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.)


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.

  • $\begingroup$ 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

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