# Covering spaces of surfaces

Let $\Sigma_g$ be a surface of genus $g\ge 2$, and let $\Sigma_k$ be an $m$-sheeted covering space of $\Sigma_g$. It is known that $k=m(g-1)+1$.

An example of such a covering space is a regular covering obtained by choosing one hole" as the center of the symmetry and take $\Sigma_k$ to have $m$-fold rotational symmetry around that chosen center (as on standard pictures in your favorite topology book).

Question: Does every finite sheeted regular covering space of $\Sigma_g$ arise in this way?

It feels like this should be known/standard, but I can't find an argument or a reference.

Edit: I agree, the question is not very precise as stated, I certainly didn't have in mind that every finite cover arises via cyclic symmetry as in the example above. The reason for the lack of precision is that I do not have a particular result in mind, but I would like to know if there is a simple geometric description of the covering map, as Misha points out, between two surfaces? The example with the cyclic group gives such a simple description of the covering map. Or, given any covering map $\Sigma_h\to \Sigma_g$ between two surfaces, is there some kind of a standard" covering $M\to \Sigma_g$, which factors through $\Sigma_h$?

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Not every finite group is cyclic, you probably had such examples in your algebra class. – Misha Aug 14 '12 at 20:55
I do not mean just cyclic groups, it is clear how this could work for other finite groups, and part of the question is: does it? – George Aug 14 '12 at 21:26
It's not clear to you what you mean by "this could work for other finite groups", but I don't think that there is any hope of doing anything like this. I recommend meditating on some more complicated finite covers (for instance, the cover corresponding to the kernel of the map $\pi_1(\Sigma_g) \rightarrow H_1(\Sigma_g;\mathbb{Z}/p)$; more general covers can be even more complicated still). – Andy Putman Aug 14 '12 at 21:30
@George: You question was about m-fold rotational symmetry around the chosen center. What could this possibly mean if the deck group in question is not cyclic? I think, you should first try to figure out what your question precisely is, before asking it. – Misha Aug 14 '12 at 22:02
If you are asking about characterization of covering spaces of the given degree rather than covering maps, then the answer to your question is trivially positive since a closed oriented surface is uniquely determined by its genus and genus of the covering surface is computed by the formula that you wrote. So, what exactly is your question: Covering spaces or covering maps? – Misha Aug 15 '12 at 15:45

Given any covering map $\Sigma_h\to\Sigma_g$ between two surfaces, is there some kind of a standard" covering $M\to \Sigma_g$, which factors through $\Sigma_h$?

In brief, the answer to this part of the question is 'You can take $M\to\Sigma_g$ to be regular, but beyond that, no.' This can already be extracted from Misha's comments above, but let me try to summarise.

First, note that there is a regular covering $M\to\Sigma_g$ that factors through $\Sigma_h\to\Sigma_g$. (Specifically, you can take $\pi_1M$ to be the intersection of all the conjugates of $\pi_1\Sigma_h$.) So, as in the earlier part of your question, you can take $\Sigma_h\to\Sigma_g$ to be regular.

You can now rephrase your question in terms of normal quotients of $\pi_1\Sigma_g$, and it becomes

Given any finite quotient $q:\pi_1\Sigma_g\to Q$, is there some kind of 'standard' finite quotient $p:\pi_1\Sigma_g\to P$ such that $q$ factors through $p$?

In particular, $P$ surjects $Q$. But any finite group can arise as $Q$ (see Misha's and algori's comments---the point is that $\pi_1\Sigma_g$ surjects a free group), so you are looking for a 'standard' family of finite groups that surjects every finite group. But there's no 'natural' definition of such a family.

Remark: Obviously, there are such families, such as $\{ Q\times\mathbb{Z}/2\}$ where $Q$ is an arbitrary finite group, but clearly this is not 'natural'.

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@HW: Just for the record, there is one more "natural" (and canonical) thing one can do, namely take the largest characteristic subgroup of finite index contained in the given finite-index subgroup. Maybe one should call this a "characteristic cover" by analogy with a regular cover. – Misha Aug 20 '12 at 20:25
@Misha - indeed, and, beyond that, you can also pass to a 'fully invariant' cover. But these are details. – HJRW Aug 21 '12 at 5:52
Ok, so this is the closest to what I had in mind and I accept this answer, but I think the actual question I am interested in (and the whole discussion above helped me to realize this as well, so thanks) is the following: given a finite covering space $\Sigma_h \to \Sigma_g$, is there a way to describe the covering map? The example with the cyclic group gives such a description. – George Aug 23 '12 at 18:00
@George - I still don't think it's clear what you mean by 'describe'. What features of the cyclic case would you like to replicate? The fact that it can be realised by rigid motions in $\mathbb{R}^3$ makes it particularly easy to picture, but of course this is very special. – HJRW Aug 25 '12 at 6:57

For an explicit example of a non-cyclic group $G$ acting freely on a surface $S$ take the union $X$ of the edges of the 3-cube $[-1,1]^3\subset\mathbb{R}^3$; set $S$ to be the boundary of a small neighborhood of $X$, and take the group generated by the rotations through $\pi$about the coordinate axes in $\mathbb{R}^3$ as $G$; note that $G$ is isomorphic to Klein's group $\mathbb{Z}/2\oplus\mathbb{Z}/2$, the "simplest" non-cyclic group there is.

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One should also note that, of course, every finite group appears as the covering group for some covering between surfaces. Say, every finite simple group appears as a covering group over genus 2 surface. – Misha Aug 15 '12 at 20:08
Misha -- re your first statement: yes, one can let a finite group act freely e.g. on the 1-skeleton of the simplex whose vertices correspond to the elements of the group, and then construct a surface as above. – algori Aug 15 '12 at 20:33
@algori: I meant an easier argument: Every genus $r$ surface group maps onto a rank $r$ free group, which then maps onto an $r$-generated finite group. For simple groups, then use the fact that they are all 2-generated. – Misha Aug 15 '12 at 20:44

The OP is asking for a classification of finite index normal subgroups of fundamental groups of closed surface. This is hairy, but algorithmic, see Gareth Jones' 1994 math Scand paper.

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Well, it's algorithmic for all finite 2-complexes, by Reidermeister--Schreier. – HJRW Aug 15 '12 at 5:59
This link has now "died"---can you please update it? – Danu Feb 25 at 10:35