Crystallography & Symmetry
patterns derived from their groups
Series

Tilings — the series

6 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. Every way regular polygons can fill a turn. The seventeen multisets of regular polygons whose interior angles add to exactly 360°, listed with the sum that qualifies each of them. They are found by a search over sizes from three upward: the largest polygon that can appear is the forty-two-gon, which needs a triangle and a heptagon beside it, and the search stops there because the smallest interior angle is a third of a turn so at most six polygons can meet. Nothing here is a table looked up — the list is the output of the search, and every count on the page downstream of it is counted from this one.

    Twenty-one vertices, eleven tilings

    Regular polygons meeting at a point must fill exactly a turn, which is a Diophantine equation with seventeen answers and twenty-one cyclic arrangements. Ten of the twenty-one tile nothing at all — and the argument that kills them counts places round a polygon rather than measuring anything.

    part 1 · classification
  2. The cell of 3.4.6.4, and the vertices in it. 3.4.6.4 drawn with the cell its own translations define. The lattice is hexagonal and the cell holds 6 vertexes, marked. Neither was chosen: the translations are the vertex-to-vertex vectors that carry every polygon of the patch onto a polygon of the patch, and the cell is the shortest independent pair of them. Expressed in that basis the vertices have coordinates that are exact and are not fractions — a vertex of this tiling sits at 1/(1 + √3) of a cell — which is why the detector that decides its group works in ℚ(√3) rather than in the rationals.

    Eleven tilings, five groups

    Hand each of the eleven uniform tilings to a detector that has never heard of tilings and ask what its symmetry is. Six of them answer p6m. Twelve of the seventeen wallpaper groups never appear at all — and the coordinates the question has to be asked in are not fractions.

    part 2 · classification
  3. 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.

    part 3 · classification
  4. a general quadrilateral tiles. A general quadrilateral — convex, with no equal sides and no parallel edges — with copies placed by half-turns about edge midpoints. The patch was checked by sampling 2000 points inside a disc: every one of them lies in exactly one tile, so there is no gap and no overlap anywhere in the region tested.

    Which shapes tile by themselves

    Every triangle tiles the plane. So does every quadrilateral, convex or not. Six sides admits three families, seven sides admits nothing at all — and the five-sided case took a hundred years and finished with a computer search. The bound at seven needs no search: it is Euler's relation with the curvature set to zero.

    part 4 · classification
  5. 3 whole-number solutions: (6, 3), (4, 4), (3, 6). Every pair of whole numbers from three to 12, with the mean face size across and the mean degree down. A square in the first colour is a pair satisfying one over p plus one over q equals a half exactly — the flat case, where a periodic net is possible — and there are 3 of them: 6 and 3, 4 and 4, 3 and 6. The lighter squares above and to the left have a sum greater than a half, which is a closed polyhedron rather than a plane tiling; the ones below and to the right have a sum less than a half and belong to a surface of negative curvature. The plane is the boundary between them and it is thin.

    Three answers in whole numbers

    One over the face size plus one over the degree equals a half. Ask for whole numbers and there are exactly three answers, which are the three nets everybody has drawn since childhood — and the pairs on either side of them are a closed polyhedron and a plane the plane has no room for.

    part 5 · classification
  6. Why a tetrahedron is not a cube cut up. The two invariants side by side. A cube's twelve right angles are each a rational part of a turn and contribute nothing; a regular tetrahedron's six edges each contribute one α, giving six. Cutting a polyhedron and rearranging the pieces cannot change the invariant, so no dissection takes one to the other however the volumes are matched. That is Hilbert's third problem, and the whole of it is one angle.

    The angle that is not a fraction of a turn

    Any two polygons of equal area can be cut into pieces that rearrange into each other. In space that fails, and the obstruction is a sum over edges of length against dihedral angle — zero for anything that fills space, and not zero for a regular tetrahedron. The whole argument reduces to one claim about one angle, and that claim is an integer computation: a sequence that is never divisible by three, when it would have to be.

    part 6 · classification

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