Seventy-three Smith forms

The denominator a group actually needs replaced the bound every account of the subject quotes, the order of the point group, with the number that actually decides how fine an intrinsic translation can be: the exponent of the cohomology group. It computed that exponent where a two-line formula applies, for a point group that is cyclic, and stopped at the classes where no formula does. The rectangular 2mm class of the plane has cohomology $\mathbb{Z}_2 \times \mathbb{Z}_2$, which was known from the plane’s extension census rather than from any formula. In space the largest exponent was “presumably six”, and the essay said plainly that “presumably” was doing work: a class whose point group holds operations of orders two, three and four might in principle need twelfths, and nothing there ruled it out.

The obstacle was never the arithmetic. It was the lack of a way to write down the cohomology group of a point group with two or three generators without enumerating a grid of candidate translations, and in space that grid is too large for the classes that matter. There is such a way, and it needs one integer matrix per class. Its Smith normal form puts the cohomology group on the diagonal, whatever the group, in whatever dimension. Run on the seventy-three arithmetic classes of space it settles the open question and shows that the finest translation in any space group is a sixth. A cubic group of order forty-eight, which the bound allows forty-eighths, needs nothing finer than halves.

A cocycle is decided on the generators #

A space group attaches a translation to each operation of its point group. Write $t(g)$ for the translation attached to $g$, taken modulo the lattice. The attachment is consistent exactly when it obeys the cocycle law

$$t(gs) \equiv t(g) + g\,t(s) \pmod{\mathbb{Z}^n},$$

which says that doing $s$ and then $g$ moves the origin by what $s$ moved it, carried by $g$, plus what $g$ moves it. Two attachments give the same group, up to where the origin was put, when they differ by a coboundary $t(g) = (g - I)v$. The cocycles modulo the coboundaries form the cohomology group, and its elements are the ways of building a space group on one arithmetic class.

The cocycle law means $t$ is fixed by its values on a set of generators. From the translations $x$ attached to the generators, $t$ can be carried to any element along any path in the Cayley graph, the graph whose vertices are the point group’s elements and whose edges are multiplication by a generator. What can go wrong is that two paths to the same element give different answers, and that is the whole of the condition a set of generator translations must satisfy.

The picture makes the recipe concrete. Choose a spanning tree of the Cayley graph rooted at the identity. Along the tree, each element’s translation becomes a definite integer combination of the generators’ translations. Every edge not in the tree closes a cycle, and the cycle’s condition is that the two combinations meeting at its end agree modulo whole lattice vectors. That is $n$ rows of integers, one for each coordinate. The cycles of a graph are generated by its non-tree edges, so these rows are a complete set of conditions: a choice of generator translations satisfying them extends to a consistent attachment on the whole group, and one violating any of them does not.

For the rectangular 2mm class the generators chosen are a mirror and the half-turn. The graph has eight edges, the tree uses three, and the other five give ten rows, of which four are not zero. Two of them say that the mirror’s translation, doubled along the line the mirror keeps, is a lattice vector, which is the familiar statement that a glide slides by half of something, arrived at as a matrix. The other two tie the mirror’s translation across its line to the half-turn’s, because the mirror times the half-turn is the second mirror and its translation is fixed by theirs.

The group is the torsion of a cokernel #

Stack the rows into a matrix $A$ with integer entries. The cocycles are the generator translations $x$ with $Ax$ an integer vector. The coboundaries include every real solution of $Ax = 0$, since a finite group’s cohomology with real coefficients vanishes, and every integer $x$, since a translation by a lattice vector changes nothing. Dividing one by the other leaves

$$H^1 \;\cong\; \big(\operatorname{im}_{\mathbb{R}} A \,\cap\, \mathbb{Z}^N\big) \big/ A\,\mathbb{Z}^m ,$$

which is the torsion of the cokernel of $A$. Bring $A$ to Smith normal form, a diagonal matrix obtained by integer row and column operations that change no lattice and lose nothing. The diagonal entries greater than one are then the invariant factors of the cohomology group. For the rectangular 2mm class the diagonal is $2, 2$, so the group is $\mathbb{Z}_2 \times \mathbb{Z}_2$. That is four ways of attaching translations, which become the three groups pmm, pmg and pgg once the change of basis swapping the two axes is taken into account.

Nothing about this uses a grid, a denominator bound or the fact that the plane is two-dimensional. The matrix has one column for each coordinate of each generator’s translation and one row for each coordinate of each cycle. A cubic class of order forty-eight with three generators gives nine columns and a few hundred rows, and its Smith form takes a fraction of a second. The column operations of the reduction also give one explicit representative translation for every class, which is what a figure or a further computation needs.

