Pythagoras, Music, Sacred Geometry and Genetic Code: Article
Pythagoras, Music, Sacred Geometry and Genetic Code: Article
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Pitkänen, M. Pythagoras, Music, Sacred Geometry and Genetic Code
Article
Abstract
A given triangle of icosahedron can contain 0, 1 or 2 edges of the cycle and the numbers of the
triangles corresponding to these triangle types classify partially the notion of harmony characterized
by the cycle. Quint cycle suggests the identification for the single edge of curve as quint interval so
that triangles would represent basic 3-chords of the harmony with 0,1, or 2 quints. Octahedron and
cube which are duals of each other and have 6 and 8 vertices respectively, and dodecahedron which
is dual of icosahedron having 20 vertices and 12 faces. Arabic music uses half intervals and scales
with 19 and 24 notes are used. Could 20-note scale with harmony defined by 5-chords assigned to the
pentagons of dodecahedron have some aesthetic appeal? The combination of this idea with the idea
of mapping 12-tone scale to a Hamiltonian cycle at icosahedron leads to the question whether amino-
acids could be assigned with the equivalence class of Hamiltonian cycles under icosahedral group and
whether the geometric shape of cycle could correspond to physical properties of amino-acids. The
identification of 3 basic polar amino-acids with triangles containing no edges of the scale path, 7
polar and acidic polar amino-acids with those containing 2 edges of the scale path, and 10 non-polar
amino-acids with triangles containing 1 edge on the scale path is what comes first in mind. The
number of DNAs coding for a given amino-acid could be also seen as such a physical property. The
model for dark nucleons leads to the vertebrate genetic code with correct numbers of DNAs coding for
amino-acids. The treatment of the remaining 4 codons and of the well-known 21st and 22nd amino-
acids requires the fusion of icosahedral code with tetrahedral code represented geometrically as fusion
of icosahedron and tetrahedron along common face which has empty interior and is interpreted as
”empty” amino-acid coded by stopping codons. In this manner one can satisfy the constraints on the
Hamiltonian cycles, and construct explicitly the icosahedral Hamiltonian cycle as (4,8,8) cycle whose
unique modification gives (4,11,7) icosa-tetrahedral cycle.
1 Introduction
The conscious experiences generated by music demonstrate a fascinating connection between algebra and
emotions. How can major and minor scale using different frequency ratios generate so different emotional
experiences. This strongly suggests the we experience music as entire time interval, 4-D patterns - rather
than time=constant snapshots. Also the ability remember the key and the tension lasting as long as the
return to the basic key has not taken place, is example of this. One of the key questions is why octaves
- that is powers of 2 of the basic note of the scale - are experienced as equivalent? One can also wonder
what is behind consonance and dissonance.
I have already earlier tried to understand music experience and considered some ideas inspired by
p-adic numbers fields - such as the idea that Pythagorean scale coming as powers of 3 for the basic note
modulo octave equivalence might relate to 3-adicity. Reading of a book titled ”Interference: A Grand
Scientific Musical Theory” by Richard Merrick [6] freely available in web (http://interferencetheory.
com/files/Interference.pdf) re-stimulated my interest. In particular, I found the idea about a connec-
tion between music scale and harmonies with Platonic solids (3-D ”sacred geometry”) as highly inspiring.
The basic question was whether the 12-tone scale could be mapped to a curve going once through each
point of icosahedron having 12 vertices and whether the 20 faces of icosahedron, which are triangles could
define the basic chords in 12-tone scale. These curves are known as Hamiltonian cycles and in the case
1 Correspondence: Matti Pitkänen http://tgdtheory.com/. Address: Köydenpunojankatu 2 D 11 10940, Hanko, Finland.
Email: matpitka@luukku.com.
of icosahedron there are 210 of them: those obtained from each other by rotation leaving icosahedron
invariant are however equivalent.
A given triangle of icosahedron can contain 0, 1 or 2 edges of the cycle and the numbers of the triangles
corresponding to these triangle types classify partially the notion of harmony characterized by the cycle.
Quint cycle suggests the identification for the single edge of curve as quint interval so that triangles would
represent basic 3-chords of the harmony with 0,1, or 2 quints.
One can make same questions also for other Platonic solids- tetrahedron (4 vertices), octahedron and
cube which are duals of each other and have (6 and 8 vertices respectively, and dodecahedron which is
dual of icosahedron having 20 vertices and 12 faces. Arabic music uses half intervals and scales with 19
and 24 notes are used. Could 20-note scale with harmony defined by 5-chords assigned to the pentagons of
dodecahedron have some aesthetic appeal? Nowadays it is possible to develop electronically music based
on this kind of scale and this kind of experimentation might be a fascinating intellectual and artistic
adventure for a young composer.
I have also played with the idea that the 20 amino-acids could somehow correspond to the 20 triangles
of icosahedron. The combination of this idea with the idea of mapping 12-tone scale to a Hamiltonian
cycle at icosahedron leads to the question whether amino-acids could be assigned with the equivalence
class of Hamiltonian cycles under icosahedral group and whether the geometric shape of cycle could
correspond to physical properties of amino-acids [4]. The identification of 3 basic polar amino-acids with
triangles containing no edges of the scale path, 7 polar and acidic polar amino-acids with those containing
2 edges of the scale path, and 10 non-polar amino-acids with triangles containing 1 edge on the scale path
is what comes first in mind.
