Crystallography & Symmetry
patterns derived from their groups
Theme

The theme: Symmetry is decidable — page 10

Unusually for a physical science, the central questions here have exact answers computable in integer arithmetic. There is no tolerance to choose and no residual to interpret.
How many components a property may have, at each rank and in each class. One row per plane point group, one column per rank of a fully symmetric property tensor, with the number of independent components in each cell — counted by averaging the tensor over the group index by index, and equal at every entry to the Molien coefficient of that degree. A symmetric property of rank r is a form of degree r, so Neumann's principle and the invariant ring are the same arithmetic in two notations. The last column is the elastic tensor, which is not fully symmetric — symmetric within each pair of indices and under exchanging the pairs — and its counts are not in the table to its left. A property with its own symmetries needs its own average, and that is why the elastic constants are not read off a degree. What symmetry decides

The parts a property splits into

A symmetric property of rank r is a polynomial of degree r wearing indices, so the number of components a class permits it is a coefficient of an invariant ring's series. The elastic tensor is not a polynomial in disguise, and its counts are not in that table — which is the most useful thing about it.

 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.

 A map from amplitudes alone, 0.59 grid steps out. The density after 150 cycles of flipping, with the atoms that produced the data drawn as rings — moved into the origin and the handedness the solution chose, because a phase set does not fix either and comparing without allowing for them measures the arbitrariness of the description. Every peak of the map is an atom and every atom has a peak. Nothing about the arrangement went into the calculation: the input was a list of amplitudes and a random set of phases. How it is known

The solver that knows no symmetry

Compute a map from amplitudes and random phases, reverse the sign of everything below a small threshold, transform back and keep the phases. Repeat. The structure appears — and so does its space group, which was never supplied.

 4mm: 17 of 25 transitions forbidden. Every pair of irreducible representations of 4mm, with the number of times the identity occurs in the product of the two with the vector operator. A zero is a prohibition: the integral that would give the transition rate vanishes for every choice of functions carrying those representations, whatever the material is made of. A positive number is a permission and nothing more. Rows are final states and columns initial ones; the labels are the dimensions of the representations, so the twos are the degenerate levels. Every one of the 17 prohibitions here was checked again against explicit polynomials. What symmetry decides

What a group forbids to happen

Two levels and a thing that might carry a crystal from one to the other. Whether it can is one sum over the group — and a zero there is a prohibition that no material, no temperature and no intensity of light gets round.

 Where time reversal does something, and what. Every wavevector of every plane group at which time reversal changes the answer, with the square of each antiunitary operator, the unitary prediction, Herring's corrected prediction and the measured degeneracies. Case (b) is Kramers' theorem in a crystal with no spin, and it happens exactly where a glide's operator squares to −1. Case (c) is a representation being carried to a different one by the antiunitary operator, so the two become one level. Nine of these rows were open before the criterion was built — three the earlier census called unaccounted and six it could not reach at all. Into space

The degeneracy time reversal forces

A crystal with a glide has levels that stick together at the edge of its zone for a reason no character table contains. The operation responsible is antiunitary, it squares to minus one, and Kramers' theorem then applies to a model with no spin anywhere in it — which closes nine rows an earlier census in this collection had to leave open.

 two sites in a hexagonal cell: 3 cutoffs, degree 3 to 12. Two atoms per hexagonal cell, at the positions graphite's carbons occupy, read as a net at a ladder of bonding cutoffs. Each row takes the cutoff just past a shell of neighbours and reports the net that results: how many edges it has, the degree of its vertices, whether its cycles generate the whole translation lattice, and the group of its own barycentric placement. The net is not in the coordinates. There is no bond in a list of positions; there is a cutoff, and moving it past a shell gives a different net from the same atoms. A row marked as a supercell is a net whose own translations turn out finer than the cell it was described in — the description was on too large a cell and the machinery says so. Symmetry at work

A net is a choice of what counts as a bond

A list of atomic positions does not contain a net. It contains distances, and somebody has to decide which of them are bonds — so the net is a fact about the cutoff as much as about the crystal, and moving the cutoff past a shell of neighbours changes the answer.

