The last two crossings fall at eighty-one and twelve thousand

A space group keeps copies of itself inside itself: subgroups that keep every rotation and mirror of the point group and give up some translations, so that the subgroup’s lattice is a sublattice of the original. Each is a superstructure the crystal could adopt without losing any of its point symmetry, and counting them by the index of the sublattice is a way of asking how much room a group leaves for ordering. In space the sublattices a group keeps grow at six speeds made that count for all seventy-three arithmetic classes, to index sixty-four for the lattices and thirty-two for the subgroups on them, and found six rates of growth. The poorer point groups grow faster, the richer ones carry more subgroups on each lattice, and a poorer group always overtakes a richer one in the end.

In the plane the overtaking happened early, every lead gone by index thirty, as the copies a group keeps grow at four speeds found. In space two did not happen inside the census. $P2/m$ grows as the square of the index and $Pmmm$ only as the index times two logarithms, yet at sixty-four $P2/m$ was still behind, 63,076 to 71,524. $P4/m$ grows as the index times a logarithm and $P4/mmm$ as the index, and at sixty-four $P4/m$ had half of $P4/mmm$’s count. The essay ended on both crossings as predictions it could not check, and on two ways to check them: continue the census, whose cost grows as the cube of the index, or find closed forms.

There is a third way, and it is the one that works. The count is multiplicative in the index, so it is fixed by its values at powers of primes, and those are either closed forms or short computations. With it the census reaches index four million, and both crossings can be read off.

A sublattice is a product of its primes #

A sublattice $L'$ of index $n$ in the lattice $\mathbb{Z}^3$ contains $n\mathbb{Z}^3$, so it is determined by its image in the finite group $\mathbb{Z}^3/n\mathbb{Z}^3$. By the Chinese remainder theorem that finite group is a product of pieces, one for each prime power dividing $n$: modulo 12, say, a vector is the same thing as a pair of vectors modulo 4 and modulo 3. A subgroup of a product of groups of coprime orders is a product of subgroups, one in each piece. So a sublattice of index 12 is exactly a pair, a sublattice of index 4 and one of index 3, and a point group keeps it exactly when it keeps both.

The subgroups on a kept lattice are counted by the cohomology $H^1(P, \mathbb{Z}^3/L')$, the same object that counts space groups over their arithmetic classes: each class of cocycle is a consistent way of attaching translations from the quotient to the point group’s operations. Cohomology turns a product of coefficient groups into a product of cohomology groups, so its order multiplies across the pieces too.

Put together, the number $c(n)$ of subgroups with the full point group on lattices of index exactly $n$ satisfies $c(mn) = c(m)\,c(n)$ whenever $m$ and $n$ have no common factor. The cumulative count to any index is then a sum of products, and every factor in it is a value at a prime power. What the census computed by enumerating sublattices one at a time is a product over primes. The multiplicativity is checked directly, against counts made by enumeration at indices 6, 12, 10 and 24 in all five classes.

Odd primes are arithmetic #

All five classes here have point groups whose order is a power of two: two for $P\bar1$, four for $P2/m$, eight for $Pmmm$ and $P4/m$, sixteen for $P4/mmm$. The cohomology of a finite group is killed by the group’s order, and on a quotient of odd order it is also killed by that odd number, so at every odd prime it is trivial. A kept lattice of odd prime-power index carries exactly one subgroup with the full point group, the obvious one, and the count at an odd prime power is just the number of kept lattices.

Those have closed forms, because at an odd prime the lattice splits into the point group’s eigenspaces. The splitting needs to divide by two, which an odd prime allows. For $P\bar1$ the point group is $\pm1$ and keeps every sublattice, so the count at $p^e$ is the number of all sublattices of that index, the sum every superstructure counted once wrote down. For $Pmmm$ the three mirrors separate the three axes, a kept sublattice is a period on each axis, and the count is the number of ways to split $e$ into three, $\binom{e+2}{2}$. For $P2/m$ the two-fold axis separates an axis from a plane on which it acts as $-1$ and keeps everything, so the count is a sum of plane counts $\sigma(p^i)$ against axis periods.

The tetragonal classes are where the arithmetic shows. Across the four-fold axis the plane is the Gaussian integers, and a sublattice of the plane kept by the quarter turn is an ideal, as an ideal across and a prime along found for the lattices alone. At a prime congruent to one modulo four, such as five, the prime splits into two Gaussian primes and there are $i + 1$ ideals of norm $p^i$; at a prime congruent to three it stays prime and there is one ideal in each even degree. $P4/mmm$ also needs its mirrors kept, which leaves only the ideals that complex conjugation fixes, one in each even degree whatever the prime. The figure shows the difference at five: $P4/m$ keeps three lattices of index five where $P4/mmm$ keeps one.

