The denominator a group actually needs

Seventeen, without a picture ends with the argument that makes its grid search legitimate:

For a finite group P acting on a module, the order of P annihilates that quotient: adding any cohomology class to itself |P| times gives zero. So every extension class is represented by translations whose coordinates are multiples of 1/|P|.

It then draws a conclusion about the seventeen: every translation appearing in any plane group is a half, a third, a quarter or a sixth, and nothing finer occurs anywhere.

The gap between those two sentences is large and the essay does not name it. The bound permits a translation of one over twelve for p6m, whose point group has order twelve, and none occurs. In space it permits one over forty-eight for the cubic groups, and the finest translation in any of the two hundred and thirty is a sixth. So the bound is true, it is quoted everywhere, and it is not the answer.

The exponent, not the order #

The quantity that gives the right answer is the exponent of the cohomology group — the smallest number of times a class must be added to itself to give nothing — rather than the order of the point group.

The exponent divides the order, which is the annihilation bound restated, and it is usually a proper divisor. For a cyclic group acting with a fixed direction the cohomology is cyclic of order n and its exponent is n, so the bound is attained: a four-fold screw does need quarters. For the same rotation acting with no fixed direction the cohomology is trivial and its exponent is one, so no fraction occurs at all and the bound is slack by its whole size.

Those two behaviours sit in one table and the difference between them is a single integer: the rank of the subgroup the rotation fixes, which is nought or one.

Why the denominators in space are 2, 3, 4 and 6 #

With the exponent in hand the fact the plane’s essay records becomes a derivation rather than an observation.

An intrinsic translation in a space group is a translation no origin removes, and the split between intrinsic and locative parts is what separates it from the part that only records where somebody put the origin. It belongs to a cohomology class, so its denominator divides the exponent of that class’s cohomology group.

For a screw axis the relevant class is the cyclic one generated by the rotation about that axis, whose exponent is the rotation’s order. So the denominators a screw can have are the divisors of 2, 3, 4 and 6 — which is eleven screws and no others, derived rather than enumerated.

For a glide plane the relevant operation has order two, so the denominator is two, which is why every glide slides by exactly half of something and why the five glide planes are five and not more.

And for everything else the exponent is one. A cubic point group of order forty-eight contains rotations of orders two, three and four and mirrors; each of those contributes its own exponent, and their least common multiple is twelve. The bound says forty-eight; the truth is twelve; and even twelve is not attained, because no single operation of a cubic group has order twelve and a translation belongs to one operation at a time.

The three, four and six in that list are the rotation orders a lattice permits, and the two is the order of a reflection. Every denominator in the two hundred and thirty is the order of a single operation that fixes something — an axis for a screw, a plane for a glide — and there is nothing else for a denominator to come from.

So the answer to “which denominators occur” is not read off the order of the point group at all. It is read off the orders of the individual operations that fix a direction, and those are the rotation orders a lattice permits.

What a change of origin can and cannot remove #

The denominator question has a second face, and it is the one a reader meets first in practice: which part of a translation is real.

An origin shift by s changes an operation’s translation by (M − I)s, so the coordinates a shift can reach are the ones in the image of M − I — and the coordinates it cannot reach are the ones in the kernel, which is the fixed subgroup again. The same subgroup decides both questions, and that is not a coincidence: a translation along a direction the operation fixes is one the operation cannot generate and one no origin can absorb.

So the exponent of the cohomology and the freedom of the origin are two readings of one integer. Where the fixed subgroup is nothing, every translation is a coboundary and every origin shift is useful; where it is a copy of the integers, the component along it survives every shift and is the intrinsic part.

That is the split between intrinsic and locative stated as a rank rather than as a construction. The construction averages a translation over the operation’s own cycle and keeps the average; the rank says in advance whether the average can be anything but nought.

Ten of thirteen, and where they are #

The plane’s census read through the exponent is a short list. Ten classes have trivial cohomology and exponent one; the rectangular mirror class has ℤ/2 and exponent two; the rectangular 2mm class has (ℤ/2)² and exponent two; the square 4mm class has ℤ/2 and exponent two.

