How large a degeneracy may be

An operator that commutes with a group cannot tell apart the members of an irreducible piece. That sentence is the reason representations matter to anyone measuring anything, and it turns a question about matrices into a question about the world: how many quantities can symmetry force to be equal?

The answer for a crystal is two in the plane and three in space, and it is not a fact about crystals so much as a fact about the numbers that describe them. The dimension of an irreducible representation cannot be chosen; it is constrained twice over, and the constraints are tight enough that the whole list of possibilities fits in a line.

The sum that does the work #

A finite group of order |G| satisfies

$$\sum_i d_i^2 = |G|$$

where the sum is over its irreducible representations and dᵢ are their dimensions. It is not an inequality with slack in it. The squares of the dimensions add to the order exactly, and the trivial representation contributes 1 to every such sum, so everything else has to fit in |G| − 1.

That alone bounds a dimension by the square root of the order. The point group 6mm has order twelve, so nothing in it can have dimension four: sixteen already exceeds twelve. It could in principle have a three-dimensional representation, since 1 + 9 = 10 leaves room for two more ones — and it does not, which the second constraint explains.

The second constraint is divisibility. The dimension of an irreducible representation divides the order of the group. For 6mm that permits 1, 2, 3, 4, 6 and 12, and the sum rule has already refused everything above 3; and 3 does not divide into a table whose remaining entries must be squares adding to 3 with six classes to fill. The two conditions together leave one possibility, and the construction finds it.

That divisibility is worth a sentence of its own, because it looks like an accident and is not. The quantity |G|/d is a sum of products of character values, and every character value of a finite group is an algebraic integer — a root of a monic polynomial with whole-number coefficients — because it is a sum of roots of unity. A sum of products of algebraic integers is another one, so |G|/d is an algebraic integer; and it is also a rational number, being a quotient of two whole numbers. A rational algebraic integer is a whole number, and there is the divisibility. The argument is the same shape as the one behind the crystallographic restriction, which observes that the trace of an integer matrix is both an integer and a sum of roots of unity and concludes that a lattice permits five orders of rotation. Both are the same manoeuvre: a quantity computed two ways is forced into the intersection of two number systems, and the intersection is small.

Why abelian groups have nothing above one #

Seven of the ten plane classes are entirely one-dimensional, and all seven of them are abelian — their operations commute. That is not a coincidence but an equivalence, and it is worth seeing why in both directions, because it is the cleanest statement in the subject.

If a group is abelian, every conjugacy class is a single element, so the number of classes is the order of the group; the number of irreducible representations equals the number of classes; and the dimension sum with that many terms forces every dimension to one. If a group has all its representations one-dimensional, then the matrices are numbers, numbers commute, and the group is abelian because the representations together separate its elements.

So the seven abelian plane classes — 1, 2, m, 2mm, 4, 3, 6 — force no degeneracy whatever. Every level of a symmetric operator on them is single, and any coincidence between two of them is something else entirely.

The three that are not abelian, and what their pair of dimensions is #

4mm, 3m and 6mm each carry exactly one or two representations of dimension two, and in every case the two-dimensional one has a concrete meaning: it is the plane itself. The pair of coordinate directions transforms into combinations of one another under a rotation of order three, four or six, and no basis makes the matrices diagonal, because a rotation of the plane through anything other than a half-turn has no real eigenvector.

That is the geometric content of the whole ladder in one sentence. A three-fold rotation mixes the two directions, so anything with a direction in it — a displacement, a field, a wave — comes in pairs that the symmetry treats as one object.

The class count, which is the constraint people forget #

The dimension sum is quotable and gets quoted. It is not sufficient on its own, and the second condition is the one that does the remaining work: the number of irreducible representations equals the number of conjugacy classes.

For a group of order eight that leaves several arithmetically possible tables. Eight can be spent as 1+1+1+1+4, which is five representations; or as 1+1+1+1+1+1+1+1, which is eight. Both satisfy the sum. Which one occurs is decided by counting classes, and the two orders-of-eight in the plane behave differently: 4mm has five classes and takes the first, while the abelian group of order eight — which is not a plane point group, but the argument is the same — would have eight and take the second.

