The half of a translation that is not a choice

P2 and P2₁ have the same point group, the same lattice, the same number of operations and the same number of atoms in a cell. They are different space groups. Asked what the difference is, most people who have met the symbols will say that in one the axis is a rotation and in the other it is a screw — which is true, and restates the question.

The useful answer is that a translation has two halves and only one of them is a fact about the group.

Moving the origin, and what it does #

Coordinates in a crystal are fractional: a point is a fraction of the way along each cell edge. Where the zero of that scale sits is a choice, and — like the choice of cell itself — it is a choice somebody made — usually to make the symbols come out tidily, sometimes because a particular atom was convenient.

Move the origin by a vector s and every operation’s translation changes. An operation is x ↦ Mx + t; describe the same operation in coordinates measured from a new origin at s, and it becomes $x \mapsto Mx + t + (M - I)\mathbf{s}$. So the translation of every operation shifts by (M − I)s, and what a change of origin can reach is exactly the image of that matrix.

For a mirror perpendicular to a, the matrix $M - I$ is diag(−2, 0, 0). Its image is the a direction alone, so an origin shift can add anything to the a-component of a mirror’s translation and nothing at all to the other two. That is why a mirror plane can be put anywhere along the direction it faces — moving the origin slides the plane — and why what it does within its own plane is untouchable.

That is already the whole result, stated geometrically: an origin shift moves an element about; it cannot change what the element does.

The split, computed #

Making it arithmetic takes one line. Apply an operation of order n to itself n times. The linear parts multiply out to the identity, so what is left is a pure translation, and one n-th of it is the part per application that no origin removes:

tintrinsic = (1/n)(I + M + M² + … + M^(n−1)) t

Everything else is location. Applying $M - I$ to the intrinsic part gives zero — the sum telescopes — so the intrinsic part is invariant under every origin shift, and the two halves partition the translation cleanly.

For the two-fold rotation of P2, with M = diag(−1, 1, −1) and t = 0, the intrinsic part is zero. For the screw of P2₁, with the same M and t = (0, ½, 0), the sum (I + M)t is (0, 1, 0) and half of that is (0, ½, 0): the intrinsic part is the whole translation, and no origin touches it.

The plane has this decomposition too, and it is nearly invisible there because the only operations with a non-zero intrinsic part are the glides — and a glide looks obviously unlike a mirror. In space there are eleven kinds of screw and five kinds of glide, and for several pairs the arithmetic is the only thing that tells them apart.

Why the axis can be anywhere and the rise cannot #

The geometric reading is worth having beside the arithmetic, because it is the one that stays.

Every operation of a space group either has a fixed point or does not. A rotation fixes its axis; a mirror fixes its plane; an inversion fixes one point. Those are the operations whose intrinsic part is zero, and for them the location part is genuinely a statement about where the element is — move the origin and the axis moves, and nothing else changes.

A screw fixes nothing, which is what puts it outside any point group and is the whole content of the screw-axis essay. Its axis is still a line, in the sense that there is a line the operation slides along without turning about anything else, but no point of that line stays put. Move the origin and the line moves; the rise does not.

This is where the International Tables’ marks come from. An axis is drawn as a polygon with as many sides as its order, filled for a rotation and open with tails for a screw, and the number of tails is the numerator of the rise. The picture is drawn from the arithmetic: a figure on this site places a mark by solving for the operation’s fixed set and reads the tails off its intrinsic part, so a mark with the wrong number of tails would require the arithmetic to be wrong rather than the drawing.

The extension, and why it is the right word #

Take a space group G. Its translations form a subgroup T, which is the lattice, and it is normal — conjugating a translation by any operation gives another translation. So the quotient G/T exists, and it is the point group P.

That is the structure the whole classification hangs on, and it is the same observation that makes a wallpaper group’s quotient by its translations finite. It is worth writing as the sequence it is: the lattice sits inside the space group, and the space group maps onto the point group, with the lattice being exactly what the map forgets. A space group is an extension of a point group by a lattice.

The question the classification answers is then: given P and T, how many groups G fit in the middle? And the answer is a count of how many consistent ways the translations can be attached.

Consistency is a real condition, not a formality. Pick a translation for each generator, and the translations of every other element follow by composition — but they must follow consistently: if two different products of generators give the same linear part, the translations they produce must agree modulo the lattice. That is the cocycle condition, and it is the reason most assignments fail.

The class 222 is the smallest place to watch that happen. Its point group has two generators, each of which may be given any of eight translations on a grid of halves, so there are sixty-four assignments to try. All sixty-four happen to close — in this class the cocycle condition turns out not to be restrictive, which is worth knowing because it is the exception rather than the rule. Eight of the sixty-four survive the quotient by the origin shift, and relabelling the axes brings the eight to four: P222, P222₁, P2₁2₁2 and P2₁2₁2₁. The counting is done twice over, in other words, and both quotients are the split this essay is about — the first divides out what an origin can change and the second divides out what a choice of which axis to call a can change.

