Two groups no single operation reveals

A space group is symmorphic when some choice of origin puts every one of its point operations through that origin with no translation attached. It is non-symmorphic when no choice does. The textbook picture of the second kind always points at one operation. A screw axis climbs as it turns, and moving the origin moves the axis but cannot stop the climb. A glide plane slides as it reflects, and no origin stops the slide. The split between intrinsic and locative parts makes that precise for a single operation. The intrinsic part of its translation, the average over its own cycle, is the part no origin removes, and a screw or a glide is an operation whose intrinsic part is not a lattice vector.

The picture suggests a theorem: a space group is non-symmorphic exactly when one of its operations has an intrinsic part that is not a lattice vector. The denominator a group actually needs leaned on something close to it, calling it “a standard fact” that restricting the cohomology to the cyclic subgroups loses nothing that matters. The theorem is false in space, in exactly two places. Across all seventy-three arithmetic classes of space, every non-zero cohomology class is revealed by some single operation except two: the non-symmorphic classes of the body-centred orthorhombic class 222 and the body-centred cubic class 23. They are the space groups I2₁2₁2₁ and I2₁3. Each of their operations, taken alone, is also a pure rotation about some line, so no operation of either group needs to be a screw. What they cannot do is put their rotation axes through one point.

What restricting to one operation means #

A cohomology class is a way of attaching translations to the operations of a point group, up to the choice of origin. Seventy-three Smith forms computed these classes for every arithmetic class of space and handed back an explicit translation for each. Given such a class, take one operation $g$ of order $k$ with its attached translation $t(g)$, and forget the rest of the group. The cyclic group $g$ generates is a small space group of its own, and the class restricts to a class of that small group.

The restricted class is zero exactly when some shift of origin removes $t(g)$ altogether. Moving the origin by $v$ changes the translation by $(g - I)v$, so the test is whether

$$t(g) \;\in\; (g - I)\,\mathbb{R}^3 + \mathbb{Z}^3 ,$$

which is the same membership test the Smith census uses for the whole group, run on a three-by-three matrix. For a two-fold rotation it asks whether the component of $t(g)$ along the axis is a whole lattice vector: if it is, the operation is a rotation about a line displaced from the origin, and if it is not, it is a screw. For a mirror it asks the same of the component in the plane. When the restricted class is not zero, the operation reveals the class. It carries an intrinsic translation that the class forces on it, and a crystallographer sees it as a screw or a glide.

The census asks this of every operation of every non-zero class of every arithmetic class of space. Sixty-one of the seventy-three classes have non-zero cohomology. Between them they carry a few hundred non-zero classes, which the normaliser then sorts into the non-symmorphic space-group types.

Nearly every class is seen by several operations #

The distribution is what the textbook picture predicts almost everywhere. A typical non-zero class is revealed by a quarter or a half of its point group’s operations, and those are the screws and glides written into its Hermann–Mauguin symbol. P2₁2₁2₁ is revealed by all three of its two-fold operations, each of which is a screw. The sixty-three non-zero classes of the orthorhombic P holohedry are each revealed by between two and six of its eight operations, in the patterns the symbols from Pmma to Pnma spell out letter by letter. No non-zero class is revealed by every operation, because the identity never reveals anything and neither does an operation that fixes no direction, such as the inversion.

Two classes are revealed by no operation at all.

I2₁2₁2₁, one operation at a time #

The body-centred orthorhombic class 222 has cohomology $\mathbb{Z}_2$. Its zero class is I222 and its non-zero class is I2₁2₁2₁. The representative translation the Smith reduction returns for the non-zero class can be read on the conventional cell, and it is instructive to take its three two-fold operations one at a time.

The two-fold along $a$ has a translation with nothing along $a$, so it is a pure rotation about a line displaced from the origin. The two-fold along $b$ has half a cell along $b$ and is a screw. Add the body-centring translation $(\tfrac12, \tfrac12, \tfrac12)$, which is a lattice vector of this lattice, and its component along $b$ becomes a whole cell: the same operation, composed with a translation the crystal already has, is a pure rotation about a different line. The two-fold along $c$ behaves like the first. So every operation of the group comes both as a screw and as a pure rotation, and taken alone each one is the operation of a symmorphic group moved off the origin.

