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 $\mathcal{W}$?


1 Answer 1


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.

  • 1
    $\begingroup$ 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
    Jun 27, 2021 at 23:59
  • 1
    $\begingroup$ 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
    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 $\endgroup$ 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
    Jun 28, 2021 at 7:22
  • 1
    $\begingroup$ Oh, I see. Yes, that makes sense, sorry I went on a wrong track. $\endgroup$
    – abx
    Jun 28, 2021 at 7:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.