## Klein's Quartic Equation

There was an interesting discussion on this newsgroup last year covering
many aspects of Klein's quartic curve. At the time, John Baez mentioned
in his "This Week's Finds" posting that the curve is the solution to the
quartic equation:

u^3 v + v^3 w + w^3 u = 0

but I couldn't quite see how to derive the topology from this. I made
some animations of the surface embedded in R^3 in various ways, but they
all started from the definition of the topology via a 14-gon in the
hyperbolic plane with some specified edge identifications.

Anyway, last weekend I sat down and really pondered the quartic equation,
and figured out a reasonably painless way to use it to get a handle on
the topology. The details (with lots of pictures) can be found at:

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

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 Greg Egan wrote: [..] > > Anyway, last weekend I sat down and really pondered the quartic equation, > and figured out a reasonably painless way to use it to get a handle on > the topology. After I read your article, I began to understand that Klein's quartic equation really does describe a 2D manifold. This was not obvious to me before, because it contains 3 complex variables (x,y,z), which can be viewed as 6 real variables. The equation itself kills off one complex variable (or 2 real ones). But now I see the "projective geometry" (dividing away by an uninteresting complex variable) kills off 2 more real variables, leaving only 2 degrees of freedom, which can be laid out on a 2D manifold! I tried using your method on some other equations, such as the simple ones: x + y + z = 0 x^2 + y^2 + z^2 = 0 Or more generally, the Fermat curves: x^p + y^p + z^p = 0 These curves should produce some nice manifolds. Your method seems to work, but I haven't got the answers yet. 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