A screw that contains its own mirror image

The same group in a bigger cell asks at which indices a plane group contains a copy of itself, and the answer turns out to be arithmetic: the index has to be a value some quadratic form represents, because a sublattice invariant under a quarter turn is an ideal of the Gaussian integers. Nothing about that argument is two-dimensional except that the plane has only one kind of direction to ask about.

In three dimensions there is a second direction, and it is the one with a surprise in it. Take a sublattice along a screw axis rather than across it, and the subgroup that comes back need not be the group it was taken from. It can be its mirror image.

The setup, and the one condition #

Write the axis as w = (R, (s/N) c), where R is a rotation of order N about c and the screw component is s/N of a period. A 4₁ axis has N = 4 and s = 1; a 6₅ axis has N = 6 and s = 5. There are fifteen kinds of axis in all, counting the pure rotations as screws with s = 0.

Now ask for a subgroup whose translation lattice is a, b, n c — the same lattice stretched by n along the axis. That subgroup contains some rotation-part-R operation, but it need not be w itself: any w′ = (R, (s/N + m) c) has the same rotation part, and w and w′ differ by a lattice translation of the parent, so both are in the parent group. Which of them is in the subgroup is what has to be decided.

The deciding condition is that the subgroup be a group. Raise w′ to the N-th power and the rotation part disappears, leaving the pure translation (s + mN) c. That translation has to lie in the subgroup’s lattice, so n has to divide s + mN — and there is such an m exactly when the greatest common divisor of N and n divides s.

That is the whole of the condition, and everything below is a consequence of it. Write s + mN = n q. Measured against the subgroup’s own period c′ = n c, the screw component of w′ is q/N, so the subgroup is an axis of type N_q — and the equation s + mN = nq read modulo N says

n q ≡ s (mod N).

One congruence, and it answers two questions at once: whether a subgroup exists at index n, and which axis it is. There is nothing geometric left to check. A sublattice of a lattice along one direction is unambiguous, the rotation is unchanged by the stretch, and the only thing that can go wrong is the arithmetic of the screw component.

Read the fourth column of that table and the result of this essay is already there. At index three, five, seven and so on, a right-handed fourfold screw contains a copy of a left-handed fourfold screw, and at index one, five, nine it contains a copy of itself. Neither of those is a matter of interpretation; the subgroup is exhibited, and its screw component in its own cell is 3/4 or 1/4 as the congruence says.

One thing to be careful about before going on: the subgroup being exhibited is a subgroup of the space group, not of the axis considered on its own. An axis is a set of operations of a space group, and stretching the lattice changes which of them survive. What the congruence tracks is the operation w′ with rotation part R that lies in the smaller group — and it is a genuine element of the parent, differing from w by a translation the parent has. Nothing has been added and nothing has been reinterpreted; a subset of the parent’s operations has been selected, and the selection happens to be a group.

The reason that selection is not obvious is that the natural guess is wrong. The natural guess is that the subgroup consists of the parent’s operations whose translation parts lie in the smaller lattice, and that set is not closed: composing the screw with itself gives translations that are fractions of the new period. The correct construction chooses the coset representative rather than filtering, which is the split between the intrinsic and location parts of a translation doing its usual work — the intrinsic part is what the congruence is about and the location part is what m absorbs.

Why the even indices vanish #

The missing entries deserve as much attention as the present ones, because they are the part a picture would get wrong.

A 4₁ axis has no isomorphic subgroup of index two along its own direction. Doubling c gives a lattice in which the screw’s translation is an eighth of a period, and an eighth is not a legal screw component for a fourfold axis — the only components a fourfold axis may have are 0, 1/4, 2/4 and 3/4, which is what the eleven screws and no others is about. The subgroup would have to be something that is not an axis at all, so it is not a subgroup of the required kind and the congruence returns nothing.

The condition gcd(N, n) | s is that statement in general. When the index shares a factor with the order of the axis, the stretched cell divides the screw component by something it cannot be divided by, unless the component was already a multiple of that factor. A 4₂ axis, whose component is 2/4, survives doubling for exactly that reason — and a 2₁ axis, whose component is 1/2, does not.

