Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
I heard of two proofs. The first one imply calculating volumes of fundamental polygon of the curve in the universal cover. The second is using Weiestrass points (which are in finite number) because the group of automorphisms acts on the Weiestrass points. –  Damien L Feb 7 '13 at 0:23
I don't know an elementary argument.To deal with higher dimension case, one can consider $\mathrm{Aut}(X)$ as an algebraic group. Its tangent space at the origin is a vector space of dimension $\dim H^0(X, T_X)$ where $T_X$ is the tangent bundle of $X$. So the whole reduces to show the vanishing of $H^0(X, T_X)$. Note that the finiteness of $\mathrm{Aut}(X)$ is not true in positive characteristic when $\dim X>1$. –  user18119 Feb 7 '13 at 21:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.