The plane is the check, and it passes class by class. The sizes of the thirteen groups agree with the extension census, which found them by searching translations on a grid of twelfths and removing coboundaries by brute force. The two computations share nothing but the cocycle law. Ten classes have trivial cohomology, the rectangular mirror class and the square 4mm class have $\mathbb{Z}_2$, and the rectangular 2mm class has $\mathbb{Z}_2 \times \mathbb{Z}_2$. The groups these carry add to seventeen, a count that belongs to the next question rather than to this one: how many groups a class carries is the number of its cohomology classes up to the symmetries of the class, and that needs the normaliser as well as the cohomology.

The number of rows is itself informative. The hexagonal 6mm class, of order twelve, imposes twenty-four rows, and its Smith form has no entry greater than one. The group is trivial, and so the most symmetric class of the plane admits no intrinsic translation at all. The classes generated by a single rotation impose none, because a rotation that fills the plane leaves nothing for its translation to be.

What the diagonal hands back #

The Smith reduction gives more than the shape of the group. Its column operations record which combination of the generators’ translations each diagonal entry refers to, and running them backwards produces one explicit translation for every class. The group is then a list of space groups rather than a list of numbers.

For the tetragonal class P4 the generator is the four-fold rotation, the Cayley graph is a cycle of four, and there is one condition: the four-fold’s translation, added to itself as the rotation carries it round, must come back to a lattice vector after four steps. The Smith diagonal is the single entry four, and the four representatives it hands back are a translation of nought, a quarter, a half and three quarters of the cell along the axis. Those are P4, P4₁, P4₂ and P4₃, read off the diagonal in order, with the screws a climb of so many quarters per turn. The hexagonal class P6 gives the single entry six and translations in sixths along its axis, the six groups from P6 to P6₅.

The square 4mm class of the plane is the more instructive case, because it shows a condition turning into a unit. Its point group contains the rectangular 2mm group, and the rectangular group’s cohomology has two independent halves: each of its two perpendicular mirrors may or may not glide. The square group adds a four-fold rotation carrying one mirror onto the other, and so it adds cycles to the Cayley graph that say the two mirrors must glide together or not at all. In the Smith form one of the two entries that were two becomes one. What is left is a single $\mathbb{Z}_2$, and its non-zero representative attaches half a cell to the four-fold rotation and half a cell to the mirror at once. That is p4g, whose glides and whose four-fold centres off the mirrors are one choice rather than two.

The reduction therefore does in a line what the plane’s census did by trying every translation on a grid of twelfths: it names the groups a class carries by their translations. The difference is that the reduction also runs in a class with three generators and forty-eight elements, where a grid would hold millions of candidates and the reduction handles a few hundred rows.

Every class of space #

The same computation runs on the seventy-three arithmetic classes that seventy-three, without a search derived from the fourteen Bravais groups. Each arrives as a list of integer matrices on a primitive basis of its lattice, which is all the computation needs.

The picture at the head of this essay is that census. Twelve classes have trivial cohomology and carry only their symmorphic group: P1, P1̄, C2, F222, P4̄, I4̄, P3̄, P6̄, R3̄, R32, R3 and F23. Fifty-three have exponent two, all sums of copies of $\mathbb{Z}_2$. They run from a single copy, as for the body-centred cubic holohedry, to six copies for the orthorhombic P holohedry, whose sixty-four classes are the sixty-four ways of choosing, for each of its three mirrors, whether it glides by half a cell along each of the two directions in its plane. Three have exponent three: P3, P321 and P312, the trigonal classes whose three-fold axis may be a screw. Three have exponent four: P4, P422 and P432. Two have exponent six: P6 and P622.

One shape in the table deserves a second look. P422 has cohomology $\mathbb{Z}_2 \times \mathbb{Z}_4$, not $\mathbb{Z}_4$ alone, and the two factors are two independent choices. The four-fold axis may climb by a quarter, a half or three quarters, which is the $\mathbb{Z}_4$. Independently of that, the two-fold axes across it may or may not become two-fold screws, which is the $\mathbb{Z}_2$. The eight elements of the product are the eight groups P422, P4₁22, P4₂22, P4₃22 and their P42₁2 counterparts. That count comes from the normaliser in the census of the distribution. What the Smith form supplies is the structure: two generators, of orders four and two.

Sixths, and never twelfths #

The question the previous essay left open can now be read off the census.

