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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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 $Y$.


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 $X\subset\mathbb{P}^n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.