While reading this paper of Kollár, the following question came up. If $f:X\to Y$ is a fiber space (i.e. surjective holomorphic map with connected fibers) with $X,Y$ smooth projective manifolds, with $K_X\sim_{\mathbb{Q}} 0$, and with $f$ a submersion everywhere, does there exist a finite étale cover $Y'\to Y$ such that $X\times_{Y} Y'$ is isomorphic to a product family (and so $K_Y\sim_{\mathbb{Q}}0$ too)?

Note that since $f$ is a submersion, every fiber $F$ satisfies $K_F\sim_{\mathbb{Q}}0$.

The case when $\mathrm{dim}X=\mathrm{dim}Y+1$ (which includes the case when $\mathrm{dim}X=2$) is easy, since then the fibers are elliptic curves, and the $j$-invariant of each fiber gives a holomorphic function on $Y$, hence constant. It then follows from Fisher-Grauert that $f$ is a holomorphic fiber bundle, and you can apply for example Lemma 17 in this paper of Kollár-Larsen to conclude.

This same argument shows that in arbitrary dimension it is enough to show that all the fibers of $f$ are biholomorphic.

Thanks to work of Kawamata we know that $\kappa(Y)\leq 0$. We can also apply Theorem 4.4 in this paper of Fujino-Gongyo and get that $-K_Y$ is semiample. So if $\kappa(Y)=0$ then we obtain that $K_Y\sim_{\mathbb{Q}}0$. At this point one can apply the arguments in Theorem 4.8 of this paper of Ambro and conclude.

So the question is reduced to the following problem: is it possible to have a holomorphic submersion $f:X\to Y$ with connected fibers, where $X,Y$ are smooth projective manifolds with $K_X\sim_{\mathbb{Q}} 0$ and $\kappa(Y)<0$ ?

If we assume that $Y$ is simply connected and that $X$ has $K_X\cong\mathcal{O}_X$ and $h^{p,0}(X)=0$ for $p=1,...,\mathrm{dim} X-1$ (so $X$ is Calabi-Yau in the stronger sense), then Corollary 2.5 in this paper of Zhang-Zuo implies that this is impossible.

In general I don't know the answer to this, but when $f$ is not assumed to be a submersion (so there might be singular fibers) then $\kappa(Y)<0$ can happen, for example when $X$ is an elliptically fibered $K3$ surface and $Y=\mathbb{P}^1$.

  • $\begingroup$ I don't know if this works, but I'd try: Use Bogomolov-Beauville to split a finite étale cover $X' \to X$ into products of tori, C-Y and HK manifolds. Then let $Y'$ be the product of factors $F$ such that the induced map $F \to X' \to X \to Y$ doesn't map everything to a point. Now hope and pray that the induced map $Y' \to Y$ is a finite étale voer. $\endgroup$ Jun 28, 2012 at 20:22
  • 1
    $\begingroup$ *voer $\to$ cover $\endgroup$ Jun 28, 2012 at 20:22
  • $\begingroup$ How do you use the assumption that $f$ has no singular fibers? $\endgroup$
    – YangMills
    Jun 28, 2012 at 21:21
  • $\begingroup$ I assume you know how to handle the case when $Y$ is a surface with $\kappa(Y)=0$. If $\kappa(Y)<0$ and $Y$ is rational then the associated (here I am using that $f$ is a submersion) map $\phi: Y\to {\mathcal M}_g$ ($g$ is the genus of fiber) is trivial, since $Y$ is simply-connected and Teichmuller space is a domain in $C^{3g-3}$. If $Y$ is ruled, $r: Y\to C_h$, then, similarly, $h\ge 1$. Again, for each fiber of $r$, its image in ${\mathcal M}_g$ is a point. Hence, $X$ admits another smooth fibration with fibers $P^1$ and base $Y'$ which, in turn, is a fibration over $C_h$ with fibers of... $\endgroup$
    – Misha
    Jul 7, 2012 at 13:50
  • $\begingroup$ of genus $g$. Since you know that $\kappa(Y')<0$, this leaves the cases $g=0$ or/and $h=0$. Maybe you can exclude these by a direct computation. $\endgroup$
    – Misha
    Jul 7, 2012 at 13:58

1 Answer 1


I am not familiar with this topic, but I remember seeing this recent paper. It claims that if the map has no singular fiber, it is a fiber bundle (all the fibers isomorphic). Moreover, if the total space is algebraic, it can be a product family after a suitable base change, as you claimed.

Edit I made my answer more informative, as the comments below suggested. I thought the abstract of the paper would answer to the question.

  • 3
    $\begingroup$ Could you please make your answer more informative? There wouldn't be much point to having MO if all the answers were "read this paper". $\endgroup$
    – Todd Trimble
    Oct 20, 2013 at 17:14
  • 1
    $\begingroup$ While I do agree that this answer should be fleshed out, I think this answer is far better than none and so I upvoted Todd's comment and also voted not to delete this answer (short answers get flagged). $\endgroup$ Oct 20, 2013 at 17:19
  • $\begingroup$ Yes, this paper answers my question. $\endgroup$
    – YangMills
    Jan 6, 2015 at 16:37
  • $\begingroup$ @YangMills Further results have been given by Gross, Tosatti, and Zhang: arxiv.org/abs/1911.07315 $\endgroup$
    – AmorFati
    Aug 19, 2020 at 23:17

Your Answer

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

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