Everything except the hexagons

Three counts of what a closed net must carry end on the same sentence, and it is an admission rather than a result. A trivalent net on a closed surface must pay a charge — twelve on the sphere, nothing on the torus, six on the projective plane — and an arithmetic that says what a net must charge does not say that a net exists.

The sphere carries the standing counterexample and it is not obscure. Twelve pentagons and a single hexagon pays the bill exactly, and no cage has it: thirteen face spirals wound up and not one of them closes, and the published enumeration agrees. So the charge is necessary and it is not sufficient, and nothing said about it so far goes further than that.

Eberhard’s theorem of 1891 says how close it comes, and the answer is stranger than either “sufficient” or “not sufficient”. Write down the numbers $p_3, p_4, p_5, p_7, p_8, \ldots$ of faces of each size other than six. If the charge

$$\sum_{k \ne 6} (6 - k)\,p_k = 12$$

comes out right, then there is some number of hexagons for which a trivalent polyhedron with those faces exists. The charge decides everything about a face vector except one entry — and that entry is precisely the one the charge cannot see.

The blind spot is not an accident of the formula #

That a hexagon charges nothing is the whole of why the theorem has the shape it has, and it is worth seeing that the zero is structural rather than a coincidence of where the six sits.

The charge comes from Euler’s relation on a net with three edges at every atom. Counting each edge from both its faces gives $\sum_k k\,p_k = 2E$, and three edges at every atom gives $2E = 3V$, and $V - E + F = 2$ closes it. What falls out is that a face of k sides contributes 6 − k, and the total is twelve. Six is the face size that makes the flat net, the honeycomb, and the charge is a measure of departure from flatness — so a hexagon is the face that departs by nothing. The zero is the whole subject rather than an artefact.

A quantity that is blind to one variable cannot constrain it. The charge holds at any number of hexagons, so if a face vector is realisable at all it is realisable at some hexagon count, and if it is unrealisable at a particular hexagon count the charge has nothing to say about that. Eberhard’s theorem states the first half — that the freedom is genuinely used — and it is not a restatement of the arithmetic, because nothing in the arithmetic says a net has to exist anywhere.

Ten vectors, every arrangement wound up #

The theorem is proved by a construction that is not reproduced here. What is computed instead is the object the theorem describes and does not give: for each face vector, which numbers of hexagons are realised.

The method is the one the projective census uses, applied to arbitrary face sizes. Peel a polyhedron’s faces off in a spiral — a first face, a ring round it, a ring round that — and write down the number of sides of each. A face vector plus a hexagon count is a multiset of face sizes, so every arrangement of that multiset is a candidate spiral, and each one is wound up: the next face is joined to the one before it and to the oldest face on the boundary still short of neighbours, until the last face closes the lid or the winding fails. The ones that close are identified up to relabelling by a canonical code read from every starting face and direction, so a solid arriving from several spirals is counted once.

Six of the ten vectors form a chain from the cube to the dodecahedron. A square charges two and a pentagon one, so trading one square for two pentagons keeps the total at twelve, and six squares, five squares with two pentagons, four with four, three with six, two with eight, one with ten and none with twelve is a walk along the charge at a fixed value. Two more carry a face larger than a hexagon, which charges a negative amount that extra pentagons have to cover, and the first is the tetrahedron’s.

What a zero costs, and the two rows that are not zeros #

A zero in that grid is worth having only if the search could have found something, so the two things that can stop it are kept apart rather than added together.

A polyhedron with no face spiral would be missed. Not every trivalent polyhedron has one; the smallest known counterexample among cages of pentagons and hexagons has three hundred and eighty atoms, and for the small vectors here the evidence that nothing is missing is that the counts reproduce the solids everybody knows and, for the twelve-pentagon row, the published enumeration.

And the planar reader can decline. Reading a wound-up adjacency back as a map on the sphere uses a fact about cages of pentagons and hexagons: two neighbouring faces share exactly two further neighbours, which is what fixes the cyclic order of the faces round a face. Three faces meeting one another pairwise without meeting at a point break that, and the reader returns nothing rather than something wrong.

The truncated tetrahedron is the case that settles the policy. It is a solid anyone can picture — four triangles, four hexagons, twelve corners — it belongs to the row for four triangles at four hexagons, and that row is declined. Reported as a zero it would be a flat falsehood; reported as a decline it is an honest silence. So every row carries the count of spirals that wound up and could not be read, and a row with any of them is a question mark rather than a dash. The limitation was found this way rather than reasoned about in advance: the four-triangle row declined at exactly one column, and the column had a famous solid in it.

