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Group $G$ acts on topological space $X$. Also, $x,x'\in X$ not in the same orbit of $G$ have open $U$, $U'$ such that $g(U)\cap U'=\varnothing$ for all $g\in G$. I have shown that $X/G$ is Hausdorff. The question asks whether the quotient map $q : X \rightarrow X/G$ a covering map.

I've worked at this for several days, but I feel like my understanding is shaky to begin with (especially since I am a beginner in both Abstract Algebra and Topology). I want to say "no", since the quotient map is not, in general, injective so it itself can't be a homeomorphism. However, a covering map has a union of disjoint open sets each mapped homeomorphically.

I can think of a (possible) example: $X=[0,1)\times[0,1]$ with the relative topology from $\mathbb{R}^2$. $G$ is the group $[0,1)\times\{0\}$ with the operation of componentwise addition modulo $1$. Then, every $[0,1)\times\{y\} \subset X$ maps to a point in $G/X$. Yet, $[0,1)\times\{y\}$ isn't a union of open sets each homeomorphic to a point.

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  • $\begingroup$ Hint: If $x\in X$, then when is the quotient map $q$ a homeomorphism close to $x$? Or for starters, when is $q$ 1-1 at $x$? How can you write this condition in terms of the action of $G$? $\endgroup$ – rfauffar Oct 28 '13 at 0:44
  • $\begingroup$ @RobertAuffarth Since the quotient map is surjective and continuous by definition, (local) homeomorphism <-> (local) injectivity. For x in open neighborhood U, this would be g*U has no intersection with U for g!=e, which is just saying that G acts evenly on X, right? $\endgroup$ – imallett Oct 28 '13 at 1:54
  • $\begingroup$ Yes, exactly. Be careful though with saying that continuous and surjective implies that local injectivity is the same as a local homeomorphism. In this case it's true that the local inverse is continuous because the projection map is open. $\endgroup$ – rfauffar Oct 28 '13 at 12:03
  • $\begingroup$ @RobertAuffarth Ah yes. I'm not sure I see how that helps though, since g(U) intersect U' being null only necessarily happened when x and x' weren't in the same orbit? $\endgroup$ – imallett Oct 28 '13 at 14:40
  • $\begingroup$ Let a topological group act on itself by translation. The quotient space $G/G$ is a singleton, so $q$ is a covering if and only if $G$ is discrete. $\endgroup$ – Daniel Fischer Oct 4 '16 at 9:03
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The following theorem is given with proof here:

Let $X$ be a topological space that is path-connected and locally path-connected. Let $G < \text{Hom}(X)$ be a group of homeomorphisms on $X$. Then the projection map $q: X \to X/G$ is a covering map if and only if $G$ acts properly discontinuously on $X$.

Properly discontinuously means that for every $x \in X$ there is a neighbourhood $U$ s.t. $gU \cap U = \varnothing$ for all $g \in G - \{e\}$.

The condition you give in the question is almost the condition that $G$ acts p.d. on $X$ so I suspect the question boils down to whether your $X$ is path-connected and locally path-connected.

Hope this helps.

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    $\begingroup$ Right, but pick $X=D(0,1)$ and $G=\mathbb{Z}_2$ given by rotation by 180 degrees. Now every pair of points that are not in the same orbit have neighborhoods that don't meet when one is rotated, but the centerpoint of the disk is fixed by $G$; thus $G$ does not act properly discontinuously. $\endgroup$ – guest May 26 '16 at 5:13

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