^{3}. H

^{3}allows an infinite number of tessellations. These consideration were inspired by the model for the recently discovered gravitational hum but they also led to a considerable progress in the understanding of the TGD inspired model of the genetic code. Here only the icosatetrahedral tessellation is discussed.

**1. Some preliminaries**

Some preliminaries are needed in order to understand Wikipedia articles related to tessellations in general.

- Schläfli symbol {p,r} (see this) tells that the possibly existing Platonic solid {p,r} has r p-polygons as faces meeting at each vertex. For instance, icosahedron {3,5} has 5 triangles as faces meeting at each vertex.
Schläfli symbol generalizes to higher dimensions. The analog of Platonic solid {p,r,q} possibly in 4-dimensions and assignable to 3-sphere has q 3-faces which are Platonic solids {p,r}. This description is purely combinatorial and is recursive. For instance, one can start from 3-D dimensional Platonic solid {p,q} with 3-D objects in dimension 4 by replacing p with {p,r}. One can also project this object to dimension 3. In this manner one obtains a projection of 4-cube (tesseract) {4,3,3} for which 3 cubes {4,3} meet at each vertex (2

^{4}=16 of them) and which has 8 3-cubes as faces as a 3-D object.In the case of hyperbolic tessellations also strange looking Schläfli symbols {(p,q,r,s)} are encountered: icosa-tetrahedral tessellation involving only Platonic solids has symbol {(3,3,5,3)}. My understanding is that this object corresponds to {3,3,5,3} as an analogue of Platonic solid associate with 4-sphere in 5-D Euclidian space and that the fundamental region of this tessellation in H

^{3}is analogous to a 3-D projection of this object. At a given vertex 3 objects {3,3,5} meet. For these objects 5 tetrahedrons meet at a given vertex. - Vertex figure is a further central notion. It represents a view of the fundamental region of tessellation from a given vertex. The vertices of the figure are connected to this vertex. It does not represent the entire fundamental region. For instance, for a cube (octahedron) it contains only the 3 (4) nearest vertices. For icosa-tetrahedral tessellation the vertex figure is icosidodecahedron (see this). The interpretation of the vertex symbol of the hyperbolic icosa-tetrahedral honeycomb (see this) is a considerable challenge.
- One cannot avoid Coxeter groups and Coxeter symbols (see this) in the context of tessellations. They code the structure of the symmetry group of say Platonic solid (tessellation of S
^{2}). This symmetry group is generated by reflections with respect to some set of lines, usually going through origin. For regular polygons and Platonic solids is its discrete subgroup of rotation group.The Coxeter group is characterized by the number of reflection hyperplanes H

_{i}and the reflections satisfying r_{i}^{2}=1. The products r_{ij}=r_{i}r_{j}define cyclic subgroups of order c_{ij}satisfying r_{ij}^{cij}=1. Coxeter group is characterized by a diagram in which vertices are labelled by i. The orders of the cyclic subgroups satisfy c_{ij}≥ 3. For c_{ij}the generators r_{i}and r_{j}commute. For c_{ij}=2 the vertices are not connected, for c_{ij}=3 there is a line and for c_{ij}>3 the number c_{ij}is assigned with the line. For instance, hyperbolic tessellations are characterized by 4 reflection hyperplanes.For instance, for p-polygon the Coxeter group has 2 generators and the cyclic group has order p. For Platonic solids the Coxter group has 3 generators and the orders of cyclic subgroups are 3, 4, or 5. For icosa-tetrahedral tessellation the order is 4.

**The most interesting honeycombs in hyperbolic 3-space**

H^{3} allows an infinite number of tessellations. There are 9 types of honeycombs. This makes 76 uniform hyperbolic honeycombs involving only a single polyhedron (see this).

4 of these honeycombs are * regular*, which means that they have identical regular faces (Platonic solids) and the same numbers of faces around vertices.
The following list gives the regular uniform honeycombs and their Schläfli symbols {p,q,r} telling that each edge has around it regular polygon {p,q} for which each vertex is surrounded by q faces with p vertices.

- H1: 2 regular forms with Schläfli symbol {5,3,4} (dodecahedron) and {4,3,5} (cube).
- H2: 1 regular form with Schläfli symbol {3,5,3}(icosahedron)
- H5: 1 regular form with Schläfli symbol {5,3,5} (dodecahedron).

From the Wikipedia article about icosa-tetrahedral honeycomb (see this) one learns the following.

