Let $X$ be a smooth projective variety of Picard number one, and let $f:X\dashrightarrow X$ be a birational automorphism which is not an automorphism.
Must $f$ necessarily contract a divisor?
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3$\begingroup$ Your question is related to Sibony-Dinh theory see link.springer.com/article/10.1007/s00208-013-0992-4 and annals.math.princeton.edu/wp-content/uploads/… $\endgroup$– user21574Nov 22, 2017 at 21:27
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5$\begingroup$ Anyway, I think their definition as " p-th relative dynamical degree" is not so fine, in fact integral must be taken on M/B and we must use fiberwise Kahler form like $\omega_{M/B}$(which is not Kahler in general) instead. to define such p-th-relative dynmical degree . See p.4 perso.univ-rennes1.fr/serge.cantat/Documents/… . But you check yourself. (this is just my opinion ) $\endgroup$– user21574Nov 22, 2017 at 21:46
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1$\begingroup$ There was an answer with an counterexample that is now deleted. The counterexample was not correct, but there was also a comment by Jason Starr on the answer giving another counterexample; I would like to know if that counterexample was correct or not. (To show my hand, I think the answer to the OP's question is yes, but I do not yet see what was the issue with Jason Starr's counterexample.) $\endgroup$– PooterNov 24, 2017 at 9:49
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$\begingroup$ Possible duplicate of Pseudo-automorphisms on Fano varieties $\endgroup$– Jérémy BlancNov 24, 2017 at 13:43
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$\begingroup$ The rational transformation in my example was actually the identity map :( $\endgroup$– Jason StarrNov 24, 2017 at 19:52
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1 Answer
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In contrast to an earlier answer on this question, I claim the answer here is yes.
In the accepted answer to this question:
Pseudo-automorphisms on Fano varieties
abx explains that for any smooth variety $X$ of Picard number 1, any pseudo-automorphism of $X$ (i.e. a birational automorphism which is an isomorphism in codimension 1) must in fact be an automorphism.
So it remains to argue that if $f$ does not contract any divisor, then it is a pseudo-automorphism. The only thing to check is that $f^{-1}$ does not contract a divisor either. You can do this by looking at a resolution of $f$ :
$$ X \leftarrow^p \widetilde{X} \rightarrow^q X $$
where $\widetilde{X}$ is smooth: the numbers of $p$-exceptional prime divisors and $q$-exceptional prime divisors must be equal, but the hypothesis that $f$ doesn't contract a divisor says that every $q$-exceptional divisor is $p$-exceptional. Hence the two sets are the same, and so $f^{-1}$ doesn't contract a divisor either.
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$\begingroup$ "So it remains to argue that if $f$ does not contract any divisor, then it is a pseudo-automorphism". It seems to me that it is essentially the definition of pseudo-automorphism: it induces an isomorphism $U\to V$ where $U,V$ are open subsets of $X$, with complements of codimension $\ge 2$. $\endgroup$ Nov 24, 2017 at 13:41
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1$\begingroup$ @JérémyBlanc: to me it seems a priori possible that $f$ does not contract a divisor but $f^{-1}$ does (in which case $f$ would not be a ps-aut.) That is the case I am trying to exclude. $\endgroup$– PooterNov 24, 2017 at 13:45