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I'm having a bit of trouble with the following exercise:

Let $G$ be a group acting properly discontinuous and continuous on a topological space $E$. Then $p:E\to G\backslash E$ is a covering. Let $N_G(H)$ be the normalizer of $H$ in $G$. Show that:
i) For all $H\subset G,p_H:H\backslash E\to G\backslash E$ is a covering.
ii) The action of $g\in G$ on E induces a deck transformation of $p_H$ iff $g\in N_G(H)$.
iii) The action in ii) induces a homomorphism $\varphi:N_G(H)\to \Delta(p_H)$ with kernel $H$, but which in general is not surjective.
iv) If $E$ is connected, $\varphi$ is surjective.

My attempt:
i) Since $H\subset G$, the action of $H$ on $E$ is properly discontinuous and continuous aswell, thus $E\to H\backslash E$ is a covering aswell. Now $H\backslash E$ is a quotient space of $E$, so applying its universal property to the covering $p$ gives us the covering $p_H$.
ii) That the action induces a homeomorphism is clear, since the action is continuous and the inverse map is given by acting with the inverse element. So it seems the normalizer property is needed to show that $p_H$ is invariant under that action. This is where I'm stuck here.
iii) Let $D_g$ be the deck transformation corresponding to $g\in N_G(H)$. Then $\varphi(gg')(e)=D_{gg'}(e)=(gg').e=g.(g'.e)=D_g(D_{g'}(e))=(D_g\circ D_{g'})(e)=(\varphi(g)\circ\varphi(g'))(e)$, thus it is a homomorphism. It is clear that $ker(\varphi)=H$ since $h.e=e\forall h\in H, e\in H\backslash E$, thus the deck transformation is the identity. I was not able to find an example for the surjectivity.
iv) I have no idea how to approach this.

I would really appreciate any criticism and/or tips.

blst
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  • Just to make sure, $G\backslash E$ is the orbit space? Or is it in some way different from $E/G$? – Daniel Fischer Jun 27 '14 at 13:02
  • It's just the quotient of $E$ by the equivalence relation $e\sim e.g$ (we use the convention of a right action). – blst Jun 27 '14 at 13:04
  • What is $\Delta(p_H)$? – Kyle Jun 27 '14 at 14:51
  • The group of decktransformations of $p_H$. – blst Jun 27 '14 at 14:51
  • Concerning (i): by this logic one could argue in the opposite direction, because equally $G \backslash E$ is a quotient and $E\rightarrow G\backslash E$ is a covering, thus getting $G \backslash E$ is a covering of $H \backslash E$, or am I missing something? – nakajuice Jul 01 '14 at 01:34
  • One requirement for the universal property is that the map we try to pass down has to be equal on elements we identify in the quotient. Every element that is identified in $H\backslash E$ is identified in $G\backslash E$ but not vice versa. – blst Jul 04 '14 at 13:37

2 Answers2

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Proof. Let's say $\phi: E \to E$ is the homeomorphism $\phi(e)=g.e$. If $\pi_H: E\to H \backslash E$ is the quotient map, then the composition $\pi_H \circ \phi: E \to H\backslash E$ induces a unique map $\overline \phi: H \backslash E \to H \backslash E$, provided that \begin{equation} \pi_H \circ \phi(e)= \pi_H \circ \phi(h.e) \quad \forall h \in H, e \in E. \tag{$*$} \end{equation} If $(*)$ holds and $\phi$ descends to a homeomorphism of $H\backslash E$, then we have \begin{equation}\tag{$**$} p_H \circ \overline \phi(H_e)=p_H(g.H_e)=G.(g.H_e)=G.H_e=p_H(H_e). \end{equation} Thus $\overline \phi$ will be a deck transformation.

