Where a gradient term survives at the edge of the zone

The cell a zone-boundary mode doubles froze order parameters that alternate from cell to cell, and found that each one halves the translations and doubles the cell. It ended on the condition that decides whether such a freezing can happen continuously at all. The free energy must contain no term linear in the gradient of the order parameter. A term of the form $\eta_i \partial_\alpha \eta_j - \eta_j \partial_\alpha \eta_i$ lowers the energy of a state in which the order parameter slowly rotates through space. When one is allowed, the uniform state at the chosen wavevector is not the minimum, and the crystal settles on a nearby wavevector that nothing pins to the lattice. That is an incommensurately modulated structure, with no larger cell to describe it.

The essay stated where this condition holds: at points of high symmetry in the zone it is satisfied, and along lines and at general points it fails. Which way the order parameter points had already computed it at the zone centre and found no Lifshitz term for any plane group. Nothing computed it anywhere else. This essay does, for every order parameter at every special point of all seventeen zones. The statement is right about lines and general points, and wrong about high symmetry. Ten order parameters at ten special points carry a Lifshitz term, and the pattern of where they sit is exact.

The count #

An order parameter at wavevector $k$ is a real, irreducible representation of the whole space group, as an order parameter is a representation set out. It carries every wavevector the point group moves $k$ to, its star, together with $-k$, because the order parameter is a real displacement or density and complex conjugation sends $k$ to $-k$. The translations that act trivially on it form a sublattice $T'$ of finite index, so the order parameter is a representation of the finite group $G/T'$. For a zone-edge point of a rectangular lattice that group has at most thirty-two elements, and for the hexagonal zone corner at most thirty-six.

A Lifshitz term is antisymmetric in the order parameter’s components and linear in the gradient. The components transform by the representation $E$, and the gradient transforms as a vector. So the terms allowed by symmetry are counted by how often the identity appears in the antisymmetric square of $E$ times the vector representation:

$$\frac{1}{|G/T'|}\sum_{h} \tfrac12\bigl(\chi_E(h)^2 - \chi_E(h^2)\bigr)\,\operatorname{tr} R_h .$$

The translations must leave the term alone too. At a point where $k$ and $-k$ agree modulo the reciprocal lattice, a term built from two copies of the order parameter carries wavevector $2k$, a whole reciprocal vector, and the translations act on it trivially. The sum over all of $G/T'$ checks that automatically.

The order parameters themselves are found without a character table. The group acting on itself from the right commutes with the group acting from the left, so a random real symmetric combination of right multiplications splits the regular representation into eigenspaces. Each eigenspace carries one real irreducible representation: either a real irrep, or a complex irrep together with its conjugate, which is exactly the pairing time reversal imposes, the same pairing that forces degeneracies in a glide group’s bands. The eigenspaces whose translations act with the characters of the star of $k$ are the order parameters at $k$. A completeness test checks that none is missed. Weighted by their dimension squared and divided by the type of each, one for real, two for complex and four for quaternionic, they must fill exactly the point group’s order times the star’s size. They do, at all ninety-two points tried.

The map #

Fifty-eight special points carry 149 order parameters between them. At the zone centre, every entry is zero, as the zone-centre essay found. At the edges and corners, 139 of the remaining order parameters are also zero. The ten that are not sit in two families.

Seven are at the zone edge of a glide group. pg has one at X and one at S. pmg has one at X and one at S. pgg has one at X and one at Y, and p4g one at X, where the star carries X and Y together. Every one is a two-component order parameter, or four-component for p4g’s two-armed star, and each carries exactly one invariant.

Three are at the corner K of a hexagonal zone, in p31m, p6 and p6m, groups that have no glide at all. p6’s corner carries two invariants for one order parameter, the others one each. The other two hexagonal groups, p3 and p3m1, have none at K. No hexagonal group has one at the edge midpoint M.

Everything else is zero. That includes every symmorphic group at every edge point, which is the case the zone-boundary essay actually froze.

One glide, and the pair it turns #

The glide groups show the mechanism most clearly, because in each case one operation does all the work.

In pg the glide reflects across a line along $x$ and then slides half a cell along it. Applied twice it is a translation by one whole cell along $x$. At X, where the wavevector has $\tfrac12$ along $x$, that translation acts on a zone-edge mode as $e^{i\pi} = -1$. So the glide, acting on the mode, squares to minus one, and its eigenvalues are $\pm i$. A glide sticks two levels together followed exactly that factor into the band structure, where it forces degeneracies no character table predicts. Here it forces the order parameter to have two real components, $\eta'$ and $\eta''$. The glide turns the pair through a quarter-turn in its own plane: $\eta'$ goes to $\eta''$ and $\eta''$ to $-\eta'$.

