When does $Aut(X)=Bir(X)$ hold? - MathOverflow

When does $Aut(X)=Bir(X)$ hold?

2013-01-31T00:37:17Z

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.

Answer by Piotr Achinger for When does $Aut(X)=Bir(X)$ hold?

2013-01-31T01:19:39Z

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

Answer by diverietti for When does $Aut(X)=Bir(X)$ hold?

2013-01-31T07:44:17Z

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).

Answer by Henri for When does $Aut(X)=Bir(X)$ hold?

2013-01-31T10:08:55Z

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.

Answer by Christian Liedtke for When does $Aut(X)=Bir(X)$ hold?

2013-01-31T11:54:39Z

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