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Jun 5, 2018 at 14:33 history undeleted S. Carnahan
Mar 14, 2018 at 19:19 history deleted user95222 via Vote
Mar 14, 2018 at 19:17 comment added user87684 I don't have a counterexample to your statement off the top of my head, but I find hard to believe that the HC for abelian varieties comes down to this.
Mar 14, 2018 at 19:14 comment added user87684 for powers of smooth projective curves. So if your question has positive answer (for every $n\ge 1$) then all Hodge classes in $C^n$ are algebraic, as they are generated by divisor classes and the cycle maps are compatible with cup products.
Mar 14, 2018 at 19:13 comment added user87684 My heuristic argument for why this really should have a negative answer: if your question had a positive answer, then it would imply the Hodge Conjecture for abelian varieties. Indeed, every abelian variety receives a surjection from the Jacobian of a smooth projective curve as long as the ground field is infinite. By Arapura's "Motivation for Hodge cycles" it is enough to check the Hodge Conjecture is true for powers of Jacobians of smooth projective curves. In turn, a sm projective curve $C$ and its Jacobian are co-motivated, so by 4.3 in loc. cit. it is enough to check the Hodge Conjecture
Mar 14, 2018 at 18:58 history asked user95222 CC BY-SA 3.0