We now show that $(*)$ holds if and only if $g$ lies in $N_G(H)$. First observe that $(*)$ is equivalent to $H.(g.(h.e))=H.(g.e)$ (for all $h$ and $e$). Now if $g \in N_G(H)$, i.e. $g H=Hg$, then \begin{align} H.(g.(h.e))&= Hg.(h.e)=gH.(h.e)=gH.e=Hg.e=H.(g.e), \end{align} as desired. On the other hand, suppose $H.(g.(h.e))=H.(g.e)$ for all $h$. Fixing $h$ now, there must exist $h' \in H$ such that $h'.(g.(h.e))=g.e$, hence $(g^{-1}h'gh).e=e$. Since $e \in E$ can only be fixed by $1 \in G$, we have $g^{-1} h' g h=1$ and thus $$g h g^{-1} = (h')^{-1}.$$ Since $ghg^{-1}$ is an element of $H$ and $h$ was arbitrary, we must have $g \in N_G(H)$. $\square$

Kyle
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  • I have one problem with this argument: Our space here is not necessarily path connected. – blst Jun 27 '14 at 14:55
  • Yeah, that's definitely a limitation. I was hoping it would be enough to inspire the more general solution. If you're familiar with the the action of a fundamental group on the fibers of a covering, perhaps it would pay to think about that. I'll give it some thought and post again when I get a chance. – Kyle Jun 27 '14 at 15:14
  • I think I might have a solution (correct me if I'm wrong): an element in $H\backslash E$ looks like $H.e$ and $p_H(H.e)={g'.(H.e),g'\in G}$. Similarly $p_H(H.(g.e))=p_H((Hg).e)={g'.((Hg).e)}$. So action by $g$ being a deck transformation is equivalent to ${g'.((Hg).e),g'\in G}={g'.(H.e),g'\in G}$. But that is equivalent to $gH=Hg$. – blst Jun 27 '14 at 16:50
  • @blst: I like that line of thought, but I don't think that $g$ being a deck transformation of $p_H : H \backslash E \to G \backslash E$ implies that $p_H(H.e)=p_H(H.(g.e))$, since $g$ needs to act on an element of $H \backslash E$. The proper statement would be $p_H(H.e)=p_H(g.(H.e))$, which doesn't uncover any new information about conjugation. – Kyle Jun 27 '14 at 17:04
  • @blst: I think I was failing to understand how you were inducing the map on $H \backslash E$. I've updated my response to account for this, and I think that your most recent comment is definitely on the right track for "Interpretation A". – Kyle Jun 27 '14 at 19:35
  • Your intepretation A is indeed what I had in mind. Also we see that $g\in N_G(H)$ is sufficient, but I'm not sure how to show it's necessary. – blst Jun 27 '14 at 20:17
  • I'm going to go ahead and clean up my answer. I'll put the "$g \in N_G(H)$ is necessary" part of the proof at the bottom so that you can wait to look at it if you want. – Kyle Jun 27 '14 at 20:53
  • @squirrel Sorry, could you explain the implication in line 4, where you from $(*)$ imply deck transformation? You use it seemingly in $G(gH_e)=G(H_e)$, but I don't see any connection behind this. – nakajuice Jul 01 '14 at 02:03
  • @haemhweg: We have $G.(g.H_e)=Gg.H_e=G.H_e$ simply because $Gg=G$, independent of $()$. The content of $()$ is simply the necessary condition for us to invoke the universal property of quotient spaces; namely that a map $f: X \to Y$ descends to a unique map $\overline{f} : X/\sim \to Y$ if we have $f(a)=f(b)$ whenever $a \sim b$. The map $ \overline{\phi}$ is defined as $H_e \mapsto g.H_e$, the map $p_H$ is defined as $H_e \mapsto G.H_e$, and $(**)$ simply unpacks such definitions. – Kyle Jul 01 '14 at 02:22
  • @squirrel Thanks a lot for a detailed explanation! So without $(*)$ statement $\bar{\phi}$ wouldn't be well-defined, right? – nakajuice Jul 01 '14 at 02:35
  • @haemhweg: Exactly. And no problem. – Kyle Jul 01 '14 at 02:52
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The example for $\text{Question (iii)} $:

Consider the space E consisted of a disjoint sequence of $S^{1}$ lying along the x-axis, each $S^{1}$ is centered at an integer point on x-axis and radius is $1/2$ . The group action G is Z, and taking H to be the trivial subgroup $\{e\}$. Consider a deck transformation from H\E to H\E by interchanging the circles that are symmetric w.r.t the origin, and this transformation cannot be represented by any element of G, hence this is a counterexample.

Kushashwa Ravi Shrimali
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user147757
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