The primes a cell can grow by

The same group in a bigger cell asks at which indices a plane group contains a copy of itself — the same group again, on a lattice that many times coarser — and finds the answer in arithmetic. For p4 the indices are the sums of two squares, because a lattice a quarter-turn carries to itself is an ideal of the Gaussian integers and its index is a norm. A screw that contains its own mirror image takes the same question up an axis and ends by naming what neither essay did: which of those copies are maximal.

The word matters because of how the copies are catalogued. A group has infinitely many copies of itself, one family for every admissible index, and the International Tables do not list them all. They list the maximal ones, the copies with no subgroup of the group lying strictly between the copy and the whole, and every other copy is reached by a chain of those. So the maximal copies are what the whole infinite family is built from, and which indices they come at is a question about primes — of the Gaussian integers for p4, of the Eisenstein integers for the hexagonal groups, and of something poorer for the groups with mirrors.

What can lie between a group and its copy #

A copy H of the group G has the same point group as G. So does any subgroup K lying between them, since K contains H’s rotations and mirrors and is contained in G’s. A subgroup with a given point group is pinned down, once one of its operations of each kind is fixed, by its translations, and K’s translations form a lattice L′ lying between H’s lattice and G’s.

Conversely, if a lattice L′ lies between the two and the point group carries it to itself, then H together with the translations of L′ is closed under composition — composing a rotation of H with a translation of L′ gives an operation of H moved by a rotated translation, which is still in L′ — and so it is a subgroup between H and G. So a copy is maximal exactly when no lattice invariant under the point group lies strictly between its lattice and the whole. The criterion is complete, not merely sufficient: every subgroup between H and G has a point group containing H’s and contained in G’s, and those are the same point group, so nothing with a different point group can slip in. That turns a question about infinite groups into one about sublattices, which can be enumerated completely.

The procedure is the one the earlier essay used, taken one step further. For every index up to thirty and every sublattice of that index, keep the sublattices the point group preserves; for each, build the subgroup by choosing cosets for the generators, closing, and handing the result to the plane-group identifier, which must return the parent’s own name. Then, for each copy found, enumerate every invariant sublattice of every proper divisor of the index and test whether it contains the copy’s lattice. A copy with no such lattice above it is maximal.

p4, and the primes of the Gaussian integers #

For p4 the invariant lattices are the ideals of ℤ[i], the multiples of some Gaussian integer a + bi, which is the fact the sublattices that stay square turns on. One ideal contains another exactly when its generator divides the other’s, and an ideal with no ideal strictly between it and the whole ring is a prime ideal. So p4’s maximal copies sit on the prime ideals of the Gaussian integers, and their indices are the norms of the Gaussian primes.

Which Gaussian primes there are is a classical fact about ordinary primes, and the same fact counts the vectors of each length in a square lattice. 2 ramifies: it is $(1 + i)^2$ up to a unit, and the prime 1 + i has norm 2. A prime congruent to 1 modulo 4 splits into two conjugate Gaussian primes, as 5 = (2 + i)(2 − i) and 13 = (3 + 2i)(3 − 2i), each of norm p. A prime congruent to 3 modulo 4 stays prime, and its norm is $p^2$.

The computation finds exactly that. p4’s maximal copies up to fifty are at indices 2, 5, 9, 13, 17, 29, 37, 41 and 49 — two, the primes one more than a multiple of four, and the squares of three and seven. At each split prime there are two maximal copies, one on each of the two conjugate prime ideals, and the picture above shows the pair at five: the lattices generated by 2 + i and by 2 − i, which are mirror images of each other. At 2, at 9 and at 49 there is one.

Everything else on p4’s row is a copy that is not maximal. Up to thirty that is 4, 8, 10, 16, 18, 20, 25 and 26 — every one of them a product of the maximal indices, and every one reached in two or more steps. The reason a given copy is not maximal is visible in the lattice of lattices.

A prime squared is sometimes maximal and sometimes not #

The diagram at the head of this essay sets p4 at index twenty-five beside p4m.

p4 has three copies of itself at index twenty-five, on the ideals generated by 5, by (2 + i)² and by (2 − i)². None of them is maximal: the first lies inside both (2 + i) and (2 − i), and each of the others inside one of them, and all of those are invariant lattices of index five. Twenty-five is on p4’s row because five is, and a copy at twenty-five is reached in two steps.

p4m’s single copy at index twenty-five is maximal. The same lattice (5) is there, and the lattices of index five that sat above it for p4 are not invariant under p4m’s point group, because a mirror exchanges 2 + i with 2 − i and so exchanges the two lattices. A mirror cannot keep either factor of a split prime without the other, so it keeps neither, and the first place p4m can grow by a multiple of five in one step is twenty-five. The two lattices (2 + i)² and (2 − i)² vanish from p4m’s diagram for the same reason: each is the mirror image of the other.