- The Schläfli symbol of icosa-tetrahedral honeycomb is {(3,3,5,3)}. This combinatorial symbol allows several geometric representations. The inner brackets would refer to the interpretation as an analogue of the Platonic solid assignable to a 4-sphere of Euclidian 5-space. At each vertex 3 objects of type {3,3,5} would meet. At the vertex of {(3,3,5)} in turn 5 tetrahedrons meet.
- Icosa-tetrahedral honeycomb involves tetrahedron {(3,3}, octahedron {(3,4)}, an icosahedron {(3,5)} as cells. That there are no other honeycombs involving several Platonic solids and only them as cells makes this particular honeycomb especially interesting. Octahedron with Schläfli symbol {3,4} can be also regarded as a rectified tetrahedron havig Schläfli symbol r{3,3}.
- The vertex figure of icosa-tetrahedral honeycomb (see this), representing the vertices a lines connecting them is icosidodecahedron (see this), which is a "fusion" of icosahedron and dodecahedron having 30 vertices with 2 pentagons and 2 triangles meeting at each, and 60 identical edges, each separating a triangle from pentagon. From a given vertex VF=60 vertices connected to this vertex by an edge can be seen. In the case of cube, octahedron, and dodecahedron the total number of vertices in the polyhedron is 2(VF+1). It is true also now, one would have 122 vertices in the basic structural unit. The total number of vertices for the disjoint polyhedra is 6+4+12= 22 and since vertices are shared, the number of polyhedra in the basic unit must be rather large.
- The numbers called "cells by location" could correspond to numbers 30, 20, and 12 for octahedrons, tetrahedrons and icosahedrons respectively inside the fundamental region of the tessellation defining the honeycomb. That the number of icosahedrons is smallest, looks natural. These numbers are quite large. The counts around each vertex are given by (3.3.3.3), (3.3.3),
*resp.*(3.3.3.3) for octahedra, tetrahedra,*resp.*icosahedra and tell the numbers of vertices of the faces meeting at a given vertex.

- The Schläfli symbol of icosa-tetrahedral honeycomb is {(3,3,5,3)}. This combinatorial symbol allows several geometric representations. The inner brackets would refer to the interpretation as an analogue of the Platonic solid assignable to a 4-sphere of Euclidian 5-space. At each vertex 3 objects of type {3,3,5} would meet. At the vertex of {(3,3,5)} in turn 5 tetrahedrons meet.
- Icosa-tetrahedral honeycomb involves tetrahedron {(3,3}, octahedron {(3,4)}, an icosahedron {(3,5)} as cells. That there are no other honeycombs involving several Platonic solids and only them as cells makes this particular honeycomb especially interesting. Octahedron with Schläfli symbol {3,4} can be also regarded as a rectified tetrahedron havig Schläfli symbol r{3,3}.
- The vertex figure of icosa-tetrahedral honeycomb (see this), representing the vertices a lines connecting them is icosidodecahedron (see this), which is a "fusion" of icosahedron and dodecahedron having 30 vertices with 2 pentagons and 2 triangles meeting at each, and 60 identical edges, each separating a triangle from pentagon. From a given vertex VF=60 vertices connected to this vertex by an edge can be seen. In the case of cube, octahedron, and dodecahedron the total number of vertices in the polyhedron is 2(VF+1). It is true also now, one would have 122 vertices in the basic structural unit. The total number of vertices for the disjoint polyhedra is 6+4+12= 22 and since vertices are shared, the number of polyhedra in the basic unit must be rather large.
- The numbers called "cells by location" could correspond to numbers 30, 20, and 12 for octahedrons, tetrahedrons and icosahedrons respectively inside the fundamental region of the tessellation defining the honeycomb. That the number of icosahedrons is smallest, looks natural. These numbers are quite large. The counts around each vertex are given by (3.3.3.3), (3.3.3),
*resp.*(3.3.3.3) for octahedra, tetrahedra,*resp.*icosahedra and tell the numbers of vertices of the faces meeting at a given vertex. - What looks intriguing is that the numbers 30, 20, and 12 for octahedrons (O), tetrahedrons (T) and icosahedrons (I) correspond to the numbers of vertices, faces, and edges for I. As if the fundamental region would be obtained by taking an icosahedron and replacing its 30 vertices with O, its 20 faces with T and its 12 edges with I, that is by using the rules
*vertex → octahedron*;*edge →I*,*face → T*. These 3-D objects would be fitted together along their triangular faces.Do the statements about the geometry and homology of I translate to the statements about the geometry and homology of the fundamental region? This would mean the following replacements:

- "2 faces meet at edge" → "2 T:s share face with an I".
- "5 faces meet at vertex" → "5 T:s share face with an O".
- "Edge has 2 vertices as ends" → "I shares a face with 2 different O:s".
- "Face has 3 vertices → "T shares a face with 3 different O:s".
- "Face has three edges" → "T has a common face with 3 I:s".

