19
$\begingroup$

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 $Y$?

$\endgroup$

2 Answers 2

37
$\begingroup$

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.

$\endgroup$
3
  • 6
    $\begingroup$ 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
  • 10
    $\begingroup$ @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
  • 6
    $\begingroup$ 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
29
$\begingroup$

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."

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.