As many heptagons as pentagons

Twelve pentagons and no way round them is Euler’s relation doing something the crystallographic restriction cannot. On a flat repeating net a five-sided face is impossible; on a closed surface of the sphere’s kind, a trivalent net must have exactly twelve of them, however large it is and however many hexagons come with them.

The argument is three lines. Euler gives V − E + F = 2; three edges at every vertex gives 3V = 2E; counting edge-sides face by face gives Σ n·pₙ = 2E. Substitute and clear denominators, and the whole of it is Σ (6 − n) pₙ = 12. A hexagon contributes nothing, a pentagon contributes one, and the number of hexagons never appears. Twelve pentagons, always.

Everything in that argument except the number two is a fact about trivalent nets, and the two is the Euler characteristic. Change the surface and the right-hand side changes with it — and the torus is where the change is most interesting, because there the answer is zero and zero does not mean “no pentagons”.

Zero is not the same as none #

On a plane the same sum comes to zero as well, and there the conclusion really is that every face is a hexagon — because a plane net is infinite, its faces cannot be counted, and the argument runs on densities where the only non-negative solution with no face below six sides is all sixes.

A torus is different in the one way that matters: it is closed, so a net on it is finite and its faces can be counted. The sum Σ (6 − n) pₙ = 0 is then a condition on a finite list of numbers, and it has many solutions. Two pentagons and two heptagons satisfy it. Ten pentagons and ten heptagons satisfy it. So do a hundred and a hundred, and so does one pentagon, one heptagon and any number of hexagons.

So the sphere fixes a count and the torus fixes a difference. That is the whole distinction, and it is worth stating in the form that makes it a prediction: a toroidal carbon cage with pentagons in it has exactly as many heptagons, and the count of either is unbounded.

The cheapest defect there is #

The claim is easy to make and worth making by construction, so here is the construction.

Wrap the honeycomb onto a torus by dividing it by a sublattice of its own translations — which is what every plane net does — and the quotient is a finite trivalent graph with 2k² vertices, 3k² edges and k² faces, all hexagons. Euler’s relation gives zero and the charge gives zero, which is consistent and uninformative.

Now rotate one bond through a right angle. The two atoms it joins keep each other and swap one neighbour each: the bond turns in place, and no vertex and no edge is created or destroyed.

That move is the Stone–Wales rotation, and it is the cheapest defect a hexagonal sheet has: no bond is broken in the counting sense, nothing is added, and the net afterwards is a perfectly good trivalent net. What it costs is four hexagons, and what it makes is two pentagons and two heptagons — the smallest possible pair of opposite charges.

The two pentagons cost two and the two heptagons pay two, so the charge is nought before and nought after, which is what a torus demands. It could not have come out otherwise, and the fact that it does is the check that the move was implemented rather than described.

What a pentagon costs, as an angle #

The charge 6 − n is a bookkeeping number and it has a geometric meaning worth having, because with it the whole result reads as a statement about curvature rather than about counting.

Three regular hexagons meeting at a point fill exactly a full turn: three times a hundred and twenty degrees is three hundred and sixty. So a trivalent net of hexagons lies flat, and every vertex of it has no angle left over.

Replace one of those hexagons by a pentagon and the angles at that vertex sum to less than a full turn — the sheet has a wedge missing and must curve towards itself to close. Replace one by a heptagon and the angles sum to more, so there is surplus material and the sheet must buckle away. The angle defect at a vertex is what is missing from a full turn, and it is positive at a pentagon’s corners and negative at a heptagon’s.

Gauss and Bonnet’s theorem for a polyhedral surface says the total angle defect over all vertices is 2π χ, and distributing each face’s contribution over its corners turns that into exactly Σ (6 − n) pₙ = 6χ. The charge is the angle defect in units of sixty degrees, and the accounting is a discrete curvature integral.

Read that way the three cases are one sentence. A sphere has positive total curvature, so it needs twelve sixty-degree wedges taken out. A torus has none, so what is taken out must be put back. A surface with more holes has negative total curvature, so material must be added, which is what a heptagon does.

