In space the sublattices a group keeps grow at six speeds

The copies a group keeps of itself grow at four speeds counted, for each of the seventeen plane groups, the sublattices its operations carry to themselves and the subgroups each such sublattice can carry, to index ninety-six. The sublattices fell into four tiers of growth, $N^2$, $N \log N$, $N$ and $\sqrt N$, set by the point group’s arithmetic. The subgroups per lattice never exceeded sixteen, so they multiplied the lattice count by a bounded factor and changed no rate. A richer group’s early lead, which the bounded factor bought, always ended by index thirty.

That essay ended by guessing what space would do. A monoclinic group keeps the sublattices a two-fold axis keeps, an orthorhombic one those three perpendicular axes keep, a tetragonal or hexagonal one an ideal across the axis times anything along it, a cubic one almost nothing. Each should grow at a definite rate, and the crossovers should again fall at small indices. Which rates occur, whether two of the seven crystal systems share one, and where the crossovers fall was the same enumeration over the sublattices of three-dimensional space.

This essay runs it for all seventy-three arithmetic classes. The guess about rates was right, and it gives six tiers. The guess about systems was half right: four systems keep one rate each and three are cut in two. The guess about crossovers was wrong. In space the richer groups carry so many subgroups on each lattice that some of their leads outlast the whole census.

What is counted #

Every sublattice of $\mathbb{Z}^3$ of index $n$ has exactly one Hermite form, a basis of three rows in triangular shape, with the diagonal entries multiplying to $n$ and each entry above the diagonal reduced below the diagonal entry beneath it. Enumerating those forms gives every sublattice once. There are $\sum d f^2$ of them at index $n$, summed over factorisations $n = a d f$, and the enumeration meets that count at every index to sixty-four: 176,874 sublattices in all.

For each arithmetic class, the point group is taken as integer matrices on its own lattice’s basis, the seventy-three built in two hundred and nineteen over seventy-three without a table. A sublattice is kept when every operation of the point group carries each of its basis rows back into it. $L(N)$ is the number kept to index $N$. On each kept lattice, the subgroups of the symmorphic space group that keep the whole point group and have exactly that lattice of translations are counted up to conjugacy. They are the first cohomology of the point group with coefficients in the finite quotient of the two lattices, the count a lattice is not a subgroup set up. That count is made to index thirty-two for every class and to sixty-four for six of them.

One correction of wording travels with it. The plane essay called these subgroups copies of the group. Most are. But the count also includes subgroups in which a mirror of the parent has become a glide of the subgroup, because the translation that accompanies the mirror is half of a new, longer period. Such a subgroup has the parent’s point group and a lattice of the parent’s shape and is not of the parent’s type. So the count is an upper bound on the copies of one type, a bounded multiple of them, and every statement about growth rates holds for both.

Six speeds #

The figure at the head of this essay shows one class from each tier, and the tiers are as distinct as the plane’s.

$N^3$. $P1$ and $P\bar1$ keep every sublattice, 176,874 to index sixty-four, with a slope of 2.97 between thirty-two and sixty-four on the logarithmic plot. The inversion is minus the identity and keeps every lattice, so it changes nothing.

$N^2$. The six monoclinic classes keep the sublattices a single two-fold axis keeps: 9,562 on the primitive monoclinic lattice and 5,482 on the centred one, slopes 2.01 and 2.02.

$N \log^2 N$. The thirteen orthorhombic classes keep those three perpendicular two-fold axes keep: 2,106 on the primitive lattice, between 958 and 1,202 on the centred ones, slopes from 1.44 to 1.48, which is the local slope of $N \log^2 N$ over this range.

$N \log N$. Thirteen classes with an axis of order three, four or six and nothing that turns the axis over keep the sublattices that are an ideal across the axis and a period along it: 330 for $P4$, $P\bar4$ and $P4/m$, 285 for $P3$ and $P\bar3$, 187 for the hexagonal classes, 214 and 174 on the body-centred and rhombohedral lattices. Slopes 1.22 to 1.31.

$N$. Twenty-four classes with such an axis and a two-fold axis or mirror through it keep only scaled and turned copies of the cross-section and a period: 210 for $P4/mmm$, 125 for $P6/mmm$, 102 for $R\bar3m$. Slopes 1.06 to 1.12.

$\sqrt[3]{N}$. The fifteen cubic classes keep the lattice scaled by $k$ and two centred versions of it, at indices $k^3$, $2k^3$ and $4k^3$: nine for the primitive and body-centred lattices to sixty-four, seven for the face-centred one. That is the collapse the richest group has the poorest arithmetic found, now placed among the other five.

Why each tier has its exponent #

As in the plane, each exponent counts free choices. A sublattice of space with no constraint has three rows, each free in direction and length, and how many sublattices there are of index $n$ grows as $n^2$, so the total to $N$ is about $N^3$.

