Timeline for Does $h^1(D)=0$ imply numerical connectedness on K3 surfaces?
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Mar 6, 2014 at 7:06 | history | edited | Roberto Pignatelli | CC BY-SA 3.0 |
removed useless $c$
|
Mar 4, 2014 at 20:20 | comment | added | rita | Ciao Pigna! welcome to MO! | |
Mar 4, 2014 at 13:18 | comment | added | rita | No. For instance if $A$ is a a rational curve with $A^2<0$, then $h^0(\mathcal O_A(-A))>0$. | |
Mar 4, 2014 at 12:08 | comment | added | Heitor | Ah OK, I see. Thank you! I thought that $h^0(\mathcal{O}_B(−A))=0$ would follow from $A$ being effective. Is this not enough? | |
Mar 4, 2014 at 11:01 | comment | added | Roberto Pignatelli | You are right, the expression for $\chi({\mathcal O}_B(-A))$ is Riemann-Roch. The injectivity gives me $h^0({\mathcal O}_B(-A))=0$, else $\chi$ could be positive. | |
Mar 3, 2014 at 19:11 | comment | added | Heitor | Just a question: I don't really see where do you use the injection $H^0(O_D)\subset H^0(O_A)$. I guess the expression for $\chi(O_B(-A))$ is just Riemann-Roch for singular curves, no? | |
Mar 3, 2014 at 17:31 | vote | accept | Heitor | ||
Mar 3, 2014 at 17:28 | history | edited | Roberto Pignatelli | CC BY-SA 3.0 |
deleted 123 characters in body
|
Mar 3, 2014 at 14:49 | history | answered | Roberto Pignatelli | CC BY-SA 3.0 |