3m1 and 31m are one class

A hexagonal lattice has two families of special directions in its plane, thirty degrees apart. One runs through the lattice points nearest the origin — the ⟨100⟩ family. The other runs between them, through the gaps — ⟨11̅0⟩.

A three-fold axis carries each family onto itself and does not exchange them. So a group built on a three-fold axis with mirrors in it has a genuine choice: the mirrors can be on the first family, or on the second, and no operation of the group turns one arrangement into the other.

As plane groups, those two arrangements are p3m1 and p31m, and this site has an essay arguing at some length that they are different groups — different patterns, different fundamental domains, different diffraction. That essay is right.

As point groups the same two objects are one class. Both are called 3m.

The two subgroups are genuinely two subgroups #

It is worth being careful, because “one class” can be read as “the distinction was never real”, and that is not what happened.

Inside the hexagonal holohedry 6/mmm there are six mirrors whose planes contain c: three with normals along the ⟨100⟩ family, three along ⟨11̅0⟩. Take the three-fold rotation about c and adjoin one mirror. Closing gives a group of order six. Adjoin a mirror from the other family and closing gives a different group of order six — different as a set of matrices, sharing only the three rotations.

So the enumeration finds two subgroups, and they are not the same subgroup.

They are also, and this is the part that makes the essay, each normal in 6/mmm. Conjugating either by any operation of the holohedry returns it unchanged. The reason is short: conjugating a mirror by a rotation turns its plane by the rotation’s angle, the six-fold turns a plane by sixty degrees, and mirrors at 0°, 60° and 120° are the same three planes as mirrors at 60°, 120° and 180°, because a plane at θ and a plane at θ + 180° are one plane.

So the orbit of each subgroup under conjugation in 6/mmm has exactly one member. There is no operation of any hexagonal lattice carrying 3m1 onto 31m.

And yet they are one class #

The relation that defines a crystal class is conjugacy in space — by any change of basis at all, not only by a symmetry of some lattice. And a rotation of thirty degrees about c carries the ⟨100⟩ family onto the ⟨11̅0⟩ family, so it carries one subgroup onto the other.

Thirty degrees is not a symmetry of a hexagonal lattice. It is a perfectly good rotation of space.

That is the whole of it, and the shape of the argument is worth extracting because it recurs: the finer relation is conjugacy inside the holohedry, and the coarser one is conjugacy in space, and the classification uses the coarser. The enumeration leans on this in both directions — it quotients by the finer relation first because that step is cheap and mechanical, and then merges what is left, and the merge is precisely where pairs like this one get identified.

The merge that identifies those two arrangements is not taken on trust. A pair of subgroups is proposed as one class when their element censuses agree — so many three-folds, so many mirrors — and a census knows nothing about how the operations compose, so agreeing on one is a proposal and not a verdict. What settles it is an explicit change of basis P with P·G·P⁻¹ = G′, constructed rather than assumed, and the enumeration refuses to return at all unless every merge it makes carries one. For this pair the matrix is the thirty-degree rotation written in the hexagonal basis, where it is a perfectly ordinary integer matrix of determinant three, and checking it is a line of arithmetic that anybody can repeat.

The bug this found #

The derivation of the symbols had this wrong, and the way it was wrong is more useful than the fix.

Symbols are read off the system’s ordered directions, and a group has to be in the right setting for the reading to give the standard name. The obvious way to find that setting is to turn the group inside its own holohedry and take the best orientation — which works for every class where turning does something.

For 3m it does nothing. The orbit has one member; the group comes back unchanged from all twenty-four conjugations; whichever of the two subgroups the enumeration happened to pick first is the one that gets read. It picked the ⟨11̅0⟩ one and derived 31m, and 31m is a real symbol for a real setting and is not on the list of thirty-two.

The assertion that the derived symbols equal the Tables’ thirty-two is what caught it, and it caught it as “missing 3m; unexpected 31m” — which names the defect exactly. The repair is that the settings search is handed every member of the merged class rather than one representative, so both subgroups are in the pool and the rule chooses between them.

A procedure that returns one answer when it should consider two is not obviously broken from the inside. Every step of the settings search ran correctly; the input was one group where it should have been two.

Why normality is the awkward property here, not a convenient one #

There is an intuition worth dislodging, because it is the one that made the bug invisible: that a normal subgroup is a well-behaved one, and that finding an orbit of size one is a simplification.

