Crystallography & Symmetry
patterns derived from their groups
Concept

Origin choice — where it appears

The convention deciding where a group's origin sits, which for some space groups has two answers and is printed twice in the tables. A structure published on one convention and read on the other is shifted by a vector nobody records.

Named by 8 essays across 4 fields — each of them below, with the objects they name alongside it.

Two cells of equal area on one rhombic lattice. One lattice — the rhombic lattice that cm sits on — with two parallelograms drawn on it: the conventional cell, and a sheared cell whose edges are the integer combinations (1, 0) and (1, 1) of it. The points are identical in both outlines; only the description changes. Each cell's contents were counted by writing every lattice point in that cell's own coordinates and sharing each one out between the cells that meet at it — a quarter at a corner, a half on an edge, one inside — and the totals come to 1 and 1, which are the determinants of the two matrices. The alternative has determinant one, so its inverse is integral and it generates exactly the same lattice; that is the whole condition, and it is why a lattice has infinitely many bases and no arithmetic can prefer one.

The cell is a choice, the lattice is not

Every lattice has infinitely many unit cells and infinitely many bases, and crystallography picks one by convention. Knowing which convention is in force is the difference between a symbol that means something and a symbol that means nothing.

lattices · Lattice One lattice, two cells, and the absences the choice creates. The same set of points described on a centred rectangular cell and on its primitive rhombic cell, with what the centred description scatters. Half the reflections vanish, and they vanish because of how the cell was drawn rather than because of anything the crystal does.

Why the bigger cell wins

A centred cell has twice the area it needs and crystallography prefers it anyway. The preference is not conservatism — it buys operations that read as whole numbers along the axes, and the price is a set of reflections that vanish for reasons having nothing to do with the crystal.

lattices · Centring Fddd has two published origins. The two conventions, computed from the operations. The International Tables place the origin at the point of highest site symmetry, and also at a centre of inversion, and for this group those are different points — so the group is printed twice, with every coordinate in the second table shifted by (-0.125, 0.125, -0.125) from the first. A structure published on one and read on the other has every atom in the wrong place by that vector, the refinement fails in a way that looks like bad data, and nothing in the symbol says which was used.

Two origins for one group

The International Tables place the origin at the point of highest site symmetry, and also at a centre of inversion. For twenty-four of the two hundred and thirty those are different points, so the group is printed twice with every coordinate shifted — and nothing in the symbol says which table a structure was written against.

space-groups · Settings The origins of p2 that change nothing. One cell of p2 with its pattern, and every point marked to which the origin may be moved without a single operation of the group changing its translation part. There are 4 of them per cell, and the count does not change when the search grid is refined, so it is a fact about the group rather than about the grid. Two coordinate lists differing by one of these vectors describe the identical arrangement, which is why no structure's coordinates are ever unique.

The same pattern, described twice

Two coordinate lists for one structure can disagree in every number and describe exactly the same arrangement, because a group does not fix its own origin. The operations that may be applied to a description without changing what it describes are its normaliser, and they can be found by looking at pictures rather than at matrices.

operations · Normalisers P222: 16 descriptions of one structure. P222 has 4 operations in a cell. 8 origins leave every one of them exactly where it was, and 8 linear parts of the lattice's holohedry normalise the group, so its Euclidean normaliser has 64 elements per cell and the index is 16. That index is the number of coordinate lists that describe one and the same arrangement of atoms. Each was applied to a motif and the resulting point sets compared: the numbers differ and the sets are identical, which is the check that makes the count mean anything.

One crystal, and sixteen coordinate lists

Two structure reports can disagree in every number and describe the same arrangement of atoms, because a space group does not fix its own origin or its own axes. How many genuinely different lists there are is the index of the group in its Euclidean normaliser — a number, computable, and the thing a structural database has to divide out before it can say two entries are the same compound.

operations · Normalisers the most consistent answer has 100 per cent of the signs. 60 runs of the sign procedure from 60 different random starts, each plotted at its self-consistency — a figure computed without any knowledge of the answer — against the fraction of its signs that are in fact right. Throwing out the 1 run that reached the uniform solution — every sign the same, perfectly consistent and physically a single peak — the highest consistency belongs to a run with 100 per cent of the signs right. The ranking works here, and the reason it works is that the cell is small. Nothing in the plot's horizontal axis knows the answer, which is the only reason a procedure of this kind is a procedure at all.

The formula that has the answer already

The tangent formula rebuilds each phase from all the others, and the true phase set is very nearly a fixed point of it — hand it the answer and it hands the answer back. Start it anywhere else and it does not arrive. Having a fixed point and finding it are different problems, and the second is where the subject spent twenty years.

diffraction · Direct methods Success of the flipping solver against how far off the subtracted lead is. The share of forty starts on the cell of nineteen carbons and one lead that the flipping solver solves, with the lead's density subtracted at a position displaced from the true one by up to sixteen grid steps, half the cell. The rate does not fall with the offset: the points scatter without a trend about the pooled rate of 40%, the band marking two binomial standard deviations, and the original solver, with nothing subtracted, solves none. Refining the position from the map does no better than holding it.

The lead is wherever it is taken out

Charge flipping solves a cell of nineteen carbons and one lead once the lead's own density is taken out before each flip and put back after. The obvious worry was precision: a lead subtracted a fraction of a grid step from where it really is leaves a residual larger than any carbon. The residual is there, larger than a carbon at a tenth of a step. It does not matter. Subtracted two steps away, or half the cell away, the lead still lets the solver succeed at the same rate, because the solver builds the whole structure round wherever it is put.

diffraction · Direct methods Success of the flipping solver against the error in the vector between two subtracted leads. The share of 80 starts on a cell of eighteen carbons and two leads that the flipping solver solves, with both leads' density subtracted, the first at its true place and the second displaced from its true place by an error in a random direction. The rate holds to a quarter of a grid step (21, 19, 18, 11, 0, 0, 0 successes at errors of 0, 0.1, 0.25, 0.5, 1, 2, 4 steps), falls by half at half a step, and is zero from one step on. With one lead the place did not matter at all; with two, the vector between them does.

Two leads need their vector to half a step

With one heavy atom in a cell with no symmetry, charge flipping did not care where the atom was subtracted: it rebuilt the structure round whatever place it was given. With two heavy atoms that freedom runs out, as predicted, because no translation can put both under wrong references at once. Subtract one of them a whole grid step from where it belongs and the solver never succeeds. But the tolerance is five times looser than the estimate it was predicted from: up to a quarter of a step the rate does not move, at half a step it halves, and the residual the solver absorbs is three carbons' worth, not one.

diffraction · Direct methods

Named alongside it

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

Direct methodsEquivalent originNormaliserSettingCentringElectron densityHeavy atomHeavy atom methodPhase problemPrimitive cellUnit cellCase analysis

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