# 5 Smooth Numbers And Geometry

 Posts 53
Casual Member
Myrtone
Casual Member
Joined: Mar 22 2010, 08:08 AM
There are three regular tilings on a plane, equilateral triangles, squares and hexagons, note that six equilateral triangles can be arranged as a regular hexagon. In any case, the number of sides of these polygons, this also being the number of corners, is always 3-smooth.

There are five platonic solids, and I have noticed that the number of edges, faces and corners of a platonic solid is always 5-smooth,* specifically, it is a factor of 5! or twelfty as Wendy calls it.

All three regular planar tilings and all five platonic solids, consist of regular polygons constructible with a compass and straight-edge.

So 5-smooth numbers seem to have some significance in geometry.

*All 3-smooth numbers are also 5-smooth, but some 5-smooth numbers are not 3-smooth.

 Posts 773
Dozens Disciple
Oschkar
Dozens Disciple
Joined: Nov 19 2011, 01:07 AM
This has to do with the fact that the symmetry group in 3 dimensions with the largest order is the icosahedral group, which has exactly order 120.

You really canâ€™t expect any primes larger than 5 in the order of a symmetry group, since its associated polyhedron should have a face with a multiple of that number of sides. Three heptagons canâ€™t meet at a vertex in spherical space, so there canâ€™t be a regular polytope of any dimension greater than 2 with heptagons for faces, which would mean that there shouldnâ€™t be any polyhedral symmetry group with 7 as a factor. Only in six dimensions and onward do we see the factor 7 in the order of a point group, but thatâ€™s because the six-dimensional simplex has 7 vertices. The same argument applies for all higher primes: there are no uniform polytope symmetry groups with prime p in their prime factorization from dimension 3 to p âˆ’ 1.

I donâ€™t know enough about this to prove that there canâ€™t be a 3D symmetry group with order, say, 1440, but at least I tried to explain why this sort of numbers have to be 5-smooth.

 Posts 53
Casual Member
Myrtone
Casual Member
Joined: Mar 22 2010, 08:08 AM
Oschkar @ May 13 2016, 07:56 AM wrote:This has to do with the fact that the symmetry group in 3 dimensions with the largest order is the icosahedral group, which has exactly order 120.
I'm not familiar with symmetry groups, has this been discussed before.

So there are five platonic solids and the symmetry group with the largest order is the factorial of the number of platonic solids.
Oschkar @ May 13 2016, 07:56 AM wrote:You really canâ€™t expect any primes larger than 5 in the order of a symmetry group, since its associated polyhedron should have a face with a multiple of that number of sides. Three heptagons canâ€™t meet at a vertex in spherical space, so there canâ€™t be a regular polytope of any dimension greater than 2 with heptagons for faces, which would mean that there shouldnâ€™t be any polyhedral symmetry group with 7 as a factor. Only in six dimensions and onward do we see the factor 7 in the order of a point group, but thatâ€™s because the six-dimensional simplex has 7 vertices. The same argument applies for all higher primes: there are no uniform polytope symmetry groups with prime p in their prime factorization from dimension 3 to p âˆ’ 1.
Okay, so this explains why the number of edges, faces and corners of a platonic solid is always 5-smooth, specifically a factor of twelfty.

But what about that fact that the two fundamental regular polygons that tessellate on a plane (six equilateral triangles meeting at a common point make a regular hexagon) both have 3-smooth numbers of sides?
Oschkar @ May 13 2016, 07:56 AM wrote:I donâ€™t know enough about this to prove that there canâ€™t be a 3D symmetry group with order, say, 1440, but at least I tried to explain why this sort ofnumbers have to be 5-smooth.
But can you, or anyone else here, prove that a regular polygon that tessellates on a plane must have a 3-smooth number of sides, specifically a factor of twelve?

Let E be the number of corners, K be the number of edges and F be the number of faces.

Apply this to the following platonic solids and the result is as follows:

*Tetrahedron - 30E+10K+30F = 360
*Cube - 15E+10K+20F and Octahedron - 20F+10K+15F both equal 360
*Dodecahedron - 6E+4K+10F and Icosahedron - 10E+4K+6F both equal 360.

wendy.krieger
wendy.krieger
The actual symmetries that occur in folded geometry are not smooth numbers at all.

All of those nifty theorems designed to describe Platonic symmetries, are eventually used to look in group theory, which at best, describe colour patterns on the infinite tilings, or polytopes that exist abstractly. The idea of 'folded geometry' is then that we can identify a cell that is a left-turning knight's move away, or something that is 2Â½ hexagons one way, and up one edge.

The different regular colourings of the square lattice, represent the gaussian integers, and one finds separate primes of the form 2, 4x+1 (eg 5, 13, 17, 29, 37, ... can divide the square of radius, without dividing the coordinates, and it is possible to divide the square tiling into tilings of these numbers).

The hexagonal tiling represents the Eisenstein integers, where the primes are of the form 3, 6x+1. One finds then 3, 7, 13, 19, 31, 37, ... in the squares of radii.

The primes that appear in the pentagon figures are 5, 10x+1, 10x+9, so one finds 11, 19, 29, 31, ... as potential numbers. Divisors 11 and 19 appear among the chords of the {3,3,5} and {5,3,3}, for example.

The groups that one finds in abstractly extending the regular definitions to 'folded geometries' (Coxeter and Moser wrote a book on this 'Generators of Finite Groups and Maps'), are numbers like 660, 1092, and their various multiples, the actual smooth numbers are not really in this series.

Coxeter describes things like the 'eleven-cell' and '57-cell', derived from the abstract painting of {3,5,3} and {5,3,5} respectively, but these do not lead to multiples of 7, the first has the same abstract group of 660 as the tiling of {5,5}_5.

One of the earliest numbers to turn up in my investigation of the heptagon, is 1079. This is 13*83, and one finds that the heptagon-numbers are those of the form 7, 14x+1, 14x-1.

Smooth numbers turn up in the general euclidean dimensions, because they involve a lot of simplex-symmetries: in fact of all of the mirror groups in all dimensions, only one does not have a simplex-symmetry for its face: {3,4,3}.