Consider a compact Riemann surface (or smooth projective algebraic curve in characteristic zero) ${X}$. One of the first facts one observes in their theory is that the group ${\mathrm{Aut}(X)}$ of automorphisms of ${X}$ is quite large when ${X}$ has genus zero or one. When ${X}$ has genus zero, it is the projective line, and its automorphism group is ${\mathrm{PGL}_2(\mathbb{C})}$, a fact which generalizes naturally to higher projective spaces. When ${g = 1}$, the curve ${X}$ acquires the structure of an elliptic curve from any distinguished point. Thus, translations by any element of ${X}$ act on ${X}$. So ${\mathrm{Aut}(X)}$ has points corresponding to each element of ${X}$ (and a few more, such as inversion in the group law).

But when the genus of ${X}$ is at least two, things change dramatically. It is a famous theorem that the number of automorphisms is bounded:

Theorem 1 (Hurwitz) If ${X}$ is a compact Riemann surface of genus ${g \geq 2}$, then there are at most ${84(g-1)}$ automorphisms of ${X}$.

I won’t write out a proof here; a discussion is in lecture 9 of the notes I’m taking in an algebraic curves class. In fact, the hard part is to show that there are finitely many automorphisms, after which it is a combinatorial argument.

This bound is often sharp. There are infinitely many genera ${g}$ together with Riemann surfaces with exactly ${84(g-1)}$ automorphisms; there are explicit constructions that I’m not very familiar with. It is false in characteristic $p$, because the Riemann-Hurwitz formula is no longer necessarily true (because of the existence of non-separable morphisms), and counterexamples are given in the notes.

What I want to describe today is that this bound is often not sharp.

Theorem 2 If ${X}$ is a compact Riemann surface of genus ${g = p+1}$ for ${p>84}$ a prime, then ${\mathrm{Aut}(X)}$ has order strictly less than ${84(g-1)}$.

I don’t remember the attribution of this result; I learned it today while paging through a book on automorphisms of Riemann surfaces.

The proof of this illustrates some useful techniques in the theory of algebraic curves. Namely, let’s suppose that ${G = \mathrm{Aut}(X)}$ reached the maximal possible, which is ${84p}$. Then we could find a ${p}$-Sylow subgroup ${T \subset G}$, which is cyclic of order ${p}$. The point is that we can take the quotient ${X/T}$ as a Riemann surface. One way to do this is to take the categorical quotient in the category of algebraic varieties (or schemes), and then get a possibly singular curve; one should normalize it then. There will be a quotient map $\displaystyle X \rightarrow X/T$

of degree ${p}$. As a result, we can compute the genus of the quotient ${X/T}$ (this is a similar technique as in the proof of Hurwitz’s theorem, in fact). By the Riemann-Hurwitz formula, it is $\displaystyle 2p = 2g(X) - 2 = p(2 g(X/T)) - 2 ) + R,$

where ${R}$ is the sum of the total ramification of the map. It follows that the genus of ${X/T}$ is at most two.

The claim is that the genus is exactly two, though. The reason is that ${R}$ is necessarily a multiple of ${p-1}$, as the fiber of the map ${X \rightarrow X/T}$ is either of cardinality ${p}$ or one at each point, by inspection. Now ${2p}$ cannot be written as either ${-2p}$ or ${0}$ plus a multiple of ${p-1}$. This means that the genus must be exactly two (and the ramification must be zero).

Now the number of ${p}$-Sylow subgroups has order ${1 + kp}$ for some ${k}$. Since any two intersect trivially, if ${p}$ is larger than ${84}$, there can be only one, so ${T}$ is necessarily normal in ${G}$. This means that ${G}$ acts on ${X/T}$. In particular, if we choose a ${\sigma \in G}$ of order seven, we get an order seven automorphism of ${X/T}$.

Now the claim is that a genus two curve such as ${X/T}$ cannot have an automorphism of order seven. If we show this, then we are done; so we have only to prove the next lemma.

\newtheorem{lemma}{Lemma}

Lemma 3 A genus two curve ${Y}$ does not have an automorphism of order seven.

Proof: Indeed, if ${Y}$ did, we could apply Riemann-Hurwitz to the map ${Y \rightarrow Y/(\mathbb{Z}/7)}$. Let’s call the quotient ${Z}$; then we have, by Riemann-Hurwitz, $\displaystyle 2 = 2g(Y) - 2 = 7 ( 2 g(Z) - 2) + R$

for ${R}$ the ramification. The ramification ${R}$ is, however, necessarily a multiple of six. This is because a fiber of ${Y \rightarrow Z}$ consists either of one point or seven, because of the definition of ${Y}$ as a quotient. So, if we reduce mod six, we find $\displaystyle 2 \equiv 2g(Z) - 2.$

We must have ${g(Z) \in \left\{0, 1\right\}}$, because again by Riemann-Hurwitz it must be less than ${g(Y)}$. Neither satisfies the congruence, though. $\Box$