That is why the ten vectors are the ten they are. Vectors with triangles and squares in quantity produce solids whose small faces meet one another three at a time, and their rows fill with declines; the pentagon-and-larger vectors produce none at all across every column shown, so their dashes are evidence.

The dodecahedron’s row, and the gap in the middle of it #

The twelve-pentagon row is the one already known in part, and it is worth seeing whole.

At no hexagons the winding finds one solid from a single arrangement, and it is the dodecahedron. At one hexagon there are thirteen arrangements, every one of them is tried, and not one closes — which is the sphere’s single exception arrived at by exhaustion rather than quoted, as it was when the gap was first reported. At two hexagons there is one solid again, at three one, at four two and at five three, and the counts go on rising.

So the row has a hole: a count that is empty with populated counts on both sides of it. That is a different object from a leading run of zeros, and the difference matters for what the theorem does and does not promise. Eberhard’s theorem says some column is populated. It says nothing whatever about which, nothing about whether the populated columns are consecutive, and nothing about whether a populated column stays populated as more hexagons are added.

And the hole is not one row’s peculiarity.

Six squares is the cube, and six squares with two hexagons is the hexagonal prism; six squares with one hexagon does not exist. Five squares and two pentagons behaves the same way. A single hexagon is the one addition that cannot be made, three times over, in vectors that have nothing else in common, and the reason is the same in all three: a lone hexagon in a net of smaller faces has six edges to place and every face it could meet already has its own quota of neighbours, so the winding runs out of boundary before the lid fits. That the sphere’s famous missing cage has two cousins among the squares is a fact nobody had asked for, and it comes out of putting the same question to a row rather than to a number.

The least count is not a function of the charge #

The clearest statement the grid supports is about what the charge determines.

Six of the ten need no hexagons at all: the tetrahedron, the cube, the dodecahedron and the solids between the last two on the trade of a square for two pentagons. One square with ten pentagons needs two. Thirteen pentagons with a heptagon need three. Fourteen pentagons with an octagon need four.

Every one of those rows charges twelve. So no arithmetic on the charge can produce the least hexagon count, because the charge takes one value across rows whose answers take four. That is the sharpest form of the theorem’s own weakness: it promises a column and cannot name it, and the reason it cannot is that the information is not in the quantity it works with.

The pattern in the four answers is suggestive and is not a rule. The vectors needing hexagons are the ones with a face that is not a pentagon — a square among ten pentagons, a heptagon, an octagon — and the odd face is one the other faces cannot surround without hexagons between. An octagon has eight edges and needs eight neighbours; pentagons brought up against it and against each other run out of charge before they run out of edges, and a hexagon is the only face that supplies an edge and costs nothing. The hexagons a vector needs are the padding its odd face requires, and the count rises with how odd the face is. Whether it rises in any stateable way is not established by four data points.

The heptagon row also shows what happens once a row starts. One solid at three hexagons, two at four, eight at five: the count rises fast, because a hexagon added to a solid that already exists can usually be inserted in several inequivalent places, and the number of places rises with the solid’s size. A row is empty, then it is populated, then it is crowded, and the interesting part of it is entirely at the left-hand end.

The same question at four bonds an atom, and on the other surfaces #

Everything above has three bonds at every atom, and that is where the six comes from. The twelve belongs to the vertex shows what happens at other degrees: with four bonds at every atom the neutral face is the square and the sphere charges sixteen, with five it is the triangle and the charge is twenty, and with six no closed cage exists at all.

Each of those has its own Eberhard’s theorem and each has its own blind spot, in the same place and for the same reason. At four bonds an atom the charge is $\sum_k (4 - k)\,p_k = 16$, a square contributes nothing, and the count of squares is the entry the charge cannot see. Eberhard’s theorem in that setting says a square count exists making the vector realisable, and the theorem is due to Grünbaum rather than to Eberhard because the four-valent case was not in the 1891 book. The structure is identical and the neutral face moves, which is the clearest evidence that the freedom is about flatness rather than about the number six.

The surfaces move it too. On the torus the charge is nothing, so a pentagon has to be paid for by a heptagon and the hexagons are free in the strong sense: the face vector permits any number of them and any balanced number of the rest. Dividing a periodic net by its translations is where such nets come from, and whether every balanced vector is realised there is the same question this page asks about the sphere, with no counterexample known either way and no census here that reaches it.

