Why should automorphism groups of compact hyperbolic curves be finite Let $X$ be a compact connected Riemann surface of genus at least two, or let $X$ be a smooth projective connected curve over an algebraically closed field of characateristic zero.
Then Hurwitz proved that the number of automorphisms of $X$ is at most $84(g-1)$.
I would like to know why Aut$(X)$ is finite (without appealing to Hurwitz' result) by an "elementary" argument.
That is, why should the automorphism group of $X$ be finite?
Once I have such an elementary argument, I believe it should be applicable to  certain higher-dimensional varieties such as varieties with ample canonical sheaf. Why should they have only finitely many automorphisms?
 A: All compact complex surfaces $X$ of general type have a finite group $Aut(X)$ of automorphism. If $X$ is minimal, even numerical bounds are known for the number $\#Aut(X)$, e.g. a bound linear in $c_1^2(X)$. Cf. "Xiao, Gang: Bound of automorphisms of surfaces of general type, I. Ann. of Math. (139) 1994, 51-77"
A: @Tom, because you are interested in a generalization to higher dimensional varieties.
To prove the finiteness of $G := Aut(X)$ for a surface $X$ of general type one can proceed as follows:
By definition $X$ has Kodaira dimension $kod(X) = 2$. The plurigenera - and the Kodaira dimension in particular - are birational invariants. A fundamental result about surfaces of general type states: The 5-canonical map
$$\phi_{\kappa_X^{\otimes 5}}: X \longrightarrow \mathbb P^N$$
maps $X$ birationally onto a normal surface $Z \subset \mathbb P^N, N = h^0(X, \kappa_X^{\otimes 5}) - 1$. 
This fact allows to represent the complex Lie group $G$ as an affine algebraic group: The natural action of $G$ on $X$ induces an action on the vector space $H^0(X, \kappa_X^{\otimes 5})$. Because $G$ stabilizes the closed subvariety $Z \in \mathbb P^N$, $G$ is a closed subgroup of the projective linear group. The latter is an affine algebraic group, hence $G$ is an affine algebraic group, too. 
If $G$ were not finite, a 1-dimensional subgroup $A \subset G, A = \mathbb G_m \ or \ A = \mathbb G_a$, would exist. The group $A$ is Abelian. The orbit space $Z/A$ exists as a variety with a rational map $f: Z \longrightarrow Z/A$. By the 1-dimensional case of the "cross-section theorem" of Rosenlicht $f$ has a rational section, which implies that $Z$ is birational to a ruled surface. Any ruled surface - and a posteriori $Z$ - has Kodaira dimension = $- \infty$. This fact contradicts $kod(X) = 2$. Hence $G$ is finite, q.e.d.   
Cf. "Rosenlicht, Maxwell: Some Basic Theorems on Algebraic Groups. American Jounal of Mathematics, Vol. 78, N0. 2 (1956), Theorem 10."
