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 $\Sigma$.

  • $\begingroup$ 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

$\Sigma'$ 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.

  • $\begingroup$ 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

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