Five solids from one inequality

The finite rotation groups of space come out of counting axes: a group of order N acting on the sphere satisfies 2 − 2/N = Σ(1 − 1/nᵢ), and the integer solutions are the cyclic groups, the dihedral groups, and three exceptions of orders 12, 24 and 60.

The five regular solids come out of the same arithmetic, and it is worth doing them separately because the two lists have different lengths. Five solids, three groups.

One expression, three signs #

Let a solid have p-gonal faces with q of them at every vertex. Counting the incidences twice gives pF = 2E = qV, and substituting into Euler’s V − E + F = 2 gives

$$V = \frac{4p}{2p + 2q - pq}, \qquad E = \frac{2pq}{2p + 2q - pq}, \qquad F = \frac{4q}{2p + 2q - pq}.$$

Three counts, one denominator. The counts are positive exactly when that denominator is, and dividing by 2pq turns the condition into

$$\frac{1}{p} + \frac{1}{q} > \frac{1}{2},$$

which for p, q ≥ 3 has five solutions and no more: (3,3), (3,4), (4,3), (3,5) and (5,3). The search that finds them is three lines, and it does not need a bound argued for — 1/6 + 1/3 is exactly a half, so nothing with p or q past six can qualify, and the search sees the failures as well as the successes.

The three signs are three geometries. Positive is a sphere and the answer is five. Zero is the plane: {3,6}, {4,4} and {6,3}, which are the triangular, square and hexagonal tilings — the three regular members of the eleven uniform tilings. Negative is the hyperbolic plane, where {3,7} and everything past it lives and the list never ends.

That is the same trichotomy the orbifold cost produces for the plane groups, and the same one that separates the seventeen from the infinitely many hyperbolic groups. It is the most reliable fact in this subject: the finite classifications sit exactly on the boundary case.

Built, not tabulated #

A list of five is easy to type in. What is not easy to type in is a construction that would break if the list were wrong, so each solid here is built from its own rotation group.

The recipe is one sentence. The vertices are the orbit of a point on an axis of order q; the face centres are the orbit of a point on an axis of order p; the edge midpoints are the orbit of a point on a two-fold axis.

The counts follow immediately, because the stabiliser of a point on an axis of order n is the cyclic group of order n:

$$V = \frac{|G|}{q}, \qquad E = \frac{|G|}{2}, \qquad F = \frac{|G|}{p}.$$

And then Euler’s formula is the axis equation. Substituting into V − E + F = 2 gives |G|(1/q − 1/2 + 1/p) = 2, which rearranges to exactly the inequality above with the excess measured rather than merely signed. Two apparently different pieces of arithmetic — a count of incidences on a polyhedron, and a count of poles fixed by rotations — are the same equation, and the construction here is required to produce the same three numbers from both. If it did not, the figure would not appear at all.

The faces are found, not assumed #

One step of the construction deserves its own paragraph because it is where the shortcut would be.

A face is the set of vertices nearest its own axis, and “nearest” means largest dot product. That is a search over the vertex list, and the number found has to be p or the drawing is refused. It nearly always is — and for the tetrahedron it is not, unless the axis is pointed the right way.

The tetrahedron’s three-fold axes are not through two faces; they are through a vertex and the face opposite it. So one end of such an axis has a single vertex nearest it and the other has three. Taking the axis with whatever sign the axis-finder returns produced a face of four vertices — the three of the real face plus the one at the far end, which ties with nothing. The fix is to try both ends and keep the one where exactly p vertices tie, which is a test rather than a rule about which solids need flipping.

That is the recurring shape of a defect in this collection: a convention that is arbitrary in most cases and load-bearing in one.

What the orbit construction quietly assumes #

Building a solid as the orbit of a point on an axis is short enough to look like a definition rather than a construction, so it is worth saying what it needs.

It needs the group to be transitive on vertices — otherwise the orbit of one point is only part of the solid — and it needs the stabiliser of a vertex to be exactly the cyclic group of order q, which is what makes the orbit |G|/q rather than something else. Both are consequences of regularity: a regular solid’s symmetry group is transitive on flags, so it is transitive on vertices, on edges and on faces separately, and the stabiliser of a vertex is the rotations about the axis through it.

For a rotation group — no reflections — the vertex stabiliser is cyclic of order q, so the orbit has |G|/q points. Include the reflections and the full symmetry group is twice as large and the stabiliser twice as large too, so the counts are unchanged. That is why the construction here uses the tetrahedral, octahedral and icosahedral rotation groups rather than the full groups: the answer is the same and the matrices are half as many.

What it does not need is any coordinates. The axes come from the group’s own elements — each rotation’s axis is its eigenvector of eigenvalue one — and the group comes from a closure of two generators. So the dodecahedron drawn on this page has never been told where a dodecahedron’s vertices are.

Five solids, three groups #

Swapping p and q swaps vertices with faces. {p, q} and {q, p} have the same denominator, and their V and F are exchanged while E is unaltered — so the two are duals, and the duality is visible in the arithmetic before any geometry is done.

