally finite, of course.

• Nice proof :-) I was aware of the notion of Dirichlet fundamental domains but if the metric was not provided, it did not occur to me that I could construct one. – Ilia Smilga Oct 8 '16 at 15:15
• I suppose you are also assuming that the action is by homeomorphisms, right? – Ilia Smilga Oct 10 '16 at 0:25
• @IliaSmilga: Not exactly, for smooth manifolds I assume diffeomorphisms and for PL manifolds I assume PL actions (for topological manifolds, I assume topological action, of course). You have to pick a category in which you are working (TOP, PL, or DIFF) and work with it. – Misha Oct 10 '16 at 5:06
• @IliaSmilga: It seems that the Dirichlet domain method can be applied in greater generality of locally path-connected locally compact spaces as they appear to admit proper geodesic metrics (by work of Bing and Moise), provided that fixed point sets of finite order homeomorphisms have empty interior. – Misha Oct 10 '16 at 5:41