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It is well know that the integral versions of the Hodge and Tate conjectures can fail. I once heard an off hand comment however that they should only fail by a "bounded amount". My question is what this means and whether this can be made more precise.

Let $\pi: X \to B$ be a smooth projective morphism of smooth varieties over $\mathbb{C}$. Assume that the (usual rational) Hodge conjecture holds for the fibre $B_x$ over every point $x \in B(\mathbb{C})$. Does there exist an integer $d$ such that for any integral Hodge class $h$ on $B_x(\mathbb{C})$, the class $dh$ is represented by an algebraic cycle for all $x \in B(\mathbb{C})$?

Of course the point of this question is whether $d$ can be chosen uniformly with respect to the family.

As a brief remark, it is quite easy to prove this when $B={\mbox{Spec}}\ \mathbb{C}$. Indeed, by assumption the algebraic classes form a subgroup of finite index inside the Hodge classes!

I am also similarly interested in the analogous question for the integral Tate conjecture for families of varieties over number fields and finite fields.

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    $\begingroup$ It seems unlikely that any of the known counterexamples to the integral Hodge conjecture would give a negative answer to your question. On the other hand, there seems to be no reason to expect a positive answer since the dimension of the space of Hodge classes could jump over infinitely many subvarieties of $B$. Interesting! $\endgroup$ – ulrich Jun 6 '14 at 9:05

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