Two is the only computation #

At two the cohomology is not trivial and the count has to be made. For each class and each power $2^e$, every Hermite form of that index is tested, the kept ones are listed, and the cohomology on each is computed. The census before this one computed it by enumerating cocycles, every assignment of a quotient vector to each generating operation, which grows as a power of the index and becomes impossible past a few hundred. Here it is read from two lattice indices instead. A cocycle is a choice of value on each generating operation satisfying one condition for each closed loop in the group’s Cayley graph; the number of cocycles and of coboundaries are both orders of finite quotient groups, and each is the index of a lattice in a space of integer vectors. Because every lattice involved contains $2^e$ times the whole space, the index can be computed by Hermite reduction with every entry reduced modulo $2^e$, so the arithmetic never leaves small integers. The result agrees with the cocycle enumeration on all 1,249 kept lattices to index sixteen in the five classes.

The table this produces is short, and its tetragonal columns are regular. $P4/m$ keeps $2e + 1$ lattices at $2^e$ and carries $24e - 16$ subgroups on them from $e = 2$ on; $P4/mmm$ keeps the same number of lattices and carries $80e - 64$. Both patterns hold at every power from four to 8,192. The orthorhombic count is quadratic in $e$, the monoclinic one roughly doubles at each step, and $P\bar1$ grows by about four each time, which are the growth rates of the six-speeds essay seen at a single prime. The largest cohomology on any one lattice is 512 for $Pmmm$, reached on the doubled cell at index eight and never exceeded, as that essay found.

Why all the multiplicity lives at two #

The split between odd primes and two is the whole economy of the method, and it has a reason worth stating. The cohomology on a kept lattice measures the ways of attaching translations to the point group’s operations that no change of origin removes, and the denominator a group actually needs found that every such translation is a fraction whose denominator divides the order of the point group. These five point groups have orders two to sixteen, so every attached translation is a fraction with a power of two below it. A quotient of odd order has no room for such a fraction: halving is invertible there, any attached translation can be absorbed by moving the origin, and the cohomology is a single class. The same fact, read the other way, is why a finite group of integer matrices survives reduction modulo three and not modulo two: two is the prime these groups are made of.

That is also why, at odd primes, the richest group has the poorest arithmetic, as that essay found for the lattices themselves: $P4/mmm$ keeps a single lattice of index five where $P4/m$ keeps three, because every extra mirror is one more condition a lattice must meet. At two the order reverses. The mirrors that make a lattice hard to keep are exactly the operations that can be given a glide of half a cell on the doubled lattice, and each one multiplies the count of subgroups there. The richer groups lose at every odd prime and win at two, and the whole contest between a poorer and a richer class is how many factors of two the indices up to $N$ contain against how many odd ones. That is the arithmetic content of the crossings: they sit where the odd primes’ gain finally outweighs two’s.

In diffraction terms the doubled cells are the superstructures whose new reflections sit at half-integer positions, the ones the reflections a superlattice adds describes, and a glide on such a cell turns some of those new reflections into systematic absences. So the multiplicity at two is a count of superstructures that differ only in which half-integer reflections they extinguish, a distinction a diffraction pattern can make and a lattice count cannot.

Two routes, one count #

The first test of the product is the census it replaces. Computed as products over prime powers, the cumulative subgroup counts to index 8, 16 and 32 for all five classes agree with the census in every entry: 43,255 for $P\bar1$ at thirty-two, 15,625 for $P2/m$, 25,023 for $Pmmm$, 701 for $P4/m$ and 1,563 for $P4/mmm$. The two computations share nothing but the definition. One lists every sublattice and every cocycle; the other lists only the sublattices of power-of-two index, counts their cohomology from lattice indices, and supplies everything else from arithmetic.

The second test is the crossings the census did see. The product puts $P\bar1$ past $P2/m$ at index 10 and past $Pmmm$ at index 20, which is where the census found them.

Eighty-one #

With the two-adic counts known to $2^9$, every index below 1,024 is exact, and the first open crossing is inside that range. $P2/m$ passes $Pmmm$ at index 81 and never falls behind again. At 64 the ratio is 0.88, as the census reported; at 81 it crosses one; at 128 it is 1.29, at 256 nearly two, and at 1,023 nearly five, with 16,269,643 subgroups for $P2/m$ against 3,319,829 for $Pmmm$.

That eighty-one is a power of three is not a coincidence of the arithmetic, but it is a coincidence of the ordering. The cumulative counts rise in steps, and the steps at indices rich in small primes are the largest; index 81 contributes $c(81)$, which for $P2/m$ is the number of plane sublattices times axis periods at $3^4$, a large term, and for $Pmmm$ only $\binom{6}{2} = 15$. The crossing lands where a large step for the poorer class meets a small one for the richer.

Twelve thousand one hundred and four #

The second crossing is much further out, and the two-adic patterns are what make it reachable. With the tetragonal classes computed to $2^{13}$, every index below 16,384 is exact. $P4/m$ passes $P4/mmm$ at index 12,104, after leading nowhere before it. Past 16,384 the counts at the higher powers of two come from the linear patterns, verified at every power from four to 8,192, and with them $P4/m$ stays ahead at every index checked to 4,194,304.

The crossing is far out because the two classes differ by a logarithm, and a logarithm is slow. The figure below shows why the separation is a logarithm and how fast it grows.