The largest exponent in space is six, reached by P6 and P622 and by nothing else, so the finest translation in any of the two hundred and thirty is a sixth. The annihilation bound allowed each class the order of its point group. The largest point group is forty-eight, and the primitive, body-centred and face-centred cubic holohedries of that order have exponent two. Their intrinsic translations are halves: the $n$-glides and $d$-type translations of Pn3̄n and its relatives. Even the quarter-cell $d$-glides of Fd3̄m are halves of a vector of the face-centred lattice. The class with the largest group needs the coarsest translations.

The worry that stopped the previous essay was that operations of orders two, three and four might combine to require twelfths without any single operation having order twelve. They do not. In every one of the seventy-three classes the exponent divides the order of some single operation of the class. So every denominator a space group needs is already the order of one of its operations, and a class needs sixths only when it has a six-fold axis.

That is weaker than it sounds, and the difference is worth stating before it is misremembered. The census shows that the exponent is always available from one operation. It does not show that every non-zero class is carried by one operation. A class of translations could need only halves and still be invisible to every single operation of the group, if the operations individually could each be written without any intrinsic translation while the group as a whole could not. The previous essay relied on the opposite as “a standard fact”: that restricting to cyclic subgroups is injective on the part of the cohomology that matters. It is not quite true in space, and two groups no single operation reveals finds the two classes where it fails.

Why the order is the wrong number #

The pattern in the exponent figure has a short explanation, and it is the one the cyclic case already gave. A class of cohomology is built from the translations of individual operations, and each operation contributes only along the directions it fixes. A rotation of order $k$ that fixes a direction contributes a $\mathbb{Z}_k$ along its axis; a reflection contributes a $\mathbb{Z}_2$ in its plane; an operation that fixes nothing contributes nothing. Adding operations to a point group cannot make a finer denominator appear. It can only impose more conditions, and each extra condition can only make the group smaller. That is why the orthorhombic P holohedry, whose three mirrors and three axes each offer a half independently, reaches $(\mathbb{Z}_2)^6$, while the cubic holohedry, whose three-fold axes tie those choices together, falls back to $(\mathbb{Z}_2)^2$ or $\mathbb{Z}_2$.

So the bound fails by its whole size in exactly the classes a reader would expect it to be tight. It counts elements, and elements in a larger group constrain one another. A group of order forty-eight has forty-eight operations and, through the cycles of its Cayley graph, a few hundred integer conditions among their translations. The Smith form is the bookkeeping of those conditions, and what survives them is a sum of copies of $\mathbb{Z}_2$.

What the census has to refuse #

The plane census is the external check: thirteen classes, sizes agreeing with an independent computation class by class. The space census has no second computation of its own groups here, but it has a count downstream: the distribution of space groups uses these representatives and arrives at 219 and 230, which would not happen if a group were wrong.

The first refusal tests the class key the census uses to tell cohomology classes apart. A cocycle moved by a change of origin, which adds $(g - I)v$ to every translation, must get the same key as before. A key that distinguished them would count every group once for each origin, which is exactly the error the coboundaries exist to remove. The second refusal is the error the whole argument since the annihilation bound has been about: the order of the point group offered as the denominator a class needs. The order-forty-eight classes refuse it, since their exponent is two.

Two conventions sit under the numbers and should be said plainly. The cohomology is taken on a primitive basis of each lattice, so a centring translation is a lattice vector and costs nothing. Taken on the conventional centred cell, the same groups would show extra halves that are centring vectors, and the exponents would look larger than they are. And the groups counted here are extension classes, not space-group types. Two classes related by an automorphism of the lattice that normalises the point group give one type, which is why $(\mathbb{Z}_2)^6$ for Pmmm is sixty-four classes and only sixteen space groups.

Still open: the normaliser, and what one operation can see #

The census gives the group for every class. Two questions are left, and each needs a computation of its own.

How the sixty-four become sixteen. The normaliser of the point group in $GL(3, \mathbb{Z})$ permutes the cohomology classes, and the space-group types are its orbits. Counting them for every class needs the normaliser acting on explicit representatives, which the column operations of the Smith reduction supply. The totals should be 219 and, with only the proper part of the normaliser, 230. Finitely many is not few asked what the distribution beneath that average looks like, and the orbits answer it.

What a single operation can see. Every exponent is the order of an operation. Whether every non-zero class is carried by one is a statement about restriction to cyclic subgroups, and it is a different statement. A group that is non-symmorphic although no one of its operations is a screw or a glide would be a group whose intrinsic translation lives in the way its operations sit relative to each other rather than in any one of them.