Every net folds onto a torus

A periodic pattern can be folded up along its own symmetries, and what is left is a smaller object that carries all the information. This collection has done it once already: two of the seventeen fold into a surface with no marked points on it at all, and those two are the ones whose groups act freely.

The same fold, performed on a net, is simpler and more useful, because a net’s translations always act freely. Divide the infinite graph by its translation subgroup and the quotient is a finite graph — the quotient graph the whole of this ladder is written in terms of — and it is not merely finite. It is a finite graph drawn on a torus, and the torus is where the accounting happens.

What the fold does to the faces #

The vertices and edges of the quotient are easy: they are the orbits of vertices and edges, and there are finitely many of each. It is the faces that make the accounting work, and they are the part a reader has to be careful about.

A plane net divides the plane into regions. Those regions are permuted by the translations, so they too fall into finitely many orbits, and each orbit becomes one face of the graph on the torus. Nothing about that requires the faces to be drawn or looked at — the orbit count is a property of the net.

And here is the point. The number of faces is not something to count. It is forced. A torus has Euler characteristic zero, so for any graph drawn on it with faces that are discs,

$$n - e + f = 0$$

and therefore f = e − n. Given the quotient graph, the number of faces per cell follows without a drawing being made. For the honeycomb that is 3 − 2 = 1: one hexagon per cell, which is right. For the kagome net it is 6 − 3 = 3: two triangles and a hexagon. For the net of squares and octagons it is 6 − 4 = 2: one square and one octagon.

One condition on that statement has to be said out loud, because it is the one that can fail. n − e + f = 0 holds for a graph on a torus whose faces are discs — every region has to be a disc, not an annulus running the whole way round. A quotient graph can fail that: take the square net and delete one of its two edges, and what is left is a graph on the torus with one vertex, one edge and one region, and the region is a cylinder rather than a disc. The subtraction would say f = 0 and the drawing has one region. What rules that case out for a net is exactly the condition the descriptions are already checked against — the cycles of the quotient must generate the whole of ℤ², and a graph whose regions wrap has cycles generating a rank-one subgroup instead. So the face count and the honesty of the description are one condition, met once.

The relation between degree and face size #

Now divide through, and something familiar appears.

Count the edge-ends. Each edge has two, so summing the degree over the vertices gives 2e; summing the face size over the faces gives 2e as well, since each edge borders two faces. Write q for the mean degree and p for the mean face size:

$$q = 2e/n \qquad p = 2e/f$$

Substitute into n − e + f = 0, divide by 2e, and

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

for every plane net there is. That is the flat case of a relation whose other cases are a sphere and a hyperbolic plane, and it is derived here with no geometry in it whatever — no angles, no lengths, no assumption that anything is regular. What went in was a graph and a translation group; what came out is a constraint on how many things can meet at a point.

There is one step in that derivation where a bug is easy and invisible, and it is the first one. Summing the degrees over the vertices gives twice the number of edges — but only if a loop is counted twice at the vertex it joins to itself, since it has two ends there. Several of the nets in this collection are written with loops: the square net is one vertex with two loops, the triangular net one vertex with three. Count a loop once and the square net comes out with mean degree two instead of four, the relation gives a mean face size of infinity, and every number in the table is still a number. So the degree sum is checked against twice the edge count for every net before the means are formed, which costs nothing and is the difference between an accounting and an arithmetic slip that looks like a result.

The averaging is the part to be careful with. The relation constrains the means, and a net with vertices of degree three and four and faces of several sizes satisfies it while having no vertex and no face of the mean size at all. Only the nets whose vertices are all alike and whose faces are all alike are constrained vertex by vertex — and those are the ones the next essay is about, where the relation has to be solved in whole numbers and has exactly three solutions.

Why the fold is legitimate #

Two conditions have to hold for the accounting to mean anything, and both are checkable.

The translations must act freely. No translation may fix a vertex, or the quotient is not a graph on a surface but something with a cone point in it. For a net this is automatic: a non-zero translation moves every vertex, because the vertices sit in a lattice-periodic arrangement and a translation that fixed one would fix the lattice.

The translation subgroup must be the whole of it. If the cycles of the quotient graph generate only a sublattice of ℤ², the fold has been performed by too small a group: the quotient is a graph on a torus that covers the real one, with twice or three times as many vertices, edges and faces as it needs. The Euler relation still holds — it holds for any torus — and every rate read off it is wrong. This is the same voltage-lattice test that keeps the descriptions honest, arriving here as a condition on a fold.