The tetrahedron is {3,3} and duals to itself. The cube {4,3} and octahedron {3,4} are a pair; the dodecahedron {5,3} and icosahedron {3,5} are a pair. Dual solids have the same symmetry group — a symmetry permutes faces and vertices together — so the five solids carry only three rotation groups: the tetrahedral one of order 12, the octahedral one of order 24 and the icosahedral one of order 60.

That is why the classification of the groups is the shorter list, and it is the reason the finite rotation groups are three exceptional cases rather than five.

Who found them, and when #

The five are the oldest classification in this collection by two thousand years. Theaetetus is credited with the proof that there are no others, in the fourth century BC; Euclid’s Elements ends with them, and Book XIII’s last proposition is precisely the statement that the list is complete — arguably the first classification theorem anybody wrote down.

The argument Euclid gives is the same one as the inequality, phrased without algebra: at least three faces must meet at a solid angle, and their angles must sum to less than a full turn. Three, four or five triangles (180°, 240°, 300°) work and six do not (360°); three squares work and four do not; three pentagons work and four do not; three hexagons already fill a turn, so no polygon with six or more sides can appear at all. Five configurations, and the enumeration is finished. What the modern version adds is not rigour but reach: written as 1/p + 1/q > ½ the same statement covers the plane and the hyperbolic case, which Euclid had no reason to look for.

Kepler’s Mysterium Cosmographicum of 1596 nested the five solids inside one another to explain the spacing of the six known planets, which is wrong and was the best-argued wrong thing of its century. He returned to the same objects in Harmonices Mundi twenty-three years later, and that is the book that also contains the eleven uniform tilings — the plane case of the same equation, drawn by the same person, at the sign he did not know he had crossed.

The same equation in four dimensions #

The Schläfli symbol extends: {p, q, r} means r of the solids {p, q} around every edge, and the condition for closure is again an inequality, this time on the angles the cells subtend. It has six solutions in four dimensions and three in every dimension above four.

Six is more than five, which is the only place in this subject where raising the dimension makes a classification larger. The extra ones are the 24-cell, the 120-cell and the 600-cell, and the last two have icosahedral cross-sections — five-fold symmetry becomes available in four dimensions in exactly the way the crystallographic restriction predicts, since the four-dimensional integer matrices include one of order five. Past four dimensions only the simplex, the hypercube and the cross-polytope survive, and the list is three for ever.

That is the same shape of result as everything else here. The interesting counts live at small dimension, and the reason is always that a constraint which is an inequality has only finitely many integer solutions until the room runs out — or, past a certain point, until it stops running out and the answer becomes boring instead.

What the round trip checked, and how #

{3,6}, {4,4} and {6,3} must fail. They make the denominator vanish, and a construction that quietly returned something for them would be returning a plane tiling as a solid. All three are refused.

{3,7} must fail. The denominator is negative and the counts would be negative, which is the hyperbolic case.

The five must not fail. A refusal test that rejected everything would pass the first four and prove nothing.

And every solid checks its own arithmetic three times over: V, E and F against the formula in p and q; V − E + F against 2; and |G| against 2E, since a rotation group of a solid has exactly two elements for every edge — the identity paired with each of the E half-turns about edge midpoints, in the count that makes the group’s order and the edge count the same measurement.

Where the exactness stops #

“Regular” here means a regular polygon at every face and the same number at every vertex. That is the Schläfli condition and it is combinatorial plus a metric condition on the faces. It excludes the Kepler–Poinsot star polyhedra, which satisfy a version of the same equation with self-intersecting faces and are four more solids by a different definition. The five is a count relative to a definition, and the definition is the one stated.

Convexity is being assumed rather than proved. The construction takes orbits of points on axes, which produces convex solids because the orbits lie on a sphere; a non-convex regular polyhedron would not be found by it. Nothing here rules them out — the definition above does.

The regular tilings appear here as failures and elsewhere as successes. {3,6}, {4,4} and {6,3} are refused here because they are not solids, and they are three of the eleven uniform tilings the classification of the plane produces. Whether a symbol is a refusal or an answer depends on which question is being asked, and the arithmetic does not distinguish them — only the sign does.

That is worth one more sentence, because it is the cleanest instance of a pattern this collection keeps meeting. One inequality, 1/p + 1/q > 1/2, has three regimes. Above the line it closes into a finite solid and there are five ways of doing it. On the line it does not close and instead tiles the plane, and the three symbols sitting exactly on it are the three regular plane tilings — which is why the same person, in the same book, wrote down both lists. Below the line it tiles the hyperbolic plane instead, and there the list is infinite, because nothing runs out. The five solids are therefore not the answer to a question about solids; they are the finite corner of a question about surfaces, and the corner is finite for the same reason every other list in this collection is: an inequality on reciprocals of integers has few solutions and then none.

