Why a character table stops where it stops

Everybody who has used a character table has noticed that it is square, and most reference books do not say why. The number of columns is the number of conjugacy classes, which is a fact about the group’s operations; the number of rows is the number of irreducible representations, which sounds like a fact about how many useful ways there turned out to be of labelling things. They are the same number, always, and the reason they are is worth having.

There is a second constraint, tighter still. The squares of the dimensions of the representations add up to the order of the group. For C2v that is $1 + 1 + 1 + 1 = 4$; for C3v it is $1 + 1 + 4 = 6$; for Td it is $1 + 1 + 4 + 9 + 9 = 24$.

Between them the two leave a group almost no freedom at all. Count the classes and the number of rows is fixed. Know the order and the dimensions are pinned down to a handful of possibilities and usually to one. A character table is not a compilation; it is the unique solution to two equations.

The first theorem is about counting twice #

The number of irreducible representations equals the number of conjugacy classes because both count the same thing from opposite ends.

A class function is a function on the group that takes the same value on conjugate elements — the same value on all three mirrors of C3v, the same value on both threefold rotations. Class functions form a vector space, and its dimension is obviously the number of classes, because a class function is determined by one number per class.

The characters of the irreducible representations are class functions, and the great orthogonality theorem says they are orthonormal under the group average. So they are linearly independent, and there are at most as many of them as there are classes. The other half — that they span the space, so there are at least as many — is the harder direction and comes from the same theorem applied to the regular representation below.

Two independent counts of one vector space, forced to agree. That is the whole of it, and it is why the table is square.

The second theorem, produced rather than quoted #

The dimension theorem is the one that can be computed with an ordinary reduction, and doing so is more convincing than stating it.

Consider the group acting on itself. Label the basis of a vector space by the group elements, and let each element act by multiplication — $g$ sends the basis vector labelled $x$ to the one labelled $gx$. That is a representation of dimension $h$, called the regular representation, and its character is trivial to write down.

The character of a permutation representation is the number of basis vectors left where they were. The identity leaves all $h$ of them alone. Every other element $g$ leaves none of them alone, because $gx = x$ would mean $g$ is the identity. So the character is

$$\chi_{\mathrm{reg}} = (h, 0, 0, \ldots, 0)$$

with the $h$ in the identity column and nothing anywhere else.

Reduce it. The reduction formula divides by the group order and sums over classes, and every class but the identity contributes zero, so the multiplicity of the representation $\Gamma_i$ comes out as $h \cdot \chi_i(E) / h = \chi_i(E) = d_i$. Each representation appears exactly as many times as its own dimension.

Now count dimensions. The regular representation has dimension $h$, and it is the sum of $d_i$ copies of each $\Gamma_i$, so

$$\sum_i d_i \cdot d_i = h$$

which is the theorem. It has been derived by running one ordinary reduction on a character with no free parameters in it.

The formula being run is the ordinary one, and it is worth saying that nothing special is done for this argument: character tables and reduction hands it the four hydrogen 1s functions of methane and gets A₁ + T₂, and the same call with $(h, 0, 0, \ldots)$ in place of that basis is what produces the theorem. Nothing in the formula knows which character it has been given.

What makes it a check rather than a restatement #

A derivation that could not fail is not evidence of anything, and this one can fail in a specific place.

The reduction formula produces multiplicities as fractions and then rounds them, and character tables and reduction records why the rounding is the check rather than a convenience: a multiplicity that is not a whole number means something upstream is wrong, and there is nothing downstream that would notice. A mistyped character in any of the nineteen tables, a class size counted wrong, a column mismatched against the generated group — every one of them produces a fraction here.

So running the regular representation through it exercises all nineteen tables at once, on a character that is not one they were built for. All nineteen return whole numbers, each representation’s multiplicity equals its own dimension, and the dimensions sum in squares to the order.

And the refusal is required as well. Handing C2v the character $(5, 0, 0, 0)$ — not the character of any representation of it — must produce a fraction and be refused, rather than being rounded to something plausible. It is.

Where the room runs out #

The practical use of the two theorems together is that they say what a table cannot contain, and this is more useful than the list of what it does.

Take Oh. It has forty-eight operations sorted into ten classes, so ten representations. Their dimensions square to forty-eight. The dimensions are 1, 1, 2, 3, 3 and then the same five again with opposite behaviour under inversion, and $2(1 + 1 + 4 + 9 + 9) = 48$ exactly.

