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The fundamental group of the Torus is $\mathbb{Z} \times \mathbb{Z}$, and unless I'm wrong all the subgroups are one of the following forms:

(Grids) $m\mathbb{Z} \times n\mathbb{Z}$ for $m, n \in \mathbb{Z}$

(Lines) $\mathbb{Z} \times (m\mathbb{Z} + b)$ for $m, b \in \mathbb{Z}$

(Trivial Group) $\{(0,0)\}$

So to find alll the covering spaces of the Torus it suffices to find a covering space corresponding to each of the groups above (thanks to the correspondence with covering spaces and subgroups of the fundamental group).

Let's get the easy ones out of the way:

$\{0\}$ corresponds to $\mathbb{R}\times \mathbb{R}$ and $S^1 \times S^1$ corresponds to $\mathbb{Z} \times \mathbb{Z}$ itself of course.

But to be honest I'm unsure about the rest.

Now for $m\mathbb{Z} \times n\mathbb{Z}$. Here we try to use $\mathbb{R}^2$ as well: Subdivide $\mathbb{R}^2$ into a grid of rectangles with length $m$ and width $n$ (vertices at integers for convenience). Then identifying the opposite sides of each rectangle in the usual way will give us our covering space. But I'm not sure if the fundamental group of this is $m\mathbb{Z} \times n\mathbb{Z}$.

As for the lines, it makes sense to me that the covering spaces look like $S^1 \times \mathbb{R}$ but I can't seem to make the idea rigorous (instead of a grid we end up with a line in $\mathbb{R}^2$ instead?)

Rotatopotato
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    The way you denoted the "lines" subgroups is somehow misleading. I would have used the notation $\langle (a,b)\rangle$ (the group generated by the element $(a,b)\in\mathbb{Z}^2$). Also, the "grids" are not necessarily aligned with the axes. For example the subgroup generated by (1,1) and (1,-1). – sss89 Oct 06 '20 at 06:23

1 Answers1

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Your list of subgroups is incomplete (and your "lines," as you've written them, aren't subgroups). The list is the following:

Rank 2: These are the lattices. They have the form $\text{span}(v, w)$ where $v, w \in \mathbb{Z}^2$ are linearly independent, and $v, w$ need not be aligned with the unit axes. Here's a picture of the lattice spanned by $v = (4, 1), w = (1, 2)$.

enter image description here

Rank 1: These are the lines. They have the form $\text{span}(v)$ where $v \in \mathbb{R}^2$ is nonzero, and again they need not be aligned with the unit axes. Your notation, as written, has the following meaning: $m\mathbb{Z} + b = \{ mx + b : z \in \mathbb{Z} \}$ (not a subgroup if $b \neq 0$), and $\mathbb{Z} \times (m\mathbb{Z} + b) = \{ (x, my + b) : x, y \in \mathbb{Z}^2 \}$ (not a subgroup if $b \neq 0$, and if $b = 0$ has rank $2$ and not $1$).

Rank 0: We only have the zero subgroup $0 = \text{span}(\emptyset)$ here.

The rank 2 subgroups are exactly the finite index subgroups, and you can show in several different ways that the corresponding covering spaces are all toruses again. The covering spaces for the rank 1 subgroups are all homeomorphic to $S^1 \times \mathbb{R}$, and of course the covering space for the zero subgroup is $\mathbb{R}^2$. This is actually important: one of the easiest ways to see what the other covering spaces are is to write them as quotients of $\mathbb{R}^2$ by the corresponding subgroups.

Qiaochu Yuan
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  • Whoops thanks about the subgroups.

    I'm not sure what you mean by seeing the other covering spaces of $\mathbb{R}^2$ as quotients of the corresponding subgroups. I know in the "artificial" construction of the universal cover as the space of (equivalence classes) of paths, we can quotient out by a subgroup of the fundamental group in a natural sense, but I'm not sure how that works for $\mathbb{R}^2$. Is it as simple as considering $\mathbb{R}^2/G$ where $G$ is one of the groups above?

     – Rotatopotato Oct 06 '20 at 11:46
  • Also I think I might need a hint to get started and understand how to frame these arguments. Would you mind helping me out how to see that the rank 2 groups correspond to toruses (tori?). Hopefully then I'll be able to figure out the rank 1 case. – Rotatopotato Oct 06 '20 at 14:41
  • @Rotatopotato: 1) yes, it's as simple as considering $\mathbb{R}^2$ mod one of the groups above. Covering space theory for topological groups is very nice! 2) Every rank $2$ lattice in $\mathbb{R}^2$ is conjugate to $\mathbb{Z}^2$ by a change of coordinates. More explicitly, if you're familiar with how $\mathbb{R}^2/\mathbb{Z}^2 \cong T^2$ corresponds to identifying sides of a square, taking $\mathbb{R}^2/\Gamma$ for $\Gamma$ a lattice corresponds to identifying sides of a parallelogram (a fundamental parallelogram of the lattice). – Qiaochu Yuan Oct 06 '20 at 17:04
  • Sorry what I meant was I don't understand why $\mathbb{R}^2/\Gamma$ should have fundamental group $\Gamma$. Thanks for your help by the way. – Rotatopotato Oct 06 '20 at 19:24
  • @Rotatopotato: if $\Gamma$ is a discrete subgroup then the map $\mathbb{R}^2 \to \mathbb{R}^2/\Gamma$ is a covering map with $\Gamma$ acting by deck transformations, and $\mathbb{R}^2$ is simply connected. – Qiaochu Yuan Oct 06 '20 at 19:26