4
$\begingroup$

The question was arisen from doing Hatcher 1.3.24 (b). I would like to know what is wrong with the following.

Let $X$ be a space that is path-connected and locally path-connected and $G$ be a group that makes $X \twoheadrightarrow X/G$ a normal covering space. Let $H_{1}, H_{2} \leqslant G$ be subgroups. Then $X \twoheadrightarrow X/H_{1}$ and $X \twoheadrightarrow X/H_{2}$ are normal covering spaces as well with $H_{j} = \text{Aut}(X \twoheadrightarrow X/H_{j})$.

Now, fix any $\phi_{1} \in H_{1}$. Consider any $x_{0} \in X$ and by normality, we have $\phi_{2} \in H$ such that $\phi_{2}(x_{0}) = \phi_{1}(x_{0})$. But then $\phi_{1}, \phi_{2} \in G$ and since $X$ is path connected, the deck transformations are uniquely determined by the image of a fixed point. Therefore, we have $\phi_{1} = \phi_{2} \in H_{2}$, showing $H_{1} \leqslant H_{2}$.

The argument can be rewritten by switching $1$ and $2$, so $H_{1} = H_{2}$.

Problem. I was supposed to show that if $X/H_{1}$ and $X/H_{2}$ are isomorphic as covering spaces of $X/G$, then $H_{1}$ and $H_{2}$ are conjugate subgroups in $G$. Clearly, I haven't used the hypothesis of the statement and the conclusion is stronger as well.

$\endgroup$
  • $\begingroup$ Wy are the coverings $X\to X/H_i$ normal? $\endgroup$ – Andreas Blass Aug 11 '13 at 15:31
4
$\begingroup$

Always good to make a diagram:

$$\begin{matrix} X & \xrightarrow{\pi_1} & X/H_1 & \xrightarrow{\psi_1} & X/G\\ & & \downarrow\iota & & \downarrow\\ X & \xrightarrow[\pi_2]{} & X/H_2 & \xrightarrow[\psi_2]{} & X/G \end{matrix}$$

So the task is to show that if there exists a homeomorphism $\iota \colon X/H_1 \to X/H_2$ with $\psi_2 \circ \iota = \psi_1$, then $H_1$ and $H_2$ are conjugate in $G$. (Note: except if $H_1 = H_2$, we do not have $\iota\circ \pi_1 =\pi_2$.)

Now, fix any $\phi_{1} \in H_{1}$. Consider any $x_{0} \in X$ and by normality, we have $\phi_{2} \in H$ such that $\phi_{2}(x_{0}) = \phi_{1}(x_{0})$.

No. The normality of $\pi_i$ says that $H_i$ operates transitively on the fibres of $\pi_i$, not on the fibres of $\pi \colon X \to X/G$. $x_0$ and $\phi_1(x_0)$ generally lie in different orbits under the action of $H_2$, and that means in different fibres of $\pi_2$, i.e. in general, there is no $\phi_2 \in H_2$ with $\phi_2(x_0) = \phi_1(x_0)$. If there always is, then $H_2 \supset H_1$.

So let's prove that if $\iota$ exists, then $H_1$ and $H_2$ are conjugate (the converse holds too, if $H_1$ and $H_2$ are conjugate, then there is such an $\iota$).

Pick any $x \in X$. Choose any $y \in X$ with $\pi_2(y) = \iota(\pi_1(x))$.

Then $\pi_G(y) = \psi_2(\pi_2(y)) = \psi_2(\iota(\pi_1(x))) = \psi_1(\pi_1(x)) = \pi_G(x)$, therefore there is a $g \in G$ such that $gx = y$.

Claim: $\pi_2 \circ g = \iota \circ \pi_1$, i.e. $g$ is a lift of $\iota$.

