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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Sign up to join this communityLet $X$ 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 $Y$?
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.
Of course, abx is completely correct in saying that the truth of the Hodge conjecture is not a birational invariant. That said, something slightly weaker is true: if $X$ and $Y$ are $K$-equivalent, then the Hodge conjecture is true for $X$ if and only if it is true for $Y$.
Here we say two smooth projective varieties $X, Y$ are $K$-equivalent if there exists a third smooth projective variety $Z$ and birational morphisms $f: Z\to X, g: Z\to Y$, such that $f^*\omega_X\simeq g^*\omega_Y$. For example, birational Calabi-Yau varieties satisfy this property. The theory of motivic integration then implies that $[X]=[Y]$ in the Grothendieck group of varieties, $K_0(\text{Var})$.
But now, this paper of Donu Arapura and Su-Jeong Kang shows that the truth of the Hodge conjecture for $X$ depends only on its class in $K_0(\text{Var})$.
So the bottom line is: no, the Hodge conjecture is not a birational invariant in general. But it is for Calabi-Yau varieties, and it is a "$K$-equivalence invariant."