Concept

Graph isomorphism — where it appears

A relabelling of one graph's vertices carrying its edges onto another's. For nets the relation is stronger, since a relabelling must also respect the voltages up to a gauge and a change of basis, and deciding it is a bounded search.

Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.

the honeycomb net, unfolded over 3×3 cells. The infinite graph the quotient graph names, drawn over 3 by 3 cells with the home cell outlined. Each edge of the quotient becomes one edge per cell, running to the cell its voltage names; the drawing adds coordinates the net does not have, and they are the placement in which every vertex sits at the average of its neighbours. two vertices, three edges, degree three — the graph of graphene and of every hexagonal mesh.

A structure with the distances thrown away

Keep which atoms are joined and throw away where they are, and what is left is an infinite graph that can be written on a postcard: a few vertices, a few edges, and a pair of integers on each. Two things about that writing-down are free, and neither of them changes the net.

applied · Nets
One vertex, two edges: one net. Three edges: no answer at all. Every net with one vertex and the stated number of edges, counted inside boxes of voltages of three sizes, up to change of basis and the sign of an edge. Two edges give one net whatever the box, and the reason is a sentence: two voltages that generate the translations are a basis of ℤ², and every basis is carried to every other. Three edges give more nets in every larger box, and that is not a failure of the search — normalise two of the voltages to a basis and the third is a free pair of integers, so the family is infinite. An enumeration inside a bound reports which of those two situations it is in rather than reporting the count it happened to reach.

Every net with one vertex, counted

A net is a few vertices, a few edges and a pair of integers on each, so a census is available: fix the numbers, bound the integers, enumerate. Two edges give exactly one net at every bound. Three give three, then nineteen, then a hundred and forty-three — and the question changes.

applied · Nets
Two vertices and three edges: two nets, at every box size tried. Every net with two quotient vertices and the stated number of edges, counted inside boxes of voltages of several sizes. One cross voltage is set to zero by the gauge — the freedom that moving one vertex into another cell gives — and the rest are drawn from the box. Each entry is the count of nets whose placement separates their vertices, plus the count of those whose does not: the first has a canonical description and stops growing, and the second does not have one and therefore keeps rising with the box. The reducible column is the descriptions thrown away for a reason the one-vertex census never had — cycles generating the whole of ℤ² and a net whose own cell holds one vertex rather than two — and it is empty at every odd edge count, because the swap that would reduce a description pairs its edges and an odd number cannot pair.

Every net with two vertices, counted

The one-vertex census could not contain the honeycomb, because the honeycomb has two vertices in its cell. Adding the second one closes a family at two nets, removes the floor of p2 entirely, makes a third of the members undrawable, and forces the census to refuse a kind of description the first one never met: an honest quotient graph written on twice the cell it needs.

applied · Nets
The sphere fixes a count; the torus fixes only a difference. Euler's relation for a trivalent net gives Σ (6 − n) pₙ = 6χ, so the surface fixes one linear combination of the face counts and nothing else. On a sphere that combination is twelve, which with no face smaller than a pentagon forces exactly twelve pentagons. On a torus it is zero, which permits any number of pentagons provided as many heptagons pay for them — and permits none at all, which is the plain hexagonal net. On a surface of two holes it is minus twelve, so heptagons become compulsory instead.

As many heptagons as pentagons

A trivalent net on a sphere must have exactly twelve pentagons. The same three lines of arithmetic on a torus give zero — which does not forbid pentagons, it makes them pay: every pentagon has to be balanced by a heptagon, and the counts are otherwise free. One rotated bond in a wrapped honeycomb makes two of each and changes nothing else.

restriction · Curvature
Every vector realised, and not at the same hexagon count. Each row is a set of faces other than hexagons whose charge — the sum of 6 − k over them — comes to twelve, which is what a closed trivalent net on the sphere must pay. Each column is a number of hexagons added to that set, and the entry is how many different solids exist with exactly those faces, found by winding up every arrangement of them into a spiral. A dash means the search found none; a question mark means the planar reader declined the row and it is not evidence either way. Every row has an entry somewhere, which is Eberhard's theorem, and the first one is at 0, 2, 3, 4 hexagons depending on the row — so the charge decides everything except the number of hexagons, and the number of hexagons is not a function of the charge.

Everything except the hexagons

Three counts of what a closed net must carry end on the same admission: an arithmetic saying what a net must charge does not say that a net exists. Eberhard's theorem says how close the charge comes to being enough, and the answer has a shape nobody would guess — it fixes every face count except the hexagons, and the hexagons are exactly the entry it cannot see.

restriction · Curvature

Named alongside it

The objects these essays reach for when they reach for this one.

Crystal netQuotient graphCensusChange of basisVoltageBarycentric placementFree actionClosed surfaceCombinatorial curvatureCoordination numberCountingCrystallographic restriction

All concepts