Why the faces are traced and not counted #

A face of a graph on a surface is not a property of the graph. The same graph embedded two ways has different faces, and “how many sides does that face have” is a question about the embedding.

What carries an embedding, combinatorially, is a rotation system: a cyclic order of the edges at each vertex, which is the order they leave it going round. Given one, the faces are found by a trace — leave a vertex along an edge, arrive at the far end, and depart along the next edge round from the one just arrived on. Repeat, and the walk closes; the walk is a face, and every directed edge lies in exactly one of them.

Doing it that way is what makes the rotation computable, and it caught an error immediately. The bond rotation is a re-linking of the rotation system and nothing else, and there is an obvious wrong way to write it: put the incoming neighbour in the slot the outgoing one has just vacated. That is not a rotation of the bond. The trace comes back with two faces of twelve sides and an Euler characteristic of minus two — which is not the surface the net is on.

The incoming neighbour has to take the slot of the neighbour that stays, displacing it round — which is what turning a bond through a right angle does to the cyclic order — and the trace then gives the two pentagons and the two heptagons.

Nothing but a trace would have said so. The wrong version produces a perfectly well-formed graph with the same vertices and the same edges; every count of those agrees; and the only thing that betrays it is that the faces it makes belong to a different surface.

What each surface charges #

Running the same three lines on a surface of genus g gives Σ (6 − n) pₙ = 6χ = 12(1 − g), and reading the ladder off is the point of having it.

At genus nought — the sphere — the charge is twelve, so pentagons are compulsory and there are twelve of them. That is the fullerene result.

At genus one — the torus — the charge is nought, so a net of hexagons alone exists. It is the only surface on which one does. That is not a fact about hexagons; it is the statement that the charge is six times the Euler characteristic, and only one closed orientable surface has characteristic zero.

At genus two and above the charge is negative, so heptagons become compulsory — twelve more heptagons than pentagons for every extra hole. A carbon structure built on a surface of many handles has to be full of seven-membered rings, and those structures have a name — schwarzites, after the minimal surfaces they are modelled on — and the arithmetic above is why nobody looking for them looks for a net of hexagons.

Pentagons close a surface and heptagons open it. A pentagon is a place where the sheet is short of material and must curve towards itself; a heptagon is a place where there is too much and it must buckle away. The charge is the total of those, and the surface’s topology is what it has to add up to.

The two counts a torus does not fix #

It is worth saying explicitly what the torus leaves free, because “the counts are otherwise unbounded” is easy to read as a technicality and is the substance of the result.

The number of hexagons is free on every surface, sphere included: it never appears in Σ (6 − n) pₙ because 6 − 6 is nought. That is why fullerenes come in a sequence — sixty atoms, seventy, seventy-six, and on upwards without limit — all with twelve pentagons and a growing number of hexagons. The sphere fixes the pentagons and says nothing about the size.

The number of pentagons is what the torus adds to that freedom. A toroidal net may have none, in which case it is a wrapped honeycomb; or two, which is one Stone–Wales rotation; or four, which is two rotations; and there is no ceiling. Each rotation adds a pentagon–heptagon pair and the accounting never notices, because the pair is neutral.

So a toroidal net has two free counts where a spherical one has one, and the extra freedom is exactly the number of neutral defects it carries. That is a statement about how much a topology constrains a structure, and the comparison is the useful part: the sphere is a strong constraint that leaves the size free, and the torus is a weaker one that leaves the size and the defect count free.

And it is why a hexagonal sheet can be rolled and a spherical one cannot be flat. A nanotube is a wrapped honeycomb with no defects, which the torus’s zero permits; its caps are hemispheres and each carries six pentagons, which is the sphere’s twelve split in half between the two ends. The accounting for the whole closed tube is the sphere’s, and the accounting for its middle is the torus’s, and the pentagons are all at the ends because that is where the curvature is.

Where the restriction is, and where it is not #

This whole family of results sits beside the crystallographic restriction and is not it, and the distinction is worth keeping.

