Which modes a site can carry

A crystal’s atoms can be displaced, and the displacements are not free. Not because anything holds the atoms in place — the question here is not about forces — but because a pattern of displacements either respects a symmetry of the crystal or does not, and the symmetric patterns available to a given set of atoms are decided before any energy is written down.

The bookkeeping is a character, and this essay computes it for every Wyckoff position of every plane group. What comes back is a table of small integers, some of them zero, and a zero is the interesting entry: a distortion those atoms cannot make, however their amplitudes are chosen.

The displacements of an orbit, as a space #

Fix a Wyckoff position — an orbit of points under the group — and let every atom of it move. If the orbit has m atoms, there are 2m numbers to choose in the plane, and the space of those choices is 2m-dimensional.

The group acts on that space. An operation moves each atom to another atom of the same orbit and rotates its displacement vector, so it permutes the components and mixes them. The space of displacements therefore carries a representation of the group — the mechanical representation of that orbit — and it splits into pieces the group keeps separate.

Splitting it is the whole of the computation, and it needs one character. For each operation, the character of the mechanical representation is

$$\chi(g) = (\text{atoms } g \text{ leaves in place}) \times \operatorname{tr}(M_g)$$

— the number of atoms that do not move, multiplied by the trace of the operation’s linear part. Atoms that move contribute nothing, because their contribution to the trace is off-diagonal; atoms that stay contribute the trace of what the operation does to a vector there.

Why atoms that move contribute nothing #

That formula looks like a trick and is not. The matrix representing an operation on the whole space of displacements is a block matrix, with one block per pair of atoms: the block in row i and column j is the operation’s linear part if g sends atom j to atom i, and zero otherwise.

A trace sums the diagonal blocks. A diagonal block is non-zero exactly when g sends an atom to itself. So the trace is the number of fixed atoms times the trace of the linear part, and every atom the operation moves is invisible to it.

That is a general fact about permutation-like representations and it is the reason this whole computation is cheap. Nothing has to be diagonalised; the character is read off by counting fixed points, which is the same count Burnside’s lemma uses for a different purpose.

The multiplicities, and what a zero means #

With the character in hand, the multiplicity of each irreducible representation is an inner product — a sum over the group, divided by its order, and required to come out a whole number. Those multiplicities are the table.

A multiplicity of zero says the atoms of that position cannot make a displacement pattern of that symmetry at all. Not that it costs energy: that no such pattern exists in the space of their displacements. An atom sitting at a fourfold rotation centre with three others at the general positions around it cannot contribute to a pattern that breaks the fourfold axis while keeping the mirrors, because there is no such motion of those four atoms.

For p4m the fourfold site supplies only the two-dimensional representation, which is the pair of translations; every one-dimensional representation has multiplicity zero there. That means a distortion of p4m carrying any of the one-dimensional symmetries needs atoms somewhere other than the fourfold centres, and a structure whose atoms are only at those centres cannot undergo that distortion at all.

Two checks, and the second is the interesting one #

The table is checked in two ways that do not overlap.

The dimensions must add up. Summing the multiplicities weighted by the dimensions of the representations must give 2m — two degrees of freedom per atom. That catches an arithmetic slip and nothing subtler; it holds on every position of all seventeen groups.

The trivial representation’s multiplicity must be the position’s free coordinates. This is the check with content. A displacement pattern transforming as the trivial representation is one that keeps every symmetry of the crystal — so after making it, the structure has the same group it started with, and the atoms have simply moved to another position of the same Wyckoff set.

The number of ways to do that is the number of free coordinates the position has: two for a general position, one for a point on a mirror, which may slide along it, and none for a point pinned at a rotation centre. That number is computed separately, from the site symmetry — how many directions the stabiliser leaves alone — and it must equal the multiplicity of the trivial representation.

It does, at every position of all seventeen groups. Two computations with nothing in common: one a character sum over the whole group, the other a count of invariant vectors under a two- or four-element stabiliser.