For this purpose it is the opposite. The settings search works by turning a group inside its holohedry and looking at what the turning produces. Its whole supply of alternatives comes from the conjugation orbit. A subgroup with a large orbit hands the search plenty to choose between; a normal subgroup hands it nothing at all, and the search then reports on the single orientation it was given as though that were the only one.

So the degenerate case — the one where the group is nicest — is the case where a procedure that samples an orbit has the least to work with and is most likely to be wrong. That is the same shape as the failure in the conjugator construction one rung down this ladder, where the group 1̅ has an intertwiner space of all nine dimensions and both systematic ways of picking an element out of it picked a singular one. A large solution space and a trivial orbit are both “the easy case”, and both are where careless picking goes wrong.

The plane group answer is the opposite, and both are right #

So why does p3m1 differ from p31m when 3m does not differ from 31m?

Because a plane group is not a point group. It is a point group together with a lattice, and the lattice is what the thirty-degree rotation destroys.

Rotating the whole configuration by thirty degrees carries p3m1’s mirrors onto p31m’s mirror directions — and carries the hexagonal lattice onto a hexagonal lattice rotated by thirty degrees, which is not the same lattice. To make the two agree the lattice has to come back, and no operation does both.

The two plane groups are therefore genuinely distinct: they have different fundamental domains, different Wyckoff positions, and — the observable difference — different systematic behaviour under diffraction, because the mirror directions sit differently relative to the reciprocal lattice.

Strip the lattice and the distinction goes with it. A point group has no lattice; it is a group of directions about a point, and there is nothing left for the thirty degrees to spoil.

Four pairs, not one #

3m is the clearest case and it is not the only one. Four of the six classes with an alternative setting are trigonal or hexagonal, and every one of them is the same situation: 32 against 312, 3m against 31m, 3̅m against 3̅1m, and 6̅2m against 6̅m2.

The pattern behind the four is a single sentence. A group whose principal axis is a three-fold has two inequivalent ways to place its secondary elements; a group whose principal axis is a six-fold has one. A six-fold rotation turns a direction by sixty degrees and so carries the ⟨100⟩ family onto itself — but a six-fold group also contains operations turning by thirty degrees when the mirrors are present, and it is the mirrors of 6mm and 6/mmm that reach both families at once. So 6mm has all six vertical mirrors and no choice to make, while 6̅2m has three mirrors and three two-folds and must decide which family gets which.

Set the two pictures side by side and the shape of the whole question is visible in one line. A setting question needs two families of directions that the group does not exchange, and it needs them to hold different things. 3m has one family carrying mirrors and one carrying nothing; 6̅2m has one carrying two-folds and one carrying mirrors. Either way there are two arrangements and no operation of the group turning one into the other, so the symbol has to choose. 6mm has mirrors on both families and therefore nothing to choose between, which is why it appears on no list of classes with alternative settings.

The two non-hexagonal cases are different in kind. mm2 against m2m is the orthorhombic system, which has no principal axis at all and where the convention is a separate clause; and 4̅2m against 4̅m2 is the tetragonal analogue, where the two secondary families are forty-five degrees apart and the four-fold rotoinversion does not exchange them.

What the two verdicts have in common #

Both essays are answering “are these the same?”, and both answer it by naming the equivalence rather than by looking. That is the habit, and it is the reason the two answers can differ without either being sloppy.

Conjugation and sameness makes the general point for the plane groups: “the same symmetry” is not a primitive notion, and every classification in this subject is a quotient by some group of changes of description. Change the group and the count changes. The seventeen plane groups are a quotient by affine equivalence; the 230 space groups are a quotient that keeps enantiomorphic pairs apart while 219 is the one that does not; and the thirty-two crystal classes are a quotient by conjugacy in space rather than in a lattice.

A count without its equivalence is not a fact. That is why this site records both numbers wherever they differ — 230 and 219, 74 two-colour subgroups against the classical 46, and now 3m as one class against p3m1 and p31m as two.

What a reader should take from the pair of essays #

The two essays disagree about two objects and agree about everything that matters, and the agreement is the transferable part.

Ask what came along with the group. p3m1 and p31m arrive with a lattice; 3m arrives with nothing. Almost every apparent contradiction in this subject’s counts is of that form — the same operations, described with or without the thing that fixes them in place. The 230 against 219 is the same question about handedness; the 74 index-two subgroups against the classical 46 two-colour groups is the same question about a change of basis; the fourteen Bravais lattices against twenty-five candidate cells is the same question about which descriptions name one lattice.

