Timeline for Reference needed for negative curves on blowup of the projective plane at generic points
Current License: CC BY-SA 3.0
12 events
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| Oct 1, 2012 at 21:50 | comment | added | Sándor Kovács | ...and I meant translate fiberwise, so you don't need an elliptic curve to act. | |
| Oct 1, 2012 at 21:47 | comment | added | Sándor Kovács | @Jim: it's funny, I had the same reaction as Artie when I first read your comment. It took me a minute to realize what you meant. Anyway, even if those 9 points are all torsion, there are other sections one may use to translate. For instance take the strict transform of a quadric through 5 of the 9 points (which also give (-1)-curves). | |
| Oct 1, 2012 at 21:07 | comment | added | Jim Bryan | @Artie: I meant "three-torsion" not "three torsion". They are the nine 3-torsion points and they are closed under multiplication. | |
| Oct 1, 2012 at 13:26 | comment | added | user5117 | Jim: I'm not sure I understand your comment, in particular, the statement that there are only 3 points. The exceptional curves give 9 points on the generic fibre of the fibration. Choosing one of them as the origin, the others generate a subgroup of the group of k(P^1)-rational points on that curve. Unless the points are in very special position (in fact, unless they are the Hesse configuration), the group they generate will be infinite | |
| Oct 1, 2012 at 7:46 | comment | added | Jim Bryan | I'm not sure what you mean by "use the group structure to translate them". I don't think we have an elliptic curve acting on this family. We can add these sections to each other, but they are closed under addition --- they are the three torsion points. Can you elaborate? | |
| Sep 30, 2012 at 17:10 | comment | added | Olivier Benoist | Yes, indeed : this is nice ! | |
| Sep 30, 2012 at 14:37 | comment | added | Sándor Kovács | ...but I think the above deformation argument fixes it. | |
| Sep 30, 2012 at 14:36 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Sep 30, 2012 at 7:03 | comment | added | Sándor Kovács | Olivier, you're right, I didn't read the question carefully | |
| Sep 30, 2012 at 6:06 | comment | added | Olivier Benoist | @Sandor : Unless I'm mistaken, through nine general points, there will be only one cubic (for instance, because the space of cubics is 9-dimensional). The construction you give applies for particular sets of 9 points : the base loci of pencils of cubics. | |
| Sep 30, 2012 at 6:02 | history | edited | Sándor Kovács | CC BY-SA 3.0 |
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| Sep 30, 2012 at 5:57 | history | answered | Sándor Kovács | CC BY-SA 3.0 |