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Octonion Mirrorhouse

 Onar Aam noticed that: 2 facing mirrors can represent the complex numbers, with 1 imaginary i which has 1 algebraic generator i and 1 dimension (the 2 mirrors (in cross-section) are parallel and make a 1-dim line). 3 facing mirrors can represent the quaternions, with 3 imaginaries i,j,k which have 2 algebraic generators i,j (since k = ij)and 2 dimensions (the 3 mirrors (in cross-section) form a 2-dim triangle). What about OCTONIONS, with their 480 different multiplications? Here are a couple of conjectures: --------------------------------Conjecture 1:------------Octonions have 7 imaginaries i,j,k,E,I,J,K which have 3 algebraic generators i,j,EThe third generator E is distinguished - we will vary it to get the 480 multiplications.  Start with a 3-dimensional tetrahedron of 4 facing mirrors. Let the floor be the distinguished third generator E and the 3 walls be I,J,K (with a specific assignment) Then, by reflection through the E floor, the reflected I J K become i j k and we now have all 7 imaginary octonions, and 1 of the 480 different multiplications.  To get another, truncate the tetrahedron. Truncation puts a mirror parallel to the floor, making a mirror roof.  Then, when you look up at the mirror roof, you see the triangle roof parallel to the floor E. The triangle roof parallel to the floor E represents the octonion -E, and reflection in the roof -E gives 7 imaginary octonions with the multiplication rule in which -E is the distinguished third generator.  Looking up from the floor, you will also see 3 new triangles having a common side with the triangle roof -E, and6 new triangles having a common vertex with the triangle roof -E.  The triangle roof + 9 triangles = 10 triangles form half of the faces (one hemisphere) of a 20-face icosahedron.  The icosahedron is only qualitatively an icosahedron, and is not exact, since the internal angle of the pentagonal vertex figure of the reflected icosahedron is not 108 degrees, but is 109.47 degrees (the octahedral dihedral angle), and the vertex angle is not 72 degrees, but is 70.53 degrees (the tetrahedral dihedral angle).  For an exact icosahedral kaleidoscope, three of the triangles of the tetrahedron should be Golden isosceles triangles. The inexactness may be related to the aperiodicity of Penrose-Wang tilings of the plane. Each of the 9 new triangles is a "reflection roof" defining another multiplication.  Now, look down at the floor E to see 9 new triangles reflected from the 9 triangles adjoining the roof -E.  Each of these 9 new triangles is a "reflection floor" defining another multiplication.  We have now 1 + 1 + 9 + 9 = 20 of the 480 multiplications.  Just as we put a roof parallel to the floor E by truncating the top of the tetrahedral pyramid, we can put in 3 walls parallel to each of the 3 walls I, J, K by truncating the other 3 points of the tetrahedron, thus getting 3x20 = 60 more multiplications. That gives us 20 + 60 = 80 of the 480 multiplications.  To get the rest, recall that we fixed the walls I, J, K in a particular order with respcet to the floor E.  There are 3! = 6 permutations of the walls I, J, K Taking them into account, we get all 6x80 = 480 multiplications.  This approach effectively fixes the 20-face icosahedron and varies the 4 faces of the EIJK tetrahedron according to the 24-element binary tetrahedral group {3,3,2} = SL(2,3) to get the 20x24 = 480 multiplications.  The truncated tetrahedron with an icosahedron at each vertex combines two types of symmetries:  tetrahedral, related to the square and the ratio sqrt(2) and the continued fraction 1 + sqrt(2) = 2 + 1/ 2 + 1/ 2 + 1/ 2 + 1/ ... which gives open systems like: an arithmetic series overtone acoustic musical scale with common difference 1/8; the Roman Sacred Cut; and multilayer space-filling cuboctahedral crystal growth.  and  icosahedral, related to the pentagon, the Golden Mean, Fibonacci sequences, and the continued fraction PHI = 1 + 1/ 1 + 1/ 1 + 1/ 1 + 1/ ... which gives closed systems like: a harmonic pentatonic musical scale; Le Corbusier's Modulor; and single-layer icosahedral crystals.  Kent Palmer has pointed out that the binary icosahedral group is isomorphic to the binary symmetry group of the 4-simplex, which he calls the pentahedron and which David Finkelstein and Ernesto Rodriguez have called the Quantum Pentacle.  The pentahedron projected into 3 dimensions looks like a tetrahedron divided into 4 quarter-tetrahedra.  If you add a quarter-tetrahedron to each truncation of a truncated tetrahedron, you get a space-filling polytope that can be centered on a vertex of a 3-dimensional diamond packing