**An attempt to understand the hyperbolic tessellations**

The following general observations might help to gain some understanding of the tessellations.

- The tessellations of E
^{3}and H^{3}are completely analogous to Platonic solis as 2-D objects. The non-compactness implies that there is infinite number of cells for tessellations. It is important to notice that the radial coordinate r for H^{3}corresponds very closely to hyperbolic angle and its values are quantized for the vertices of tessellation just like the values of spherical coordinates are quantized for Platonic solids.The tessellations for E

^{3}are scale covariant. For a fixed radius of H^{3}characterized by Lorentz invariance cosmic time this is not the case. One can however scale the value of a. - What distinguishes between regular tessellations in E
^{3}and H^{3}is that the metric of H^{3}is non-flat an H^{3}has negative curvature. H^{3}is homogeneous space meaning that all points are metrically equivalent (this is the counterpart of cosmological principle in cosmology). Since both spaces have rotations as symmetries, this does not affect basic Platonic solids as 2-D structures assignable with 2-sphere if the edges are identified as geodesic lines of S^{2}. Quite generally, isometries characterize the tessellations, whose fundamental region corresponds to coset space of H^{3}/Γ by a discrete group of the Lorentz group acting as isometries of H^{3}. - The modifications induced by the replacement E
^{3}arrow H^{3}relate to the 3-D aspects of the tessellation. This because the metric is non-flat in the radial direction. The negative curvature implies that the geodesic line diverge. One can use counterpart of the standard spherical coordinates and in this coordinates the solid angles assignable to the vertices of Platonic solid are smaller than in E^{3}. Also the hyperbolic planes H^{2}emerging from edges of the tessellation of H^{3}diverge in normal direction the angles involved are smaller.

- Important constraints come from the topology and combinatorics. Basic equations for the numbers V ,E, and F for number of vertices, edges an faces are purely topological equation V E+F=2, and the equation pF=2E=qV. Manipulation of these equations gives 1/r+1/p= 1/2+1/E implying 1/r+1/p>1/2. Since p and q must be at least 3, the only possibilities for {p,q} are {3, 3}, {4, 3}, {3, 4}, {5, 3}, and {3, 5}.
- The angular positions of the vertices at S
^{2}are basic angle variables. In H^{3}hyperbolic angle assignable to the radial coordinate is additional variable of this kind analogous to the position of the unit cell in the E^{3}tessellation. The cosmological interpretation is in terms of redshift. - There is the Euclidian angle φ associated with the vertex of the face given by π/p. Here there is no difference between E
^{3}and H^{3}. - The angle deficit δ associated with the faces meeting at a given vertex due to the fact that the faces are not in plane in which case the total angle would be 2π. δ is largest for tetrahedron with 3 faces meeting at vertex and therefore with the sharpest vertex and smallest for icosahedron with 5 triangles meeting at vertex. This notion is essentially 3-dimensional, being defined using radial geodesics, so that the δ is not the same in H
^{3}. In H^{3}δ is expected to be larger than in E^{3}. - There is also the dihedral angle θ associated with the faces as planes of E
^{3}meeeting at the edges of the Platonic solid. θ is smallest for tetrahedron with 4 edges and largest for dodecahedron with 20 edges so that the dodecahedron is not far from flat plane an this angle is not far from π. The H^{3}counterpart of θ is associated faces identified as hyperbolic planes H^{2}an is therefore different. - There is also the vertex solid angle Ω associated with each vertex of the Platonic solid {p,q} given by Ω =qθ -(q-2)π. For tessellations in E
^{3}the sum of these angles is 4π. In H^{3}it Euclidian counterpart is larger than 4π. - The face solid angle is the solid angle associate with the face when seen from the center of the Platonic solid. The sum of the face solid angles is 4π. For Platonic solid with n vertices, one has Ω= 4π/n. The divergence of the geodesics of H
^{3}implies that this angle is smaller in H^{3}: there is more volume in H^{3}than in E^{3}.

^{3}allows only single regular tessellation having cube as a unit cell. H

^{3}allows cubic and icosahedral tessellations plus two tessellations having dodecahedron as a unit. Why E

^{3}does not allow icosahedral an dodecahedral tessellations and how the curvature of H

^{3}makes them possible? Why the purely Platonic tetra-icosahedral tessellation is possible in H

^{3}?

The first guess is that these tessellations are almost but not quite possible in E^{3} by looking the Euclidian constraints on various angles. In particular, the sum of dihedral angles θ between faces should be 2π in E^{3}, the sum of the vertex solid angles Ω at the vertex should be 4π. Note that the scaling of the radial coordinate r decreases the dihedral angles θ and solid angles Ω. This flexibility expected to make possible so many tessellations and honeycombs in H^{3}. The larger the deviation of the almost allowed tessellation, the larger the size of the fundamental region for fixed a.

