## Hurwitz’s Theorem on Automorphisms

This is this blog’s 100th post. Now, not quite my hundredth, nor nearly the hundredth with actual math content, but still, it’s a number which, when expressed in base ten, happens to have some zeros. More importantly, however, tomorrow marks one year since my first post went up. Now that that’s done, we’re going to spend today looking at other numbers. Specifically, the number of elements of any finite group acting on a complete curve.

Fix a complete curve $C$ and a finite group $G$. Then we say that $G$ acts on $C$ if we have a map $G\times C\to C$ denoted by $(g,p)\mapsto g\cdot p$ such that $(gh)\cdot p=g\cdot (h\cdot p)$ and $e\cdot p=p$ for all $g,h\in G,p\in C$. We call an action effective if the condition $g\cdot p=p$ for all $p$ implies that $g$ is the identity element.

Now, we can take quotients here. We’ll think analytically for a moment. $C$ is a Riemann surface, and $G$ acts on it effectively. Then we have the space $C/G$ with elements the sets $\{g\cdot p|g\in G\}$ for each $p$, which were a partition of $C$. So $C/G$ is defined as a set. In fact, it’s defined as a topological space by saying that a set is open if and only if its preimage is. Even better, we can actually get a complex structure on it to make it into a Riemann surface, which means that the quotient is again an algebraic curve. The quotient map $\pi:C\to C/G$ has degree $|G|$ and at any point $p\in C$, we have $mult_p \pi$ is equal to the order of the stabilizer of $p$, that is, the number of group elements fixing the point $p$.

Now we want to apply Hurwitz’s formula to this map. If we have $k$ branch points $y_1,\ldots,y_k$ in $C/G$, then over each of them there are $|G|/r_i$ ramification points, each having multiplicity $r_i$ (it all has to be the same, because the group action lets us permute them) so Hurwitz tells us that we have $2g(C)-2=|G|(2g(C/G)-2)+\sum_{i=1}^k \frac{|G|}{r_i}(r_i-1)$, which can be simplified a bit to $2g(C)-2=|G|\left(2g(C/G)-2+\sum_{i=1}^k (1-1/r_i)\right)$. We set $R=\sum_{i=1}^k (1-1/r_i)$. From here on, we assume that $C$ has genus $\geq 2$.

So now we break it into cases. If $g(C/G)\geq 1$, we have two.
If $R=0$, then there’s no ramification, and so $g(C/G)\geq 2$. So $|G|=(2g-2)/(2g(C/G)-2)=(g-1)/(g(C/G)-1)$. As $g(C/G)-1\geq 1$, we then have $|G|\leq g-1$. Now if $R\neq 0$, it must automatically be greater than $1/2$. Checking this (and a couple of other claims) are elementary and tedious, so we’ll leave those to be done in private, rather than here. So then $2g(C/G)-2+R\geq 1/2$, and so $2g-2\geq |G|/2$, which tells us that $|G|\leq 4(g-1)$. But this isn’t the case that gives the universal bound which is valuable.

Now, we look at the case where $g(C/G)$ is zero. Then Hurwitz’s formula becomes $2g-2=|G|(R-2)$. As $2g-2>0$, we must have $R>2$. It is again nothing but a tedious check to see that if it is greater than 2, it is greater than $2+\frac{1}{42}$. So then $2g-2\geq |G|(2+\frac{1}{42}-1)$, and so $84(g-1)\geq |G|$.

This gives us Hurwitz’s Theorem on Automorphisms, that a curve of genus $g\geq 2$ has at most $84(g-1)$ automorphisms. This follows from the fact that the automorphism group is finite and must act effectively.

Now for a bit of fun. We call a group a Hurwitz group if it is of order $84(g-1)$ for some $g$ and is also the automorphism group of an algebraic curve of that genus. Now, these groups are interesting in and of themselves, algebraically. They’re groups that can be generated by two elements $g,h$ such that $g^2=h^3=(gh)^7=e$. Of course, there are quite a few famous groups which arise this way, not the least of them is the Monster. However, the details of this have been done better than I could do them elsewhere, so I’ll just link to Lieven le Bruyn and Marni Dee Sheppeard.

Hurwitz’s theorem on automorphisms has another good consequence. If we denote by $M_{g,n}$ the functor of flat families of genus $g$ curves with $n$ marked points (that is, specified sections of the family) then for any $g$, we can fix $n$ big enough that curves with this many marked points don’t have any automorphisms, because an automorphism would take $C\to C$, but would also have to just permute the marked points. So then for $g\geq 2$ and $n$ sufficiently large, we’ll actually get a fine moduli space representing this functor, and it won’t just be a scheme, but it will be a variety! However, if we don’t want to be marking points, because that increases the dimension of the moduli space and can alter it enough that we lose some good geometric data, we still only get a coarse moduli space, because though the automorphism groups are finite, they aren’t zero.

Well, that’s all this week, next week we continue on through some more topics in the theory of curves.