When does a projective algebraic surface have an infinite automorphism group? Is there a simple criterion, or at least a sufficient condition?


2 Answers 2


Well, let us start with a baby case. Consider a projective smooth curve $C$ over an algebraically closed field $K$. Let $T_C$ be the tangent bundle on $C$. The following facts are well known.

  • If $C=\mathbb P^1$, then $\mathrm{Aut}(C)=\mathrm{PGL}(2,K)$ is an algebraic group of dimension $3$. On the other hand, $\dim_K H^0(C, T_C)=3$.

  • If $C$ has genus $1$, then $\mathrm{Aut}(C)$ is an algebraic group whose connected component is $C$ (once an origin is chosen), hence has dimension $1$. On the other hand, $\dim_K H^0(C, T_C)=\dim_K H^0(C, O_C)=1$ ($T_C\cong O_C$ in this case).

  • If $C$ has genus $g\ge 2$, then $\mathrm{Aut}(C)$ is a finite (algebraic) group, hence has dimension $0$. On the other hand $\dim_K H^0(C, T_C)=0$ because $\deg T_C=2-2g <0$.

In higher dimension the picture is a little more complicate. Let $X$ be a projective variety over $K$. The group $\mathrm{Aut}(X)$ is the points of a group scheme over $K$. This is proved using Hibert schemes and by viewing an automorphism as a closed subscheme of $X\times X$ via its graph. See e.g. the excellent book of Kollár, "Rational curves on algebraic varieties", I.2.10. The group scheme is locally of finite type and the tangent space at the origin has dimension $\dim_K H^0(X, T_X)$ (Exercice I.2.16.4, use Thm I.2.16).

  • 1st conclusion: if $K$ has characteristic $0$, as any group scheme locally of finite type is automatically smooth, the dimension of $\mathrm{Aut}(X)$ is $\dim_K H^0(X, T_X)$. In positive characteristic, the former is bounded by the latter.

  • In characteristic $0$, in order for the automorphisms group to be finite, $H^0(X, T_X)$ must vanish. The converse is not true because $\mathrm{Aut}(X)$ can be a discrete infinite group.

This of course is not a very satisfactory answer to your question. But:

  • If $X$ is of general type (curves of general type are those of genus $\ge 2$), then $\mathrm{Aut}(X)$ is always finite. This results from a theorem of Kobayashi-Ochiai in characteristic zero (generalized by Deschamps-Ménégaux in positive characteristics). This theorem is a generalization to higher dimension of de Franchis theorem for morphisms of curves.

There is a lot of work (over $\mathbb C$) on the bound of the order of $\mathrm{Aut}(X)$ when $X$ is of general type. But this is another story.

  • $\begingroup$ So if the global sections of the tangent bundle have dimension n, then the automophism group is an algebraic group of dimension n? $\endgroup$
    – arsmath
    Sep 13, 2013 at 12:45
  • $\begingroup$ Yes, you are essentially correct. $\endgroup$
    – Cantlog
    Sep 13, 2013 at 20:03
  • $\begingroup$ Nice answer, Cantlog. $\endgroup$ Sep 13, 2013 at 21:16

Although it's a bit late, I can't resist adding the following comment to Cantlog's excellent answer.

As mentioned there, looking at vector fields doesn't suffice, because there is the possibility that $\mathrm{Aut}(X)$ can be an infinite discrete group.

In the latter case, however, we can get some information by looking at the natural representation of $\mathrm{Aut}(X)$ on the Néron–Severi group $NS(X)_{\mathbf{R}}$ (meaning divisors on $X$ modulo numerical equivalence, tensored with the real numbers). The homomorphism $\mathrm{Aut}(X) \rightarrow GL \left(NS(X)_{\mathbf{R}} \right)$ always has kernel an algebraic group, so in this case it will be finite. So the image will be an infinite subgroup of the general linear group.

Now there's a natural convex-geometric object acted on by $\mathrm{Aut}(X)$ via this representation, namely the cone of curves $C(X)$ of $X$. A rational polyhedral cone which is fixed by an infinite subgroup of integer matrices cannot have finitely many extremal rays. So the conclusion is that if $C(X)$ has finitely many extremal rays, then $\mathrm{Aut}(X)$ cannot be infinite discrete. This is useful because in practice, one can often calculate $C(X)$.


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