The entries above one are worth a sentence. At index two a 4₂ axis contains both 4₁ and 4₃, because the congruence 2q ≡ 2 (mod 4) has the two solutions q = 1 and q = 3. A 4₂ axis is not chiral — it is its own mirror image — and it contains both hands as subgroups, in the same cell, at the same index. So handedness is not something that is inherited downward and it is not something that is conserved; it is a property of a particular axis in a particular cell, and the same object looked at through a longer period has a different one.

There is a second way to see why the even indices fail, and it is the one to keep. A 4₁ axis carries a point one quarter of a period along c per quarter turn. Double the period and the same operation carries it one eighth of the new period, and an eighth is not a quarter, a half or three quarters. The subgroup would need an operation that is a fourfold rotation combined with an eighth of its own period, and no such operation exists in three dimensions — the screw axis is the essay that says which components are legal and why. So the failure is not that the subgroup is hard to find; it is that the object it would have to be is not on the list of objects.

The whole table #

Doing this for all fifteen axis types at once produces a pattern that is easier to see than to state.

Three readings of that grid are worth having. The first is that the pure rotation rows are solid: with s = 0 the congruence is n q ≡ 0, which q = 0 always solves, so an axis with no screw component contains itself at every index. That is the two-dimensional answer, and it is the only case in which the axial direction behaves like the directions across it.

The second is that each screw row is periodic with period N, since the congruence depends on n only through its residue. So the whole of an infinite question is decided by at most six cases, and the table above is a picture of N columns repeated rather than of twenty different situations.

The third is the density. A chiral screw keeps exactly the indices coprime to N, so it keeps a fraction φ(N)/N of them — a half for the twofold, two thirds for the threefold, a half for the fourfold, a third for the sixfold.

Getting Euler’s totient out of this is not a coincidence and it is not deep. The indices that work for a chiral screw are the ones invertible modulo N, and the count of those is what the totient is. It is worth naming because the same function turns up in the crystallographic restriction for an entirely different reason — there it bounds the degree of a cyclotomic polynomial and so the dimension a rotation of order n needs — and the two appearances have nothing to do with each other beyond both being about residues.

Periodicity in n has a practical consequence worth stating separately. Because the congruence sees only n modulo N, and because every index that works is a product of prime indices that work, the whole infinite lattice of isomorphic subgroups along an axis is generated by the behaviour at the primes not dividing N. That is the same shape as the chain of maximal subgroups takes elsewhere: an infinite family described by what happens one prime at a time, with composites following.

Eight axes and their mirrors #

The eight chiral axes are the interesting rows, and collecting them makes the result of the essay a single line.

The mirror of a screw of component s/N is the one of component (N−s)/N, so a mirroring index is an n with n(N−s) ≡ s (mod N), which is n s ≡ −s. When s is invertible modulo N that is n ≡ −1, and the smallest such index is N − 1. When it is not — 6₂ and 6₄, whose component shares a factor of two with six — the congruence is weaker and the smallest index is N/gcd(s, N) − 1, which is two. That is the one place in this table where the obvious answer is wrong, and it is wrong on a quarter of the rows.

What makes this worth an essay rather than an exercise is what it says about handedness. The law that hides handedness is about a measurement that cannot see the difference between the two members of a pair; this is about the pair itself. No operation of any crystal, and no change of coordinates preserving the cell, takes P4₁ to P4₃ — the groups one hand may sit in depends on that. But P4₁ contains P4₃ as a subgroup, and so P4₃ contains P4₁, and each contains the other infinitely often.

That is not a contradiction and it is worth saying why. Being a subgroup is not being the same group, and neither containment gives an isomorphism of the two as space groups in a fixed cell, which is what the enantiomorphic distinction is about. What it does mean is that the distinction is not hereditary: a property that some subgroup of a group must share is not a property that separates P4₁ from P4₃, so no argument of that shape can prove them different. That rules out a whole class of would-be proofs, which is the practical use of the result.

