The screw a dimension does not have

Seventeen, without a picture counts the plane groups as extensions and ends by saying what the machine is for:

The extension count runs in any dimension, and the interesting question is what it does when the answer is not already known: the same computation applied where it is not a check but a prediction.

It also says why the plane was worth doing first: “A method whose only test is a computation nobody can check is not a method, and the seventeen are the one case where every step can be verified against a list arrived at four other ways.”

So a prediction has to be earned. What earns it here is a closed form: for a point group that is cyclic — generated by one rotation — the extension count is not a search over a grid of translations at all. It is two lines of linear algebra over the integers, exact in any dimension, and it can be checked against the two tables that exist before it is asked about a dimension where none does.

The formula, and where its answer comes from #

A cocycle for a cyclic group is determined by the translation attached to the generator, and closure forces one condition: composing the generator with itself n times gives the identity, so the sum of the rotated copies of that translation must be a lattice vector. That makes the cocycles the vectors the norm N = I + A + A² + ⋯ + Aⁿ⁻¹ sends into the lattice, and the coboundaries the vectors (A − I)s. Dualising gives the form above: the fixed vectors of A, over the image of N.

The numerator is where the whole result comes from, and it is decided by one question: does A have an eigenvalue equal to one? If it does, there is a direction A leaves alone and the fixed vectors are a copy of the integers along it. If it does not, there are no fixed vectors at all, the numerator is nothing, and the cohomology is trivial.

A rotation of order m needs a lattice of dimension at least φ(m) — which is where five-fold becomes legal, since φ(5) is four — and in exactly that dimension its characteristic polynomial is the m-th cyclotomic, which has no root equal to one. So a rotation filling its own smallest lattice fixes nothing.

Why the numerator is where everything happens #

The formula has a numerator and a denominator and it is worth saying why only one of them ever decides anything.

The denominator is never the obstacle. The norm of a rotation acting with no fixed vector is the zero matrix — the sum of all the powers of a matrix whose eigenvalues are the primitive m-th roots of unity is the value of a cyclotomic sum, which is nought — so the image of the norm is nothing. Nothing divided by nothing is nothing, and the cohomology is trivial because the numerator was.

And where there is a fixed direction the denominator is exactly right. The fixed vectors are a copy of the integers along the axis; the norm acts on that copy by multiplication by n, since each of the n powers of the rotation leaves it alone; so the quotient is the integers modulo n. The denominator does the work of turning an infinite group into a finite one and does no more.

So the whole content of the computation is the rank of the fixed subgroup, which is nought or one and is decided by whether the dimension exceeds the degree of the cyclotomic polynomial. One integer, and it settles the answer in every dimension.

That is worth contrasting with what the general extension count does. There the cocycles and the coboundaries are both large and the answer is their quotient, so both halves matter and neither can be read off in advance — which is why the plane’s census computes them on a grid and has to argue that the grid is fine enough. Here the grid question does not arise, because nothing is searched.

Checked against the two tables that exist #

The formula is asked first where an answer is already known from elsewhere.

In the plane the answer is one, at every order. A two-, three-, four- or six-fold rotation of the plane acts on a two-dimensional lattice with no fixed direction, so the cohomology is trivial, so p2, p3, p4 and p6 have no partners. That is the fact the extension census records as ten of the thirteen classes having nothing to choose, arrived at from the structure of the action rather than from a grid search.

In space the answer is the order. A rotation of space about an axis leaves that axis alone, so the fixed vectors are a copy of the integers, the norm multiplies that copy by n, and the quotient is cyclic of order n. The four-fold gives four — P4, P4₁, P4₂ and P4₃ — the three-fold three and the six-fold six, which are exactly the counts eleven screws and no others enumerates by asking which fractions of a turn a lattice can absorb.

Eight numbers, all of them in a table, all of them reproduced. The formula has earned a question nobody has a table for.

In four dimensions the five-fold has no screw #

A five-fold rotation is an integer matrix in four dimensions and in no fewer. Its characteristic polynomial there is $\Phi_5 = x^4 + x^3 + x^2 + x + 1$, whose roots are the primitive fifth roots of unity and do not include one. So it fixes no lattice vector, the numerator of the cohomology is nothing, and the class carries one space group: the symmorphic one, with no screw in it.