Counting classes is not a formality. Two operations are in the same class when some operation of the group carries one to the other, and deciding that is a computation rather than an inspection: the two mirrors of 2mm look alike and are in different classes, because nothing in 2mm turns one into the other; the four mirrors of 4mm fall into two classes of two, because the four-fold rotation carries axial mirrors to axial mirrors and diagonal ones to diagonal ones and never mixes the two kinds. The class structure is what fixes the shape of the table before a single character is computed.

Three dimensions, where the ceiling rises to three #

Running the same construction on the thirty-two crystal classes of space changes nothing about the method — the matrices are three by three, and the rest of the file is the same file — and it changes the answer by exactly one.

The cubic classes are the ones with more than one axis of order three, and the three-dimensional representation is again the space itself: three mutually perpendicular directions that the body-diagonal rotations cycle. A cubic crystal’s three principal directions are not three independent things; they are one three-dimensional object, which is why a cubic crystal has an isotropic dielectric constant and a single elastic modulus in place of three.

Nothing in space reaches four. The largest crystallographic point group has order forty-eight, so the sum rule permits up to six in principle; divisibility and the class count remove it. The enumeration confirms it on all thirty-two rather than arguing it in general, which is this collection’s habit with a bounded list.

Why five is impossible, and where the reader has met that before #

A five-fold degeneracy would need a five-dimensional irreducible representation, which needs a group of order at least twenty-five with five dividing its order. The crystallographic point groups have orders 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48, and five divides none of them — which is the crystallographic restriction arriving in a place it was not sent.

The restriction is usually stated about rotations: a lattice permits orders one, two, three, four and six and no others, because the trace of an integer matrix is an integer. Its consequence for the orders of the point groups is immediate, and its consequence for the degeneracies follows from that. So a crystal cannot have five levels forced together for the same reason it cannot have a five-fold axis, one implication further along.

The doubling that is not in the table #

The figure above is the exception every account of this subject has to state, and stating it here rather than later avoids a false impression.

4, 3 and 6 are abelian, so by the argument above they force no degeneracy at all. Their character tables contain complex numbers, though — the four-fold rotation is represented by i in one of them — and a physical operator with real matrix elements is unchanged by complex conjugation, which is a symmetry no group of motions contains. Conjugation carries each complex representation to its conjugate partner, so the two levels it labels cannot be separated by a real operator, and they are observed as one doubled level.

This is the reason a textbook on molecular vibrations prints certain pairs of one-dimensional representations bracketed together as a “separably degenerate” pair. The bracket is not a group-theoretic statement; it records the presence of a symmetry — the reality of the operator, or equivalently the reversibility of time — that the group of motions never contained. The next rung separates that case from a genuine coincidence, and the reciprocal-space ladder meets it again as the pairing of a wavevector with its negative.

What the bound does and does not say #

Three statements, in decreasing strength.

A degeneracy larger than the bound is never forced by the point group. That is a theorem and the enumeration confirms it on every class.

A degeneracy larger than the bound can occur. Nothing forbids two levels of different symmetry from coinciding, and the previous paragraph gives a systematic reason for one such case. The bound is a floor on what symmetry guarantees, not a ceiling on what happens.

A degeneracy smaller than the bound is impossible where the bound applies. If a quantity transforms as a two-dimensional representation, its two components are equal in value, and no measurement respecting the symmetry finds otherwise. That is what makes the count useful before anything is measured — the same argument as Neumann’s principle, which decides how many independent components a property has before anybody builds an apparatus.

What a bound of two costs, and what it buys #

It is worth being concrete about what the ceiling means for a measurement, because “at most two” sounds like a small statement and is not.

A property with a direction in it — a vector — has two components in the plane and three in space. In a class whose largest representation is one-dimensional, those components transform separately, so a crystal can respond differently along each axis and a measurement has to find each number independently. In a class carrying a two-dimensional representation, the components in the plane are one object, and a single number describes both: measuring along one direction is measuring along the other.

That is the whole of why a hexagonal crystal is optically uniaxial while an orthorhombic one is biaxial, and it is decided by the dimension of a representation rather than by anything about the atoms. The property essays count independent components a different way — by a character sum over the tensor’s own representation — and the two counts have to agree, which they do because they are two arrangements of the same orthogonality relation.

The cost is the mirror image of the benefit. A class with a two-dimensional representation cannot be persuaded to have different values along the two directions, however the crystal is prepared, so a material with an intrinsically anisotropic response in the plane cannot sit in such a class. Symmetry gives a count for nothing and takes the freedom to disagree with it.

