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Sep 12, 2010 at 5:01 vote accept David Eisenbud
Sep 12, 2010 at 5:01 vote accept David Eisenbud
Sep 12, 2010 at 5:01
Aug 5, 2010 at 14:02 comment added Tom Goodwillie Answering a question that was not asked, I believe it is easy to give examples over $mathbb Z$ of bundles $E$ such that these two symmetric squares of $E$ are nonisomorphic, where the argument does not rely on proving that the bundles are not isomorphic in characteristic $2$.
Aug 4, 2010 at 15:05 answer added Tom Goodwillie timeline score: 14
Aug 4, 2010 at 13:21 answer added Torsten Ekedahl timeline score: 28
Aug 4, 2010 at 12:34 comment added Torsten Ekedahl Yes, not only do they have the same Chern classes they give the same element in the Grothendieck group.
Aug 4, 2010 at 11:09 comment added Tom Goodwillie These two bundles always appear the same in the Grothendieck group, right?
Aug 4, 2010 at 8:37 comment added algori Torsten -- yes, it is invariant under twisting. My bad.
Aug 4, 2010 at 7:48 comment added Torsten Ekedahl @algori: It is almost true, up to a twist by a line bundle every vector bundle (on some quasi-projective variety) is the pullback of the tautological bundle on some Grassmannian. The problem is invariant under twisting by line bundles. So yes the problem is equivalent to the same problem for the tautological bundles.
Aug 4, 2010 at 7:18 comment added algori Angelo -- could you please explain why: as opposed to the topological case, not all algebraic vector bundles are induced from the tautological bundles over Grassmannians.
Aug 4, 2010 at 6:41 comment added Angelo Hi David! The examples to try would be tautological bundles on grassmannians (if the isomorphim exists for those, then it should always exist, right?). Have you tried with tangent bundles to projective spaces?
Aug 4, 2010 at 0:37 comment added algori I've added the "Algebraic geometry" tag.
Aug 4, 2010 at 0:36 history edited algori
added ag tag
Aug 3, 2010 at 23:27 history asked David Eisenbud CC BY-SA 2.5