The same holds for the eight-, ten- and twelve-fold rotations, which are also four-dimensional and also fill their lattices. Every rotation acting on the smallest lattice that holds it admits exactly one group.

And add one more dimension for the rotation to leave alone, and the count becomes five: a five-fold rotation and four five-fold screws, in five dimensions. A screw needs one more dimension than its rotation does, whatever the order — the rotation needs φ(m) dimensions to exist as an integer matrix and one more to have somewhere to climb.

That is a statement about every dimension at once, and it is short enough to check: a screw is a rotation with a translation along its own axis, an axis is a direction the rotation fixes, and a rotation fixes a direction exactly when the dimension exceeds the degree of its cyclotomic polynomial.

What the prediction is worth, and what it is not #

Three things should be said plainly about a computation nobody here can check.

It is a prediction about one class, not about a dimension. The four-dimensional classification has seven hundred and ten arithmetic classes and four thousand seven hundred and eighty-three groups. What is computed above is the cohomology of one class — the cyclic group generated by one five-fold rotation on the lattice it fills — and it says that class carries one group. It says nothing about the classes containing a five-fold rotation together with other operations, which are the ones a four-dimensional table would mostly consist of.

The computation is not being extended; it is being restricted. The general extension count enumerates translations on a grid and quotients by coboundaries and by a normaliser, and none of that runs in four dimensions here. The cyclic case avoids all of it by having a closed form, which is why it is the case a prediction can be made in.

The four-dimensional entry is not a number anybody can look up here. Every other count on this page is in a printed table and was compared against it; this one is the formula’s own output, and the only evidence for it is that the formula is right in the two dimensions where evidence exists. That is what a prediction is, and saying so is the difference between a prediction and a claim.

And the answer is one a specialist would already expect. That a screw needs an axis is not a surprise to anybody who has met the space groups, and the four-dimensional literature does not contain a five-fold screw. What is worth having is not the fact but the route: the same machine that produced seventeen, run without modification, gives it — and the reason it gives it is a statement about a cyclotomic polynomial rather than about a picture of an axis.

The rule, stated for every dimension at once #

The census is short enough to read as one sentence, and the sentence is worth extracting because it settles a family of questions rather than one.

A rotation of order m on a lattice of dimension φ(m) admits one group. On a lattice of dimension φ(m) + 1 with the extra direction fixed, it admits m.

That covers every order at once and it explains the pattern in the space groups without enumerating anything. In space a three-, four- or six-fold rotation needs two dimensions to exist and has a third to climb, so it has screws, and the number of them is its order. In the plane the same rotation has no third dimension, so it has none. The seventeen have no screws for a reason of arithmetic rather than of drawing, and it is the same reason that gives the two hundred and thirty theirs.

It also settles the two-fold, which is the case easiest to get wrong by intuition. A half-turn is minus the identity in two dimensions and fixes nothing, so p2 has no partner; in three dimensions a two-fold axis fixes its axis, and the class carries two — P2 and P2₁. Same operation, same argument, two answers, decided by a dimension.

And it says what a screw is, in a form that never mentions a screw. The fixed direction is the whole of it: a cocycle is a translation attached to the generator, a coboundary is what an origin shift can remove, and an origin shift can remove any component in a direction the rotation moves. What it cannot touch is the component along a direction the rotation leaves alone, because shifting the origin along that direction changes nothing about where the rotation sends it. So the surviving classes are exactly the values that component can take, and on a lattice they are the $m$ multiples of $1/m$ of the repeat. The screw fraction is not an extra piece of geometry added to a rotation; it is the residue of an origin shift that had nowhere to go.

Where the cyclic case stops #

The closed form is cheap because the group has one generator, and almost no arithmetic class does.

The thirteen classes of the plane include 2mm on a rectangular lattice, whose point group has two generators and whose cohomology is (ℤ/2)² — four extension classes and three groups, the one place the plane’s two counts differ. A group with two generators has a cocycle condition for each and a compatibility condition between them, and no two-line formula.

