Crystallography & Symmetry
patterns derived from their groups
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The theme: Generated, not drawn — page 4

Every pattern on this site is the orbit of a motif under a group, and the group is then rediscovered from the drawing. A picture that was drawn by hand can have symmetries nobody intended.
the honeycomb net: cmm against p6m. the honeycomb net drawn twice. On the left a placement chosen by hand, whose symmetry group is cmm of order 4; on the right the placement in which every vertex sits at the average of its neighbours, whose group is p6m of order 12. The graph is identical in the two — the same vertices joined the same way — so every symmetry of the left-hand drawing is a symmetry of the net and the right-hand drawing has them all. Each detected operation is then required to carry every edge of the quotient graph to an edge, which is what makes it a symmetry of the net rather than of the point set. Symmetry at work

The placement nobody chose

A net has no coordinates, so drawing one means inventing them. There is exactly one way to invent them that involves no choice: put every vertex at the average of its neighbours. The drawing that results has the largest symmetry group the net admits, and this site's own detector finds it.

 the kagome net: one of 1 mechanism. An infinitesimal mechanism of the kagome net, drawn as a velocity at every joint. The vector is an exact solution of the rigidity matrix — a set of joint velocities and a rate of change of the cell's metric under which no bar's length changes to first order — with the two rigid translations projected out so that what is left is a motion rather than a shift. Whether it continues into a finite motion is a separate question that a first-order calculation cannot answer, and this collection answers it for one framework by constructing the motion explicitly. Symmetry at work

A fold that keeps its symmetry

The kagome framework has exactly one mechanism, and it does not stop at first order. Every triangle turns, alternate ones the other way, the cell shrinks to half its size, and not one bar changes length — and the count that found the mechanism cannot see how many there really are.

 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. Symmetry at work

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.

 A screw dislocation of Burgers vector 1, after 40 steps of growth. The height of a growing surface, light for low and dark for high, over a patch 25 cells across with a screw dislocation at its centre. Growth is an integer rule — a site rises when it has a neighbour a layer higher — and the only thing that makes this patch different from a flat one is a branch cut along which the comparison is offset by the Burgers vector. The step winds round the centre instead of running out: after 40 steps the centre has climbed 10 layers and the surface is still growing at 110 sites a step. The shading is normalised to the patch's own range, so the shape is the steady state the mechanism predicts and is the same at every step count; the numbers at the foot are what changes, and they are what the claim of unending growth is actually about. Symmetry at work

The step that never runs out

A perfect crystal face cannot grow: an atom arriving on a flat plane touches it on one side and leaves again. Faces grow anyway, and the reason is a defect — a screw dislocation puts a step on the surface that winding round it never consumes.

 Four of the 980 piles in a three-cube box. A stack of unit cubes in the corner of a box, seen down the body diagonal. Every visible face is one of three rhombi and the picture is a tiling of one fixed hexagon — the same hexagon for every pile, because a pile in an a×b×c box always shows ab+bc+ca faces however it is stacked. The four here are taken at even intervals through the enumeration, from the empty box to the full one. Order without repetition

A facet with no energy in it

Stack cubes into the corner of a box and look down the body diagonal: the pile is a tiling of a hexagon by three rhombi, and the number of piles is a product MacMahon wrote down in 1916. Because the count is exact, so is the average pile — and the average has a flat corner meeting a rounded middle, which is the shape of an equilibrium crystal, arrived at by counting with no surface energy anywhere in the argument.

 The domain is a polygon, and its edges are elements. The Dirichlet domain of a point whose stabiliser is trivial: the set of points at least as close to it as to any other point of its orbit. It is a convex polygon, it is a fundamental domain, and each of its edges lies on the bisector of the base point and one image of it — so each edge already carries the element that produced it, with no search. Edges are drawn by kind: paired with another edge, fixed pointwise by a reflection, or folded in half by a half turn. Operations

The relations a polygon dictates

Poincaré's theorem has two halves. The walls of a fundamental domain name the generators, which is the half this collection already computes; walking round its corners names the relations, which needs a domain with corners rather than a domain made of pixels. Building the Dirichlet polygon exactly gives a presentation of each of the seventeen — and coset enumeration says every one of them is right.

