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Let $R^2$ be the plane, and let G act on it with orientation preserving homeomorphisms, and assume that

  • every orbit of G is a discrete subset in $R^2$
  • G acts freely: $(\forall g \in G, g \neq e)$, $(\forall x \in R^2)$ $xg \neq x$.

Is it true that $R^2/G$ 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.).

  • By fixed point free do we mean no element $g$ of $G$ fixes any point, or that there is no point fixed by all of $G$? – Chris H May 31 '19 at 06:19
  • The former: $(\forall g \in G, g \neq e)$, $(\forall x \in R^2)$: $xg \neq x$. I edited the question. –  May 31 '19 at 06:41
  • The answer here may help, since I think your claim is false without the "properly discontinuous" requirement needed for a hausdorff quotient. https://math.stackexchange.com/questions/1082834/properly-discontinuous-action-equivalent-definitions – Chris H May 31 '19 at 07:27

1 Answers1

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The point is that we may quickly prove that the action is properly discontinuous, that is for every compact $K \subset \mathbb R^2$, $I(K)=\{g \in G \mid gK\cap K \neq \emptyset \}$ is finite. To see this choose an alleged counterexample $K$ ( i.e., $K$ is a compact subset of $\mathbb R^2$ and $I(K)$ is infinite )and note that the set $$\bigcup_{g \in I(K)}gK$$ has diameter at most $3$ times the diameter of $K$ and contains $I(K)k$ for every $k \in K$. Then for any $k \in K$ we have that $I(K)k$ is (by freeness) an infinite bounded subset of $\mathbb R^2$ and thus must have a limit point. This contradicts the discreteness of orbit hypothesis.

Greg Conner
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  • Thank you for your answer! I think you assumed that G acts with isometries of the plane. In my question we only can assume that G acts by homeomorphisms. –  Jun 26 '19 at 15:14