The hexagonal groups repeat the pattern in the Eisenstein integers ℤ[ω], where ω is a cube root of unity and primes behave according to their residue modulo three. The lattices a three-fold turn preserves are the ideals of that ring, and their indices are the values of $a^2 - ab + b^2$, the Löschian numbers the sublattices that are the same shape list. 3 ramifies, with the Eisenstein prime 1 − ω of norm three. A prime congruent to 1 modulo 3 splits, as 7 = (3 + ω)(2 − ω). A prime congruent to 2 modulo 3 stays prime, with norm $p^2$.

So p6’s maximal copies up to fifty are at 3, 4, 7, 13, 19, 25, 31, 37 and 43, with two at each split prime, and p3’s are the same. Forty-nine — seven squared — is not among them, because seven splits, and the diagram shows the three copies at forty-nine sitting under the two ideals of index seven exactly as p4’s copies at twenty-five sat under the ideals of index five. p6m’s maximal copies are at 3, 4, 25 and 49: the ramified prime, and the square of every other prime, for the same reason as p4m.

The ring counts the copies as well as placing them #

Every ideal of ℤ[i] of norm n carries a copy of p4, and every copy sits on one, so the number of lattices carrying a copy at index n is the number of ideals of that norm — and that number has a closed form. Add up, over the divisors d of n, +1 for each d one more than a multiple of four, −1 for each d one less, and nothing for the even ones. At twenty-five the divisors 1, 5 and 25 give three, the three lattices at the foot of the head diagram. At ten, 1 and 5 count and 2 and 10 do not, giving two, the ideals (1 + i)(2 + i) and (1 + i)(2 − i). At nine, 1 and 9 count +1 and 3 counts −1, giving one. The same sum with residues modulo three counts p3’s and p6’s copies, and built one by one up to index thirty, the count found agrees with the sum at every index for all three groups.

The sum also says why a split prime gives two maximal copies and an inert one gives none at p and one at $p^2$. At a prime the divisors are 1 and p, so the sum is two when p is +1 and zero when it is −1. At $p^2$ it is three for a split prime and one for an inert prime, and that single ideal is (p) itself, which is prime — so its copy is maximal. p4m and p6m carry exactly one copy at each index on their rows, because a mirror keeps only the ideals that are their own mirror images: those generated by an ordinary integer, by the ramified prime 1 + i or 1 − ω, or by a product of those. Eighteen is (3)(1 + i) and twelve is (2)(1 − ω); neither group has anything at ten or at seven.

All seventeen #

The groups with no rotation beyond a half-turn have the plainest rows, and the rows explain themselves. p1 and p2 have maximal copies at exactly the primes, and at each prime p there are p + 1 of them: every sublattice of prime index is maximal, since a lattice of prime index has nothing between it and the whole, and a half-turn preserves every lattice. There are p + 1 sublattices of each prime index, which is the whole count.

pm and pmm keep two at each prime, the two rectangular lattices that stretch the cell along the mirror or across it. pg and pmg keep two at each odd prime and only one at two. The lost one is the doubling along the glide line: two glides make the translation along the line that the glide’s half-step squares to, and a cell doubled in that direction would have to leave that translation out. The centred lattice of index two fails for the same reason, and the doubling across the glide line is all that remains.

cm and cmm lose the prime two. Their copies skip every index two more than a multiple of four, and their maximal copies are at the odd primes only, two at each. Both groups have a copy at four, on the lattice doubled in both directions, and it is never maximal: a lattice of index two lies above it that the mirror preserves. That lattice carries a subgroup too — every invariant lattice does — but not a copy of cm, which has none at two, so the subgroup in between is a different group and still stops the copy at four from being maximal. pgg loses every even index, and its maximal copies are the odd primes, two at each.

p4m keeps 2, 9, 25 and 49 up to fifty, since the ideal 1 + i is carried to itself by a mirror and the copy on it is p4m again. p4g has copies at 9, 25 and 49 and nowhere else up to fifty, and all three are maximal: a lattice between would have index three, five or seven and would have to be preserved by both a quarter-turn and a mirror, and no lattice of those indices is.

p3m1 and p31m have copies at 4, 9, 16, 25, 36 and 49, and are maximal at 4, 25 and 49. Neither has a copy at three. The ideal 1 − ω is invariant under both groups’ point groups, but the lattice it generates is turned through thirty degrees against the original, and a turn of thirty degrees exchanges the two ways three mirrors can sit on a hexagonal lattice. So the subgroup on it comes back as the other member of the pair — p31m inside p3m1 and p3m1 inside p31m, the klassengleiche relation of index three that the two ways down finds. That subgroup still sits above the copies at nine, which is why nine is on both rows and is not maximal on either. p6m, whose mirrors run both ways, is unchanged by the turn and keeps three.

