No cell joins the seven halves

One cell reaches every Wyckoff set ended with a fact about positions that had taken a good deal of checking. The affine normaliser of a space group can mix axes without limit in the low-symmetry systems, and it identifies positions that look different. Yet for every one of the 219 types there is one cell on which those same identifications are made by isometries. In the triclinic group $P\bar1$ the normaliser even contains a cycle of seven half-shifts, which looks as if it should join things a cell cannot. It turns out not to matter, because a plain shift of the origin had already joined them all.

That essay’s last paragraph pointed sideways. Three of them, and they are equivalent asked which subgroups a normaliser fuses and could ask only in the plane, because the normalisers of the space groups had not then been computed. With all 219 of them in hand the same comparison is a loop over what is already computed. The natural expectation is that it will come out the same way, and that the fact about positions was really a fact about normalisers.

It very nearly does. The exception is sharp, it has a single cause, and it turns out to be the crystallographic restriction itself, showing up where nothing would have suggested looking for it.

The simplest subgroups a group has #

A subgroup of index two is the place to start, because here the enumeration needs no search. Every such subgroup is normal, so conjugacy within the group does nothing to them. And each one is the kernel of a map from the group onto two elements, one where half of the operations go to $+1$ and the other half to $-1$.

Such a map sends every square to $+1$. Twice a lattice vector is the square of that vector, so the map vanishes on the sublattice $2L$, and it therefore factors through the finite group $Q = G/2L$. Take the subgroup $K$ of $Q$ generated by all squares. Then $Q/K$ has every element of order at most two and is commutative, so it is a vector space over the field of two elements, of some dimension $r$. The maps to two elements are the $2^r - 1$ non-zero linear functionals on it, and each one is a subgroup of index two. So the subgroups of index two of any space group are a finite list, with no search and no choice of bound.

Only one convention needs stating. A subgroup is a set of actual motions, including translations, on a fixed lattice. So the sublattice of even-$x$ points and the sublattice of even-$y$ points are two different subgroups of $P1$, even though they look alike, and the question throughout is what motion makes them equal.

The list contains both kinds of subgroup, as the two ways down calls them: those that keep all the translations and halve the point group, and those that keep the point group and halve the lattice. Across the 219 types there are 1,813 subgroups, the quotient reaching rank six at most. Two types have none at all, the cubic $F23$ and $P2_13$: in those, every operation is a product of squares.

Three ways to make two subgroups the same #

Now compare three levels of sameness, the same three the earlier essay used for positions.

The first is an origin shift: a translation of the whole picture that carries the group onto itself. It is available on every cell, for any metric at all, so anything it joins is joined always.

The second is the whole affine normaliser: any change of basis, with any shift, that carries the group onto itself. That is what the International Tables mean by an equivalence of descriptions, and it treats two subgroups as the same whenever a linear map of space turns one into the other, whether or not that map preserves any lengths.

The third is the best cell. Choose a metric, and keep only the normaliser elements that are isometries of it. Different metrics allow different elements, and the question is how few classes any one metric can achieve.

The enumeration keeps those three numbers for every type. Summed, they are 1,392 classes under shifts, 1,159 under the affine normaliser, and 1,161 on the best cell. The last two differ by two, and the whole of that difference is two groups.

For anything past the monoclinic system, the equality of best cell and affine has a one-line reason, and it is the one the earlier essay used. The normaliser’s linear parts form a finite group, so averaging any metric over that group gives a metric they all preserve. So one cell has them all as isometries, and it makes every identification the normaliser does. That half of the table is no surprise.

The monoclinic row needs work, because its normaliser is infinite, and no averaging can be done over all of it. The computation tries each element of finite order with the group it generates, and that gives only an upper limit for the best cell. Since the best cell can never do better than the normaliser, an upper limit that meets the affine count is exact. It meets it in all thirteen monoclinic types. That is a measurement rather than a proof that it must, and the bound on the matrix entries (at most one on a primitive basis) is a stated limit of the search.

