Curvature — the series · Crystallography & Symmetry

60 vertices, 12 pentagons. A closed net with three edges at every vertex: 60 vertices, 90 edges and 32 faces, of which 12 are pentagons and 20 are hexagons. The pentagons are picked out in the second colour. Their number is not a property of this cage — it is twelve for every closed trivalent net of pentagons and hexagons, at any size, and the hexagon count is free.

Twelve pentagons, and no way round them

The crystallographic restriction forbids a five-fold face in a flat repeating net. Curve the net into a closed cage and the same three lines of arithmetic require exactly twelve of them — at any size, with the hexagon count free. What a lattice forbids, closing up compels.

part 1 · restriction

The sphere fixes a count; the torus fixes only a difference. Euler's relation for a trivalent net gives Σ (6 − n) pₙ = 6χ, so the surface fixes one linear combination of the face counts and nothing else. On a sphere that combination is twelve, which with no face smaller than a pentagon forces exactly twelve pentagons. On a torus it is zero, which permits any number of pentagons provided as many heptagons pay for them — and permits none at all, which is the plain hexagonal net. On a surface of two holes it is minus twelve, so heptagons become compulsory instead.

As many heptagons as pentagons

A trivalent net on a sphere must have exactly twelve pentagons. The same three lines of arithmetic on a torus give zero — which does not forbid pentagons, it makes them pay: every pentagon has to be balanced by a heptagon, and the counts are otherwise free. One rotated bond in a wrapped honeycomb makes two of each and changes nothing else.

part 2 · restriction

The same accounting, at every coordination number. One row per number of edges at a vertex. The bill a sphere charges is 2dχ; the face worth nothing is 2d/(d − 2), which is a whole number at three, four and six and is 10/3 at five; the faces that can pay are those with fewer sides than that; and the last column is every way of paying the whole bill with faces of a single size. At three edges a vertex there are three such ways and twelve pentagons is one of them. At six there are none, which is the statement that six-fold coordination belongs to the plane and to no closed surface at all.

The twelve belongs to the vertex

Twelve pentagons is read as a fact about closing a surface. It is not: it is a fact about three edges meeting at a point. Let four edges meet instead and the sphere charges eight triangles; let five meet and it charges twenty; let six meet and it cannot be paid at all.

part 3 · restriction

Every closed surface, and the two that charge nothing. The same accounting indexed by Euler characteristic rather than by genus. An orientable surface has χ = 2 − 2g, so it only ever occupies an even row; a non-orientable one has χ = 2 − k and occupies every row from one downwards. The odd rows therefore belong to surfaces that cannot be oriented and to nothing else — and the first of them, the projective plane, charges six. Six pentagons is a bill no orientable surface presents.

The surfaces a count by genus skips

A count indexed by genus steps in twelves and lands only on even numbers. A closed surface can have any characteristic at or below two, and the odd ones belong to the surfaces that cannot be oriented — where the projective plane charges six pentagons, a bill no orientable surface ever presents.

part 4 · restriction

Where the sphere and the projective plane have no net. The number of different closed nets with three bonds at every atom and faces that are pentagons and hexagons only. On the sphere, with twelve pentagons and k hexagons for k up to 12, every count has at least one net except k = 1. On the projective plane, with six pentagons and h hexagons, each count sits under the sphere count it lifts to, since every hexagon of a projective net becomes two on the sphere. The projective counts for h = 0 to 6 are 1, 0, 0, 1, 1, 3, 3, so the projective plane has no net at h = 1 or 2: two gaps where the sphere has one. Every sphere count was found by enumeration and agrees with the published one.

A gap the sphere does not have

A net of pentagons and hexagons on the projective plane must have six pentagons, and the count permits any number of hexagons. Not every number happens. Lifting each net to the sphere turns the question into one about which cages have a centre — and the answer leaves two gaps where the sphere has one.

part 5 · restriction

A centre at every other ring

A census cannot settle an infinite row, and the construction proposed to settle it was a tube capped at both ends, lengthened a ring at a time. Carried out, it alternates: a centre appears at every other ring and never between, the two families it permits reach two arithmetic progressions rather than a row, and the first of them opens with exactly the cage the census found could not halve.

part 6 · restriction

Everything except the hexagons

Three counts of what a closed net must carry end on the same admission: an arithmetic saying what a net must charge does not say that a net exists. Eberhard's theorem says how close the charge comes to being enough, and the answer has a shape nobody would guess — it fixes every face count except the hexagons, and the hexagons are exactly the entry it cannot see.

part 7 · restriction

The fewest contacts twelve pentagons can have, by size. For every cage of pentagons and hexagons up to forty-four atoms, the number of pairs of pentagons sharing a bond. The lower line is the fewest any cage of that size achieves — 30, 24, 21, 18, 17, 15, 14, 12, 11, 10, 9, 8 — the upper line the most, and the dashed line the bound that counting edges gives: the twelve pentagons carry sixty edges between them, a contact uses two and an edge to a hexagon uses one, so the contacts cannot fall below 30 − 3h with h hexagons. The bound is attained while the hexagons are few and goes loose at five, after which each extra hexagon removes about one contact rather than three. The number of cages at each size is printed beneath, and it is the least rather than the average that the bound is about.

How close the twelve must be

The charge fixes twelve pentagons and says nothing about where they go, because it is a sum over faces and cannot see which face touches which. What it cannot see is a graph on twelve points, and the fewest edges that graph can have falls from thirty to eight over the cages a census reaches — then keeps falling at a rate that puts its first zero exactly where the truncated icosahedron is.

part 8 · restriction