About tesselations in hyperbolic 3-space and their relation to the genetic code

The post represents new results about the tessellations of H³. H³ 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.

There are 9 types of honeycombs. This makes 76 uniform hyperbolic honeycombs involving only single polyhedron (see this). 4 of these are regular, which means that the have identifical 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.

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

There is a large number of uniform honeycombs involving several cell types. There exists however a “multicellular” honeycomb, which is completely unique in the sense that for it all cells are Platonic solids. This icosa tetrahedral (or more officially, tetrahedral-icosahedral) tessellation has tetrahedrons, octahedrons, and icosahedrons as its cells. All faces are triangles. The tetrahedral-icosahedral honeycomb is of special interest since it might make possible the proposed icosa tetrahedral realization of the genetic code (see this). Octahedron can be seen also as a rectified tetrahedron r{3,3}.

From the Wikipedia article about tetrahedral-icosahedral honecomb one learns following.

1. Its Schläfli symbol is {(3,3,5,3)}. Unfortunately, there was now explanation for the meaning of the symbol. I manage to find only the definition of Schläfli symbol {p,q,r}. Could the symbol mean that at each edge corresponds to 1 tetrahedral , 1 icosahedral and 1 octahedral cell with Schläfli symbol {3,3} resp. {3,5} resp. r{3,3}={3,4} respectively (note that polygon {p,q} has regular p-sided faces at each vertex (see this). The rectified tetrahedron has Schläfli symbol r{3,3} and is octahedron having Schläfli symbol {3,4}. This interpretation conforms with the fact that the angle defined by the edge is smallest for tetrahedron and largest for icosahedron.
2. The vertex figure of tetrahedral-icosahedral honeycomb (see this), representing the vertices a lines connecting them is icosidodecahedron (see this), is a “fusion” of icocahedron 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, an dodecahedron the total number of vertices in 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 is 6+4+12= 22 and since vertices are shared, the number of polyhedra in the basic unit must be considerably higher than 1+1+1=3.
3. 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 for by (3.3.3.3), (3.3.3) and (3.3.3.3) or octahedra, tetrahedra and icosahedra and tell the numbers of vertices of the faces meeting at a given vertex.

An attempt to understand the hyperbolic tesselations

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

1. The tessellations of E³ and H³ 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 H³ 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 E³ are scale covariant. For a fixed radius of H³ 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 E³ and H³ is that the metric of H³ is non-flat an H³ has negative curvature. H³ 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 S². Quite generally, isometries characterize the tessellations, whose fundamental region corresponds to coset space of H³/Γ by a discrete group of the Lorentz group acting as isometries of H³.

The modifications induced by the replacement E³arrow H³ 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 E³. Also the hyperbolic planes H² emerging from edges of the tessellation of H³ diverge in normal direction the angles involved are smaller. It is useful to start from the description of Platonic solids. They are characterized combinatorially by integers and by geometrically by various kinds of angles. Denote by p the number of vertices/edges of the face and by q the number of faces meeting at vertex.

1. 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}.
2. The angular positions of the vertices at S² are basic angle variables. In H³ hyperbolic angle assignable to the radial coordinate is additional variable of this kind analogous to the position of the unit cell in the E³ tessellation. The cosmological interpretation is in terms of redshift.
3. There is the Euclidian angle φ associated with the vertex of the face given by π/p. Here there is no difference between E³ and H³.
4. 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 H³. In H³ δ is expected to be larger than in E³.
5. There is also the dihedral angle θ associated with the faces as planes of E³ 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 H³ counterpart of θ is associated faces identified as hyperbolic planes H² an is therefore different.
6. 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 tesselations in E³ the sum of these angles is 4π. In H³ it Euclidian counterpart is larger than 4π.
7. 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 H³ implies that this angle is smaller in H³: there is more volume in H³ than in E³.

E³ allows only single regular tesselation having cube as a unit cell. H³ allows cubic and icosahedral tessellations plus two tessellations having dodecahedron as a unit. Why E³ does not allow icosahedral an dodecahedral tessellations and how the curvature of H³ makes them possible? Why the purely Platonic tetra-icosahedral tessellation is possible in H³?

The first guess is that these tesselations are almost but not quite possible in E³ by looking the Euclidian constraints on various angles. In particular, the sum of dihedral angles θ between faces should be 2π in E³, 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 tesselations and honeycombs in H³. The larger the deviation of the almost allowed tesselation, 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 E³ and H³.