The seven ways to halve a lattice #

That leaves the two triclinic groups. Take $P1$. It is a lattice and nothing else, so its subgroups of index two are the sublattices of index two. There are exactly seven, one for each non-zero linear functional on a three-dimensional space over two elements: keep the vectors whose $x$ is even, or whose $y$ is, or whose $z$ is, or $x + y$, or $x + z$, or $y + z$, or $x + y + z$.

The normaliser of $P1$ is the group of all integer matrices of determinant $\pm 1$, because any change of basis of a lattice is a redescription. Reduced modulo two, those matrices give the whole of the group of invertible three-by-three matrices over the field of two elements, which has 168 elements and acts transitively on the seven functionals. So, under the affine normaliser, all seven halves of a lattice are one thing. That is plain to anyone who has re-chosen a basis: any sublattice of index two becomes “the even-$z$ points” in a suitable basis, by choosing a basis that makes it so.

An origin shift does nothing here, since a shift commutes with translations and leaves each sublattice alone. So the shift column for $P1$ reads seven. In the earlier essay, the shift column for positions in $P\bar1$ had already joined everything, and the infinite cycle had nothing left to do. Here the shift has done nothing, and the entire identification falls on the linear part.

Now ask what one cell can do. Choose any metric at all. Its isometries form a finite group of integer matrices, and that group’s image modulo two permutes the seven. Which finite groups can arise? Every finite group of integer matrices lies inside one of the fourteen lattice groups that the fourteen Bravais lattices are named for, up to a change of basis, so the search has an end.

The fewest orbits any lattice group achieves is two. A cubic cell with body centring splits the seven as three and four, and a cubic cell with face centring as one and six. Primitive cubic gives three orbits. No cell makes the seven one, so on no cell at all are the seven sublattices of a lattice equivalent by isometry. The best-cell number for $P1$ is two against the normaliser’s one, and $P\bar1$ repeats the difference, three against two. Its fifteen subgroups of index two are seven that keep the inversion’s translations in a halved lattice, and eight that drop the inversion and keep a whole lattice or a centred copy; the seven are the ones the normaliser joins and no cell can.

Why it could not have come out otherwise #

That result looks like an accident of a list of fourteen. It is not, and the reason takes two lines.

A group acting transitively on seven things has order divisible by seven, since the orbit has size seven and an orbit’s size divides the group’s order. A finite group of order divisible by seven contains an element of order seven, by Cauchy’s theorem. That element is an integer matrix of size three by three. And the crystallographic restriction says that no three-dimensional integer matrix has order seven, since the smallest dimension where a rotation of order seven can act on a lattice is six.

So the fact that a cell cannot join the seven halves of a lattice is the crystallographic restriction, reappearing as a statement about sublattices. The number seven is the number of ways to halve a lattice in space, and it is also an order the lattice cannot have. The first is $2^3 - 1$. The second comes from the totient function being larger than three at seven. Nothing connects these two facts, and here they collide.

The plane is the comparison, because it shows that the collision is a matter of arithmetic and not of principle.

In the plane a lattice has $2^2 - 1 = 3$ halves. Three is an order the plane allows. A hexagonal cell’s three-fold rotation permutes the three, and so in the plane one cell joins them, as the earlier essay on subgroups found in general. In space the corresponding number is seven, and in four dimensions it is fifteen, which is $3 \times 5$, both of which a four-dimensional lattice can have.

The same seven appeared without being recognised in one cell reaches every Wyckoff set, as a cycle in $P\bar1$’s normaliser joining the half-shift centres. A matrix of order seven modulo two exists, and lifts to an integer matrix of infinite order, and that matrix is what the earlier essay called a trap. For positions it was a false alarm, since half-shifts joined the same points regardless. For subgroups it is not a false alarm. Shifts do not touch translation subgroups, the infinite-order element really is the only thing that can do the job, and that is why this is the case where a cell fails.

