Three answers in whole numbers

Folding a plane net onto its own torus produced one equation. If q is the mean number of edges at a vertex and p the mean number of edges round a face, then

$$\tfrac{1}{p} + \tfrac{1}{q} = \tfrac{1}{2}$$

for every periodic graph in the plane, with no assumption that anything is regular and no length anywhere in the derivation.

Now ask for the cases where the means are not merely means. A net in which every vertex has the same degree and every face the same size must satisfy the equation in whole numbers, and whole-number solutions of an equation like that are countable.

The three #

Rearranged, the equation says p = 2q/(q − 2), and p is a whole number for only three whole values of q above two.

q = 3, p = 6. Three edges at every vertex, six edges round every face: the honeycomb. It is the net of graphite, of a beehive’s cross-section, and of every hexagonal mesh ever woven.

q = 4, p = 4. Four and four: the square net, which is the square lattice read as a graph.

q = 6, p = 3. Six edges at every vertex, three round every face: the triangular net.

And that is the list. Not three that anybody has found so far — three, with nothing else possible, by an argument that is two lines of arithmetic on a relation that was itself two lines of arithmetic on a fold.

The three are worth taking one at a time, because the quotient graph each of them is written as gets smaller as the degree goes up, and that is the opposite of what a reader expects.

The honeycomb is the largest of the three descriptions and the sparsest of the three nets. Its quotient needs two vertices because no single vertex with loops can have odd degree — a loop contributes two — and three is odd. That is a small piece of arithmetic with a visible consequence: every net of odd degree needs at least two vertices in its quotient, so the honeycomb’s description could not have been shorter however it was written.

Two integers per loop, four integers in all, and the object described is the square lattice with every nearest-neighbour pair joined. The face count is 2 − 1 = 1, and the face is the square. Add a third loop and the degree goes to six, which is the last value the relation permits — so the square net and the triangular net are adjacent descriptions, one loop apart, and there is nothing beyond the third loop that stays flat.

Why this is not the same count as the seventeen #

It is worth being careful about what has been enumerated here, because this collection contains several counts that sound alike and are not.

The seventeen is a count of symmetry groups. The eleven uniform tilings is a count of tilings by regular polygons with all vertices alike, and its argument is about angles adding to 360°. This is a count of combinatorial types of regular net, and its argument has no angles in it at all: it never asks what a face looks like, only how many edges bound it.

The three counts are related and none of them implies the others. In particular the three regular nets here are the three regular tilings that the uniform-tiling count also finds — but the uniform count reaches them through a statement about the interior angles of regular polygons, which is a fact about Euclidean geometry, and this one reaches them through a statement about the Euler characteristic of a torus, which is a fact about topology. Two arguments, sharing nothing, arriving at the same three.

The derivation, in full #

The rearrangement is short enough to write out, and writing it out shows exactly where the finiteness comes from.

Start from 1/p + 1/q = 1/2 and solve for p:

$$\tfrac{1}{p} = \tfrac{1}{2} - \tfrac{1}{q} = \frac{q-2}{2q}$$

$$p = \frac{2q}{q-2}$$

Now q must be at least three, since a vertex of degree one or two in a periodic net either dangles or is a point on an edge. And p must be at least three, since a face bounded by one or two edges is a loop or a pair of parallel edges rather than a region.

Write 2q/(q − 2) as 2 + 4/(q − 2). For p to be a whole number, q − 2 must divide four — so q − 2 is one, two or four, and q is three, four or six. Substituting gives p as six, four and three.

The finiteness comes from a divisibility condition on the number four, and that is the whole reason the list is short. The same shape of argument runs through this collection: only five rotation orders survive because an integer in an interval of length four is one of five numbers, and only five plane lattices exist for a reason of the same kind. A classification is finite when it comes down to a small integer having few divisors.

The near misses on both sides #

The relation is an equality only for the plane, and the pairs that fail it are not failures — they are the other two geometries.

Sum greater than a half. Then the accounting is that of a sphere, and the object closes up: it is a finite polyhedron rather than an infinite tiling. The pairs (3, 3), (3, 4), (4, 3), (3, 5) and (5, 3) are the five Platonic solids, and there are five for exactly the reason there are three here — an inequality in two whole numbers with very little room in it.