A quarter-turn preserves the antisymmetric combination $\eta'\,d\eta'' - \eta''\,d\eta'$, because it preserves orientation in the plane of the pair. The glide leaves the $x$-derivative alone, because it reflects $y$ and not $x$. So $\eta' \partial_x \eta'' - \eta'' \partial_x \eta'$ is invariant under the glide, and it is invariant under translations and the identity. That is the one Lifshitz term the census finds for pg at X, and its direction is the glide’s direction. At Y the wavevector has $\tfrac12$ along $y$ and nothing along $x$. The glide squared is a translation along $x$, which now acts as $+1$. Its eigenvalues are real, the order parameter has one component, and an antisymmetric square of one component is zero.

What the term does is visible from its form. Write the pair as one complex amplitude, $\eta' + i\eta'' = A\,e^{iqx}$, a mode sitting at X with a slow extra twist along $x$. The stiffness of the order parameter costs energy in proportion to $A^2 q^2$. The Lifshitz term, with coefficient $c$, contributes in proportion to $c\,A^2 q$, linear in $q$ and of either sign. A quadratic plus a linear term in $q$ has its minimum away from $q = 0$ whenever $c$ is not zero. So the mode’s wavevector leaves X by an amount set by a ratio of two coefficients, along the glide, and the period of the resulting structure owes nothing to the lattice.

The same reasoning predicts every other entry in the rectangular groups, and each prediction was checked against the census rather than assumed. A glide is active at a point when its square, a whole translation along the glide, acts there as $-1$, so that the point has $\tfrac12$ along that glide’s direction. pmg has one glide, along $x$, and a mirror; its entries are pg’s, at X and S. pgg has a glide along $x$ and another along $y$. At X only the first is active and the invariant runs along $x$; at Y only the second is active and it runs along $y$.

At S both are active, and the invariant disappears. The second glide reflects $x$, so it reverses $\partial_x$. Being active, it also acts on the pair as a quarter-turn and so preserves the antisymmetric combination. The product changes sign, and the term that the first glide allowed the second forbids. The same happens along $y$ with the roles exchanged. At X, by contrast, the second glide is inactive, and it acts on the pair as a reflection, which reverses the antisymmetric combination while reversing $\partial_x$. The two signs cancel and the term survives. p4g has the same pair of perpendicular glides as pgg, and its corner M has none for the same reason. So the census draws a line the zone-boundary essay did not: one active glide allows a gradient term along itself, and two crossed ones cancel it.

A corner with no glide #

The hexagonal corner is the case the glide story cannot explain, and it is why the refusal at the end of this essay exists.

The corner K is not half a reciprocal vector. It is a third of one, so $K$ and $-K$ are different points of the zone, and $K + (-K) = 0$ is what makes a product of a $K$ mode and a $-K$ mode survive the translations. The three-fold rotation fixes $K$ and multiplies a mode there by a cube root of unity, $\omega^a$. The gradient has two circular components, $\partial_x \pm i\partial_y$, which the three-fold multiplies by $\omega$ and $\omega^{-1}$. A term $\eta_K\,(\partial_x + i\partial_y)\,\eta_{-K}$ is invariant under the three-fold exactly when the phases of $\eta_K$, $\eta_{-K}$ and the gradient multiply to one.

In p3 the only thing relating $\eta_{-K}$ to $\eta_K$ is complex conjugation, so $\eta_{-K}$ carries $\omega^{-a}$. Their product carries no phase, the gradient carries $\omega$, and nothing is invariant. p3m1 adds mirrors, and each of them carries $K$ to $-K$. But a mirror reverses the three-fold’s sense, so the mirror image of $\eta_K$ again carries $\omega^{-a}$, and the count stays zero.

In p6 a half-turn carries $K$ to $-K$, and the half-turn commutes with the three-fold. So $\eta_{-K}$, defined as the half-turn’s image of $\eta_K$, carries the same $\omega^a$ as $\eta_K$, and the product carries $\omega^{2a}$. For $a = 1$ that is $\omega^2 = \omega^{-1}$, which cancels the gradient’s $\omega$. The census finds two invariants for that order parameter, one for each circular component, and none for $a = 0$.

p31m reaches the same result by a different route. Its mirrors do not move $K$ at all: they fix it, so the operations keeping $K$ form the group $3m$. That group has a two-dimensional representation whose two circular components turn oppositely under the three-fold, by $\omega$ and by $\omega^{-1}$. A $K$ component turning by $\omega^{-1}$, paired with a $-K$ component that also turns by $\omega^{-1}$, carries $\omega^{-2} = \omega$, and together with the other circular component of the gradient the product is invariant. The census finds one invariant for that order parameter and none for the one-dimensional ones. p6m has both the half-turn and the mirrors that fix $K$, and it has one.

