Timeline for Negativity of contraction
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2010 at 0:28 | vote | accept | Carlos | ||
| Aug 25, 2010 at 2:42 | answer | added | Paul Hacking | timeline score: 5 | |
| Aug 11, 2010 at 18:10 | comment | added | Henri | Ok, I didn't notice the irreducibility of the divisor $E$, thanks! | |
| Aug 11, 2010 at 17:23 | comment | added | Donu Arapura | To be more explicit, $E$ is supposed to be irreducible and $C\subseteq E$, which implies $E=C$, when $X$ is a surface. So $C^2 <0$ by Mumford. | |
| Aug 11, 2010 at 16:38 | comment | added | Donu Arapura | Mumford's ^result (I wish I could edit comments). | |
| Aug 11, 2010 at 16:37 | comment | added | Donu Arapura | Yes $E_1\cdot E_2=0$, but the assumptions of the question are not satisfied here. In fact, the question does have an affirmative answer for surfaces using Mumford's about negative definiteness of the intersection matrix. | |
| Aug 11, 2010 at 16:17 | history | edited | Charles Matthews | CC BY-SA 2.5 |
upper case title
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| Aug 11, 2010 at 16:01 | comment | added | Henri | In the case where X is $P^2$ blown up in two points, the two exceptional divisors $E_1$ and $E_2$ satisfy your hypothesis but as they are distinct curves, you have necessairly $(E_1 \cdotp E_2) \geq 0$. Or am I wrong? | |
| Aug 11, 2010 at 13:49 | history | asked | Carlos | CC BY-SA 2.5 |