The number of DNAs coding for a given amino-acid [5] could be also seen as such a physical property.
The model for dark nucleons leads to the vertebrate genetic code with correct numbers of DNAs coding
for amino-acids. It is not however clear how to interpret DNA codons geometrically.
It however turns out that one can understand only the role of 60 codons in the icosahedral framework.
The treatment of the remaining 4 codons and of the well-known 21st and 22nd amino-acids requires
the fusion of icosahedral code with tetrahedral code represented geometrically as fusion of icosahedron
and tetrahedron along common face which has empty interior and is interpreted as ”empty” amino-acid
coded by stopping codons. In this manner one can satisfy the constraints on the Hamiltonian cycles, and
construct explicitly the icosahedral Hamiltonian cycle as (4,8,8) cycle whose unique modification gives
(4,11,7) icosa-tetrahedral cycle.
The well-known but story (good story but √ probably not true) tells that a pupil of Pythagoras demon-
strated that the diagonal of unit square ( 2) cannot √ be rational number and had to pay with his life
for the discovery. Pythagoras himself encountered 2 through music theory. He asked what is the note
exactly in the middle of the of the scale. Modern mathematician would answer half of octave corre-
sponding to the frequency ratio 21/2 . Algebraic numbers did not however belong to the world of order of
Pythagoras and he obtained to a non-satisfactory rational approximation of this number. This was very
natural since only rational approximations of algebrais
√ are possible in the experimental approach using
only strings with rational number valued lengths. 2 represents the interval C − F# known as tritone and
this
√ this interval was associated with devil and its use was denied also by church. Only after reformation
2 was accepted and this interval appears repeatedly in the compositions of Bach.
The amazing connections between evolution of mathematics and evolution of the religious beliefs
inspires the question whether the evolution of consciousness could at basic level correspond to th evolution
of the complexity of the number field behind the dynamics underlying consciousness. For instance, in
TGD framework the vision about physics as generalized number theory allows one can to ask whether
the mathematical evolution could have meant quite concretely the emergence of increasingly algebraic
extensions of rationals for the coefficients of polynomials describing space-time surfaces serving as space-
time correlates of consciousness.
1. Since 12-note scale is used in practically all classical western music and even in atonal music based
on 12-note scale, the natural question is whether 12-note scale could be mapped to a connected,
closed, non-self-intersecting path on icosahedron going through all 12 vertices and consisting of edges
only. Closedness would mean that base note and its octave are identified by octave equivalence.
2. This mathematical problem is well-known and curves of this kind are known as Hamilton cycles
and can be defined for any combinatorial structure defined by vertices and faces. Hamilton proved
that Hamiltonian cycles (possibly identifiable as 20-note scale) at dodecahedron is unique module
rotations and reflection leaving dodecahedron invariant. Also in the case of tetrahedron and cube
the Hamiltonian cycle is unique.
3. For octahedron and icosahedron this is not the case [3] and there are both cycles containing only
faces with at least 1 edge of the path and also cycles containing no faces containing no edges of the
path. Numerical experimentation is rather straightforward manner to determine Hamiltonian cycles
and H = 210 = 1024 cycles can be found. The number of topologically non-equivalent cycles (not
transformable to each other by the isometries of icosahedron) is factor of this number. The group of
orientation preserving isometries of icosahedron is the alternating group A5 of 60 even permutations
of five letters. The full group of isometries is G = A5 × Z2 containing N = 120 elements.
4. Some subgroup of G leaves given path invariant and its order must be factor M of N so that
topological equivalence class of cycles contains R = N/M elements. The number of topologically
non-equivalent cycles in given class with H(top) elements is Ntot = H(top)/R so that R must be a
factor of H(top).
Before continuing it is good so summarize the geometry of icosahedron shortly. There are 20 faces
which are triangles, 12 vertices, and 30 edges. From each vertex 5 edges. Therefore the construction
of Hamiltonian cycles means that at each vertex on path one must select between four options edges
since one cannot return back. This gives 412 = 224 ∼ 1.6 × 107 alternatives to be considered. Therefore
the numerical search should be relatively easy. Keeping account of the points already traversed and not
allowing self intersections, the actual number of choices is reduced. The construction requires labeling of
the vertices of the icosahedron by integers 1,...,12 in some manner and defining 12 × 12 matrix A(i, j)
whose element equals to 1 if vertices are neighbours and 0 if not. Only the edges for with A(i, j) = 1
holds true are allowed on the path. A concrete representation of icosahedron as a collection of triangles in
plane with suitable identifications of certain edges is needed. This helps also to visualize the classification
of trianbles to three types discussed below. This can be found in the Wikipedia article.
1. Proceeding in Pythagorean spirit, the neighboring points would naturally correspond to progression
by fifths - that is scalings by powers of 3/2 or in equal temperated scale by powers of 27/12 . This
would mean that two subsequent vertices would correspond to quint.