The three classes of the plane with two generators are also the three the series of invariant sublattices thins most for, which is a coincidence of small numbers rather than a connection: a class with two generators has more conditions in every direction at once.

So the honest scope of this page is: the cyclic classes, in any dimension, exactly. That is a small fraction of any classification and it is a fraction with a clean answer, and the clean answer is the thing worth carrying — a rotation’s screws are counted by whether it fixes a direction, and by nothing else about it.

Where the exactness stops #

Computed here. For each rotation order whose cyclotomic polynomial has degree at most four: the companion matrix, its order, the norm, the fixed subgroup as a kernel over the integers, the image of the norm expressed in that subgroup’s basis, and the invariant factors of the quotient by Smith normal form. And the same with one extra dimension the rotation leaves alone. Four checks that can fail, including the counts in the plane and in space that the Tables record.

Every number is an integer computation. No grid of translations is searched anywhere, so the grid caveat the plane’s extension count carries — that a search on a grid can quietly decide an answer — does not apply. What replaces it is the closed form, which is derived and not checked except through its answers.

The four-dimensional answer is not checked against anything. It is the formula’s output in a dimension where no independent list is available here, which is exactly what the extension count was asked for and exactly why it should be read as a prediction. A four-dimensional table would check it in a line.

The rotation is the whole point group. Every class computed is generated by one rotation and nothing else. A class containing a five-fold rotation together with a reflection, or with a second rotation, is a different module and a different cohomology, and none of them is computed.

And “one group” means one up to the equivalences the classification uses. The cohomology counts extension classes; turning those into groups needs a further quotient by the normaliser, which for a cyclic class with trivial cohomology has nothing to act on. Where the cohomology is non-trivial — the axial cases — the normaliser can merge classes, and the merging of P4₁ with P4₃ under a reflection is exactly the eleven enantiomorphic pairs question. The counts above are before that quotient.

The first refusal is a five-fold screw axis, in four dimensions or any other where the rotation fills its lattice. It is the prediction, stated as something the computation must turn away, which is the only form in which a prediction can be tested by a machine that has no table.

Who runs the machine in four dimensions #

Hans Zassenhaus gave the algorithm in 1948 and the four-dimensional count was produced by Harold Brown, Rolf Bülow, Joachim Neubüser, Hans Wondratschek and Hans Zassenhaus in 1978, running it by machine: seven hundred and ten arithmetic classes and four thousand seven hundred and eighty-three groups. The five-fold classes are among them and their entries are what they are.

The four-dimensional groups have a use, which is the reason anybody produced them. A quasicrystal’s diffraction pattern is indexed by more integers than space has dimensions, and the structure is described by a periodic object in the higher space cut by a three-dimensional slice — so the symmetry of an incommensurately modulated crystal is a four-dimensional space group, and the tables of them are consulted by people refining real structures.

That is where the prediction on this page would be used, if it were needed. A modulated structure whose average has a five-fold axis would have a four-dimensional group, and the question of whether that group could have a five-fold screw is the question this page answers. Nobody has needed to ask it, because five-fold axes do not occur in average structures — which is the crystallographic restriction, keeping the question hypothetical.

Still open: the classes with two generators #

The obvious next question is what the machine does on a class whose point group is not cyclic, in a dimension without a table.

The two-generator case has a shape and not a formula. For a point group generated by two rotations the cocycles satisfy one condition per generator and one for the relation between them, so the cohomology is a kernel modulo an image again — but of a larger matrix, assembled from the group’s presentation. That is a computation rather than a closed form, and it would run in four dimensions as readily as in two.

What it would need is the arithmetic classes, and enumerating the seven hundred and ten is the step nothing here can make. The cyclic ones can be written down without enumerating anything, which is why they are the ones this page reaches.

The narrower question worth asking first is about the dihedral classes — a rotation and a reflection — where the cohomology has a short description too, and where the answer in space is the difference between a screw and a glide. Whether the same statement holds there, that the count is decided by what the operations fix, is a computation of the same size as this one and is not made here.