Timeline for The existence of primitive and sufficiently ample line bundles on K3 surfaces?
Current License: CC BY-SA 2.5
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| Dec 17, 2009 at 4:00 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Dec 17, 2009 at 3:53 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Oct 25, 2009 at 2:56 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Oct 22, 2009 at 14:45 | comment | added | Jorge Vitório Pereira | @Andrea: I agree that it is never a square, but I don't see how it implies the primitiviness. If the primitive curve has self-intersection N then its multiples will have self-intersection N k^2. | |
| Oct 22, 2009 at 12:43 | comment | added | Andrea Ferretti | I hope it's readable; line breaks were lost :-( | |
| Oct 22, 2009 at 12:42 | comment | added | Andrea Ferretti | More simply, the only nontrivial common factor between 2k and 2k+3 can be 3. So if the product is a square, either both are squares (in which case 2k+1, so no solution here) or else 2k = 3 a^2 2k +3 = 3 b^2 But then 6k and 6k + 9 are both squares, which has no solution again. So I would say that number is never a square. | |
| Oct 22, 2009 at 12:25 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Oct 21, 2009 at 17:05 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Oct 21, 2009 at 16:35 | comment | added | Andrea Ferretti | There is a slight problem, which I should have pointed out before if comments could be larger. In Beauville's construction you can always take the genus g curve to be primitive, and it will remain primitive when you take the deformation. But your construction does not yield a primitive vector, since O(k) = k O(1). | |
| Oct 21, 2009 at 15:16 | history | edited | Jorge Vitório Pereira | CC BY-SA 2.5 |
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| Oct 21, 2009 at 13:54 | comment | added | Andrea Ferretti | First you can show the existence of algebraic K3 surfaces such that the generic hyperplane section is a smooth curve of genus g. This is done in Beauville's book on algebraic surfaces Then you study deformations using the local Torelli theorem. Deformations which keep a given cohomology class of type (1,1) are locally an hypersurface in the deformation space. So you can keep the class of your curve of type (1,1) and kill all other cohomology classes, if any. Use exponential sequence and Riemann-Roch to check that this is still the class of a curve, so the Picard group has now rank 1. | |
| Oct 20, 2009 at 1:58 | history | answered | Jorge Vitório Pereira | CC BY-SA 2.5 |