Two forty-sixes that do not match

Two counts of the subperiodic groups end on the same number. Eighty is seventeen, seventeen and forty-six found that a layer group is a plane group with a sign on each operation, plus for an operation that keeps the sheet’s two faces and minus for one that exchanges them, and that the signs come in three ways: all plus, doubled by a mirror in the sheet, or a two-colouring of the plane group. There are forty-six two-colourings, so forty-six layer groups of the third kind. Seventy-five is twenty-nine and forty-six found that a rod group is a finite group of the plane across the rod extended by a group of motions of the axis, and that the axis group comes in two kinds: translations only, for twenty-nine rods, and translations with reversals, for forty-six.

That essay closed on the question of whether the second forty-six is the first in disguise, a way of reading a two-coloured plane pattern as a rod that turns its axis over, or a coincidence of small numbers. It named the first test: whether the two lists have the same distribution over a natural invariant. This essay runs that test on six invariants, and the answer is the same on all six. The two lists do not match, and the ways they fail to match say why the totals agree only by accident.

Making the two lists comparable #

The test needs something to compare, and the two lists are made of different things. One is a list of plane patterns with two colours, the other a list of threads in space. They become comparable once each entry is read as a pair: a group and a subgroup of index two.

A two-colour plane group is a plane group $G$ together with its colour-keeping half $H$, the operations that send black to black. $H$ has index two in $G$, and the colouring is the pair, considered up to the affine changes of description that carry $G$ to itself. Seventy-four colourings, forty-six groups found seventy-four such halves among the seventeen plane groups and forty-six classes of them.

A rod that turns its axis over is a rod group $R$ together with its axis-keeping half $R^+$, the operations that send the rod’s upward direction to itself. $R^+$ has index two in $R$, and it is one of the twenty-nine rods over translations alone.

Each member of each pair has a point group in the plane. For a colouring it is the point group of $G$ or of $H$, read off the linear parts of their operations. For a rod it is the group of in-plane parts of the operations of $R$ or of $R^+$: what the rod shows when looked at end on, with every operation’s action along the axis forgotten. Both are among the ten crystallographic point groups of the plane, 1, 2, 3, 4, 6, m, 2mm, 3m, 4mm and 6mm, which the plane’s finite groups leaves once the restriction has had its say.

So both lists carry the same invariants: the point group of the whole, the point group of the half, the pair of them, the order of the whole’s point group, whether whole and half share a point group, and whether the whole has a mirror. A correspondence between the lists that preserved any one of these would need the two lists to have the same distribution over it. If the distributions differ, no such correspondence exists. That is a test with only one way to pass and many ways to fail.

The whole #

The first invariant is the one the earlier essay proposed, and it fails at once.

The colourings pile onto the rectangular mirror groups: seventeen on 2mm, ten on m, eight on 4mm, and none at all on 3. The rods pile onto the hexagonal and square ones: eleven on 6mm, nine on 4mm, seven on 2mm, and one on 3. The two distributions overlap only at the ends, with one each on the trivial group. To make them equal, seventeen of the forty-six entries would have to move from one point group to another, which is more than a third of the list.

The shapes are nearly opposite, and the reason is visible in what each list is made of. A two-colouring of a plane group needs a subgroup of index two, and there are two places one can come from. It can halve the point group, keeping the rotations and colouring the mirrors, or keeping one mirror and colouring the other. It can halve the lattice, colouring alternate rows or alternate cells. The rectangular mirror groups offer both in abundance. Their point groups have several subgroups of index two, and their lattices have three halvings, all invariant under perpendicular mirrors, along each axis and along the diagonal of the cell.

The hexagonal groups offer almost nothing. Their lattices have three halvings too, but a three-fold rotation permutes the three, so no single halving is kept by the group, and none can be the colour-keeping half. Their point groups 3 and 6 have few subgroups of index two, and 3 has none. That is why p3 has no two-colouring at all, and why the five hexagonal groups between them carry six of the forty-six.

The rods are fed from a different source. What makes a rod over the reversing axis group distinct from another is the turn of its step and the operations at its reversal points, and the step’s turn is a screw: eleven screws and no others are the possible turns, and the six-fold and four-fold groups have the most of them, in both hands. A rod with six-fold screws and two-fold axes across the rod shows 6mm end on, because the in-plane part of a two-fold axis across the rod is a reflection. Six of the rods’ eleven on 6mm are of that kind, $622$ and the five screws $6_1 22$ to $6_5 22$, and the rest add mirrors to them. The plane has no screws to spend. Its only translational freedom is in the lattice, and the hexagonal lattice spends none of it on two colours.