What the check would catch #

A check that has never failed is worth examining for what it could catch, and this one could catch the commonest error in the subject: using the wrong coset representative for a non-symmorphic group.

The character formula asks how many atoms an operation leaves in place, and for a glide the answer depends on which representative of the glide’s coset is used — the operation itself, or the same operation composed with a lattice translation. Pick the wrong one and the character is wrong, the multiplicities come out as fractions or as negative numbers, and the inner product refuses to divide.

That refusal is the site’s standing pattern: a multiplicity is a count, so a remainder is a bug and not a rounding error. Here it is also the only warning available, because a wrong multiplicity that happened to be a whole number would produce a table that looks exactly like a right one.

Where the count comes from geometrically #

There is a second way to see the same numbers, and it is the one that makes them memorable.

The displacements of an orbit are determined by the displacement of one atom, transported around by the group — except that the transport must be consistent: if two operations send the representative atom to the same place, they must send its displacement to the same vector. That consistency is a condition only at atoms with a non-trivial stabiliser, and it is exactly the condition that the displacement at such an atom be invariant under its own site symmetry.

So the space of displacement patterns of an orbit is the space of vectors at one atom that its site symmetry permits, induced up to the whole group. An atom on a mirror may only carry a displacement along the mirror in the fully symmetric pattern; an atom at a rotation centre may carry none.

The word for that construction is induction, and the character formula above is the induced character written out. This collection has met the same construction in the other direction — restricting a representation of a big group to a small one — and the two are adjoint operations in a sense that is not developed here.

A structure’s modes, and where they come from #

Adding up over all the positions a structure occupies gives the modes of the whole structure, and the sum decides what distortions the structure can make.

Two structures in the same space group but with atoms at different Wyckoff positions have different mode tables, so they permit different distortions and can undergo different transitions. That is a statement about structures rather than about groups, and it is the reason this table is per-position rather than per-group.

It also gives a rule with practical bite: a structure whose atoms all sit at maximally symmetric positions has very few modes available, and every one of those modes is a translation or a strain. To break a symmetry at all, a structure needs atoms with room to move — which is to say, atoms at positions with a small site symmetry.

A worked position: the atom on a mirror #

Take p4m and an atom on one of its mirror lines. Its orbit has four members, so eight numbers describe their displacements, and the table’s row for that position reads one of each of the four one-dimensional representations and two of the two-dimensional one — eight in total, as it must.

Reading that row as a description of motions: one of the eight combinations keeps every symmetry, and it is the atom sliding along its own mirror with its three images sliding along theirs. That is the position’s single free coordinate, and moving it produces a structure that is still p4m with the atoms at a different place in the same Wyckoff set.

Three combinations keep some symmetries and break others: the atom moving off its mirror one way while its images move correspondingly, breaking either the mirrors or the fourfold axis and keeping the rest. And four more make up the two copies of the two-dimensional representation, which are the patterns whose symmetry is lowest.

Every one of those eight motions is available to that atom. What the table adds to the picture is the sorting, and the sorting is what makes the count of eight into a statement about which distortions of the crystal are possible.

Every group, in one sweep #

Running the computation over all seventeen groups gives 105 positions, and both checks hold on every one of them: the weighted multiplicities sum to twice the atom count, and the trivial multiplicity is the number of free coordinates.

The sweep is worth doing rather than checking a few cases, because the failure it is guarding against is not uniform. A group with a glide has coset representatives that can be chosen wrongly; a group with two distinct classes of mirror — p3m1 and p31m are the standard pair — has two positions that look alike and are not; a group with a threefold axis has representations whose characters are complex, so the inner products run through cyclotomic arithmetic rather than integers.

Each of those is a place a plausible wrong answer could appear, and the sweep visits all of them. It is the same argument this site makes about drawing every figure the essays ask for rather than only the defaults: a claim that holds in the cases somebody thought to check is a claim about those cases.

The same table under another name #

Spectroscopy does this computation constantly and calls it the factor group analysis: which vibrational modes a crystal has, sorted by symmetry, from the Wyckoff positions its atoms occupy. The tables printed in that literature are these tables, in three dimensions, with the acoustic modes subtracted.

