Fermat Quartic Tiling I have been reading about the Fermat quartic $Q \subset \mathbb{P}^{2}$, defined in homogeneous coordinates as $X^{4}+Y^{4}+Z^{4}=0$. This is the second most symmetric non-hyperelliptic surface of genus 3, after of course the Klein quartic. 
We know that $Q$ has a symmetry group $\mathrm{PSL}(\mathbb{Z}/8)$ of order 96 and a quotient that is the $(2,3,8)$-orbifold. Furthermore, this curve is tiled by 12 octagons glued together such that 3 meet at each vertex. I want to know how to express points on these octagons in terms of $X,Y,Z$, in particular centers and vertices, which I suspect are Weierstrass points. Could anyone show how this is done? Am I missing something obvious? Thanks very much.
 A: Any Riemann surface mapping "nicely" into a $(2,m,n)$-orbifold will have its vertices, edge centers, and face centers mapping to the critical points of the orbifold; when the map into the orbifold is a quotient by a group action, these will be the points fixed by an automorphism.
An order-$8$ automorphism, which should thus preserve the center of an octagon, can be expressed explicitly on the Fermat quartic as follows:
$$X \mapsto Y, Y \mapsto iX, Z \mapsto Z.$$
Since this is a projective curve, we must solve, simultaneously,
$$X=\alpha Y, Y=i\alpha X, Z=\alpha Z$$
for some $\alpha \in \mathbb{C}$.  First, note that $X=\alpha(i\alpha X)$, so either $\alpha^2=-i$ or $X=Y=0$, and the latter would not lie in the quartic, so WLOG we can say $\alpha=(1-i)/\sqrt{2}$, and so the relevant point is
$$\big[1:\frac{1+i}{\sqrt{2}}:0 \big]$$
and the other octagon centers can be described as the image of that point under the symmetry group, which turns out to be the group generated by permuting coordinates and multiplying any of them by fourth roots of unity.
I leave vertices and edge centers as an exercise.
