# Complex fibration over complex torus

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 $$\mathcal{W}$$?

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.

• 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? Commented Jun 27, 2021 at 23:59
• A priori you have only a map from the torus to the (coarse) moduli space. How do you lift it to Teichmüller space?
– abx
Commented Jun 28, 2021 at 3:37
• @abx, thanks, I missed multiple fibers, But in case the bundle is a smooth fibration I think it works. I'll correct the answer Commented Jun 28, 2021 at 6:49
• That does not completely answer my question: how do you define the map from the torus to the Teichmüller space?
– abx
Commented Jun 28, 2021 at 7:22
• Oh, I see. Yes, that makes sense, sorry I went on a wrong track.
– abx
Commented Jun 28, 2021 at 7:53