The centring is what makes this possible. In a primitive lattice a screw’s half-cell climb cannot be cancelled, because no lattice vector has a half along the axis. In a body-centred lattice the centring vector has a half along every axis at once, so it turns every two-fold screw into a two-fold rotation about a parallel line, and every two-fold rotation into a screw. I222 has two-fold screws for exactly the same reason, since its symbol names the rotations and not the screws that come with them. The symbol I2₁2₁2₁ names the screws of a group that has rotations too, and I222 names the rotations of a group that has screws too. The two symbols list the same kinds of operation, and the difference between the groups is somewhere else.

The axes that never meet #

It is in where the rotation axes lie.

The picture at the head of this essay is that difference. In I222 the pure two-fold axes along $a$ sit at heights nought and a half, the axes along $b$ at heights nought and a half, and the axes along $c$ pass through the corners and edge-centres of the cell, so axes of all three directions pass through the origin together. Put the origin there and every operation of the point group goes through it with no translation. That is the definition of a symmorphic group.

In I2₁2₁2₁ the axes along $a$ sit at heights a quarter and three quarters, the axes along $b$ at nought and a half, and the axes along $c$ pass through the quarter-points of the cell’s face. Two axes of different directions meet only if they share the coordinate along the third direction, and in every pair the census checks, the two coordinates are a quarter out of step. No two rotation axes of I2₁2₁2₁ meet anywhere in the crystal. There is therefore no point that every operation can be made to pass through, and so no origin that makes the group symmorphic, although every operation passes through plenty of points on its own. The failure is not in any one operation. It is in how the operations sit relative to each other, and that is exactly the part of a cohomology class that restriction to one cyclic subgroup cannot see.

Why a quarter, and why the centring #

The distance between the axes is not arbitrary, and a short piece of geometry fixes it. Two half-turns about perpendicular lines compose to a half-turn about the line perpendicular to both — when the lines meet. When they are a distance $d$ apart along their common perpendicular, the composite is a half-turn about that perpendicular direction followed by a translation of $2d$ along it, which is a screw that climbs $2d$. The product of the two-folds along $a$ and $b$ is always the two-fold along $c$, so how far apart the $a$ and $b$ axes sit decides how far the $c$ operation climbs.

In I222 the axes meet, $d = 0$, and the product is a pure rotation. In I2₁2₁2₁ they sit a quarter of the cell apart along $c$, so the product climbs half a cell: a two-fold screw along $c$. A primitive lattice would make that screw an intrinsic translation of the group, and the group would be P2₁2₁2₁’s relative with an operation that reveals it. The body-centred lattice already contains a translation with half a cell along $c$, so the screw composed with the centring is a pure rotation about a parallel line. The skew of the axes is absorbed by the centring one product at a time, and never absorbed all at once. Each pair of directions satisfies the relation with its own absorbing centring translation, and no single origin satisfies all three pairs, because each pair’s axes are a quarter out of step in a different coordinate.

A quarter is also the only skew that works. A distance of a third or a sixth would give a climb of two thirds or a third, which no lattice vector of a body-centred cell cancels, and the operations would compose to a screw the group could not absorb. So the arithmetic permits exactly two arrangements on this lattice, axes meeting and axes a quarter apart. That is the $\mathbb{Z}_2$ the Smith form reports for the class.

The cubic case is the same construction with a three-fold axis added.

I23 contains I222 and I2₁3 contains I2₁2₁2₁, each with the three-fold rotations along the body diagonals added. The three-folds themselves reveal nothing: a three-fold’s cohomology on this lattice is trivial, and the census finds no class of the cubic body-centred 23 revealed by any three-fold operation. The two-fold axes do what they did in the orthorhombic group, crossing in I23 and never meeting in I2₁3. That the census finds exactly these two classes and no others, among a few hundred non-zero classes on sixty-one arithmetic classes, says that the combination is rare. Both have a body-centred lattice, whose centring vector has a half along every axis at once, and a point group with no mirror and no four-fold operation. Every other body-centred class has one or the other, and in every one of them the census finds each non-zero class revealed by some operation.