Nothing here is crystallography yet. These are finite groups, and a crystal’s point group must additionally preserve a lattice. Applying the restriction to the three groups keeps T and O and deletes I, because the icosahedral group has five-fold axes and no lattice admits one. So of the five solids, three have crystallographic symmetry and two do not — the dodecahedron and the icosahedron are shapes a crystal may not have as its symmetry, however often mineral collections seem to show them.

What the icosahedral group is doing in a crystallography collection #

The group the restriction deletes is the one that turns out to matter most in this collection’s later essays, and the reason is worth stating here where its construction is.

An icosahedral group cannot be the point group of a periodic crystal. It is the point group of a great many other things: virus capsids, boron clusters, the C₆₀ molecule, and — most consequentially — the alloy Shechtman measured in 1982, whose diffraction pattern has icosahedral symmetry and sharp peaks at once. That combination was thought impossible because sharp peaks were taken to imply a lattice, and a lattice forbids the group.

The resolution is that sharp peaks imply a Fourier module rather than a lattice, and a module can have icosahedral symmetry in three dimensions if it is the shadow of a lattice in six. The sixty rotations built here — as explicit matrices, from the icosahedron’s own vertices — are the same sixty that act on that six-dimensional lattice, which is why this construction gets reused far from the solids it was written for.

The pyrite exception, which is not one #

Pyrite grows in shapes that look like dodecahedra, and they are not. Its faces are pentagons, but irregular ones — the crystal is cubic, its group is one of the crystallographic classes, and the “pyritohedron” is a distortion of the dodecahedron with the five-fold axes broken into two-folds and three-folds.

That distinction is decidable and this collection has the machinery for it: the shape does not name the class, and a pentagonal face is no more evidence of five-fold symmetry here than a pentagonal tile is in the plane. What settles it is the group, and the group is measured.

Where the ladder goes next #

Every argument here assumed that a finite group has a fixed point to take axes through, and that a group preserving a lattice preserves a metric. Both are true, both are used constantly in this collection, and neither has been shown. They are the same trick applied twice, and it is the next rung.

Relax the hypothesis and the list grows twice #

Five is the answer to a question with two conditions in it — all faces the same regular polygon, the same number at every vertex. Weakening the first while keeping the second is the standard next question, and its answer has the same shape as the plane’s.

Allow the faces to be regular polygons of different kinds, and require only that every vertex have the same arrangement around it. The five stay on the list, and thirteen more appear — the truncated solids, the cuboctahedron and icosidodecahedron, the rhombi- and snub families. Thirteen, and no fourteenth.

Two infinite families come with them, and they are the reason the count is quoted as thirteen rather than as some larger number. A prism with regular polygon ends and square sides satisfies every condition, for every polygon; so does an antiprism. Neither family is finite, and both are excluded by convention rather than by arithmetic, which is worth knowing before comparing anybody’s count with anybody else’s.

The plane’s version of this question is the eleven, and the parallel is exact: uniform tilings by regular polygons are the flat case of the same relaxation, and they behave the same way — a small finite list, arrived at by enumerating what can surround a vertex and then asking which of those arrangements extends.

And the same trap sits in both. One solid satisfies “every vertex is surrounded alike” without being vertex-transitive: the elongated square gyrobicupola, obtained by twisting one cap of the rhombicuboctahedron through a right angle. Its vertices all have a triangle and three squares around them, and no symmetry of the solid carries a vertex of one half to a vertex of the other. Whether it belongs on the list depends on which of the two conditions is taken as the definition, and the two conditions are not the same condition — which is precisely the distinction the eleven tilings turn on.

The other side of the inequality #

The positive case gives five, the zero case gives three, and the negative case is dismissed above as “infinitely many”. It is worth saying what infinitely many amounts to, because the objects on that side are the reason this inequality is studied outside geometry.

Every pair $(p, q)$ with $\tfrac1p + \tfrac1q < \tfrac12$ gives a tiling of the hyperbolic plane by regular $p$-gons, $q$ at each vertex. The smallest is $\{7, 3\}$ — heptagons, three at a corner — and after it the list never stops.

The symmetry groups of those tilings are infinite, so they are not directly comparable with the five. What is comparable is what happens when the hyperbolic plane is folded up into a closed surface: the tiling descends to a finite tiling of a surface of some genus, and its symmetry group becomes finite.

How large can that finite group be? Hurwitz answered it with this inequality. A surface of genus $g \ge 2$ has at most $84(g - 1)$ orientation-preserving symmetries, and the bound is attained exactly when the surface carries a $\{7, 3\}$ tiling — the value 84 is $\tfrac{2}{1/2 - 1/7 - 1/3}$, read straight off the same expression this essay is about.

The smallest case is famous. Genus three gives a bound of 168, attained by Klein’s quartic curve, whose 168 symmetries make it the most symmetric surface of its genus. The group is the second-smallest non-abelian simple group, and it arrives here from the same arithmetic that produced the icosahedron on the other side of the sign.

So the three signs are not three cases of a classification. They are three different subjects, and the inequality is the only thing they have in common.