Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a projective complex manifold. Under what condition do we have the equality $Aut(X)=Bir(X)$? Here $Aut(X)$ denotes the group of holomorphic automorphisms of $X$ and $Bir(X)$ the group of birational morphisms of $X$.

I am interested in the case when $\dim_{\mathbb{C}}X=2,3$. Maybe there are not universal criteria, so I would appreciate your providing me with any examples for which the equality holds.

share|improve this question
4  
There is a large body of work on this question for Fano varieties beginning with the theorem of Iskovskikh and Manin proving equality for smooth quartic threefolds. (Note that Fano varieties are covered by rational curves so these examples are of a very different nature from the ones mentioned in the answers so far.) –  ulrich Jan 31 '13 at 13:12
3  
Just to follow up the important point raised by ulrich: the keywords here are "birational rigidity", "superrigidity", and "Sarkisov program". –  Jason Starr Jan 31 '13 at 15:07
    
Thank you for the useful comments. Do you know anything about Calabi-Yau case? K3 surface is minimal and thus the equality holds (thanks to Christian'sanswer below). What about dimension 3 case? –  Koopa Feb 1 '13 at 1:10

6 Answers 6

To complete the answer of Divierietti and the comment of Roy Smith, here is a statement which might interest you:

Theorem If $X,Y$ are varieties over a field $k$, assume $X$ is smooth and $Y$ proper containing no rational curves. Then any rational map $X\dashrightarrow Y$ is everywhere defined.

You can find that statement in Debarre's book Higher Dimensional Geometry, Corollary 1.44 p.31.

In particular, if $X$ is smooth projective and contains no rational curves, then its automorphism group is equal to the group of its birational endomorphisms.

share|improve this answer
    
$Y$ containing no rational curves ? or $X$? –  diverietti Jan 31 '13 at 10:28
    
Yes, Y. Consider the two projection $p_1$, $p_2$ from the graph to $X$ and $Y$. We know that the exceptional locus of $p_1$ contains rational curves $C \subset X\times Y$. By assumption, $p_2$ must contract $C$, so that $C$ is contracted by $p_1$ and $p_2$, which is absurd. –  Henri Jan 31 '13 at 10:50
1  
Yes, of course, I misunderstood what you wrote! Cheers ! –  diverietti Jan 31 '13 at 13:49

It also holds for minimal surfaces of Kodaira dimension $\kappa\geq0$.

share|improve this answer
3  
True, essentially for the uniqueness of the minimal model. It is worth pointing out that in this case $X$ can contain rational curves, in particular it is not necessarily Kobayashi hyperbolic. –  Francesco Polizzi Jan 31 '13 at 12:21
2  
even worse: a general (algebraic) K3 surface contains infinitely many rational curves, giving examples very away from being Kobayashi hyperbolic –  Christian Liedtke Jan 31 '13 at 15:16

One example: This holds for abelian varieties, because a rational map to an abelian variety is always regular.

share|improve this answer
    
Thank you for the example, Piotr. Do you know how to prove the fact about Abelian varieties? –  Koopa Jan 31 '13 at 2:44
4  
maybe something to do with the absence of rational subvarieties? –  roy smith Jan 31 '13 at 3:03
    
Milne's notes on abelian varieties, p. 15. –  Piotr Achinger Jan 31 '13 at 17:14
    
Thanks. I will take a look at it. –  Koopa Feb 1 '13 at 0:57

I realize that no one addressed the question of the OP about Calabi-Yau manifolds. Two remarks:

1) The equality $\mathrm{Bir}(X)=\mathrm{Aut}(X)$ holds if $K_X$ is nef and $\mathrm{Pic}(X)=\mathbb{Z}$ (see for instance the introduction of this paper). This applies in particular to Calabi-Yau complete intersections (of dimension $\geq 3$).

2) There are examples of birational automorphisms of holomorphic symplectic manifolds which are not biregular, see §6 of this paper.

share|improve this answer

A large class of compact complex manifolds for which (more generally) $$ \operatorname{Aut}(X)=\operatorname{Bim}(X) $$ holds is given by Kobayashi hyperbolic compact complex spaces. Here $\operatorname{Bim}(X)$ is the group of bimeromorphic automorphism.

A compact complex space $X$ is Kobayashi hyperbolic iff there is no non-constant holomorphic map $f\colon\mathbb C\to X$. For instance, by Liouville's theorem, a compact complex space $X$ is hyperbolic as soon as its universal cover is a bounded domain in $\mathbb C^n$. Other examples are given by compact complex manifolds whose cotangent bundle is Griffiths positive (or, more generally, with ample cotangent bundle).

If $X$ is moreover projective, it is conjectured by Lang that being hyperbolic should be equivalent to have only subvarieties of general type.

This latter class of projective manifolds (of general type, with all subvarieties of general type) have indeed the property your are asking for, too. This is because the indeterminacy locus of a birational map is covered by rational curves (and cannot be of general type, nor hyperbolic).

share|improve this answer
1  
Sorry for a (potentially) stupid question, but could you give me an example of compact $X$ such that $Bir(X)\ne Bim(X)$? –  Koopa Feb 1 '13 at 1:04
1  
Hi Koppa. Of course if X is projective then the rational function field equals the meromorphic function field. From this it follows easily that any meromorphic mapping $f\colon X\to Y$ between complex projective varieties is indeed rational. When I wrote "more generally", I meant that I was looking at general abstract compact complex manifolds, where the notion of rational mapping is not even defined... –  diverietti Feb 1 '13 at 7:15
    
I see. Thank you for clarifying the point. –  Koopa Feb 1 '13 at 21:48

Let $X$ be a variety with at most canonical singularities and ample canonical divisor $K_X$. Then $Aut(X) = Bir(X)$.

The canonical ring $R$ of $X$ is finitely generated. Furthermore $$X\cong Proj(R)=X_{can}$$ since $K_X$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\rightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular.

More generally, if $X,Y$ are projective varieties with $K_X$, $K_Y$ ample and at most canonical singularities, then any birational map $f:X\rightarrow Y$ is indeed biregular.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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