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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.)

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2 Answers 2

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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.

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  • $\begingroup$ 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, 2021 at 10:36
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    $\begingroup$ @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, 2021 at 11:04
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    $\begingroup$ Oh yes, I was silly indeed, and I missed $y_0$. What a rookie mistake! Thanks! $\endgroup$
    – pbelmans
    Oct 22, 2021 at 11:08
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Let $C$ be a general curve of genus $g$ and take $\pi: Sym^g(C) \to Jac(C)$.

Then brill-noether says that that fibers of dimension $r$ arise from $h^0=r+1$ line bundles which form a $g-h^0 h^1$ dimensional subspace of $Jac(C)$.

We have $h^0-h^1 = 1$ for our degree $g$ and so generically there is one point in the fiber i.e birational.

But take $h^0 \cdot h^1 = g, h^0 = 1+h^1$ so roughly $h^0, h^1 = \sqrt{g}$ then there are finitely many points with fiber roughly $\sqrt{g}$ dimensional.

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