The six regular polytopes

Every dimension has its most symmetric shapes — the regular ones. In the plane they are the regular polygons, in space the five Platonic solids. In four dimensions there are six, and one of the six is unlike anything in any other dimension. In this chapter we will work out why there are exactly that many, build each one from coordinates, and fold up a tesseract from its net — the very net Salvador Dalí painted.

Polygons and the five solids

A regular polygon is a convex polygon with all sides equal and all angles equal. For every $p \ge 3$ there is one: place $p$ points around a circle at equal angles of $360°/p$ and join neighbours. There are infinitely many, and they are written simply $\{p\}$: $\{3\}$ is the equilateral triangle, $\{4\}$ the square, $\{5\}$ the regular pentagon. The angles of a $p$-gon add up to $180°(p-2)$, so each angle of a regular $p$-gon is $180° - 360°/p$: 60° for the triangle, 90° for the square, 108° for the pentagon, 120° for the hexagon.

A regular polyhedron is a convex polyhedron whose faces are all the same regular polygon, with the same number of faces at every vertex. If the faces are $p$-gons and $q$ of them meet at each vertex, the polyhedron is written $\{p, q\}$. There are five of them, known since antiquity: Euclid’s Elements ends by constructing them and proving there are no others. They are the tetrahedron $\{3,3\}$, the cube $\{4,3\}$, the octahedron $\{3,4\}$, the dodecahedron $\{5,3\}$ and the icosahedron $\{3,5\}$.

Why only five? It all comes down to one simple observation about vertices.

Intuitively the lemma is obvious: cut a cardboard corner along one edge and flatten it on a table — the faces will not close up, a gap remains. If there is no gap, the angles add up to exactly 360° and lie flat, and you do not get a corner.

The equality $(p-2)(q-2) = 4$ is interesting too: $\{3,6\}$, $\{4,4\}$ and $\{6,3\}$ — six triangles, four squares or three hexagons around a point. The angles add up to exactly 360°, and instead of a solid we get an infinite tiling of the plane. There are three regular tilings, and below we will meet their four-dimensional relatives.

We will also need the dihedral angle of a polyhedron — the angle between two faces along their common edge. It is convenient to compute from $\sin(\delta/2) = \cos(\pi/q) / \sin(\pi/p)$; for the cube, for example, $\cos 60° / \sin 45° = \tfrac{\sqrt2}{2}$, so $\delta/2 = 45°$ and $\delta = 90°$, as it should be.

The Schläfli symbol

The notations $\{p\}$ and $\{p,q\}$ are the start of a chain devised by the Swiss mathematician Ludwig Schläfli. Its logic is simple: each new number says how many of the previous shapes meet around the next kind of element.

• $\{p\}$ is a polygon: $p$ edges, two at each vertex.
• $\{p,q\}$ is a polyhedron: faces $\{p\}$, $q$ around each vertex.
• $\{p,q,r\}$ is a four-dimensional solid: its three-dimensional cells are polyhedra $\{p,q\}$, $r$ of them around each edge.

The four-dimensional counterpart of a polyhedron is called a polychoron, or more often simply a 4-polytope: it is bounded by three-dimensional cells, neighbouring cells are glued along two-dimensional faces, faces along edges. A polytope $\{p,q,r\}$ is regular if all its cells are identical regular polyhedra $\{p,q\}$ and the same number $r$ of them surround every edge.

The symbol has a second meaning as well. Slice a corner off the polyhedron $\{p,q\}$ close to a vertex: the cut is a regular $q$-gon, the vertex figure. For a polytope $\{p,q,r\}$ the cut near a vertex is three-dimensional, and it is the polyhedron $\{q,r\}$. In the tesseract $\{4,3,3\}$, for instance, four edges leave each vertex, and their far ends are all equally far from each other: they form a regular tetrahedron $\{3,3\}$. So the symbol reads from both ends: the left tells you what the cells are made of, the right what the neighbourhood of a vertex looks like.

This also explains duality. Put a point at the centre of every cell and join the centres of neighbouring cells: you get a new regular polytope, and its symbol is the old one read backwards, $\{p,q,r\} \to \{r,q,p\}$. In three dimensions this turns the cube $\{4,3\}$ into the octahedron $\{3,4\}$ and the dodecahedron into the icosahedron.

Why exactly six

For polyhedra, the proof rested on the vertex: the faces around it, flattened out, must leave a gap. In four dimensions the same idea works one level up — not for faces around a vertex, but for cells around an edge.