2. The twenty triangles of the icosahedron would naturally correspond to 3-chords. Triangles can
contain either 0, 1, or 1 edges of the 12-edge scale path. The triangle containing 3 edges is not
possible since it would reside on a self-intersecting path. A triangle containing one edge of path the
chord would contain quint which suggest a chord containing basic note, quint and minor or major
third. The triangle containing two edges would contain subsequent quints - CDG is one possible
example by octave equivalence. If the triangle contains no edges of the path one can say that the
chord contains no quints.
The numbers of triangles classified according to the number of path edges contained by them serves
as the first classification criterion for a given harmony characterized by the Hamiltonian cycle (note
that one cannot exclude the possibly of non-closed paths since Pythagorean construction of the
scale by quints does not yield quite precisely octave as outcome).
Figure 1: There 3 different types of triangles characterized by the number of edges contained by them.
This predicts chords with 0,1 or 2 quints.
1. The topologically equivalent cycles must have same numbers of faces containing 0, 1, or 2 edges of
the Hamiltonian path since isometries do not change these numbers. Let us denotes these numbers
by n0 , n1 and n2 . The total number of faces is 20 so that one has
n0 + n1 + n2 = 20 .
Furthermore, each of the 12 edges on the path is contained by two faces so that by summing over
the numbers of edges associated with the faces one obtains twice the number of edges:
0 × n0 + 1 × n1 + 2 × n2 = 2 × 12 = 24 .
n0 = n2 − 4 , n2 ≥ 4 ,
n1 = 24 − 2n2 , n2 ≤ 12 .
If these integers characterize the topological equivalence completely and if the allowed combinations
are realized, one would have 12-4=8 topologically nonequivalent paths. The actual number is
Ntot = 2k , k ≥ 7, so that the integers cannot characterize the topology of the path completely.
2. The number of Hamiltonian paths on icosahedron is known to be 2560 [1]. Numerical calculations
[2] suggest that the number of Hamiltonian cycles is 210 = 1024. This would mean that the sum
over the numbers N (n2 ) if cycles associated with differ values of n2 satisfies
12 X
X
N (n2 , i)) = 210 .
n2 =4 i
N (n2 , i) is the number of paths of given topology with fixed n2 . The numbers N (n2 , i) are integers
which are factors of N = 120 of the order of the isometry group of the icosahedron. The average of
N (n2 , i) is 27 = 128.
Figure 2: The edge of the cycle belongs to 2 triangles, which as chords can correspond to 1 resp.2 , 1
resp. 1 and 2 resp. 2 quints.
since the cycle contains 12 edges. There are p12 +2p11 = n1 1-quint triangles and (p12 +2p22 )/2 = n2
2-quint triangles (note double counting responsible for division by two). Altogether this gives
2. These three Diophantine equations are for integers and would allow for real numbers only single
solution and for integers it in the generic case there are no solutions at all. Situation changes if
the equations are not independent which can happen if the integers ni satisfy additional conditions.
By subtracting first and second and second and third equation from each other one obtains the
consistency condition
n1 = 24 − 2n2 .
This condition is however second of the conditions derived earlier so that only two equations, say
the first two ones, are independent.
n1
p22 = p11 + n2 − 2 ,
p22 = n2 − p212 .
giving
One must have 0 ≤ pij ≤ 12 and p12 ≤ n1 from p11 = (n1 − p12 )/2. Here one has p12 ∈
{0, 2, ...M in{12, 2n2 , n1 } so that M in{7, n2 + 1, [n1 /2] + 1} solutions are possible. The condition
that the cycle has no self-intersections can forbid some of the solutions.
3. The first guess for the ”biological harmony” possibly associated with amino-acids would be (n0 , n1 , n2 ) =
(3, 10, 7): this if one neglects the presence of 21st and 22th amino-acid also appearing in pro-
teins. It turns out that a more feasible solution fuses tetrahedral code and icosahedral codes with
(n0 , n1 , n2 ) = (4, 8, 8) giving (n0 , n1 , n2 ) = (4, 11, 7) for icosatetrahedral code.
For instance, (n0 , n1 , n2 ) = (3, 10, 7) would give p12 ∈ {0, 2, 4, 6, 8, 10}, p11 ∈ {5, 4, 3, 2, 1, 0}, p22 ∈
{7, 6, 5, 4, 3, 2} so that one has 6 alternative solutions to these conditions labelled by p12 . The
number of neighboring triangles containing single quint is even number in the range [0, 10]: this
brings in mind the possibility that the neighboring single quint triangles correspond to major-minor
pairs. Clearly, the integer p12 is second topological invariant characterizing harmony.
manners. The remaining 2-2 edges can be chosen only in one manner. This gives altogether
To summarize, one obtains large number of notions of harmony are possible although one cannot
expect that the absence of self-intersections does not allow all topologies for the cycle.
