A three-fold axis puts a grid point on itself

The domain in reciprocal space treated the irreducible wedge of the Brillouin zone as a fundamental domain, the piece of reciprocal space a point group repeats, and measured its share on grids of wavevectors through the zone centre. The share always came out above one over the order of the group. A wavevector on a mirror or an axis is carried to itself by some operation, so its star is short, and a grid with points on symmetry elements has more stars than its size divided by the group’s order.

That essay said two things about the excess that pull against each other. In one place it observed that a calculation choosing its own grid can offset it from the zone centre, and that an offset grid can avoid the special points entirely, which makes every star full length. In another it said that the count’s excess over the area is something nothing a finite calculation does reaches. If an offset grid avoids every special point, a finite calculation reaches the area exactly. If no offset grid can, the first sentence is wrong. Neither was checked, and in three dimensions the answer turns out to depend on the group in a way the plane essay could not have guessed.

This essay checks both, for every Laue group on every lattice it occurs on. Four groups have a grid that avoids every special point, and for those the ideal share is reached exactly, at a grid as small as the group. The rest do not, and for every group with a three-fold axis the reason is a theorem rather than a failed search.

A grid, an offset, and what free means #

A grid of wavevectors is a finite set in the torus of reciprocal space, the zone with its opposite faces identified. The grids a calculation actually uses come in two kinds. A grid through the zone centre is a finite subgroup of that torus, and it is exactly the set of wavevectors a periodic box holds, the object crystallography in a box counted and a whole star needs a cubic box sized star by star. An offset grid is the same set moved by a fixed vector, the kind of sampling Monkhorst and Pack made standard.

A grid keeps a group when every operation carries the grid onto itself. For an offset grid that is a condition on the offset as well as on the box, and it is a finite one: because time reversal sends every wavevector to its negative, the group always contains the inversion, and an invariant offset must then be a half-step of the grid. So for a box of $n$ cells the offsets that could possibly work lie on a grid of spacing $1/2n$, and the search over them is exhaustive rather than sampled.

A grid is free when no point of it is fixed by any operation but the identity: no point on a mirror, on a rotation axis, or at one of the eight time-reversal-invariant momenta. A free grid’s points fall into stars of full length, so it has exactly its size divided by the group’s order in stars, the area’s share with no boundary correction. And a free grid’s size must be a multiple of the group’s order, since its stars partition it into pieces of that size. That last fact makes the search cheap, because only sizes that are multiples of the group’s order need trying.

The convention is the one a band-structure calculation without spin–orbit coupling uses: the symmetry of the grid is the Laue group, the point group with the inversion added, because time reversal makes the energy at a wavevector equal to the energy at its negative, the same reason the symmetry diffraction adds is an inversion. Each lattice is taken on its primitive basis and each Laue group as the integer matrices that represent it there, the seventy-three classes of two hundred and nineteen over seventy-three with the inversion adjoined, which gives twenty-four distinct pairs of a lattice and a Laue group.

Four groups go free #

The figure at the head of this essay is the census. Free grids exist for exactly four Laue groups: the triclinic $\bar1$, the monoclinic $2/m$, the orthorhombic $mmm$ on all four of its lattices, and the tetragonal $4/m$ on both of its. In every case the smallest free grid has exactly as many points as the group has operations, two, four, eight and eight, which is the least the divisibility allows. A grid of eight wavevectors in an orthorhombic crystal, offset by a quarter of the reciprocal cell along each axis, keeps all eight operations of $mmm$ and is a single star: every wavevector is carried to every other by some operation, and nothing is fixed.

For those four groups the second sentence of the earlier essay is simply false. The count’s share of the zone equals the area’s share exactly, on a grid of eight points, and stays exact on the free grids of every multiple of the order the search found: sixteen points with two stars, thirty-two with four. A calculation on a crystal of one of those groups can sample its zone with no point on a symmetry element and no star shorter than the group, at any size divisible by the group’s order.

For the other twenty pairs the first sentence is false. Every Laue group with a three-fold axis, $\bar3$, $\bar3m$, $6/m$, $6/mmm$, $m\bar3$ and $m\bar3m$, on every lattice that carries one, has no free grid at any size searched, and the tetragonal group with mirrors, $4/mmm$, has none either, on either lattice, to sixty-four points and, in a separate run, to a hundred and twenty-eight. No offset avoids the special points in the most symmetric crystals, which are the ones where the saving the wedge offers was largest.

