Timeline for sections of the cotangent bundle of elliptic surfaces
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2013 at 23:04 | comment | added | Qing Liu | @rvarma: yes, if $\Omega_{X/C}$ is torsion free (equivalently, all fibers of $f$ are reduced), your isomorphism is correct. But this isomorphism holds under a slightly more general situation, it is enough that there is no multiple fibers (this is proved in the paper with T. Saito in any characteristic, but probably there is a simpler proof in characteristic $0$). | |
| Feb 17, 2013 at 20:13 | vote | accept | rvarma | ||
| Feb 17, 2013 at 19:32 | comment | added | rvarma | Oh yes $\theta$ could be $0$ as in the isotrivial case. | |
| Feb 17, 2013 at 19:28 | comment | added | rvarma | So as long as $\Omega_{X/C}$ is torsion free, we get $K_C \cong f_*(\Omega_{X/k})$ which I never knew. Tell me If I was wrong somewhere. Once again thank you. | |
| Feb 17, 2013 at 19:27 | comment | added | rvarma | hi prof Liu, thanks for the answer. The thought of using $f_*$ did not cross my mind :) . Anyway but in my case since $C$ is a smooth curve and $\Omega_{X/C}$ generically restricts to the canonical sheaf of the elliptic curve (say F), in particular $h^0(\Omega_{X/C}\mid_F) = 1$, the sheaf $f_*(\Omega_{X/C}$ splits $L_0 \oplus T$ where $L_0$ is a line bundle and $T$ is a sky-scrapper sheaf. Now on the other hand $f_{*1}(O_x) \otimes \Omega_{C/k}$ is locally free as well. So the kernel of $\theta$ has to be $T$. | |
| Feb 17, 2013 at 17:41 | history | answered | Qing Liu | CC BY-SA 3.0 |