1
$\begingroup$

Let $\Sigma$ be a closed orientable surface of genus $g \geq 2$. Suppose that $\Sigma = \partial H$ where $H$ is a handlebody. We then have the subgroup $N$ which is the kernel of the inclusion $\pi_1(\Sigma) \to \pi_1(H)$. I know that this map is surjective on $\pi_1$ and since $\pi_1(H)$ is a free group of rank $g$, $N$ must be a countably generated free group.

What is the topology of the surface $\Sigma'$ covering $\Sigma$ corresponding to $N$? I imagine that since the group must be free, there can not be any genus. A reference for the result would be wonderful.

I know that orientable open surfaces are classified by their genus (either finite or infinite) and their ideal boundary. More generally I would love to know which of these types of surfaces can occur as (regular and/or irregular?) coverings of $\Sigma$.

$\endgroup$
  • $\begingroup$ Since you appear to be looking for a reference, I have added the appropriate tag to your question. Please feel free to rollback if this is not what you want. $\endgroup$ – Xander Henderson Mar 10 '18 at 4:44
0
$\begingroup$

$\Sigma'$ is homeomorphic to the sphere minus a Cantor set.

Edited to address comments: One way to prove this is to prove that the universal cover $\tilde H$ of the handlebody $H$ is homeomorphic to the 3-ball $B^3$ minus a Cantor subset $C \subset \partial B^3 = S^2$, and that the restriction of the universal covering map $\tilde H \to H$ to the boundaries $\partial\tilde H \to \partial H = \Sigma$ is a regular covering map with deck transformation group $N=\pi_1 H$, hence $\Sigma' \approx S^2 - C$.

And if you want to know why $\tilde H$ has the description that I say it has, a good way is to use 3-dimensional hyperbolic space $\mathbb{H}^3$. Using the ping-pong argument, there is an action of $F_g$ on $\mathbb{H}^3$ and a Cantor subset $C$ in the sphere at infinity $S^2_\infty = \partial \mathbb{H}^3$ (the set $C$ is called the "limit set" of the action) such that the restriction of $F_g$ to $\overline{\mathbb{H}}^3 - C \approx B^3 - C$ is free and properly discontinuous, and the quotient of this restricted action is homeomorphic to the genus $g$ handlebody.

Also, just because one is removing a Cantor subset $C \subset S^2$ it does not follow that $S^2 - C$ is homotopy equivalent to an uncountable wedge of circles. In fact $S^2-C$ is homotopy equivalent to a countable wedge of circles.

$\endgroup$
  • $\begingroup$ Thanks, for the response! Do you know how this is proven? Also wouldn't such a space have an uncountably generated free group (since it is homotopy equivalent to an uncountable wedge of circles) which contradicts being a subset of $\pi_1(\Sigma)$. I am guessing you are correct and I am wrong in thinking that the sphere minus a cantor set this way. $\endgroup$ – user101010 Mar 11 '18 at 21:22

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.