From the 120-cell to E8 (Part 2)

So far we have 120 points on the unit sphere in 4 dimensions, with coordinates as follows (since the last section, I've swopped two axes):

(±1, 0, 0, 0) and permutations thereof (8 points)
(0, ±1/2φ, ±1/2, ±φ/2) and even permutations thereof (96 points)
(±1/2, ±1/2, ±1/2, ±1/2) (16 points).

Each point has 12 nearest neighbours, all at distance 1/φ. By joining up all the nearest neighbours we obtain the skeleton of a regular polytope in 4-dimensions, having 600 tetrahedral-cells. The 120 points may also be interpreted as the cell-centres of the 120-celled polytope, whose cells are dodecahedra.

Now consider the lattice generated by the position vectors of the 120 points. Each lattice-point is a linear combination of the 120 vectors, with integer coefficients.

Adding a point to an almost opposite one we get the difference vector between nearby points, which, as we've seen, will be one of the original points, divided by φ (=1.61803...). So the lattice will contain a scaled-down copy of the original 120 points. But it will also contain the difference vectors between points of the scaled-down copy which will lead to a second copy, scaled down still further - and so on. A tiny sphere, centred on the origin, will contain an infinity of lattice-points; and the same goes for any other point of the lattice.

In a manner of speaking, there are too many lattice-points to fit into 4 dimensions. The result - the lattice-points pile up almost on top of one another. The remedy - to move into more than 4 dimensions!

The coordinates of the 120 points (and hence of all the lattice-points) are a mixture of rational and irrational parts. Suppose we separate them out and assign the irrational parts to 4 new dimensions? There are many ways one could do this, but in the present case a very simple method presents itself. All the coordinates arising in the lattice consist of a rational number plus a rational multiple of φ (1/φ is no problem because it can be written φ-1). In other words, they can be written as a+bφ, with a and b rational. In fact, things will work out a bit neater if we use as our 'fundamental' irrational not φ, but ψ = φ-1. Each coordinate can be written a+bψ, and to it we can assign the pair of rational numbers (a,b). Thus

φ = 1+ψ goes to (1,1)
1 goes to (1,0)
1/φ = ψ goes to (0,1) and
1/φ² = 1-ψ goes to (1,-1).

To the point (0, 1/2φ, 1/2, φ/2), for example, we assign the rational 8-tuple

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2).

The even permutations of the coordinates then give rise to even permutations of the pairs

(0, 0), (0, 1/2), (1/2, 0), (1/2, 1/2).

If any pair is replaced by its negative, we get another point belonging to the set of 120. The point (1/2, 1/2, 1/2, 1/2) translates as

(1/2, 0, 1/2, 0, 1/2, 0, 1/2, 0).

Subtracting this from the point already considered we get

(-1/2, 0, -1/2, 1/2, 0, 0, 0, 1/2),

which does not belong to the original set because of the pair (-1/2, 1/2) in second place (this pair corresponds to the real number -1/2φ²). Similarly, we obtain all even permutations of the pairs

(0, 0), (0, 1/2), (1/2, 0), (1/2, -1/2),

and any of these pairs can be replaced by its negative. Taking the difference of the points

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2) and (0, 0, 0, -1/2, 1/2, 0, 1/2, 1/2)

we obtain

(0,0,0,1,0,0,0,0).

Clearly that 1 could be in any of the 8 places, which implies that any 8-tuple of integers belongs to the lattice. Finally, we note that the sum of

(0, 0, 0, 1/2, 1/2, 0, 1/2, 1/2)

and its even permutation

(1/2, 1/2, 1/2, 0, 0, 1/2, 0, 0)

is

(1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2),

from which we could if we wanted subtract

(1/2, 0, 1/2, 0, 1/2, 0, 1/2, 0)

to obtain

(0, 1/2, 0, 1/2, 0, 1/2, 0, 1/2).

From all of this it follows that any 8-tuple of the following form belongs to the lattice: start with an arbitrary 8-tuple of integers (possibly negative), then add a half-integral part taking one of the following forms:

0000, 1111, 2222, 3333, 0123 or any even permutation of 0123.

Here 0 stands for the pair (0,0), 1 for (0, 1/2), 2 for (1/2, 0), 3 for (1/2, 1/2).

