It seems that this is possible. The toric harmonies have by following argument 12 DNA doublets which each code single amino-acids. In icosahedral model one has 10 doublets. This corresponds to the almost exact U ↔ C and A ↔ G symmetries of the genetic code. In the following I give the argument in detail.
1. Some basic notions
First some basic notions are in order. The graph is said to be equivelar if it is a triangulation of a surface meaning that it has 6 edges emanating from each vertex and each face has 3 vertices and 3 edges (see this) Equivelarity is equivalent with the folllowing conditions;
- Every vertex is 6-valent.
- The edge graph is 6-connected.
- The graph has vertex transitive automorphism group.
- The graph can be obtained as a quotient of the universal covering tesselation (3,6) by a sublattice (subgroup of translation group). 6-connectedness means that one can decompose the tesselation into two disconnected pieces by removing 6 or more vertices
- Edge graph is n-connected if the elimation of k<n vertices leaves it connected. It is known that every 5-connected triangulation of torus is Hamiltonian (see this). Therefore also 6-connected (6,3)p=2,q=2 tesselation has Hamiltonian cycles.
- The Hamiltonian cycles for the dual tesselation are not in any sense duals of those for the tesselation. For instance, in the case of dodecahedron there is unique Hamiltonian cycle and for icosahedron has large number of cycles. Also in the case of 6,3) tesselations the duals have different Hamilton cycles. In fact, the problem of constructing the Hamiltonian cycles is NP complete.
Can one say anything about the number of Hamiltonian cycles?
- For dodecahedron only 3 edges emanates from a given vertex and there is only one Hamiltonian cycle. For icosahedron 5 edges emanate from given vertex and the number of cycles is rather large. Hence the valence and also closely related notion of n-connectedness are essential for the existence of Hamilton's cycles. For instance, for a graph consisting of two connected graphs connected by single edge, there exist no Hamilton's cycles. For toric triangulations one has as many as 6 edges from given vertex and this favors the formation of a large number of Hamiltonian cycles.
- Curves on torus are labelled by winding numbers (M,N) telling the homology equivalence class of the cycle. M and M can be any integers. Curve winds M (N) times around the circle defining the first (second) equivalence homology equivalence class. Also Hamiltonian cycles are characterized by their homology equivalence class, that is pair (M,N) of integers. Since there are only V=12 points, the numbers (M,N) are finite. By periodic boundary conditions means that the translations by multiples of 2e1+2e2 do not affect the tesselation (one can see what this means geometrically from the illustration at this). Does this mean that (M,N) belongs to Z2× Z2 so that one would have 4 homologically non-equivalent paths.
Are all four homology classes realized as Hamiltonian cycles? Does given homology class contain several representatives or only single one in which case one would have 20 non-equivalent Hamiltonian cycles?
3. It is possible to find the Hamiltonian cycles
It turned out that there exist programs coding for an algorithm for finding whether given graph (much more general than tesselation) has Hamiltonian cycles. Having told to Jebin Larosh about the problem, he sent within five minutes a link to a Java algorithm allowing to show whether a given graph is Hamiltonian (see this): sincere thanks to Jebin! By a suitable modification this algorithm find all Hamiltonian cycles.
- The number NH of Hamiltonian cycles is expected to be rather large for a torus triangulation with 12 vertices and 24 triangles and it is indeed so: NH=27816! The cycles related by the isometries of torus tesselation are however equivalent. The guess is that the group of isometries is G= Z2,refl⋌ (Z4,tr⋌ Zn,rot). Zn,rot is a subgroup of local Z6,rot. A priori n∈{2,3,6} is allowed.
On basis of the article about toric tesselations (see this) I have understood that one has n=3 but that one can express the local action of Z6,rot as the action of the semidirect product Z2,refl× Z3,rot at a point of tesselation. The identity of the global actions Z2,refl× Z3,rot and Z6,rot does not look feasible to me. Therefore G= Z2,refl⋌ (Z4,tr⋌ Z3,rot) with order ord(G)=24 will be assumed in the following (note that for icosahedral tesselation one has ord(G)=120 so that there is symmetry breaking).
Z4 would have as generators the translations e1 and e2 defining the conformal equivalence class of torus. The multiples of 2(e1+e2) would leave the tesselation invariant. If these arguments are correct, the number of isometry equivalence classes of cycles would satisfy NH,I≥ NH/24=1159.
