### Carbon Beauty III

Observe how the buckyball trinity builds the prime $p = 11$ from lower primes, particularly the prime 5. Not only do we use the Hecke group $H5$, but the 11 buckyballs (truncated icosahedra) defining the genus 70 curve each have 60 vertices, which come from 5 copies of the 12 vertex truncated tetrahedron, which has 4 hexagonal and 4 triangular faces. Another choice for the Euler structure $(V,E,F)=(12,18,8)$ is the 4 pentagon and 4 square faces of the third ball in the ternary geometry of the cube. And since buckyballs are mixtures of pentagons and hexagons, the buckyball trinity averages these geometries.

Note also that the $p = 7$ truncated cube has $(V,E,F)=(24,36,14)$, which is the same Euler structure as the permutohedron. Thus all three truncated Platonic geometries have a double Euler structure. The $3 \times 8$ splitting of the 24 vertices of the permutohedron is like the pairing of squares to form three cubes in M Theory, or the three squares of the associahedron, which was associated to the crossings defining the trefoil knot in $\mathbb{R}^{3}$, the complement of which has the cover of the modular group, namely the braid group on three strands, as fundamental group. Gee, if I repeat that a lot it's because I find it very, very interesting!

Note also that the $p = 7$ truncated cube has $(V,E,F)=(24,36,14)$, which is the same Euler structure as the permutohedron. Thus all three truncated Platonic geometries have a double Euler structure. The $3 \times 8$ splitting of the 24 vertices of the permutohedron is like the pairing of squares to form three cubes in M Theory, or the three squares of the associahedron, which was associated to the crossings defining the trefoil knot in $\mathbb{R}^{3}$, the complement of which has the cover of the modular group, namely the braid group on three strands, as fundamental group. Gee, if I repeat that a lot it's because I find it very, very interesting!

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