1. For a proper choice of the mapping of the 12-note scale to the surface of icosahedron the 20 triangles
could correspond to 20 amino-acids analogous to 3-chords and that the 3 types of 3-chords could
correspond to 3 different classes of amino-acids. One can of course consider also the mapping
of amino-acids to a unique sequence of 20 vertices of dodecahedron representing 20-note scale or
20-chord scale and replacement of the 3-chords defining the harmony with 12 5-chords.
2. Amino-acids are characterized by the non-constant side chain and these can be classified to three cat-
egories: basic polar, non-polar, and polar (http://en.wikipedia.org/wiki/Amino-acids). The
numbers of amino-acids in these classes are a0 = 3, a1 = 10, a2 = 7. Could these classes correspond
to the numbers ni characterizing partially some topological equivalence classes of Hamiltonian paths
in icosahedron? There is indeed a candidate: a0 = n0 = 3, a1 = n1 = 10, a2 = n2 = 7 satisfies
the conditions discussed above. 3 basic polar amino-acids would correspond to the triangles with
no edges on the Hamiltonian cycle, 10 non-polar amino-acids to triangles containing one edge, and
7 acidic polar and polar amino-acids to those containing two edges. One can criticize the combi-
nation of polar and acidic polar amino-acids in the same class. One can also classify amino-acids
to positively charged (3), negatively charged (2) and neutral (15) ones. In this case the condition
is however not satisfied. Thus the proposal survives the first test - assuming of a course that these
Hamiltonian cycles exist! This has not been proven and would require numerical calculations.
3. As found Hamiltonian paths have also other topological characteristics and they could correspond
to physical characteristics and it would be interesting to see what theyh are. To proceed further
one should find the total number of the Hamiltonian paths with n2 = 7 and identify the isometries
of different topological equivalence class having n2 = 7.
Amino-acid sequences would correspond to sequences of 3-chords. The translation of mRNA of gene
to amino-acid sequence would be analogous to the playing of a record. The ribosome complex would be
the record player, the amino-acid sequence would be the music, and mRNA would be the record. Hence
genes would define a collection of records characterizing the organism.
1. Could different DNA triplets coding for the same amino-acid correspond to various octaves of the
chord? The most natural expectation would be that the number of octaves so that one would have 3
DNAs would code single amino-acid and stopping codon would correspond to 4 DNAs. It is difficult
to understand why some 3-chords could correspond to 6 octaves and one of them only one.
2. Could the degeneracy correspond to the ordering of the notes of the 3-chord? For the 3-chords there
are 6 general orderings and 3 cyclic orderings modulo octave equivalence and characterizing by the
choice of the lowest note. The simplest assumption would be that the allowed orderings - degenera-
cies - are characterized by a subgroup of the cyclic group S3 yielding the allowed permutations of
the notes of the chord. The subgroup orders for S3 are 1,2,3, and 6. The allowed degeneracies are
6,4,3,2, and 1 so that this identification fails for D = 4.
3. Could the different correspondences between DNA codons and amino-acids correspond to the dif-
ferent topological equivalence classes of n2 = 7 Hamiltonian cycles. This does not seem to be the
case. The number of different DNA-amino-acid correspondences obtained by choosing one rep-
resentative from the set of DNAs coding for a given amino-acid (and not stopping sign) is the
product of the numbers D(ai ) coding amino-acid ai . From the table below this number is given by
63 × 45 × 31 × 29 × 12 = 34 × 221 and clearly much larger than N = 210 .
4. Could the different codons coding for codon code for some additional information so that amino-
acids would in some aspect differ from each other although they are chemically identical? Here the
magnetic body of amino-acid is a natural candidate. This would suggest that the folding pattern of
the protein depends on what DNA sequence codes it. This information might be analogous to the
information contained by notes besides the frequencies. Durations of notes corresponds is the most
important information of this kind: the only candidate for this kind of information is the value of
hef f = n × h associated with the amino-acid magnetic body determining its size scale. Magnetic
fields strength could be also code by DNA codon besides amino-acid.
d 6 4 3 2 1
N 3 5 2 9 2
Table 1. The number of amino acids N associated with a given degeneracy d telling the number of
DNA triplets mapped to the amino acid in the genetic code. The degeneracies are always smaller than 7
as predicted by the proposed explanation of the Genetic Code.
Second question concerns genetic code itself. Could the DNA degeneracies D(ai ) (number of DNAs
coding for amino-acid ai ) be understood group theoretically in terms of icosahedral geometry? The
triangles of the icosahedron are mapped the triangles under the isometries.
1. One can start by looking the Table 1 for the genetic code telling the number N (d) of amino-acids
coded by d DNA codons. One finds that one can divide DNAs to three groups containing n = 20,
n = 20, resp. n = 21 codons.
(a) There are 3 amino-acids codes by 6 codons and 2 amino-acids coded by 1 DNA: 3×6+2×1 = 20
codons altogether.
Note: One could also consider 1 amino-acid coded by 2 codons instead of 2 coded by 1 codon
3 × 6 + 1 × 2 = 20.
(b) There are 5 amino-acids coded by 4 codons making 5 × 4 = 20 codons altogether.
(c) There are 9 amino-acids coded by 2 codons and 1 by 3 codons making 9 × 2 + 1 × 3 = 21
codons.
