Finding all connected covering spaces of the circle $S^1$

Question

I want to solve this problem that I found in a qualifying topology exam:

"Let $S^1$ be the unit circle in the complex plane. How many isomporphism classes of connected covering spaces of $S^1$ exist? Construct a representant of each class."

The following theorem is taken from Hatcher's Algebraic Topology:

So, this theorem could help to find the set of basepoint-preserving isomorphism classes of path-connected covering spaces $p:(\widetilde{X},\widetilde{x}_0)\to (S^1,1)$ where $S^1$ is the unit circle and $1=(1,0)$, and we have $\pi_1(S^1,1)=\Bbb Z$, but how can we know what $\pi_1(p_*(\widetilde{X},\widetilde{x}_0))$ is if we don't know $(\widetilde{X},\widetilde{x}_0)$? Also, this theorem helps to find path-connected covering spaces, but the problem is asking for connected covering spaces, so, is this theorem useful to solve this problem or not? And finally, how can the covering spaces be represented? I guess it's by a permutation because that's the next topic in Hatcher after this theorem, but how can we do that?

Answer 1

For your first question: as the theorem says, one side of the bijection is "the set of subgroups of $\pi_1(X,x_0)$". So for any subgroup $H < \pi_1(X,x_0)$, there is a corresponding covering space $p : (\widetilde X,\tilde x_0) \to (X,x_0)$, where the correspondence is given by $H = p_*(\pi_1(\widetilde X,\tilde x_0))$. The proof of the theorem tells you how to find the covering space that corresponds to $H$.

For your second question, $X$ is locally path connected, so all of its covering spaces are also locally path connected. Now you can apply a theorem from topology: assuming a space $Y$ is locally path connected, it follows that $Y$ is connected if and only if $Y$ is path connected. And, of course, like all manifolds the space $X=S^1$ is locally path connected.

For your final questions, as I said earlier, the representation of the covering space is given in the proof of the theorem in Hatcher.

Finally, let me address the the qualifying exam problem. If you know the fundamental group of $\pi_1(S^1)$, then you will be able to use your knowledge of algebra to list, explicitly, every single one of its subgroups. And now your job is to find the covering space corresponding to each subgroup $H < \pi_1(S^1)$. One can find that covering space, in general, by following the proof of the theorem, but perhaps your knowledge about $S^1$, which is a pretty simply space, will aid you in finding that covering space directly.

Answer 2

The good news is that we do know a universal cover of $S^1$, namely $\mathbb{R}\to S^1$ given by $x\mapsto e^{ix}.$ It is easy enough to see that this is a covering map, and we know that $\mathbb{R}$ is simply connected. We also know that the fundamental group $\pi_1(S^1,1)\cong \mathbb{Z}$ acts on $\mathbb{R}$ by deck transformations, namely topological automorphisms of $\mathbb{R}$ that preserve the projection $p:\mathbb{R}\to S^1$.

Now, another important fact here is that $S^1$ is a smooth manifold, and the covering spaces of smooth manifolds are also smooth manifolds. On a smooth manifold, the topology is regular enough that connectedness is equivalent to path connectedness. So, the path connected covering spaces of $S^1$ are exactly the connected covering spaces of $S^1$. In particular, the theorem applies.

We know the subgroups of $\mathbb{Z}$, they are of the form $n\mathbb{Z}$ for $n\in \mathbb{N}$. Because $\mathbb{Z}$ is Abelian, the conjugacy classes of subgroups are in bijection with the subgroups themselves. The connected covers corresponding to each subgroup are obtained by quotienting $\mathbb{R}$ by the suitable subgroup under the $\pi_1(S^1,1)$ action. So, if we take $n\mathbb{Z}$, the quotient $\mathbb{R}/n\mathbb{Z}\cong S^1$ (topologically) and the associated covering map is $q_n: S^1\to S^1$ given by $q_n(\theta)=\theta^n$, where the elements of $S^1$ are viewed as a elements of $\mathbb{C}$.