A lattice is not a subgroup

The primes a cell can grow by ends with a caveat it does not develop:

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.

That is a real gap between what has been computed here and what the International Tables print. The Tables list subgroups; every count so far has been a count of lattices. The criterion for maximality is about lattices and is correct as stated, but the number beside an entry in the Tables is not the number of lattices, and the difference is not a constant.

The smallest case, and why nothing repairs it #

pm has a mirror every half a cell — the mirror lines of a plane group with a mirror are twice as dense as the lattice — and a copy of pm on a lattice doubled across them keeps every other one. There are two sets of every other one, and both are genuine copies of pm with that lattice and that point group.

The parent could carry one onto the other by a translation, and the translation that would is a translation by half the new cell, which is a translation of the old lattice. It is in the parent. But it is not in either copy, and conjugating one copy by it gives the other copy rather than itself — which is exactly what it means for the two to be different subgroups rather than one subgroup described twice.

So the two copies are not conjugate in either of them and are conjugate in the parent, and the Tables’ convention is to count them as two subgroups. The same distinction has been met here from the other side: two mirrors a coset cannot tell apart is the observation that a coset of a plane group holds every mirror of a family, and telling them apart needs more than the group modulo its translations.

What a copy on a fixed lattice is #

The general statement takes four lines and is the same composition arithmetic the groups are built with.

The parent’s operations are pairs (M, t) and a copy with the same point group and the sublattice’s translations has operations (M, t + v) for some v in the lattice, read modulo the sublattice — one v per element of the point group. Composing two of them and asking for closure gives v(gh) = M(g) v(h) + v(g), the cocycle condition, with the coefficients living in the lattice modulo the sublattice rather than in the lattice itself. And moving the origin by s changes v(g) by (M(g) − I)s, which is a coboundary, and produces a copy conjugate to the first.

So the copies on one lattice, up to conjugacy in the parent, are the cocycles modulo the coboundaries — the first cohomology of the point group with coefficients in the finite quotient.

That is a different object from the one the seventeen are counted by. There the coefficients are the lattice itself, the cocycles are the ways of attaching translations to a point group, and the group is a second cohomology. Here the coefficients are a finite quotient and the group is a first. Two cohomologies, two questions, and only the vocabulary is shared.

The two counts, for all seventeen #

With the computation stated, the gap can be measured rather than described.

p1’s two counts agree everywhere, and they have to: its point group is trivial, so there is one element to assign a shift to and the cocycle condition forces that shift to be zero. A group with no symmetry carries exactly one copy of itself on each of its lattices, which is the case the lattice count was silently assuming.

Every other group carries more somewhere. pm at index two carries two copies on each of its three lattices. p2 at index four carries two on most of its lattices and four on one. p3 at index three carries three on its single lattice, because the cocycles there number three and the coboundaries number one — the rotation acts trivially on a quotient of order three, so nothing can be removed by moving the origin.

Over the indices two to six the totals are: p1, thirty-two lattices and thirty-two copies; p2, thirty-two and fifty-six; pm and pg, eighteen and thirty-four; cm, ten and fifteen; pmm, pmg and pgg, eighteen and seventy-two; cmm, ten and twenty-six; p4, four and six; p4m and p4g, two and eight; p3, two and four; p6 and p6m, two and two.

The correction is not a constant and it is not monotone in anything obvious. p4m has two lattices and eight copies, a factor of four; p6m has two and two, a factor of one. The factor is the order of a cohomology group, and cohomology groups of different point groups acting on different quotients have no reason to be comparable.

p3 at index three, in full #

One case is small enough to do completely, and doing it makes the cohomology concrete.

p3 has a single invariant sublattice at index three: the hexagonal lattice scaled by the ramified Eisenstein prime, turned through thirty degrees. Its point group is the three-fold rotation, with three elements, and the lattice modulo the sublattice has three elements. So an assignment of shifts is a triple, there are twenty-seven of them, and the cocycle condition cuts that to three.

