Is there a self-homeomorphism of a genus 2 (closed, orientable) surface, which has finite order and exactly 5 fixed points?


Of course, the same question can be asked replacing 2 by $g$ and $5$ by any number $k$. An upper bound for possible values of $k$ is (generalized Lefschetz fixed point theorem) $2g+2$. 


For $g=0$, a sphere, only $k=0,2$ are possible.


For $g=1$, the torus, $k=0,1,2,3,4$ are all possible. For example, the map $(x,y) \mapsto (y,-x-y)$ is of order 3 and has exactly 3 fixed points.



Automorphism of genus 2 surface with 5 fixed points
Although Jason and Dylan seemed to have answered this question in the comments, I decided to work out what the generalized version of this kind of statement is.


Let $\sigma$ be an automorphism of a Riemann surface. Any automorphism that is orientation-reversing with a fixed point has a fixed curve, so let's assume $\sigma$ is orientation-preserving.


By Rieman-Hurwitz, if $\sigma$ has order $k$ and $n$ fixed points than:


$$ 2k -n (k-1) \geq 2 - 2g$$


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


Moreover if $k$ has order $2$ then the number of fixed points is congruent to $2g+2$ modulo $4$. So this already rules out many possibilities.


In particular, for $g=2$ this rules out $5$. 


There are other congruence conditions. If $k$ is a power of a prime $p$ then all the terms in the Riemann-Hurwitz formula are multiples of $p$ except for the $-n(k-1)$ and $2-2g$ so we get $n \equiv 2-2g $ modulo $p$.


so the list of possibilities becomes something like:


$n \leq 2g+2$ and congruent to it mod $4$ (order $2$)


$n \leq g+2$ and congruent to it mod $3$ (order $3$)


$n \leq 2g/3+2$ and even (order $4$)


$n \leq g/2+2$ and congruent to $2-2g$ modulo $5$ (order $5$)


$n \leq g/5+2$ with no congruence condition (order $6$)


I think you can show that almost all of these are actually achieved, by solving the Riemann-Hurwitz equation, drawing the right orbifold, and using the formula for the orbifold fundamental group.


