13.5.1 Invariants and the Inner Product on Class Functions
Proposition 13.5.1 (Averaging projection). Let be a representation of , and let Then is a projection onto : its image is , and it is the identity on . Consequently,
Proof. If , every term equals , so . For any and , since runs over as does; so . Thus and . Then : every is with , and a vector in both summands satisfies . In a basis adapted to this decomposition, has matrix with ones, so . By linearity of the trace, . ∎
Corollary 13.5.2 (Dimension of ). For representations and of ,
Proof. By Proposition 13.1.10, . Apply Proposition 13.5.1 to the representation , and use (Proposition 13.4.3). ∎
This formula suggests an inner product. Let denote the complex vector space of class functions on . A class function is determined by its values on the conjugacy classes, and these values can be chosen freely, so the indicator functions of the conjugacy classes form a basis of . In particular, is the number of conjugacy classes of .
Definition 13.5.3 (Inner product on class functions). For class functions and on , let
This is a positive-definite Hermitian inner product on . It is conjugate-linear in the first variable and linear in the second, the convention of Chapter 9; it satisfies ; and for . Grouping the terms by conjugacy classes, where runs over the conjugacy classes of and denotes the value of on . This is the form in which we compute with character tables.
Theorem 13.5.4 (Orthogonality of characters). For representations and of , . In particular, if and are irreducible, then So the characters of pairwise non-isomorphic irreducible representations are orthonormal with respect to .
Proof. The first statement is Corollary 13.5.2, and the second follows from it by Corollary 13.3.2. ∎
Corollary 13.5.5 (At most as many irreducibles as conjugacy classes). Up to isomorphism, has finitely many irreducible representations, and their number is at most the number of conjugacy classes of .
Proof. Orthonormal vectors are linearly independent: if , applying gives . So the characters of pairwise non-isomorphic irreducible representations form a linearly independent subset of , and there are at most of them. ∎
From now on, denote the irreducible representations of , one from each isomorphism class, with characters and dimensions . So far we know that is at most the number of conjugacy classes; Theorem 13.6.2 below shows that the two numbers are equal.
Corollary 13.5.6 (A representation is determined by its character). Let be a representation of . In any decomposition of into irreducible subrepresentations, the number of summands isomorphic to is So , and the multiplicities do not depend on the decomposition. Two representations with the same character are isomorphic.
Proof. If (Corollary 13.2.3), then by Proposition 13.4.3, and by Theorem 13.5.4. So the are determined by , and two representations with the same character have the same multiplicities, hence are isomorphic. ∎
Corollary 13.5.7 (Irreducibility criterion). If , then . In particular, is irreducible if and only if .
Proof. By orthonormality, . This sum of squares of nonnegative integers equals exactly when one is and the others are , that is, when for some . ∎
Note. The number is a quick count of the irreducible pieces of . For small values it says:
| is |
|---|
| irreducible; |
| a direct sum of two non-isomorphic irreducibles; |
| a direct sum of three pairwise non-isomorphic irreducibles; |
| a direct sum of four pairwise non-isomorphic irreducibles, or for an irreducible . |
Twisting by a -dimensional representation is a cheap way to produce new irreducible representations from old ones.
Proposition 13.5.8 (Twisting by a -dimensional representation). Let be a -dimensional representation of , given by a homomorphism . For every representation of , , and is irreducible if and only if is.
Proof. Fix a basis vector of . The map , , is a linear isomorphism, and So, transported by , acts on as the operator acts on ; taking traces gives . A subspace of is invariant under all the operators if and only if it is invariant under all , because the are nonzero scalars. Hence matches the subrepresentations of with those of , and one is irreducible exactly when the other is. ∎
13.5.2 The Regular Representation
Definition 13.5.9 (Regular representation). The regular representation of is the permutation representation for the action of on itself by left multiplication: has basis , and .
Proposition 13.5.10 (Decomposition of the regular representation). Each irreducible representation occurs in with multiplicity : .
Proof. By Proposition 13.4.3(v), is the number of with , which is if and otherwise. By Corollary 13.5.6, the multiplicity of in is ∎
Corollary 13.5.11 (Sum of squares of the dimensions). .
Proof. Compare dimensions in Proposition 13.5.10: . ∎