Timeline for Self-intersection of the diagonal on a surface
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2023 at 21:24 | comment | added | Marco Golla | @R.vanDobbendeBruyn: thanks, I have to get this misconception (or terminology mix-up) out of my system. | |
| Oct 29, 2023 at 13:24 | vote | accept | Tim | ||
| Oct 29, 2023 at 0:11 | answer | added | R. van Dobben de Bruyn | timeline score: 4 | |
| Oct 28, 2023 at 15:41 | comment | added | R. van Dobben de Bruyn | @MarcoGolla the OP did not specify that the Cartier divisor should be effective. @ Tim, somehow the moving lemma says that this is always possible, but I'm not sure you'll find a clean recipe that works for all curves, even of a given genus... | |
| Oct 28, 2023 at 15:34 | comment | added | Tim | @DonuArapura To better understand and give another example, if $X$ is defined over the algebraic closure of a finite field, can one move $\Delta$ using geometric Frobenius (at the cost of dividing by $q$ eventually), say? | |
| Oct 28, 2023 at 15:07 | review | Close votes | |||
| Nov 2, 2023 at 3:03 | |||||
| Oct 28, 2023 at 14:45 | comment | added | Donu Arapura | For $X=\mathbb{P^1}$, $X\times X$ is a quadric in $\mathbb{P^3}$, so take $D= X\times X\cap H$ for $H$ a general hyperplane. For $g=1$, you can translate $\Delta$ using the group law on $X$. | |
| Oct 28, 2023 at 14:18 | comment | added | Tim | @MarcoGolla What about when $g=0$ or $g=1$? Even for $X=\mathbf{P}^1$, I'd be interested in constructing an explicit $D$ that is itself a prime divisor if possible. Thanks for commenting | |
| Oct 28, 2023 at 13:56 | comment | added | Marco Golla | Since the self-intersection is negative as soon as $g>1$, the diagonal is rigid and you can't find such a divisor. | |
| Oct 28, 2023 at 13:15 | history | asked | Tim | CC BY-SA 4.0 |