Classifying Covering Spaces using First Cohomology

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?

• Maybe I'm wrong, but I'm not sure that any finite covering space is a $G$-principal bundle. To be on the safe side I'd require the additional property of being a regular covering space. – Riccardo Feb 13 '16 at 8:49
• Yes, I think that is correct. Thank you. – user39598 Feb 13 '16 at 8:56
• Well we have a natural isomorphism of functors $H^1(-,\mathbb{Z}/k\mathbb{Z})\cong [-,K(\mathbb{Z}/k\mathbb{Z},1)]$, and then notice that $K(\mathbb{Z}/k\mathbb{Z},1)$ is a $B(\mathbb{Z}/k\mathbb{Z})$, therefore it classifies $\mathbb{Z}/k\mathbb{Z}$-bundles, which are not all the $k$-coverings. This is what I'd say – Riccardo Feb 13 '16 at 9:48
• you have all the rights to be confused, this is somewhat delicate if seen for the first time. I'd suggest you to check the nice book of tom Dieck Algebraic Topology, at page 328-329. He is very clear. Roughly speaking, we require the presence of an action of $\mathbb{Z}/k\mathbb{Z}$ on the bundle, (compatible with the structure) and such an action turns out to be transitive and free, by axioms. So for discrete groups, a $G$-bundle is something more than a $|G|$-covering, the latter has an obvious action of the deck transformation group, which is always free, but not necessarily transitive – Riccardo Feb 13 '16 at 9:58
• $S^1$ is a $B\mathbb{Z}\simeq K(\mathbb{Z},1)$, so $H^1(X;\mathbb{Z})\cong[X,B\mathbb{Z}]\cong[X,S^1]$ classifies fiber bundles with structure group $\mathbb{Z}$, hence regular coverings with countably infinite fibers. The "universal guy" of such is the standard covering $\mathbb{Z}\rightarrow\mathbb{R}\rightarrow S^1$. – archipelago Feb 13 '16 at 10:14

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 $$X$$.

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.