The twelve belongs to the vertex

Twelve pentagons and no way round them proves its number in three lines, and as many heptagons as pentagons changes one integer in those lines and gets a different answer. Between them they make the twelve look like a fact about surfaces: the sphere charges twelve, the torus charges nothing, a surface with two holes charges twelve the other way.

That reading is half right, and the half it leaves out is the half a crystallographer needs. The second essay says so in its own closing section, under what it does not own: “Every relation above uses three edges at every vertex, and a net with vertices of mixed degree obeys a longer identity in which the vertex degrees appear. Carbon sheets are trivalent and that is why the arithmetic is this short; a four-coordinated net has its own version and it is not this one.”

This is that version, and the twelve does not survive it. Silicon is four-coordinated. A metal is twelve-coordinated. Carbon is trivalent in a sheet and tetrahedral in diamond, and the same element changes the answer by changing how many bonds leave each atom. So the interesting question is not what the sphere charges but what the vertex charges, and the twelve turns out to be one row of a ladder that has four more.

The three lines, with one letter left alone #

The derivation is worth rerunning rather than cited, because exactly one substitution changes and everything else is identical.

Euler’s relation is unchanged: V − E + F = χ. Counting edge-sides face by face is unchanged: Σ n·fₙ = 2E. The middle line is the one carrying the assumption. It was 3V = 2E, three edges at every vertex and two ends to each edge; with d edges at every vertex it is d·V = 2E, which is the same sentence with a different number in it.

Substituting and clearing denominators gives

$$\sum_n \bigl(2d - n(d-2)\bigr)\,f_n = 2d\,\chi.$$

At d = 3 the coefficient is 6 − n and the right side is 6χ, which is the sum the previous two rungs are about. Nothing in the derivation was ever about three. It was about the fact that a number appears, and three is the number carbon happens to supply.

So a face of n sides carries a charge that depends on the vertex it is attached to, and the useful thing about writing it that way is that the charge is now a function of two variables rather than one.

The face that costs nothing, and the three times it exists #

Setting the coefficient to zero asks which face a net may have any number of, and the answer is n = 2d/(d − 2).

At three edges a vertex that is six, which is why a fullerene may be any size: hexagons are free, and only the pentagons are on the bill. At four edges a vertex it is four, so a four-coordinated net may carry unlimited squares. At six edges a vertex it is three, so triangles are free.

Those three pairs — hexagons at three, squares at four, triangles at six — are the three regular tilings of the plane, and they are here for a reason that does not mention flatness anywhere. A plane net is one whose every face is free, because the plane’s characteristic is zero and a sum of terms that must vanish, with no term allowed to be negative, is a sum of zeros. So the question which regular tilings are there and the question which faces cost nothing are the same question, and the second is easier.

The value at five edges a vertex is ten thirds, and that absence is the most informative entry in the table. There is no face size a five-valent net can carry freely, so every face of every five-valent net is charged, so no five-valent net of the plane exists. The missing regular tiling {n, 5} and the missing free face are one fact, and neither is a coincidence about small numbers: 2d/(d − 2) is an integer only when d − 2 divides 2d, which is to say when d − 2 divides four, which happens at d = 3, 4, 6 and nowhere else.

That is a divisibility argument of exactly the kind the crystallographic restriction turns on, arriving in a question about coordination rather than about rotation. Five is excluded in both places and for arithmetic that has nothing in common — there the trace of an integer matrix must be an integer, here d − 2 must divide four — which is worth noticing precisely because it is not evidence of a deeper connection. Two unrelated divisibilities both happen to exclude five, and reading that as one phenomenon is the error the coincidence invites.

What each degree charges, and what can pay it #

A sphere presents a bill of 2dχ, which is 4d, and the faces that can pay it are those whose charge is positive — the ones with fewer sides than the free face.

At three edges a vertex the bill is twelve and three face sizes can pay: a pentagon is worth one, a square two, a triangle three. That is why the fullerene result has room in it, and why the same twelve can be spent in degrees rather than in pentagons alone.

At four edges a vertex the bill is sixteen and exactly one face size can pay: the triangle, worth two. So every four-coordinated closed net has eight triangles’ worth of charge and no alternative denomination whatever, which is a much tighter statement than the trivalent one. The octahedron is the case with nothing else in it; the cuboctahedron is the case with squares added, and the squares are free.

At five edges a vertex the bill is twenty and again only the triangle pays, this time worth one, so twenty triangles.

At six the bill is twenty-four and nothing can pay it. Every face has three sides or more, every charge is zero or negative, and a sum of non-positive terms cannot come to a positive number. There is no closed surface of positive characteristic with six edges at every vertex — no such cage, at any size, with any faces.

Five divisions, and the five solids come out #

The last column of the ladder is an enumeration rather than a description, and it produced the one result in this essay that was not expected before it was computed.