The same arithmetic, three times over #

It is worth collecting how often one small piece of arithmetic has now decided a count in this collection.

Rotation orders. The trace of an integer matrix is an integer, so 2 cos(2π/n) is an integer, so n is 1, 2, 3, 4 or 6. That is the restriction in one line.

Point groups. The orders permitted are then 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 — and the thirty-two classes are what closure over those orders permits in space.

Degeneracies. The dimensions of the irreducible representations of a group of one of those orders divide the order and have squares summing to it, which leaves 1, 2 and 3 and nothing else.

Each step is elementary and each is exact. The chain is why a crystallographer can say, with no measurement at all, that a cubic crystal’s three principal refractive indices are equal, that a tetragonal crystal’s are two rather than three, and that no crystal has a level five deep because a lattice would not permit the rotation that would force it.

What the ceiling does to a level that arrives too large #

The bound is stated as a prohibition, and its most visible consequence is not a prohibition at all: it is what happens to a degeneracy that was larger before the crystal arrived.

A free atom is spherically symmetric, and its levels are degenerate to the dimensions the full rotation group provides — one, three, five, seven, and upwards without limit. A d level is five-fold degenerate, and five is above every ceiling computed on this page. So a d level cannot survive intact in any crystal: the five states must split into pieces whose sizes are dimensions the point group actually has.

For a site of cubic symmetry the answer is 3 + 2. The five states break into a three-dimensional piece and a two-dimensional one, and the split is forced by arithmetic before any statement about energies is made — nothing here says which of the two lies lower, or by how much, only that the five must become a three and a two. Lowering the symmetry further splits them further: at a site with a single four-fold axis the three breaks again, since no representation of an abelian group has dimension above one.

That is the crystal-field splitting, and setting it beside the ceiling is the cleanest way to see what the ceiling is for. The bound is not chiefly a statement that no crystal has a five-fold degeneracy. It is a statement that a five-fold degeneracy arriving from somewhere else must break, into pieces the group’s own table lists, in a pattern the table fixes completely.

The degeneracy the table cannot see #

There is a two-fold degeneracy that no point group forces and no point group can remove, and it is worth naming because it is a genuine exception to the reading the largest representation is the largest degeneracy.

Time reversal is not a spatial operation and does not appear in any of the tables here. For a system with an odd number of electrons it is an operation whose square is −1, and that single sign has a consequence: every level is at least two-fold degenerate, in every crystal, at every site, whatever the point group is — including the trivial one. This is Kramers’ theorem, and the degeneracy it forces sits on top of whatever the spatial symmetry provides.

So the honest statement of the ceiling has a qualifier attached. The point group forces degeneracies of one, two or three, and those are the degeneracies of the spatial part of a state. A system carrying half-integer spin has each of them doubled, giving observed degeneracies of two, four and six — and the doubling is invisible in the character table because it comes from an operation the table does not contain.

The repair, where it is needed, is to work with the double group, in which a rotation by a full turn is a distinct operation from the identity and the extra representations account for the doubling. That is a larger machinery than this page uses, and the boundary is worth stating plainly: everything computed here is about the thirty-two ordinary point groups acting on ordinary tensors, and it is exactly right for a vibration, a tensor property or an orbital, and incomplete for a state with spin.

Both of the paragraphs above are the same observation about what a bound is a bound on. The ceiling of three is a statement about the irreducible representations of a finite group of matrices, and it binds anything those matrices act on. A degeneracy arriving from a larger symmetry — the sphere’s, in the crystal-field case — must break down into pieces the ceiling permits. A degeneracy arriving from an operation that is not one of those matrices at all — time reversal, in Kramers’ case — is not bounded by them and multiplies whatever they give. Neither case contradicts the count, and both are outside it, which is the usual position of a theorem meeting a measurement.

Where this ladder goes #

The next rung takes the third statement above and tests it: a degeneracy that symmetry forces cannot be moved by anything respecting the symmetry, so moving the weights of an operator and watching what survives separates a forced degeneracy from a coincidence. And the shell that splits applies the decomposition to the object a crystallographer actually has — a shell of neighbours around an atom — where the multiplicities say which combinations of displacements exist before any of them is computed.