Sum less than a half. Then the plane has no room and the object belongs on a surface of negative curvature. Seven triangles at a vertex, or five squares, or four pentagons: each is a perfectly good combinatorial arrangement that cannot be drawn in a flat plane and can be drawn in a hyperbolic one. Those are the hyperbolic tilings, of which there are infinitely many, and Escher’s circle prints are pictures of them.

So the three answers of this essay sit on a boundary with a finite list on one side and an infinite one on the other, and the boundary is thin: it is the only place where the arrangement can be both unbounded and flat.

Relaxing the regularity #

Requiring every vertex to be alike and every face to be alike is a strong condition, and almost no real structure meets it. What happens when it is dropped is the reason the relation is worth having.

The equation still holds, with p and q as means. So a net with vertices of degree three and four has a mean degree strictly between them, and its mean face size is forced to be whatever makes the sum a half — which is a number no individual face need have. The star net has mean degree three and mean face size six, and has faces of three and of twelve and none of six.

That is the sense in which the relation constrains a real structure: not by telling it what to be, but by fixing one number once the other is known, whatever the mixture.

Two useful consequences fall out immediately. A plane net’s mean degree is at most six, because faces have at least three sides. And a plane net whose vertices all have degree five must have faces of several sizes, since ten-thirds is not a number of sides — but such nets exist in quantity, and one of them is in this collection.

The half-truth about pentagons #

The relation is often quoted in the form the plane cannot be tiled by regular pentagons, and that statement is true and is not what the arithmetic says.

What the arithmetic says is that a net cannot have every face of five sides and every vertex of the same degree, because the equation would need q = 10/3. Faces of five sides are not the problem at all: fifteen convex pentagons tile the plane, and the Cairo tiling is a pentagonal tiling anybody can find on a pavement. What those tilings do is have vertices of two different degrees — some of three and some of four — so the mean falls between and the relation is satisfied.

The word regular is doing all the work, and dropping it turns a theorem into a false statement about pentagons. It is worth the pedantry, because the false version is repeated constantly.

The three are also told apart by a measure that has nothing to do with the relation, and it is worth quoting because it is so plain. Counting outwards from any vertex, the honeycomb finds 3, 6, 9, 12 vertices at successive distances, the square net finds 4, 8, 12, 16 and the triangular net 6, 12, 18, 24. Each is exactly the degree times the distance, with no correction term at any distance — which is not true of most nets and is not implied by regularity, and which is the cleanest evidence that these three are the simple cases rather than merely the ones an equation happened to select.

Two nets that are not the same net and satisfy the same pair #

Solving the equation gives a pair of numbers, and a pair of numbers is not a net. It is worth exhibiting the gap.

Take degree four with faces of four. The square net satisfies it. So does the net obtained by writing the square net on a doubled cell and joining the corners differently — and so, more interestingly, does the kagome net, whose mean degree is four and whose mean face size is four, made up of triangles and hexagons in the ratio two to one. Three genuinely different objects, one pair of numbers.

Only one of them has every face of four sides, and that is the extra condition the regular case imposes. But the fact that the pair (4, 4) is realised by several nets is the point: the equation is a constraint on a net and never a description of one, and any argument that treats a solution of it as an identification has skipped the step where the net is actually determined.

That step is what the rest of this ladder is for. A coordination sequence separates the three; a canonical placement separates them and names their groups; and the pair of numbers separates nothing.

What the three have that the others do not #

There is one more property of the three, and it is the one that makes them the right answer to a question rather than the solutions of an equation.

Each of the three is the most symmetric net of its degree. That is not automatic — a net’s group is whatever its equilibrium placement turns out to have, measured on the basis its own edges ask for — and it is checkable: the honeycomb and the triangular net both come out at p6m, the square net at p4m, and no other net in this collection of the same degree does better. The skew net has degree six like the triangular net and its group is pmm, of order four rather than twelve.

So regular in the combinatorial sense and most symmetric in the geometric sense agree here. They need not have: they are different conditions, one on a graph and one on a drawing, and the fact that they pick out the same three nets is a small piece of evidence that the combinatorial notion is the right one.

Where the three turn up outside this collection #

A list of three is short enough to be met everywhere, and it is worth noticing how often the same three arrive from different directions.

