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 $\mathcal{W}$?



Complex fibration over complex torus
**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.


