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Let $ f: \mathbb{P}^n \dashrightarrow (\mathbb{P}^1)^n , [x_0:\dots :x_n] \mapsto ([x_0,x_1],[x_0,x_2], \dots ,[x_0,x_n])$ a birational map.

In particular, if $X$ is the blow-up of $\mathbb{P}^n$ at $r+n-1$ points and $X'$ is the blow-up of $(\mathbb{P}^1)^n$ at $r$ points. Consider $\phi : X \dashrightarrow X'$ the birational map that makes the diagram with the blow-up's and $f$ commute, is $\phi$ a Crepant birational map?

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  • $\begingroup$ Maybe you don't believe me, but I still contend that $f$ cannot be a birational map. It is projective: factor as the graph of $f$ followed by the projection onto the second factor. $f$ being separated, means that the graph of $f$ is a closed immersion. Thus, $f$ is projective and therefore proper. Kontsevich proved that if $X$ and $Y$ admit a proper birational map, then the hodge numbers must be the same. In the previous question, I show that the hodge numbers are not the same: mathoverflow.net/questions/123700/crepant-birational-map/… $\endgroup$
    – Andrew Stout
    Mar 6, 2013 at 7:14
  • $\begingroup$ Even if you believe $f$ is a birational morphism, since blow-ups are projective, we have that $X$ has the same hodge polynomial of $\mathbb{P}^n$ and $X'$ has the same hodge polynomial of $(\mathbb{P}^1)^n$, but I have already computed these hodge polynomials in the last question. They are not equal. Therefore, there is no crepant resolution of $X \rightarrow X'$ by Batyrev's or Kontsevich's theorem because $H(X,u,v) \neq H(X',u,v)$. $\endgroup$
    – Andrew Stout
    Mar 6, 2013 at 8:00
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    $\begingroup$ @Andrew: doesn't $f$ restrict to an isomorphism between the obvious open subsets isomorphic to $\mathbb{A}^n$? $\endgroup$
    – quim
    Mar 6, 2013 at 8:22
  • $\begingroup$ yes, this is true, $f$ is birational, but it is not universally closed. I made an error. $\endgroup$
    – Andrew Stout
    Mar 6, 2013 at 8:38

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Suppose for the moment that $n=2$. Then the birational map $f\colon \mathbb{P}^n\dashrightarrow(\mathbb{P}^1)^n$ described in the question, which is $f\colon [x:y:z] \dashrightarrow([x:y],[x:z])$ has two base-points, $[0:0:1]$, $[0:1:0]$ and the inverse is $f^{-1}\colon ([a:b],[c:d]) \dashrightarrow[ac:bc:ad]$ and has one base-point, namely $([0:1],[0:1])$. If $X\to \mathbb{P}^2$ is the blow-up of $[0:0:1]$, $[0:1:0]$ and $X'$ is the blow-up of $(\mathbb{P}^1)^2$ at $([0:1],[0:1])$, the corresponding map $X \dashrightarrow X'$ is an isomorphism, and thus a "crepant" birational map. If you want to blow-up more points on $X$ and $X'$, which corresponds via this isomorphism you again get an isomorphism. If you blow-up other points you get a birational map which is not an isomorphism and thus not crepant (between two smooth projective surfaces crepant=isomorphism).

For $n>2$, the birational map $f$ has more that only isolated base-points, so you need to blow-up more complicated stuff, then contract, and maybe blow-up again... and can get a sequence of simple birational maps (blow-ups or inverse of blow-ups) from $\mathbb{P}^n$ to $(\mathbb{P}^1)^n$. In any case, you cannot get morphisms if you only blow-up points of $\mathbb{P}^n$.

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  • $\begingroup$ @Jérémy: just a LaTeX suggestion: if you use $\dashrightarrow$ it looks much better. Cheers! $\endgroup$
    – Sándor Kovács
    Mar 6, 2013 at 11:53
  • $\begingroup$ of course, yes, segre embedding is a blow up of one point of $\mathbb{P}^2$. The exception divisors below should line up in the formula I presented. $\endgroup$
    – Andrew Stout
    Mar 6, 2013 at 12:03
  • $\begingroup$ @Sándor Kovács. Normally I use $\dasharrow$ and it works in LaTeX but not on this website, thanks for the hint. I fixed it. You can also use \ mapstochar\ dashrightarrow in fact. @Andrew: By Segre embedding you probably mean from $\mathbb{P}^1\times \mathbb{P}^1$ to the quadric in $\mathbb{P}^3$. But in this case it is the blow-up of two pts of $\mathbb{P}^2$, followed by the contraction of the line passing through the two points. $\endgroup$
    – Jérémy Blanc
    Mar 6, 2013 at 12:11
  • $\begingroup$ In fact $\mapstochar$ does not work on MO. $\endgroup$
    – Jérémy Blanc
    Mar 6, 2013 at 12:13

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