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Let $X$ and $Y$ be two birational smooth projective varieties over the complex numbers. Assume $X$ satisfies the Hodge conjecture.

Is it known that the Hodge conjecture holds for $Y$?

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1 Answer 1

No. If you blow up a smooth subvariety $X$ of $\mathbb{P}^n$, the Hodge conjecture for the resulting variety is equivalent to the Hodge conjecture for $X$. So the Hodge conjecture for rational varieties (= birational to $\Bbb{P}^n$) implies the Hodge conjecture in general.

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On the other hand, the weak factorization theorem implies that the Hodge conjecture is a birational invariant of smooth projective varieties of dimension $\leq 4$. –  Olivier Benoist Jan 26 at 23:19
@abx: I don't want to be nitpicking, but I had to reread what you wrote to understand what you're saying. Perhaps, saying "blow up $\mathbb P^n$ along a smooth subvariety $X$" would be a clearer way to put it. Then again, it might be just me... Cheers! (and +1) –  Sándor Kovács Jan 27 at 0:50
In fact, the Hodge conjecture, the Tate conjecture, and the various standard conjectures hold for all varieties if they hold for all rational varieties (and they do all hold for $\mathbb{P}^n$). See: Tankeev, S. G. Monoidal transformations and conjectures on algebraic cycles. (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 71 (2007), no. 3, 197--224; translation in Izv. Math. 71 (2007), no. 3, 629--655 –  abz Jan 27 at 4:43

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