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Let $M$ be a 3-dimensional complex manifold, and $\Lambda$ a discrete lattice in $\mathbb C^2$. Suppose there is a holomorphic submersion $f:M\to\mathbb{C}^2/\Lambda$ with fibers given by 1-dimensional compact complex manifolds. And these fibers form the leaves of a 1-dimensional holomorphic foliation $\mathcal{W}$.

Then can we have a more specific description of $M$? Is $M$ a fiber bundle or even a trivial bundle over $\mathbb{C}^2/\Lambda$? Can we say something about the leaves of $\mathcal{W}$?

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Added. Thanks to the comment of abx below I understood that there was a big gap in the reasoning, and $M$ doesn't need to be a fiber bundle.

Example. I'll construct an example $S$ of a complex surface that admits a submersion to an elliptic curve $E$ but that is not a fiber bundle over $E$. Then $M$ can be taken as $S\times E'$, where $E'$ is any elliptic curve (and $M$ admits a submersion to $E\times E'$).

So, let's start with the elliptic curve $E$ and any complex curve $C_1$ (say of genus $>2$) that admits a fixed point free involution $\sigma_1$. Now take any degree two cover $C_2$ of $E$ with ramifications (so that the genus of $C_2$ is $>1$). Let $\sigma_2$ be the corresponding involution of $C_2$. Finally consider the quotient of $C_1\times C_2$ by $\mathbb Z_2 $ that is acting on the $C_1$ factor by $\sigma_1$ and on the $C_2$ factor by $\sigma_2$. The resulting suface $(C_1\times C_2)/\mathbb Z_2$ admits a submersion to $E$.

(corrected) Old answer. The fact that such a manifold is a submersion implies that all fibers are smooth. As the above example shows, the submersion doesn't need to be a fiber bundle, but in case it is a fiber bundle something can be said.

Namely, one can say that the fiber bundle is isotrivial, i.e. all fibers are isomorphic curves. Indeed, we can associate to such a manifold a holomorphic map from $\mathbb C^2$ to the corresponding Teichmüller space, and since the latter space is a bounded domain, the map is constant.

At the same time we can't claim that the fibration is trivial, it can be isotrivial. Indeed, one can find curves $\Sigma$ that admit a non-trivial holomorphic action of $\mathbb Z^4$ on them. Then we can take a quotient of $\Sigma\times \mathbb C^2$ by such an action. It should not be hard to classify all such actions for small $2g-2=\chi(\Sigma)$, but I guess that for larger $g$ this might be difficult.

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    $\begingroup$ A minor wording suggestion: since you are discussing triviality or not for a bundle, maybe something like "difficult" rather than "non-trivial" as the final word? $\endgroup$
    – LSpice
    Commented Jun 27, 2021 at 23:59
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    $\begingroup$ A priori you have only a map from the torus to the (coarse) moduli space. How do you lift it to Teichmüller space? $\endgroup$
    – abx
    Commented Jun 28, 2021 at 3:37
  • $\begingroup$ @abx, thanks, I missed multiple fibers, But in case the bundle is a smooth fibration I think it works. I'll correct the answer $\endgroup$
    – Dmitri Panov
    Commented Jun 28, 2021 at 6:49
  • $\begingroup$ That does not completely answer my question: how do you define the map from the torus to the Teichmüller space? $\endgroup$
    – abx
    Commented Jun 28, 2021 at 7:22
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    $\begingroup$ Oh, I see. Yes, that makes sense, sorry I went on a wrong track. $\endgroup$
    – abx
    Commented Jun 28, 2021 at 7:53

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