[Math] Abelian Covering Space of Surface of genus $g$

algebraic-topologycovering-spaces

I am doing Exercise 1.3.19 from Hatcher's Algebraic Topology 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 $g > 3$?

Thanks.

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

Best Answer

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.

Related Question