# Klein's quartic

1. Jan 15, 2016

### Lapidus

Can someone tell me how I have to glue together the hyperbolic plane
so that I get the surface (Klein's quartic) shown to the left. I found this picture on the net, but without a desription how to glue.
thanks!

2. Jan 15, 2016

### mathwonk

this is a fun question. i don't know the answer but it is fun staring at the pictures. since the klein quartic is a surface of genus 3, one must mnake a minimum of 6 loop cuts to render it topologically a disc as in th picture. but there are 14 edges to the disc in the picture. this has something to do with th symmetries of the klein quartic which has symmetries of orders 2,3, 7, so the decomposition is done in a way that respects these symmetries. e.g. rotation of order 7 about the center gives one of the automorphisms of the quartic, but i don't quite see how to glue it back together.

3. Jan 15, 2016

### Staff: Mentor

Oh brother. Beyond my pay grade - so John Baez on understanding part of this problem
http://www.tony5m17h.net/JBGEKQ.html

Neat movie.

4. Jan 15, 2016

### mathwonk

this is a nice link with a description of how to glue, but it puzzles me since again making more than 6 loops cuts would separate the disc, so somehow those cuts are not all loop cuts?

5. Jan 21, 2016

### zinq

Since the genus of a g-holed torus is g, the Klein quartic has genus 3, and hence Euler characteristic 2 - 2g = -4.

After identifying the edges in pairs, there will be 1 2-cell, 7 edges, and say V vertices.

The alternating sum 1 - 7 + V is the Euler characteristic, so we have

1 - 7 + V = -4​

and so

V = 2.​

The reason that the number of cuts seemed too many is that there will be 2 vertices after identification, not just 1.

6. Jan 22, 2016

### mathwonk

equivalently the last cut along the edge joining the two distinct vertices is not a loop cut.

7. Jan 22, 2016

### zinq

In fact, all 7 "cuts" are between the two vertices. Though it is better to call them edges.

Last edited: Jan 22, 2016