 One group refuses two colours and three refuse three. The two counts side by side, with the rows that refuse a number of colours marked. p3 is the only group with no two-colouring; p4, p4m and p4g are the only ones with no three-colouring. Neither list is a subset of the other and both come from the same arithmetic — a rotation order that divides nothing the symmetric group has. The classification

What a half-turn does to three colours

Ten of the seventeen plane groups have no three-colouring, because a half-turn cannot permute three colours cyclically — that is the first rung of this ladder and it is true. Drop the word cyclically and the answer changes completely: a half-turn permutes three colours perfectly well by swapping two and fixing one, and only the three four-fold groups refuse three colours at all.

 Perfection is a rank, and most lattices do not reach it. For each lattice, the rank of the matrices vvᵀ built from its shortest vectors, against the dimension of the space of symmetric matrices those live in. Reaching it means the shortest vectors pin the form down completely: no deformation keeps every one of them at its length. Falling short means there is a direction left to move in, and the lattice is not a local maximum of density. Lattices

One perfect form in space

Which lattice packs spheres most densely is a question about a maximum over a continuum, and Voronoi turned it into a rank calculation and a sign check. A lattice is a local maximum exactly when its shortest vectors pin its shape down completely and its inverse can be written over them with positive coefficients. Searching every reduced integer form of minimum two finds one such lattice in the plane and one in space.

 Thirty-two classes, from fourteen Gram matrices. The five hundred and ten subgroups sorted by how many operations of each kind they contain — a determinant and a trace decide which of the ten kinds a matrix is. Thirty-two answers come out, and they are the thirty-two crystal classes: matched against the construction elsewhere in this collection by signature rather than by name, since nothing here names a point group. What symmetry decides

Thirty-two from fourteen matrices

Write the fourteen Bravais lattices as Gram matrices, ask each one which integer matrices preserve it, and take every subgroup of every answer: five hundred and ten of them. Sort those by how many operations of each kind they contain — which a determinant and a trace decide — and thirty-two answers come out. They are the crystal classes, from a construction in which no point group is ever named.

 Four vectors summing to zero, and six numbers on the edges. A superbasis is the three basis vectors together with their negated sum, so the four sum to nothing and their pairwise products sit on the six edges of a tetrahedron. Selling's rule is: while any edge is positive, apply one transformation. The right panel is the same lattice reduced, with the vanishing parameters marked — and a vanishing parameter is a face the Voronoi cell does not have. Lattices

A reduction with one rule

Niggli's reduction is eight numbered conditions with sub-cases, applied in order until none applies. Selling's is a single rule on four vectors that sum to zero: while any of six numbers is positive, do one thing. It terminates sooner, its termination is a quantity that visibly falls, and when it stops the six numbers are the Voronoi cell — the pattern of which ones vanish gives Fedorov's five solids and nothing else.

 Twenty-two halvings the fourteen lattices permit. Every lattice has exactly seven subgroups of index two, whatever its shape. The third column is how many of the seven the lattice's own group carries onto themselves, and the fourth is how many of those survive as distinct types once a change of basis within the type is allowed to identify them. The running total ends at twenty-two, which with the fourteen grey lattices is the thirty-six magnetic Bravais lattices — and the row that ends at zero is the face-centred cubic lattice. Lattices

The halving a lattice will not permit

Admit time reversal and a lattice splits into points that leave the moments alone and points that reverse them. The second set is a coset of a subgroup of index two, and every lattice has exactly seven of those, whatever its shape. What differs is how many of the seven the lattice's own symmetry survives — and the face-centred cubic lattice survives none of them.

 The kinds of line defect each breaking allows. The eleven proper crystal classes, each with the order of the binary group that covers it, the number of conjugacy classes of that group other than the identity — which is the number of kinds of line defect — and whether the group commutes. 6 of the eleven do not, and in those media two defect lines cannot pass through each other without leaving a third line behind. Symmetry at work

The defect that needs two laps

Which defects a medium can have is not a fact about the medium. It is a fact about the space its order parameter lives in, and for a rotational symmetry broken down to a point group that space has a fundamental group twice the size of the point group. The kinds of line defect are its conjugacy classes — and in six of the eleven cases they do not commute, which means two defect lines cannot pass through each other.

 Seventy-three arithmetic classes, from fourteen groups. Every subgroup of every lattice's own group, split by whether the subgroup's own Bravais group is that lattice's. The ones that are not belong to a lower lattice and are counted there, which is what stops the same class being counted twice. The running total ends at seventy-three, and no conjugacy in GL(3, ℤ) was ever decided. What symmetry decides

Seventy-three, without a search

The unit a space group is built from is a point group together with the lattice it acts on, and there are seventy-three of them. Getting there looks like it needs conjugacy in GL(3,ℤ), which is a search this collection tried and abandoned. It does not: every finite group of integer matrices carries a canonical larger group that says which lattice it belongs to, and once that is computed the search has nothing left to do.