They are the three regular tilings of the plane, which is the geometric statement. They are the three lattices whose Voronoi cells are regular polygons, which is a statement about the cell nobody chose. They are the three two-dimensional structures whose atoms all have identical environments and whose bonds are all identical — the condition a chemist would call uninodal and edge-transitive. And they are the three ways of dividing a torus into congruent regular pieces.

Each of those is a different property and each picks out the same three objects. That is the situation this collection is always looking for: a small count reached along several routes, none of which is a rewording of another. When it happens, the count is usually saying something about the plane rather than about the definition that produced it.

The contrast is with counts that are artefacts. If the definition had required faces to be regular polygons rather than to have the same number of sides, the answer would be the same three; if it had required only that the vertices be alike, the answer would be eleven; and if it had required nothing at all it would be infinite. Which count a number belongs to is the first thing to establish about it.

What is not settled #

The three are combinatorial types, not structures. A net of degree three with hexagonal faces might be built of atoms at any spacing, in any cell shape the hexagonal metric allows, and the equation says nothing about which.

Nothing here counts nets. The equation constrains a pair of numbers; it does not enumerate the nets that realise a given pair, and there is no reason to expect that count to be small. There are many nets of degree four besides the square one, and telling them apart is what the coordination sequence and the canonical placement are for.

Nor does the equation know about symmetry. Two nets can satisfy the same pair, have the same coordination sequence and still differ; and two nets that satisfy different pairs can have the same plane group, which several rows of the registry do. The equation is an accounting identity and accounting identities do not identify.

And the three-dimensional analogue does not exist in this form. The Euler relation on a three-torus gives n − e + f − c = 0 with cells as well as faces, which is one equation in three unknowns rather than two, and it constrains far less. The tidiness of this argument is a two-dimensional tidiness.

Why the list has the symmetry it has #

Three answers, and they are not three unrelated numbers: the list has a structure, and naming it explains why the middle case is the odd one out.

The equation 1/p + 1/q = 1/2 is symmetric in p and q, so if (p, q) is a solution then so is (q, p). The three come out as one self-paired case and one exchanged pair — (4, 4), and (6, 3) with (3, 6) — which is why a list of three looks like it should have been even.

That exchange is not a formal symmetry of the equation but a real relation between the objects: it is duality. Put a vertex in the middle of every face and join two whenever their faces share an edge, and a net of degree q with faces of p sides becomes one of degree p with faces of q sides. The honeycomb and the triangular net are dual to each other; the square net is dual to itself.

So the three regular nets are two objects, one of which is its own partner. And every count in this ladder that comes out at three has that shape underneath it — the eleven uniform tilings and their eleven duals are the same statement made where the pairing is not so nearly trivial.

The same relation, one vertex at a time #

The equation above is global: it holds for a whole periodic net and says nothing about any particular vertex. There is a local version, it is what the global one is a sum of, and it is more useful for irregular structures than the mean-value reading.

Assign to each vertex the quantity

$$1 - \frac{q_v}{2} + \sum_{f \ni v} \frac{1}{p_f},$$

where q_v is the vertex’s degree and the sum runs over the faces meeting it. That is the combinatorial curvature at that vertex, and summing it over a fundamental domain gives exactly the relation this essay is about — zero for a periodic plane net, positive for a closed cage, negative in the hyperbolic case.

The local form does two things the global form cannot. It applies to a single vertex, so a net with a mixture of vertex types can be examined one vertex at a time and the deficits and excesses read off directly. And it turns the classification into an accounting: a net with a few positively curved vertices must have negatively curved ones to pay for them, which is the constraint an irregular net actually satisfies.

It also has consequences the mean-value version does not reach. A net every one of whose vertices has strictly positive combinatorial curvature must be finite — it closes up, and cannot be a periodic plane net at all — which is the same statement as twelve pentagons on a cage generalised past regularity. The averages hide that, because a mean can be zero while individual terms are not, and the whole interest of an irregular structure is in the individual terms.

Where this goes #

The arithmetic here is the flat member of a family whose other members are already in this collection: twelve pentagons on a closed cage is the spherical one, and five solids from one inequality is the same inequality solved for finite objects. Reading the three side by side is the clearest way to see that a single accounting is doing all of it.

And the constraint that runs in parallel to it — the crystallographic restriction proved about graphs — has the same shape: a local count constrained by a global fact about ℤ², with no geometry in the argument.