The half #

The second invariant is the point group of the half, and the comparison is worse.

The colourings’ halves are mostly m and 2mm, twelve each, since colouring a rectangular mirror group usually leaves it a mirror group of its own. The rods’ halves are mostly rotation groups, with eight on 6, six on 4 and five each on 2 and 3. That is also what the construction predicts. The axis-keeping half of a rod with two-fold axes across it loses those axes, and with them the reflections they showed end on, so a rod that shows 6mm whole usually shows 6 in its half. The colour-keeping half of a colouring loses an index-two piece of its operations. When that piece is a translation coset, the point group does not change at all.

The pair #

The pair of whole and half together is the finest invariant of the six, and it separates the lists most.

Twenty-one pairs occur between the two lists. Several occur on one side only: a rod can show 3 whole and 3 in its half, or 6 whole and 6 in its half, and no colouring can, since p3 has no colouring and p6’s one colouring halves its rotations. A colouring can show m whole and m in its half seven times, and a rod only twice. The finest invariant moves nineteen of the forty-six, which is the answer one would expect from two lists built by unrelated rules, not from two descriptions of one list.

There is one more invariant worth naming, the order of the whole’s point group, since it is the coarsest and the one most likely to agree by accident. It does not. The colourings have twelve of order two and nineteen of order four, and the rods six and ten, while the rods have eleven of order twelve to the colourings’ three. Fifteen entries would have to move. And whether the whole has a mirror at all, the coarsest test of the six, gives forty against thirty-five.

Where the two extremes come from #

The two lists are most different at their extremes, and each extreme has a short explanation.

The plane’s empty term is p3. A two-colouring of p3 would need a subgroup of index two, and there are only two places to find one. The point group 3 has three elements and no subgroup of index two, since three is odd. The lattice could be halved instead, colouring alternate rows. The lattice vectors modulo twice the lattice form a group of four elements, three of them non-zero, and each non-zero one names a halving: the vectors whose class is zero or that one. A third-turn of the hexagonal lattice permutes the three non-zero classes in a cycle, so it carries each halving to a different one and keeps none. A subgroup of p3 of index two would have to contain the third-turns, since they have odd order, and so it would have to be kept by them. No halving is, and p3 has no two-colouring at all. The same cycle explains why the other hexagonal groups have so few. Each of their colourings has to come from the point group, because the lattice offers nothing a three-fold rotation will keep.

The rods’ largest term is the trivial kernel, with thirteen. A rod whose kernel is trivial has no operation fixing its axis pointwise, so all its structure is in the step and the reversal points, and that structure is plentiful. Three of the thirteen have the same operation at every reversal point, a mirror across the axis, a centre or a two-fold axis across the rod. Four alternate two kinds of reversal point. Six are screws with two-fold axes across them, turning by a third, a quarter or a sixth between one reversal point and the next, in both hands. The plane has no analogue of any of these, because a plane group has no axis to reverse and no screw to turn. A colouring of p1, the plane’s trivial point group, can only halve the lattice, and up to the affine normaliser there is one way to do it.

So the two lists are not merely distributed differently. Each is largest where the other has nothing to offer. The plane’s variety lives in its lattice, and the rods’ in their axis.

The nearest miss #

One invariant comes within one of agreeing, and it is the most interesting of the six because of what it means on each side.

For a colouring, whole and half share a point group exactly when some translation changes colour, since two operations with the same linear part differ by a translation, and if one keeps colour and the other changes it, that translation changes it. These are the colourings in which the colour is carried by the lattice rather than by the point operations, a chessboard rather than a mirror pattern. There are twenty of them. That the equivalence holds is checked colouring by colouring rather than assumed: the computation reads off both conditions independently and requires them to agree.

For a rod, whole and half share a point group end on when every operation that turns the axis over has an in-plane part already shown by the half. The simplest case is a mirror across the axis, whose in-plane part is the identity. There are twenty-one of them.

Twenty against twenty-one is close enough to be tempting and far enough to settle the question. A correspondence keeping this invariant would need exactly equal counts, and one entry has nowhere to go. The breakdown makes the miss worse rather than better. The colourings’ twenty sit mostly on m and 2mm, where the chessboard colourings of rectangular patterns live, while the rods’ twenty-one spread over all ten point groups with no more than three on any.

Two sums that meet #

If no invariant matches, the agreement of the totals has to be explained as arithmetic rather than structure, and the way to see that is to write each total as the sum it is.

