Quotient of the hyperbolic plane with respect to commutator group of $\pi_1(\Sigma_g)$

Question

Let $\Sigma_g$ be a Riemann surface of genus $g\geq 2$ and $G=\pi_1(\Sigma_g)$. Let $\pi\colon \mathbb{H}\to \Sigma_g$ be the universal covering map. What kind of surface is $\mathbb{H}/[G,G]$?

Moreover, what is $[G,G]$; e.g. if $g=2$?

Answer 1

Let me start by interpreting the question "What kind of surface is $S$?" in the case of a general connected oriented topological surface (without boundary). (I am considering only oriented surfaces just for simplicity of discussion.) If $S$ had finite complexity, i.e., would be homeomorphic to the interior of a compact oriented surface, you probably would be satisfied by the answer of the type "$S$ is has $n$ ends and genus $g$", since this provides a complete set of topological invariants. Surfaces of infinite complexity are also classified by a certain set of invariants:

1. Its set of ends (regarded as a topological space).

2. Its genus.

3. Its set of ends with positive genus.

You can find more details and references in this MO post.

If you look closely at the surface you are interested in, $H^2/[G,G]$, you realize that its invariants are:

1. The surface is 1-ended (simply because the abelian group $G/[G,G]$ is 1-ended).

2. It has infinite genus (this is easy to see and is explained in Sam's answer).

3. In particular, its only end has positive genus.

To summarize: Your surface is the unique connected oriented topological surface of infinite genus and one end. If you are looking for a different answer, you should clarify what does your question really mean.

Answer 2

$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = \Sigma_2$ be the genus two surface. In this case, $\ZZ^4$ is the deck group of the desired covering. Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$ parallel to the four coordinate axes. This gives a Cayley graph for $\ZZ^4$.

Next, starting at every vertex of the graph we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses). This is possible because the boundary word describes a closed loop in the graph. Let $S'$ be the resulting two-complex. Every edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells. The eight corners give the vertex a disk neighborhood in $S'$.

Thus $S'$ is a surface. Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$. By the Galois correspondence, $S'$ is the desired covering space. Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended. The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point. Thus $S'$ has genus, and so has infinite genus.

This construction works in any genus. When $g = 1$ the construction produces the universal cover.

Answer 3

Consider the Abel-Jacobi map $\mu : \Sigma_g \to J(\Sigma_g )=\mathbb{C}^g/\Lambda$. Then take the lift $\widetilde{\mu}: \mathbb{H} \to \mathbb{C}^g$ from the universal covering $\mathbb{H}$ of $\Sigma_g$. It seems to me that the image $X := \widetilde{\mu}(\mathbb{H})$ is the surface $\mathbb{H}/[G,G]$.