A single two-fold axis splits space into a line and a plane perpendicular to it, and a kept sublattice is nearly a plane sublattice times a period along the line, glued at worst by halves. The plane sublattice is still free, and the plane’s own count grows as its index; summed against the divisors along the line, the total grows as $N^2$. Three perpendicular axes leave only a period along each, three divisors of the index, and the number of ways to write $n$ as a product of three factors grows on average as the square of its logarithm.

An axis of order three or more pins the cross-section to an ideal of the Gaussian or Eisenstein integers, as the sublattices that are the same shape found in the plane, and the ideals of norm up to $x$ grow as $x$; times a period along the axis this is an ideal across and a prime along, which grows as $N \log N$. Something that turns the axis over, a two-fold rotation or a mirror containing it, pins the angle within the ideal to the parent’s own or one turned direction, leaving a scale and a period, and $\sum_c \sqrt{N/c}$ grows linearly. The cube pins everything but the scale.

The glue never changes an exponent. Between the split sublattices and the rest lie centred ones, such as the body-centred sublattice of a primitive tetragonal lattice or the rhombohedral sublattices of a hexagonal one. They add at most a bounded number of sublattices for each split one. That is why $P3$ keeps 285 sublattices to sixty-four and $P6$ only 187: a three-fold axis alone admits the rhombohedral sublattices of the hexagonal lattice, and the six-fold’s half-turn refuses them, but the two counts grow at the same rate.

Three systems on two rates #

The plane essay’s question, whether two crystal systems share a rate, has a sharper answer than it expected. Four systems keep one rate each, and no other system shares it: triclinic $N^3$, monoclinic $N^2$, orthorhombic $N \log^2 N$, cubic $\sqrt[3]{N}$. The other three, tetragonal, hexagonal and the rhombohedral classes, share two rates between them, and every one of the three is cut in two.

The cut is by a simple property of the point group: whether it is abelian. A point group with one axis of order three or more and nothing that turns that axis over consists of rotations about the axis, the inversion and the mirror across it, and all of these commute. Add a two-fold axis or a mirror containing the principal axis and the group stops commuting, and that is exactly the operation that pins the angle. So $P4/m$ grows as $N \log N$ and $P4/mmm$ as $N$, while their lattice, their system and their four-fold axis are the same.

The measured slopes say the same thing without any classification. Every class’s slope between thirty-two and sixty-four lies inside its tier’s band, and the bands do not overlap: 2.97, 2.01 to 2.02, 1.44 to 1.48, 1.22 to 1.31, 1.06 to 1.12, and below 0.4. Two closed forms are met index by index as well: the triclinic count against $\sum d f^2$, and the primitive cubic count against one sublattice at each cube, twice a cube and four times a cube.

The subgroups on each lattice #

The multiplicities grow with the point group, as in the plane, and much faster. $P1$ has one subgroup on every lattice. $P\bar1$ averages 1.9 and has at most eight. $P2/m$ averages 6.6 with up to sixty-four, and $Pmmm$ averages thirty-two, with 512 on the doubled cell $2\mathbb{Z}^3$, where each of its three mirrors can be kept, moved by half a parent cell, or given a glide of half a parent cell in either direction within it, eight choices for each mirror, and every one of the $8^3$ combinations closes into a group. $P4/mmm$ averages sixteen with up to sixty-four. The cubic groups, with only nine lattices, have up to eight on one.

The ceiling has the plane’s reason. The cohomology that counts them is killed by the order of the point group and determined by its values on at most three of the group’s generating operations, each a vector of the finite quotient, so its size is bounded by the point group alone, whatever the index. The bound is reached early: $Pmmm$’s 512 appears at index eight and is never exceeded to sixty-four. The subgroup count is the lattice count multiplied by a factor between one and 512, and so it grows at the lattice count’s rate in every class.

What a crystal can do without losing an operation #

The tiers are also a statement about which superstructures a crystal can form while keeping every operation of its point group, which is the situation the reflections a superlattice adds describes from the diffraction side: new reflections appear, and no reflection that the point group relates to another becomes inequivalent.

A cubic crystal that orders and stays cubic in its point group can do so only on a cell multiplied by a cube, or on the centred versions of such a cell at twice or four times a cube. To index sixty-four that is nine possible lattices in all, and the first three, at indices two, four and eight, are the familiar ones: a primitive cubic lattice thinned to a face-centred or body-centred sublattice, or doubled along every edge. Every other superstructure of a cubic crystal lowers its point group. A tetragonal crystal with mirrors through its axis has 210 such lattices to sixty-four, a tetragonal crystal without them 330, an orthorhombic one 2,106 and a monoclinic one nearly ten thousand. The richer the point group, the narrower the menu of superstructures that keep it, and in space the menu narrows by six orders of growth rather than four.