The same accounting, in three places #

This is the third time this collection has run an Euler argument, and the three are worth setting side by side because the differences are as instructive as the similarity.

On a sphere. Twelve pentagons and no way round them: a closed cage of hexagons and pentagons must have exactly twelve pentagons, whatever its size, because the characteristic is two rather than zero and the surplus has to be paid for somewhere.

On a quotient of the plane. The two that fold into a surface: a wallpaper group acting freely folds the plane onto a torus or a Klein bottle, and there are exactly two such groups because there are exactly two flat closed surfaces.

On a torus, for a graph. This essay: the characteristic is zero, so the faces are determined by the vertices and edges, and the mean degree and mean face size satisfy one equation.

The common structure is that a global topological invariant — a single integer attached to a surface — constrains a local count that a reader would expect to be free. Nothing about a vertex knows how many edges the rest of the net has; the accounting nevertheless forbids most combinations.

The fourth case is the one this collection does not draw and it completes the pattern. A surface of genus two or more has negative Euler characteristic, so the same substitution gives 1/p + 1/q < 1/2 — and every pair of whole numbers with a sum below a half is available, which is why there are infinitely many regular tilings of the hyperbolic plane and only three of the Euclidean one. Read in that order the three cases are one statement about a sign: a positive characteristic forces a surplus that pentagons must pay, a zero characteristic forces an exact balance and admits three solutions, and a negative one imposes no upper bound at all. The plane is the boundary case, and boundary cases are thin. That thinness is what the next essay is about, and it is why so much of crystallography is a search among a handful of possibilities rather than among many.

The fold, and what a fold is for #

It is worth saying what is gained by folding, because the answer is not simply “a smaller object”.

An infinite graph has infinitely many vertices, and a question about all of them is a question with no finite answer unless something reduces it. The fold reduces it exactly: a property that is invariant under the translations is a property of the quotient, and everything in this ladder — the degrees, the faces, the coordination sequence, the group, the rigidity — is invariant under the translations, because the translations are symmetries.

That is a general principle and it is the same one that made the seventeen countable. A wallpaper group has infinitely many elements; the classification is possible because the interesting structure lives in the quotient by the translations, which is a point group of order at most twelve. The fold performed here is the same fold performed on a graph rather than on a group, and it does the same work: it turns an infinite question into a finite one without losing anything.

The price is that the quotient has to be read with its voltages. Two graphs on a torus can be the same abstract graph and different nets, and the difference is entirely in how the edges wrap — which is the whole content of the two freedoms that leave a net unchanged and the one piece of information that must not be dropped.

What a face is, when there is no drawing #

There is a subtlety worth facing directly, because the essay has been using the word face as though a net had faces and a graph does not.

A graph has faces only once it is drawn on a surface, and different drawings can give different numbers of them. What rescues the argument is that a plane net comes with a drawing: it is a graph in the plane, and its regions are determined. The fold then carries those regions to the torus and the count is the count of their orbits.

So the accounting is about a net as a plane graph rather than about the abstract graph underneath. That is a genuine restriction and it is worth stating, because the abstract quotient graph of the honeycomb — two vertices, three parallel edges — can also be drawn on a sphere, where it has three faces rather than one and the relation gives 2 rather than 0. The voltages are what pick out the torus, and without them the question has no answer.

What the relation forbids, in ordinary words #

The equation is short and its content is easier to feel in a list than in symbols.

Three edges at every vertex asks for faces of six sides: the honeycomb. Four edges at every vertex asks for faces of four: the square net. Six edges asks for three: the triangular net. Those are the three that come out exactly, and everything else has to be a mixture.

Five edges at every vertex would ask for faces of ten-thirds of a side, which is not a number of sides anything can have. So a net cannot have every vertex of degree five and every face the same size — although it can perfectly well have every vertex of degree five, with faces of several sizes averaging out, and one such net appears in this collection. That distinction is easy to lose and it is exactly the distinction between a constraint on averages and a constraint on individuals.

Seven edges at every vertex would ask for faces of fourteen-fifths, which fails the same way and fails harder: the mean degree cannot exceed six with faces of at least three sides, because 1/3 + 1/q = 1/2 gives q = 6 and any larger q makes the sum too small. A plane net’s mean degree is at most six, and it is exactly six only when every face is a triangle. It is the same bound the densest packing of the plane runs into from the other side, where six is how many equal discs can touch one. That is a bound obtained by arithmetic on two integers, and it holds for every periodic graph in the plane whatever it looks like.

