6
$\begingroup$

Is there an example of a birational morphism of smooth complex projective varieties $f\colon X\to Y$, that cannot be factored into a chain $X\to X_1\to\cdots\to X_n\to Y$ of blow-down along smooth centers?

(By weak factorization theorem, we know in general that $f$ can be factorized into a zig-zag of blow-ups and blow-downs along smooth centers.)

$\endgroup$
11
$\begingroup$

Let $X \subset \mathbb{P}^2_{x_i} \times \mathbb{P}^6_{y_j}$ be given by the equations $$ x_1y_1 + x_2y_2 + x_3y_3 = x_1y_4 + x_2y_5 + x_3y_6 = 0. $$ It is smooth because its projection to $\mathbb{P}^2$ is a $\mathbb{P}^4$-fibration. This also implies that the rank of the Picard group of $X$ is 2. Now let $$ f \colon X \to \mathbb{P}^6 $$ be the projection. It is a birational morphism, and if it is a sequence of smooth blowups, it is itself a smooth blowup (because the difference of the Picard ranks is 1). But it is not a smooth blowup, because $f$ has 1-dimensional fibers over a codimension 2 subvariety of $\mathbb{P}^6$ and 2-dimensional fiber over a point.

$\endgroup$
3
  • $\begingroup$ Possibly silly question, but which point has a 2-dimensional fiber? Such $[y_1:\ldots:y_6]\in\mathbb{P}^6$ would have to make the equations identically zero, but at least one term of them will always be non-zero I think. $\endgroup$
    – pbelmans
    Oct 22 at 10:36
  • 1
    $\begingroup$ @pbelmans: In $\mathbb{P}^6$ we have seven homogeneous coordinates, and so there is a unique point which has $y_1 = \dots = y_6 = 0$. $\endgroup$
    – Sasha
    Oct 22 at 11:04
  • $\begingroup$ Oh yes, I was silly indeed, and I missed $y_0$. What a rookie mistake! Thanks! $\endgroup$
    – pbelmans
    Oct 22 at 11:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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