MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '11 at 17:35

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

share|cite|improve this answer
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 '11 at 4:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.