Crystallography & Symmetry
patterns derived from their groups
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60 vertices, 12 pentagons

60 vertices, 12 pentagons
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.

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.

11 essays call closed-net. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about. Every one of this site's 393 essays names its parameters at the call site, which the standard pass of 2026-08-09 established and param-floor holds.

Where it is called

Changing this generator changes every one of these figures.

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. What a lattice forbids

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.

 orders 5 and 7 reach a site of symmetry 1 and no more. A molecule whose only symmetry is one n-fold axis, and the highest site symmetry it may occupy in any of the 45 space groups this site builds. The site's symmetry has to be a subgroup of the molecule's, so the site's order must divide n and the site group must be cyclic. Orders 1, 2, 3, 4 and 6 reach a site of their own order. Orders 5 and 7 reach one, because no site symmetry in any space group contains an operation of order five or seven — the orders available are 1, 2, 3, 4, 6, computed by asking every operation of every group whether it moves a point. A five-fold molecule keeps its axis; the crystal simply has no use for it. What a lattice forbids

What a molecule gives up to sit in a crystal

A molecule brings its own symmetry. A crystal offers sites with symmetries of their own, and the two have to be compatible — the site's symmetry must be a subgroup of the molecule's. So a molecule may always keep more than its site offers, and a molecule with a five-fold axis may sit only where the crystal offers nothing at all.

 11 nets, and one accounting. Every plane net folds onto a torus when its own translations are divided out, and a torus has Euler characteristic zero — so the quotient's vertices, edges and faces satisfy n − e + f = 0 and the number of faces is not something to be counted off a drawing but e − n. Dividing through gives one over the mean face size plus one over the mean degree equal to a half, which is the same relation that forbids a plane tiling by pentagons, reached here with no geometry in it at all. It holds for every net in the table. The classification

Every net folds onto a torus

Divide a plane net by its own translations and the quotient is a finite graph drawn on a doughnut. A doughnut has Euler characteristic zero, so the number of faces is not something to count — it is forced, and with it a relation between how many edges meet at a vertex and how many bound a face.

 3 whole-number solutions: (6, 3), (4, 4), (3, 6). Every pair of whole numbers from three to 12, with the mean face size across and the mean degree down. A square in the first colour is a pair satisfying one over p plus one over q equals a half exactly — the flat case, where a periodic net is possible — and there are 3 of them: 6 and 3, 4 and 4, 3 and 6. The lighter squares above and to the left have a sum greater than a half, which is a closed polyhedron rather than a plane tiling; the ones below and to the right have a sum less than a half and belong to a surface of negative curvature. The plane is the boundary between them and it is thin. The classification

Three answers in whole numbers

One over the face size plus one over the degree equals a half. Ask for whole numbers and there are exactly three answers, which are the three nets everybody has drawn since childhood — and the pairs on either side of them are a closed polyhedron and a plane the plane has no room for.

 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. What a lattice forbids

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.

 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. What a lattice forbids

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.

 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. What a lattice forbids

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.

 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. What a lattice forbids

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.

 A centre at every other ring, and never between. Two families of closed cage, each a tube of hexagons closed at both ends by a cap of six pentagons, taken from no rings of hexagons to 8. The top row has five faces to a ring and a pentagon at each pole; the bottom row has six and a hexagon. Each box holds the cage's number of atoms with its number of hexagons beneath, and a box is drawn solid with a dot under it when the cage has a symmetry that reverses orientation and fixes nothing — a centre, which is what lets the cage halve onto the projective plane. The five-family has one at even numbers of rings and the six-family at odd ones, so their hexagon counts are 0, 10, 20, 30 … and 8, 20, 32, 44 … — two arithmetic progressions rather than two rows. What a lattice forbids

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.

 Every vector realised, and not at the same hexagon count. Each row is a set of faces other than hexagons whose charge — the sum of 6 − k over them — comes to twelve, which is what a closed trivalent net on the sphere must pay. Each column is a number of hexagons added to that set, and the entry is how many different solids exist with exactly those faces, found by winding up every arrangement of them into a spiral. A dash means the search found none; a question mark means the planar reader declined the row and it is not evidence either way. Every row has an entry somewhere, which is Eberhard's theorem, and the first one is at 0, 2, 3, 4 hexagons depending on the row — so the charge decides everything except the number of hexagons, and the number of hexagons is not a function of the charge. What a lattice forbids

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.

 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. What a lattice forbids

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.

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