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

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

Theorem 1 (Hurwitz)If is a compact Riemann surface of genus , then there are at most automorphisms of .

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 together with Riemann surfaces with exactly automorphisms; there are explicit constructions that I’m not very familiar with. It is false in characteristic , 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 2If is a compact Riemann surface of genus for a prime, then has order strictly less than .

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 reached the maximal possible, which is . Then we could find a -Sylow subgroup , which is cyclic of order .

The point is that we can take the quotient 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

of degree . As a result, we can compute the genus of the quotient (this is a similar technique as in the proof of Hurwitz’s theorem, in fact). By the Riemann-Hurwitz formula, it is

where is the sum of the total ramification of the map. It follows that the genus of is at most two.

The claim is that the genus is exactly two, though. The reason is that is necessarily a multiple of , as the fiber of the map is either of cardinality or one at each point, by inspection. Now cannot be written as either or plus a multiple of . This means that the genus must be exactly two (and the ramification must be zero).

Now the number of -Sylow subgroups has order for some . Since any two intersect trivially, if is larger than , there can be only one, so is necessarily normal in . This means that acts on . In particular, if we choose a of order seven, we get an order seven automorphism of .

Now the claim is that a genus two curve such as 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 3A genus two curve does not have an automorphism of order seven.

*Proof:* Indeed, if did, we could apply Riemann-Hurwitz to the map . Let’s call the quotient ; then we have, by Riemann-Hurwitz,

for the ramification. The ramification is, however, necessarily a multiple of six. This is because a fiber of consists either of one point or seven, because of the definition of as a quotient. So, if we reduce mod six, we find

We must have , because again by Riemann-Hurwitz it must be less than . Neither satisfies the congruence, though.

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