Stand at the midpoint of an edge and look in every direction perpendicular to it. In four-dimensional space there are three such directions: the perpendiculars to a line form a three-dimensional space. In that space each cell containing the edge appears as a flat wedge — its cross-section by the plane perpendicular to the edge. The angle of the wedge at the edge is exactly the cell’s dihedral angle $\delta$. There are $r$ cells around the edge, so $r$ wedges, and they close up into a convex polyhedral angle.

The theorem says “at most six”. In fairness, two remarks. First, there is a second necessary condition, at the vertex: the vertex figure $\{q,r\}$ must itself be a Platonic solid. The six satisfy it — their vertex figures are the tetrahedron, octahedron, icosahedron, tetrahedron, cube and tetrahedron. That is no accident. Substitute the formula for the dihedral angle: $r\delta < 2\pi$ means $\delta/2 < \pi/r$, that is, $$\sin\frac{\pi}{p}\,\sin\frac{\pi}{r} > \cos\frac{\pi}{q}.$$ The inequality is symmetric in $p$ and $r$, so the edge condition for $\{p,q,r\}$ is the same as the edge condition for its dual $\{r,q,p\}$, and checking the cases shows the vertex condition removes nothing beyond it. For instance, $\{3,5,3\}$ is not ruled out by its vertex — its vertex figure $\{5,3\}$, a dodecahedron, is perfectly fine — but by the fact that three icosahedra do not fit around an edge: $414.6° > 360°$.

Second, the inequality is only a necessary condition. That all six really exist is proved by construction: below each one has vertex coordinates, and the showcase widget builds cells, faces and edges from these coordinates and counts them itself.

The six

Here they all are. The numbers are the vertices $V$, edges $E$, two-dimensional faces $F$ and cells $C$.

The card under the showcase is not copied from the table: the widget builds the polytope from its vertex coordinates (cells are the sets of vertices that are extreme in the direction of a normal, faces are where neighbouring cells meet) and counts everything from the resulting structure — including how many cells meet at an edge and how many edges leave a vertex.

The 5-cell: a four-dimensional tetrahedron

Five vertices, each joined to every other: 10 edges, 10 triangles, 5 tetrahedra — any two, three or four of the five vertices. The simplest coordinates live in five-dimensional space: the points $(1,0,0,0,0)$, $(0,1,0,0,0)$, …, $(0,0,0,0,1)$ are all $\sqrt2$ apart and lie in the four-dimensional hyperplane $x_1 + \dots + x_5 = 1$. This is the simplex; there is one in every dimension, and the $n$-dimensional one has $n + 1$ vertices.

The tesseract and the 16-cell

The tesseract’s vertices are the sixteen points $(\pm1, \pm1, \pm1, \pm1)$, and its eight cubic cells lie in the hyperplanes $x = \pm1$, $y = \pm1$, $z = \pm1$, $w = \pm1$. The centres of those cells are the eight points $(\pm1, 0, 0, 0)$, $(0, \pm1, 0, 0)$, $(0, 0, \pm1, 0)$, $(0, 0, 0, \pm1)$ — the vertices of the dual 16-cell. It is the four-dimensional octahedron: each vertex is joined to all others except its opposite (24 edges), and the 16 tetrahedral cells come from taking one vertex from each of the four axes, $2^4 = 16$ ways. Four cells meet around each of its edges — the only one of the six with $r = 4$. If you think of points of four-dimensional space as quaternions, the 16-cell’s vertices are $\pm1, \pm i, \pm j, \pm k$, and they form a group under multiplication.

The 24-cell: the one without a partner

Twenty-four vertices: all permutations of $(\pm1, \pm1, 0, 0)$ — two of the four places for the ones, $\binom42 = 6$ ways, and four choices of signs. The cells are 24 regular octahedra, three around each edge; the vertex figure is a cube. Its dual is itself (turned by 45°). All simplices are self-dual too, but the 24-cell is the only regular self-dual polytope in any dimension that is neither a polygon nor a simplex. It has no regular counterpart in three dimensions, nor in five or more; its closest three-dimensional relative is the cuboctahedron, which has two kinds of faces.

You can build it from a tesseract. On each of the eight cubic cells of the tesseract $(\pm1, \pm1, \pm1, \pm1)$ erect a four-dimensional pyramid with its apex at $(\pm2, 0, 0, 0)$ and so on. That gives 24 vertices, 16 old and 8 new. Each square face of the tesseract now carries two square pyramids, one from each of the two cubes that share it, and these two lie in the same hyperplane: for the face $(1, 1, \pm1, \pm1)$ the apexes $(2, 0, 0, 0)$ and $(0, 2, 0, 0)$ satisfy the same equation $x + y = 2$. The two pyramids fuse into a regular octahedron with edge 2. The tesseract has 24 square faces — hence 24 octahedral cells. In three dimensions the same trick (six square pyramids on a cube) gives the rhombic dodecahedron, which is not regular.

