Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In the Wikipedia article on the Hodge Standard Conjecture it is written (note: it has since been fixed):

In characteristic zero the Hodge standard conjecture holds, being a consequence of Hodge theory. In positive characteristic the Hodge standard conjecture is known only for surfaces and abelian varieties.

I have three questions:

(1) Is the characteristic 0 version the Hodge Index Theorem?

(2) If so, what is a good reference for an algebraic geometry proof? I know it can be proved for a surface using the Riemann-Roch Theorem. Does this continue to be true for higher dimensions?

(3) Is the conjecture really only known in positive characteristic for surfaces and abelian varieties? Surely it should be possible to at least compute a result for projective $n$-space.

share|cite|improve this question
Actually the Hodge standard conjecture is not even known in positive characteristic for abelian varieties --- it is only known that it is implied by the Hodge conjecture for complex CM abelian varieties (see my 2002 Annals paper). Of course, the conjecture is trivial for projective n-space. A good reference for these things is Kleiman's article in the proceedings of the Motives conference, Seattle 1991, published by AMS 1994. – JS Milne Jun 12 '10 at 1:13

1 Answer 1

Probably too late, but...

(1) Not quite. The Hodge index theorem only works for $\mathbb{C}$. To extend it to all characteristic $0$ ground fields, you need the Lefschetz principle and the comparison theorem.

(2) As J. S. Milne mentioned, the wikipedia article was mistaken (it has since been fixed). It isn't known how to extend Segre-Grothendieck to higher dimension varieties.

(3) It seems that only surfaces are known so far. The Lefschetz conjecture for projective spaces and Grassmannians follows from intersection theory, but I'm not aware of such results for the Hodge standard conjecture.

share|cite|improve this answer

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.