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 $Y$?
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 $Y$? 


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 CalabiYau 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 SuJeong 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 CalabiYau varieties, and it is a "$K$equivalence invariant." 

