# Why do people think that abelian varieties are the hardest case for the Hodge conjecture?

Today, I heard that people think that if you can prove the Hodge conjecture for abelian varieties, then it should be true in general. Apparently this case is important enough (and hard enough) that Weil wrote up some families of abelian 4-folds that were potential counterexamples to the Hodge conjecture, but I've never heard of another potential counterexample.

Anyway, in short:

1) Does the Hodge Conjecture for abelian varieties imply the full Hodge conjecture?

2) If not, is there an intuitive reason why abelian varieties should be the hardest case?

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This reminds me of a comment of JS Milne: “the Tate conjecture (including num=hom) implies that the category of motives over finite fields is generated by abelian varieties, and so the standard conjectures for abelian varieties over finite fields then implies it for all varieties over finite fields”. The Hodge conjecture being a caracteristic 0 analogue of the Tate conjecture, the case of abelian varieties should be of prime interest. I wonder how Milne's statement is proved though and what is different over a non finite field. – AFK Mar 3 '10 at 23:27
Haha, that's exactly the quote I was going to post. – Harry Gindi Mar 3 '10 at 23:31
YBL and Harry, have you learned anything on this since you commented? Could anybody else please comment on the difference between finite and characteristic 0 fields? Also, how do abelian varieties over finite fields generate the category of motives over finite fields? I suppose those are pure motives, and how is "generate" to be taken, as tensor categories, abelian categories, tannakian categories? Please, any comment. – plm Mar 19 '12 at 9:27

I would say the answer to both questions is no. In fact, abelian varieties should be an "easy" case. For example, it is known that for abelian varieties (but not other varieties), the variational Hodge conjecture implies the Hodge conjecture. It is disconcerting that we can't prove the Hodge conjecture even for abelian varieties, even for abelian varieties of CM-type, and we can't even prove that the Hodge classes Weil described are algebraic. So if the Hodge conjecture was proved in one interesting case, e.g., abelian varieties, that would be a big boost.

Added: As follow up to Matt Emerton's answer, a proof that the Hodge conjecture for abelian varieties implies the Hodge conjecture for all varieties would (surely) also show that Deligne's theorem (that Hodge classes on abelian varieties are absolutely Hodge) implies the same statement for all varieties. But no such result is known (and would be extremely interesting).

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I was about to post a quote from you, but the real thing is certainly better. – Harry Gindi Mar 3 '10 at 23:25
Ok, it appears that I must have misunderstood. Thanks for clearing this up! (And everyone else, too, lots of good answers) – Charles Siegel Mar 4 '10 at 3:34
The Hodge conjecture is true in dimension at most 3 for "almost trivial" reasons (it's true for divisors and there is a duality trick). Is it still open for abelian varieties of dimension 4? For CM abelian varieties of dimension 4? – Kevin Buzzard Mar 4 '10 at 11:34
It is still open in for abelian varieties of dimension 4. And actually in dimension 4, it are the CM abelian varieties that are difficult. The others were dealt with by Moonen–Zarhin, “Hodge classes on abelian varieties of low dimension”. – jmc Jan 22 '14 at 14:34

The class of Abelian varieties is the simplest class of varieties where the Hodge conjecture is not known to be true. So naturally a certain amount of effort is directed toward them. However, it's not clear me that one can make an obvious reduction from more general smooth projective varieties to Abelian varieties. The reason I'm skeptical is because the Hodge structures for such varieties need not lie in the tensor category generated by Hodge structures of Abelian varieties.

One last thing. The Hodge conjecture is false for compact Kaehler manifolds, and in fact for complex tori! Cf. Voisin, IMRN vol 20 (2002). There is also an older example due to Zucker.

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"The Hodge conjecture is false....for complex tori." Right. My point (or rather E. Izadi's point) is that nevertheless it is true for generic complex tori in every dimension. – Pete L. Clark Mar 4 '10 at 5:42
Sorry if my last comment came across as a veiled criticism of your post. Since you mentioned the Kaehler version, I wanted to make it clear to people where things stood. It's interesting how close to the boundary of truth and falsity this conjecture is. But I agree with your intuition or Elham's about the generic case being easier. In particular, for a generic torus there should be no Hodge cycles at all. – Donu Arapura Mar 4 '10 at 13:17
No apologies necessary. Thanks for coming to MO -- we can benefit from your expertise. – Pete L. Clark Mar 4 '10 at 14:06

My next door neighbor (in the math department) is a Hodge theorist, and I have never heard her say that abelian varieties are the hardest case of the Hodge conjecture.

However, they are certainly a demonstrably "rich" case of the Hodge conjecture. My neighbor did tell me once that the Hodge conjecture is presumably true "most of the time" even for compact Kahler manifolds, because a generic Kahler manifold doesn't have enough nontrivial cohomology groups to make the Hodge conjecture an interesting statement. On the other hand, an abelian variety of large dimension has lots of large dimensional cohomology groups, and there are many known families of abelian varieties with sufficiently small Mumford-Tate group so that the Hodge conjecture cannot be verified just by intersecting divisors together.

For a brief exposition around the Hodge conjecture and abelian varieties, you may consult

http://math.uga.edu/~pete/mtnotes.pdf

Caveat lector: I am not an expert on this subject.

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Can the phrase "generic Kahler manifold" be made to have any sense? (I don't mean that as a flame --- it is an honest question.) – Ravi Vakil Jul 26 '11 at 17:02
@Ravi: I don't know. Anyway, I don't mean "generic" in any of the technical senses of algebraic geometry. Perhaps a better word would be "typical". – Pete L. Clark Jul 26 '11 at 18:58

Related to Jim Milne's answer, one might mention that Deligne proved that for abelian varieties, all Hodge cycles are "absolutely Hodge" (i.e. when you think of them embedded diagonally inside the product of the algebraic de Rham cohomology and $\ell$-adic cohomology (for every $\ell$) and apply an automorphism of $\mathbb C$, the resulting cycles are again diagonally embedded rational cycles, and are in fact again Hodge). Note that if the Hodge conjecture holds, then this is certainly true (since the conjugate under any automorphism of $\mathbb C$ of an algebraic cycle is again an algebraic cycle).

On the one hand, this is much more than is known about the Hodge conjecture for more general classes of varieties.

On the other hand, one can't immediately extend this to other classes of varieties because the motives of abelian varieties don't generate all motives over a field of char. 0 (in fact, far from it, as far as I know), a fact already brought up in Donu Arapura's answer.

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