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

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