# When does $\operatorname{Aut}(X)=\operatorname{Bir}(X)$ hold?

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

• 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.)
– naf
Jan 31, 2013 at 13:12
• Just to follow up the important point raised by ulrich: the keywords here are "birational rigidity", "superrigidity", and "Sarkisov program". Jan 31, 2013 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? Feb 1, 2013 at 1:10

To complete the answer of Diverietti 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 Algebraic 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.

• $Y$ containing no rational curves ? or $X$? Jan 31, 2013 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. Jan 31, 2013 at 10:50
• Yes, of course, I misunderstood what you wrote! Cheers ! Jan 31, 2013 at 13:49

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

• 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. Jan 31, 2013 at 12:21
• even worse: a general (algebraic) K3 surface contains infinitely many rational curves, giving examples very away from being Kobayashi hyperbolic Jan 31, 2013 at 15:16
• @ChristianLiedtke: I hope this comment will be noticed even if the question is from a while ago. You answer is about birational morphisms, is it true for birational maps? if $X$ is a k3 surface, for example, is it true that a birational map $p: X \dashrightarrow X$ is everywhere defined? May 28, 2017 at 18:29

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 Chen - Rational self maps of Calabi–Yau manifolds). 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 Beauville - Some remarks on Kähler manifolds with $$c_1 = 0$$.

A simplest example is a curve $$C$$, then $$\operatorname{Bir}(C)\cong\operatorname{Aut}(C)$$.

A careful note that $$\operatorname{Bir}(X)$$ is not a group scheme in general. Moreover if $$X$$ and $$Y$$ are birational then $$\operatorname{Bir}(X)$$ is not isometric with $$\operatorname{Bir}(Y)$$

So we need to add some condition on $$X$$ in the sense of Minimal Model Program such that in scheme group theoretic sense $$\operatorname{Bir}(X)$$ behaves well since we have lack of well defined multiplication rule. In fact if $$X$$ is a minimal model with terminal singularities it is a theorem that $$\operatorname{Bir}(X)$$ is a group scheme as an Abelian variety. In general $$\operatorname{Aut}^o(X)\cong\operatorname{Bir}^o(X)$$

where $$\operatorname{Aut}^o$$ is the identity connected component. See Hanamura - On the birational automorphism groups of algebraic varieties.

Let $$X$$ be a projective variety. Then we have the following isomorphism from MMP:

$$\operatorname{Aut}(X_\text{min})\cong\operatorname{Bir}(X_\text{min})$$ where $$X_\text{min}$$ is the minimal model of the projective variety $$X$$.

For any surface of non-negative Kodaira dimension, we have $$\operatorname{Aut}(X)\subset\operatorname{Bir}(X)\cong\operatorname{Bir}(X_\text{min}) \cong\operatorname{Aut}(X_\text{min}).$$

If $$X$$ is a Fano variety of $$\dim\geq 5$$ and its Picard group is generated by anticanonical divisor of the variety $$X$$, then it is conjectured (see Chel'tsov - Birationally rigid Fano varieties) that $$\operatorname{Aut}(X)\cong\operatorname{Bir}(X)$$.

Moreover for moveable log pair $$(X,M_X)$$ which is log Calabi–Yau pair, with at worst canonical singularities then $$\operatorname{Aut}(X)\cong\operatorname{Bir}(X)$$.

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

• Thank you for the example, Piotr. Do you know how to prove the fact about Abelian varieties? Jan 31, 2013 at 2:44
• maybe something to do with the absence of rational subvarieties? Jan 31, 2013 at 3:03
• Milne's notes on abelian varieties, p. 15. Jan 31, 2013 at 17:14
• Milne's notes on Abelian varieties referenced by @PiotrAchinger. Jul 14, 2022 at 22:07

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

The canonical ring $$R$$ of $$X$$ is finitely generated. Furthermore $$X\cong\operatorname{Proj}(R)=X_\text{can}$$ 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.

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).

• Sorry for a (potentially) stupid question, but could you give me an example of compact $X$ such that $Bir(X)\ne Bim(X)$? Feb 1, 2013 at 1:04
• 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... Feb 1, 2013 at 7:15
• I see. Thank you for clarifying the point. Feb 1, 2013 at 21:48