Timeline for About the isotriviality of pencils of plane curves
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2018 at 12:11 | comment | added | Alan Muniz | @JasonStarr Then a pencil with a node not in the base locus will give a nonisotrivial fibration, right? Could you give any good reference to study the Kodaira-Spencer map with reasonable examples? | |
| Jun 3, 2018 at 15:11 | comment | added | Jason Starr | For a transverse crossing at $t_0\in C$, the first-order deformation to the versal deformation space of the node $x_0\in f^{-1}(t_0)$ is a nonzero quotient of the Kodaira-Spencer map $T_B\to h^1(R\pi_*\textit{RHom}(L_f,\mathcal{O}_X))$. So the Kodaira-Spencer map is nonzero. For a family of hypersurfaces, that target is locally free (on a neighborhood of $t_0$), so the family is nonisotrivial. | |
| Jun 2, 2018 at 12:57 | comment | added | Alan Muniz | @JasonStarr, could you please elaborate this or point out a reference? | |
| Jun 2, 2018 at 9:35 | comment | added | Jason Starr | Isotrivial pencils are everywhere non-transversal to the discriminant divisor, which is much stronger than being non-Lefschetz (i.e., that the pencil is somewhere non-transversal). | |
| Jun 2, 2018 at 2:58 | history | answered | HYL | CC BY-SA 4.0 |