It is worth being precise about what this says of real crystals, because it is easy to overstate. A crystal in $P1$ has a metric, and that metric decides which of the seven halves are congruent as point sets. For a general cell they are seven different sublattices with seven different shapes, and only the abstract description makes them one. So in a triclinic superstructure of index two, where atoms order on one of the seven sublattices, the seven possibilities are seven physically different orderings. To call them “the same subgroup” is a statement about bookkeeping, about what a table of subgroups would list, and it is never a symmetry the crystal can have. That a pseudo-cubic triclinic crystal could make three of them nearly degenerate, or four of them, is possible, and the fact here is that it can never make all seven so. No distortion of a crystal, however special, brings every index-two ordering of a lattice to the same energy by symmetry.

Where the fusing happens #

The rest of the table says something about where the affine normaliser earns its keep.

In the orthorhombic system the shifts take 825 subgroups to 627, and permutations of axes take those to 444. That is the largest fall of any system, and it comes from the same freedom that gives six ways to name one group. In the hexagonal and rhombohedral systems the affine normaliser adds nothing to the shifts at this index: 134 and 134, 17 and 17. Tetragonal falls only a little and cubic by two classes. The pattern is a simple one. Fusion happens where the lattice leaves room for it, and index two in a highly symmetric system is a small list already sorted by the group’s own structure. Shifts alone account for about two-thirds of the whole fall, 421 of the 654 classes lost, because they exchange subgroups that differ only in which of the group’s centres they keep. The affine normaliser’s linear parts account for the remaining third, and nearly all of that is in the orthorhombic and monoclinic systems.

So the summary is that normalisers do their real work in the low-symmetry systems, and that is the same place a cell’s power to imitate them is smallest. The triclinic case is where those two trends finally cross.

What the census does not say #

The result is a statement about index two. It gives no information about higher index. How many subgroups of index three showed how quickly the plane’s counts rise, and for a higher prime there are non-normal subgroups in space, with conjugacy to handle and a search to bound. Index three may have failures of its own. In three dimensions a lattice has $3^2 + 3 + 1 = 13$ sublattices of index three, and thirteen is another prime no lattice in space admits, so $P1$ fails again in the same way, which can be claimed from the argument alone without a census. What happens in the other low-symmetry groups at index three is unknown here.

A subgroup here is a subset of a group’s motions on a fixed lattice, and the question has been about equivalence as descriptions. Two subgroups that are equivalent in that sense are of the same affine type; the converse is false. Two subgroups can be the same type and not be joined by any element of the parent’s normaliser, which the plane essay found in some of its groups. The comparison in space would need each subgroup’s type identified, and it is left out here.

Then there is the monoclinic row’s limit. The search tries matrices with entries at most one, as the earlier essay did, and it finds the same count on the affine and best-cell sides. A larger bound could not reduce the best-cell count below an affine count it already matches, but it could conceivably reduce the affine count below it. The earlier essay reported that the action on positions did not depend on the bound, and the same check here is less complete, since it uses the one bound.

What the census had to pass #

The list of subgroups rests on two facts that can both fail visibly. The quotient by the squares must have order exactly a power of two, and the computation stops if it does not. And a normaliser element must act on that quotient as a map, carrying each coset to a coset; one that does not is discarded rather than trusted. The check on the whole is that $P1$ has seven subgroups, the count any reader can verify by hand, and that $P2/m$ has rank five. The claim that fails, and has to, is that one cell joins $P1$’s seven. It was tried on all fourteen lattice groups rather than deduced, and the deduction above came afterwards as the explanation.

Still open: four dimensions, where fifteen is allowed #

In four dimensions a lattice has $2^4 - 1 = 15$ halves, and $15 = 3 \times 5$. Both three and five are orders a four-dimensional lattice can have, so the argument above does not forbid a cell joining them, and the obstacle disappears. A five-fold lattice has one free number has both factors available in one group: the 240 symmetries of $A_4$ have elements of both orders. Whether any finite group of four-by-four integer matrices acts transitively on all fifteen halves is a question about the list of maximal finite groups that the four-dimensional essays have already used. It is a small computation, and its answer would say whether the plane’s success or space’s failure is the general pattern. Five dimensions gives $31$, a prime that is again forbidden, and six gives $63$, composite. If the answer tracks only whether $2^n - 1$ has a large prime factor, that would make the failure here an accident of the number three, one the restriction would then have supplied by coincidence rather than for a reason. It has not been computed here.