Re: Klein's Quartic Equation

Greg Egan

In article <44D7B4B7.2050904@xs4all.nl>, Gerard Westendorp
<westy31@xs4all.nl> wrote:

[snip]

> Before struggling on, I thought maybe someone might give
> some useful hints on this. After all, there are so many
> pictures of Platonic solids on the web, and so few of Platonic
> tilings of hyperbolic surfaces!
>
> Gerard

Have you seen Don Hatch's beautiful page of tesselations of the
hyperbolic plane?

http://www.hadron.org/~hatch/HyperbolicTesselations/

BTW, I've added a new movie to my page on KQE:

http://gregegan.customer.netspace.ne.../KleinQuarticE
q.html

which shows the curves you get by rotating a certain
three-real-dimensional surface through C^3 and plotting the intersection
with Klein's quartic curve.

Gerard Westendorp

Greg Egan wrote:


[..]

> Have you seen Don Hatch's beautiful page of tesselations of the
> hyperbolic plane?

Yes, its a cool page. But I still want to understand a bit better the
relation with algebraic curves.
I made some progress her myself.

For example, the surface associated with

u + v + w = 0

Is the Riemann sphere!

To see this, we can write the solution as:

(u,v,w) = (1 , z , -(1+z)) /R

Where R is a normalization factor. I will leave out the normalization
factor in the remaining.

So the solution surface is parametrised by the complex number z, it
is locally a copy of the complex plane. But when z gets very large,
for example, lets fill in z=1000, we get:

(u,v,w)= (1,1000,-999) ~ (0,1,-1).


It is not so hard to see that at |z| -> inf, all points on C map
to the single point (0,1,-1). This is just the Riemann sphere, which
has genus 0. It is also possible to check the genus by looking at
the Euler characteristic of a tiling. A tiling on a bounded sheet of
paper has Euler characteristic 1. Around the boundary, there is a polygon
of N vertices and N edges. As we fuse the entire boundary into a single
vertex, we will change the Euler characteristic with N-N+1, giving 2.
This means genus 0.

Now we can also figure out the genus of any Fermat curve:

u^p + v^p + w^p = 0

Write the solution as:

(u,v,w) = (1 , z , [-(1+z^p]^(1/p)) /R

Once again we have a surface parametrized by a complex number(z).
But because of the p-th root, we get p copies of Riemann spheres
instead of just 1. A special case occurs when w=0, then all
p Riemann spheres share a single point, joining them together.
There are p such points, corresponding to the z^p=-1

If we put tilling on the p Riemann spheres, they will have
Euler characteristic 2p. Then we glue them together at the p
points. At each point, we turn p vertices into 1 vertex, reducing
the Euler characteristic by p-1.

So the Euler characteristic will be

chi = 2p-p(p-1) = 3p-p^2

genus = 1-chi/2 = (p-1)(p-2)/2

Here is a table of the first 5:

p genus
1 0
2 0
3 1
4 3
5 6

So Fermat's quartic has the same genus as Klein's quartic.

Gerard