This site tests consistency the blunt way, and the bluntness is deliberate. Rather than deriving the cocycle condition and checking it, each candidate assignment is simply closed — compose the generators until nothing new appears — and kept only if the result has exactly the order the point group has. An inconsistent assignment produces a closure that keeps generating new elements, running away from the right size, and gets discarded. That the closure can be run at all is what the three-dimensional machinery bought. That test is equivalent to the condition and it cannot be got subtly wrong, which a hand-derived condition can.

What splitting means #

Among all the ways the translations can be attached there is always one trivial one: attach nothing. Give every generator a translation of zero, and the closure is a copy of the point group sitting inside the space group, fixing the origin.

A group that can be written that way — at some origin — is symmorphic. Since “attach nothing” is a single answer, there is exactly one symmorphic group per arithmetic crystal class, and so seventy-three of them.

The definition needs care in a way that is easy to get wrong, and this site got it wrong first. A group with a screw in it can still be symmorphic. C2 has a two-fold rotation and, because the C centring translation is (½, ½, 0), it also has that rotation composed with the centring — which is a 2₁ screw. So C2 contains a screw axis and is nonetheless symmorphic, because the rotation is also there and the origin can be put on it.

An early version of the test here asked whether any operation of the group had a non-zero intrinsic part, and answered “non-symmorphic” when one did. That called C2, R3, Fm3̅m and Im3̅m non-symmorphic — four errors from one line, and all four are groups a reader would know were wrong. The test now asks the question the definition asks: is there an origin at which some operation in each coset of the translation subgroup loses its translation? Which is a search, and the search is what runs.

The grid the search runs on is twenty-fourths, not twelfths, and that detail is not arbitrary either. A shift s changes a translation by (M − I)s, whose entries can be twice s — so a translation of a twelfth is reachable only from a shift of a twenty-fourth. A grid of twelfths would report some symmorphic groups as non-symmorphic, and again it would fail in the safe-looking direction, by finding nothing.

The same operation, at eleven different places #

There is a trap in all of this that catches every diagram drawn by hand, and it follows directly from the split.

A group’s operation list is the list modulo lattice translations. P2₁/c has four entries: identity, screw, inversion, glide. But the cell contains four 2₁ axes, not one — at x = 0 and x = ½, and at z = ¼ and z = ¾ — and they all belong to the single screw entry, differing by lattice translations.

Where they sit is not obvious, and this is the part that goes wrong. Compose the screw with the translation a and the result is another screw; its axis is not one cell along, it is a half cell along, because the axis moves by (I − M)⁻¹a rather than by a. For a two-fold that inverse halves; for a fourfold it halves the diagonal; for a threefold it produces thirds.

Every element in every plan diagram in this field is placed by solving (I − M)x = t − tintrinsic for each translate of each operation, which is an exactly solvable and exactly singular system: singular because the solution set is the whole axis rather than a point, so a solver that assumes invertibility cannot place a single one of them.

The consequence a chemist meets #

The split between symmorphic and non-symmorphic is not a taxonomic nicety, and there is a clean way to say what it costs.

In a symmorphic group there is a point of the crystal whose site symmetry is the entire point group — a place where all of it acts at once. An atom put there is fixed by every operation, so it contributes a single atom to the cell rather than a whole orbit. In a non-symmorphic group there is no such place, and the highest site symmetry available is a proper subgroup of the point group.

That has a direct consequence for structure solving. The smallest number of formula units a cell can contain is set by the multiplicities the group makes available, and a non-symmorphic group’s smallest orbit is larger — so a molecule crystallising in P2₁/c must contribute at least two copies per cell unless it happens to sit on an inversion centre, which requires the molecule itself to be centrosymmetric. Roughly a third of all published organic structures are P2₁/c — a concentration with no analogue in the plane’s seventeen, where no group dominates, and that constraint is part of the reason: molecules that are not centrosymmetric are common, and a group with plenty of general positions and few special ones suits them.

Where the exactness stops #

The split is exact, and two things about it are not.

The intrinsic part is defined modulo the lattice. A screw with a rise of ½ and one with a rise of 3/2 are the same operation, because the difference is a lattice vector, and the lattice is what everything here is measured against. Everything here reduces into [0, 1), which is correct and means the “intrinsic translation” is a coset rather than a vector. It has to be — otherwise 4₁ and a 4₁ composed with the c translation would be different operations, and they are not.