 One number, and it is the fraction. Data simulated from crystals that are nought, a quarter, a half, three quarters and wholly inverted, each fitted for the single parameter. The fitted values sit on the diagonal to better than five parts in a hundred, which is what makes the parameter a measurement of composition rather than a test of a hypothesis: a crystal is allowed to be part one hand and part the other, and a value near a half is a real answer about the specimen rather than a failure of the determination. How it is known

How much of it is the other hand

A crystal of one enantiomer is a hypothesis, not an observation. What the diffraction actually measures is a fraction — how much of the specimen is the inverted structure — and the useful part of that measurement is the uncertainty on it.

 Where the axes are free, they move. Five different invariant tensors of each of three classes, with the trace of each tensor's principal axes on the page. In the orthorhombic class every sample gives the same three directions: the axes are the two-fold axes and symmetry has fixed them. In the monoclinic class one direction is common to every sample and the other two rotate freely in the plane across it. In the triclinic class nothing is common at all. Each sample stands for a different material, or the same material at a different wavelength — which is what makes the middle picture the dispersion of the optic axes. What symmetry decides

The axes a class pins down

A property tensor has a shape and an orientation, and symmetry treats them differently. Three principal directions fixed for ever in an orthorhombic crystal; one in a monoclinic one, with the other two turning as the wavelength changes.

 11 frameworks, 3 where the count is wrong. Every net in this collection read as a framework of rigid bars and free joints, with the cell free to change shape. Maxwell's count and the number of mechanisms agree on most of them and not on all: a framework with a state of self-stress has a bar the count treats as removing a freedom that the others had already removed, and it has a mechanism the count cannot see. Here that is fes, snb, ring5, where the count says 2, -1, -4 and the rank says 3, 0, 9. The identity Maxwell is always right about — count equals mechanisms minus self-stresses — holds on every row. Symmetry at work

The count that promises a mechanism

Count the joints, count the bars, subtract. The number that comes out promises rigidity when it is small and a mechanism when it is large, and it is wrong in both directions — because it assumes every bar removes a freedom the others have not already removed.

 Two populations, and neither of them empty. The reflections of a structure in which three quarters of the atoms are paired by a half-cell shift, sorted by the parity of h + k and each class scaled by its own mean. The two histograms have the same shape, which is the point: each class on its own is an ordinary acentric distribution. What differs is the scale — the odd class is a sixth of the even one on average — and no odd reflection is absent, so no extinction rule fires and nothing about the space group is affected. How it is known

A translation that is nearly there

Half a structure copied onto the other half by a half-cell shift, with nothing exact about it. No reflection vanishes, so no extinction rule fires — and the test for a centre of symmetry answers yes about a structure that has none.

 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.

 The kagome net's level that does not move. Three levels of the kagome net across the zone, one of them flat. The reason is drawn beside it: a state that alternates in sign round one hexagon and vanishes everywhere else is an exact eigenvector of the adjacency operator at −2, because every site outside the hexagon that touches it touches exactly two of its vertices and those two carry opposite signs. The check is integer arithmetic in a supercell of 27 sites, with a residual of exactly zero. A state confined to one hexagon has no wavevector, and a level made of such states cannot depend on one — which is what a flat line across a zone means. Symmetry at work

The level that does not move

Three levels cross the kagome net's zone and one of them is a horizontal line. The reason is a state that alternates in sign round a single hexagon and is exactly zero everywhere else — a solution with no wavevector in it at all, which is why no wavevector can move it.

 Two plane structures with one Patterson. Two arrangements of 4 atoms on a 4 by 4 torus. No translation and no half-turn carries one onto the other, so they are different structures; every interatomic vector occurs the same number of times in both, so no measurement of intensities distinguishes them. The plane case is not the chain case with an extra index: a mirror in the plane sends a vector set to its mirror image rather than to itself, which is the one place the analogy with a cycle breaks. How it is known

Two structures on a torus, and one Patterson

Homometry was settled here on a ring of positions, which is a crystal in one dimension. Moving the same exhaustive search to a torus asks whether the coincidence is commoner or rarer when the vectors have a plane to land in — and the honest answer is that dimension is not what decides it.

