## Re: Klein's Quartic Equation

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?

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.

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 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

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