I am trying to find all possible covering spaces of the 2-torus $T = S^1 \times S^1$ that is finitely-sheeted of degree $n$, that is, each point $x \in T$ has a neighborhood $U_x$ whose preimage under the covering map is the union of precisely $n$ disjoint open sets which are homeomorphic to $U_x$, up to isomorphism.
I used the classification theorem for covering maps to put the total number of isomorphism classesinto one-to-one correspondence with conjugacy classes of subgroups of $\pi_1(T) = \mathbb{Z} \times \mathbb{Z}$, we know from algebra there are 3 such classes. I also know that the following spaces are covering spaces for $T$:
- The torus $T$ itself
- The open cylinder $S^1 \times(0,1)$
- The open square $(0,1) \times (0,1)$
Which are not equivalent since (1) is compact and (3) is simply connected. It is clear that (1) is a covering space of itself with 1 sheet, and it can be made a covering space of precisely $n$ sheets by reparameterizing itself as a torus that spins around "$n$ times as fast". However, when it comes to spaces (2) and (3), I can only think of covering maps of infinite degree. Are there finitely-sheeted covering maps for (2) and (3)? If not, why?
What's a geometrically intuitive way of approaching this kind of problem?
Every finite cover of a compact space is compact, and in this case every finite cover of a torus is in fact another torus. This is because every finite index subgroup of the fundamental group $\mathbb{Z}^2$ is another copy of $\mathbb{Z}^2$ sitting discretely and cocompactly inside $\mathbb{R}^2$ (a lattice in $\mathbb{R}^2$), so the quotient by such a subgroup is just another torus.
Every lattice is generated by two integer vectors $(a, b), (c, d) \in \mathbb{Z}^2$ which are linearly independent over $\mathbb{R}$, meaning that $ad - bc \neq 0$, and corresponds to an $n$-sheeted cover where
$$n = \left| \det \left[ \begin{array}{cc} a & c \\ b & d \end{array} \right] \right| = |ad - bc|$$
is the area of a fundamental parallelogram for the lattice. These are the subgroups of $\mathbb{Z}^2$ of index $n$, and it's an interesting question to count them exactly; the answer turns out to be the divisor function $\sigma_1(n) = \sum_{d \mid n} d$.