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Is every pseudo-automorphism (self-birational map which does not contract any hypersurface) of a smooth Fano variety of Picard rank $1$ equal to a biregular automorphism?

Remark: For $\mathbb{P}^n$, the answer is yes, and easy: every birational map of degree $>1$ contracts a hypersurface, given by its Jacobian. Same for any projective surface (because birational maps are sequence of blow-ups and blow-downs).

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  • $\begingroup$ Can't you just apply Hartog's theorem / S2 extension to sections of (positive) tensor powers of the anticanonical bundle? $\endgroup$ Aug 22, 2014 at 1:40
  • $\begingroup$ When you write "birational map of degree > 1", what precisely do you mean? Do you mean "rational self-map" rather than a birational map? $\endgroup$ Aug 22, 2014 at 2:48
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    $\begingroup$ @JasonStarr: "Degree" probably refers to the induced map on the Picard group, not the degree of the extension of function fields. $\endgroup$
    – naf
    Aug 22, 2014 at 4:29
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    $\begingroup$ Yes, the degree of a birational map $\mathbb{P}^n\dashrightarrow \mathbb{P}^n$ is the degree of the polynomials (given without common factors), and is also the induced map on the Picard group. @JasonStarr, Could you explain more what you mean with Hartog's Theorem / S2 extension in this context? I google it and did not understand exactly what you meant. $\endgroup$ Aug 22, 2014 at 6:38
  • $\begingroup$ Can the strict transform of an ample divisor under a non-automorphism pseudoautomorphism still be ample? I'd expect it gets a base locus along the indeterminate subvarities. $\endgroup$
    – user47305
    Aug 22, 2014 at 7:31

3 Answers 3

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You don't need your variety, say $X$, to be Fano, only $\mathrm{Pic}(X)=\mathbb{Z}$. A pseudo-automorphism $u$ of $X$ induces an automorphism of $\mathrm{Pic}(X)$, which must be the identity. Let $L$ be a very ample line bundle on $X$; since $u^*L\cong L$, $u$ induces an automorphism of $H^0(X,L)$ (here you use Hartogs theorem, as Jason pointed out). Then $u$ induces an automorphism of $\mathbb{P}(H^0(X,L))$ which preserves the image of $X$, hence an automorphism of $X$.

Note that if $K_X\geq 0$, any birational map is a pseudo-automorphism, and therefore biregular. Of course this doesn't hold for Fano varieties.

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  • $\begingroup$ The part "and therefore biregular" is not correct, in dimensions $3$ and higher $\endgroup$
    – YangMills
    Apr 4 at 17:56
  • $\begingroup$ @YangMills: This is under the previous hypothesis, namely $\operatorname{Pic}(X)=\mathbb{Z} $. $\endgroup$
    – abx
    Apr 4 at 18:25
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This is also true for every smooth Fano variety $X$, with any Picard number. One can see it using Mori dream spaces: $X$ is a Mori dream space (by BCHM), and hence has (up to isomorphisms) only finitely many "small $\mathbb{Q}$-factorial modifications" (SQM) = $f\colon X$-->$Y$ birational, isomorphism in codimension one, with $Y$ projective, normal, and $\mathbb{Q}$-factorial. These SQMs correspond to the chambers in the decomposition of the cone of movable divisors in $\mathcal{N}^1(X)$, the chamber corresponding to $f$ being $f^*\text{Nef}(Y)$. For arbitrary Mori dream spaces, it can happen that some $Y$ is isomorphic to $X$, and then $X$ has a pseudo-automorphism. But if $X$ is Fano this is impossible, because the anticanonical class is in the interior of the chamber $\text{Nef}(X)$, so it will not be ample on any other model $Y$.

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It seems to me that the Fano variety $X$ does not need to be smooth. It is enough to have that $-K_X$ is $\mathbb{Q}$-Cartier. Indeed, in this case we can embed $X$ in a projective space $\mathbb{P}^n$ with the linear system $|-mK_X|$ for $m\gg 0$.

For any pseudo-automorphism $\phi:X\dashrightarrow X$ we have that $\phi^{*}(-mK_X) = -mK_X$, and hence $\phi$ induces an automorphism of $\mathbb{P}^n$ stabilizing $X\subset\mathbb{P}^n$.

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