Crystallography & Symmetry
patterns derived from their groups
Concept

Coordination number — where it appears

How many nearest neighbours an atom has in a structure. It is the same number for both close packings of equal spheres, which is why a density and a coordination cannot separate them and a space group can.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

One layer, and the two ways to sit on it. A close-packed layer — large pale discs, each touching six others, which is as tight as one layer of equal spheres can be. Its hollows come in two sets, marked in the two smaller colours, and a second layer must take one set or the other; the two choices are mirror images and equally good. The third layer then faces the same choice again, and this time the two answers are genuinely different: over the first layer, or over the hollows the second did not use. That single decision, repeated, is the whole of close packing — and nothing in the geometry prefers either answer, because both give the same density and the same number of touching neighbours.

Two stackings, one density

Stack spheres as tightly as they will go and the third layer has a free choice. Both answers fill exactly the same fraction of space and give every sphere the same twelve neighbours — and their space groups are Fm3̅m and P6₃/mmc, which is the only thing that tells them apart.

applied · Packing The whole space of plane lattices, and its corner. Every plane lattice appears exactly once in this picture. Scaling changes no density, so the leading coefficient is fixed at one; reduction then confines the other two to 0 ≤ b ≤ 1 ≤ c, and every lattice has exactly one reduced form. The curves are the levels of constant density, which are parabolas — a density d needs 4c − b² to equal (π/2d)². They crowd toward the corner b = c = 1, which is the hexagonal lattice at π/√12 ≈ 0.9069; the square lattice sits on the left edge at π/4 ≈ 0.7854. The picture is a search over a region rather than over a list, which is what makes the answer a decision: there is nowhere else for a lattice to be.

The densest lattice in the plane

Which arrangement of equal discs covers the most floor is a question about infinitely many lattices, and reduction turns it into a question about a two-parameter region with a corner. The answer is at the corner, and the argument finishes.

applied · Packing Cube, octahedron, rhombic dodecahedron — from connectivity alone. Every form of index two or less, classified by how many chains lie inside it: two or more and the face is flat, exactly one and it is stepped, none and it is kinked. The number beside each flat form is how many chains it contains, which is the rule's own tie-break — a face with three chains is flatter than one with two. Nothing about interplanar spacing enters, and the three structures are told apart by their bonds.

Which faces are flat

Bravais ranks a crystal's faces by how far apart their planes lie. Hartman and Perdok classify them by how many uninterrupted chains of bonds run inside them, which uses no spacing at all — and on the three cubic structures the two rules put the same face first every time. Then the second rule's power turns out to live entirely in where the chain list is cut off.

applied · Growth 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 How far apart points on a sphere can be kept. For each number of points, the largest smallest angle a search could find between any two of them. Two unit spheres touching a third do not overlap exactly when their contact points are 60° or more apart, so the largest count whose best arrangement still clears 60° is the kissing number. Twelve clears it with three degrees to spare and thirteen falls short by more than three. The circle column is the same problem in the plane, where the answer is exactly 360/n and needs no search at all.

The room a thirteenth sphere would need

Twelve equal spheres touch one, and whether a thirteenth could was argued in 1694 and settled in 1953. The reason it took so long is measurable: the twelve leave three and a half degrees of slack, which is enough room to look promising and not enough to use — and in the plane, where the same question has no slack at all, nobody ever argued.

applied · Packing The same accounting, at every coordination number. One row per number of edges at a vertex. The bill a sphere charges is 2dχ; the face worth nothing is 2d/(d − 2), which is a whole number at three, four and six and is 10/3 at five; the faces that can pay are those with fewer sides than that; and the last column is every way of paying the whole bill with faces of a single size. At three edges a vertex there are three such ways and twelve pentagons is one of them. At six there are none, which is the statement that six-fold coordination belongs to the plane and to no closed surface at all.

The twelve belongs to the vertex

Twelve pentagons is read as a fact about closing a surface. It is not: it is a fact about three edges meeting at a point. Let four edges meet instead and the sphere charges eight triangles; let five meet and it charges twenty; let six meet and it cannot be paid at all.

restriction · Curvature The fewest contacts twelve pentagons can have, by size. For every cage of pentagons and hexagons up to forty-four atoms, the number of pairs of pentagons sharing a bond. The lower line is the fewest any cage of that size achieves — 30, 24, 21, 18, 17, 15, 14, 12, 11, 10, 9, 8 — the upper line the most, and the dashed line the bound that counting edges gives: the twelve pentagons carry sixty edges between them, a contact uses two and an edge to a hexagon uses one, so the contacts cannot fall below 30 − 3h with h hexagons. The bound is attained while the hexagons are few and goes loose at five, after which each extra hexagon removes about one contact rather than three. The number of cages at each size is printed beneath, and it is the least rather than the average that the bound is about.

How close the twelve must be

The charge fixes twelve pentagons and says nothing about where they go, because it is a sum over faces and cannot see which face touches which. What it cannot see is a graph on twelve points, and the fewest edges that graph can have falls from thirty to eight over the cages a census reaches — then keeps falling at a rate that puts its first zero exactly where the truncated icosahedron is.

restriction · Curvature

Named alongside it

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

Close packingCrystal netPacking fractionCombinatorial curvatureCrystallographic restrictionEnumerationKissing numberPolyhedronBasis reductionBfdhCensusCounting

All concepts