by wendy » Sat Dec 02, 2017 12:53 pm
The process of 'isomorphism' is where one cycles through the roots of various equations. For the class-2 systems, this usually means replacing sqrt(5) as +2.236067 or -2.236067. A similar scheme exists in all other polygon systems, where one transforms various equation-solutions.
If one takes a graph of rays, say the ray graph of eight radiating lines, the steps in order (0,1,2,3,4,5,6,7), give a regular octagon, while (0,5,2,7,4,1,6,3) gives an octagram. If one supposes that 0 points to 1,0 and 2 to 0,1 (in clock-face style), then 1 is at ½r2, ½ r2 (where r2 is sqrt(2) ). But changing sqrt(2), replaces r2 by -r2, and this swaps 1 and 5, and 3 and 7, while leaving 0,2,4,6 alone. This is equivalent to multiplying the point number by 5, modulo 8. The order (0,1,2,3,4,5,6,7) now gives the octagram, while 0,5,2,7,4,1,6,3 gives an octagon.
A clock-face with the numbers running from 0 to 9, can be used to draw a pentagon (0,2,4,6,8), a pentagram (0,4,8,2,6), a decagon (0,1,2,3,4,5,6,7,8,9) and a decagram (0,3,6,9,2,5,8,1,4,7). If you multiply the coordinates by 3 or 7, pentagons and pentagrams swap, and decagons and decagrams swap. It's a bit more complex than just +sqrt(5) <> -sqrt(5), but this is indeed the heart of the matter.
The thing can then be represented as a kind of dual-projection of something in 2n space, where in one case, an octagon appears, and in the other an octagram appears.
The relative importance of all of this, is that if a figure is finite, it must have finite conjugates. This means that while we could put three decagrams around a corner (as x10/3o3o) the conjugate must also be of the same nature (ie x10o3o as three decagons, does not exist), and so this can not be finite. The first is an infinitely wound polyhedron, and the second is a hyperbolic tiling. Information about both tells us a lot more than about either alone. We can see that x10o3o can not contain x10o6o, because the conjucate of the first is a spherical polyhedron, while the second is a dense hyperbolic tiling. Since the conjugates can not contain each other, nor can the original set.
One should also note that every tiling that solves a finite polynomial, with a unit first term, has solutions that are "algebraic integers", every algebraic integer necessarily divides a real integer. Such is usually the unit term on the equation, or the determinate of the defining matrix.
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the dream we dream together is reality.