Note: One could also consider the decomposition 8 × 2 + 2 × 1 + 1 × 3 = 21 codons implied if
1 amino-acid is coded by 2 codons in the first group.
This makes 61 codons. There are however 64 codons and 3 codons code for stopping of the translation
counted as ”empty” amino-acid in the table.
1. This would suggest the division to 60 + 4 codons. The identification of additional 4 codons and
corresponding amino-acids is not so straightforward as one might first think. 3 of the 4 additional
codons could code for ”empty” amino-acid (Ile) and 1 of them to Ile (empty amino-acid).
2. What suggests itself strongly is a decomposition of codons in 3 different manners. 3 groups of 6
codons plus 2 groups of 1 codon (1 group of 2 codons), 5 groups of 4 codons, and 10 groups of 2
codons (9 groups of 2 codons plus plus 2 groups of 1 codon).
This kind of decompositions are induced by the action on the triangles of icosahedron by thee subgroups
of the isometry group A5 ×Z2 of the icosahedron having 120 = 2×2×2×2×3×5 elements and subgroups
for which number of elements can be any divisor of the order. The orbit associated with a subgroup with
n elements has at most n triangles at its orbit. This allows immediately to deduce the values of n possibly
explaining the genetic code in the proposed manner.
1. The 3 amino-acids coded by 6 codons must correspond to n = 6. This subgroup must have also two
1-element orbits (1 2-element orbit): in other words, 2 triangles must be its fixed points (form its
orbit).
(a) The non-abelian group S3 permuting the vertices of is the first candidate for the subgroup in
question. The triangles at the opposite sides of the icosahedron remain invariant under these
permutations. S3 however has two orbit consisting of 3 triangles which are ”wall neighbours”
of the triangles which remains fixed.
(b) Second candidate is the abelian group Z̃2 × Z3 . Here Z3 permutes the vertices of triangle
and Z̃2 is generated by a reflection of the triangle to opposite side of icosahedron followed by
a rotation by π. This group has 3 orbits consisting of 6 triangles and 1 orbit consisting of
2 triangles (the triangles at opposite side of icosahedron). This group seems to be the only
working candidate for the subgroup in question.
2. The 5 amino-acids coded by 4 codons must correspond to n = 4 and therefore to Z̃2 × Z2 . This is
indeed subgroup of icosahedral group which permutes triangles at the vertices of inscribed tetrahe-
dron. Now all orbits contain 4 triangles and one must have 5 orbits, which are obtained by acting
on the 5 triangles emanating from a given vertex. Note that also Z5 is subgroup of icosahedral
group: this would give a variant of code with 4 amino-acids coded by 5 codons if it were possible
to satisfy additional consistency conditions.
3. Consider next the group consisting of 9 amino-acids coded by 2 codons and Ile (”empty” amino-
acid) coded by 3 codons. Since only the Z̃2 × Z3 option works, this leaves 9 amino-acids coded by 2
codons and 2 amino-acids coded by 1 codon. The subgroup must correspond to n = 2 and thus Z2
acting on fixed triangle and leaving it and its Z̃2 image invariant. One has 9 2-triangle orbits and
two single triangle orbits corresponding to the triangles at the opposite sides of the icosahedron.
The 9 amino-acids coded by 2 codons are all real or 8 of them are real and 1 corresponds to ”empty
amino-acid” coded by two codons.
3-element orbits are lacking and this forces to consider a fusion of of icosahedral code with tetrahe-
dral code having common ”empty-acid” - common triangle of icosahedron and tetrahedron) coded by
2 icosahedral codons and 1 tetrahedral codon. Ile would be coded by 3 codons assignable to the orbit
of Z3 subgroup of tetrahedral symmetry group S3 and would be associated with the tetrahedron. This
would predict 2 additional amino-acids which could be understood by taking into account 21st and 22nd
amino-acid (Sec and Pyl [4]).
The Hamiltonian cycle is not explicitly involved with the proposed argument. Some property of the
cycle respected by the allowed isometries might bring in this dependence. In Pythagorean spirit one might
ask whether the allowed isometries could leave the Hamiltonian cycle invariant but move the vertices along
it and induce a mapping of faces to each other.
The amino-acid triangle at given orbit cannot be chosen freely. The choices of amino-acid triangles
associated with the three groups of 20 DNAs must be different and this gives geometric conditions for the
choices of the three subgroups and one can hope that the assignment of amino-acid toa given triangle is
fixed about from rotational symmetries.
3.2.1 Does the understanding of stopping codons and 21st and 22nd amino-acids require
fusion of tetrahedral and icosahedral codes?
Several questions remain. Could one also understand the additional 4 DNA codons? Could one understand
also how one of them codes amino-acid (Ile) instead of stopping codon? Can one related additional codons
to music?
1. Something is clearly added to the geometric structure, when 4 additional DNA codons are brought
in. They could represent orbits of faces of Platonic solid with 4 faces. The four faces are most
naturally triangles and actually must be so since tetrahedron is the only Platonic solid having 4
faces and its faces are indeed triangles. Tetrahedron has symmetry group S3 containing Z3 and
Z2 as subgroups. Z3 leaves one of the tetrahedral triangles invariant so that one has two orbits
consisting of 1 and 3 triangles respectively.