However, this set of 8-tuples turns out to be closed under addition, which implies that the lattice consists exactly of this set of 8-tuples. Why it should be closed under addition is not obvious to me - perhaps it is to you? It's all a question of those half-integer parts. When we add them, we make take 1/2+1/2 = 1 to be equivalent to 0, since we are not worried about the integer parts. So we have the addition table for pairs

 | 0 1 2 3
---------
0| 0 1 2 3
1| 1 0 3 2
2| 2 3 0 1
3| 3 2 1 0

(Klein's four-group). Note that the rows are all even permutations of one another, which implies that when 1111 (or 2222 or 3333) is added to 0123 we get an even permutation of 0123. We don't need the complete addition table for pair-quadruples, but note that if the column under 0000 is

0000
1111
2222
3333
0123
0312
0231
1032
1203
1320
2301
2130
2032
3210
3021
3102

then the column under 0123 (obtained by adding 0123 to the previous column) will be

0123
1032
2301
3210
0000
0231
0312
1111
1320
1203
2222
2013
2130
3333
3102
3021.

Note that the origin has 240 nearest neighbours in the lattice, all at (8-dimensional) distance 1. These are

(±1, 0, 0, 0, 0, 0, 0, 0) and permutations (16 points)
(±1/2, 0, ±1/2, 0, ±1/2, 0, ±1/2, 0) (16 points)
(0, ±1/2, 0, ±1/2, 0, ±1/2, 0, ±1/2) (16 points)
(0, 0, 0, ±1/2, ±1/2, 0, ±1/2, ±1/2) and all even permutations of the four pairs (192 points).

From the 120-cell to E8 (Part 1)

Take three golden rectangles made of cardboard. Make a slot down the centre of each one, parallel to the longer side, and equal in length to the shorter side. Insert the second rectangle into the slot in the first one, thus:

Finally, insert no. 3 into the slot in no.2, at the same time contriving that no.1 goes through the slot in no.3 (!).

Then the 12 vertices of the 3 rectangles mark out the 12 vertices of a regular icosahedron.

To see this, let the short side of the rectangles be 1, the long side φ (the golden number, such that φ² = φ+1). We shall show that the length of the vector joining a vertex of rectangle no.1 to a nearby vertex of no.2 is 1. In a suitable coordinate system, the coordinates of the first point are (φ/2,0,1/2), and of the second, (1/2, φ/2,0); so the coordinates of the vector are (-(φ-1)/2, φ/2, -1/2) or (-1/2φ, φ/2, -1/2). So the square of the length of the vector is

(1/φ² + φ² + 1)/4 = (φ² - 2φ + 1 + φ² + 1)/4 = 1,

as was to be shown. Thus the length of the vector is the same as the short side of a rectangle. Consequently, any vertex is surrounded by five nearby vertices all at distance 1. Moreover, each of the five is separated from its neighbours by the same distance. Thus the vertices enclose 20 equilateral triangles, and thus mark out the vertices of a regular icosahedron.

Note that the coordinates of the 12 vertices are (±φ/2,0,±1/2) , and all even permutations of that form (i.e. permutations obtained by an even number of exchanges: the even permutations of abc are bca - obtained by swopping the first two elements, then the second two - and cab).

Ready for four dimensions? Consider the set of points

(±1, 0, 0, 0) and permutations thereof
(±φ/2, ±1/2, 0, ±1/2φ) and even permutations thereof
(±1/2, ±1/2, ±1/2, ±1/2).

There are 8 of the first sort, 8 times 12 of the second, 16 of the third: altogether, 8 times 15 or 120. From the calculation we already did for the icosahedron, we see that each of these 120 points lies at distance 1 from the origin: in other words, they all lie on a (hyper)sphere of radius 1.

Now consider the nearest neighbours of (1,0,0,0). These are (φ/2, 1/2, 0, 1/2φ) and 11 other points obtained by evenly permuting the last three coordinates, and plus-or-minus-ing them. Components of the vector:

(-(2-φ)/2, 1/2, 0 , (φ-1)/2).

Observe that 2-φ is the square of φ-1 and so equals 1/φ²; so the vector is equal to

(-1/2φ, φ/2, 0, 1/2)

divided by φ. So its length is 1/φ. (In fact, we note that the vector times φ is actually the position vector of one of the original 120 points.)

It is clear that the distance between neighbours among the 12 points at height φ/2 is also 1/φ.

We won't go through the calculations, but it is straightforward to show that each and every one of the 120 points has 12 nearest neighbours at distance 1/φ. The points are thus disposed in a highly symmetrical manner on the (3-dimensional) 'surface' of the hypersphere, and are in fact the vertices of 600 tetrahedra, which fit together to form a 600-celled polytope, which is the 4-dimensional analogue of the 3-dimensional icosahedron.

