TITLE: Why do people think that abelian varieties are the hardest case for the Hodge conjecture?
QUESTION [25 upvotes]: 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?

REPLY [13 votes]: 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.