No extinction can tell them apart #

The consequence a crystallographer meets first is in the diffraction pattern. Systematic absences are the reflections a group extinguishes for every arrangement of atoms, and each one comes from a single operation with an intrinsic translation. A screw along $c$ extinguishes the odd reflections along $c^*$; a glide extinguishes a zone. The absences read off intrinsic translations one operation at a time, which is exactly the information restriction to cyclic subgroups keeps.

So a class that no single operation reveals cannot produce an absence of its own. Computed from the groups themselves, by summing the structure factor of a general orbit at two unrelated positions and keeping the reflections where both sums vanish, I222 and I2₁2₁2₁ extinguish the same reflections: the half of them with $h + k + l$ odd, which the body centring alone extinguishes, and nothing else. I23 and I2₁3 do the same. The International Tables list both pairs among the space groups that systematic absences cannot distinguish, and there the fact is recorded rather than explained.

The census runs the same comparison on every non-zero class of every arithmetic class of space. Each class is set against the symmorphic group on its own arithmetic class, and the two sets of absent reflections with indices up to three are compared. Every other non-zero class changes the absences: it extinguishes reflections the symmorphic group keeps, because some operation carries an intrinsic translation and extinguishes a row or a zone of reflections. The two that change nothing are the two no operation reveals. In space, being invisible to every single operation and being invisible to every extinction rule turn out to be the same condition. The first is a statement about cohomology and the second about structure factors, and the census finds that they pick out the same two groups. The direction from the first to the second is the argument just given, since an extinction needs an intrinsic translation. The converse is what the comparison adds: every class some operation reveals leaves a trace in the absences, so a revealing operation never has its intrinsic translation cancelled out of the diffraction pattern by the rest of the group. The explanation is the one above: the groups differ by a cohomology class that every cyclic subgroup kills, and an extinction rule is a statement about cyclic subgroups.

That makes these two pairs different in kind from the other ambiguities of space-group determination. Most groups sharing absence conditions share them because they differ in something the absences never see, such as the presence of a centre of symmetry or a mirror with no glide component, which statistics or anomalous scattering can then detect. I222 and I2₁2₁2₁ differ in something the absences cannot see even in principle, and they have the same point group, the same lattice and the same kinds of operation. What separates them in a structure determination is whether the refined atoms sit in one arrangement of axes or the other, which is a question about the whole structure rather than about any reflection.

The plane has nothing of the kind #

In the plane the textbook picture is exactly right. The rectangular mirror class, the rectangular 2mm class and the square 4mm class carry five non-zero classes between them, and a reflection reveals every one as a glide. So each of the four non-symmorphic plane groups, pg, pmg, pgg and p4g, has a glide in it, and the glide is the whole reason it is non-symmorphic. The plane cannot build the space examples, because it has no body-centred lattice with a half along two independent axes at once. Its centred rectangular lattice has a centring vector with a half along both edges, but the only point groups on it, m and 2mm, have trivial cohomology there.

What the census has to refuse #

The census depends on two computations agreeing: the Smith forms, which say which classes exist, and the restriction test, which says which operations reveal each. A mistake in either would show up somewhere the result can be checked.

The totals are the strongest check available. The same representatives and the same class keys, taken up to the normaliser, give 219 space-group types and 230 with enantiomorphs kept apart. That would fail if the key confused two classes or split one. The axes test is geometric and independent of the cohomology: it locates every pure two-fold axis of the two pairs of groups in the conventional cell and asks whether any two of different directions meet. The refusal is the sentence from the earlier essay. It survives as a statement about denominators, since every exponent in space divides the order of an operation, and fails as a statement about classes, in two groups.

Still open: which dimension first, and how often #

The plane has no class that every cyclic subgroup kills, and space has two, in two arithmetic classes out of seventy-three. Whether the phenomenon grows with dimension, becoming common in four dimensions, where body-centred lattices are more varied and point groups larger, is a count nobody here has made. The census method runs unchanged on any finite group of integer matrices, and the four-dimensional classes are the obvious place to point it.

There is also a question about the diffraction consequence. The argument shows that a hidden class produces no absence. It does not show that two groups with identical absences always differ by a hidden class, and they do not, since a centre of symmetry changes no absence. The finer statement would sort every pair of space groups with the same point group, lattice and absences by why they coincide, and it would say which of the ambiguities in space-group determination are statistical and which are structural. It has not been done here.