4
$\begingroup$

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.

enter image description here

$\endgroup$

1 Answer 1

0
$\begingroup$

I noticed that instead of considering the cone with 8 edges, I can just use the area formed by the edges, because they are homeomorphism.
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.

I painted the inside (bottom side) by glay and the outside (top side) by light blue.
The order to patch them is:
$P_1\to P_7\to P_5\to P_3$ 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 $Q_3,R_3,Q_7,R_7$,
in the direction from top to bottom.

At first I couldn't image that it can be realized in 3-dimensional space so I made a model by paper. I observed it is actually a surface with genus 3.

enter image description here

enter image description here

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .