The offset a quarter turn allows lies on a mirror

A band-structure calculation samples the zone of wavevectors on a finite grid, and the cheapest grid is one that keeps the crystal’s whole point group and puts no wavevector on any mirror, rotation axis or point where time reversal folds the zone onto itself. Such a grid is called free here, and on a free grid every star of wavevectors has full length, so the calculation reaches the irreducible wedge’s share of the zone exactly. A three-fold axis puts a grid point on itself ran the census for every Laue group on every lattice it occurs on. Four groups had free grids, as small as the group itself: the triclinic, the monoclinic, the orthorhombic, and the tetragonal group $4/m$. Every group with a three-fold axis had none, and the essay proved why. One case was left as a census and nothing more: the tetragonal holohedry $4/mmm$, which had no free grid on either tetragonal lattice to a hundred and twenty-eight points, while its subgroup $4/m$ had one at eight.

That essay put the question in the right place. The four-fold rotation alone can be dodged, the mirrors alone can be dodged, and so the obstruction, if there is one, must use the two together. It suggested two possible shapes for it: a count of fixed points modulo some prime, or the diagonal mirror lines on the torus, rational lines that every square-symmetric grid might be forced to meet. The second guess is the right one, and the reason is shorter than either guess suggested. The quarter turn decides where the grid’s offset may sit, and that one place is always on a mirror.

Where a quarter turn lets a grid sit #

The plane is enough to see the whole argument, and space adds only one step, so the plane comes first. The square reciprocal lattice has the point group $4mm$: four rotations about the zone centre and four mirrors, two along the axes and two along the diagonals. A grid of wavevectors is a finite set in the torus of reciprocal space, the zone with opposite edges glued. The grids a calculation uses are the wavevectors a periodic box holds, as crystallography in a box set out, possibly all moved by one fixed offset. So a grid is a coset: an offset $w$ plus a lattice $\Lambda$ of wavevectors, where $\Lambda$ contains the reciprocal lattice and is finer than it.

For the grid to keep the four-fold rotation $r$, two things are needed. The lattice $\Lambda$ must be carried to itself by the quarter turn, and the rotated offset must differ from the offset by a vector of $\Lambda$. The second condition says that $w$, read on the torus $\mathbb{R}^2/\Lambda$, is a point the quarter turn fixes. A rotation fixes a point of a torus when $r w - w$ is a lattice vector, and the number of such points is the absolute value of the determinant of $I - r$, a close relative of the trace that the crystallographic restriction turns into its list of allowed orders. For a quarter turn that determinant is two. So on any torus the quarter turn fixes exactly two points, the origin and one other, and every grid that keeps the four-fold rotation has its offset at one of them.

The figure in the next section draws both. The origin is the zone centre, and an offset there puts a grid point on the rotation axis itself, which ends the question at once: the four-fold rotation fixes it. The other fixed point is the second four-fold centre of the grid’s own lattice, the point at the middle of each of its square cells when the cell is drawn with lattice points at the corners. That is the only place an offset that keeps the quarter turn can avoid the axis, and the only grids the question is about are the ones offset there.

The census before this one had already found the consequence without naming it. The free grid of $4/m$ at eight points is offset by a quarter of the reciprocal cell along both axes across the four-fold, and a quarter is exactly the middle of a cell of the half-spaced lattice that grid is built on. No other offset was ever going to work.

Two orientations, and the prime that ramifies #

The next question is which lattices $\Lambda$ the whole square group keeps, mirrors included. The rotations alone allow many: any lattice that a quarter turn carries to itself. Writing the reciprocal plane as the complex numbers, with the zone centre at nought, a lattice the quarter turn keeps is a set closed under multiplication by $i$, which makes it a module over the Gaussian integers, the numbers $a + bi$ with $a$ and $b$ whole. Every such module is a single multiple $\gamma \mathbb{Z}[i]$ of the Gaussian integers, because they have unique factorisation, and counting them by index is the plane’s version of how many sublattices a lattice has, cut down to the square ones. The sublattices that stay square used exactly that fact to decide which indices a square sublattice can have.

A mirror adds a condition. The mirror across the real axis is complex conjugation, so the lattice must also equal $\bar\gamma \mathbb{Z}[i]$, which means $\bar\gamma$ is $\gamma$ times a unit. Factor $\gamma$ into Gaussian primes and the condition sorts them. A prime that lies over an ordinary prime congruent to one modulo four, such as $2 + i$ over five, has a conjugate $2 - i$ that is a different prime, so the two must appear equally often, and together they make the ordinary prime five. A prime congruent to three modulo four stays prime and is its own conjugate. And the prime two is the odd one out: it is $-i(1+i)^2$, the square of a single Gaussian prime up to a unit, so $1 + i$ is its own conjugate up to a unit and may appear once without a partner. Two is the only ordinary prime that ramifies in the Gaussian integers.

So a lattice the square group keeps is $\gamma \mathbb{Z}[i]$ with $\gamma$ either a real number or a real number times $1 + i$. The first is a square lattice aligned with the mirrors, its rows along the axes. The second is a square lattice turned by half a right angle, its rows along the diagonals. There is no third orientation, and the census checks it: of the thirteen square-symmetric boxes up to sixty-four points, and of every one up to four hundred, each is aligned with $k^2$ points or turned with $2k^2$ points, and none is anything else.

Now the second four-fold centre can be written down. The middle of the cell of $\gamma\mathbb{Z}[i]$, spanned by $\gamma$ and $i\gamma$, is $\gamma(1+i)/2$. If $\gamma$ is real, that point is a real multiple of $1 + i$: it lies on the diagonal through the origin. If $\gamma$ is a real multiple of $1 + i$, the point is a real multiple of $(1+i)^2 = 2i$: it lies on the vertical axis through the origin. The figure shows both cases. The ringed point sits on a mirror line, exactly, not modulo anything.

Every grid is caught #

The argument is now complete. Take any grid that keeps $4mm$. Either its offset is at the zone centre, and the grid point there sits on the four-fold axis, or its offset is at the second four-fold centre $\gamma(1+i)/2$, which lies on a mirror line through the zone centre. A point on a mirror line is fixed by that mirror. So every grid that keeps the square group, of any size and any shape, has a point fixed by some operation other than the identity, and no grid of $4mm$ is free. The proof uses nothing about the size of the grid, so the census’s limit of a hundred and twenty-eight points stops mattering.

The figure at the head of this essay draws the three ways the catch happens. When the grid is aligned and its side is odd, the offset is a half-integer multiple of a lattice step that happens to land a grid point on a rotation centre of the reciprocal lattice, and the rotation catches it before any mirror is needed. When the side is even, no rotation fixes anything, which is precisely why the group $4$ alone has free grids, and the diagonal mirror catches the column of points along the diagonal. When the grid is turned, the offset lies on the vertical axis and the axial mirror catches it.

The census sorts all thirty-four grids up to four hundred points into those three outcomes, and they fall exactly where the argument puts them.

The pattern in the table is sharper than the theorem needs. On aligned grids of even side, only the diagonal mirrors fix anything; the axial ones fix nothing, because the grid’s coordinates are odd multiples of half a step and never land on a cell edge. On turned grids of even $k$ the opposite holds, and only the axial mirrors fix anything. Each orientation has its own class of mirror. That is why neither class alone is an obstruction. The rectangular group $2mm$, which has only the axial mirrors, has free grids, and they are aligned ones. A square group with only the diagonal mirrors would have free grids too, the turned ones. The square group with its four-fold rotation has both classes, and the rotation is what removes every other choice of offset, leaving the two orientations, each of which meets one class.

The same argument in space #

A crystal is three-dimensional, and the census that left the question open was about the tetragonal lattices in space. The argument lifts with one extra step.