The colourings are $1 + 2 + 5 + 2 + 3 + 5 + 5 + 2 + 5 + 2 + 5 + 3 + 0 + 1 + 1 + 1 + 3$ over the seventeen plane groups in their usual order. The rods are $13 + 9 + 7 + 3 + 3 + 3 + 3 + 3 + 1 + 1$ over their kernels, the operations that fix the axis pointwise. The first sum is spread evenly and its largest terms come from the groups with two perpendicular mirrors. The second is steep and its largest term comes from the kernel with nothing in it, where all the variety is in the screws and the reversal points.

Neither sum is a rearrangement of the other, and neither is a refinement of the other in any sense the invariants above could detect. They meet at forty-six because two different collections of small numbers sum to the same small number, which is what small numbers do. The layer groups’ forty-six and the rods’ forty-six are both counts of index-two halves of a family of infinite groups. That is the only thing they share, and it is a description of the question each answers, not a map between the answers.

The pairing that does exist is the one already known, and it runs between different lists. The forty-six colourings are forty-six layer groups, by the sign on each operation: a colour change becomes a flip of the sheet. Two lists of forty-six that are one list read twice can be told apart from two lists that merely share a total, because the first kind agrees on every invariant. Run on the colourings and on the layer groups that turn the sheet over, every invariant above agrees, by construction. Run on the colourings and the rods, none does.

What the test cannot show #

A failed test of this kind is conclusive in one direction only. Each invariant rules out a correspondence that preserves that invariant. It cannot rule out a correspondence that preserves none of them, and a bijection between two lists of forty-six always exists, since any two sets of the same size can be paired. The question was never whether a bijection exists but whether a natural one does, one that respects the structure both lists are made of. The six invariants are the structure: point groups, their orders, the relation between a group and its half. A correspondence that ignored all six would be a renumbering, not an explanation.

Nor does the test say that the two lists have nothing in common. Both arise from the same arithmetic of a subgroup of index two, and two colours and a swap is about exactly that arithmetic in the plane. What the test says is that the arithmetic is fed differently in the two cases, by lattice halvings in the plane and by screws along a line, and that the difference reaches every invariant that could carry a correspondence.

The convention matters for the counts, though not for the conclusion. The colourings are counted up to the affine normaliser, which identifies colourings related by any change of description. The rods are counted up to motions of space that keep handedness, which keeps the two hands of each screw apart. Counting the rods with reflections allowed would merge the four pairs of hands among them and give forty-two, and counting the colourings up to the smaller Euclidean normaliser could only split classes, never merge them, and so could only give more. The equality of the totals depends on both conventions. That is one more sign that it is not structural, since a correspondence of structure would not depend on a choice of how to count each side.

Who counted them #

The forty-six two-colour plane groups were enumerated in the mid-twentieth century, in the tradition of black-and-white symmetry that Heesch, Shubnikov and Belov developed, and they appear as the magnetic layer and plane groups in the tables Daniel Litvin compiled. The seventy-five rod groups and their split by the axis action appear in volume E of the International Tables, edited by Vojtěch Kopský and Litvin. The comparison here uses neither table. The colourings come from the plane groups’ own index-two subgroups, identified under a searched affine normaliser, as seventy-four colourings, forty-six groups computed them. The rods come from kernels and reversal points, as the essay before this one built them, and their count was checked there cell by cell against the rods counted class by class.

Five tests and one refusal #

The first two tests check the inputs: forty-six on each side, and the equivalence between a colour-changing translation and a half that keeps the point group, read independently on every colouring. The next three record the comparisons. The refusal is the hypothesis the earlier essay raised, and it is refused on six counts rather than one. A single invariant that matched would have reopened it, since agreement on a coarse invariant is weak evidence for a correspondence and disagreement on any invariant is decisive against one that keeps it.

Still open: the rods’ own colourings #

The comparison here set the plane’s colourings beside the rods that turn their axis over, and the two are fed by different sources. The rods have two-colourings of their own, the black-and-white rod groups of the magnetic tables, counted by the same rule as the plane’s: a rod group with a half of index two, up to the changes of description that carry the rod to itself. A rod has two kinds of index-two half to offer, one from its point operations and one from its translations along the axis, where alternate repeats can be coloured alike, and its screws give it a third, since a screw’s step can change colour while its repeat keeps it.

Which of the two shapes found here those colourings take is a question the same computation could answer. If they pile onto mirrors as the plane’s do, the source of a colouring is the point group and the screws only add to it. If they pile onto the six-fold and four-fold groups as the reversing rods do, the screws dominate. Each answer says something about where index-two halves come from in a group that repeats in one direction only. The table has not been drawn.