4
$\begingroup$

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.).

$\endgroup$
4
  • $\begingroup$ I have already asked this question on MSE math.stackexchange.com/questions/3246060/… $\endgroup$
    – user143512
    Aug 28, 2019 at 15:46
  • $\begingroup$ "Is a discrete group of" is not very precise. For instance, being a discrete subgroup of the homeomorphism group is not enough to act properly. So it might mean that the action is proper. Your question amounts to ask whether a free action with discrete orbits implies that the action is proper (it would sound natural to first assume closed discrete orbits, not just discrete). $\endgroup$
    – YCor
    Aug 28, 2019 at 16:12
  • $\begingroup$ "Your question amounts to ask whether a free action with discrete orbits implies that the action is proper" Yes, that is an equivalent, and probably better form of my question. If I understand correctly, in the quoted statement, by "discrete group" they a mean group where every orbits are discrete subsets of $R^2$. $\endgroup$
    – user143512
    Aug 28, 2019 at 16:44
  • $\begingroup$ But do you confirm they mean "discrete" and not "closed discrete"? (There are easy examples of (non-free) actions with discrete orbits, but not all closed.) $\endgroup$
    – YCor
    Aug 28, 2019 at 17:48

1 Answer 1

5
$\begingroup$

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).

$\endgroup$
2
  • 1
    $\begingroup$ Where does the second example fail to be a covering map? (Also I think F is homeomorphic to a Möbius-band, and is not homeomorphic to P so I might not get it.) In the first example, the quotient space $E/G$ fails to be Hausdorff at the images of (0,1) and (1,0). That is what I needed. Thank you both! $\endgroup$
    – user143512
    Aug 29, 2019 at 6:41
  • $\begingroup$ @MacskaBonifác: You are right, I was careless. I corrected the statement. $\endgroup$
    – Misha
    Aug 29, 2019 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.