Swapping the species does not halve the count

Counting what a group cannot tell apart put two species on the sixteen sites of a four-by-four block with the symmetry of p4m and found 805 structures among 65,536 arrangements. It named three directions it did not take, and one of them was to let the species themselves be interchangeable: to count a structure and the structure with every A turned into B and every B into A as one. That is the natural count whenever the two species are two states of one thing — spin up and spin down, one orientation of a molecule and its reverse, occupied and vacant with the roles of site and gap symmetric.

The obvious answer is half: every structure has a partner, the arrangement with the species exchanged, and the two merge. Half of 805 is 402.5, which is not a count, and the correct number is 433.

The gap is the structures that are their own partner. Exchange the species in one of them and the result is not a new structure but the same one, turned or moved by an operation of the group. There are 61 such structures on this block, and $\tfrac12(805 + 61) = 433$. What they are, how many there are and where they sit is the whole of this essay.

The lemma, applied to a group twice as large #

Burnside’s lemma counts orbits as the average number of arrangements each operation of the group leaves alone. Letting the species be exchanged doubles the group: every operation $g$ of the plane group, on its own or followed by the exchange. The count up to that larger group is the average over both halves.

The first half is the old sum, whose average is the 805. The second half is new, and it is where all the interest is. An operation $g$ followed by the exchange leaves an arrangement alone exactly when every site $x$ holds the species opposite to the one at $g$’s image of $x$. Follow a site round the cycle $g$ makes of it: the species must alternate A, B, A, B round the cycle and come back to where it started. That is possible only if the cycle has even length, and then it can be done in two ways, starting with A or with B.

So an operation with even one cycle of odd length contributes nothing to the new half. In particular, any operation that fixes a site contributes nothing, because a fixed site is a cycle of length one and would have to hold the opposite of itself. An operation whose cycles are all even contributes two raised to the number of its cycles.

On the four-by-four block of p4m, 95 of the 128 operations have only even cycles. Three translations cut the sixteen sites into eight pairs, twelve into four cycles of four; twelve half-turns about points that are not sites cut them into eight pairs; sixteen quarter-turns into four cycles of four; and the reflections and glide reflections that avoid every site do the same in three patterns. The contributions add to 7,808, and divided by 128 that is 61 — the number of structures carried onto their own complement.

The formula is general, and it says two things at once. The count with the species interchangeable is

$$\frac12\left(N + S\right), \qquad S = \frac{1}{|G|}\sum_{g \text{ with only even cycles}} 2^{c(g)},$$

where $N$ is the ordinary count and $c(g)$ the number of cycles. And $S$ is itself a count: the number of structures that some operation reverses. Burnside’s lemma applied to the larger group has delivered a second number for free, and that second number is the one with a meaning.

The smallest block, by hand #

The formula is easy to trust on a block small enough to list. Take four sites in a two-by-two block with only translations — p1 — so the group has four operations. There are sixteen arrangements. The identity fixes all sixteen; each of the three other translations pairs the sites into two cycles of two and fixes four. The ordinary count is $(16 + 12)/4 = 7$: all A, all B, one A, one B, a pair of A in one row, a pair in one column, and a pair on the diagonal.

With the exchange admitted, the new half of the sum is the three translations again, because each has only even cycles, and each contributes $2^2 = 4$. So $S = 12/4 = 3$, and the count up to exchange is $(7 + 3)/2 = 5$. The list confirms it. All A and all B merge; one A and one B merge; and the row pair, the column pair and the diagonal pair are each carried onto their own complement by a single step along the block — a row pair shifted by one row is the other row, which is the complement. The three self-complementary structures are the three ways of halving a two-by-two block, and they are the stripe patterns and the chessboard that every account of a two-sublattice antiferromagnet starts from.

That small case already has the whole shape of the large one. The self-complementary structures are the balanced ones some operation reverses; the others pair off; and the reversing operation here is a translation every time, because p1 has nothing else, which is exactly the situation of p1 on the four-by-four block, where all sixty are reversed by translations.