Every non-trivial exponent in the plane is two, and every one of them comes from a reflection. So the finest translation in any of the seventeen is a half, and the bound of twelve that p6m’s order permits describes nothing — p6m’s cohomology is trivial and the class admits one group.

That is the sharpest form of the gap this page is about. The bound at its weakest is twelve to one, in the group with the most symmetry, and the group with the most symmetry is the one with nothing to bound.

The bound that is not about the lattice #

There is a reason the bound is so much weaker than the answer, and it is worth stating because it is a general shape rather than an accident of these groups.

The annihilation bound is a theorem about any finite group acting on any module: the order kills the cohomology, always. It uses nothing about the module — not its rank, not whether the action has fixed vectors, not whether it is faithful. It is a statement about the group alone.

The answer uses the module. Whether the rotation fixes a direction is a fact about how it acts on the lattice, not about how large it is, and it is what the whole answer turns on. So the bound is weak for exactly the reason it is general.

That is also why the bound is the right thing to quote when it is being used as a bound. The plane’s extension count needs it to justify searching a grid of denominator twelve, and for that purpose a weak bound is as good as a sharp one — it only has to be finite. A bound used to terminate a search and a bound used to predict an answer are different objects, and the same inequality serves one well and the other badly.

What the plane does with it #

The plane is the case where the bound is furthest from the answer, and the reason is the cyclic closed form.

In the plane a rotation fills its lattice, so it fixes no direction, so its cohomology is trivial and its exponent is one. Ten of the thirteen classes have a cyclic point group or a rotation filling the lattice, and every one of them carries a single group with no intrinsic translation anywhere in it.

The three exceptions all have a mirror, and a mirror does fix a direction — the line it lies on. So the intrinsic translations of the plane are all glides, all of denominator two, and the bound of twelve that p6m’s order permits is not merely unattained but unattainable: p6m’s cohomology is trivial, so it has no intrinsic translation at all.

Four of the seventeen are non-symmorphic — pg, pmg, pgg and p4g — and every one of them owes its intrinsic translation to a reflection rather than to a rotation. The half-cell in a glide is the only fraction the plane has.

The same slackness, in the search it licenses #

It is worth following the bound into the place it is actually used, because there its weakness has a price and the price is measurable.

The plane’s extension count searches translations on a grid of denominator twelve, which is the least common multiple of the orders the point groups take. The bound guarantees that nothing finer is needed. For ten of the thirteen classes the grid holds a hundred and forty-four points per coordinate pair and the answer is that every one of them is a coboundary — a hundred and forty-four candidates to establish that there is nothing to choose.

With the exponent in hand the same ten classes need no search at all: the fixed subgroup is nothing, so the cohomology is trivial, so the class carries one group. One rank computation replaces the grid.

That is not a criticism of the search. A search that runs on any class, cyclic or not, is worth a great deal more than a formula that runs on some of them, and the plane’s census is right to use one. What the comparison says is where the effort goes in higher dimensions: the classes with a cyclic point group and a rotation filling the lattice are free, and the effort is entirely in the others. Of the seven hundred and ten classes in four dimensions, how many are free by that criterion is a count nobody here has made and one that would say how much of the 1978 computation was avoidable.

Where the exactness stops #

Computed here. For a rotation of each order in the smallest dimension that holds it and in one more, the invariant factors of the second cohomology by the closed form the cyclic case establishes, and from them the exponent. The bound, which is the order. And the ratio of the two, which is the slack.

The closed form is for cyclic groups. Every exponent computed is the exponent of a cyclic class. The claim that the denominators occurring in a general space group are the divisors of the exponents of its cyclic subclasses is a statement about how the cohomology of a group relates to the cohomology of its cyclic subgroups, and it is used above rather than proved — the restriction to a cyclic subgroup is injective on the part of the cohomology that matters here, which is a standard fact and is not derived.