Where it stops being flat #

The relation 1/p + 1/q = 1/2 is an equality for a plane net and it is worth asking what a departure from it would mean, because the answer organises three geometries into one line.

If 1/p + 1/q > 1/2 the accounting is that of a sphere: the graph is a polyhedron, it closes up, it is finite. If 1/p + 1/q < 1/2 the accounting is hyperbolic: the graph has more room than the plane can supply, and it can only be drawn on a surface of negative curvature. The plane is the boundary case, and it is the only one of the three that is periodic.

That is a strong statement about what a crystal net can be. A net that wanted seven hexagons round a vertex is not a net that is hard to draw — it is a net the plane has no room for, and the arithmetic says so before any drawing is attempted. The next essay solves the equation in whole numbers and finds the three that survive.

An accounting with no drawing at any point #

Every number in this essay was produced without a picture being made, and it is worth tracing the chain once to see how little was needed.

The input is a list of edges with voltages. The vertex count is the length of a list. The edge count is the length of another. The face count is a subtraction. The mean degree and mean face size are two divisions. The relation is checked by adding two reciprocals. At no point is anything placed anywhere, and at no point is a length used.

That is the sense in which this is the cleanest result in the ladder. The placement needed a linear solve; the rigidity needed a metric and a rank; the coordination sequence needed a search over an infinite graph. The Euler accounting needs six integers and two divisions, and it constrains every net there is.

It is also, for the same reason, the weakest. A relation that costs nothing decides very little: it rules out a great many combinations and identifies none.

Two things this does not say #

It does not say a net is determined by its face and degree counts. Two nets can agree on every number in the table above and be different nets — the relation is one equation and a net is a great deal more information than that.

It does not say the faces are the chemistry. In a real framework the rings that matter are the ones a guest molecule can pass through, and those are not always the faces of the plane graph: a large face may be subdivided by bonds that lie outside the plane, and in three dimensions the notion of a face disappears entirely and is replaced by the far more delicate business of counting rings. The two-dimensional case is clean precisely because it is two-dimensional.

The faces are cycles, and two cycles are not faces #

The face count came out of the Euler relation as e − n. It can be got a second way, from the quotient graph’s cycles alone, and the second route says which cycles the faces are and which they are not.

The cycles of a graph form a vector space, and its dimension is e − n + 1 — one independent cycle for each edge beyond a spanning tree. That number is one more than the face count, and the discrepancy is exactly two once the relation among the faces is accounted for: the face boundaries sum to zero, so they span a subspace of dimension f − 1, and (f − 1) + 2 = e − n + 1.

So two independent cycles of the quotient graph do not bound faces, and they are the two that carry the periodicity — the ones whose voltages generate ℤ². That is the same pair the index check is about, seen from the other side: a quotient graph whose cycles fail to span the translations is one whose two non-bounding cycles are not independent, and the fold was performed by the wrong group.

The consequence is a floor. A two-periodic net needs at least two independent non-bounding cycles, so e − n + 1 ≥ 2 and therefore f ≥ 1: every plane net’s quotient has at least one face, which is why the smallest quotient in the registry — one vertex, two loops — has exactly one.

What a face is doing in an object with no drawing #

The essay is careful to say that a graph has faces only once it is drawn. The cycle argument above makes that precise rather than merely cautious, and it says exactly how much of the drawing is being used.

A cycle is combinatorial: it is a closed walk in the quotient graph, and it exists whether or not anything is drawn. A face is not: it is a cycle chosen as bounding a region, and the choice comes from the embedding.

What the essay’s accounting relies on is that the embedding is fixed by the net rather than by anybody’s drawing — the net is a plane graph, so its faces are the regions of the plane, and folding carries them to the quotient’s faces. The freedom a general graph has, of being drawable on several surfaces with different face counts, is not available here, and that is the whole of what the plane periodicity buys.

So the relation 1/p + 1/q = 1/2 is a statement about plane nets and would fail immediately for the same abstract graph drawn elsewhere. A quotient graph with two vertices and three edges can be drawn on a torus with three faces — the honeycomb — and on a surface of higher genus with fewer, and only the first of those is a net.

Where this goes next #

The next essay takes the relation and asks for solutions in integers, which is where the three regular nets come from and where the other two geometries appear as the near misses.

Two rungs away, the same quotient graph gives a different constraint: the crystallographic restriction, which turns out to be a theorem about periodic graphs with no lengths in its proof either — and, like this one, it constrains a local count by way of a global fact about ℤ².