to form a Dirichlet domain of the 3-dimensional diamond packing.  A Dirichlet domain of a vertex in a packing is the set of points in the space in which the packing is embedded that are nearer to the given vertex than to any other.  The 4 most distant vertices of the Dirichlet domain polytope are vertices of the dual diamond packing in 3-dimensional space. Therefore, the OCTONION MIRRORHOUSE shows a TETRAHEDRAL DIAMOND packing network of ICOSAHEDRA, or, equivalently, of PENTAHEDRA.  A pentahedron has 5 vertices, 10 edges, 10 areas, and 5 cells. The 10 areas of a pentahedron correspond to the 10 area faces of one hemisphere of an icosahedron.  Pentahedra could be used to define the 4-dimensional spacetime lattice for the 4-dimensional HyperDiamond Feynman Checkerboard version of the D4-D5-E6 physics model. The closed nature of the Golden Mean pentahedral symmetry corresponds to the role a spacetime as a passive fixed stage.  The symmetries of the tetrahedra that were truncated might then describe the internal symmetries of the 4-dimensional HyperDiamond Feynman Checkerboard. The open nature of the sqrt(2) tetrahedral symmetry corresponds to the role of particles and fields as active players.  --------------------------------Conjecture 2:------------The imaginary octonions i j k E I J K can be considered as an associative (quaternion) triangle ijk plus a coassociative square EIJK As in Conjecture 1, let EIJK be the faces of a tetrahedron.  Then truncate the tetrahedron as in Conjecture 1.  Reflections then produce an icosahedron for each of the 4 faces E I J K  The binary icosahedral symmetry group {5,3,2} = SL(2,5) of each icosahedron has 120 elements. The 480 multiplications are the 4x120 elements of the 4 binary symmetry groups of the 4 icosahedra.  Each of the 4 binary icosahedral symmetry groups can be represented as the faces of a 120-cell in 4-real-dimensional space.  A pair of the 4 binary icosahedral symmetry groups can be represented as the 240 faces of a Witting polytope in 4-complex-dimensional space.  All 4 of the 4 binary icosahedral symmetry groups can be represented as the 480 faces of a Witting polytope truncated at each of its 240 vertices by its self-dual polytope in 4-complex-dimensional space.  This approach effectively fixes the 4 faces of the EIJK tetrahedronand varies the icosahedron (or pentahedron) according to the 120-element binary icosahedral group {5,3,2}=SL(2,5) to get the 4x120 = 480 multiplications.  Since the Witting polytope lives in 4 complex dimensions, its lines are complex, and therefore contain complex phase information that is useful in implementing a Many-Worlds sum-over-histories quantum theory. --------------------------------  THE OCTONION MIRRORHOUSE and the 3-dim DIAMOND PACKING:  -------------------------------- DIAMOND PACKING FROM THE ICOSAHEDRAL POINT OF VIEW: The reflection pattern of octonion truncated tetrahedra has icosahedra at the truncations. If you truncate the icosahedra, they do not fill space, but they do fit together to make a 3-dimensional diamond packing:
There are two distinct dual 3-dimensional diamond packings.  Their sum is a bcc lattice, denoted D*3 because it is the dual of the D3 lattice. Note that in 4 dimensions, D*4 = D4, but that in 3 dimensions,D*3 is a bcc lattice and D3 is an fcc lattice, and two fcc lattices make a diamond packing.  There is a 3-dimensional representation of a D*3 bcc lattice by the centers of equal numbers of space-filling cuboctahedra and octahedra:
