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I wonder what could be a Kähler surface, which is a total space of principal elliptic bundle over a curve. I believe that there is a classification and that it must be pretty simple, but I cannot find it in the literature.

Fo example, though there exist a full and complicated classification of elliptic surfaces, I didn't manage to find the list of all elliptic surfaces with no singular fibers.

Is it true, for example, that any of such surface is covered by the product of the base and the fiber? In [Höfer, Thomas, Remarks on torus principal bundles, J. Math. Kyoto Univ. 33 (1993), no. 1, 227–259.] it is proved, that such a fibration must be flat (in the following sence: it is covered by a flat principal $\mathbb{C}^*$-bundle), so it is enough to show that it has finite monodromy...

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  • $\begingroup$ I don't think so. See my answer to this mathoverflow.net/questions/257147/… $\endgroup$
    – Donu Arapura
    Commented Dec 14, 2016 at 21:05
  • $\begingroup$ @DonuArapura so my conjecture is true if one asks the total space to be projective (instead of Kähler)? Thank you very much, I liked the example. $\endgroup$
    –  V. Rogov
    Commented Dec 15, 2016 at 2:40
  • $\begingroup$ Yes, I think that it would be true in the projective case. $\endgroup$
    – Donu Arapura
    Commented Dec 15, 2016 at 3:25
  • $\begingroup$ > ...which is constant on the orbits $\{y+h(\gamma)\}$... I don't really see it. The monodromy somehow acts on meromorphic functions... $\endgroup$
    –  V. Rogov
    Commented Dec 15, 2016 at 4:29
  • $\begingroup$ The complete classification of fiber and principal bundles of elliptic curves over a smooth projective complex curve is given in Barth, Peters and Van den Ven, Compact Complex Surfaces, p. 143-147. Over a projective line base, the Kaehler ones are trivial bundles. With elliptic curve base, they are complex tori. Over any higher genus curve, there is a finite unramified covering by a trivial bundle. $\endgroup$
    – Ben McKay
    Commented Dec 15, 2016 at 8:32

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A (partial) answer was given by Donu Arapura at Are most Kähler manifolds non-projective?

Let $C$ be a smooth projective curve of genus $g>0$, $\Gamma =\pi_1(C)$, and $\tilde{C}$ the universal cover. Choose an elliptic curve $E$ and a group homomorphism $h\colon \Gamma \to E$. Define an action of $\Gamma$ on $\tilde{C} \times E$ by $\gamma(x,y)=(\gamma x,y+h(\gamma))$, and let $S$ be the quotient.

$S$ is Kähler. If hh has infinite image, then $S$ is not algebraic.

Proof. $S$ is Kähler because $\tilde{C}\times E$ has an invariant Kähler metric. For the second statement, assume that $h$ has infinite image. Projection on the first factor gives a holomorphic map $f \colon S \to C$. The fibres of $f$ can be identified with $E$. Restricting a meromorphic function $F$ on $S$ to a fibre gives a meromorphic function on $E$ which is constant on the orbits $\{y+h(\gamma)\}$ and therefore constant. Therefore $F$ comes from $C$. This shows that transcendence degree of the field of meromorphic functions on $S$ is 1.

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