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Can $\mathbb P \times \mathbb P \times \mathbb P$ be obtained from $\mathbb P^3$ by a finite succession of blow-ups and blow-downs along non-singular centers?

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    $\begingroup$ In characteristic zero, any birational map is the composition of a finite succession of blow-ups and blow-downs along non-singular centers. This is a famous theorem of Wlodarczyk. $\endgroup$
    – abx
    Mar 8, 2015 at 9:10

2 Answers 2

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Yes it can. This is a particular case of the weak factorization theorem of Abramovich, Karu, Matuski and Włodarczyk (Torification and factorization of birational maps, J. Amer. Math. Soc. 15 (2002), no. 3, 531–572). The proof of that theorem uses resolution of singularities hence requires characteristic zero (at least, for the moment).

Actually, your example is of an equivariant birational map between toric varieties, and this case had been treated earlier by Włodarczyk (Decomposition of birational toric maps in blow-ups & blow-downs*, Trans. Amer. Math. Soc. 349 (1997), no. 1, 373–411) in any characteristic (the result is equivalent to a purely combinatorical statement on fans).

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  • $\begingroup$ Any explicit constructions of blow-ups and blow-downs between $\mathbb P \times \mathbb P \times \mathbb P$ and $\mathbb P^3$ is known? $\endgroup$
    – Creg
    Mar 8, 2015 at 9:58
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    $\begingroup$ In the projective toric case, it's about cutting off corners of the two relevant polytopes -- the cube and the 3-simplex -- to get an answer in common. Here, you cut off three corners of the simplex making one of the faces into a hexagon (blow up three points). Then shrink the old edges of that hexagon to singular points (blow down the three original lines connecting them). Then pull that resulting triangle out a bit, creating three new edges (blow up the three singular points to $\mathbb P^1$s, differently). Keep pulling it out, to get a point (blow down its $\mathbb P^2$). Now it's a cube. $\endgroup$ Mar 8, 2015 at 10:46
  • $\begingroup$ To avoid blowing down to singular things: blow up three general lines in $\mathbb P^3$, and there's an obvious map to $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$ given by the pencils of planes containing the lines. What it does is contract the unique quadric containing the lines to a rational curve in $\mathbb P^1 \times \mathbb P^1 \times \mathbb P^1$. I guess the image is the diagonal, up to choice of coordinates. (This is a different map from the toric example, since you can't find three disjoint torus invariant lines without blowing up points first.) $\endgroup$
    – user47305
    Mar 9, 2015 at 16:38
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One can obtain $\overline{M}_{0,n}$ (the moduli space of $n$-pointed rational curves) in at leat two ways:

  1. Blowing-up $n-1$ points $\{p_1,...,p_{n-1}\}$ in $\mathbb{P}^{n-3}$ and then all the strict transforms of linear spaces generated by subsets of $\{p_1,...,p_{n-1}\}$ in order of increasing dimension.
  2. Blowing-up, in order of increasing dimension, suitable smooth subvarieties of the diagonals of $(\mathbb{P}^1)^{n-3}$.

The first construction is due to Kapranov, you can find it for instance in Section 6.2 "B. Hassett, Moduli spaces of weighted pointed stable curves, Advances in Mathematics 173 (2003) 316–352". The second one is due tue Keel, see Section 6.3 of the same paper.

Putting togheter 1 and 2 you see that $(\mathbb{P}^1)^{n-3}$ can always be obtained from $\mathbb{P}^{n-3}$ by a sequence of blow-ups and blow-downs along smooth centers.

In your case, $n =6$, Kapranov constrction is as follows:

  • begin with $p_1,...,p_5\in\mathbb{P}^{3}$ in general position;
  • blow-up the points $p_{1},...,p_{5}\in\mathbb{P}^{3}$;
  • blow-up the strict transforms of the lines $\left\langle p_{i},p_{j}\right\rangle$, $i,j=1,...,5$.

In this way you get a morphism $\phi:\overline{M}_{0,6}\rightarrow\mathbb{P}^3$.

While Keel construction is as follows:

  • Let $X_1$ be the blow-up of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$ in $p_1 = ([0:1],[0:1],[0:1])$, $p_2 = ([1:0],[1:0],[1:0])$, and $p_3 = ([1:1],[1:1],[1:1])$.
  • Consider the projections $\pi_i:\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3\rightarrow\mathbb{P}^1_i$, and define $F_0 = \bigcup_{i=1}^3\pi_i^{-1}([0:1])$, $F_1 = \bigcup_{i=1}^3\pi_i^{-1}([1:0])$, $F_{\infty} = \bigcup_{i=1}^3\pi_i^{-1}([1:1])$. Let $\Delta_2$ be the union of the $2$-dimensional diagonals of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$. Then we have $X_2$ the blow-up of $X_1$ along the strict transform of $\Delta_2\cap (F_0\cup F_1\cup F_{\infty})$.
  • Finally, the blow-up $X_3$ of $X_2$ along the strict transform of the $1$-dimension diagonal $\Delta_1$ of $\mathbb{P}^1_1\times\mathbb{P}^1_2\times\mathbb{P}^1_3$. This is $\overline{M}_{0,6}$.

Now, let $\psi:\overline{M}_{0,6}\rightarrow (\mathbb{P}^1)^3$. You can take your sequence of blow-ups and blow-downs as $f = \phi^{-1}\circ \psi$.

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