Ask, at each degree, for the ways of paying the whole bill with faces all of one size. The number of faces is 2dχ divided by that size’s charge, and the pair counts only when the division comes out whole. Running it over every degree and every face size from three to twelve gives five solutions in the entire ladder — and the five are the five Platonic solids.

Four triangles is the tetrahedron. Six squares is the cube. Twelve pentagons is the dodecahedron. Eight triangles at four edges a vertex is the octahedron, and twenty at five is the icosahedron. The accounting has produced the classification of the regular solids without ever asking for regularity.

That is worth being careful about, because it is not quite the theorem it resembles. What the division establishes is that the face vector of a single-size closed net must be one of five; it does not establish that the net is regular, that its faces are congruent, or that its edges are the same length. A cage with twelve pentagons of five different shapes satisfies the arithmetic perfectly and is not a dodecahedron. What the enumeration bounds is the combinatorics, and the geometry is a separate question that the inequality answers by a different route.

Read the other way, though, the two arguments are one. And that is the connection this rung exists for.

The charge and the excess are the same number #

The coefficient factors, and the factorisation is the whole of it:

$$2d - n(d-2) \;=\; 2(n+d) - nd \;=\; 2nd\left(\frac{1}{n} + \frac{1}{d} - \frac{1}{2}\right).$$

The bracket is the quantity in the Platonic inequality, which sorts the pairs {n, d} into solids where it is positive, plane tilings where it is zero and hyperbolic tilings where it is negative. The factor in front is 2nd, which is positive for every pair anybody would write down. So a face pays into a sphere’s bill exactly when its Schläfli symbol closes into a solid, is free exactly when the symbol tiles the plane, and costs exactly when the symbol is hyperbolic.

This is the surprising part and it is worth stating in the form that makes it useful. The fullerene sum and the classification of the regular solids are not two results about polyhedra that happen to both come from Euler. They are one expression read at two scales: the inequality asks about a single vertex figure and answers with a sign, and the sum adds that same quantity over all the faces of a net and answers with a total. A local sign and a global total, and the local quantity is what is being totalled.

Which explains something the two earlier rungs left unexplained. They observed that the arithmetic on a sphere gives twelve and on a plane gives zero, and treated the two as different cases of one identity. In this reading the plane case is not a case at all — it is the boundary of the inequality, the set of pairs where the local excess vanishes, and a net of such faces has nothing to add up.

Why the five come in pairs #

The expression is symmetric in n and d, and that symmetry is not decoration.

2(n + d) − nd is unchanged when the two letters are exchanged, so the pair {n, d} and the pair {d, n} carry the same charge and stand or fall together. Reading the five solutions with that in mind, they are not five unrelated rows: six squares at three edges a vertex and eight triangles at four are the same pair of numbers written in the two orders, and so are twelve pentagons at three and twenty triangles at five. The tetrahedron, at three and three, is fixed by the exchange.

What the exchange does to a net is take its dual — one vertex per face, one face per vertex — which turns a d-valent net with n-gonal faces into an n-valent net with d-gonal faces. So the cube and the octahedron are one entry in the ladder seen twice, the dodecahedron and the icosahedron are another, and the tetrahedron is its own. That the regular solids come in dual pairs is usually shown by construction; here it is a property of an algebraic expression, and the construction is what makes it about polyhedra.

The symmetry also explains a feature of the ladder that would otherwise look arbitrary. The free faces are the pairs where the charge vanishes — {6, 3}, {4, 4}, {3, 6} — and that list is closed under the exchange, with the square self-paired. The three regular tilings of the plane are a dual pair and a self-dual case, which is the same three-and-not-four structure the solids have, one row lower.

And it says why d = 5 has no plane case in a way the divisibility argument does not. A plane tiling at five edges a vertex would need a face size n with {n, 5} on the boundary, and its dual would be a five-sided face at n edges a vertex — so the missing tiling and the missing free face are the same missing pair approached from the two sides, and neither is available because the boundary of the inequality passes between the integers there rather than through one.

Vertices that are not all the same #

Real nets are not of uniform degree. A zeolite framework has bridging oxygens with two bonds and silicons with four; a defective sheet has a vertex where three bonds should be and two are; and a net with mixed degrees obeys neither the trivalent identity nor any of the rows above.

The general form comes out of the same substitution with the middle line left as a sum. Writing v_d for the number of vertices with d edges:

$$\sum_n (6-n)\,f_n \;+\; 2\sum_d (3-d)\,v_d \;=\; 6\chi.$$

The second sum is zero exactly when every vertex has three edges, which is why the short form looked complete for two whole essays. It is not a correction to the trivalent result — it reduces to it — and it is not a generalisation for its own sake either, because the term it adds is large.

The cuboctahedron is the case worth carrying. Its face term is thirty-six — eight triangles at three each, six squares at nothing — and thirty-six is not twelve and is not sixteen. A reader who applied the fullerene sum to it would get an answer three times too large and would have no way of knowing, because thirty-six is a perfectly ordinary-looking number and the sum has no built-in objection to it. What says otherwise is the vertex term, which is minus twenty-four and is not optional.

