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I am doing Problem 1.3.19 of Hatcher and I come to this part of the problem:

For $n = 3$ and $g\geq 3$, describe a normal covering space $\tilde{X}$ of $X=M_g$, the surface of genus $g$ explicitly with deck transformation group $G(\tilde{X})$ consisting of translations isomorphic to $\Bbb{Z}^3$.

Now in the case that $g = 3$ we can get such a covering space from the topological quotient map $p : \Bbb{R}^3 \to \Bbb{R}^3/\Bbb{Z}^3$, that is we quotient out $\Bbb{R}^3$ by the action of $\Bbb{Z}^3$ on $\Bbb{R}^3$ defined as follows. For $x \in \Bbb{R}^3$ and $(a,b,c) \in \Bbb{Z}^3$.

$$(a,b,c) \cdot x = \text{translation by the vector $(a,b,c)$}.$$

The orbit space is isomorphic to $M_3$ and so $\Bbb{R}^3$ is the desired covering space. However for $g > 3$ I am finding it hard to visualize such a covering space. How can I find one for $g > 3$?


Edit: I just realised the orbit space is $S^1 \times S^1 \times S^1$ and so the example above fails.

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up vote 10 down vote accepted

Let's focus on the surface $\Sigma_3$ of genus $3$ just to be concrete.

Here's the geometric idea: picture

The basic idea is that the lattice $\mathbb{Z}^3$ has to act by deck transformations on the covering space. So we have to have three curves on $\Sigma_3$ which "unwind" to the grid pictured on the right in the covering space. From there it's not a stretch to the cut-and-paste operation in the picture.

Formally - without exhibiting explicit formulae - I think of this two ways.

  1. Three disjoint essential simple closed curves on $\Sigma_3$, $\alpha_1,\alpha_2,\alpha_3$, generate a copy of $F_3$, the free group on three generators, in $\pi_1\Sigma_3$. Associated to $F_3$ is a covering $\widehat{\Sigma_3}\to \Sigma_3$, with $F_3$ acting by deck transformations on $\widehat{\Sigma_3}$ so that $\widehat{\Sigma_3}/F_3 = \Sigma_3$. Since $Z(F_3)<F_3$, we have a quotient cover $\widehat{\Sigma_3}/Z(F_3)\to \Sigma_3$, with the deck group $F_3/Z(F_3)\cong \mathbb{Z}^3$. Geometrically, take the "asterisk" fundamental domain (the one you get by cutting along the red curves) and glue it so it looks like the Cayley graph of $F_3$. Then quotient by the action of the commutators of the generators.

  2. The fundamental group $\pi_1(\Sigma_3)$ projects onto the first homology group $H_1(\Sigma_3;\mathbb{Z})$. Associated to $H_1(\Sigma_3;\mathbb{Z})$ is an abelian cover $A(\Sigma_3)\to \Sigma_3$. The images of $\alpha_1,\alpha_2,\alpha_3$ generate a copy of $\mathbb{Z}^3$ in $H_1$, which corresponds to a $\mathbb{Z}^3$-covering space of $\Sigma_3$.

I think there should be a nice picture of a fundamental domain of this cover in $\mathbb{H}^2$ by stringing together right-angled dodecagons, but I don't quite see it.

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Consider the space:

$$M = (\mathbb{R} \times \mathbb{Z} \times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{R} \times \mathbb{Z}) \cup (\mathbb{Z} \times \mathbb{Z} \times \mathbb{R}).$$ This looks like a $1$-dimensional "grid" inside $\mathbb{R}^3$. Now, it's easy to see that you can fatten up $M$ a little bit, and delete its insides, so that it becomes hollow; this way, you get something that looks like a "torus" grid inside $\mathbb{R}^3$. Call this surface $\widetilde{M}$.

