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Daniel Loughran
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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 Hodgealgebraic classes form a subgroup of finite index inside the algebraicHodge 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.

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 Hodge classes form a subgroup of finite index inside the algebraic 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.

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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Daniel Loughran
  • 21.3k
  • 3
  • 46
  • 136

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 Hodge classes form a subgroup of finite index inside the algebraic 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.

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.

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

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 Hodge classes form a subgroup of finite index inside the algebraic 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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Daniel Loughran
  • 21.3k
  • 3
  • 46
  • 136

Bounding failures of the integral Hodge and Tate conjectures

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.

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