Timeline for symplectic classes on rational surfaces.
Current License: CC BY-SA 3.0
9 events
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May 22, 2011 at 23:32 | comment | added | Dmitri Panov | Paul, yes, indeed $C\cdot X$ must be positive, the point is that if you we blow up $\mathbb CP^2$ in $n\ge 9$ generic points the number of rational $-1$-curves is infinite, and the structure of this set infinite is quite involved. If you blow up $\mathbb CP^2$ in $8$ points the number of $-1$ curves is around 250 (I am lazy now to make the precise calculation) | |
May 22, 2011 at 20:05 | comment | added | Paul | OK, I get it, its the converse that is tricky. | |
May 22, 2011 at 16:44 | comment | added | Paul | @Dmitri: I didn't think the question was that involved: Isn't it true that if $C$ is a symplectic class, then $C^2>0$ (the symplectic class is a volume form) and $C\cdot X>0$ for any complex curve $X$ (since every complex curve is a symplectic submanifold). So $C$ has to satisfy $C^2=a^2-\sum b_i^2 >0$ and $aA-\sum b_iB_i>0$ when $X=AH+\sum_i B_iE_i$ is represented by any complex curve. Maybe I'm missing something obvious. | |
May 22, 2011 at 13:36 | comment | added | Dmitri Panov | You are very welcome :) ! | |
May 22, 2011 at 13:22 | vote | accept | Yunhyung Cho | ||
May 22, 2011 at 9:51 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
added 6 characters in body
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May 22, 2011 at 9:35 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
improved exposition
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May 22, 2011 at 0:22 | history | edited | Dmitri Panov | CC BY-SA 3.0 |
added 202 characters in body
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May 22, 2011 at 0:06 | history | answered | Dmitri Panov | CC BY-SA 3.0 |