- The actual number is obtained as sum of cycles characterized by groups H⊂ Z12 leaving the cycle invariant and one can write NH,I= ∑H (ord(H)/ord(G)) N0(H) , where N0(H) is the number of cycles invariant under H.
4. What can one say about the symmetry group H for the cycle?
Simple arguments below suggest that the symmetry group of Hamiltonian cycles is either trivial or reflection group Z2,refl.
- Suppose that the isometry group G leaving the tesselation invariant decomposes into semi-direct product G= Z2,refl⋌ (Z4,tr⋌ Z3,rot), where Z3,rot leaves invariant the starting point of the cycle. The group H decomposes into a semi-direct product H=Z2,refl ⋌ (Zm,tr× Z3,rot) as subgroup of G=Z2,refl ⋌ (Z4,tr× Z3,rot).
- Zn,rot associated with the starting point of cycle must leave the cycle invariant at each point. Applied to the starting point, the action of H, if non-trivial - that is Z3,rot, must transform the outgoing edge to incoming edge. This is not possible since Z3 has no idempotent elements so that one can have only n=1. This gives H=Z2,refl ⋌ (Zm,tr. m=1,2 and m=4 are possible.
- Should one require that the action of H leaves invariant the starting point defining the scale associated with the harmony? If this is the case, then only the group H=Z2,refl would remain and invariance under Zrefl would mean invariance under reflection with respect to the axis defined by e1 or e2. The orbit of triangle under Z2,refl would consist of 2 triangles always and one would obtain 12 codon doublets instead of 10 as in the case of icosahedral code.
If this argument is correct, the possible symmetry groups H would be Z0 and Z2,refl. For icosahedral code both Zrot and Z2refl occur but Z2,refl does not occur as a non-trivial factor of H in this case.
The almost exact U ↔ C and A ↔ G symmetry of the genetic code would naturally correspond to Z2,refl symmetry. Therefore the predictions need not change from those of the icosahedral model except that the 4 additional codons emerge more naturally. The predictions would be also essentially unique.
- If H is trivial Z1, the cycle would have no symmetries and the orbits of triangles would contain only one triangle and the correspondence between DNA codons and amino-acids would be one-to-one. One would speak of disharmony. Icosahedral Hamiltonian cycles can also be of this kind. If they are realized in the genetic code, the almost exact U ↔ C and A ↔ G symmetry is lost and the degeneracies of codons assignable to 20+20 icosahedral codons increase by one unit so that one obtains for instance degeneracy 7 instead of 6 not realized in Nature.
5. What can one say about the characer of toric harmonies?
What can one say about the character of toric harmonies on basis of this picture.
- It has been already found that the proposal involving three disjoint quartets of subsequent notes can reproduce the basic chords of basic major and minor harmonies. The challenge is to prove that it can be assigned to some Hamiltonian cycle(s).
The proposal is that the quartets are obtained by Z3rot symmetry from each other and that the notes of each quartet are obtained by Z4,tr symmetry.
- A key observation is that classical harmonies involve chords containing 1 quint but not 2 or no quints at all. The number of chords in torus harmonies is 24 =2× 12 and twice the number of notes. The number of intervals in turn is 36, 3 times the number of the notes. This allows a situation in which each triangle contains one edge of the Hamiltonian cycle so that all 3-chords indeed have exactly one quint.
- By the above argument harmony possesses Z2 symmetry or no symmetry at all and one has 12 codon doublets. For these harmonies each edge of cycle is shared by two neighboring triangles containing the same quint. A possible identification is as major and minor chords with same quint. The changing of the direction of the scale and the reflection with respect to the edges the Hamiltonian cycle would transforms major chords and minor chords along it to each other and change the mood from glad to sad and vice versa.
The proposed harmony indeed contains classical chords with one quint per chord and for F,A,C # both minor and major chords are possible. There are 4 transposes of this harmony.
- Also Hamiltonian cycles for which n triangles contain two edges of Hamiltonian path (CGD type chords) and n triangles contain no edges. This situation is less symmetric and could correspond to a situtation without any symmetry at all.
- One can ask whether the classical harmonies corresponds to 24 codons assignable to the toric harmony and to the 24 amino-acids being thus realizable using only amino-acids. If so, the two icosahedral harmonies would represent kind of non-classical exotics.
For a summary of earlier postings see Latest progress in TGD.
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