Let $G$ acts on space $X$, We say the action of $G$ on $X$ is proper discontinuous, if for every $x\in X$, there is an open set $U$ containing $x$ such that $gU\cap U=\phi$ for all $g\in G, g\neq e$

We say the action is $\textit{covering space action}$ if $G$ acts by homeomorphism and action is proper discontinuous. We know if action of $G$ on $X$ is a covering space action, $$X\rightarrow X/G$$ is a covering map.

Let $H$ be a subgroup of $G$, Is it true that $$X/H\rightarrow X/G$$ is also a covering map? How to prove it.

$\textbf{Edits}$ I have tried the following approach, Kindly correct me if there is a mistake.

Let $x\in U\subset X/G$ be an open set, since $X\xrightarrow{p} X/G$ is a covering map, $p^{-1}(U)$ is disjoint union of open set in $X$, that is $p^{-1}(U)=\sqcup V_{\alpha}$,

Now Using the fact that covering map is open, that is, It takes open sets to open sets, We conclude that $q(V_{\alpha})$ is a open set in $X/H$ [Since $X\xrightarrow{q}X/H$ is a covering map].

Let's denote $f:X/H \to X/G$. I guess there is no problem in concluding that $f^{-1}(U)=\sqcup q(V_{\alpha})$, hence showing that our map is a covering map.

Is there any relation between $[G:H]$ and the above covering covering map ?

  • 1
    $\begingroup$ If $f:X/H \to X/G$ is indeed a covering map, and all is well in this world, then $[G:H]$ should be the number of points of $f^{-1}(x)$ for any $x \in X/G$. $\endgroup$ – Arthur Nov 28 '16 at 6:22
  • $\begingroup$ @Arthur Oh! Right. My bad $\endgroup$ – Tensor_Product Nov 28 '16 at 6:24
  • $\begingroup$ @Tensor_Product, Is $H$ any subgroup of $G$? I think one can prove that $X/H\to X/G$ is a covering map when $H$ is the normal subgroup generated by elements of $G$ having fixed points. For any $H$ I'm not so sure. $\endgroup$ – R_D Nov 28 '16 at 7:08
  • $\begingroup$ @R_D Yes, H is any subgroup of G $\endgroup$ – Tensor_Product Nov 28 '16 at 7:19
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    $\begingroup$ So far, so good, but incomplete, you are not quite done yet. $\endgroup$ – Moishe Kohan Nov 28 '16 at 21:13

I will give you a sketch of a proof assuming that $H$ is a normal subgroup of $G$. (This is essentially so that $G/H$ becomes a group)

We need to prove that $X/H\to X/G$ is a covering map. First notice a few things -

  1. $G/H$ has a natural action on $X/H$ given by $gH\cdot [x]=[g\cdot x]$ where $[x]$ is the orbit of $x$ under the $H$ action on $X$ and $g\cdot x$ is the action of $G$ on $X$.

  2. $(X/H)/(G/H)$ is homeomorphic to $X/G$

From (2) it is enough to prove that $G/H$ acts on $X/H$ by homeomorphisms and the action is properly discontinuous.

Proving any $gH:X/H\to X/H$ is a homeomorphism : bijectivity is straight forward, bicontinuity can be done by a little diagram chasing (showing inverse image of an open set is open and open set gets mapped to an open set) $\require{AMScd}$ \begin{CD} X @>g>> X\\ @V p V V @VV p V\\ X/H @>>gH> X/H \end{CD} Here the vertical $p$ is the quotient map which is open (since it is given by a group action) and continuous.

Proving $G/H$ action on $X/H$ is properly discontinuous :

Let $[x]\in X/H$. Let $U$ be an open set in $X$ containing $x$ such that $gU\cap U=\emptyset$ for all $g\neq e$. Then $p(U)$ is an open set in $X/H$ containing $[x]$.

Suppose there is a $[y]\in gH\cdot p(U)\cap p(U)$. Then there is $y'\in U$ such that $[y]=gH\cdot[y']=[g\cdot y']$. So there is a $h\in H$ such that $y=h\cdot g\cdot y'\implies y\in hg\cdot U\cap U\implies hg=e\implies g=h^-1\in H$. Thus $gH=H$ is the identity element of $G/H$.

We are done. $\square$


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