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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 $Y=\mathbb{P}^1$.

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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. –  Gunnar Þór Magnússon Jun 28 '12 at 20:22
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*voer $\to$ cover –  Gunnar Þór Magnússon Jun 28 '12 at 20:22
    
How do you use the assumption that $f$ has no singular fibers? –  YangMills Jun 28 '12 at 21:21
    
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... –  Misha Jul 7 '12 at 13:50
    
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. –  Misha Jul 7 '12 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.

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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". –  Todd Trimble Oct 20 '13 at 17:14
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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). –  Karl Schwede Oct 20 '13 at 17:19

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