Timeline for Del Pezzo surfaces of degree $2$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 13, 2016 at 13:41 | vote | accept | Daniel Loughran | ||
| Jun 11, 2016 at 5:34 | answer | added | Sasha | timeline score: 3 | |
| Jun 9, 2016 at 20:05 | comment | added | Daniel Loughran | This looks great! But there is still something I'm confused about: how does one view $\sqrt{B^2 - 4AC}$ (in Jason's notation) as a global section of $-2K_S$? It looks horribly undefined as written. | |
| Jun 9, 2016 at 15:08 | comment | added | Martin Bright | Beaten to it! Yes, what Jason Starr said. | |
| Jun 9, 2016 at 15:07 | comment | added | Jason Starr | @MartinBright. That is what I was thinking as well. | |
| Jun 9, 2016 at 15:07 | comment | added | Martin Bright | I guess you can find $f$ up to constants because it cuts out the ramification locus of the anticanonical map. View $q$ as a quadratic form in $s,t$ with coefficients involving $x,y,z$; then its discriminant is a polynomial of degree $4$ in $x,y,z$, which is your $f$. | |
| Jun 9, 2016 at 15:04 | comment | added | Jason Starr | Is it just the following: write $q(s,t,x,y,z) = A(x,y,z)s^2 + B(x,y,z)st + C(x,y,z)t^2$ and let $f(x,y,z)$ be $B^2-4AC$? | |
| Jun 9, 2016 at 14:23 | history | asked | Daniel Loughran | CC BY-SA 3.0 |