The same question in the plane #

The plane shows the same division where it can be drawn. A rectangular lattice with its two mirrors, $2mm$, has a free grid of four points, offset by a quarter cell along both axes; the square lattice with only its four-fold rotations, the group $4$, has one too. Add the diagonal mirrors and make it $4mm$, and no symmetric grid to thirty-six points is free; the best still puts points on the diagonal mirrors. The hexagonal groups $6$ and $6mm$ have no free grid either, and the grids of $6$ always have a point on a six-fold centre.

So in both dimensions the same two things forbid a free grid: a rotation of order three or six, and a four-fold rotation with mirrors through it. Each has a different reason, and only the first is understood here.

The three-fold theorem #

Every Laue group with a three-fold axis contains the rotoinversion $\bar3$, the three-fold rotation followed by the inversion. Across the axis, on the two-dimensional torus of wavevectors perpendicular to it, $\bar3$ acts as a rotation by a sixth of a turn. A sixth of a turn has a special property among plane rotations: the determinant of the identity minus the rotation is one. So the equation for a fixed point of the sixth of a turn, with whatever translation the grid’s offset attaches to it, has exactly one solution on the torus. The census checks it on every translation of a grid of twelfths: always one fixed point, never zero and never more.

That fixed point lies under the grid. The grid projects onto the transverse torus as another symmetric grid, and the unique fixed point of an invertible-over-the-integers affine map that keeps a coset of a finite group must lie in that coset, since solving for it inside the coset has a unique solution there too. So there is always a column of grid points standing above the point $\bar3$ fixes.

Along the axis, $\bar3$ acts as a reflection, $y \mapsto c - y$, since the three-fold rotation does nothing along its own axis and the inversion reverses it. Done twice, that reflection does nothing at all. But $\bar3$ done twice is the three-fold rotation itself, as an operation of the whole space. So the three-fold rotation fixes every grid point in that column. Every grid that keeps a group with a three-fold axis has a point on the axis, whatever the box and whatever the offset, and no such grid is free. The census confirms the theorem directly: every one of the 70 invariant grids of $\bar3$ on the rhombohedral lattice to eighteen points has at least one point a three-fold rotation fixes.

The argument also says why four-fold and two-fold axes behave differently. A quarter of a turn has determinant two in the same place, and a half-turn four, so the corresponding equations have two or four solutions on the transverse torus, and an offset can be chosen to put none of them under the grid. Only a sixth of a turn pins its fixed point to the grid, and every Laue group with a three-fold axis contains one, because it contains the inversion and the inversion times the three-fold rotation is $\bar3$.

The four-fold with its mirrors #

The tetragonal group with mirrors is the case the theorem does not reach. In $4/m$ a free grid of eight points exists, offset by a quarter along every axis, and the four-fold rotations, mirrors across the axis and the inversion all miss it. Adding the vertical and diagonal mirrors of $4/mmm$ destroys every such grid the search can find: to a hundred and twenty-eight points on either tetragonal lattice, and in the plane, where $4mm$ is the corresponding group, to thirty-six.

A reason is easy to half-see. Projected onto the plane across the four-fold axis, a free grid of $4/mmm$ would have to be a free grid of $4mm$, since every operation of $4mm$ does nothing along the axis and a point fixed in the plane would be fixed in space. The plane search finds none. But why a square group with its mirrors cannot act freely on any symmetric set of plane wavevectors is something the search establishes only up to its limit; the determinant argument gives two fixed points for each four-fold rotation, enough freedom to dodge in principle, and the obstruction must come from the mirrors and the rotations together. No proof is offered here, and the statement for $4/mmm$ is a census, not a theorem.

How far from the ideal the rest stay #

For the groups that cannot go free, the useful question is how close a grid can come, and the census answers it size by size by taking every symmetric box and every invariant offset and keeping the grid with fewest stars. The tetragonal group with mirrors does best: at thirty-two and sixty-four points its best grids have one and a half times the ideal number of stars, half of their points on some mirror. The rhombohedral group with mirrors stays between about two and a half and four times the ideal, the hexagonal holohedry at four, and the cubic holohedry at three to eight on these sizes.

