On Jul 16, 2015, at 3:02 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Veit Elser <ve10@cornell.edu> [Jul 16. 2015 08:15]:
There’s something to be said for having conventions, but you also lose sight of interesting things when you are rigid in your ways.
The Galois automorphism sqrt(5) <-> -sqrt(5) has geometrical consequences. Perhaps you’ve encountered the construction of the 2D Penrose tiling starting from a lattice in 4D? Or its counterpart in 3D derived from a lattice in 6D? Basic to those constructions is an orthogonal decomposition of the 4D or 6D space by a pair of irrational subspaces. The automorphism has the effect of swapping those subspaces. So in addition to those obvious symmetries of the tiling, e.g. the pentagon (or icosahedron), there is the less obvious symmetry associated with changing the sign of sqrt(5). I’m actually surprised the term “Penrose involution” doesn’t bring up any hits.
Could you point me to any reference regarding this (especially 6D-->3D)?
The book by Marjorie Senechal is pretty good: http://www.amazon.com/Quasicrystals-Geometry-Marjorie-Senechal/dp/0521575419 But it might be quicker if you follow these steps: 1) Label half the vertices of the regular icosahedron 1,…,6 in some arbitrary way, just so no two labels are opposite the origin. 2) Construct a 6x6 “golden matrix” G by these rules: a) all the elements are -1/2 or +/12 b) diagonal elements are + c) off-diagonal elements are + if the corresponding icosahedron vertices form an acute angle at the origin, - otherwise 3) Check that G^2 = G+1. 4) Check that det G = -1. Together with 3) this implies that G has eigenvalues {tau,tau,tau,-1/tau,-1/tau,-1/tau}. You may therefore write G = tau P + (-/tau) P’, where P and P’ are projections into orthogonal 3-spaces. 5) Using P+P’=1 and the G you constructed in 2), use G = tau P + (-/tau) P’ to solve for P. 6) Now you can project things from 6D into 3D using P. The “checkerboard lattice" in 6D (points with integer coordinates that have even sum), called D_6, is a natural place to start because G acts as a bijection on D_6. Example: applying P to the 12 points {[2,0,0,0,0,0], … , [0,0,0,0,0,-2]} gives you the icosahedron. You can also try things like selecting points of D_6 that are close to the origin of one of the 3-spaces, by using P’, and project those into the other 3-space using P. That’ll give you something like a 3D Penrose tiling. -Veit
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