Timeline for Calabi-Yau fiber space without singular fibers implies finite quotient of product?
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| Jan 6, 2015 at 16:37 | vote | accept | YangMills | ||
| Oct 20, 2013 at 16:55 | answer | added | user25340 | timeline score: 5 | |
| Jul 28, 2012 at 23:10 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jul 28, 2012 at 23:03 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jul 18, 2012 at 1:15 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jul 11, 2012 at 3:06 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jul 7, 2012 at 15:16 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jul 7, 2012 at 15:13 | comment | added | YangMills | Dear Misha, thanks for your comment. However, since I assume that $K_X$ is torsion, the same is true for the fibers (since $f$ is a submersion). Therefore if the fibers are curves they have genus $0$. I gave a complete argument for this case in my question (using the lemma from Kollar-Larsen). So the case I don't know how to do is when dim$X$>dim$Y$+1. | |
| Jul 7, 2012 at 13:58 | comment | added | Misha | 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. | |
| Jul 7, 2012 at 13:50 | comment | added | Misha | 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... | |
| Jul 7, 2012 at 2:36 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jun 29, 2012 at 15:22 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jun 28, 2012 at 21:21 | comment | added | YangMills | How do you use the assumption that $f$ has no singular fibers? | |
| Jun 28, 2012 at 20:22 | comment | added | Gunnar Þór Magnússon | *voer $\to$ cover | |
| Jun 28, 2012 at 20:22 | comment | added | Gunnar Þór Magnússon | 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. | |
| Jun 28, 2012 at 18:30 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jun 28, 2012 at 3:03 | history | edited | YangMills | CC BY-SA 3.0 |
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| Jun 27, 2012 at 21:19 | history | asked | YangMills | CC BY-SA 3.0 |