The hexagonal figures say something stronger than the excess. At twenty-four and at forty-eight points, the best grid of $6/mmm$ has every one of its points on some mirror or axis: there is no grid of those sizes with even one wavevector in general position. The same is true of the best cubic grids the census found. For those groups the trade-off the earlier essay described, between sampling the special points and avoiding them, does not exist at the sizes a first calculation uses: every symmetric grid samples them, and the only choice is which ones.

What the census must refuse #

The refused claim is the earlier essay’s sentence about offsets, in its general form: that an offset grid can always avoid the special points. It holds for four of the eleven Laue classes and fails for the other seven, among them every hexagonal, trigonal and cubic crystal.

What the pictures cannot show is the three-dimensional grid itself. The plane panels draw the same question where a cell is a square or a rhombus and every point is visible, and the theorem’s figure draws its argument in steps; a grid of forty-eight points in a three-dimensional zone, with the mirrors and axes it touches, is a picture in which the property that matters, whether a point lies exactly on a plane, cannot be read by eye. The census reads it from integer arithmetic instead: a point is fixed when an operation sends it to itself modulo the reciprocal lattice, exactly.

What this means for a calculation #

The practical sampling literature has long distinguished grids through the zone centre from offset grids, and has known that offset grids sometimes break a crystal’s symmetry, so that a hexagonal crystal is often sampled on a grid centred on the zone centre. The census turns that experience into a classification. An offset that keeps the group exists for every Laue group; an offset that also avoids every special point exists for exactly four of them, and for a crystal with a three-fold axis it never exists, whatever the offset and whatever the grid. The excess the earlier essay measured is therefore a fixed cost of symmetry in seven Laue classes and an avoidable one in four.

The theorem also tells a calculation where its unavoidable special points are. In a crystal with a three-fold axis, every symmetric grid has wavevectors on that axis, the line through the zone centre that the tables call $\Lambda$ or $\Delta$ according to the lattice. On a rhombohedral crystal that line ends at the point T, which the rhombohedral box census found held whole by a box of two cells. Those wavevectors have the largest little groups the grid contains along that line, and their bands are the ones in which symmetry forces degeneracies. A calculation that wanted to avoid them entirely would have to give up the symmetry of the grid, which the reduction to the wedge depends on.

Where the census stops #

The search runs to forty-eight points for most groups and further for the tetragonal ones, and for the groups with a three-fold axis it does not matter where it stops, because the theorem covers every size. For $4/mmm$ it matters, and the census is evidence rather than proof: a free grid of some large size divisible by sixteen is not excluded. The cubic groups are covered by the theorem, which is as well, because their invariant boxes are so few, as the three that stay cubic found, cubes and twice and four times cubes, that the first size divisible by forty-eight at which one exists is four hundred and thirty-two.

The census also assumed time reversal. A magnetic crystal, or a calculation with spin–orbit coupling and a magnetic field, has a smaller group with no inversion forced on it, and then offsets need not be half-steps and the counting changes. Whether a three-fold axis without the inversion still forces a grid point onto itself is a different question, since a third of a turn has determinant three where a sixth has one, and the argument above does not apply to it.

Monkhorst, Pack, and the points that cannot be dodged #

Special-point sampling began with Baldereschi in 1973 and with Chadi and Cohen in the same year, and Monkhorst and Pack’s grids of 1976 became the standard way of choosing wavevectors. That offset grids can lose symmetry on hexagonal lattices, and that grids through the zone centre are used there for that reason, is common working knowledge in the field. What the census adds is the exact boundary: which Laue groups admit a grid that keeps their symmetry and avoids every special point, the size of the smallest, and a one-line reason why no group with a three-fold axis ever does.

Still open: why the mirrors forbid what the four-fold allows #

The four-fold axis alone permits a free grid and the four-fold axis with its mirrors does not, in the plane and in space, as far as the search reaches. An explanation would have to be an obstruction that uses the mirrors and the rotation together, since each alone can be dodged: a counting of fixed points modulo some prime, perhaps, or a reduction to the diagonal mirrors’ lines on the torus, which are rational lines of slope one and minus one that every square-symmetric coset might be forced to meet. Whether a proof exists, or whether a free grid of $4/mmm$ waits at some large size, is the one case in the census that is still a question. It has not been settled here.