No glide is involved. These are symmorphic groups, and their zone corners carry a Lifshitz term because of how the point group acts on the three-fold’s phases, not because of any fractional translation. The census therefore refuses the tempting generalisation of the glide groups, that only a non-symmorphic operation puts a Lifshitz term at a special point. In the plane, three symmorphic groups contradict it.

Off the special points, always #

The zone-boundary essay’s other claim is right, and the census shows why. At a point on a line of the zone, or a general point, $k$ is not equal to $-k$, and an operation that fixes $k$ fixes its direction. So the component of the gradient along $k$ itself is invariant under every operation that keeps $k$. A term pairing the $k$ mode with the $-k$ mode, times that component of the gradient, survives. It amounts to letting the wavevector drift along its own direction. At $(\tfrac14, 0)$ and at $(\tfrac16, \tfrac13)$ every order parameter of every group has at least one invariant. The count grows with the order parameter’s size and falls with the symmetry that fixes $k$: two for p1’s two-component order parameter on a line, one each for the two-component order parameters of pm, pmm and pmg there, one for each four-component order parameter of p4m, and two for the eight-component one at its general point.

That is the sense in which only special points can hold a continuous commensurate transition. What the census adds is that being special is necessary and not sufficient. Of the special points of the plane, ten fail through an active glide or through the corner’s three-fold, and a transition there goes the way a general point’s would.

What this changes about the zone-boundary transitions #

The transitions the cell a zone-boundary mode doubles froze all live in symmorphic groups at edge points, where every entry of the map is zero. Its enumeration stands. What changes is the reading of the glide groups. A structure with pg, pmg, pgg or p4g symmetry has order parameters at its zone edge that the Landau argument cannot freeze continuously into a doubled cell. The two-component order parameter at pg’s X is exactly the kind whose degenerate levels the glide forces together. It comes with a gradient term, so its continuous instability is toward a modulation along the glide, not toward a cell twice as long.

That is a statement about the continuous route only, and about the gradient term only. Landau’s other condition, that no cubic invariant may exist, is separate and is not tested here. A first-order transition can still jump to the doubled cell, and a modulated phase can lock into a commensurate one at lower temperature when higher-order terms favour the lattice’s own period. The satellites that need a second integer and superspace groups in the plane describe what the modulated phase then looks like. The census only says which order parameters must pass through such a phase if they order continuously.

What the census does not decide #

The count is of terms allowed by symmetry. It does not say their coefficients are large, and a small coefficient produces a modulation so long that a real crystal, with its finite size and its defects, may never show it. It also treats one order parameter at a time. When two order parameters at the same point are close in energy, terms coupling them can matter, and none of those are counted.

The plane is also a restriction. In space the same computation runs over 219 groups with many more special points, and the glide mechanism has screws alongside it: a screw applied $n$ times is a translation along its axis, and at a zone-boundary point that translation can act as $-1$ as well. Whether the rule found here survives there — one active operation of that kind allows a term, two crossed ones cancel it — is not settled by anything computed in this essay.

What the census has to refuse #

Two of the tests guard the computation itself. The completeness test would fail if an eigenspace mixed two order parameters or if the translations were read wrongly, because the dimensions would no longer add up. The two counts, one from characters and one from the representation matrices element by element, would disagree if the squares $h^2$ were composed wrongly. Two guard the claims inherited from earlier essays: none at the zone centre, as which way the order parameter points computed, and at least one everywhere off the special points, as the zone-boundary essay stated. The refusal is the generalisation the glide groups invite and the hexagonal corner contradicts.

Still open: screws, and the corners of space #

The natural next computation is the same census in space, where the special points include the corners and face centres of fourteen kinds of zone, and screws join glides as operations that square to a translation. The glide rule found here makes a definite prediction to test there: an order parameter at a zone-boundary point where exactly one screw or glide is active should carry a Lifshitz term along that operation’s direction, and crossing a second should remove it. The hexagonal corner makes a second prediction: in space the three-fold’s phases at the corner of a hexagonal zone should admit a gradient term in some symmorphic groups and not others, by the same arithmetic of cube roots of unity. Both are finite computations of the size already done here.