1. 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 E³. In E³ r=4 cubes can meet at the edge. In H³ 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 tesselation in H³.

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 H³. 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 tesselation in E³. In E³ This is true only for 8 cubes per vertex (Ω= π/2) so that the cubic honeycomb is the only Platonic honeycomb in E³. The minimal number of cubes per vertex is 9 in H³. 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 H³.

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, 3^1/2φ, (3-φ)^1/2φ] ,

[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 , 20φ³/(3-φ), 20φ²/3]

≈ [.942809,8,3.771236,61.304952,17.453560] . What one can say about icosa tetrahedral tessellation?

1. Consider first the dihedral angles θ. The values of dihedral angles associated T, O, and I in H³ is reduced from that in E³ so that their sum in E² sene must be larger than 2π. Therefore at least one of these cells must appear twice in H³. 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 E³. 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 H³.

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)] propto [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]

New results about the relation of the icosa tetrahedral tessellation to the dark genetic code

How could the icosa tetrahedral tessellation relate to the proposed dark realizations of the genetic code cite{btart,bioharmony2020,TIH}? Consider first the realizations of the dark genetic codes in terms of dark proton triplets at monopole flux tubes parallel to the ordinary DNA and to the realization in terms of dark photon triplets.

1. The TGD based inspired model of the dark photon genetic code assumes that the dark realization of genetic code involves 3 icosahedral Hamiltonian cycles giving rise to 20+20+20 dark DNA codons and the unique tetrahedral Hamiltonian cycle giving the remaining 4 codons.
2. In the dark proton realization a given codon would correspond to a selected triangular face of I or T carrying dark protons at the vertices of this face. The dark 3-proton states would correspond to 64 codons.

A possible problem is that one obtains only 32 states for dark proton triplets. One bit must be introduced somehow. Dark protons can be effectively neutrons as far as charge is considered. This is possible if the bonds connecting the dark protons can be both neutral and negatively charged. Weak interactions are as strong as electromagnetic interactions in a given biological scale (such as DNA scale) if the dark Compton length proportional to h_eff is larger than this scale and the weak transitions change the dark protons to effective dark neutrons.

Dark photon triplets would correspond to the representation of codons as frequency triplets represented by the realization of icosahedral and tetrahedral Hamiltonian cycles as frequency triplets. The 3 frequencies would be associated with the cyclotron transitions of dark protons of dark proton triplet.

Dark photon triplets would make possible communications between identical codons by using 3-resonance. Also dark genes as sequences of N dark codons could act as a single quantum coherent unit and 3-N resonances between identical dark genes would become possible. The mechanism is very similar to that used in the computer language LISP. The modulation of the frequency scale by modulating the thickness of the monopole flux tubes would make possible coding of the signal and it would be transformed to a sequence of resonance pulses at the receiving end.

The proposal is that the icosa tetrahedral honeycomb at the light-cone proper time a= constant surfaces identifiable as hyperbolic 3-space H³ allows to realize the dark genetic code. This raises several questions.

The icosa tetrahedral honeycomb is the unique honeycomb, which involves only Platonic solids. This inspires the question whether genetic code could be universal and realized in all scales by induction, which means that the tessellation of H³ induces tessellation of 3-surface X³⊂ H³ by restriction. Also the induction to H³(a) projection of X⁴ makes sense.

The TGD view of holography indeed predicts that the space-time surfaces in H=M⁴× CP₂ as analogs of Bohr orbits go through H³(a_n)⊂ M⁴⊂ H, where a_n corresponds to a root of the polynomial with integer coefficients determining to a higher degree a given region of the space-time surface by M⁸-H duality cite{btart/M8H1,M8H2}. The fundamental region of the icosa tetrahedral tessellation contains 30 octahedrons, 20 tetrahedrons, and 12 icosahedrons and the cautiously proposed interpretation is that the cells meeting at each edge of the tesselation have either the cyclic structure TOTI or OTOI, and each vertex involve 3 O:s, 2 T:s and 1 I. Could one interpret this in terms of the dark icosahedral realization of the genetic code?