We know that $\pi_2(gx) = \iota(\pi_1(x))$. Let $z \in X$ arbitrary, and choose a path $\gamma$ connecting $x$ and $z$. Since paths can always be lifted, there is a path $\beta$ in $X$ starting at $y$ such that $\pi_2 \circ \beta = \iota \circ \pi_1 \circ \gamma$. Now,

$$\pi_G \circ \beta = \psi_2\circ \pi_2 \circ \beta = \psi_2 \circ \iota \circ \pi_1 \circ \gamma = \psi_1 \circ \pi_1 \circ \gamma = \pi_G \circ \gamma,$$

that is, $\beta$ is a lift of $\pi_G \circ \gamma$ starting in $y$. But $g\cdot \gamma$ is also such a lift of $\pi_G \circ \gamma$, and hence $\beta = g\cdot\gamma$ by the uniqueness of lifts, and thus $\beta(1) = g\gamma(1) = gz$, whence $\pi_2(gz) = \pi_2(\beta(1)) = \iota(\pi_1(\gamma(1))) = \iota(\pi_1(z))$.

$\pi_2\circ g = \iota \circ \pi_1$ is a different way of writing $H_2g \supset gH_1$ or $gH_1g^{-1} \subset H_2$.

We have not yet used that $\iota$ was supposed to be a homeomorphism, just that it was a continuous map with $\psi_2\circ \iota = \psi_1$. The same construction for $\iota^{-1}$ choosing $y = gx$ and $x$ as starting points shows $\pi_1 \circ g^{-1} = \iota^{-1}\circ \pi_2$ or $g^{-1}H_2 \subset H_1g^{-1}$. Together $H_2 = gH_1g^{-1}$.

$\endgroup$
  • $\begingroup$ Thank you. I have followed different books from Hatcher since when I posted this question and now this started to make sense (maybe Hatcher was too difficult for me to start with this subject). $\endgroup$ – Gil Oct 15 '13 at 14:54
  • $\begingroup$ @DanielFischer Can you please explain why is $\pi_2\circ g=i\circ \pi_1$ is same as saying that $H_2g\supseteq gH_1$? Thank you. $\endgroup$ – caffeinemachine May 29 '16 at 11:18
  • $\begingroup$ @caffeinemachine Pick $\xi \in X$ and $h_1 \in H_1$. Then we have $\pi_1(h_1\xi) = \pi_1(\xi)$ by definition of $\pi_1$, and consequently $\iota(\pi_1(h_1\xi)) = \iota(\pi_1(\xi))$. Now using $\iota\circ\pi_1 = \pi_2(g)$ we obtain $\pi_2(gh_1\xi) = \pi_2(g\xi)$, that is, $g\xi$ and $gh_1\xi$ lie in the same fibre of $\pi_2$, i.e. there is a $h_2\in H_2$ with $h_2(g\xi) = gh_1\xi$. The deck transformations $h_2g$ and $gh_1$ agree at one point ($\xi$), so they are the same deck transformation (since $X$ is connected), i.e. $gh_1 = h_2g$. That means $gh_1 \in H_2g$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22
  • $\begingroup$ Since $h_1\in H_1$ was arbitrary, we have $gH_1\subseteq H_2g$. This is the direction I use, and I didn't really intend to state equivalence. But well, I sort of did, so: Conversely, if we have $gH_1\subseteq H_2g$, the covering $\pi_2\circ g\colon X\to X/H_2$ factors through $X/H_1$, i.e. there is a unique covering map $\theta\colon X/H_1\to X/H_2$ with $\pi_2\circ g=\theta\circ\pi_1$, and one verifies that $\psi_2\circ\theta=\psi_1$. By definition of $g$, we have $\iota(\pi_1(x)) = \pi_2(gx) = \theta(\pi_1(x))$, so the two covering maps $\theta$ and $\iota$ agree at the point $\pi_1(x)$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22
  • $\begingroup$ Locally, we can write both of $\iota$ and $\theta$ in the form $\psi_2^{-1}\circ \psi_1$ (with an appropriate branch of $\psi_2^{-1}$), hence [since the $\psi_k$ are local homeomorphisms], the sets $A = \{p \in X/H_1 : \theta(p) = \iota(p)\}$ and $B = \{ p \in X/H_1 : \theta(p) \neq \iota(p)\}$ are both open. By connectedness, one of the two is empty. Since $\pi_1(x) \in A$, it follows that $B = \varnothing$, i.e. $\theta = \iota$. $\endgroup$ – Daniel Fischer May 29 '16 at 13:22

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.