A multiplicative count has a Dirichlet series, a product over primes of the series at each prime, and the growth of the cumulative count is read from the series’ pole at one. For $P4/mmm$ the odd primes contribute one lattice in each even degree of the cross-section times any period along the axis, which multiplies to $\zeta(s)\zeta(2s)$ with the factors at two removed, and the factor at two is the series of $c(2^e)/2^{es}$, which the linear pattern sums to 101 at $s = 1$. A simple pole: the count to $N$ is about $N$ times $\zeta(2) \cdot \tfrac12 \cdot \tfrac34 \cdot 101 = 62.30$, and the computed counts divided by $N$ settle on that number. For $P4/m$ the cross-section contributes every Gaussian ideal, whose series is $\zeta(s) L(s, \chi_4)$, and with the axis this is $\zeta(s)^2 L(s, \chi_4)$: a double pole, a count of about $N$ times a constant times $\log N$. The constant is $L(1, \chi_4) = \pi/4$ times a quarter times the two-adic factor 35, which is 6.87, so the count divided by $N$ climbs by $6.87 \log 2 = 4.76$ each time the index doubles, as the figure shows.

So the contest is a constant, 62.3, against a quantity that starts near fifteen and climbs by 4.8 a doubling. It needs about seven and a half doublings past index sixty-four to catch up. The richer group’s larger multiplicity on its doubled cells is what delays it: the two-adic factor 101 for $P4/mmm$ against 35 for $P4/m$ is three times larger, and it multiplies everything.

What it means for a supercell search #

The practical reading of the six-speeds essay was that a search over supercells for an ordering that keeps a crystal’s whole point group pays for the richer group’s multiplicity at the sizes such searches use. The crossings say how far that reaches. For an orthorhombic crystal with mirrors against a monoclinic one, the richer group carries more candidate subgroups to index eighty, comfortably beyond any search that enumerates supercells. For a tetragonal crystal with mirrors against one without, it carries more to index twelve thousand, which is beyond any search there is. At every size a calculation could use, a tetragonal crystal with mirrors offers more ways to order without losing a symmetry than one without them, and the asymptotic statement that the order eventually reverses has no consequence a calculation would meet.

The arithmetic also says where the subgroups are. At odd indices every kept lattice carries exactly one subgroup, so all the multiplicity, the factor between lattice counts and subgroup counts, comes from the powers of two dividing the index. A supercell of odd index keeps exactly the symmetry-preserving orderings its lattices allow; one of even index has many more, and the excess is concentrated on the doubled cells, where mirrors can become glides by half a parent cell.

What the counts do not settle #

The patterns past 8,192 are observed, not proved. The tetragonal two-adic counts are linear in $e$ at every power from four to 8,192, and the crossing at 12,104 uses nothing beyond that range; but the claim that $P4/m$ stays ahead to four million uses the linear patterns past it. A proof would describe the kept lattices at $2^e$ for the tetragonal classes, $2e + 1$ of them, and their cohomology, in closed form, which the table invites and nothing here does.

Subgroups, not types. As in the census before it, a subgroup here is counted whether or not it has the same space-group type as its parent: on the doubled cells some of $Pmmm$’s 512 have glides where $Pmmm$ has mirrors. The count is of subgroups with the full point group, which is the quantity the rates and crossings are about, not the number of isomorphic subgroups in the tables’ sense.

What the pictures cannot show is a single subgroup. Every figure is a count, and the counts are products of numbers no figure draws. The cohomology on one kept lattice, the 512 ways of moving three mirrors on the doubled cell, is the object being counted, and it is described in words in the census before this one and drawn nowhere.

Who counted what #

The multiplicativity of sublattice counts, through the Dirichlet series $\zeta(s)\zeta(s-1)\zeta(s-2)$ for all sublattices of $\mathbb{Z}^3$, is classical, and Dirichlet series of sublattices kept by a finite group belong to the theory of zeta functions of groups and orders developed by Solomon, Bushnell and Reiner and, for crystallographic groups, by du Sautoy and others. That cohomology vanishes at primes not dividing the group’s order is a standard fact. What the computation adds is the use of both to carry the count of subgroups with the full point group for five classes far past any enumeration, the two-adic tables that are the only part needing computation, and the two crossings the census left open.

What the counts must refuse #

The refused claim is the premise of the census before this one: that the count of subgroups is a sum over indices that has to be enumerated index by index, at a cost that grows with the cube of the index. It is a product over the primes of each index, and only the prime two needs any enumeration.

Still open: the two-adic patterns, proved #

The tetragonal counts at $2^e$ are $24e - 16$ and $80e - 64$ from $e = 2$, and the kept lattices number $2e + 1$ in both classes. A lattice kept by $P4/m$ at $2^e$ is an ideal of the Gaussian integers across the axis, a power of $1 + i$ since two ramifies, glued to a period along the axis, with a centred gluing when the two can be matched; counting those and their cohomology would turn the patterns into theorems and the crossing’s stability past 16,384 into a proof. The same analysis at two for $Pmmm$, where the count is quadratic in $e$, and at three for the hexagonal classes, where three divides the point group’s order and the same split into a closed form and one computed prime applies, would place the crossings among the hexagonal and trigonal classes too. None of it has been done.