 Both sides of the transformation, on five lattices. A Gaussian of width set by t on every point of a lattice, summed; and the same sum over the dual lattice with the width inverted and the covolume divided out. The two agree to the last bit a double carries, at every t and on lattices with no symmetry in them, so nothing here is a coincidence of parameters. The identity is exact and the reason to have it is that the two sides do not cost the same. Lattices

The sum that turns a lattice into its dual

Put a Gaussian on every point of a lattice and add them up. The answer equals the same sum over the dual lattice with the width inverted and the covolume divided out — exactly, to the last bit a double carries, on lattices with no symmetry in them. The identity is free and the reason to have it is that the two sides do not cost the same: at one end of the range the direct sum needs forty thousand terms and the dual sum needs a hundred and twenty-five.

 The same terms, added in two shapes. Partial sums of the alternating 1/r sum over the simple cubic lattice, taken over expanding cubes and over expanding spheres. The terms are identical and only the order differs. The cubes creep towards 1.747565 — 1.7258 by the last point drawn — and the spheres do not settle at all, landing at -3.527 after passing through values on both sides of it. A conditionally convergent sum has no value until the order is named. Lattices

The sum whose answer depends on the shape

Give the points of a cubic lattice alternating signs and add up one over the distance. Added over expanding cubes the total creeps towards 1.747565; added over expanding spheres it does not converge at all, landing on both sides of that number and never settling. The terms are identical and only the order differs. Splitting the sum in two with the theta transformation gives it a value — ten decimal places from a few thousand terms.

 Three variants, and not one undistorted plane. The three tetragonal variants a cubic parent produces, with the principal stretches of each. Every one of them has the same three numbers in a different order, and the middle one is not one — so none of the three leaves any plane undistorted, and none of them can meet the parent phase across an interface. That is the difficulty the whole of the crystallographic theory of martensite exists to resolve, and it is visible in one column. Symmetry at work

The plane a deformation leaves alone

Two differently deformed regions can meet across a plane only if that plane is deformed identically from both sides — which forces the two deformations to differ by a rank-one term. Multiplying each side by its own transpose removes the rotation and leaves a condition on a signature: one positive eigenvalue, one negative, one exactly zero. In that form the classical rule that the middle principal stretch must be one is not quoted but derived, and it says that no single variant of a cubic-to-tetragonal transition can meet its parent at all.

 Where the laminate's middle eigenvalue crosses zero. The middle eigenvalue of FᵀF − I for the average deformation of a twinned laminate, against the volume fraction of one variant. At both ends the laminate is a single variant and the value is well away from zero; in between it crosses, twice, and each crossing is a volume fraction at which the laminate can meet the parent phase across a plane. The two roots are complementary, which is the same plate with the two variants exchanged. Symmetry at work

The plate that only fits when it is twinned

No single variant of a cubic-to-tetragonal transition can meet its parent across a plane. A fine mixture of two variants can, because its average deformation carries a free parameter — the volume fraction — and that parameter passes through the compatibility condition twice. Sweeping it gives the two fractions, the two habit planes, and one inequality: the plate exists exactly when the two principal stretches satisfy η₁² + η₃² ≤ 2.

 Three places to put the boundary. A large site and a small one, with three candidate boundaries between them. Halfway is the ordinary Voronoi cell and it cuts through the large sphere. Splitting in the ratio of the radii is the natural repair and its surfaces are not planes, so the cells do not fit together. The power plane sits where the tangent lengths agree, which is further from the large site than halfway and is still a plane — and being a plane is the whole reason the construction works. Lattices

Where the boundary goes when the atoms differ

Assigning each point of space to the nearest atom is the right rule only when every atom is the same size. Splitting the distance in the ratio of the radii is the obvious repair and it produces curved faces that do not fit together. The repair that works measures to a sphere rather than to a point: the boundary stays a plane, the cells still tile exactly, and a small enough atom loses its cell altogether — at a radius ratio of exactly one in eight.

