One cell reaches every Wyckoff set

The same site under two names sorted the seventy-two positions of the seventeen plane groups by whether a change of description can exchange them. A motion in a group’s normaliser rewrites a report without changing what it reports, so two positions it exchanges are one position written from two origins, and the orbits are called Wyckoff sets. The essay ended on a fact it was careful about. The widest possible normaliser, the affine one, which carries the group onto itself whether or not it preserves lengths, makes exactly the identifications that isometries of some single cell make. It called that a fact about the plane, checked group by group, and “not one to carry into three dimensions by analogy”, because in space the affine normaliser of a monoclinic or triclinic group mixes the cell’s axes in ways no metric obviously accommodates.

The check can now be made, and its first result is a warning of a different kind. The obvious place to look for a failure is the triclinic system, where the normaliser is as large as it can possibly be, and there is an object there that looks exactly like a counterexample. It is not one, and the reason it is not is worth more than the answer. On one cell for each space-group type, the Euclidean sets are the affine sets. The plane’s tidy answer holds in space, and past the monoclinic system it needs no search at all, since it follows from averaging a quadratic form. The monoclinic groups, where no such argument exists, turn out to need a special cell in exactly one case out of thirteen.

Finding the positions, and two numbers to check them by #

The positions have to exist before they can be sorted. The points a group treats differently defined them in the plane, and the definition carries over unchanged: a point’s site symmetry is the set of motions that fix it, written out in full with whole-cell translations included, and two points are in one position when their stabilisers are conjugate in the group. Done on a grid of twenty-fourths of the conventional cell, that is a little over thirteen thousand points for each type, and the types are the 219 of the census over the seventy-three arithmetic classes.

Such a list could easily be slightly wrong, so the external check matters. The International Tables print the total number of Wyckoff positions across the 230 groups, which is 1,731. Counted over the 219 affine types with each enantiomorphic pair twice, the stabilisers here give 1,731. Nothing was fitted to reach that.

The normaliser is then found in two pieces. A change of basis comes from the finite group of what a normaliser can do to the axes past the monoclinic system, and from the integer matrices with small entries below it. An origin shift for each is searched on the same grid, and the pair is kept when conjugating by it carries the group onto itself. The orbits on the positions then give affine sets, and their total over the 219 types is 1,128, the same total as the published count of types of Wyckoff set. That number is quoted rather than read from a table of sets here. Agreement between a stabiliser search and a normaliser search with both published figures is the reason to believe what they say next.

The picture at the head of this essay is the plane’s table, one dimension up. In the plane the totals ran seventy-two, fifty-three, fifty-one, fifty-one. In space they run 1,694, 1,182, 1,128, 1,128. Most of the fusion, a third of all the positions, happens on the general cell, through the origin shifts. And the third and fourth numbers agree, which is the claim. What matters is that it holds type by type and partition by partition rather than only in total, and the comparison was made on the sets themselves.

Almost all of the table is one column. Tetragonal, trigonal, hexagonal and cubic groups have the same number of sets on the general cell as under the affine normaliser, and so do the triclinic ones. A cell can only matter where the affine normaliser contains something the general cell does not make an isometry, and for all those systems it contains nothing of the kind. A tetragonal group’s general cell already has the four-fold axis and every mirror the normaliser could use.

Where the cell matters #

Twenty-seven types depend on the cell. Twenty-six of them are orthorhombic, out of fifty-nine in that system, and all twenty-six behave in the same way, which is the way $pmm$ behaved in the plane. Three inequivalent axes make three separate families of positions that a change of basis could exchange if only the axes had equal length. No shift can turn a mirror normal to $a$ into a mirror normal to $b$.

$Pmmm$ is the clearest case, because it is where the numbers are largest.

It has twenty-seven positions: eight points where three mirrors meet, twelve lines where two meet, six mirror planes, and everything else. On a general orthorhombic cell the shifts by half a cell along each edge carry the eight points onto one another, so those are one set, and they carry each mirror plane onto the parallel one at the other height. That is eight sets out of twenty-seven before anything else is considered, and nothing here depended on the lengths of the axes. Only the relabelling of axes is left, and that does depend on them. A structure with its atoms on mirrors normal to the short axis and one with its atoms on mirrors normal to the long axis are different crystals, since the distances between atoms differ. On a cubic cell they are the same crystal, turned, and the eight sets become four.