The topmost point, (1,0,0,0), lies 'inside' the 12 points at level φ/2, which mark out the vertices of an icosahedron. Joining up vertices, we get 20 tetrahedra all sharing the topmost point as vertex. 120 times 20 makes 2400, but each tetrahedron has 4 vertices so we divide by 4 to obtain 600 as the number of tetrahedra.

The 'dual' of the 600-cell (each of whose cells is a tetrahedron) is the 120-cell (each of whose cells is a dodecahedron). To see this, consider the 20 tetrahedra which meet at a given vertex of the 600-cell. The centres of the tetrahedra are symmetrically disposed, and mark out the vertices of a dodecahedron. The geometrical centre of the dodecahedron does not quite coincide with the given vertex, but lies on the radius joining the origin to the vertex.

A neighbouring vertex will give rise to a second dodecahedron. But its cluster of 20 tetrahedra will have 5 in common with the cluster of the first vertex. This implies that the two dodecahedra have a pentagon in common: their pentagonal faces will touch. The dodecahedra will be 'glued' together.

In the same way, each vertex of the 600-cell gives rise to a dodecahedron; and the 120 dodecahedra are glued together, thus forming the cells of a second regular polytope, the 120-cell.

What we like about the Cosmic PDS Hypothesis

We must recognize that this whole PDS business may turn out to be moonshine - without basis in fact. If that's the case, then, with heavy heart, we will have to turn away from it, and look to pastures new. Meanwhile, there's no reason why we shouldn't take a look at some of the reasons we would like it to be true.

Firstly, there is the long-standing 'mysticism of the dodecahedron'.

***Plato, in his Timaeus, says of it (after describing the other four so-called 'Platonic' solids):

There still remained a fifth construction, which the God used for embroidering the constellations on the whole heaven.

The other four solids were assigned to the four terrestrial elements (fire, air, water and earth). The dodecahedron was set apart and associated with the heaven, i.e. with the sphere of the fixed stars, thought of as an outermost skin or shell of the (physical) world.

***Proposition 17 of book 13 of Euclid's Elements, the second last proposition of the whole book, explains how to construct a dodecahedron inside a sphere (touching it in 20 points). Its position in the book makes it hard not to think of this construction as a sort of grand finale of the whole work.

***Johannes Kepler was extremely susceptible to this sort of mysticism. He said that the greatest moment of his life was when he realized, in a moment of dazzling illumination , that the orbits of the planets were determined by celestial polyhedra, nested inside one another, as detailed in Kepler's famous diagram:

***Maybe Salvador Dali had the last word on dodecahedron-mysticism, in the shape of his painting of 1955, entitled 'Sacrament of the Last Supper'. The painter described it as an "arithmetic and philosophical cosmogony based on the paranoiac sublimity of the number twelve".

Secondly, there is the curious coincidence that if Roukema is right, our sphere of visibility, here and now (limited by our 'cosmological horizon') has almost the same diameter as the cosmic dodecahedron. If we had occurred much earlier in the history of the cosmos, our visibility sphere would have been much smaller than the dodecahedron; much later, and it would have been much greater than the dodecahedron - big enough, in fact, to contain many copies of the dodecahedron.

To some minds, this coincidence is so absurd that it seems to be a reason to reject the hypothesis. Myself, I relish the absurdity, the unlikeliness of it. It's as good as the fact that we live in an era when the moon is just big enough in the sky to blot out the sun when she stands in front of him. In times to come, the moon will have receeded still further from the earth and will be too small to blot out the sun, and there will no total eclipses (just annular ones).

Thirdly, there is the close link between Poincaré Dodecahedral Space and a beautiful mathematical object which exists in four dimensional space: the 120-cell or dodecaplex. In four dimensions there exist highly symmetrical solids (or rather, hypersolids) which are the analogues in 4 dimensions of the regular polyhedra in 3 dimensions. Their king, the 120-cell, will be the subject of future posts.

The 120-cell is linked in turn to the remarkable sphere-packing lattice in 8 dimensions known as E8, which is linked in turn to the biggest of the exceptional Lie groups, also called E8. Some people think that this group may hold the key to particle physics (see, for example, Garrett Lisi's Exceptionally Simple Theory of Everything). Clearly, it would be 'kinda neat' if the particles and forces of the universe turned out to be modelled on the 'ultimate' Lie group - and even better if this was somehow tied in with the topology of the universe.

The first link in this chain (120-cell to E8) will be the subject of my next post.