There is no eleventh row and no larger dimension. A four-dimensional representation would contribute sixteen on its own, and forty-eight minus sixteen leaves thirty-two to be made up by nine other representations, of which two must be one-dimensional — the totally symmetric one and its partner — leaving thirty for seven, which cannot be done with squares while keeping ten rows in total.

This is what degeneracy is a group theorem rests on, and it puts a hard ceiling under it. An octahedral molecule cannot have four orbitals at one energy, for any reason a symmetry argument accepts, because the largest dimension in its table is three and the sum of squares leaves no room for a fourth. A fourfold degeneracy in a real octahedral system is therefore an accident of the parameters, a spin degeneracy, or a mistake — and the three possibilities are distinguishable.

A group with few classes has large representations #

The two theorems together produce a rule of thumb that is exact rather than approximate: the fewer classes a group has for its order, the larger its representations must be.

D2h and Td illustrate it from opposite ends. D2h has order eight and eight classes — every operation is its own class, because the group is abelian — so it has eight representations, and eight squares summing to eight forces every one of them to be one-dimensional. An abelian point group has no degeneracies at all, and this is the whole proof.

Td has order twenty-four and only five classes. Five squares summing to twenty-four cannot all be small: the only solution with five positive integers is $1 + 1 + 4 + 9 + 9$. The large dimensions are forced by the shortage of classes, and the shortage of classes is a statement about how much conjugation mixes the operations up.

The comparison is sharper with the order held fixed, because then nothing but the class count can be doing the work. D6h has the same twenty-four operations as Td and sorts them into twelve classes rather than five.

So a threefold degeneracy in a tetrahedral molecule and its absence in a hexagonal one of the same group order are not two facts about the molecules. They are one fact about how conjugation sorts twenty-four operations, read twice.

So the threefold degeneracies of a tetrahedral molecule — the t₂ set of hydrogen orbitals in methane, the t₂ vibrations, the p orbitals themselves — are not three separate observations. They are one arithmetic fact about a group with twenty-four operations in five classes, and no tetrahedral system can avoid them.

The consequence shows up wherever a tetrahedral molecule is counted. Methane’s nine vibrations reduce to A₁ + E + 2T₂ — computed in character tables and reduction — and six of those nine coordinates are in threefold sets, which is not a fact about C–H bonds. Td’s table has two three-dimensional rows and could not have had fewer, so any nine-coordinate basis on a tetrahedral molecule will put most of them into threes.

The classes are counted from the molecule, which is where this could go wrong #

Everything above takes the number of conjugacy classes as given, and it need not be given: it can be computed from the molecule’s own operations, by conjugating each one by every other and collecting what lands on what.

That is the step where the two sides of the square could part company. If the conjugation tolerance were too loose, two genuinely distinct classes would merge and the count of columns would fall below the count of rows; if it were too tight, one class would split and the count would rise above. Either failure produces a table that is not square, and the reduction formula immediately produces fractions for every basis it is handed.

The tolerance is a thousandth, and the reason it is safe is a fact about point groups rather than about arithmetic: the closest two distinct operations any of these nineteen groups holds differ by a rotation of forty-five degrees. There is nothing to confuse at a thousandth. That margin is what lets the class count be treated as an integer produced by a search rather than as a parameter, and the squareness of every generated table is then evidence that the search worked.

It is worth noticing what would happen without it. A molecule whose coordinates are slightly off — a structure taken from a measurement, or built by a minimiser that stopped early — has operations that are slightly wrong, and conjugating them accumulates the error. This site’s answer is to refine every operation into the exact orthogonal matrix realising its atom permutation before any product is taken, so that a chain of conjugations is bit-identical rather than drifting. The counting theorems then check the refinement from a completely different direction: if refinement had failed, the tables would not be square.

The four relations the tables are verified against #

The nineteen tables here are tabulated data — the one tabulated ingredient in the whole symmetry argument — and they are verified before use against four internal relations, of which the two counting theorems are the outermost.

Rows are orthonormal. Two different representations’ characters, averaged over the group with class sizes as weights, give zero; a representation with itself gives one. That is the great orthogonality theorem in the form a table can be checked against, and it is checked for every pair in every table.

Columns are orthogonal too. The same statement read down instead of across, which is a separate arithmetic fact and catches a different kind of typing error.

The class sizes sum to the order, which is trivial and catches the case where a column has been left out.

The dimensions square to the order, which is this essay’s second theorem.

A table failing any of them should be rejected rather than used, and the reason for that severity is that a wrong character produces a wrong multiplicity, a wrong multiplicity produces a plausible level diagram, and nothing further downstream asks whether the diagram is right. The checks are cheap and they are the only thing standing between a mistyped digit and a wrong picture with a confident caption.