Consider now the constraints on the basic parameters of the Platonic solids (see this) in E^{3} and H^{3}.

- The values of didedral angle for tetrahedron, cube, octahedron, docecahedron, and icosahedron are
[θ (T), θ (C), θ (O), θ(D),θ(I)]≈ [70.3

^{°}, 90^{°}, 109.47^{°}, 116.57^{°}, 138.19^{°}] .Note that r=5 tetrahedra could almost meet at single edge in E

^{3}. In E^{3}r=4 cubes can meet at the edge. In H^{3}r should be larger. This is indeed the case for the cubic honeycomb {4,3,5} having r=5 .For r=3 icosahedrons the the sum dihedral angles exceeds 2π which conforms with the that {3,5,3} defines an icosahedral tessellation in H

^{3}.For r=4 docecahedra meeting at the edge the total dihedral angle is larger than than 360

^{°}: r=4 is therefore a natural candidate in H^{3}. There are indeed regular dodecahedral honeycoms with Schläfli symbol {5,3,r}, r=4 and r=5. Therefore it seems that the intuitive picture is correct. - The values of the vertex solid angle Ω for cube, docecahedron, and icosahedron are given by the formula Ω =qθ -(q-2)π giving
[Ω (C), Ω (D), Ω (I)] ≈ [1.57080, 2.96174,2.63455].

The sum of these angles should be 4π for a tessellation in E

^{3}. In E^{3}This is true only for 8 cubes per vertex (Ω= π/2) so that the cubic honeycomb is the only Platonic honeycomb in E^{3}. The minimal number of cubes per vertex is 9 in H^{3}. It is convenient to write the values of the vertex solid angles for D and I as[Ω(D),Ω(I)]= [0.108174, 0.209651]× 4π .

The number of D:s

*resp.*I:s must be at least 10*resp.*5 for dodecahedral*resp.*icosahedral honeycombs in H^{3}. - The basic geometric scales of the Platonic solids are circumradius R, surface area A and volume V. The circumradius is given by
R=(a/2)×tan(π/q)tan(θ/2), where a denotes the edge length. The surface area A of the Platonic solid { p,q} equals the area of face multiplied by the number F of faces: A=(a/2)
^{2}Fpcot(π/p). The volume V of the Platonic time is F times the volume of the pyramid whose height is the length a of the face: that is V= FaA/3.Choosing a/2 as the length unit, the circumradii R, total face areas A an the volumes V of the Platonic solids are given by

[R(T),R(C),R(O),R(D),R(I)]= [3

^{1/2}/2, 3^{1/2},2^{1/2}, 3^{1/2}φ, (3-φ)^{1/2}φ] ,[A(T),A(C),A(O),A(D),A(I)]= [ 4×3

^{1/2}, 24, 2×3^{1/2}, 12×(25+10×5^{1/2})^{1/2}, 20× 3^{1/2}] ,and

[V(T),V(C),V(O),V(D),V(I)]≈ [8

^{1/2}/3, 8, 128^{1/2}/3 , 20φ^{3}/(3-φ), 20φ^{2}/3]≈ [.942809,8,3.771236,61.304952,17.453560] .

- Consider first the dihedral angles θ. The values of dihedral angles associated T, O, and I in H
^{3}is reduced from that in E^{3}so that their sum in E^{2}sene must be larger than 2π. Therefore at least one of these cells must appear twice in H^{3}. It could be T but also O can be considered.For 2T+O+I and T+2O+I the sum would be 388.26

^{°}*resp.*427.43^{°}in E^{3}. 2T+O+I*resp.*T+2O+I could correspond 4 cells ordered cyclicaly as ITOT*resp.*IOTO. - The values of the vertex solid angle Ω for tetrahedron, octahedron, and icosahedron are given by
[Ω(T), Ω(O),Ω(I)] =[0.043870, 0.108174, 0.209651]4π
If the numbers of T, O and I are [n(T),n(O),n(I)], one must have [n(T)Ω(T), +n(O)Ω(O) +n(I)Ω(I)>4π in H
^{3}.If the number of the cells for the fundamental domain are really [N(T), N(O), N(I)]=[30,20,12], the first guess is that [n(T), n(O), n(I)] ∝ [N(T), N(O), N(I) is approximately true. For [n(T), n(O), n(I)]=[2,3,1]n(I), one obtains Ω= n(T)Ω(T)+n(O)Ω (O)+ n(I)Ω (I)= n(I)× .629 × 4π. This would suggest n(I)=2 giving [n(T), n(O), n(I)]=[4,6,2]

For a summary of earlier postings see Latest progress in TGD.

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