Why is $\mathbb{Z}$ not a Kähler group?

Question

Is there some simple proof that $\mathbb{Z}$ is not isomorphic to the fundamental group of any compact Kähler manifold? This follows from the main result of https://arxiv.org/abs/0709.4350 which states that any $3$-manifold group which is Kähler must be finite. The simplest non-finite 3-manifold group is $\mathbb{Z} = \pi_{1}(S^{2} \times S^{1})$. So I was wondering, is there an argument to rule out this particularly simple group, without applying to a lot of machinery? (also historically when were we first able to rule this group out?)

Answer 1

If $X$ is a compact Kähler manifold, then $h^{p,q}(X) = h^{q,p}(X)$ and $b_k(X) = \sum_{p+q=k}h^{p,q}(X)$, so in particular, $b_1(X) = h^{1,0}(X) + h^{0,1}(X) = 2h^{1,0}(X)$ is even. Now,

$$b_1(X) = \operatorname{rank} H^1(X; \mathbb{Z}) = \operatorname{rank} \operatorname{Hom}(\pi_1(X), \mathbb{Z}).$$

If $\pi_1(X) = \mathbb{Z}$, $b_1(X) = 1$ which is not even, so $X$ is not Kähler. More generally, if $\pi_1(X)$ is abelian, then it must have even rank.

Note, there are non-compact Kähler manifolds with fundamental group $\mathbb{Z}$, for example $X = \mathbb{C}^*$.

Answer 2

Every free group is not Kähler (fundamental group of compact Kähler manifold). In particular, a free group of rank 1 is not Kähler.

This follows since the rank of the abelianization of a Kähler group is even, even rank free groups contain finite index subgroups of odd rank (see here for example), and finite index subgroups correspond to covers of Kähler manifolds which are also Kähler.

This and other interesting examples can be found in Fundamental Groups of Compact Kähler Manifold by Amorós, Burger, Corlette, Kotschick, and Toledo.