Symmorphic is not the same as “no screws”. It is the sentence above, and the C2 example is why the distinction has to be maintained. A group can contain screw axes, glide planes, and a full complement of both, and still split. The test is about whether they can all be avoided at one origin, not about whether any exist.

And the point group is not a subgroup. For a non-symmorphic group, the sentence “the point group of P2₁/c is 2/m” is true in the sense that dropping every translation leaves 2/m, and false in the sense that there is no copy of 2/m sitting inside P2₁/c. The map from the space group to the point group has no section. That is what “does not split” means and it is the reason the word extension is the right one — the group is built on top of the lattice in a way that cannot be undone.

This has a consequence worth carrying into the diffraction essays: the point group is what a diffraction pattern’s intensities show, and it is recoverable. The extension — whether the translations are attached — is what the absences show, and it is recoverable separately. Two independent readings of one experiment, and they answer the two halves of this essay’s split.

Where this goes next is the counting. Attaching translations consistently, quotienting by the origin and then by relabelling the axes, is the whole procedure — and running it on the class with the most answers is the next essay, where the number sixteen becomes ten for a reason that is worth an argument of its own.

The name the split has elsewhere #

Everything above is one computation performed twice — once to strip out what an origin shift can change, and once to check that what is left can be attached consistently. That pair of operations is a standard object outside crystallography, and knowing its name makes the classification’s shape easier to see.

The assignment of translations to point operations is a cocycle. Write $\tau(g)$ for the translation attached to the point operation $g$. Composing two operations composes their translations in the way the matrices dictate, and the requirement is

$$\tau(gh) = \tau(g) + M_g\,\tau(h) \pmod{\text{lattice}}.$$

That is the consistency condition the enumeration tests, written once instead of case by case.

An origin shift is a coboundary. Moving the origin by $\mathbf{s}$ changes every $\tau(g)$ by $(M_g - I)\mathbf{s}$, and an assignment of that form is exactly what carries no information — it is the trivial assignment seen from somewhere else.

So the classification is a quotient: consistent assignments, divided by those an origin shift produces. That quotient is the second cohomology group of the point group with coefficients in the lattice, and its elements are the genuinely different ways of building a space group on a given point group and lattice.

Read the census through it and the numbers stop being a list. A class whose cohomology is trivial admits one group, and that group is symmorphic — everything can be pushed to zero by a single choice of origin. A class whose cohomology is non-trivial admits as many groups as it has elements, and all but one of them are non-symmorphic. In the plane the cohomology is trivial for thirteen of the seventeen; in space, seventy-three of the two hundred and thirty are symmorphic, so the remaining hundred and fifty-seven are counted by the non-trivial elements.

This is not a reinterpretation after the fact. It is how classifications beyond three dimensions are actually done: the four-dimensional space groups — 4,783 of them — were enumerated by computing these cohomology groups mechanically, because case analysis of the kind that produced the 230 by hand does not survive the increase in dimension. Zassenhaus’s algorithm is that computation, and it takes as input precisely the two ingredients this essay separates.

What a non-symmorphic group does to the levels #

The split is algebra, and it has a consequence that a measurement sees directly — one which is unavailable to any symmorphic group and which is the reason solid-state physicists care about the distinction at all.

A state in a crystal is labelled by a wavevector, and the operations that fix that wavevector act on the states carrying that label. In a symmorphic group those operations form an ordinary group — the point operations, with their translations set to zero by a suitable origin — and the possible degeneracies are the dimensions of that group’s representations, exactly as a degeneracy is bounded by a representation.

In a non-symmorphic group the intrinsic translations survive, and they do not vanish from the calculation. Composing two operations then picks up a phase factor $e^{i\mathbf{k}\cdot\boldsymbol{\tau}}$ that depends on the wavevector, so the operations act by a projective representation rather than an ordinary one — a representation in which composition holds only up to a phase.

At the zone centre the phases are all one and nothing has changed. At the zone boundary they need not be, and where they are not, the projective representations can be forced to have larger dimension than any ordinary representation of the same group. The consequence is a degeneracy that no material parameter can lift.

That is band sticking. In a crystal with a glide plane, bands arrive at the relevant zone-boundary plane in pairs and touch there — not because the two happen to have the same energy, but because the group permits no state at that wavevector that is not part of such a pair. Change the composition, the pressure, the temperature; the two bands still meet at the boundary. Nothing analogous happens in a symmorphic group.

So the intrinsic half of the translation is measurable, which is the strongest possible answer to the question this essay opens with. P2 and P2₁ differ by something with no fixed point, invisible in the crystal’s shape, invisible in the point group, and invisible in any table of atom positions read without its symmetry. It is nevertheless the difference between a band structure whose degeneracies are optional and one whose degeneracies are compulsory — and a spectroscopy that finds an unliftable degeneracy at a zone boundary has found the half of the translation that is not a choice.