The rotation acts trivially on the quotient — every element of the lattice is carried to itself modulo the sublattice, which is the fact that makes the sublattice invariant in the first place — so the condition reduces to v(gh) = v(g) + v(h), and the cocycles are the homomorphisms from a cyclic group of order three to a group of order three. There are three.

The coboundaries are the shifts (M − I)s, and with a trivial action M − I is zero on the quotient, so there is exactly one coboundary and it is trivial. Three cocycles, one coboundary, three copies.

Geometrically the three copies are the three positions a set of three-fold centres can take relative to the parent’s. The parent’s centres sit at three positions in its cell; a copy on the coarser lattice keeps one of them and drops the other two, and which one it keeps is the choice. No translation of the parent carries one choice onto another, because the translations that would are the ones the coarser lattice has lost.

What a subgroup count is conjugate up to #

The Tables’ counts and a computation’s counts can differ for a reason that has nothing to do with lattices, and it is worth separating from the multiplicity above.

Two copies that no translation of the parent carries onto one another may still be carried onto one another by something outside the parent — a motion normalising the parent without belonging to it. Three of them, and they are equivalent is that distinction made on the subgroup tables directly: three subgroups of one type may be three copies the group itself cannot tell apart and the normaliser can, and the tables mark which.

Everything on this page counts up to conjugacy in the parent, which is the finer count. The coarser one — up to conjugacy in the normaliser — is what a reader sometimes wants and is a quotient of it by an action this computation does not perform. So the number here is an upper bound on the count a reader may have in mind, and how much of an upper bound depends on how much freedom the normaliser has in each case.

That is a second correction on top of the first, in the opposite direction, and the two are independent: the multiplicity raises the lattice count and the normaliser lowers the copy count.

What this does and does not change #

The three counts this one corrects are not wrong, and it is worth saying precisely which of their statements survive.

Maximality is untouched. The criterion is that no invariant lattice lies strictly between, and it is a statement about lattices with a complete argument behind it. Every index called maximal here is maximal, and every index not called maximal, is not. The multiplicity changes how many subgroups sit at a maximal index, not which indices those are.

The Euler product is untouched as a statement about lattices and becomes a different statement about subgroups. The number of lattices is multiplicative, checked at every index to thirty; whether the number of copies is multiplicative is a question about how the cohomology behaves under a product of coprime quotients, and it is not asked here.

And the plane’s rows of indices are untouched, since an index carries a copy exactly when it carries a lattice that carries one, and the multiplicity only says how many.

What does change is the reading of a number in the Tables. An entry saying a group has two maximal isomorphic subgroups at index five is, for p4, two lattices with one copy each; an entry saying a group has some number at index four may be one lattice with several. The two readings coincide only for p1, and a reader translating between the Tables and a lattice count needs the multiplicity to do it.

The groups that carry nothing extra #

Three of the seventeen carry exactly as many copies as lattices over the range measured, and they are worth naming because the reason differs in each case.

p1 carries one per lattice because its point group is trivial, so there is nothing to twist. That is the structural case and it is the only one that holds at every index.

p6 and p6m carry one per lattice over indices two to six because their invariant lattices in that range are few and the quotients they leave are ones the point group acts on with no invariants to spare — a six-fold rotation moves almost everything, so almost every twist is a coboundary and is removed by a change of origin. That is a coincidence of the range rather than a theorem, and a computation reaching further would be the way to find out whether it lasts.

And p31m carries one per lattice where p3m1 carries two, at the same indices, on lattices of the same shape. The two groups differ only in where their mirrors sit relative to the three-fold centres — the pair that is the standing example that position is not a detail — and the multiplicity is another place the difference shows. A count that could not tell them apart would be a count that had thrown away the thing the pair exists to illustrate.

Where the exactness stops #

Computed here. For each of the seventeen and each index from two to six: every sublattice of that index the point group preserves; for each, the cocycles with coefficients in the lattice modulo the sublattice, found from their values on a generating set and checked on every pair of elements; the coboundaries, enumerated from the origin shifts; and the quotient. Four checks that can fail, including that every quotient is a whole number and that p1 carries exactly one copy on each lattice.