And the count is worthless without the equivalence. A reader who has been told “there are thirty-two crystal classes” and “p3m1 and p31m are different groups” has been told two true things that appear to conflict, and no amount of staring at either will resolve it. What resolves it is that they are quotients by different relations, and both relations are worth having.

What a reader should do with a symbol that has two forms #

The practical consequence, for anybody reading a structure report rather than building a classification, is small and easy to get wrong in a specific way.

Meeting 31m in a paper means the authors have used the setting with the mirrors on the ⟨11̅0⟩ family. It does not mean a different crystal class from 3m, and it does not mean anything is unusual about the material. The setting is a statement about how the axes were chosen, which is a fact about the report rather than about the crystal.

Where it does matter is in comparing two reports. Two structures described in different settings have Miller indices, atomic coordinates and extinction conditions that do not line up, and reading one against the other without transforming between them produces disagreements that look like disagreements about the structure. That is the ordinary form of the problem and it is a bookkeeping hazard rather than a conceptual one.

Where it matters much more is one level down. As space groups, P3m1 and P31m really are two different groups, with different Wyckoff positions and different systematic absences, and there the setting question has become a classification question because the lattice has arrived. A reader who has internalised “3m1 and 31m are the same thing” from the point-group level and carries it to the space-group level will conclude that two different space groups are one.

The rule that transfers is to ask what came with the group — and to notice that the answer changes between the two levels rather than assuming it does not.

Where the exactness stops #

“Conjugate in space” here means conjugate by a rational matrix, which is what the enumeration constructs. That is the right relation for crystal classes. It is not the finest one anybody uses: requiring the change of basis to carry one lattice onto another gives the 73 arithmetic crystal classes, and in that classification 3m on a hexagonal lattice really does split — into 3m1 and 31m — because the lattice is back in the picture.

So the honest statement is that these two objects are one geometric crystal class and two arithmetic ones, and which is meant depends on whether the lattice is part of the question. This site computes the geometric classification and does not compute the 73.

Nothing here is measured. The two subgroups are integer matrices, the thirty-degree rotation is exhibited as a rational conjugator in the lattice basis, and the verdict is an equality of matrix sets. There is no tolerance in it and no picture is being trusted.

Producing the witness, and what its absence would mean #

The verdict here rests on exhibiting a thirty-degree rotation that conjugates one subgroup onto the other. That is the right shape of evidence, and it is worth saying what the other verdict would have to look like, because the asymmetry between the two is the whole reliability of the method.

A positive answer is a matrix. Hand over the conjugator, check that it carries every operation of one group to an operation of the other, and the claim is settled — by anybody, in a line of arithmetic, with no trust in the search that found it.

A negative answer is a search. Establishing that no conjugator exists means ruling out every invertible rational matrix, which is an infinite set. A bounded search over matrices with small entries can fail to find one and has proved nothing: the conjugator might have entries of a thousand.

That asymmetry is why the merges in this collection’s enumerations are always accompanied by their conjugating matrices and the non-merges are not accompanied by anything. A pair reported as distinct is reported on the strength of an invariant — the element signature, or a character — which proves a difference outright. A pair reported as the same is reported on the strength of a witness.

Neither kind of evidence is available for the other verdict, which is the point the collection keeps returning to: an invariant can only separate, and a construction can only identify, and a classification needs both because it makes both kinds of claim.

What the two arithmetic classes buy #

The last section’s conclusion — one geometric class, two arithmetic ones — sounds like a bookkeeping refinement. It has a consequence that is neither small nor abstract, and it is visible in the list of two hundred and thirty.

An arithmetic class is a point group together with the lattice it acts on, and it is the input to the space-group enumeration. So the two arithmetic classes here produce two separate families of space groups, and both families are in the tables: P3m1 and P31m, P3̄m1 and P3̄1m, P3c1 and P31c, and so on down the trigonal section.

Those are genuinely different groups, with different absences, different site symmetries and different diffraction. A reader who took the geometric merge to mean the distinction had evaporated would expect one of each pair and find two.

So the merge and the distinction sit at two levels and both survive. At the level of point groups there is one class, because space has a thirty-degree rotation. At the level of space groups there are two of everything, because a hexagonal lattice does not. The classification is not a single verdict but a chain of them, and each link is taken relative to its own equivalence.

Where this goes #

The classification is finished and the classes are named. What they decide comes next, and the first thing is not a property of a material at all — it is a limit on what an experiment can report. Eleven of the thirty-two contain the inversion, and a diffraction pattern cannot see past that, so a structure determination begins with an eleven-way answer to a thirty-two-way question.