 How many reflections the centre test needs. The error rate of two tests for a centre of symmetry against the number of reflections used, measured on 60 centrosymmetric and 60 non-centrosymmetric structures at each point. The moment test — the one in every textbook, comparing ⟨|E|² − 1⟩ to its two theoretical values — reaches one error in twenty at 160 reflections and one in a hundred at 320. A likelihood ratio, which uses each reflection's own value instead of one average, reaches the same at 40 and 80. The gap is the price of summarising a distribution by its mean, and it is about a factor of four. How it is known

How many reflections it takes to know there is a centre

The test for a centre of symmetry compares one average of the intensities against two theoretical values a quarter apart. Whether that is a measurement depends on how many reflections went into the average, and the only honest way to find out is to run the test on structures whose answer is already known and count the mistakes.

 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.

 6°: a boundary with a dislocation every 9.5 cells. Two crystals of the same lattice, each turned by half of 6 degrees in opposite senses, meeting on the dashed line. Almost everywhere along it the atoms of one side face the atoms of the other at very nearly the right distance — the boundary is good crystal — and at the marked places the misfit has accumulated to a whole lattice vector and an extra half-plane has to be inserted. Those are the edge dislocations, and they are 9.5 cells apart against the 9.6 that Frank's formula gives. Symmetry at work

A small angle is a row of dislocations

Turn one crystal a degree against another and the coincidence arithmetic says they share almost nothing. The boundary between them is nevertheless nearly perfect crystal, and both statements are true: the misfit stays small for a long way and then, all at once, needs an extra half-plane.

 Which descents change the shape of the cell, and into how many shapes. Each descent the modes produced, with the number of independent strain components the parent class permits and the number the child permits. A transition is ferroelastic exactly when the second is larger — the child leaves alone a distortion the parent moves — and the difference is a spontaneous strain the crystal acquires without being pushed. The count of distinct shapes is the orbit of that strain under the parent, which can be smaller than the number of domains: two domains may differ in something a change of shape cannot show. Every count here is a rank of an averaged set of quadratic forms, computed twice — once by averaging, once from a character. Symmetry at work

The strain that arrives with the transition

A crystal that loses symmetry usually changes shape, and whether it does is a subtraction: how many strain components the child permits, minus how many the parent did. The difference is a distortion nobody applied, and it is what makes a domain visible in a microscope.

 3m → 1: the two directions a wall between domains may take. The difference between the strains of two domains, sampled around a circle of directions: the first colour where that direction is stretched, the second where it is compressed. The two solid lines are the directions where it is neither, and those are the only orientations a straight wall between the two domains can take without straining itself — Sapriel's condition, one dimension down from the planes it is usually written for. There are exactly two, and that is not luck: the two domains are images of one another under the parent group, so their strains have the same area change and their difference changes no area at all. A form that changes no area takes both signs, and its zero set is a pair of directions. Symmetry at work

The walls a strain permits

Two domains of different shape can only meet along a line neither of them stretches. That condition is a quadratic in a direction, so a pair of domains has exactly two permissible walls — and the reason there are always two rather than sometimes none is that their strains differ by no area at all.

 A mode with no dipole, landing in a phase that may have one. Two marks per row: the first is filled when the mode itself carries a dipole — the displacements, weighted by charge, summing to something other than zero — and the second when the class of the phase it produces permits a polarisation at all. A row with the first empty and the second filled is an improper case: nothing about the transition was about becoming polar, and the phase that results may be polar anyway, so a polarisation appears as a side effect at second order in an order parameter that is about something else. The zone-boundary rows are where these occur; at the zone centre in the plane there are none, because the only two-dimensional order parameters available there are the polarisation itself. Symmetry at work

The polarisation nobody asked for

A mode whose displacements cancel exactly can still leave a phase whose class permits a polarisation. The crystal then becomes polar as a side effect of a transition that was about something else — and in the plane, at the zone centre, the arithmetic says this cannot happen at all.