2. One amino-acid is coded by 3 rather than only 2 codons. One can indeed understand this symmetry
breaking geometrically. If the tetrahedron is attached on icosahedron along one of its triangular
faces and this icosahedral face corresponds to amino-acid (Ile) belonging to a 2-triangle orbit under
icosahedral Z2 one obtains 3 codons coding for Ile! I have actually proposed earlier a model involving
attachment of tetrahedrons to icosahedron as a model for genetic code.
3. Tetrahedron should bring in three additional amino-acids. ”Empty” amino-acid could correspond to
either one of them or to the common base triangle which is indeed geometrically in unique position.
One could even demand that this triangle is ”empty” so that icosatetrahedron would be non-singular
continuous manifold. This would mean exchange of the roles of ile and empty amino-acid both being
coded by three codons. The 3-triangle orbit outside the icosahedron would correspond to Ile and
base triangle to empty amino-acid or vice versa. Base triangle would be coded by 1 tetrahedral
codon plus 2 icosahedral codons.
One of the outsider triangles corresponds to ”empty” amino-acid or ile but two other triangles to two
new exotic amino-acids. In some species there indeed are 21st and 22nd amino-acids (selenocysteine
and pyrrolysine, http://en.wikipedia.org/wiki/Amino_acid) with sulphur replaced with selene.
This modification does not change the polarity properties of cys and lys: cys is non-polar and lys
basic polar implying (n0 , n1 , n2 ) → (4, 11, 7).
4. The naive assumption (n0 , n1 , n2 ) = (3, 10, 7) before the modification need not be correct and
the numerous futile attempts to construct this cycle and the argument below for construction of
icosahedral Hamilton cycle suggests that (n0 , n1 , n2 ) = (4, 8, 8) is more feasible option.
5. The two outsider tetrahedral triangles could correspond to the orbits of Z2 subgroup of S3 acting
as reflection with respect to media of the bottom triangle. Outside faces form orbits consisting of
1 triangle and 2-triangles. These orbits could correspond to 21st and 22nd amino-acids coded by 1
and 2 exotic variants of stopping codons respectively.
The 2 exotic amino-acids are however coded by codons which are usually interpreted as stopping
codons. Something must however distinguish between standard and exotic codings. Is it ”con-
text” giving different meaning for codons and perhaps characterized by different magnetic bodies
of codons?
The base triangle is counted neither as amino-acid nor as a part of the ”icosatetrahedron”. The
interpretation is natural since the resulting topological structure is non-singular and defines a continuous
albeit not differentiable manifold topology (sphere).
Figure 4: Icosatetrahedron is obtained by attaching tetrahedron along one of its faces to icosahedron.
The resulting structure is topological manifold if the common face is replaced with empty set and it is
natural to identify it as ”empty” aminoacid.
1. One should be able to modify the icosahedral Hamiltonian cycle so that the numbers (n0 , n1 , n2 )
charactering icosahedral cycle change so that they conform with the properties of the two exotic
amino-acids. Selenocystein (Sec) is nonpolar like cys and pyrrolysine (Pyl) basic polar like Lys so
that (4, 11, 7) seems to be the correct characterization for the extended system. One must have
(n0 , n1 , n2 ) → (4, 11, 7).
2. One must visit the additional vertex, which means the replacement of one edge from the base triangle
with wedge visiting the additional vertex. There are several cases to be considered depending on
whether the base triangle is 1-quint triangle or 2-quint triangle, and what is the type of the edge
replaced with wedge. One can even consider the possibility that the modified cycle does not remain
closed.
If the icosahedral cycle has (n0 , n1 , n2 ) = (3, 10, 7), the value of n2 is not changed in the construction.
For a closed cycle edge is replaced with wedge and the only manner to preserve the value of n2 is
that the process producing 1 tetrahedral 2-quint triangle transforms 1 icosahedral 2-quint triangle
identified as base triangle to 1-quint triangle. If the replaced edge of base triangle is of type 2-1, one
has n1 → n1 + 1 since one icosahedral 1-quint triangle disappears and 2 tetrahedral ones appear.
Icosahedral n0 increases by 1 units. Hence the condition (3, 10, 7) → (4, 11, 7) would be met. It
however seems that (4, 8, 8) is more promising starting cycle as the argument below shows.
3. The number options is at most the number n2 of 2-quint triangles serving as candidates for ”empty”
amino-acid. An additional condition comes from the requirement that replaced edge is of type 2-1.
Figure 5: The modification of (4, 4, 8) icosahedral Hamiltonian cycle consistent with the constraints that
icosatetrahedral cycle corresponds to (4, 11, 7) consistent the classification of amino-acids in three classes.
satisfies (n0 , n1 , n2 ) = (3, 10, 7), and the following proposal starts from different icosahedral cycle. The
following is just a trial, which should be checked by a direct calculation.
1. The most obvious guess for the cycle to be modified to cycle at icosatetrahedron having (n0 , n1 , n2 ) =
(4, 11, 7) (the triangle corresponding to ”empty” amino-acid is not counted) is (n1 , n2 , n3 ) = (3, 10, 7).