The restriction forbids a five-fold rotation of a lattice, and its proof is that the trace of an integer matrix is an integer. It is about symmetry.

The counting here forbids or compels five-sided faces, and its proof is Euler’s relation. It is about topology, and it never mentions a symmetry: a fullerene’s twelve pentagons are there whether or not the cage has any symmetry at all, and the icosahedral ones are the ones somebody chose to build.

The two meet in the flat case and agree, which is why they are easy to confuse. A periodic trivalent net in the plane has charge zero and therefore all hexagons; a periodic pattern in the plane has no five-fold rotation. Both say “no fives”, from different premises, and on a torus they part company: the torus has charge zero like the plane, and it admits pentagons, because it is closed and the plane is not.

Who noticed, and in what order #

The sphere’s twelve is old — it is in Euler’s own relation and was written down for polyhedra long before anybody wanted a carbon cage — and the torus’s zero is a corollary nobody had a use for until there were structures to have it about.

What made it a subject was the fullerene, in 1985, and then the nanotube. Both are the sphere’s arithmetic: a cage has twelve pentagons and a capped tube has twelve split between its two ends. The toroidal case arrived immediately afterwards as a question — can a carbon torus exist? — and the answer the counting gives is yes, with any number of pentagons provided the heptagons match.

The Stone–Wales rotation is older than either, proposed in 1986 as the cheapest way for a fullerene to rearrange itself, and it is the move that makes the balance visible. Applied to a sphere it turns twelve pentagons into a different arrangement of twelve; applied to a flat sheet or a torus it makes a pentagon–heptagon pair out of nothing. The same move does different things on different surfaces, and the accounting is what says which.

The lesson generalises past carbon and is worth carrying: a local move that is neutral under a topological invariant can be applied anywhere, and a move that is not is confined to the surfaces whose invariant permits it. That is why the pentagon–heptagon pair is the universal defect of hexagonal sheets and a lone pentagon is not a defect at all — it is a change of surface.

What the trace refuses #

The last is the one that makes the torus’s balance a statement. If a sphere could also have equal counts, “as many heptagons as pentagons” would be an identity about counting rather than a fact about the surface — and it cannot, because the sphere’s charge is twelve and twelve is not nought. The test enumerates the sphere’s solutions and requires that none of them balance.

The second and third are a pair, and the pair is what the implementation is checked by. A move that leaves the counts alone but makes the wrong faces is the failure the wrong re-linking produced; a move that makes the right faces by adding a vertex somewhere would be a different move with the same name.

Where the exactness stops #

Computed here: the honeycomb wrapped onto a k by k torus as a graph with a rotation system read once off the geometry; the faces of it, by tracing darts; the vertex, edge and face counts and the Euler characteristic from that trace; the bond rotation as a re-linking of the rotation system; the faces again; and the charge Σ (6 − n) pₙ before and after.

Trivalent throughout. 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.

A net on a torus, not a torus of atoms. Nothing here says a toroidal carbon cage is stable, or that a Stone–Wales rotation is energetically cheap in a real sheet — it is famously the cheapest topological defect and its energy is a chemical quantity this collection does not compute. What the counting settles is which face vectors are possible, which is a smaller and firmer claim.

Genus, and orientability. The ladder above is for closed orientable surfaces. A Klein bottle also has Euler characteristic zero and also admits a hexagonal net, and it is not on the ladder because the drawing of the ladder is in genus and a Klein bottle has none. The arithmetic is the same and the bookkeeping of non-orientable surfaces is not done here.

Where the ladder goes next #

Back, to the sphere: twelve pentagons and no way round them, where the same three lines give twelve rather than nought, and to every net folds onto a torus, where dividing a plane net by its own translations is what puts it on the surface used here.

Sideways, to the symmetry version of the same prohibition: the crystallographic restriction, which forbids a five-fold rotation for a reason that has nothing to do with Euler, and where five-fold becomes legal, where more dimensions relax it.

Onward, to the surfaces a group can wrap onto: the two that fold into a surface, where p1 gives a torus and pg a Klein bottle, and ten ways for space to be flat, which is the same question one dimension up.