The split in the middle of that list is worth reading physically. A tetragonal crystal whose point group has mirrors through the four-fold axis can keep them on a supercell only if the new cell’s cross-section is the old one scaled, or scaled and turned through forty-five degrees, the $\sqrt2 \times \sqrt2$ cell that tetragonal and perovskite-derived structures so often adopt. A crystal of class $4/m$ has no such mirrors to keep, and it may adopt any cross-section that is an ideal of the Gaussian integers: a cell turned through the angle whose tangent is a half, at index five in the cross-section, keeps all of $4/m$ and none of the mirrors of $4/mmm$. That is the same distinction the sublattices that are the same shape drew in the plane, between the similar sublattices of a square lattice and the ones that also keep its mirrors, and in space it is what separates two tiers.

The three-fold axis shows the glue at work. A hexagonal lattice has, besides the sublattices that split into a cross-section and a period, the rhombohedral sublattices of index three that stack their layers in a spiral. A three-fold axis keeps them and a six-fold axis does not, because the half-turn contained in a six-fold sends the obverse stacking to the reverse one. So $P3$ keeps 285 sublattices to index sixty-four and $P6$ keeps 187, and the subgroups follow: $P3$ carries 725 and $P\bar3$, whose inversion constrains the shifts further, only 400. Both pairs grow at the same rate. The glue is a factor and never a rate, and this is the clearest instance of it in the census.

The richer groups lead for longer #

A bounded factor cannot change a rate, but it can delay a crossing, and in space the factors are thirty times the plane’s. The plane’s three contests were settled at indices ten, eleven and twenty-seven. In space $P\bar1$ passes $P2/m$ between index nine and sixteen and passes $Pmmm$ at index twenty, after which it is never caught: by sixty-four it has 334,570 subgroups to $Pmmm$’s 71,524.

Two contests are not settled at all within the census. $P2/m$ grows as $N^2$ and $Pmmm$ as $N \log^2 N$, so $P2/m$ must pass it eventually; but at index sixty-four it has 63,076 against 71,524. The ratio has risen from sixty-two per cent at thirty-two to seventy-six at forty-eight and eighty-eight at sixty-four, so the crossing is near, but it is not inside the census. $P4/m$ against $P4/mmm$ is further off: 1,676 against 3,326 at sixty-four, a ratio that has moved only from forty-five to fifty per cent since thirty-two, because the logarithm that separates their rates grows slowly and $P4/mmm$’s multiplicity is three times $P4/m$’s. The plane’s conclusion that every lead ends by index thirty does not survive into space. The leads all end, by the argument about rates, but some of them end beyond the indices any supercell search uses.

That matters for the practical reading the plane essay gave. A search over supercells for an ordered arrangement that keeps a crystal’s whole point group works within a few tens of cells. In the plane the richer groups were expensive only at the smallest sizes. In space an orthorhombic or tetragonal group with mirrors carries more candidate subgroups than its mirror-free relative across the whole of that range, and the budget of such a search is set by the multiplicity rather than by the lattice count.

Counted twice, refused twice #

The first refused claim is the one a reader carries over from the systems: that each crystal system has one rate of growth. Tetragonal, hexagonal and rhombohedral classes each grow at two. The second is that a cubic class can keep a sublattice that is not a scaled primitive, body-centred or face-centred cell of its own lattice; every cubic count to sixty-four sits at $k^3$, $2k^3$ or $4k^3$, which is why only three stay cubic seen as a growth rate.

What the figures cannot show is the rate itself. A slope measured between thirty-two and sixty-four is a slope, not an exponent, and the distinction between $N \log^2 N$ and $N^{1.5}$ is invisible on any range a census can reach. The tiers are identified by the counting argument above, which says which choices a kept sublattice has; the census confirms that the measured slopes fall where those arguments put them and that no class falls between. Nor do the subgroup counts distinguish types: as stated above, they include subgroups with glides where the parent has mirrors.

Who counted what #

Counting the sublattices of $\mathbb{Z}^3$ by index, through the Dirichlet series $\zeta(s)\zeta(s-1)\zeta(s-2)$, is classical; counting the sublattices a finite group of integer matrices keeps is a question about its integral representations, studied for crystallographic groups by Zassenhaus and later by Plesken among others. The isomorphic subgroups of each space group are listed in the International Tables by series with their index formulas, following work by Bertaut, Billiet and Müller. What the census adds is the comparison across all seventy-three classes at the same indices, and the measurement of how long a richer group’s multiplicity keeps it ahead.

Still open: where the last two crossings fall #

$P2/m$ should pass $Pmmm$ somewhere beyond index sixty-four and $P4/m$ should pass $P4/mmm$ much further out. Both follow from the rates, and neither has been seen. Locating them needs either the census continued, which grows as the cube of the index for the classes that keep the most lattices, or closed forms for the subgroup counts, sums over the kept lattices of cohomology orders. For $Pmmm$ such a form would be a sum over factorisations of the index into three periods, weighted by how many of the three are even; the same structure as every superstructure counted once but with a different weight. Writing it down and checking it against the census would place both crossings exactly. It has not been done here.