2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Just to make sure, $G\backslash E$ is the orbit space? Or is it in some way different from $E/G$? $\endgroup$ Jun 27 '14 at 13:02
  • $\begingroup$ It's just the quotient of $E$ by the equivalence relation $e\sim e.g$ (we use the convention of a right action). $\endgroup$
    – blst
    Jun 27 '14 at 13:04
  • $\begingroup$ What is $\Delta(p_H)$? $\endgroup$
    – Kyle
    Jun 27 '14 at 14:51
  • $\begingroup$ The group of decktransformations of $p_H$. $\endgroup$
    – blst
    Jun 27 '14 at 14:51
  • $\begingroup$ 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? $\endgroup$
    – nakajuice
    Jul 1 '14 at 1:34
2
$\begingroup$

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 $g \in N_G(H)$. $\square$

$\endgroup$
11
  • $\begingroup$ I have one problem with this argument: Our space here is not necessarily path connected. $\endgroup$
    – blst
    Jun 27 '14 at 14:55
  • $\begingroup$ 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. $\endgroup$
    – Kyle
    Jun 27 '14 at 15:14
  • $\begingroup$ 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 $gH=Hg$. $\endgroup$
    – blst
    Jun 27 '14 at 16:50
  • $\begingroup$ @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. $\endgroup$
    – Kyle
    Jun 27 '14 at 17:04
  • $\begingroup$ @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". $\endgroup$
    – Kyle
    Jun 27 '14 at 19:35
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.