Classifying Covering Spaces using First Cohomology

Question

I am familiar with the classification of covering spaces of a space $X$ in terms of subgroups of $\pi_1(X)$ (up to conjugation). However, if $X$ is a manifold, I know that $H^1(X; G)$ classifies G-bundles over $X$ (using Cech cohomology here). I think finite regular covering spaces are $\mathbb{Z}/k \mathbb{Z}$-bundles; regular means that the deck transformations act transitively on the fiber (and regular covers correspond to normal subgroups of $\pi_1(X)$). Does this mean that $H^1(X; \mathbb{Z}/k\mathbb{Z})$ is in bijection with k-sheeted regular covering spaces over $X$. I could not find such a statement anywhere and so am a bit suspicious.

Also, if this is correct, what does $H^1(X; \mathbb{Z})$ classify? I'm not sure what a $\mathbb{Z}$-bundle is - what has automorphism group equal to $\mathbb{Z}$? Also, $H^1(X; \mathbb{Z}) = [X, S^1]$ so if $H^1(X; \mathbb{Z})$ classifies some kind of bundles, there should be universal bundle over $S^1$ which pulls back to these bundles. What is this bundle?

Answer 1

It's somewhat delicate here to get all the details right. First, $G$-bundles for a finite group $G$ are not required to be connected, so the relevant version of the classification of covering spaces is the disconnected version, which goes like this: the category of covering spaces of a nice connected space $X$ with basepoint $x$ is equivalent to the category of $\pi_1(X, x)$-sets.

More explicitly, $n$-sheeted covers (possibly disconnected) are equivalent to actions of $\pi_1(X, x)$ on $n$-element sets, or even more explicitly to conjugacy classes of homomorphisms $\pi_1(X, x) \to S_n$. Said another way, $n$-sheeted covers, possibly disconnected, are classified by the nonabelian cohomology set

$$H^1(X, S_n).$$

Among these, the connected covers correspond to the transitive actions, which are classified by conjugacy classes of subgroups of $\pi_1(X, x)$ of index $n$. Among these, the regular covers correspond to normal subgroups.

Now, for $G$ a finite group, a $G$-bundle is more data than a $|G|$-sheeted cover: the fibers are equipped with a free and transitive right action of $G$ and everything has to be compatible with this. Said another way, $G$-bundles are equivalent to actions of $\pi_1(X, x)$ on $G$ regarded as a right $G$-set, or more explicitly to conjugacy classes of homomorphisms $\pi_1(X, x) \to G$ (thinking of $G$ as a subgroup of $S_{|G|}$ to make the connection back to covers).

Given a finite regular $n$-sheeted cover $Y \to X$ with corresponding normal subgroup $H = \pi_1(Y, y)$ of $\pi_1(X, x)$, we can think of this cover as a $G = \pi_1(X, x)/H$-bundle, but not all $G$-bundles arise in this way (the monodromy map $\pi_1(X, x) \to G$ is not required to be surjective in general), and we only know that $G$ is some finite group of order $n$. Moreover, the data of a $G$-bundle includes the data of an isomorphism between $G$ and this quotient; it's not enough just to know that it exists.

So we can find a finite regular $n$-cover which is not a $\mathbb{Z}/n\mathbb{Z}$-bundle, even up to isomorphism of covers, by finding a group $\pi_1(X, x)$ with a normal subgroup $H$ of index $n$ such that the quotient is not $\mathbb{Z}/n\mathbb{Z}$. A simple example is $X = T^2, \pi_1(X, x) \cong \mathbb{Z}^2$; take $H = 2 \mathbb{Z}^2$, so that the quotient is $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

And we can find a $\mathbb{Z}/n\mathbb{Z}$-bundle which is not a finite regular $n$-cover in the usual sense, again even up to isomorphism of covers, by finding a disconnected such bundle; for example, $X \times \mathbb{Z}/n\mathbb{Z}$ for $n \ge 2$ and any $X$.

Answer 2

Regular cyclic $q$-coverings are classified by normal subgroups $$N\hookrightarrow \pi_1X \to \pi_1X/N\cong \mathbb Z/q,$$

hence there is a natural isomorphism to $Hom(\pi_1X,\mathbb Z/q)$ mod automorphism of $\mathbb Z/q$.

Note that $K_q=K(\mathbb Z/q,1)$ is universal in the sense that the map $$[X,K_q] \to Hom(\pi_1X,\mathbb Z/q) = H^1(X,\mathbb Z/q),$$ $$ [f]\mapsto f_*$$ is an isomorphism. In particular you see that every cyclic $q$-covering is obtained by pulling back the universal cover of $K_q$.

Corollary If $X$ is a smooth orientable manifold, then all infinite cyclic covers (or cyclic which factor through $\mathbb Z$) are obtained by cutting and gluing along a hypersurface.