The D*3 bcc lattice is the sum of two cubic Z3 lattices, as can be seen by letting the cuboctahedra represent one Z3 and the octahedra represent the other Z3. The cuboctahedra can be transformed into octahedra, and vice versa, by the Fuller jitterbug transformation or by the Fuller tensegrity transformation The tensegrity transformation uses 6 struts, 3 pairs of struts, each pair on the X, Y, Z axes:the octahedron, each pair together making one axis; the icosahedron, each pair making a golden rectangle; and the cuboctahedron, each pair making a square.  The intermediate stage in both transformations is the icosahedron. As can be seen from the figure, the triangular faces of the intermediate icosahedron are twisted with respect to the triangular faces of the octahedron and cubocatahedron by a golden ratio twist.  Here is another way to visualize the Golden twist: Again, consider the face/base shared by the small tetrahedron in the space-filling pattern and the icosahedron in the reflection pattern.  Now, consider the other 3 faces of the small tetrahedron. They are on the far half of the icosahedron, you do not see them in the first-order reflections, only in the second-order reflections.  Each of the other 3 faces of the small space-filling tetrahedron should (if you were looking directly at them) produce an icosahedron. These 3 icosahedra are not the same as the icosahedron you see directly in the reflection pattern, but are twisted by a Golden ratio.  To see this, try to INSCRIBE a single icosahedron inside the small tetrahedron of the space-filling pattern.  You can do this, but the 4 triangular faces of the icosahedron that lie within the 4 faces of the small tetrahedron are twisted:  A /\ / \ / \ X / \ / p \ / \Z / r \ / q \ B/________________\C Y  If ABC is the triangle face of the small tetrahedron, and if X, Y, and Z divide AB, BC, and CA in the Golden ratio, then, if p is the intersection of XC and AYq is the intersection of AY and BZr is the intersection of BZ and XC  the triangle face of the inscribed icosahedron is pqr . So the second-order icosahedra are twisted with respect to the original icosahedron by a Golden ratio.  By taking opposite Golden ratios, (where BX is the short length, etc), you will get another oppositely oriented inscribed icosahedron corresponding to the opposite mirror image space-filling pattern.  Therefore, you get 2 different icosahedra for each of the 3 other faces of the small tetrahedron of the space-filling pattern. These 6 correspond to the other 6 E8 lattices that are distinct from each other and from the E8 lattice corresponding to the original icosahedron. This accounts for all 7 distinct E8 lattices.  The Golden twist is a manifestation of octonion non-associativity.  -------------------------------- DIAMOND PACKING FROM THE TETRAHEDRAL POINT OF VIEW: The 3-dimensional diamond packing is the space-filling pattern of truncated tetrahedra PLUS small tetrahedra at each truncation, with each small tetrahedron made up of 4 little tetrahedra.  There are two distinct dual 3-dimensional diamond packings.  The oppositely oriented dual mirror image packing of truncated tetrahedra is made up of oppositely oriented dual truncated tetrahedra.  A truncated tetrahedron is a representation of the Schlafli double-6  There is a tensegrity transformation from the tetrahedron to the truncated tetrahedron to the icosahedron:  -------------------------------- FULLER TENSEGRITY TRANSFORMATIONS IN 3-DIM are 3-DIM PROJECTIONS of transformations in 4-DIM and 8-DIM  To see this, compare the structures described in this section with those used to construct the HyperDiamond Feynman Checkerboard.  Particularly, notice that: Fuller's Vector Equilibrium, the cuboctahedron, has 4 axes that are 3-dim projections of the 4-dim coordinate axes of the 4-dim 24-celland has many interesting 3-dim transformations that are inherited from the 24-cell. The 24-cell can be projected into 3-dim to 2 octahedra plus a central cuboctahedron, (a cuboctahedron being made up of 6 half-octahedra plus 8 tetrahedra, just as a large tetrahedron is made up of 4 small tetrahedra plus an octahedron); and an icosahedron can be made from Golden Ratio points on the edges of an octahedron, while a 120-vertex 600-cell (4-dim hyper-icosahedron) can be made from vertices and Golden Ratio points on the edges of a 24-cell. The 600-cell
is dual to the 120-cell
 
-------------------------------- References (including some illustration sources) for this page include:   Some very nice graphics of high-dimensional geometry, including semi-regular polytopes derived from the 120-cell and the 600-cell, are on the Kyoto WWW page http://kojigen.gaia.h.kyoto-u.ac.jp/homepage_e.html Professor Koji Miyazaki at Kyoto has done much work on golden isozonahedra and related geometric structures.   Coxeter - Complex Regular Polytopes, 2nd ed - Cambridge 1991; Fuller - Synergetics - Macmillan 1975; Holden - Shapes, Space, and Symmetry - Dover 1971; and Kappraff - Connections - McGraw-Hill 1991.   


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