The convention this rests on, said plainly #

Two conventions are doing work above and neither is visible in the arithmetic.

A face is a face of an embedding, not of a graph. The counts fₙ are read off a rotation system — the cyclic order of the edges at each vertex — exactly as the torus rung reads them, and the same abstract graph drawn on two surfaces has different faces. Everything here is a statement about an embedded net.

Degree means graph degree, not chemical coordination. They are the same for a framework whose bonds are the edges, and they part company the moment a structure is described with a bond nobody would draw or a contact nobody would call a bond. A twelve-coordinated close-packed metal has no natural net of degree twelve in this sense, because its contacts do not form a planar embedding at all — and the accounting above says nothing whatever about it. This is the point at which the arithmetic stops applying to a large class of real crystals, and it stops for a reason of definition rather than of difficulty.

Where the numbers were checked and where they were not #

Computed here: the identity’s coefficient at every degree from three to eight; the free face size as an exact fraction rather than a decimal; the paying face sizes; the single-size solutions by division; the face census and charge of the dodecahedron, the octahedron, the icosahedron and the cuboctahedron, each built from a rotation group and each measured off its own faces; and the mixed-degree identity on two of them.

The charge is unreduced, and it matters once. The four-valent charge can be written 4 − n against a bill of 4χ, which is the same statement divided by two, and mixing the reduced form with the unreduced bill gives every four-valent answer off by a factor of two. The first version of this file did exactly that and the octahedron caught it: a measured charge of sixteen against a predicted eight is not a subtle disagreement, and no argument would have found it because both numbers are plausible.

Nothing here says a net exists. The identity is necessary and not sufficient, which is the same limitation the trivalent case records: a face vector satisfying the sum may still have no net, and the trivalent ladder has a known exception at a single hexagon. Whether the four-valent and five-valent ladders have exceptions of their own is not settled here and is not obviously easier.

And the ladder is a ladder of degrees, not of surfaces. Every bill above is a sphere’s. Combining the two directions — degree four on a torus, degree five on a surface with two holes — is a matter of putting a different χ into the same expression, and the arithmetic permits it in every combination; whether the nets exist is again a separate question with no general answer here.

What a chemist reads off the ladder #

The rows are not equally populated in nature, and the reason they are not is the useful end of all this.

Three is the row of the sheets. Graphite, boron nitride, the whole family of trivalent carbon cages. The bill is twelve and there are three denominations, so the structures are various and the pentagons are the interesting part of every one of them.

Four is the row of the frameworks. Silica, the zeolites, the silicon and germanium networks. The bill is sixteen, there is exactly one denomination, and it is a triangle — which is why a four-connected framework built from tetrahedra has to reach its curvature through three-rings, and why three-rings are strained, and why so much of that chemistry runs on rings of four and larger with the curvature supplied elsewhere. The accounting says the bill must be paid; the chemistry says the only currency is expensive; and the resolution is that most such frameworks are not closed at all.

Six is the row that is only ever flat. A close-packed layer, a triangular lattice, a kagome net’s parent. No closed cage exists with six edges at every vertex, so a six-coordinated structure that curves has to have vertices of other degrees in it, and those are its defects. That is the disclination arriving from the accounting side rather than from the lattice side.

And five is the row with no flat case. No plane tiling, and on a sphere only the icosahedron. A five-coordinated net is curved wherever it exists, which is a strong statement and an accurate one — the twelve five-coordinated vertices of a fullerene’s dual are exactly the places its surface is curving, and there is nowhere for a five-valent region to be flat.

Where this ladder goes: the surfaces the genus skipped #

The ladder above indexes by degree and leaves the surface fixed at a sphere. The other direction was fixed in the previous rung, which indexed by genus and left the degree at three, and drew a table whose rows were the orientable surfaces.

That table skips every odd number, and the numbers it skips are not gaps. A closed surface has any integer characteristic at or below two, and the odd ones belong to the surfaces that cannot be oriented — where the accounting still holds, still gives a whole number, and gives one no orientable surface can ever demand. The projective plane charges six, which is half of the sphere’s twelve for a reason that can be built rather than argued about: it is the sphere with antipodal points identified, and the dodecahedron divided by its own centre is a net of six pentagons on it.

Sideways from here sit the two rungs this one leans on. Twelve pentagons is the d = 3 row read alone, and five solids from one inequality is the same coefficient read one face at a time — which, as of this rung, is a statement about the same expression rather than a family resemblance between two arguments.

And underneath both sits the question of which of these nets a crystal may actually contain. The crystallographic restriction forbids a five-fold rotation to a lattice and says nothing about coordination; the ladder here forbids a six-valent cage and says nothing about symmetry. Neither implies the other, both exclude something at five, and what a net’s own symmetry can be is where the two constraints are finally applied to one object.