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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 $g=2$?

Sam Nead
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Mohammad Farajzadeh-Tehrani
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    The commutator subgroup of a surface group is free. See corollary 4 at http://chiasme.wordpress.com/2014/08/27/on-subgroups-of-surface-groups/ In particular, it can be deduced that $\mathbb{H} / [G,G]$ is not compact from the abelianization of $G$. – Seirios Oct 18 '14 at 03:50
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    I would run that argument the other way: $\mathbb{H}/[G, G]$ is an infinite covering space of $\Sigma_g$, hence it is noncompact. As a noncompact surface, it's homotopy equivalent to a $1$-dimensional CW complex, hence its fundamental group is free. – Qiaochu Yuan Oct 18 '14 at 05:10
  • Is there any particular dictionary for surfaces you would like to use, Mohammad? The surface has fairly direct descriptions in most ways I can think of thinking of surfaces. – Ryan Budney Oct 18 '14 at 05:36
  • Thanks Seirios: But what is rank of it? Can we say that? at least for g=2? – Mohammad Farajzadeh-Tehrani Oct 18 '14 at 11:22
  • Of course it is infinite, but it should be still describe-able in terms on number of ends and genus; e.g. a pair of pants is genus zero with 3 ends. In this case, I feel it should be a genus 0 curve with infinity many ends dictated by the type of free graph. – Mohammad Farajzadeh-Tehrani Oct 18 '14 at 11:37
  • I think the following describes the generators:http://mathoverflow.net/questions/38413/commutator-subgroup-of-a-surface-group – Mohammad Farajzadeh-Tehrani Oct 18 '14 at 14:04
  • The classification of noncompact boundaryless surfaces is more complicated than that. If orientable, they are classified genus (maybe infinite) and (homeo class of) a pair of compact totally disconnected metrisable spaces of ends $(E\supset F)$, with $F$ the subspace of nonplanar ends. This is due to Kerekjarto (1923). In your case, I would guess that $E$ is a Cantor and $F$ is empty, but this is only a guess. – BS. Oct 19 '14 at 08:21
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    It seems I was wrong. Kerekjarto says (if I correctly understood, p. 183 of "Vorlesungen uber Topologie) that this "homology covering" of a genus $g\geq 2$ surface has infinte genus and only one end (so $E=F$=point). So it is homeomorphic to the half-infinite "Jacob ladder". – BS. Oct 19 '14 at 15:06
  • @ BS: interesting, I had never heard of such terminologies. Is there any somewhat recent book discussing this sort of things! – Mohammad Farajzadeh-Tehrani Oct 20 '14 at 00:26
  • For fun: Finite Jacob ladder can be seen here http://www.cambridgeblog.org/2013/01/into-the-intro-games-and-mathematic/. – Mohammad Farajzadeh-Tehrani Oct 20 '14 at 00:45
  • Unfortunately I didn't found any recent textbook on this subject. By the way, the classification theorem proves that the semi-infinite Jacob ladder is homeomorphic to the "Loch Ness monster"! http://en.wikipedia.org/wiki/Loch_Ness_monster_surface – BS. Oct 20 '14 at 10:23

3 Answers3

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$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$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.

Sam Nead
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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.

Community
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Misha
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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 $\mathbb{H}/[G,G]$.

Holonomia
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