Automorphism of genus 2 surface with 5 fixed points 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.
 A: 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.
A: There is an orbifold quotient of an orientable genus two surface that is a sphere with five cone points of order two. However, covering group here is $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ so a single homeomorphism will generate this group. 
The existence of the quotient is as follows. First, a bit of notation. Following Thurston, we will use $F(n_1, .. n_m)$ to indicate an orbifold with underlying space $F$ and $m$ cone points, where the order of each cone point is given by $n_j$.
Then the Riemann-Hurwitz formula for orbifold euler characteristic is: 
$$\chi(F(n_1, .. n_m)) = \chi(F) - \sum_{j=1}^m(1-\frac{1}{n_j}). $$
Note, using this formula we see the only possible orientable orbifold quotient of a genus two surface with five cone points is $S^2(2,2,2,2,2)$. As noted above this quotient can be realized and in fact this is the only case we need to consider. While $\chi(S^2(2,2,2,2,3)) = -2/3$, any three fold cover of this orbifold will have cone points of order 2. All other cases can be dismissed by considering the minimal manifold covering degree in a similar fashion.
From the analysis of the torus, we know that a a torus covers the ``pillowcase'' the orbifold $S^2(2,2,2,2)$. At points other than the cone points this is a 2-1 covering. Thus, we have the following covering $p: T^2(2,2) \to S^2(2,2,2,2,2)$ (forgive the rough sketches). Cone points of order 2 are indicated by dots and the axis of symmetry is indicated by the dashed line. 


One can do a case analysis to show that a cyclic covering from the genus 2 surface to $S^2(2,2,2,2,2)$ does not exist. By looking for subgroups of $\pi_1^{orb}(S^2(2,2,2,2,2)$ that are isomorphic to the fundamental group of the genus two surface.
The following MAGMA code shows does this case analysis and shows that all quotient groups are $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$.
> G := Group<a,b,c,d,e|a*b*c*d*e, a^2,b^2,c^2,d^2,e^2>;
> L := LowIndexSubgroups(G,<4,4>);  
for> if AQInvariants(l) eq [ 0, 0, 0, 0 ] then
for|if> print l, AQInvariants(quo<G|l>);         
for|if> end if;
for> end for;

