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Suppose that over an algebraically closed field $K$ of finite characteristic the numerical equivalence of cycles relation (for algebraic cycles of smooth projective varieties) coincides with the homological one (with respect to etale $\mathbb{Q}_l$-adic cohomology). Does the Hodge standard conjecture follow? Moreover, does it follow from the existence of a 'nice' motivic $t$-structure for Voevodsky's motives over $K$ (i.e. we assume that this $t$-structure 'splits' Chow motives into 'numerical pieces', and that mixed motives have filtration 'by these pieces'). If the answer (even) to the second question is 'no', then why does one usually believe in the HSC?

flag	asked Jan 23 2011 at 17:17 Mikhail Bondarko 5,876●1●4●21

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I would be surprised if the Hodge standard conjecture would follow from the others (which is not to say that it doesn't). This conjecture is true in char $0$ by Hodge theory. Perhaps that and a general sense of optimism, at the time, were sufficient, but I'm just speculating. – Donu Arapura Jan 23 2011 at 17:35

SGP · Answer 1 · 2011-01-24 03:04:38Z UTC

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Somewhat relevant to your question: Milne has shown that the usual Hodge conjecture for complex abelian varieties of CM type implies the Hodge standard conjecture for all abelian varieties over the algebraic closure of any finite field. See Milne's paper

answered Jan 24 2011 at 3:04

SGP
2,614●4●14

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The Tate conjecture for varieties over finite fields implies that the motives of abelian varieties generate all motives over finite fields. Thus the Hodge standard conjecture (over all fields) is implied by the Hodge conjecture for complex abelian varieties + the Tate conjecture for varieties over finite fields. That may, or may not, be reason for believing it. – mephisto Feb 21 2011 at 4:59

If the numerical equivalence of cycles coincides with the homological one, does the Hodge standard conjecture follow?

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