The 24-cell also has a property it shares, among the six, only with the tesseract and the 16-cell: identical copies of these shapes, placed cell to cell, fill the whole of four-dimensional space without gaps, as cubes fill ours. Quaternions again give a neat description: scaled down by half, its vertices are $\pm1, \pm i, \pm j, \pm k$ together with the sixteen numbers $\tfrac12(\pm1 \pm i \pm j \pm k)$ — the twenty-four Hurwitz units, which also form a group.

The 600-cell and the icosians

One hundred and twenty vertices on the unit 3-sphere: 8 points like $(\pm1, 0, 0, 0)$, 16 points $(\pm\frac12, \pm\frac12, \pm\frac12, \pm\frac12)$ and 96 points $\frac12(\pm\varphi, \pm1, \pm\varphi^{-1}, 0)$ with all even permutations of the coordinates, where $\varphi = \frac{1+\sqrt5}{2}$ is the golden ratio. There are 600 cells, all tetrahedra, five around each edge and twenty around each vertex (the vertex figure is an icosahedron). As quaternions these 120 points are the icosians, and they are closed under multiplication: a group of 120 elements, the binary icosahedral group. Each of the 60 rotations that carry an icosahedron onto itself corresponds to two of its quaternions, $\pm q$ — the ones that give that rotation by $v \mapsto q v \bar q$ (see the chapter on rotation).

The 120-cell and its rings of dodecahedra

It is the dual of the 600-cell: its 600 vertices are the centres of the 600 tetrahedra, and its 120 dodecahedral cells correspond to the 120 vertices of the 600-cell. That is exactly how our showcase builds it. The angle between neighbouring cells is 144°, so the centres of adjacent dodecahedra are $180° - 144° = 36°$ apart as seen from the centre, and exactly ten dodecahedra stacked face to face close up into a ring along a great circle. All 120 cells split into 12 such rings of 10, and the rings are interlinked just like the circles of the Hopf fibration. The 600-cell has a similar picture: its 600 tetrahedra stack into 20 twisted helices of 30. The “Hopf” colouring on the showcase paints each ring its own colour.

Euler in four dimensions

Work out the alternating sum $V - E + F - C$ from the table:

$$5 - 10 + 10 - 5 = 16 - 32 + 24 - 8 = 8 - 24 + 32 - 16 = 24 - 96 + 96 - 24 = 0,$$ $$600 - 1200 + 720 - 120 = 120 - 720 + 1200 - 600 = 0.$$

For polyhedra $V - E + F = 2$, but for 4-polytopes the same sum is zero — and not only for regular ones but for every convex one. This is a case of the Euler–Poincaré formula: the boundary of a convex $n$-dimensional solid is a sphere of dimension $n - 1$, and the alternating sum depends only on that sphere. For the 2-sphere it is 2, for the 3-sphere 0. Schläfli stated the formula for every dimension in that same work, but his proof was incomplete; Henri Poincaré gave a rigorous one in the 1890s. Why zero, specifically, can be seen from duality. In the dual polytope vertices and cells swap places, as do edges and faces, so the sum becomes $C - F + E - V$ — it changes sign. If it is the same for all convex 4-polytopes, it equals minus itself, and so it is zero. In three dimensions duality swaps only $V$ and $F$, the sum $V - E + F$ keeps its sign, and nothing stops it from being 2.

Five dimensions and up: only three

The same test works one more level up. A regular five-dimensional solid $\{p,q,r,s\}$ is made of four-dimensional cells $\{p,q,r\}$ — necessarily one of our six — with $s$ of them around each two-dimensional face. Perpendicular to a two-dimensional face in five dimensions there are again three directions, so again we need $s\delta < 360°$, where $\delta$ is now the angle between neighbouring cells of the four-dimensional solid. Those angles are $\arccos\frac14 \approx 75.52°$ for the 5-cell, 90° for the tesseract, 120° for both the 16-cell and the 24-cell, 144° for the 120-cell and $\approx 164.48°$ for the 600-cell.

• 5-cell: $s = 3$ and $s = 4$ pass ($226.6°$ and $302.1°$), $s = 5$ does not. That gives $\{3,3,3,3\}$ and $\{3,3,3,4\}$, the five-dimensional simplex and “octahedron”.
• Tesseract: $s = 3$ gives the five-dimensional cube $\{4,3,3,3\}$; $s = 4$ is exactly 360°, a honeycomb of tesseracts.
• 16-cell and 24-cell: $s = 3$ is exactly 360° — two more honeycombs, $\{3,3,4,3\}$ and $\{3,4,3,3\}$. These are the fillings of four-dimensional space mentioned above.
• The 120-cell and the 600-cell do not pass at all.