The product table is bounded by the same arithmetic #

Multiplying two representations gives a representation, which reduces to a sum of irreducible ones, and the reduction is bounded by the dimensions: the product of a $d_1$-dimensional and a $d_2$-dimensional representation has dimension $d_1 d_2$, and whatever it reduces to must add up to that.

For Td, T₂ × T₂ has dimension nine and reduces to A₁ + E + T₁ + T₂, whose dimensions are $1 + 2 + 3 + 3 = 9$. The check is arithmetic and it is made every time a product is computed here.

The rule that matters for spectroscopy falls out of the same place. A transition is allowed when the product of the two states’ representations with the operator’s contains the totally symmetric representation, and it contains it at most once — because the totally symmetric representation appears in a product exactly when the two factors are the same representation, and then exactly once. Selection rules are one theorem uses that fact throughout; the counting theorems are why it holds.

Every product in Td reduces that way, and each one is checked the same way it is computed: the dimensions of what it reduces to must sum to the product of the two factors’ dimensions, which is arithmetic on integers and either holds or does not. That check is what the products are drawn with rather than a property claimed for them afterwards.

The infinite case, where both theorems still hold and neither helps #

A linear molecule has infinitely many operations — every rotation about the axis — and the two theorems survive the limit in a form that is true and useless.

C∞v has infinitely many classes, one for each rotation angle, so it has infinitely many representations: Σ⁺, Σ⁻, Π, Δ, Φ and onwards without end. The dimension theorem becomes a statement about an infinite sum that does not converge in any useful way, since the group order is infinite too.

An infinite group, worked in a finite one works out what is done in practice: the calculation is carried out in a finite subgroup, C2v or D2h, and the answer is translated back. What the counting theorems say about that trade is exactly the cost. C2v has four representations; C∞v has infinitely many. The correspondence must therefore be many-to-one, and it is — Π and Φ and every other odd-indexed species map to the same pair of C2v labels, and a calculation done in the subgroup cannot tell them apart.

That is a cleaner statement of the trade than “the finite group is an approximation”. Nothing is approximated. Information is discarded, and the counting theorems say precisely how much: the number of distinct answers available drops from infinite to four.

Two conditions, and what each one catches #

The table being square and the dimensions summing to the order are two separate constraints, and it is worth noticing that they catch different mistakes — so a table satisfying one and failing the other is wrong in a locatable way.

The count of representations against the count of classes catches a mis-sorted group. If two operations have been put in the same class that should be in different ones, or the reverse, the class count changes and the two sides stop matching. That is a mistake about the group’s structure rather than about its representations.

The sum of squared dimensions against the order catches a missing or spurious representation. A row left out reduces the sum below the order; an invented one pushes it above. That is a mistake about the table’s contents rather than about the group.

So the pair localise a fault. A table with the right number of rows and the wrong sum has a row with the wrong dimension in it; one with the right sum and the wrong number of rows has been built on a bad class structure; and one failing both has more wrong with it than either check can say.

Both are integer conditions on small numbers, both take a moment, and neither requires knowing what any of the representations mean — which makes them the right first thing to do with any character table that has been typed in by hand.

What is left unproved here #

Two things in this essay are quoted rather than derived, and both are worth naming.

The great orthogonality theorem is the source of everything above and is not proved here. It is a statement about matrix elements averaged over a group, it takes a page, and the counting theorems are its corollaries rather than the other way round.

That the characters span the class functions — the harder half of the first theorem — follows from the regular representation containing every irreducible representation, which is exactly what the reduction above demonstrated. So that half is in fact covered here, in the section that looked like it was only about dimensions. The regular representation is doing two jobs at once, which is why it appears in every proof of both results.

The rest of it is arithmetic, checked in nineteen places, and the arithmetic is what makes a character table a closed object rather than a list somebody stopped adding to.

Where the counting theorems earn their keep #

The two theorems are stated here for the nineteen groups tabulated, and their real use is for groups without a table.

Every group a molecule can fall to computes the subgroups of a point group by closing subsets of its operations, and finds thirty below methane’s Td of which eight have no table here.

For each of the eight without one the search still returns an order and a class count — and the two theorems then say, without any further work, how many representations the missing table has and what its dimensions must be. A group of order eight with eight classes is entirely one-dimensional. A group of order twelve with four classes has dimensions squaring to twelve in four terms, which is $1 + 1 + 1 + 9$ and nothing else.

That is not the whole table, since the characters themselves still have to be found. But it is the shape of it, obtained from two integers, and it is the reason a missing table is a bounded gap rather than an open question.