# is the Hodge conjecture birationally invariant?

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 $$\mathbb{P}^n$$ 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.

• On the other hand, the weak factorization theorem implies that the Hodge conjecture is a birational invariant of smooth projective varieties of dimension $\leq 4$. Jan 26, 2014 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) Jan 27, 2014 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, 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."