# Why is Hodge more difficult than Tate?

There are strong connections between the Hodge and the Tate conjectures, mainly at the level of similarities and analogies. To quote from an answer of Matthew Emerton on MathOverflow:

"[...] we also have a natural abelian (in fact Tannakian) category in play: in the complex case, the category of pure Hodge structures, and [when the field of definition $K$ is finitely generated over its prime subfield], the category of $\ell$-adic representations of $G_K$ (the absolute Galois group of $K$) (for some prime $\ell$, prime to the characteristic of $K$ in the case when $K$ is a finite field).

Now taking cohomology gives a functor from the category of smooth projective varieties to this latter category (via Hodge theory in the complex case, and the theory of étale cohomology in the other cases). The Hodge conjecture (in the complex case) and the Tate conjecture (in the other cases) then says that this functor is fully faithful.

Moreover, it is known that the Hodge conjecture for CM abelian varieties over $\mathbb{C}$ implies the Tate conjecture over finite fields (this was proven by J. Milne), and that the Tate conjecture for abelian varieties over finitely generated fields implies the Hodge conjecture over $\mathbb{C}$ (this was proven by P. Deligne I. Piatetski-Shapiro - see anon's answer) (see this workshop summary). However, I remember that during a seminar at my university, someone said that the Hodge conjecture is expected to be more difficult to solve than the Tate conjecture. Also, the Hodge conjecture (H), unlike the Tate conjecture (T), is part of the Millenium problems, which could suggest that H is more difficult/deeper that T. My question is thus the following: are there any reason why H should be more difficult than T?

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I wholly support this question, but that opinion is subjective. I recently spoke with a number theorist who believes the Tate conjecture is more difficult. –  Brandon Carter Apr 29 '12 at 20:32
@BrandonCarter Fair enough. I slightly changed my question to make it look less contingent on the opinion of one person in one seminar. However, I would be very curious to hear the arguments of your number theorist. –  M Turgeon Apr 29 '12 at 20:52
Well, Ray Hudson, in a (soccer) match a few months ago shown on Gol TV, said of the other team "They are trying to solve the problem that is Barcelona. They might as well try to solve the Hodge Conjecture." Email to the station resulted in no explanation of the comment. See my question at goltv.tv/en/blog/caffeine-free-clasico –  Will Jagy Apr 29 '12 at 21:44

## 2 Answers

Here is an argument that Tate is harder than Hodge:

• We know the Hodge conjecture in the codimension one case (this is the Lefschetz $(1,1)$ Theorem). On the other hand, the Tate conjecture remains open even in codimension one except in some very special cases. Also, those special cases have often been proved by reducing to the Hodge conjecture.

Here is an argument that Hodge is harder than Tate:

• The Tate conjecture (over either finite fields or number fields) involves contexts (motives over finite fields or number fields) which one might hope to fit into the context of the Langlands program, and in particular of Shimura varieties (where one can at least imagine describing the Galois-invariant cohomology classes, and relating them to actual cycles). On the other hand, the Hodge conjecture is about all varieties over $\mathbb C$, and most of these don't fit into the highly structured context of the Langlands program.

The fact that both arguments seem reasonable shows (in my view) that the question doesn't have a good answer. Indeed, with our current state of knowledge being what it is (pretty poor for either conjecture), it is hard to compare the two --- they both seem like hard problems! They also seem to be closely coupled (and not just at the level of analogies).

One thing that the second point suggests is that it could be important to reduce the Hodge conjecture from the case of general complex varieties to the case of varieties over number fields (or, equivalently, varieties over $\overline{\mathbb Q}$). (I first learned this suggestion from Langlands.) That is probably also a very hard problem, but more accessible than the Hodge conjecture itself.

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It isn't --- the Tate conjecture is harder than the Hodge conjecture. For example, the Tate conjecture implies that all absolutely Hodge classes are algebraic. Thus, in the presence of Deligne's conjecture that all Hodge classes are absolutely Hodge (which is known for abelian varieties), the Tate conjecture implies the Hodge conjecture. There is no similar statement going the other way. Another reason that Tate gives is that the Tate conjecture doesn't say which cohomology classes are algebraic --- it only gives their $Q_l$ span. Incidentally, it was Piatesky-Shapiro who first proved that the Tate conjecture for abelian varieties implies the Hodge conjecture for abelian varieties, not Deligne. Added: See Pjateckiĭ-Šapiro, I. I. Interrelations between the Tate and Hodge hypotheses for abelian varieties. (Russian) Mat. Sb. (N.S.) 85(127) (1971), 610--620.

Concerning the last part of Matt E's answer: "Assume that the Hodge conjecture is known for varieties $X_\bar{Q}$ defined over $\bar{Q}$ and (weakly) absolute Hodge classes on them. Then the Hodge conjecture is true for (weakly) absolute Hodge classes." Voisin, Claire Hodge loci and absolute Hodge classes. Compos. Math. 143 (2007), no. 4, 945--958.

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Dear anon, As an incidental question, is there a reference for Piatetski-Shapiro's argument? Best wishes, –  Matt E May 1 '12 at 0:36
Dear anon, Thanks for the additions to your answer. Best wishes, –  Matt E May 1 '12 at 14:47