Nine is one of the steps p4’s row is built from #

It is tempting to say each row of copies is built from the primes that appear on it, and for p4 to name two and the primes congruent to one modulo four. That leaves out nine, which is on p4’s row — p4 contains a copy of itself on the lattice of multiples of 3, at index nine — and is a product of neither. The steps the row is built from are not the prime numbers that happen to be indices; they are the norms of the primes of the ring, which include the squares of the primes that stay prime. Every copy is a chain of maximal copies, and a chain of lattice inclusions multiplies indices, so every index on the row is a product of maximal indices — and for p4 the maximal indices are 2, the split primes, and the inert primes squared.

The same correction applies to the hexagonal rows, where four is one of the steps and two is not an index at all, and to p4g, whose row up to fifty consists of nothing but steps.

What the International Tables are listing #

The Tables give, for every space group, its maximal subgroups of the kind that keep the point group — the klassengleiche ones — and among those the isomorphic ones, usually as the first few members of infinite series indexed by primes. The descent with no shortcut explains why maximal subgroups are the natural thing to tabulate: every chain of subgroups from a group down to any subgroup passes through them, one step at a time.

What the computation above adds is the reason the series look the way they do. An entry of index p for every prime p ≡ 1 mod 4, two subgroups each, is the statement that such a prime splits in ℤ[i]; an entry of index $p^2$ for the other odd primes is the statement that they stay prime; and a series that appears for a group without mirrors and is missing for the same lattice with mirrors is the statement that complex conjugation exchanges the two factors of a split prime. The Tables’ lists are the primes of a ring, written as subgroups.

The convention, and what was not computed #

The indices run to thirty for every group, and to fifty for p3, p4, p4m, p4g, p3m1, p31m, p6 and p6m. The arithmetic predicts every row at every index; the computation confirms p4’s, p3’s and p6’s predictions exactly where it reaches.

The numbers count lattices, not subgroups. One lattice can carry more than one copy. pm’s lattice doubled across its mirrors carries a copy whose mirrors lie on the original mirror lines at even steps and another whose mirrors lie on the odd ones — mirrors that are not conjugate in the group, the distinction two mirrors a coset cannot tell apart draws. Maximality depends only on the lattice, by the criterion above, so every copy on a large dot is maximal; the count beside the dot is of lattices.

The plane only. In space the lattices invariant under a point group are modules over larger rings, the rotations along different axes constrain one lattice at once, and the screw axes add the congruences the screw essay worked out. Nothing here computes that.

The checks, and what they refuse #

The second refusal carries the result’s structure in it: the sublattice with basis vectors (25, 0) and (0, 1) has the right index and is not carried to itself by a quarter-turn, so it cannot carry p4, while the three that are carried to themselves each carry a copy, and none of the three is maximal.

Who worked out the series #

Gauss classified the primes of ℤ[i] in the course of his work on biquadratic reciprocity, and Eisenstein did the same for ℤ[ω] and cubic reciprocity; which primes split, ramify or stay prime in each ring was settled by the middle of the nineteenth century. The maximal isomorphic subgroups of the space groups were worked out more than a century later, by Yves Billiet in the 1970s and with Erwin Bertaut at the end of that decade, as part of tabulating every maximal subgroup of every space group, and their infinite series appear in the volume of the International Tables on subgroups. The two lists are one list, and the diagrams above are where they meet.

Where this goes: the rings in space #

The next step is to ask the same question of the space groups with a single principal axis of order three, four or six, where the lattice across the axis is a module over ℤ[i] or ℤ[ω] and the lattice along it is a module over ℤ. A copy can then grow across the axis by a prime ideal, along it by a prime, or both at once, and the screw axes forbid some of the combinations. Whether the maximal copies of P4₁ are exactly the products of p4’s prime ideals with the primes the screw congruence allows, or whether the two constraints interact, is a question the plane’s answer suggests and does not decide.