There is a physical reading of the mirroring result which is worth setting down carefully, because it is easy to overstate. A crystal in P4₁ really does contain, as a sublattice of its own symmetry, a P4₃ group — but that group has a cell three times as long, and every operation in it is an operation of the original crystal. Nothing has changed hands physically: the same atoms sit in the same places, and the same helices wind the same way. What has changed is the description, and the description’s handedness is defined relative to a period. A helix that advances a quarter of a period per quarter turn advances three quarters of a longer period per quarter turn, and calling that left-handed rather than right-handed is a statement about which period is being counted against.

That is why the result rules out a class of arguments rather than a class of crystals. It does not say the two members of an enantiomorphic pair are secretly the same; the round trip in space and the diffraction argument both keep them apart. It says that any proof of their difference has to use the cell, because a proof that only used the subgroup structure would prove too much.

Which orders can be handed at all #

The twofold row is the case that makes the counting honest. A 2₁ screw is chiral in the sense that its trajectory is a helix and a helix has a hand — and it is not chiral in the sense that matters here, because the mirror of component 1/2 is component 1/2. Turning the helix over gives the same axis back. That is the whole reason P2₁2₁2₁ has no enantiomorphic partner while P4₁2₁2 does, and it comes out of the congruence as s ≡ N − s rather than out of a picture of a helix.

Counting the eight also settles a small confusion. There are eleven enantiomorphic pairs of space groups and eight chiral kinds of axis, and the two numbers are not the same because a space group can be handed by having several axes rather than by having one of these — and because two space groups can share an axis type and differ elsewhere. The relationship between the two counts is the space group enumeration’s business rather than this table’s.

What the congruence has to refuse #

The first refusal is the one that would be easiest to write into a table by hand and hardest to notice. Every odd index works for 4₁, and a reader building the list from the pattern rather than from the congruence would very likely write “every index” and be wrong on half of them. The check that catches it is that the solution count agrees with the greatest common divisor in every one of the fifteen rows and every index, which is a statement about the congruence rather than about any particular axis.

The fourth refusal is the one this essay was written wrong on first. n ≡ −1 is the right residue when the screw component is invertible modulo the order, which is six of the eight cases, and it is not when it is not; asserting N − 1 everywhere passes six rows and fails 6₂ and 6₄. An assertion that names the general form rather than the common case is the difference between a check and a coincidence, and this one caught a sentence that had already been written into the prose.

The last refusal is worth a word too. It is a claim the table makes false by having eight rows in it, and it is included because a version of this computation that lost the distinction between an axis and its mirror — by comparing screw components up to sign, say, which is a very natural simplification — would still produce a plausible table with all the same indices in it. What it would lose is the only interesting column.

One more consequence is worth drawing because it is the sort of thing a table of subgroups is actually used for. If a structure is refined in P4₁ and a referee suspects the true cell is three times longer, the group they should be testing against is P4₃ and not P4₁ — the congruence says so before any data is looked at. Getting that the wrong way round produces a refinement that cannot converge for a reason no residual will explain, and the diagnosis is arithmetic rather than crystallographic. A bigger cell and sometimes the mirror is the plane version of the same trap, where the extra thing a longer cell can bring back is a reflection.

Where this stops #

The directions across the axis are the two-dimensional problem again, and this collection has them: the index must be a value of a quadratic form, the same forms whose values are the lengths a lattice has. A general isomorphic subgroup of a space group combines a sublattice in the plane with a stretch along the axis, so its index is a product of one of those values with one of these residues, and both factors have to be admissible independently.

What is not done here is the enumeration of maximal isomorphic subgroups, which is the form the tables in the literature take. A subgroup at a composite index is generally not maximal — it factors through the subgroups at its prime factors — so the useful list is the one at prime index, and turning this congruence into that list means checking, for each prime, whether the subgroup at that index is contained in anything else. That is a question about the whole space group rather than about one axis, and it needs the other directions and the other generators. The congruence here settles the axis, which is the part with the handedness in it.