I have not found cycle with these characteristics.
2. It seems however possible to find cycle with (n1 , n2 , n3 ) = (4, 8, 8). From this can obtain the
desired kind of extended cycle if the ”empty” triangle is 2-quint triangle and the edge replaced with
the wedge is of type 2-2. The replacement of icosahedral edge eliminates two icosahedral 2-quint
triangles and generates 1 tetrahedral 2-quint triangle giving n2 → n2 − 2 + 1 = n2 − 1 = 7. The
disappearance of the icosahedral edge generates two icosahedral 1-quint triangles of which second
one corresponds to empty amino-acid and is not counted and 2 tetrahedral 1-quint triangles giving
n1 → n1 + 3 = 11.
The figure below represents the construction of cycle (4, 8, 8, ). The icosahedron is constructed from
regions P (I) glued to the triangle t along one edge each. The arrows indicate that the one pair of edges
of type 1 and 2, 1 and 3 and 3 and 2 are identified. Also the long edges I of T are identified with pairs
of subsequent edges of P (I) as the arrows indicate.
Figure 6: A proposal for a Hamilton cycle realizing bio-harmony (n1 , n2 , n3 ) = (4, 8, 8) allowing extension
to cycle (3, 11, 7) on tetraicosahedron. Circled ”0”, ”1” and ”2” indicates whether a given small triangle
is 0-, 1-, or 2-quint triangle. It is relatively easy to verify that the condition (n1 , n2 , n3 ) = (4, 8, 8) for
bio-harmony is satisfied.
3.3 How could one construct the Hamiltonian cycles on icosahedron with a
minimal computational work?
Although the construction of Hamiltonian cycles is known to be an NP hard problem for a general graph,
one can hope that in case of Platonic solids having high symmetries, a direct construction instead of
straightforward numerical search might work. The following is a proposal for how one might proceed. It
relies on paper model for icosahedron.
1. The basic observation about one can get convinced by using paper model is following. One can
decompose the surface of icosahedron to three regions P (I), I = 1, 2, 3, with pentagonal boundary
and containing 5 triangles emanating from center vertex plus one big triangle T containing 4 pen-
tagonal triangles and one lonely small triangle t opposite to it. These 5 regions span the surface of
icosahedron. There is clearly a symmetry breaking and there is great temptation to assume that t
corresponds to the triangle along which the tetrahedron is glued to the icosahedron in the model of
genetic code realizing the modification of icosahedral cycle for which the first guess is (3, 7, 10) but
for which also (4, 8, 8) can be considered.
2. The Hamiltonian cycle must visit at the centers of each P (I): one enters pentagonal region P (I),
I = 1, 2, 3 along one of the five interior edges beginning at pentagonal vertex aI,i , i = 1, .., 5 and
leaves it along second edge ending at vertex bI,j , j 6= 5. One can call these edges interior edges.
The edges at boundaries of P (I) can be called boundary edges. Interior edge can correspond to
|i − j| = 0, 1 or i − j > 1. For |i − j| = 1 the interior edge gives rise to 2-quint triangle. For i − j = 0
there is no boundary edge after b( I, j).
3. Pentagonal boundary edges come in three types. 2 of them are shared with T , 1 with t opposite to
it, and 2 with another pentagonal region P (I). One can label P (i) in such a manner that the P (I)
shares two boundary edges with P (I + 1).
The boundary edges of small and big triangle are boundary edges of the 3 pentagonal regions so
that they are not counted separately.
4. One can assume that the cycles begins from a vertex of T . Since the cycle is closed it returns back
to this vertex. The last edge is either at the boundary of T or goes through one or two edges of the
small interior triangle of T so that this triangle is either 0-,1- or 2-quint triangle.
t can be 0-, 1-, or 2-quint triangle.
5. The total number of the interior edges inside the 3 pentagonal regions is 3 × 2 = 6 so that 6
remaining edges must be boundary edges associated with P (I) and interior edges of T : otherwise
one would visit some pentagonal center twice and self-intersection would occur. The boundary edges
associated with t and T are boundary edges of P (I), I = 1, 2, 3
6. At the vertex b( I, j) of pentagonal region one must turn right or left and move along the boundary
edge. One can move at most nI = 4 − j boundary edges along the pentagonal boundary in clockwise
direction and nI = j − 2 edges in counterclockwise direction (clockwise is the direction in which the
index labelling 5 vertices grows). The maximum number of boundary edges is 3 and obtained for
j − i ± 1.
P
7. The condition P nI + n(T ) = 6, where n(T ) = 1, 2 is the number of interior edges of T , holds true
so that one has n(I) ≡ ntot ∈ {4, 5}. The numbers and types (shared with pentagon, T , or t)
of the boundary edges of P (I), the differences ∆(I) = jI − iI , the number of edges in t and the
number of interior edges of T characterize the Hamiltonian cycle besides the condition that it is
closed. The closedness condition seems possible to satisfy. One must enter big triangle through one
of the vertices of T and this vertex is uniqely determined once the third pentagon is fixed. One can
therefore hope that the construction gives directly all the Hamiltonian cycles with relatively small
amount of failed attempts, certainly dramatically smaller than n = 224 ∼ 107 of blind and mostly
un-succesful trials.
