# Explicit triangulation of a genus 3 surface $X^4+Y^4+Z^4=0$

I tried to describe the Riemann surface $$X^4+Y^4+Z^4=0$$ as a surface in real 3-dimensional space (without self-intersection) with triangulation.

I picked $$12$$ points on the surface: $$P_{a} = [0:1:\zeta^a], Q_{a}= [\zeta^{a}:0:1], R_{a} = [1:\zeta^a:0]$$, where $$a \in \{1,3,5,7\}, \zeta = (1+i)/\sqrt2$$.

I join $$P_a$$ $$Q_b$$ by the parametlized curve: $$[X:Y:Z] = [ζ^b t: ζ^{-a} \sqrt[4] {1-t^4}:1]$$. Similarly I join $$Q_aR_B$$ and $$R_aP_b$$.

Then I found following triangles are contractible:
$$\triangle P_1Q_1R_5, \triangle P_1Q_1R_7, \triangle P_1Q_3R_3, \triangle P_1Q_3R_5, \triangle P_1Q_5R_1, \triangle P_1Q_5R_3, \triangle P_1Q_7R_1, \triangle P_1Q_7R_7$$
$$\triangle P_3Q_1R_3, \triangle P_3Q_1R_5, \triangle P_3Q_3R_1, \triangle P_3Q_3R_3, \triangle P_3Q_5R_1, \triangle P_3Q_5R_7, \triangle P_3Q_7R_5, \triangle P_3Q_7R_7$$
$$\triangle P_5Q_1R_1, \triangle P_5Q_1R_3, \triangle P_5Q_3R_1, \triangle P_5Q_3R_7, \triangle P_5Q_5R_5, \triangle P_5Q_5R_7, \triangle P_5Q_7R_3, \triangle P_5Q_7R_5$$
$$\triangle P_7Q_1R_1, \triangle P_7Q_1R_7, \triangle P_7Q_3R_5, \triangle P_7Q_3R_7, \triangle P_7Q_5R_3, \triangle P_7Q_5R_5, \triangle P_7Q_7R_1, \triangle P_7Q_7R_3$$
(I found this by integrating $$3$$ holomorphic differentials which I previously described here)

Thus I have $$12$$ points and $$48$$ edges, and $$32$$ triangles on the surface. (The Euler characteristic shows that this is a genus 3 surface.)

So I tried to choose $$12$$ points on a sphere with $$3$$ handles, and draw $$48$$ edges ($$8$$ edges for each point) to triangulate, satisfying the relations above. But it was really difficult and I couldn't succeed. Is this actually possible (and can you draw it)?

What I tried else: I considered 4 cones with:
Apex:$$P_1$$, 8-edged bottom $$Q_1R_5Q_3R_3Q_5R_1Q_7R_7Q_1$$
Apex:$$P_3$$, 8-edged bottom $$Q_1R_3Q_3R_1Q_5R_7Q_7R_5Q_1$$
Apex:P_5, 8-edged bottom Q_1R_1Q_7R_5Q_5R_7Q_3R_7Q_1
Apex:$$P_5$$, 8-edged bottom $$Q_1R_1Q_3R_7Q_5R_5Q_7R_3Q_1$$
Apex:P_7, 8-edged bottom Q_1R_1Q_3R_5Q_5R_3Q_7R_3Q_1
Apex:$$P_7$$, 8-edged bottom $$Q_1R_1Q_7R_3Q_5R_5Q_3R_7Q_1$$
Then patch together with the blue edges or pink edges respectively. But it seems to me that it is not possible in 3-dimensional space without self-intersection (but I am not sure).

EDIT I tried again with more symmetric shape and succeeded so I have posted it as an answer.

I used more symmetric 4 shapes each has $$8$$ points and $$8$$ edges with the relation described in the question. The shapes are shown in the attached pictures.
$$P_1\to P_7\to P_5\to P_3$$ in the upper area containing $$Q_1,R_1,Q_5,R_5$$,
$$P_5\to P_3\to P_1\to P_7$$ in the lower area containing $$Q_3,R_3,Q_7,R_7$$,