Let $\mathbb{R}^2$ be the plane, and let a group $G$ act on it with orientation preserving homeomorphisms, and assume that
- every orbit of $G$ is a discrete subset in $\mathbb{R}^2$
- $G$ acts freely: $(\forall g \in G, g \neq e)$, $(\forall x \in \mathbb{R}^2)$ $xg \neq x$.
Is it true that $\mathbb{R}^2/G$ is a manifold with the factor topology, and $G$ determines a covering to it?
In EMS: Geometry II$^1$, it is stated in a slightly more general way:
If $\Gamma$ is a discrete group of orientation-preserving homeomorphisms of a
surface $X$, then the mapping it: $\pi: X \rightarrow X/\Gamma$ is a ramified covering
(Kerekjarto [1923]$^2$)
So the statement may be true. But the source is a German textbook. Can anyone prove it, and/or give English sources, or provide a counter example?
1: Gamkrelidze, R. V. (ed.); Vinberg, E. B. (ed.), Geometry II: spaces of constant curvature. Transl. from the Russian by V. Minachin, Encyclopaedia of Mathematical Sciences. 29. Berlin: Springer-Verlag. 254 p. (1993). ZBL0786.00008.
2: von Kerékjártó, B., Vorlesungen über Topologie. I.: Flächentopologie. Mit 80 Textfiguren., Berlin: J. Springer, (Die Grundlehren der mathematischen Wissenschaften. Bd. 8.) VII u. 270 S. gr. $8^\circ$ (1923). ZBL49.0396.07.).
Best Answer
There are examples of free actions on $\mathbb{R}^2$ where every orbit is discrete and closed but the action is not properly discontinuous and the quotient is non-Hausdorff. The example is rather standard. I will use $G\cong {\mathbb Z}$. Let its generator act on the punctured plane $P:=\mathbb{R}^2 - \{(0,0)\}$ via $$ (x,y)\mapsto (2 x, 2^{-1} y). $$ This defines an action of $G$ on $P$. It is an easy exercise to verify that the action is free and every orbit is closed and discrete. But the action is not properly discontinuous. (Consider $G$-images of any compact subset $K\subset P$ whose interior has nonempty intersection with both coordinate lines.)
In order to get a similar example of action on a plane, remove from $P$ the half-line $L=\{(x,0): x< 0\}$. The resulting subset $E\subset P$ is homeomorphic to $\mathbb{R}^2$ and is $G$-invariant. The action of $G$ on $E$ again fails to be properly discontinuous. Moreover, the quotient $E/G$ is non-Hausdorff, hence, is not a manifold (in the usual sense). However, the quotient map $E\to E/G$ is a covering map.
One can modify this example so that the quotient by $G$ is not a covering map. Namely, in $P$ consider the positive quadrant $$ Q=\{(x,y): x> 0, y> 0\}. $$ Next, take $T:= P \setminus Q$. Define $F$ to be the quotient space of $T$ where we identity boundary arcs via $(x,0)\sim (0,y)$ ($x>0, y>0$) via the map $(x,0)\mapsto (0, x^{-1})$. The action of $G$ on $P$ preserves $T$ and descends to the quotient space $F= T/\sim$. It is easy to check that every $G$-orbit in $F$ is discrete, closed, but the map $F\to F/G$ is not a covering map (say, at the equivalence class of (0,1)). In this example, $F$ is homeomorphic to the open Moebius band. To obtain an example of an action on a simply-connected space, remove from $F$ the projection of the half-line $L$.
More interestingly, there are examples of smooth free actions of ${\mathbb R}$ on $\mathbb{R}^2$ such that the quotient is Hausdorff, each orbit is closed but the action is not proper. The quotient is homeomorphic to $[0,1)$.
Lastly, in the cited book (Vinberg et al) they deal with isometric group actions; for those it is easy to prove that every free isometric action with discrete and closed orbits (on a metric space) is properly discontinuous, hence, a covering action. You do not even need local compactness of the space. Few years ago I wrote a note discussing issues related to different definitions of proper discontinuity (here).