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.