Timeline for Curves with negative self intersection in the product of two curves
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Oct 23, 2009 at 21:35 | history | edited | Ilya Nikokoshev | CC BY-SA 2.5 |
corrected
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| Oct 23, 2009 at 21:29 | comment | added | Dmitri Panov | If x>0, y>0, then x^2 (C1)^2+y^2(C2)^2+2xy= 0 + 0 + 2xy >0, so positive. If you take a generic product C1xC2, then all curves on it will be numerically equivalent to xC1+yC2 with x, y>0, so the square is always positive. Of course I forgot to say, that I am looking for irreducibe and reduced curves --- i.e. of multiplicity one. If a product C1xC2 satisfying conditions that I want exists, that will be something extremely rigid... | |
| Oct 23, 2009 at 21:15 | history | edited | Ilya Nikokoshev | CC BY-SA 2.5 |
add info
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| Oct 23, 2009 at 21:02 | comment | added | Dmitri Panov | Sorry I don't quite get your question:) If we consider C1xC1 as a topological 4-fold, there will many 2-surfaces inside with negative self intersection, for example the diagonal. But overwise a curve with negative self intersection has negative expected dimension, so on a generic product C1xC2 all complex curves have postive self-intersection | |
| Oct 23, 2009 at 20:51 | history | answered | Ilya Nikokoshev | CC BY-SA 2.5 |