Timeline for A K3 cover over a Del Pezzo surface
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 4, 2022 at 18:10 | comment | added | Basics | You're right. Thanks for the answer. | |
| Jul 3, 2022 at 19:31 | comment | added | Evgeny Shinder | I think $S$ has no embedding as a quartic in $\mathbb{P}^3$: we are looking for an integral vector $v = zh - xe_1 - ye_2$ such that $v$ intersects $e_1$, $e_2$, $h - e_1 - e_2$ positively and $v^2 = 4$. From the positivity of the intersection we get $x, y > 0$, $z > x+y$, so that $z^2 - x^2 - y^2 > 2xy$ and from the square $4$ condition $2xy \le z^2 - x^2 - y^2 = 2$, so $xy < 1$ and we don't get any solutions. | |
| Jul 2, 2022 at 21:28 | comment | added | Basics | There are many (probably, infinitely many) elements of square 4. For example, $210 h - 183 E_1 - 103 E_2$ is of squre 4 but the condition 3 doenot hold for it. I would like to find some geometric methods, not investigating diophantine equations. | |
| Jul 2, 2022 at 21:14 | comment | added | Evgeny Shinder | Then after changing the basis, the intersection form seems to be diagonal $2(x^2 - y^2 - z^2)$, and one needs to solve $x^2 - y^2 - z^2 = 2$ to find elements of square $4$. One obvious element of square $4$ is $2h - E_1 - E_2$ (pullback from the del Pezzo surface multiplies the degree by two), however it contracts the other $(-2)$-curve (pullback of $h - E_1 - E_2$). | |
| Jul 2, 2022 at 18:11 | history | edited | Basics | CC BY-SA 4.0 |
added 8 characters in body
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| Jul 2, 2022 at 17:15 | comment | added | Basics | Yes, they form a basis. | |
| Jul 2, 2022 at 16:24 | comment | added | Evgeny Shinder | $S$ has three $(-2)$-curves obtained as preimages of lines on the del Pezzo surface. It seems that these curves form a basis of $NS(S)$? | |
| Jul 1, 2022 at 23:05 | history | asked | Basics | CC BY-SA 4.0 |