So the cell’s whole contribution to a Wyckoff set in space is to make the axes equal. Twenty-six of twenty-seven cases are that one thing.

A proof that takes one line #

Why should it always work? Past the monoclinic system there is a short answer that makes the plane’s group-by-group check look like more work than it needed to be.

A normaliser must permute the axes proved that past the monoclinic system a space group’s affine normaliser has only finitely many linear parts. Take any finite group of integer matrices $M$, and form the sum of $M^{T}M$ over the group. That sum is a positive definite symmetric matrix, and because multiplying by another group element only reorders the sum, every $M$ preserves it. It is a metric, and on the cell with that metric, every element of the affine normaliser is an isometry. So a single cell reaches the affine sets for every type past the monoclinic, by construction. There is nothing to search for. The computation here formed that average for each of those types and compared the partitions, as a check on the arithmetic rather than on the argument, and all 204 agree. For the twenty-six orthorhombic ones the averaged form has forty-eight symmetries: it is a cell in which all three axes are equal.

This is a much older idea than crystallography, the same averaging that shows every finite group of matrices is conjugate to a group of orthogonal ones. Its role here is to explain why the plane’s result was not luck. In the plane every affine normaliser with something to fuse was finite as well, once the shifts had done their part, and the essay’s careful check was verifying a theorem it had no need to doubt.

It also says exactly where the doubt was justified. The argument fails if the normaliser’s linear part is infinite, and that happens in the triclinic and monoclinic systems, where the normaliser only acts modulo two and an infinite group has no average. Those fifteen types are where a counterexample could live.

The trap in $P\bar1$ #

The largest normaliser of all belongs to $P\bar1$, whose linear part is every integer matrix of determinant plus or minus one. Its positions are the eight inversion centres, at the corners, edge midpoints, face centres and body centre of the cell, and the general position. Those eight centres look like the perfect place for the plane’s answer to fail.

Set the origin on one of them and the other seven are the seven non-zero halves of the lattice. The integer matrices permute these through their reductions modulo two, and the group of three-by-three matrices over two elements has order 168, which contains an element of order seven. So there is a change of basis that carries one half-shift through all seven in a single cycle.

That matrix cannot be an isometry of any cell. An isometry of a lattice has finite order, and in three dimensions no integer matrix has order seven, which is the crystallographic restriction. This one is the companion matrix of $x^3 - x - 1$, whose largest root is the plastic number, 1.3247, the same number that governs the smallest inflation factor on the aperiodic side of this subject. Its powers grow without end, their entries running 1, 1, 1, 2, 2, 3, 4, 5.

And no finite group can do the job this matrix does by being larger, either.

A group transitive on seven things has order divisible by seven, and no finite group of integer matrices in three dimensions does. Sorting the seven by each of the fourteen Bravais lattices shows what the best that can be done looks like: face-centred cubic gives an orbit of six and leaves one over, body-centred cubic splits them four and three, and the others do less. That is a striking table. Here is a set of seven points, permuted transitively by an affine normaliser, which no cell can make equivalent. It looks like exactly the failure the essay warned of.

It is not one, because the seven were never separate. They are origin shifts, and a shift by half a cell is already in the normaliser of $P\bar1$ on every cell, since an inversion at the origin and one at any half-point generate the same group with the lattice. So all eight centres are one Wyckoff set on the most general cell there is, and the matrix joins nothing that the shifts had not. The computation says this directly: nine positions, two sets, on the triclinic cell and under every change of basis. The affine normaliser is infinite, and all of that infinity is spent moving positions within a set they already share.

This was the first version of the idea behind this essay, and it was wrong. A claim that the seven centres cannot be joined on any cell was going to be the headline until the computation came back with one set rather than two. What the seven-cycle really shows is about descriptions rather than positions: the coordinate lists of one $P\bar1$ structure related by such a change of basis cannot be related by any isometry of any cell, and that is a real statement about what a change of basis can mean. It is just not a statement about sets.

