TITLE: Why is Hodge more difficult than Tate?
QUESTION [22 upvotes]: 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?

REPLY [13 votes]: 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.