


By the Hurwitz's automorphisms theorem there is an upper bound $|\text{Aut}(C)|\leq 84(g-1)$ for all Riemann surfaces $C$ with $g(C)\geq 2$, but it is not sharp if $g=2$. What is the sharp upper bound for genus $2$ Riemann surfaces? Also, I'm asking for the simple proof of the fact that there is no genus $2$ Riemann surfaces with $|Aut(C)|=84$.


2 Answers


One proof uses the fact that every genus $2$ Riemann surface is hyperelliptic.

If $C$ is a hyperelliptic Riemann surface, then there is a unique degree $2$ map to $\mathbb P^1$, which defines a homomorphism $Aut(C) \to Aut(\mathbb P^1)$ whose kernel is of order $2$, generated by the hyperelliptic involution. This gives a finite subgroup of $Aut(\mathbb P^1) = PGL_2(\mathbb C)$ which preserves some set of exactly $2g+2$ points.

A finite subgroup of $PGL_2(\mathbb C)$ must lie in its maximal compact subgroup, which is $SO(3)$. We can use the classification of finite subgroups of $SO(3)$ to find the possible automorphism groups. We need to find the largest finite subgroup of $SO(3)$ that preserves a set of $6$ points on the sphere.

We use the fact that all subgroups are either cyclic, dihedral, $A_4$, $S_4$, or $A_5$. A cyclic subgroup that fixes $6$ points is no larger than $C_6$. A dihedral subgroup that fixes $6$ points is no larger than $D_6$. $A_5$ has no orbits of fewer than $12$ points. $S_4$ is the largest remaining possibility, and preserves $6$ points (the $6$ vertices of an inscribed regular octahedron), so the largest possible value for $|Aut(C)|$ is $2|S_4|=48$.

5 should be of interest to you.


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