# Why it stinks living in five dimensions

I’ve been skimming through Coxeter’s Regular Polytopes. I nearly understand how to enumerate regular polytopes in n dimensions, and thought I’d put my paraphrase here.

There is one number that characterizes a regular (convex) polygon (convexity assumed throughout this): The number of edges (p). So a regular polygon can be described by {p}. For example, a square is {4}.

There are two numbers that characterize regular (convex) polyhedra: the number of edges on each face (p) and the number of faces around each vertex (q), so a regular polyhedron can be described by {p, q}. For example, a cube is {4, 3} (3 squares – that is, 3 {4} — at each vertex) while an octahedron is {3, 4} (4 triangles — 4 {3} — at each). {4, 4} is not a polyhedron — it’s a honeycomb: a square infinite lattice. But you can consider that a kind of degenerate polyhedron.

Consult Euclid if you want to know how to enumerate the regular polyhedra. There are five: {3, 3}, {3, 4}, {4, 3}, {3, 5}, and {5, 3}.

The vertex figure of a polyhedron is the polygon whose vertices are the center points of the edges that meet at a vertex of the polyhedron. Since each of the faces of the polyhedron that meet at a vertex includes two of the edges at that vertex, it follows that an edge of the vertex figure lies in each face of the polyhedron, so the number of edges in the vertex figure equals the number of faces surrounding each vertex. That means that for a polyhedron (or honeycomb) {p, q}, the faces are {p} and the vertex figures are {q}.

In n dimensions, an n-polytope is characterized by an (n–1)-polytope (the cell figure) and the number of cells meeting around each (n–2)-dimensional hyperplane. In 4 dimensions, if the cell figure is {p, q} and there are r cells around each edge, then the polytope can be denoted {p, q, r}. For example, a hypercube has 3 cubes meeting at each edge. A cube is {4, 3}. So a hypercube is {4, 3, 3}. In 5 dimensions a polytope is denoted {p, q, r, s} (s {p, q, r} around each face), and so on.

The vertex figure of an n-polytope is an (n–1)-polytope whose vertices are the midpoints of the edges meeting at a vertex. Each of the (n–1)-polytopes meeting at a vertex has its own (n–2)-polytope vertex figure which is a cell of the vertex figure of the n-polytope. So for example, for a 4-dimensional hypercube, the vertex figure is a polyhedron each of whose cells (i.e. faces) is the vertex figure of a cell of the hypercube; that is, the vertex figure of a cube — which is a triangle. There are 3 cubes around each edge, so 3 triangles around each vertex in the vertex figure: a tetrahedron. In numbers, a hypercube is {4, 3, 3}; its cell is {4, 3} and there are 3 at each edge; the vertex figure of a cell is {3}, 3 of which around each vertex form the vertex figure of the hypercube, {3, 3}. More generally a {p, q, r} has cells {p, q} and vertex figures {q, r}. And even more generally, in n dimensions, a {p, q, r, … y, z} has cells {p, q, r, … y} and vertex figures {q, r, … y, z}.

So right away, we can limit the possible regular 4-polytopes {p, q, r} to ones where both {p, q} and {q, r} are regular polyhedra. That is, we can consider: {3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 3}, {4, 3, 3}, {4, 3, 4}, {4, 3, 5}, {3, 5, 3}, {5, 3, 3}, {5, 3, 4}, and {5, 3, 5}, and nothing else.

Now comes a lot of trigonometric rigamarole, too much to go into here but not hard to follow; it’s in Coxeter section 7.7. It leads to Schäfli’s criterion, a necessary (but not sufficient) condition for the existence of a regular polytope in the form of an inequality relating the integers p, q, r, … y, z. In particular, in 3 dimensions,

1/p + 1/q ≥ 1/2

which is satisfied, of course, by all the regular polyhedra (with >) and the one regular 3-honeycomb (with =). In 4 dimensions,

sin π/p sin π/r ≥ cos π/q

and in 5 dimensions,

(cos2 π/q) / (sin2 π/p) + (cos2 π/r) / (sin2 π/s) ≤ 1

So what does this tell us? Using the criterion for 4 dimensions, we find that the allowed polytopes are:

{3, 3, 3}, {3, 3, 4}, {3, 3, 5}, {3, 4, 3}, {4, 3, 3}, and {5, 3, 3}

The allowed honeycomb is:

{4, 3, 4}

The following are eliminated:

{4, 3, 5}, {3, 5, 3}, {5, 3, 4}, and {5, 3, 5}

Then for 5 dimensions we can combine pairs of 4-polytopes and apply the criterion; we get the following allowed polytopes:

{3, 3, 3, 3}, {3, 3, 3, 4}, and {4, 3, 3, 3}

Allowed honeycombs:

{3, 3, 4, 3}, {3, 4, 3, 3}, and {4, 3, 3, 4}

Excluded:

{3, 3, 3, 5}, {4, 3, 3, 5}, {5, 3, 3, 3}, {5, 3, 3, 4}, {5, 3, 3, 5}

But now in 6 dimensions, the only possible polytopes generated by pairs of regular 5-polytopes are:

{3, 3, 3, 3, 3}, {3, 3, 3, 3, 4}, and {4, 3, 3, 3, 3}

and the only regular honeycomb is

{4, 3, 3, 3, 4}.

And clearly in all higher dimensions, all we can get are polytopes {3, …, 3}, {3, …, 4}, and {4, …, 3} and honeycomb {4, …, 4} (where the ellipses are all 3s). The polytopes are respectively simplexes, cross polytopes, and measure polytopes (analogs of tetrahedra, octahedra, and cubes) which are evidently existent in all dimensions. The 4-polytopes {3, 3, 5}, {3, 4, 3}, and {5, 3, 3} have to be constructed to show they exist, and they do. (That’s in Coxeter too, and is the hard part.)

So: the number of regular polytopes in n dimensions is:

infinite (n = 2)

5 (n = 3)

6 (n = 4)

3 (n ≥ 5)

How boring, to live in a number of dimensions with only 3 regular polytopes!