They all sit at half composition #

An operation of the group moves sites and keeps the number of sites of each species. So a structure carried onto its own complement has as many A as B, and a self-complementary structure on sixteen sites has exactly eight of each. Nothing needs computing to see that; what the census adds is how large a share of the balanced structures it is.

Pólya’s refinement splits the 805 by composition: one with no A, one with one, five with two, and so on, rising to 153 at eight of each and falling symmetrically. Of the 153 balanced structures, 61 are self-complementary. The other 92 are balanced by count and not by symmetry: exchanging their species gives a different structure that happens to have the same composition, and in the count up to exchange each of them merges with its partner.

That distinction is invisible in a composition. A chemist given a formula AB with eight of each on a sixteen-site cell cannot tell from the formula whether the structure is one of the 61 or one of the 92, and yet the two kinds behave differently under the operation that matters most when the species are states rather than atoms.

Read as spins: a zero moment forced, or a zero moment by count #

Call A spin up and B spin down, and the exchange becomes time reversal: running a current loop backwards reverses its moment and moves nothing, which is the operation that reverses time. A structure with equal numbers up and down has zero net moment. The census separates two ways of having it.

In a self-complementary structure the zero is forced. Some operation of the crystal carries the arrangement onto its time reverse, so the magnetisation is carried onto minus itself by a symmetry and must vanish. That is an antiferromagnet in the strict sense, and the zero survives any perturbation that respects the symmetry. In the other 92 the zero is a coincidence of count. Nothing relates the up sites to the down ones, and a perturbation that changes the moment on one kind of site more than on the other — a field, a strain, a different neighbour — produces a net moment immediately. That is a compensated ferrimagnet, balanced at one composition and one temperature and nowhere else.

The operations that carry a self-complementary structure onto its complement, together with the ones that fix it, form a group in which half the elements reverse the species. That is a two-colour group — the kind seventy-four colourings, forty-six groups counts — and each self-complementary structure is a pattern whose full symmetry is one of them.

Reversed by a translation, or by a turn #

Which operation reverses the species is not a detail. If a pure translation does, the structure read as spins has a translation that reverses every moment, so the magnetic cell is larger than the chemical one and magnetic reflections appear where the chemical ones are absent. If only a rotation or reflection does, the magnetic cell is the chemical cell.

The first kind is also the kind that forms domains of a particular sort. A translation that reverses the structure relates two arrangements that differ by nothing but a shift, so a crystal can hold both at once with a boundary between them where the shift changes — an antiphase boundary, across which the pattern is continuous and the species are exchanged. A structure reversed only by a turn has no such boundary available: its two versions differ by an orientation, and the domains between them are twins rather than antiphase domains.

For p4m, 23 of the 61 are reversed by some translation and 38 only by rotations and reflections. For p2, 48 and 78. For p1 all 60 are reversed by translations, because p1 has nothing else. And for five groups on this block — p1, cm, p3, p3m1 and p31m — not one self-complementary structure is reversed by a point operation, for the reason the lemma already gave: every rotation and reflection those groups have on this block fixes a site, and an operation that fixes a site cannot reverse the species on it. A three-fold rotation has cycles of length one and three and never qualifies. A mirror through a row of sites fixes the row.

That is the counting version of a rule magnetic crystallography states about sites: an atom sitting on an element that reverses time must carry a moment that the element leaves alone, and for a moment that can only point up or down, no such moment exists. The census reaches it without mentioning moments, through the parity of a cycle.

The seventeen #

Across the seventeen the self-complementary share runs from about one structure in seventy for p1 (60 of 4,156) to one in thirteen for pmm, pmg and pgg (115 of 1,459), and the count up to exchange is never half. Every figure in the first three columns is computed twice: by the Burnside sums, and by generating all 65,536 arrangements, marking each orbit, and asking whether the complement of a representative lies in its own orbit. The two agree for every group.

