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Let $X$ be a smooth projective variety over the complex numbers, and $\text{Hdg}^p(X)_{\mathbf{Q}}$ the abelian group of Hodge classes in $H^p(X,\mathbf{Q}(p))$.

Denote by $\text{Hdg}^*(X)$ the subring of $H^*(X,\mathbf{C})$ generated by $\bigoplus_p\text{Hdg}^p(X)_{\mathbf{Q}}$.

Suppose $X$ is the $n$-th product power of a smooth projective curve $C$, ie. $X = C^n$.

Is $\text{Hdg}^*(X)$ generated in degree $1$? (in other words, is it generated under cup products and sums, by Hodge classes in $\text{Hdg}^1(X)_{\mathbf{Q}}$?)

Is there an explicit example in which it is not?

(My expectation is that it isn't)

Improve this question
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  • $\begingroup$ My heuristic argument for why this really should have a negative answer: if your question had a positive answer, then it would imply the Hodge Conjecture for abelian varieties. Indeed, every abelian variety receives a surjection from the Jacobian of a smooth projective curve as long as the ground field is infinite. By Arapura's "Motivation for Hodge cycles" it is enough to check the Hodge Conjecture is true for powers of Jacobians of smooth projective curves. In turn, a sm projective curve $C$ and its Jacobian are co-motivated, so by 4.3 in loc. cit. it is enough to check the Hodge Conjecture $\endgroup$
    – user87684
    Commented Mar 14, 2018 at 19:13
  • $\begingroup$ for powers of smooth projective curves. So if your question has positive answer (for every $n\ge 1$) then all Hodge classes in $C^n$ are algebraic, as they are generated by divisor classes and the cycle maps are compatible with cup products. $\endgroup$
    – user87684
    Commented Mar 14, 2018 at 19:14
  • $\begingroup$ I don't have a counterexample to your statement off the top of my head, but I find hard to believe that the HC for abelian varieties comes down to this. $\endgroup$
    – user87684
    Commented Mar 14, 2018 at 19:17

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