1. Why only I and T would contribute to the code. The reason could be that all faces of O are shared by either I or T so that they do not contribute.
2. What could the role of O:S be? There are 3 O:s per single I in vertex and one T per single I in edge for OTOI option. For the TOTI option there are two T:s per I.
   1. OTOI option is slightly favored by the fact the number of octahedrons is largest. For the OTOI option there are 4 independent T codons and the total number of codons is 60+4=64 if either T or I codon but not both is realized. Why this?
   2. TOTI option is favoured if I-T behaves as a single unit in the isohedral code: either I codon or T codon is selected but not both. For both T:s, one T codon is however also an I codon. This would leave 3+3 independent T codons giving 60+ 3+3 =64+2 codons.

There indeed exist 2 exotic amino acids: selenocysteine and pyrrolysine. Could they correspond to the 2 additional T codons? I have proposed an alternative explanation of exotic amino acids in terms of failure of tRNA-amino acid correspondence to be completely unique. This would reflect the incompleteness of the chemical representation of the dark code.

One can also consider an alternative interpretation explaining the problem that the number of dark proton triplets is 32 in the simplest model. The additional tetrahedral bit appearing in the TOTI option would double the number of dark proton codons from 32 to 64. Second T with 2 free faces would define a separate strand and a selection of T would correspond to this bit.

Bioharmony involves 3 icosahedral Hamiltonian cycles. All the combinations of the 3 -cycles with symmetries Z₆, Z₄ and Z₂ predict the same code. These bioharmonies are interpreted as correlates for emotional states appearing already at the basic bio-molecular level. The motivation comes from the fact that the icosa tetrahedral harmony emerges as a geometric model for the music harmony and music indeed both creates and expresses emotions.

Could icosahedral tessellation allow us to understand the presence of these 3 icosahedral Hamiltonian cycles? Could one assign a dark DNA strand with both O:s and I-T units in the case of T-O-T-I option? The choice of O at the vertex of the dark DNA strand as a path inside the tessellation would select at each step one of the icosahedral Hamiltonian cycles allowed by a given bioharmony.

One can criticize the assumption that there is only a single codon per single I and T. I:s could in principle carry several codons. This however gives a restriction that the codons inside given I and T are different and restricts the representative power of the code if it involves more than 2 strands. This restriction is however automatically satisfied for the base-paired codon and anticodon in the DNA double strand! The most ambitious model would describe the entire DNA double strand and relate the model bioharmony to the properties of the icosa tetrahedral tessellation. There are however many questions remaining.

1. Single DNA and RNA strand would correspond to a “half realization” for which the T and I cells would contain only single codon. The splitting of DNA could have a geometric interpretation as an effective replication of the induced tessellation to two tessellations to RNA type tessellations.
2. There are 20 amino-acids and an icosahedron involves 20 faces. Is this a mere accident? Could icosahedral honeycomb describe amino-acid sequences geometrically. tRNA-amino-acid pairing would involve pairing icosa tetrahedral and icosahedral tessellations.
3. DNA has a helical structure. Helical tessellations are known to exist (see this). If icosa tetrahedral tessellation is induced, the helical structure would most naturally reflect the dynamics of the corresponding space-time surface.

This could also explain why only a sequence of Is: is selected from the set of 12 I:s from a given fundamental region of the icosa tetrahedral tessellation. 12 DNA codons would correspond to the distance 10.952 nm by using the rise/base pair equal to 0.332 nm for the B form of DNA (see this). This makes 9.96 Angstrom per codon. A full turn for the DNA helix corresponds to an integer number of codons and to the length 11 nm very near to the distance assignable to 12 DNA codons.

Does this mean that the diameter of the fundamental region of the icosa tetrahedral tessellation is 11 nm? This is not far from the “nominal value” of the p-adic length scale L(151)= 10 nm. The definition of this scale is subject to uncertainties and does not exclude L(151)= 11 nm.

One can of course ask whether the helical structure could be an inherent property of the icosa tetrahedral tessellation. This does not look plausible to me.

One can also ask, whether the helical structure of a double DNA strand combined with the coiling in the scale of 10 nm:s giving chromosomes could allow for the DNA strand to visit maximal number of I:s inside the fundamental regions to guarantee maximal packing efficiency. See the article Overall View about Models of Genetic Code and Bio-harmony or the chapter with the same title.

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

Source: http://matpitka.blogspot.com/2023/07/about-tesselations-in-hyperbolic-3.html