Timeline for sections of the cotangent bundle of elliptic surfaces
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2013 at 20:13 | vote | accept | rvarma | ||
| Feb 17, 2013 at 20:01 | answer | added | Sándor Kovács | timeline score: 1 | |
| Feb 17, 2013 at 17:41 | answer | added | Qing Liu | timeline score: 3 | |
| Feb 17, 2013 at 13:38 | comment | added | Ariyan Javanpeykar | @rvarma. You're right I was thinking about the smooth case. | |
| Feb 17, 2013 at 13:38 | comment | added | Ariyan Javanpeykar | @Damian My apologies. I was implicitly thinking about the smooth case. | |
| Feb 16, 2013 at 11:35 | comment | added | rvarma | @Damian yes that is what I think too, infact I believe it will torsion free only when the fibers are reduced. Otherwise as I commented above won't it be of the form $\omega_f \otimes I_Z$ ?. | |
| Feb 16, 2013 at 11:20 | comment | added | Damian Rössler | @Ariyan: I dońt understand your exact sequence: $\omega_f$ should be the sheaf of differentials of $f$. This will not be a line bundle unless $f$ is smooth everywhere. | |
| Feb 16, 2013 at 5:12 | comment | added | Will Sawin | Note that this statement is still true if you remove the singular fibers, and then you don't have to worry about the singularity at all. Adding back in the singular fibers only requires you to check that if it's $0$ at the nonsingular fibers then it's $0$ at the singular ones, which is true as long as you're torsion-free. This removes the singular fibers from consideration and frees you to focus on the monodromy, which is the interesting part, but I don't fully see how to complete the argument. | |
| Feb 16, 2013 at 5:09 | comment | added | Will Sawin | This proof has to take advantage of the fact that the family is not isotrivial, because the theorem is false there. One way to do this is using Hodge theory to compare sections of the tangent bundle to a cohomology group, and the Leray spectral sequence to compute the cohomology group. But presumably that's just a version of the dimension argument you don't like. So you need some more clever way to take advantage of that. As Ariyan points out, the key fact is that any section of $\Sigma$ must map to $0$ on $\omega_f$ and thus arise from $K$. | |
| Feb 16, 2013 at 3:20 | history | edited | rvarma | CC BY-SA 3.0 |
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| Feb 16, 2013 at 2:38 | comment | added | rvarma | @ariyan I think if I assume further there are no multiple fibers (which I should have mentioned above) only then the quotient sheaf $\Omega/ f^*K$ is torsion free. In that case I thought this quotient should be of the form $\omega_f \otimes I_Z$ where $I_Z$ is the ideal sheaf defining the singular points in the fibers. | |
| Feb 15, 2013 at 18:17 | comment | added | Ariyan Javanpeykar | You could write down what the entire exact sequence should be. If I'm not mistaken, it should be $0\to f^\ast K\to \Omega\to \omega_f\to 0$, where $\omega_f$ is the relative dualizing sheaf. This is a line bundle on $X$. It is zero because $\omega_f$ reduces to the canonical sheaf on the generic fibre of $X\to C$ , and the generic fibre is an elliptic curve, right? If you don't like "relative dualizing sheaves", then just note that the quotient of $\Omega$ by $f^\ast K$ reduces to the canonical sheaf on the generic fibre. So your "surjective map" should be an isomorphism. | |
| Feb 15, 2013 at 13:45 | history | edited | rvarma | CC BY-SA 3.0 |
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| Feb 15, 2013 at 11:00 | comment | added | rvarma | hi yes I was trying that .. :) | |
| Feb 15, 2013 at 10:59 | history | edited | rvarma | CC BY-SA 3.0 |
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| Feb 15, 2013 at 10:53 | comment | added | Serge Lvovski | Please fix the markup. | |
| Feb 15, 2013 at 10:49 | history | edited | rvarma | CC BY-SA 3.0 |
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| Feb 15, 2013 at 10:45 | history | edited | Ariyan Javanpeykar | CC BY-SA 3.0 |
Fixed the LateX code
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| Feb 15, 2013 at 10:36 | history | edited | rvarma | CC BY-SA 3.0 |
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| Feb 15, 2013 at 10:32 | history | edited | Sasha | CC BY-SA 3.0 |
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| Feb 15, 2013 at 10:26 | history | asked | rvarma | CC BY-SA 3.0 |