A few of the rows repeat each other: pm and pg give the same counts on this block, and so do pmm, pmg and pgg. That is a property of the four-by-four torus, whose small size lets a glide and a mirror act on the sixteen sites with the same cycle structure, and not a statement that the groups are alike; on a larger block they part. Counts on one block are counts of one quotient of the group, which is the limitation every count of this kind carries and which the table of marks makes explicit.

The block must also have even side for any of this to happen. On a three-by-three block every translation has cycles of length one or three, and so does every other operation of p3; the census finds 32 structures and not one self-complementary. Nine sites cannot be split evenly, and the lemma says the same thing through cycles.

Three species, and any relabelling #

The same enlargement works with more species. With three, and any permutation of them allowed, the group is the plane group times the six permutations, and a pair made of an operation and a relabelling $\sigma$ fixes an arrangement exactly when, round each cycle of length $\ell$, the species at the start is one that $\sigma^\ell$ leaves alone. Each cycle contributes the number of such species. That is de Bruijn’s extension of Pólya’s theorem, from 1959.

For p4m there are 359,955 structures of three species on the block and 60,742 up to relabelling, against a naive 59,992.5. The excess is again the structures some relabelling carries onto themselves, now in more varieties: a structure fixed by exchanging two species, and one fixed by cycling all three, which needs every cycle of the operation to have length divisible by three. On the smallest block the count is checked against direct enumeration of all eighty-one arrangements; on the four-by-four block it rests on the formula and on its divisibility, which fails if a single term is wrong.

What the numbers depend on #

The block is a torus of side four. Every count here is of arrangements on a four-by-four block with its edges identified, which is a finite quotient of the infinite pattern: a structure with a larger period is invisible, and two structures that differ only beyond the block are one. That is the standard device and the standard cost of counting on a finite block.

The convention is that exchange is allowed. Whether it should be depends on the physics. For two chemical species it is not — copper and gold are not interchangeable — and the ordinary count stands. For two states of one thing it is, and then the ordinary count double-counts every structure that is not its own complement. The number to quote depends on which, and quoting 805 for a spin system or 433 for an alloy is an error of convention rather than of arithmetic.

No figure here shows a symmetry. The pairs of blocks show an arrangement and its complement; that the complement is the same structure is a claim that some operation of the group maps one onto the other, which the census checks by trying every operation and the picture can only suggest.

Where the count meets the colour groups #

Two colours and a swap classifies the groups that exchange two colours: the counterchange patterns, whose symmetries include operations that reverse black and white. This census comes at the same objects from the other side. It does not classify groups; it counts arrangements, and among them it finds the ones whose full symmetry, once exchange is admitted, is one of those groups. Sixty-one of the p4m structures on this block are counterchange patterns, and the lemma found them by nothing more than asking which operations have only even cycles.

The two routes meet in one fact: an operation that reverses the colours must move every site, because a site it fixed would have to be both colours. For a group that is the statement that the colour-reversing elements have no fixed points on the coloured set; for a count it is the statement that only even cycles contribute. The halving a lattice will not permit meets the translation half of the same fact, where an antitranslation is a lattice vector that must move every site to one of the other sublattice.

Still open: the census by two-colour group #

The census sorts the self-complementary structures by whether a translation reverses them, which is one bit of information about their two-colour group. The full statement would sort them by the group itself — which of the two-colour plane groups each one has — and that needs the structure’s complete symmetry with exchange computed and identified, not merely the kinds of operation that reverse it. On the four-by-four block that is 61 identifications for p4m and a few hundred across the seventeen, each a small computation of a stabiliser and a comparison against the forty-six.

The other question is the spin version with more than two states. A moment that can point in three directions — the three-state Potts arrangement, or a triangular antiferromagnet’s three sublattices — is three species with the relabellings of a cyclic group rather than all six, because time reversal does not permute three directions arbitrarily. Which subgroup of relabellings is physical, and how the count changes with it, is a question de Bruijn’s formula answers once the subgroup is named, and naming it is a question about the spins rather than about counting.