 One curve falls and the other does not. The boundary's share of a ball, against the radius, for a plane group and for the free group on two generators. The plane group's falls like one over the radius and goes to zero; the free group's rises to two thirds and stays. A group with no sequence of regions whose boundary becomes negligible has no shape-independent average, and that is not a difficulty in the analysis — it is a property of the group. Operations

The boundary a growing region forgets

Quoting a density assumes the region it was averaged over does not matter, and that assumption is a property of the group of translations rather than of the crystal. A ball in a plane group grows like R² and its boundary like R, so the edge becomes negligible — and where that fails, the average genuinely moves. The free group on two generators keeps two thirds of itself on the boundary forever, and a slab seven layers deep is wrong by exactly one seventh however wide it is made.

 The table of marks of 4mm. Every conjugacy class of subgroup of 4mm, against every other. The entry is the number of cosets of the column's subgroup that the row's subgroup holds still. The first row is the identity, which fixes everything, so it is the size of each coset space; the last column is the whole group, whose only coset is fixed by everybody. What symmetry decides

The table that decides every action

Burnside's lemma counts orbits and stops there — two completely different actions with the same orbit count are indistinguishable to it. The object that settles the whole question is a square table whose entries count fixed cosets: lower triangular because a subgroup fixes no coset of anything smaller, positive on the diagonal because it fixes its own, and therefore invertible. Inverting it turns a list of fixed-point counts back into the orbits themselves.

 The five, as generator counts. Each of the five convex bodies that tile space by translation, built as the set of combinations of a handful of vectors with coefficients between zero and one. Three generators give a cube, four give either a hexagonal prism or a rhombic dodecahedron depending on whether three of them are coplanar, five give the elongated dodecahedron and six the truncated octahedron. The last column is what the same number of generators would give in general position, and the shortfall is the number of faces lost to coplanarity. The classification

Every parallelohedron is a shadow of a cube

Take a few vectors and form every combination of them with coefficients between zero and one. All five of the convex bodies that tile space by translation come out of that recipe, from three vectors, four, four, five and six — and since the recipe is exactly the image of a cube of that many dimensions, the truncated octahedron is a three-dimensional shadow of a six-dimensional cube. The five are not the generic answers: they are the degenerate ones, and the degeneracy is what the tiling demands.

 Five solids, twice each. Each Platonic solid as a framework of rods hinged at the corners, and again with its faces made rigid by adding their diagonals. The rank of the rigidity matrix reaches 3V − 6 exactly when the framework cannot move; the shortfall counts the ways it can. Three of the five are rigid as rods and all five are rigid as plates, which is Cauchy's theorem in the form a rank computation can see. Symmetry at work

The polyhedra that can flex

A cube of rods folds and a cube of cardboard does not, and the difference is a rank. Cauchy proved in 1813 that a convex polyhedron with rigid faces is rigid; the rank of a rigidity matrix sees it directly, and it also sees where the hypothesis is doing the work. Drop convexity and an octahedron flexes — followed here for forty steps with every edge length held to five parts in a thousand million million.

 Seven fields and the number that counts each one. The degree of each field, computed by triangulating the sphere drawn round the defect, mapping every vertex, and adding the signed areas of the image triangles. The total is 4π times the degree, and the integral column is that total divided by 4π before rounding. Each is read on three successively finer meshes and required to give the same integer on all three, because a mesh too coarse for its field does not produce a noisy answer — it produces a confident wrong one. Symmetry at work

The point defect whose charge has no sign

A line defect is read on a loop; a point defect is read on a sphere, and the number that comes off the sphere is a degree. In a nematic that degree is an integer whose sign depends on a choice nobody can make — and the media where no such number exists at all are exactly the ones whose residual symmetry is a crystal class.

 One determinant, three dimensions. The Cayley–Menger determinant of a set of squared distances, at three sizes. Its value is the squared content of the simplex those distances describe, times a factor that alternates in sign with the dimension. At three points it is Heron's formula rewritten; at four it gives a tetrahedron's volume from its six edge lengths with no coordinates anywhere. The alternating sign is not a convention — a value of the wrong sign means the distances belong to no set of points at all. Symmetry at work

What six lengths decide and nine do not

A tetrahedron's volume is a determinant in its six edge lengths, with no coordinates anywhere. Add a fifth vertex and the lengths stop deciding: two shapes with identical edges and identical faces have volumes in the ratio 2.6. What survives is that the possibilities are finite — which is the whole reason a flexing polyhedron cannot change its volume.

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