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 $Y$?


2 Answers 2


No. If you blow up $\mathbb{P}^n$ along a smooth (closed) subvariety $X$ of codimension $\geq 2$, 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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    $\begingroup$ On the other hand, the weak factorization theorem implies that the Hodge conjecture is a birational invariant of smooth projective varieties of dimension $\leq 4$. $\endgroup$ Jan 26, 2014 at 23:19
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    $\begingroup$ @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) $\endgroup$ Jan 27, 2014 at 0:50
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    $\begingroup$ 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 $\endgroup$
    – abz
    Jan 27, 2014 at 4:43

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."


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