TITLE: Automorphism Groups of Projective Algebraic Surfaces
QUESTION [7 upvotes]: When does a projective algebraic surface have an infinite automorphism group?  Is there a simple criterion, or at least a sufficient condition?

REPLY [4 votes]: 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 $C(X)$.