Project the grid onto the plane across the four-fold axis. The projection of a coset is a coset, the projected lattice is one the square group keeps, and the four-fold rotation still carries the projected grid to itself, so the projected offset is again either on the axis or at the second four-fold centre of the projected lattice. Some point of the grid projects onto that place exactly, because projection maps the coset onto the projected coset. If it is the axis, that grid point lies on the four-fold axis in space. If it is the second centre, that grid point lies in the vertical mirror plane through the diagonal or through an axis, because a vertical mirror plane is exactly the set of points whose projection lies on its line. A vertical mirror fixes every point of its own plane. Either way a grid point is fixed.

Two features of the space version are worth stating. The step uses the vertical mirrors only. The horizontal mirror and the inversion play no part, and the census confirms the sharper statement: of the 1,456 grids $4/mmm$ keeps on the primitive tetragonal lattice to a hundred and twenty-eight points, and the 412 on the body-centred one to sixty-four, a vertical mirror alone fixes a point of every one. And the projection works on the body-centred lattice as well as the primitive one, although its reciprocal lattice is face-centred and its layers across the axis are staggered. The argument never measures anything modulo the reciprocal lattice. It finds a point lying exactly in a mirror plane, and a point lying exactly in a mirror plane is fixed by that mirror whatever the lattice is.

The free grid that the mirrors end #

A proof that rules out too much is wrong, and this one has a test case built in. The group $4/m$ has the four-fold axis and the inversion and the horizontal mirror, and lacks the vertical mirrors. The argument needs a vertical mirror at its last step, so it says nothing about $4/m$, and $4/m$ does have a free grid: eight wavevectors, offset by a quarter along every axis, two layers along the four-fold of four points each. The census finds it again here, and the figure draws it.

Drawn down the axis, the eight points of that grid sit at the four points a quarter of the cell from the centre along each diagonal. Those are exactly the second four-fold centres the argument names, and every one of them lies on a diagonal line. In $4/m$ there is no mirror on that line, so nothing fixes them. In $4/mmm$ there is, and all eight are fixed at once. The free grid of $4/m$ is free because of where its offset sits, and the same position is what destroys it when the mirrors arrive. The two outcomes of the earlier census, free for one group and never free for the other, are one fact about one point.

What the argument does not use #

Three things the earlier census assumed turn out to be unnecessary, and saying which is part of the result.

The inversion is not used. The census included time reversal, which adds the inversion to every group a band calculation sees for the reason the symmetry diffraction adds is an inversion, and it was the inversion that forced every symmetric offset onto a half-step of the grid, which is what made the search finite. The argument here never mentions it. So the statement holds for the polar group $4mm$ in space as well as for $4/mmm$. A magnetic crystal or a calculation with spin–orbit coupling in a field can have a group with no inversion, and the essay that left this open noted that its own three-fold argument needed the inversion through the rotoinversion. The four-fold argument does not, so for a grid that keeps a four-fold axis and a mirror containing it, nothing about time reversal matters.

The size of the grid is not used. The argument is about the torus and its fixed points, and it is the same at eight points and at eight million.

The metric is not used beyond the square symmetry. The tetragonal lattice’s two cell lengths, along the axis and across it, can be anything, and the reciprocal lattice’s shape along the axis never enters, because the projection discards it. What the argument does use is the arithmetic of the plane across the axis: that the square group’s lattices are the ideals of the Gaussian integers that conjugation keeps, and that those come in only two orientations. The obstruction is the ramified prime two, the one place where the Gaussian integers allow an ideal that is its own mirror image without being an ordinary integer. If there were a third orientation, an ideal kept by conjugation with its second centre off every mirror line, a free grid of $4mm$ could exist. There is not, because two is the only prime that ramifies in $\mathbb{Z}[i]$.

Compared with the three-fold proof #

The three-fold argument and this one are worth reading side by side, because they look alike and work differently.