8. Each P (I) containing boundary edges gives rise to least 2 2-quint triangles associated with bI (I)
and aI+1 .
If all 3 P (I) have |i − j| > 1, one has n2 = 3 × 2 = 6. The contribution of regions P (I) is larger
if some pentagon interiors have |∆(I)| = |j(I) − i(I)| = 1. |j(I) − i(I)| = 1 gives ∆n2 (I) = 1 and
∆n1 (I) = 0 since 2 1-quint triangles are replaced with single 2-quint triangle.
The interior of the T can give 1 2-quint triangle.
(a) Each pentagonal interior edge pair leading from a(I, j) to b(I, j) contributes 2 1-quint triangles
for ∆(I) 6= ±1, otherwise one obtains only 1 2-quint triangle. This would give maximum
number of 6 1-quint triangles associated with the interior edges of 3 pentagons.
(b) P (I) pentagonal boundary edges contribute 2 × (P (I) − 1) additional 1-quint triangles.
(c) T contributes at most 4 1-quint triangles.
(d) t can correspond 1-quint triangle and would do so if the interpretation of extended code is
correct.
10. The construction also breaks the rotational symmetry since the decomposition of icosahedron to
regions is like gauge fixing so that one can hope of obtaining only single representative in each
equivalence class of cycles and therefore less than 210 . By the previous argument related to icosate-
trahedral code, t and the triangle opposite to it cannot however correspond to amino-acids coded
by 1 codon as one might guess first. Rather, t corresponds to ”empty” aminoacid and to 1-quint
triangle belonging to Z2 orbit.
The number of cycles should be 210 . One can try to estimate this number from the construction.
Each bI,j can be chosen in 4 manners at the first step but at later steps some vertices of the neighboring
pentagon might have been already visited and this reduces the available vertices by n + 1 if n subsequent
edges are visited. At each vertex bI,j one has 4 options for the choice of the boundary edges unless some
boundary edges of pentagon (shared with other pentagons) have been already visited. It is also possible
that the number of boundary edges vanishes. One can start from any vertex of triangle. This gives the
upper bound of 24 choices giving N < 212 paths going through 4 pentagon-like regions. The condition
that the path is closed, poses constraints on the edge path assignable to T but the number of choices is
roughly 24. The condition that path goes through all vertices and that no edge is traversed twice must
reduce this number to 210 .
The numerical construction of Hamiltonian cycles should keep account about the number of vertices
visited and this would reduce the number of candidates for b(I, j) and for the choices of P (I) for I > 1
as well as the number of edge paths associated with T .
4 Other ideas
The book of Merrick discusses also other ideas. The attempts to understand music in TGD framework
relate to these ideas.
1. Music metaphor is key element of TGD inspired view about biology and neuroscience. In particular,
TGD based view about dark matter leads to the proposal that bio-photons are ordinary photons
resulting as transformations of dark photons with large Planck constant hef f = nh to ordinary
photons. The further hypothesis is that the energy spectrum of bio-photons is universal and contains
visible photons and UV photons, which defined transition energies of biomolecules. This hypothesis
follows if the value of hef f assignable to a magnetic flux tube characterizes ion and is proportional
to its mass number. The notion of gravitational Planck constant identified as hgr = GM m/v0 ,
where v0 is a velocity parameter assignable to the two-particle system can be identified in the case
of elementary particles and ions with hef f and predicts also the universality of bio-photon spectrum.
2. In this framework bio-photons would represent music as light inducing molecular transitions. Notes
that is different energies of bio-photons would correspond to different magnetic field strengths at
magnetic flux tubes as was proposed much earlier in the quantum model of hearing [10]. Could the
biochemical and physiological aspects involved with the generation of music experience be realized
in terms of bio-photon emission induced by the listening of music?
vacuum extremals for which the light velocity at the space-time surface corresponds to arbitrarily low
velocity at the level of imbedding space.
Acknowledgements: I want to thank Tommi Ullgren for directing my attention to the book of Richard
Merrick as well as for fascinating discussions about music. I am also grateful for Salla Vasenius for drawing
the illustrations.
References
Mathematics
[1] Icosahedral graph. Wolfram MathWorld. http://mathworld.wolfram.com/IcosahedralGraph.
html.
[2] Why are there 1024 Hamiltonian cycles on an icosahedron? http://mathoverflow.net/questions/
37788/why-are-there-1024-hamiltonian-cycles-on-an-icosahedron.
[3] B. Hopkins. Hamiltonian parths on Platonic graphs. IJMMS. http: // www. emis. de/ journals/
HOA/ IJMMS/ 2004/ 29-321613. pdf , 30:1613–1616, 2004.
[10] M. Pitkänen. Quantum Model for Hearing. In TGD and EEG. Onlinebook. http://tgdtheory.
fi/public_html//tgdeeg/tgdeeg/tgdeeg.html#hearing, 2006.