The Octagon (CLP-1)

Let’s start with an equation: y²=x⁸-1. Solving for y is easy, because for each x we appear to have just two choices for the sign, good and evil. If we do this in the complex plane, the set of solutions therefore looks like two copies of the x-plane. There is a little problem at the eighth roots of unity, because there, good and evil coalesce. 

 

Octagon 01

A good way to imagine this is to think about the (extended) complex plane as two disks, and of each disk as a regular octagon, with vertices at these eighth roots of unity. Then it takes four such octagons to build the solution space of the equation y²=x⁸-1, and we need to have four octagons at each vertex coming together, alternating between good and evil. Luckily, this can be done in the hyperbolic plane, using a tiling by regular right-angled hexagons.To get an idea how these are glued together, it helps to think about the equation in the form x⁸=y²+1. This represents the same solution space as 16 copies of a single triangle, with vertices at the octagon centers as shown above. Thus the entire solution space can also be obtained by gluing together the edges of the 16-gon above, where the identifications are indicated by the (extended) edges of the central octagon.

Wouldn’t it be nice if we could visualize this in ℝ³? This is indeed possible if we are willing to conformally bend our octagon a little so that every other edge becomes a straight segment, and the other edges lie in planes that meet the octagon orthogonally along that edge.

Octagon clp

This allows to extend the octagon by rotating and reflecting about its edges like above, which shows four such hexagons, i.e. the entire solution space. If you do this right, you get one of the many views of the CLP surface of Hermann Amandus Schwarz. CLP stands for crossed layers of parallels. This is once again a triply periodic minimal surface. Here is another translational fundamental piece that corresponds to the 16-gon:

Clp 0

Let’s begin to rotate through the associate family. For angle π/16, we see how the touching vertices are being separated.

Clp 16

At π/4, we get a nice symmetric piece, but translational copies will intersect so that the surface will not remain embedded.

Clp 4

At π/2 we meet the conjugate surface of the CLP surface. The amusing point here is that it is congruent to the CLP surface, a feature it shares with the Enneper surface and one surface in the family of Riemann’s minimal surfaces.Clp 2

 

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