Infinite cyclic cover corresponding to non-zero cohomology class $\alpha \in H^1(x,\mathbb Z)$

Question

I want to understand the following sentence:

Let X a compact (complex) manifold which has a non-zero cohomology class $\alpha \in H^1(X,\mathbb Z)$. Let $\pi: \bar X\to X$ be the corresponding infinite cyclic covering.

What does this mean? It seems that an infinite cyclic covering is a cover with fiber $\mathbb Z$.
But why does such a covering exist, and how is it related to the cohomology class?

Answer 1

The way this works is that in the universal coefficient theorem, $$0\to \operatorname{Ext}^1(H_0(X),\mathbb{Z}) \to H^1(X;\mathbb{Z}) \to \operatorname{Hom}(H_1(X),\mathbb{Z}) \to 0,$$ the $\operatorname{Ext}$ group is trivial, so the cohomology is isomorphic to the group of homomorphisms $H_1(X)\to \mathbb{Z}$. By the Hurewicz theorem, this is isomorphic to the group of homomorphisms $\pi_1(X)\to\mathbb{Z}$ since $H_1(X)$ is the abelianization of $\pi_1(X)$. One could also use $H^1(X;\mathbb{Z})=[X,K(\mathbb{Z},1)]=\operatorname{Hom}(\pi_1(X),\mathbb{Z})$.

Given a map $f:\pi_1(X)\to \mathbb{Z}$, there is a normal covering space corresponding to $\ker f$. The infinite cyclic group $\mathbb{Z}$ is the group of deck transformations of this covering space, and so it is called an infinite cyclic cover.

(For non-normal covers, say for a subgroup $G\subset \pi_1(X)$, there is a map $\pi_1(X)\to \operatorname{Sym}(\pi_1(X)/G)$ of permutations of the fiber, where cosets of $G$ are in one-to-one correspondence with points of that fiber. If $G$ is a normal subgroup then the image of this map has a group structure.)

Concretely, given $\alpha\in H^1(X;\mathbb{Z})$, the corresponding homomorphism $f:\pi_1(X)\to\mathbb{Z}$ "integrates $\alpha$ along a path." This function measures how many sheets up or down one travels in the infinite cyclic cover when walking along that path.

Since $X$ is a compact complex manifold, it is orientable. Thus, Poincare-Lefschetz duality applies and gives $H^1(M)\cong H_{n-1}(M,\partial M)$. In certain situations, the corresponding homology class to $\alpha$ might be an embedded $(n-1)$-manifold (for example $n=2$, or if $X$ is merely orientable but not complex, $n=3$). One can produce an infinite cyclic cover by cutting $X$ along this submanifold, which should be thought of as doing a branch cut, then taking one copy of this for each $n\in\mathbb{Z}$, and finally gluing the $n$th copy to the $(n+1)$th copy along the cut for all $n$.