Exercise I.F-8 from "Arbarello, Cornabla, Griffiths, Harris: Geometry of algebraic curves" states that for a complex algebraic genus $g$ curve and its automorphism $\varphi$ of order $n$ the number of fixed points $\alpha$ of $\varphi$ satisfies

$\alpha\leq \frac{2g-2+2n}{n-1}$

with inequality only if the quotient $C/\varphi$ has genus $0$ (i.e. $\mathbb{P}^1$).

So the inequality itself is an easy part (it follows from Riemann-Hurwitz). But I can't see why the number of fixed points is exactly that number if $g(C/\varphi)\ne 0$. Can you help me?

Edited: in fact this number can be non-integer and it's confusing.


The map $\pi:C\to C/\varphi$ is totally ramified over the fixed points of $\varphi$. So if equality holds in the given inequality, we obtain that the degree of the ramification divisor of $\pi$ is $\geq \alpha.(n-1)=2g-2+2n$. On the other hand, it is an easy observation that for a degree $n$ morphism of curves $f:X\to Y$, the degree of the ramification divisor is maximal if $Y=\mathbb{P}^1$ and in this case is equal to $2g-2+2n$.

So, $\mbox{deg} R=2g-2+2n$ and by Riemann Hurwitz $g(C/\varphi)=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.