Now consider the (covering) action generated by the translations: $$T_1(x, y, z) = (x+1, y, z),~~T_2(x, y, z) = (x, y+1, z),~~T_3(x, y, z) = (x, y, z+1).$$ The quotient $\widetilde{M}/\langle T_1, T_2, T_3 \rangle$ is a genus $3$ surface. To see this, think of a fundamental region for the quotient; this should look like three tubular segments of length $1$ having one common intersection. Each of these tubes has its edges identified, so you get $3$ "holes" (sorry for not providing pictures, but I think you'll be able to see this). This is your surface for $g = 3$. If you now consider, more generally, the action generated by $$T_1(x, y, z) = (x+(m-2), y, z),~~T_2(x, y, z) = (x, y+1, z),~~T_3(x, y, z) = (x, y, z+1),$$ then $\widetilde{M}/\langle T_1, T_2, T_3 \rangle$ should give you the surface of genus $2(m-2)+1$, for $m \geq 3$. So this takes care of all coverings of surfaces having odd genus.

Can you get the even ones?

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How did you come up with such a covering space? I am having a problem even visualizing $\tilde{M}$. – user38268 Nov 9 '12 at 5:31
@BenjaLim can you visualize $M$? If I'm not mistaken, you can picture $\widetilde{M}$ as being the set of points $x \in \mathbb{R}^3$ such that $d(x, M) = 1/10$, where $d$ is just the euclidean distance. Any very small distance should do the job. The idea here is to convert line segments inside $\mathbb{R}^3$ into small cylinders. This gives you a surface. Now, in any cylindrical segment (that is, finite-length cylinder), if you identify the opposite edges, you get a torus. Does that help? – student Nov 9 '12 at 6:03
I have posted an answer below. Please see it. – user38268 Nov 9 '12 at 7:13

My understanding of student's answer is as follows. Take the subspace of $\Bbb{R}^3$ that is

$$M = (\text{lattice in $x - y$ plane} \times \Bbb{R}) \cup (\text{lattice in $y - z$ plane} \times \Bbb{R}) \cup (\text{lattice in $x - z$ plane} \times \Bbb{R}).$$

Now if we let $\Bbb{Z}^3$ act on this plainly, the orbit space of this will just be $S^1 \vee S^1 \vee S^1$. What do we do? Well $M$ above is the space of all lines in $\Bbb{R}^3$ parallel to either the $x$,$y$ or $z$ axis such that each line has to pass through only lattice points.

We let $\tilde{M}$ be the space of all "hollow pipes" or "tubes" that are parallel to either the $x$, $y$ or $z$ axis such that each of these tubes or pipes has to pass through lattice points. The orbit space of this under the usual action of $\Bbb{Z}$ looks like the unit cube whose edges are pipes:

enter image description here

such that the "beginning" and end of the pipes are made the same. In other words, the orbit space is $M_3$.

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Perhaps it would help to visualize what sort of family of curves in $M_g$ could possibly generate a subgroup of $\pi_1M_g$ isomorphic to $\mathbb{Z}^3$. This is a family of $3$ disjoint curves. Then cut along three disjoint circles transverse to each of these curves, make infinitely many copies, and glue subject to the commutation reltaions. This is where the "lattice of tubes" comes from. – Neal Nov 9 '12 at 11:02
For precision at the expense of clarity: if $\gamma_j$ are disjoint circles generating $\mathbb{Z}^3$ in $\pi_1$, choose three disjoint circles $\alpha_i$ so that the geometric intersection number $\iota(\alpha_i,\gamma_j)=\delta_{ij}$ and pick a circle in each homotopy class which realizes these numbers. Then cut, etc. ... – Neal Nov 9 '12 at 11:51
@Neal Thanks for your feedback. Are you saying that my interpretation of the student's answer above is correct? Also how do we get those covering spaces for $g$ even? – user38268 Nov 9 '12 at 12:02
@BenjaLim You're on the right track, but the way you wrote the set $M$, it seems that it contains planes, when in fact it's just a bunch of lines, as you correctly state afterwards. The region you drew is not really the quotient; you have to reduce it some more because the edges in that picture are still identified. Instead of thinking of a "cube" as in your picture, think about a common intersection of three tubes. – student Nov 9 '12 at 16:50
@BenjaLim I have written an answer. I'm sorry to take so long to get to it. – Neal Nov 15 '12 at 5:13

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