Non-algebraic Kähler threefolds with abelian $\pi_1$ of arbitrarily large rank

Question

Do there exist non-algebraic Kähler threefolds with abelian $\pi_1$ of arbitrarily large rank?

Answer 1

Let's construct such a Kahler $3$-fold $X$. It will be obtained as an elliptic fibration over a projective surface $S$ with abelian fundamental group $\mathbb Z^{2g}$.

Construction. Recall first that the space of principally polarised Abelian varieties of dimension $g$ has dimension $g(g+1)/2$. Let us consider the space of all complex tori of dimension $g+1$ that are obtained as an extension of a principally polarized Abelian variety $A_g$ by an elliptic curve $E$

$$E\to T_{g+1}\to A_g.$$

It is not hard to see that (up to natural equivalence) this space has dimension $g(g+1)/2+1+g$. Here $g(g+1)/2$ stands for the dimension of the space of ppavs, $1$ for elliptic curves, and $g=2g-g$ - for the dimension of extensions of a fixed Abelian variety by a fixed elliptic curve modulo the natural equivalence .

Since $g(g+1)/2+1+2g=(g+1)(g+2)/2$, we see that the space of such tori has the same dimension as the moduli space of principally polarized Abelian variteties of dimension $g+1$. However, not all constructed tori are projective*, and so we can pick one of such $T_{g+1}$ that is not an Abelian variety.

Now, take a very ample line bundle $L$ on the corresponding $A_g$ and consider a compete intersection $S\subset A_g$ of dimension $2$. Finally we take $X\subset T_{g+1}$ as the induced elliptic fibration

$$E\to X\to S.$$

Since $\pi_1(S)\cong \mathbb Z^{2g}$, and the elliptic fibration is a topologically trivial bundle, we have $\pi_1(X)\cong \mathbb Z^{2g+2}$. So we only need to prove that $X$ is not projective.

(non) Projectiveness of $X$. Assume by contradiction that $X$ is projective, I want to deduce then that $T_{g+1}$ is projective as well. To do this, I believe it is enough to construct a divisor $D$ on $T_{g+1}$ that intersects positively the fibre of the fibration. To construct it, take a curve $C\subset S$ such that $C+C+\ldots + C$ ($g$ times) is $A_g$. Let $E\to Y\to C$ be the restriction of the bundle to $C$. Since $Y$ is a projective surface, we can take a curve $C'\subset Y$ that projects to $C$ with positive degree. Finally take $g$-fold sum $C'+\ldots +C'=D\subset T_{g+1}$. This is the desired divisor (since by construction it projects to the whole $A_g$).

Once we have $D$, we can construct a divisor in $T_{g+1}$ whose $g+1$ power is positive, hence it is ample. This contradicts the fact that $T_{g+1}$ is not projective.

*I think this can be deduced, for example from Poincare reducibility, as follows. Let $T_{g+1}$ be one of constructed complex tori. Assume that it is an Abelian variety. Then the dual $T_{g+1}^*$ is also an Abelian variety. Moreover, it contains a sub-torus, and so thanks to Poincare reducibility it is isogenous to a product of two tori. Hence the same holds for $T_{g+1}$. But this is not the case for a general member of the constructed family.