Three solids remain: the simplex, the cube and the cross-polytope (the generalised octahedron). All three exist, and from here on the picture repeats in every dimension: the dihedral angles of these three grow with the dimension, and nothing new can be assembled from them. So four dimensions is the richest case: it has more regular solids than any other dimension except two.

Unfolding: Dalí’s cross

A cube can be cut along its edges and unfolded into the plane — you get a cross of six squares; a cube has 11 different nets in all. A tesseract can likewise be cut along its square faces and unfolded into our space: eight cubes glued face to face. The tesseract has 261 different nets, as Peter Turney counted in 1984. The most famous is a column of four cubes with one more cube on each of the four sides of the second from the top. Salvador Dalí painted it in Crucifixion (Corpus Hypercubus) of 1954: the cross on which Christ hangs is a tesseract net.

It has to be folded in the fourth dimension. In our space the “arm” cubes would jam into each other. In four dimensions each cube turns about the square that glues it to its neighbour — in the plane spanned by that square’s normal and the $w$ direction. The square itself stays put, like a door hinge (in four dimensions a rotation happens about a plane, not an axis; more on that in the chapter on rotation). When every hinge has turned by 90°, the 64 vertices of the eight cubes merge into the 16 vertices of the tesseract, and the 48 squares into 24: seven pairs of faces were already glued in the net, and seventeen more pairs meet during folding.

Who found them

All six regular polytopes — and also the fact that in five or more dimensions there are only three regular solids — were found by Ludwig Schläfli in a long treatise, Theorie der vielfachen Kontinuität (“Theory of multiple continuity”), written in 1850–1852. It never appeared in full in his lifetime: it was printed in 1901, six years after his death. Meanwhile the polytopes were rediscovered: in 1880 the American mathematician Washington Irving Stringham described them independently, and his paper contained some of the first published pictures of four-dimensional figures. The word “tesseract” was coined by Charles Hinton in 1888. Reinhold Hoppe introduced the German word Polytop in 1882, and it was carried into English by Alicia Boole Stott, daughter of the logician George Boole: without any mathematical training, she found and drew the three-dimensional sections of all six (in a paper of 1900; more in the chapter on cross-sections). The story was summed up by H. S. M. Coxeter in Regular Polytopes (1948), still the standard reference on the subject.

Summary

• The Schläfli symbol $\{p,q,r\}$ describes a regular polytope: cells $\{p,q\}$, $r$ around each edge; the vertex figure is $\{q,r\}$ and the dual is $\{r,q,p\}$.
• There are five Platonic solids because the face angles at a vertex must add up to less than 360°: $(p-2)(q-2) < 4$.
• There are six regular polytopes because the dihedral angles of the cells around an edge must also add up to less than 360°: three cases with tetrahedra pass, and one each with the cube, the octahedron and the dodecahedron.
• The 24-cell is self-dual and has no regular counterpart in other dimensions; the vertices of the 600-cell are the 120 icosians, a group under quaternion multiplication; the 120 dodecahedra of the 120-cell form 12 rings of 10.
• For every convex 4-polytope $V - E + F - C = 0$.
• From five dimensions on there are three regular solids: the simplex, the cube and the cross-polytope.
• The tesseract has 261 nets; Dalí’s cross folds into a tesseract by turning its cubes in the fourth dimension about their shared faces.

The polytopes live on the three-dimensional sphere — all their vertices lie on it. Next comes the hypersphere: how that sphere itself is built and why it falls apart into linked circles.

Sources

• H. S. M. Coxeter. Regular Polytopes. 3rd ed. New York: Dover, 1973 (1st ed. London: Methuen, 1948).
• L. Schläfli. Theorie der vielfachen Kontinuität. Denkschriften der Schweizerischen Naturforschenden Gesellschaft 38 (1901).
• W. I. Stringham. Regular figures in n-dimensional space. American Journal of Mathematics 3 (1880), 1–14.
• P. Turney. Unfolding the tesseract. Journal of Recreational Mathematics 17 (1), 1984–85, 1–16.
• Salvador Dalí. Crucifixion (Corpus Hypercubus), 1954. The Metropolitan Museum of Art.
• Euclid. Elements, Book XIII.
• Wikipedia: Regular 4-polytope, 24-cell, 600-cell, 120-cell, Icosian.