Crystallography & Symmetry
patterns derived from their groups
Concept

Duality — where it appears

The exchange of a tiling's tiles for vertices and its vertices for tiles, joining two new vertices when their tiles shared an edge. Applied to the eleven uniform tilings it gives eleven tilings by a single repeated shape, three of which are pentagons.

Named by 9 essays across 4 fields — each of them below, with the objects they name alongside it.

3.4.6.4 and its dual. The tiling in pale outline with its dual drawn over it: one dual vertex at the centre of every tile, one dual edge across every shared edge, and one dual tile round every vertex. 3.4.6.4 has 3 kinds of tile and one kind of vertex; its dual has one kind of tile and 3 kinds of vertex, and the congruence of those tiles is checked rather than eyeballed — every dual face presents the same cyclic sequence of squared edge lengths, compared exactly. That swap is what the eleven duals are for: read one way the list classifies tilings with all vertices alike, read the other it classifies tilings with all tiles alike.

Eleven duals, one tile each

Swap the vertices of a uniform tiling for its tiles and the eleven come back as eleven tilings by a single repeated shape. Three of those shapes are pentagons — which is worth pausing over on a site whose other essays prove that five-fold symmetry cannot exist.

classification · Tilings Which Schläfli symbols close. Every {p, q} with p polygons round each face and q faces round each vertex, from three to six of each. A solid exists only when 2p + 2q − pq is positive, which is the same statement as 1/p + 1/q > ½; the five that qualify carry their vertex, edge and face counts, and the three on the diagonal where the expression vanishes are the three regular tilings of the plane. Past them the expression is negative and the answer is the hyperbolic plane, where the list never ends. The five, the three and the infinity are one inequality read at its three signs.

Five solids from one inequality

Five families of rotation group in space, five regular solids, three regular tilings of the plane and an endless supply of hyperbolic ones — all of it is 1/p + 1/q compared with a half, read at its three signs.

restriction · Finite groups hexagonal: 0.5 and 0.577. The hexagonal lattice with both radii drawn together: the small circles are the largest that do not overlap and the large ones the smallest that leave no gap. The line runs from a lattice point to the deepest hole, which is a corner of the cell around it, and its length is the covering radius 0.5774 against a packing radius of 0.5. The deep hole was found by search on a grid of 24 and then refined, and checked afterwards against an independent grid.

Covering and packing want different lattices

A lattice has two natural radii — the largest spheres on its points that do not overlap, and the smallest that leave no gap — and both are radii of the same Voronoi cell. In the plane one lattice is best at both. In space the best packer and the best coverer are different lattices, and they are duals of one another.

lattices · Wigner–Seitz cells 60 vertices, 12 pentagons. A closed net with three edges at every vertex: 60 vertices, 90 edges and 32 faces, of which 12 are pentagons and 20 are hexagons. The pentagons are picked out in the second colour. Their number is not a property of this cage — it is twelve for every closed trivalent net of pentagons and hexagons, at any size, and the hexagon count is free.

Twelve pentagons, and no way round them

The crystallographic restriction forbids a five-fold face in a flat repeating net. Curve the net into a closed cage and the same three lines of arithmetic require exactly twelve of them — at any size, with the hexagon count free. What a lattice forbids, closing up compels.

restriction · Curvature 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 Every arrangement on two tori, by how it winds. The ice arrangements on tori eight and nine vertices across, each placed at the pair of numbers it winds by — how much the height on the faces gains going once round the torus across and once round it down — with a disc whose area grows with the number of arrangements there. On the even torus every winding is even and the arrangements crowd the centre; on the odd torus every winding is odd and there is no centre to crowd, so they sit at plus and minus one. Solid discs are the sectors whose windings are both multiples of three, which are exactly the arrangements that are also three-colourings. On the even torus that is exactly a third of everything; on the odd torus it is the four sectors at plus and minus three, and they hold about one arrangement in eighty.

A winding does not dilute

Three-colourings of an odd torus are a small and shrinking fraction of the ice arrangements on it, and the natural guess is that the fraction goes to nothing. It does not. It settles near one in twenty-eight, because the windings that decide it are a property of the whole surface rather than of its edge — and on even tori the same windings put exactly a third of the arrangements in one class, which in the limit is Jacobi's theta identity.

aperiodic · Entropy Face- and body-centred cubic change places at t = 1. The Gaussian sum of the face-centred cubic lattice minus that of the body-centred cubic lattice, both at unit volume, across widths from 0.35 to 2.8. Below t = 1 the difference is positive, so the body-centred lattice has the smaller sum; above, it is negative and the face-centred lattice does. At t = 1 the two are equal to the last digit a computer carries, and not by accident: at unit volume each lattice's sum at width t is the other's at width 1/t times a known factor, and at t = 1 the factor is one. Neither lattice is best at every width, and no other lattice can be either.

No lattice in space wins at every width

Spread a Gaussian over every point of a lattice and add them up. In the plane one lattice gives the smallest sum at every width — the hexagonal one — and that is a theorem. In space no lattice does, and the proof is two lines: at large widths the winner must be the densest lattice, at small widths it must be the densest lattice's dual, and in space those are face- and body-centred cubic. They change places at exactly the width where each is the other's dual.

lattices · Lengths The exact third on every even rectangle. Square ice on twelve tori with both sides even, eight of them rectangles that are not squares, up to 6 × 12 and 8 × 8: the number of ice arrangements, the number whose two windings are both divisible by three, and their ratio. It is one third exactly, as a ratio of integers, on every one.

The third survives a rectangle

On every even square torus, exactly a third of the square-ice arrangements have both windings divisible by three, and the explanation offered was a theta function that is its own dual. A test was proposed: on a torus twice as long as it is wide the two windings have different stiffnesses, the explanation gives a different number, and the exact third should fail. It does not fail on any of eight rectangles. But the explanation, worked through properly for a rectangle, never predicted that it would. The two directions of a rectangle are each other's duals, and their factors cancel to a third at every shape.

aperiodic · Entropy E₈ has the smallest sum at every width. For four eight-dimensional lattices of unit covolume — D₈, its dual D₈, the cubic lattice Z⁸ and two copies of D₄ — the ratio of their Gaussian sums, less the constant term, to E₈'s, at widths from 0.2 to 5, logarithmic. Every ratio exceeds one everywhere: E₈ wins at every width against every rival. The closest approach is at t = 1, where D₈ and its dual are equal, a factor of 1.31 above E₈. D₈ is the better of the pair at wide widths and D₈ at narrow ones, and neither comes close to E₈ at either end.

In eight dimensions one lattice wins at every width

In space no lattice has the smallest Gaussian sum at every width, because the densest lattice wins when the Gaussian is narrow and its dual wins when it is wide, and fcc is not bcc. The argument fails exactly when the densest lattice is its own dual. In eight dimensions it is: E₈, with 240 shortest vectors, beats D₈, its dual, the cubic lattice and a sum of two four-dimensional lattices at every width tried, by a factor of at least 1.31, and its sum is mirror-symmetric in the logarithm of the width.

lattices · Lengths

Named alongside it

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

Crystallographic restrictionEnumerationTheta seriesClose packingThe Euler characteristicArchimedean tilingCountingDual latticeEntropyHeight functionResidual entropyTorus

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