Timeline for Isomorphism problem for commutative algebras and schemes.
Current License: CC BY-SA 2.5
12 events
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| Apr 21, 2010 at 12:08 | comment | added | naf | I think the analogue of the Narasimhan-Nori theorem (only finitely many polarisations of a given degree upto automorphism) might be known for K3 surfaces; perhaps this follows from the "Kawamata-Morrison conjecture" on the structure of the nef cone of Calabi-Yau's which is known for K3 surfaces. However, I do not know if generators and relations for the automorphism group can be computed explicitly and even if they could it is not clear how one could use that to check for isomorphisms. | |
| Apr 21, 2010 at 11:02 | history | edited | damiano | CC BY-SA 2.5 |
removed incorrect answer
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| Apr 20, 2010 at 19:20 | history | edited | damiano | CC BY-SA 2.5 |
added 145 characters in body
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| Apr 20, 2010 at 14:41 | comment | added | naf | Boundedness is OK for Enriques surfaces but it is not clear to me why that is enough since an isomorphism, if it exists, needn't preserve the sextic structure (as far as I can see). | |
| Apr 20, 2010 at 14:38 | comment | added | naf | For the abelian case, the Kummer is a quartic in P^3 only in the principally polarised case. However, one can probably compute the endomorphism ring of an abelian variety and then also all the polarisations, so I agree that this case is probably OK. | |
| Apr 20, 2010 at 14:34 | comment | added | naf | I didn't understand the details of your argument for K3's at all, so didn't comment earlier. Firstly, I do not see how you get an isomorphism of the resuctions mod p of the two surfaces if you know that the (rational) Picard lattices are isomorphic. Secondly, an isomorphism mod p need not lift to an isomorphism so I do not see how you can use this. Perhaps you can expand your answer a little? | |
| Apr 20, 2010 at 14:28 | comment | added | Donu Arapura | This looks interesting. I agree, however, that the K3 surface case needs some clarification. To apply Torelli in a more conventional approach, I think you would need to compare the transcendental lattices. Somehow you're avoiding this issue, but I'm not quite sure how. | |
| Apr 20, 2010 at 12:59 | comment | added | damiano | (I would like to emphasize further that the case that worries me most is the case of K3 surfaces.) | |
| Apr 20, 2010 at 12:58 | comment | added | damiano | For abelian surfaces, construct the Kummer variety, realize the Kummer as a quartic in P^3, and project away from a node. You get a double cover branched along six lines that are tangent to the same conic: I believe that the Jacobian of the genus two curve that you obtain by considering the double cover of the conic branched along the six tangency points is at least isogenous to the initial Abelian surface. This should bound things enough to make the construction computable. | |
| Apr 20, 2010 at 12:58 | comment | added | damiano | I've been a little sloppy: Enriques surfaces are probably ok, since they are all sextics in $P^3$ with double lines along the edges of a tetrahedron; this gives you the boundedness that seems needed. (Abelian surfaces in another comment, since they do not fit here!) | |
| Apr 20, 2010 at 12:05 | comment | added | naf | 1) Are you sure that the ample divisors on Enriques surfaces are constant? (The fact that the Picard rank is constant does not impy this in general.) 2) Only principally polarised abelian surfaces correspond to genus 2 curves and there exist non-isomorphic genus 2 curves which have isomorphic Jacobians. | |
| Apr 20, 2010 at 11:15 | history | answered | damiano | CC BY-SA 2.5 |