Six is where it stops, and the reason is the enumeration. The cocycles are found by trying every assignment on a generating set, so the work grows as the index raised to the number of generators; at index six with two generators that is thirty-six candidates per lattice, and it grows quickly.

A copy here has the parent’s point group and the sublattice’s translations. That is the right definition for an isomorphic subgroup and it does not check that the result is the parent’s own type — in the plane it always is, for the reason the rows themselves give, and along an axis in space it is not.

The generating set is found rather than given. A cocycle is determined by its values on generators, and the enumeration uses a greedy search for the smallest generating set of the point group’s matrices. Every candidate is then checked on every pair of elements rather than only on the generators, so a wrong generating set would produce too few cocycles and not too many — and p1’s answer of one copy per lattice, which is forced, is the control that would catch it.

And conjugacy is taken in the parent. Two copies are counted as one when a translation of the parent carries one onto the other. The Tables’ convention is conjugacy in the parent too, but their count of “how many subgroups of this type at this index” sometimes means conjugacy classes and sometimes means subgroups; the two differ by the index of a normaliser, and that difference is not computed here.

The refusal is pm’s doubled lattice offered as carrying one copy, which is the case this whole page exists for. It is a good refusal because the wrong answer is the one a reader arrives with: the lattice is one lattice, the group is one group, and nothing about the picture suggests there is a choice left to make. A test that only ever confirmed the lattice count would pass for a computation that had not noticed the distinction at all.

Where the copies sit in the descent #

A subgroup relation is not a chain but a lattice of routes, and the multiplicity is what makes the routes branch.

Going down from a group to a copy of index four can pass through two different copies of index two, and if each index-two step has two copies on its lattice then the number of routes multiplies rather than adding. Every way down and no way round shows the corresponding structure for lattices alone — an infinite tree in which every vertex has p + 1 neighbours and no path comes back — and the copies sitting on each lattice thicken it.

So a diagram of the isomorphic subgroups of a plane group has more vertices than a diagram of its invariant lattices, by exactly the factors on this page, and the two diagrams have the same shape only for p1. That is worth knowing before reading either: a subgroup diagram drawn from the lattice poset is the right picture of which indices occur and the wrong picture of how many objects there are.

Why the caveat sat there through two counts #

It is worth being honest about why a stated caveat went undeveloped, because the reason is structural rather than an oversight.

The maximality criterion is about lattices, and it is complete: the argument that a subgroup between two isomorphic subgroups is determined by its lattice needs nothing about which subgroup sits where. So a question asking which indices are maximal can be answered entirely in the lattice poset, and the multiplicity never arises. It arises the moment the question changes from which indices to how many subgroups, which is the question a table answers and a criterion does not.

That is a general shape and it is worth carrying. A classification and a census are different pieces of work, the first is usually the harder one, and the second is where the discrepancies with a printed table come from. The classification was done three counts ago and the census is here.

Still open: whether the multiplicity is multiplicative too #

The lattice count is multiplicative and the row has an Euler product. Whether the copy count does is a question about the first cohomology of a point group with coefficients in a quotient, and it has a plausible answer that is not checked.

The quotient factors. If the index is a product of coprime numbers, the lattice modulo the sublattice splits as a direct sum of two pieces of coprime order, and cohomology takes direct sums to direct sums — so the cocycles and the coboundaries should each factor, and the multiplicity with them. That would make the subgroup count multiplicative as well and give the Tables’ series an Euler product of their own.

What could break it is the lattice count and the multiplicity not factoring compatibly: a lattice of index fifteen corresponds to a pair of lattices of indices three and five, and the copies on it should correspond to pairs of copies on those, which needs the correspondence between lattices and pairs to be the one the cohomology sees. That is a check a computation to index fifteen would make and this one does not reach.

The other direction worth following is the axial case. There a lattice carries the parent’s type or the enantiomorph’s, and the multiplicity question becomes two questions — how many copies of each type sit on one lattice — which is a first cohomology with a twist in it that the plane has no room for.