The character table is almost unitary except that we have to weight thecolumns by the sizes of the conjugacy classes. The transpose of a unitary matrixis also unitary, so the columns of a character table are orthogonal (for thehermitian inner product) and have norms given by |G|/size of conjugacy classwhich is just the order of the centralizer of an element of the conjugacy class.The orthogonality of characters makes it very easy to work with representations.For example:• We can count the number of times an irreducible representation occurs insome representation by taking the inner product of their characters.• A representation is irreducible if and only if its character has norm 1.• Two representations are isomorphic if and only if they have the same char-acter. (The analogue of this fails in cases when we do not have completereducibility, such as modular representations of finite groups.)We can use this to decompose the regular representation: its character is |G|at the identity and 0 elsewhere. So its inner product with the character of anyirreducible representation V is dim(V ), so V occurs dim(V ) times. In particular|G| =Pdim(V )2, and we can use this to check that a list of irreducibles iscomplete.Theorem 267 The number of irreducible characters of a finite group is equalto the number of conjugacy classes, and the irreducible characters form an or-thonormal basis for the class functions.Proof The irreducible characters are orthogonal class functions, so it is suf-ficient to show that the number of conjugacy classes is at most the number ofirreducible representations. The key point is to observe that any class functionis in the center of the group ring and so by Schur’s lemma acts as a scalar on anyirreducible representation. If the number of conjugacy classes were greater thanthe number of irreducibles, we could therefore find a non-zero class functionacting as 0 on all irreducibles, and therefore as 0 in the regular representation,which is nonsense. It is natural to ask if there is a canonical correspondence between irreduciblesand conjugacy classes. At first glance, the answer seems to be obviously no. Forexample, for infinite compact groups the number of conjugacy classes cane un-countable while the number of irreducibles is countable, and for even the cyclicgroup of order 3 there is no canonical way to match up the irreducibles withelements of the group. However a close look show that in many cases there doesindeed seem to b e some sort of natural correspondence. For example, for sym-metric groups the conjugacy classes correspond to partitions, and we will latersee the same is true for their representations. An even deeper look shows thatrepresentations of semisimple Lie groups (and automorphic forms) correspond toconjugacy classes of their “Langlands dual group” though this correspondenceneed not be 1:1 in general. This is closely related to Langlands functoriality: ifthe correspondence were 1:1 (which it is not in general) then a homomorphismof Langlands dual groups would induce a map between representations or au-tomorphic forms on different groups. This is expected to hold and is known asLanglands functoriality.103We can describe the irreducible representations of a group most convenientlyby giving their character tables: These are just square matrices giving the valuesof the characters on the conjugacy classes.One reason why character tables are useful is that they are usually easy tocompute. (This only applies to complex character tables: modular charactertables are far harder to compute.)Examples: We compute the character tables of S3, S4, S5using ad hoc meth-ods. (In fact we will see later how to compute the characters of all symmetricgroups in a uniform way.)We can guess the character table of SU(2). has an obvious 2-dimensionalrepresentation. Its conjugacy classes correspond to diagonal matrices with en-tries u, u−1for u of absolute value 1, except that we can exchange the entries byconjugating by (0 ii 0) (generating the Weyl group: note that there is no matrix oforder 2 in SU(2) swapping the entries). The trace of this on the 2-dimensionalrepresentation is just u + u−1. We can also take the representation consistingof polynomials of degree n on this 2-dimensional representation. This has abasis xn, xn−1y, . . . , ynon which the trace is un+ un−2+ · · · u−n, so this isthe character of this n + 1-dimensional representation. So the characters aregiven by (un+1− u−n−1)/(u − u−1), which will turn out to be a special case ofthe Weyl character formula. We will soon see that these are all the irreduciblerepresentations.What about orthogonality of characters? The characters are orthogonal evenfunctions on the unit circle if we change the measure by a factor of (u − u−1)2.Where has this funny-looking factor come from? The answer is that we shouldreally be integrating over the whole of SU(2), not just over the torus. Whilethe torus does contain a representative of each conjugacy class, we cannot justchange integrals of class functions on the group to integrals over the torus,because some conjugacy classes are in some sense bigger than others. Thefactor (u − u−1)2accounts for the fact that some conjugacy classes are bigger,and is essentially the Weyl integration formula for