Regular covering corresponding to a kernel

Question

I am reading about the Burau representation of Braid groups from this thesis, and there's a line where it says

Let $\tilde{D}_n$ be the regular covering space of $D_n$ corresponding to the kernel of φ.

Here, $D_n$ is the $n$-times punctured disc and the homomorphism φ takes each loop to its total winding number (sum of winding numbers with respect to each puncture).

I don't understand what it means for a covering space to correspond to a kernel. At first, I thought thay it could be that the group of deck transformations was isomorphic to this kernel, but I think that's not the case, because, as it says, the group of deck transformations is $\mathbb{Z}$.

Afterwards, the author geometrically explains the covering space, but I can't find any relationship between the covering space and the kernel (the loops with $0$ total winding number).

Edit: In this particular case, there is a representation where the total winding number represents a slice of the covering space (each positive wind moves upwards and each negativ wind moves downwards). Therefore, the kernel corresponds to staying at the level $0$, but does this work in general for any "covering space corresponding to a kernel"?

Answer 1

This is a core idea in the theory of covering spaces. Allen Hatcher's Algebraic Topology, Chapter 1, is a good place to start.

Here's an example to build intuition (and your intuition is already correct in your "Edit"): Let $X$ be the unit circle. It's fundamental group is $\mathbb{Z}$. Every subgroup is normal (since the group is abelian) and so is the kernel of some homomorphism. To the kernel of the map which reduces an integer modulo $n > 1$ corresponds the $n$-fold covering of $X$ by itself, i.e. $p: X \to X$ is given by $p(z) = z^n$, where $X$ is identified with the unit complex numbers. Every loop in the total space is mapped to a loop which winds around the origin zero times (modulo $n$) in the base space. So, elements of the fundamental group of the total space are mapped to elements in the kernel $n \mathbb{Z}$. Conversely, a loop whose homotopy class is in the kernel lifts to a loop in the total space. (In general, if you choose any loop, you do not expect it to lift to a loop, but rather just to a path.)

In general, elements of a given subgroup of the fundamental group of the base space are represented by loops which lift to loops in the corresponding covering space. And loops which do not lift to loops (but rather the lift has distinct endpoints) do not belong to the given subgroup.

To do this correctly and rigorously, it is essential to keep track of the basepoints. However, if the given subgroup is normal, then you can more or less ignore the basepoints since the covering space is "regular" (sometimes called normal or Galois). There are other subtleties as well, especially the existence of coverings and the correct notion of what it means for two (based) covering spaces to be equivalent so that there is indeed a 1-1 correspondence between subgroups and covering spaces.)

Answer 2

I think I've finally found it.

In Hacther's Algebraic Topology there's a table of examples of covering spaces for $S^1\lor S^1$ labeling the loops as $a$ and $b$, respectively. For each free group generated by combinations of $a$ and $b$ (and their inverses), a covering space is given. One can see these groups the groups generated by $\rho_*(\pi_1(\tilde{X},\tilde{x}_0))$ in $\pi_1(X,x_0)$, where $\rho$ is the covering map and $\tilde{x}_0\in\rho^{-1}(x_0)$ (a different $\tilde{X}$ for each covering).

In this case, the kernel is a group. Hence, we can assign a covering space of $D_n$ according to this group. The covering given is the corresponding to the kernel since, fixing a base point $d_0\in D_n$, any loop in $\tilde{D}_n$ based at a fixed $\tilde{d}_0\in\rho^{-1}(d_0)$ goes upwards with every positive wind and downwards with every negative wind. Thus, to end up in $\tilde{d}_0$ again, it has to have total wind number $0$. Since this loop is a lifting from a loop at $d_0$, the total winding number is the same, hence the loop is in the kernel (this fact is geometrically evident).