 Σ5: three lattices in one picture. Two copies of the square lattice turned by 36.87 degrees against one another — one drawn pale, one drawn in the second colour — with the points they share ringed. The fine dots are the lattice generated by both together, the DSC lattice, which contains each crystal with index 5 exactly as the coincidences sit inside each crystal with index 5. Three lattices nested at the same index, and the middle one is the crystal. Symmetry at work

The dislocations a boundary allows

Two crystals meeting at a coincidence angle share one lattice and generate another. The second is where a boundary's own defects live, its shortest vector is one over the square root of the index, and a dislocation's energy is the square of that.

 Every wall is a frieze. Each ferroelastic descent, with the frieze group of each of the two walls its domains permit. A wall is periodic along its length and bounded across it, so its symmetry group is one of the seven — the classification this collection derived early as the same argument on a strip, arriving here as a fact about interfaces. The last column counts the operations in the wall's group that exchange the two domains rather than fixing them: a wall is unchanged by having its sides swapped, so those belong to it, and they are why a wall is often more symmetric than either domain. Symmetry at work

The wall has a group of its own

A boundary between two domains is periodic along its length and bounded across it, so its symmetry is a frieze. The seven, derived here early on as an exercise on a strip, turn out to be the classification of interfaces.

 A hundred and thirteen orbit types, merged into shapes. The three counts, and what stands between them. 113 is the number of kinds of form this site publishes: one for every stabiliser a face can have, in every class. Allowing a stratum to change shape along its own family raises it to 164. Merging entries that are the same solid with the same symmetry, wherever they occur, brings it down to 48 — 30 that enclose a volume and 18 that do not, which is the count every mineralogy text prints, with the dome and the sphenoid kept apart rather than merged. The last line is the warning: throwing away the symmetry of the solid and keeping only its combinatorial type leaves 35, because a rhombic dipyramid, a tetragonal dipyramid and an octahedron are one and the same arrangement of eight triangles. Symmetry at work

A hundred and thirteen orbits, and forty-eight shapes

This collection reports 113 kinds of crystal form and every mineralogy text reports 47. That difference was explained here in a paragraph and never computed, which means nobody had checked it. Computing it needs a definition of *shape* a program can decide, and the definition turns out to be the interesting part.

 The figure of merit a supercell always beats. One line list, indexed on the true cell and on five multiples of it. The mean discrepancy is not merely similar down the column, it is identical to every digit: the supercell's grid of allowed Q values contains the true cell's grid exactly, so each line lands on precisely the same place and misses by precisely the same amount. Any figure of merit built on agreement alone therefore returns one number for the whole family, and cannot prefer the true cell. What falls is the last column, and the only thing in it that the fit does not already contain is the count of lines the cell says should have been seen. Symmetry at work

The figure of merit a supercell always beats

Indexing a powder pattern returns a ranked list rather than an answer, and the ranking needs a number. The obvious number — how well the cell accounts for the lines — is exactly the number a supercell cannot lose on, because the supercell's grid contains the true cell's grid and the discrepancies are identical to every digit. What has to be paid for is the lines nobody saw.

 How densely each shape packs, by translation alone. The densest lattice packing of each shape, as its area over the critical determinant of its difference body. The two that tile the plane by translation reach one and must, which is a check on the search rather than a result of it. The triangle reaches exactly two thirds because its difference body is a hexagon. The many-sided approximation to a circle reaches π/√12, which this collection computes a completely different way. And the pentagon is the worst of them, which is where the search is doing work nobody could do by inspection. Symmetry at work

The densest packing of a shape that is not a disc

Which lattice packs equal discs most densely has a proof that finishes. Replace the disc with a pentagon and the same question has no closed form, but it does have a reduction: translates overlap exactly when the difference of their positions lies inside the shape minus itself, so the question becomes the smallest determinant a lattice can have while avoiding one convex body — and that is a search with a resolution attached.

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