The one monoclinic case #

The same thing happens in twelve of the thirteen monoclinic types. The normaliser is infinite, and the general cell’s sets already equal the affine ones, so its extra elements have nothing to fuse. The one exception is $C2/c$.

$C2/c$ has six positions: four kinds of centre on or off its two-fold axes, a line of axis points, and the general position. On a general cell they fall into four sets, and under the affine normaliser into three. The extra identification joins the two inversion centres at $(0, 0, 0)$ and $(0, \tfrac12, 0)$ with the two at $(\tfrac14, \tfrac14, 0)$ and $(\tfrac14, \tfrac14, \tfrac12)$. No shift does it, and a change of basis does.

Since the normaliser is infinite, averaging over it is not available, but something narrower is: find one element of finite order that makes the identification. The group it generates with the point group is finite, so that group can be averaged, and on the resulting cell the element is an isometry. The search finds an element of order two, and the averaged form has sixteen symmetries, so it is a tetragonal cell. $C2/c$ on a tetragonal cell has three sets, the affine number. A structure in $C2/c$ whose lattice happens to be tetragonal to within error can be reported with its atoms on either pair of centres, and those reports describe one crystal.

This is the only case in the census where a monoclinic group’s cell can change what its positions mean, which is a small and definite thing to know about the most commonly reported family of crystal structures.

What this does not show #

That one cell reaches the affine sets does not make them the right sets for a crystal. The distinction the plane’s essay drew still holds and is if anything sharper here. An orthorhombic crystal with three different cell edges and atoms on mirrors normal to one of them is not the same crystal as one with atoms on mirrors normal to another. Identifying them because a cubic cell would is the false positive. The result here is that the affine classification has a geometric meaning, and it should not be read as licence to use it on a measured cell.

The monoclinic search is bounded. Candidate matrices have entries of at most one in a primitive basis, and the claim that this generates all the identifications rests on the normaliser acting only modulo two there and on the agreement with 1,128. It is not an independent enumeration of an infinite group.

The 1,128 is quoted from memory of the literature on Wyckoff sets, not from a table consulted during this computation, and the agreement is of a total rather than of the individual sets.

The small squares of the $Pmmm$ figure are a bookkeeping picture. They show how many positions there are and which are grouped. They do not show where a position is in the cell, and two squares in one shade may be points as far apart as the cell allows.

How the sets were checked, and what they had to refuse #

Each motion is required to carry every position onto one with the same site order, and the computation stops if one does not. That check is the one that makes the partitions trustworthy, because a wrong shift would usually fail it. Shifts are kept only if conjugating the operation list returns the list exactly, and translations must go to translations before any shift is considered.

Numbers from outside are the main check, and there are two, each of which a mistake in the corresponding half of the computation would have changed: 1,731 on the stabilisers and 1,128 on the normalisers. Beneath those, $Pmmm$ has to have twenty-seven and $P2/m$ fifteen. A shift that is not in the normaliser has to be turned away. Offered to $Pmmm$, a quarter-cell shift would carry mirrors at height zero to heights that hold no mirror, and it is not found.

And the headline this essay nearly carried has been refused by name. That the seven half-shifts of $P\bar1$ are positions no cell can join is what the figures alone would suggest, and the test is that they are one set before any matrix is applied.

Where the question goes next #

The argument depends on one property of three dimensions and of two: every affine normaliser with something to fuse beyond the shifts has a finite subgroup that does the fusing. Whether that survives in four dimensions is a real question, since there the crystallographic restriction permits orders that make the seven-cycle’s trick available to other primes, and the normalisers of low-symmetry groups are again infinite. In the plane and in space a shift always got there first. Whether a four-dimensional group exists whose positions are joined only by an element of infinite order, so that the affine set is reached on no cell at all, would need the positions of a four-dimensional type list, and that has not been built.

Sideways, the same positions now make other questions possible. Three of them, and they are equivalent asked which subgroups a normaliser fuses and could only ask it in the plane, and it named screws as the case to watch. With the normalisers of all 219 types in hand, the same comparison for subgroups of small index in space is a loop over what is already computed.