The three-fold argument used the rotoinversion $\bar3$, which acts across its axis as a sixth of a turn, with the determinant of $I$ minus the operation equal to one. A determinant of one means a single fixed point on the torus whatever offset the grid has, so the operation pins a column of the grid by itself, and squaring it turns that column into points on a three-fold axis. No mirror was needed and none was used. That is why every group with a three-fold axis fails, with or without mirrors: $\bar3$ alone is enough.

The quarter turn has determinant two. It does not pin the grid; it leaves two places, and the grid may sit at either. One of them, the axis, is fatal. The other is not fatal to the rotation, which is why $4$ and $4/m$ go free, and it becomes fatal only because a mirror passes through it. The half-turn, with determinant four, leaves four places, and for the groups with only two-fold axes and mirrors there is always a place that no element passes through, which is why $2/m$ and $mmm$ go free. So the census’s division, four free groups and seven that are never free, is a division by how tightly each group’s rotations pin an offset and whether a mirror passes through the places they leave.

That gives a short account of the whole table. A group is never free when its rotations leave no offset off a symmetry element. A sixth of a turn leaves none at all. A quarter turn leaves one, which a vertical mirror covers. A half-turn leaves several, and the groups with only half-turns cannot cover them all.

What the pictures cannot show #

The figures draw the plane, and one grid in space seen two ways. They cannot draw the general statement, which is about every coset of every lattice the square group keeps, and they cannot show the step that makes it general: that the Gaussian integers have unique factorisation, so that every such lattice is a single multiple of them. That is a fact about arithmetic the pictures take on trust. The census checks its consequence, that only two orientations of box occur, on every box to four hundred points in the plane, and the space census checks the conclusion directly on the 1,868 grids it reaches. Beyond those sizes the statement rests on the proof, which is short enough to check by hand.

The figures also draw grid points as dots on a mirror line, and a dot can look as if it lies on a line when it is a pixel away. The census does not look: a point is on a mirror when the mirror sends it to itself modulo the reciprocal lattice, in integer arithmetic on coordinates measured in units of half a grid step. The one place the check uses something stronger is the witness, the point reported lying exactly on a line through the origin before any reduction, and that is checked as an equality of integers too.

Who did what #

The arithmetic of the Gaussian integers is Gauss’s, from 1832, and the fact that two is the only ramified prime is as old. The use of offset grids for sampling the zone is Monkhorst and Pack’s, from 1976, after Baldereschi, Chadi and Cohen, and that offset grids can lose symmetry on some lattices is working knowledge in the field. The domain in reciprocal space asked whether an offset grid can avoid the special points entirely, and the census before this one answered it for every Laue class except one. The argument here closes that one, and the census of grids it reports was run for this essay.

What the census must refuse #

The refused claim is the one a careless version of the argument would make: that the four-fold axis by itself prevents a free grid. It does not. The group $4/m$ has a free grid of eight points, found by the census and drawn above, and the argument correctly says nothing about it, because its last step needs a vertical mirror. A proof of the result for $4/mmm$ that also excluded $4/m$ would be a wrong proof, and the refusal is there to show that this one is not.

Still open: groups that turn the axis over #

With $4/mmm$ settled, every Laue group’s answer is a theorem rather than a census, and the earlier essay’s table of free and never-free groups is complete for grids that keep the whole group with time reversal included. Two questions remain that the argument here does not reach. One is the magnetic groups, where an operation can combine a spatial symmetry with time reversal: a primed mirror reverses time as well as reflecting, and whether a point it fixes counts as special depends on what is being sampled, so the question of a free grid changes meaning rather than answer. The other is the grids a calculation uses when it gives up part of the symmetry on purpose, keeping only a subgroup so that a free grid exists, and paying for it in a larger wedge. Which subgroup of $4/mmm$ gives the cheapest free grid, and whether it is always $4/m$, is a finite comparison that has not been made.