Where the exactness stops #

Computed here. For each of ten face vectors and each hexagon count from none to five: every arrangement of the faces, wound up; the ones that close, read as maps on the sphere; the distinct solids among them by canonical code; and, for each solid found, a check that it has 2F − 4 atoms, 3F − 6 bonds and three bonds at every atom. The counts of spiral arrangements tried are printed beside the rows, and the largest is the twelve-pentagon row at five hexagons.

Not proved. Eberhard’s theorem itself. Nothing above establishes that every charge-twelve vector is realised somewhere; what is established is that ten of them are, at the counts shown. A construction proving the general statement is Eberhard’s and is not reproduced.

The spiral assumption, restated where it bites. A dash means no spiral arrangement closed. A polyhedron with no face spiral at all would produce no arrangement that closes and would not be seen, and the search cannot distinguish that from absence. For the twelve-pentagon row the counts agree with an enumeration made by a method that assumes no spiral, which is the only place that assumption is independently checked.

A decline is not a zero. Two rows in the grid carry question marks, and they are rows the planar reader could not report on. The truncated tetrahedron sits in one of them.

And five is where the search stops. Every row is taken to five hexagons and no further, because the number of arrangements grows as a binomial and the twelve-pentagon row at six hexagons is already past the bound the search is allowed. Nothing here says what any row does beyond five.

The solids are combinatorial, not convex. Everything found is a pattern of contacts, and whether it can be drawn with flat faces is a separate question with a separate theorem behind it — Steinitz’s, which says a three-connected planar graph is a convex solid and which nothing here verifies. The same distinction runs through the construction that finds a centre in a cage, where a cage with a centre is a combinatorial object and the centre is an involution rather than a point.

The first refusal is the one that keeps the theorem’s hypothesis honest. Eleven pentagons charges eleven, not twelve, so Eberhard’s theorem promises nothing about it — and the search finds nothing at any hexagon count to three, which is what a vector outside the theorem’s reach should look like. Without that row the grid would be a table of successes with no control in it.

Who proved it, and what it cost him #

Victor Eberhard published the theorem in 1891, in a book on the theory of convex polyhedra, and the proof is a construction: it starts from a solid and adds faces in a way that changes one entry of the face vector at a time, so that any admissible vector is reached from a known one. It is long, it is elementary, and it produces a hexagon count without any control over how large that count is — which is the feature every account since has had to apologise for.

Branko Grünbaum gave the modern treatment in the 1960s and it is his framing the statement above uses. He also proved the sharper result the twelve-pentagon row needs: with Theodore Motzkin in 1963, that a trivalent solid with twelve pentagons and k hexagons exists for every k except one, which is the whole of the sphere’s row and is far more than Eberhard’s theorem gives for that vector. The gap between the two results is the gap between “some hexagon count” and “these hexagon counts”, and it has been closed for a handful of vectors and for no general family.

What Eberhard had no way to anticipate is where the theorem would be needed. A trivalent solid whose faces are pentagons and hexagons is a carbon cage, and the question of which face vectors are realised became a question about which molecules could exist ninety-four years after he answered it. The rows with heptagons and octagons in them are the defective cages, where the heptagon is a hole in the lattice that a disclination describes from the other side.

Still open: the count the theorem will not name #

Eberhard’s theorem says a column exists and every account of it says the same thing about the proof: the hexagon count it produces is not bounded by anything usable. For the ten vectors here the least count is four or less, which is not evidence about vectors in general — it is evidence about ten small ones, chosen partly because they could be searched.

Two questions follow and neither is answered above. How does the least count grow? A vector with several large faces has more to pad and presumably needs more hexagons, and whether the growth is linear in the charge deficit of the large faces, or in the number of them, or in neither, is not something four values can decide. And is a row eventually always full? Every row here that starts, continues, and the counts rise; the one hole found sits at one hexagon in three rows and nowhere else. Whether a hole can occur at a larger count, in any vector at all, is not known here and would be the first thing to look for in a search that could afford to go further.

The row that would settle the second question fastest is not one of these ten. It is a vector whose least count is large — an octagon or a nonagon among pentagons — where the first populated column is far enough out that there is room for a hole behind it, and the search that would reach it is the one this page’s binomial bound is in the way of.