Subtracting the acoustic modes is worth a sentence, because it is the one step this essay has not taken. Three of the modes of any structure — two in the plane — are rigid translations of the whole crystal, which change nothing. They transform as the vector representation, so the count of genuine internal modes is this table with one copy of the vector representation removed.

That subtraction is where a mode table meets a measurement: what a spectrum shows is the internal modes, and which of them are visible depends on a further symmetry question — whether the mode’s representation is one the measurement couples to. This collection computes the first half and takes no position on the second, which is a fact about the interaction of light with matter rather than about the crystal.

The modes that break something #

Everything above counts; the essays after this one use the counts.

A distortion that lowers the symmetry is a mode transforming as a non-trivial representation, and the table says which positions can supply one of each kind. Freezing such a mode into a structure produces a new structure whose group is a subgroup, and the next ladder does exactly that — builds the displacement pattern, moves the atoms, and hands the result to the detector.

What connects the two is that the mode has to come from somewhere. An order parameter transforming as a given representation is only available to a crystal whose atoms supply a mode of that symmetry, and the multiplicity in this table is the number of independent ways they can supply one. Where it is zero, the transition is not available to that structure however the energies fall.

The subtraction, and the two lists it produces #

The essay stops one step short of the table a spectroscopist reads, and taking the step is worth doing because it turns a count of modes into a prediction about which of them a measurement can see.

Of all the modes at the zone centre, two are not vibrations at all: the two translations, in which every atom moves the same way and nothing is stretched. They transform as the vector representation, they are always present, and they are the acoustic modes. Subtracting them from the total leaves the optic modes, which are the ones with a frequency.

The optic modes then divide by what they couple to, and the rule is one line in each case.

A mode is infrared-active when it transforms as a vector. Absorption of infrared light is a coupling to the electric field, which is a vector, so a mode can absorb only if its own symmetry is one a vector has. Which representations those are is a character this collection already computes — it is the same one that decides the polar classes.

A mode is Raman-active when it transforms as a symmetric rank-two tensor. Raman scattering couples to the polarisability, which is such a tensor, so the permitted modes are those appearing in its decomposition — and that character is the one behind the optical indicatrix.

So one mode table and two characters give three lists: how many modes of each symmetry there are, which of them absorb infrared, and which of them scatter. All of it before any frequency is computed and before the material is named, which is why a vibrational spectrum is assigned by symmetry first and by chemistry second.

The exclusion a centre imposes #

There is a consequence of those two rules that is sharp enough to be a diagnostic, and it follows from a parity argument this collection has made three times in other contexts.

A vector is odd under inversion and a symmetric rank-two tensor is even. So in a centrosymmetric crystal every representation is either odd or even, the infrared-active modes are the odd ones and the Raman-active modes are the even ones, and no mode can be both.

That is the rule of mutual exclusion, and it is used the way every prohibition in this collection is used: in the negative direction. A material whose spectra show a line at the same frequency in both the infrared and the Raman is a material without a centre of symmetry — and that is a conclusion about the structure drawn from two spectra and no model whatever.

The converse fails, as always. Two spectra with no coincidence are consistent with a centre and do not establish one, since a mode can be inactive in both for reasons of its own. The rule forbids, and forbidding is the direction in which a permission-based table says something no measurement will contradict.

What the table does not say #

It says nothing about frequencies. Two modes with the same symmetry can have any energies whatever; symmetry sorts them into kinds and does not order them. A soft mode — one whose restoring force vanishes — is a mode with a particular symmetry and a particular energy, and only the first is in this table.

It says nothing about which mode a crystal chooses. That needs an energy, and the whole of this collection’s position is that symmetry supplies the list and not the choice.

And it is a zone-centre table. Every mode counted here has the same displacement pattern in every cell. A crystal can also distort with a pattern that alternates from cell to cell, and those modes belong to a different group — the little group of a wavevector — with a different table. The cell-doubling essay is where that case is computed, and its tables are larger.