“No denominator finer than a sixth” is a fact about two and three dimensions. In four dimensions the eight-, ten- and twelve-fold rotations exist and a lattice of five dimensions gives them axes, so denominators of eight, ten and twelve occur there. The statement that the finest is a sixth is the crystallographic restriction, not the cohomology.

Nothing here counts groups. The exponent says what denominators can occur; how many classes there are at each is the order of the cohomology, which is a different number. A cohomology of (ℤ/2)² has exponent two and order four, and the two facts answer different questions — one about how fine a translation gets and one about how many groups the class carries.

And the mirrors are not computed. The glide’s denominator of two is read off the order of a reflection; the cohomology of a dihedral class, which is where a plane group’s intrinsic translations actually live, is the computation this page does not make and the one the cyclic case names as what it would take next.

The second refusal is the whole page: the order of the point group offered as the denominator a group needs. It is a natural mistake because the bound is stated in exactly that form in every account of the subject, and because in the one case anybody computes by hand — a screw axis — the two happen to agree.

A bound, a prediction and a search are three uses of one inequality #

The pattern this page is about recurs, and naming it is worth a paragraph because it is not special to cohomology.

A bound that terminates a search only has to be finite. Minkowski’s divisibility on the order of a finite group of integer matrices is the same shape: it allows twenty-four in the plane where the largest group is twelve, and the slackness costs nothing because its job is to let an enumeration stop.

A bound offered as a prediction has to be sharp, and neither of those is. Quoting the annihilation bound as the denominators a group needs, or Minkowski’s bound as the size a point group reaches, is using an inequality for a job it was not proved for.

And a bound read as a description of the answer is worse still, because a reader takes the slack for content: twenty-four suggests a group of order twenty-four might exist, and one over forty-eight suggests a translation of a forty-eighth might. Neither does.

The tell in both cases is the same. A bound that mentions only the group and not what the group acts on cannot know the answer, because the answer depends on the action. Minkowski’s bound is a product over primes and knows nothing about lattices; the annihilation bound is the order and knows nothing about the module; and both are exactly as weak as that description suggests.

Who proved the bound, and what it was for #

The annihilation of group cohomology by the order of the group is one of the first results in the subject, from Eilenberg and Mac Lane’s papers of the 1940s, and it is proved in three lines by averaging a cochain over the group — the same averaging trick the finiteness of the classification rests on, applied to functions on the group instead of to a quadratic form.

Its use in crystallography is Zassenhaus’s, and it is the step that makes his algorithm finite: without a bound on the denominators there is no finite set of candidate cocycles to enumerate, and the classification in any dimension is not a computation. The bound is what makes the four-dimensional count possible, and its slackness costs only time.

That is worth keeping in view against everything above. The bound is quoted as though it described the answer and it does not; but it was never proved in order to describe the answer. It was proved to make a search terminate, it does that, and a sharper bound would have made the search faster and the theorem harder.

Still open: the exponent of a class with two generators #

The exponent is computed here only where the closed form applies, and the interesting classes are the others.

The rectangular 2mm class of the plane has cohomology (ℤ/2)², whose exponent is two — so its denominators are halves, which is what pmm, pmg and pgg show. That one is known from the census rather than from a formula, and it is the only non-cyclic exponent available here.

In space the largest exponent is presumably six, since a six-fold screw is the finest translation in the two hundred and thirty, but “presumably” is doing work in that sentence: the exponent of a class whose point group contains operations of orders two, three and four could in principle be their least common multiple of twelve, and what stops it is that no operation has order twelve. Whether that argument is complete — whether a cohomology class can require a denominator no single operation’s order supplies — is a question about how the restrictions to cyclic subgroups combine, and it is not answered here.

The test that would settle it is small and is not run: compute the exponent of each of the seventy-three arithmetic classes in space and compare the largest with six. Six of the seventy-three are enumerable here, and their intrinsic translations run to quarters and thirds; the classes with the largest point groups are the ones the question is about and the ones the enumeration cannot reach.