We know that are orthonormal in , so they are linearly independent and is at most the number of conjugacy classes. We now prove that they also span , so that equals the number of conjugacy classes. The proof uses a weighted version of the averaging projection.
13.6.1 Weighted Averages
Proposition 13.6.1 (Weighted averaging operators). Let be a class function on and a representation of , and let Then is a homomorphism of representations, and .
Proof. For , where we substituted , which runs over as does, and used that is a class function. So for every . The trace formula follows from linearity of the trace. ∎
For the constant function , is the averaging projection of Proposition 13.5.1. We also note two simple facts. If is a direct sum of subrepresentations, then preserves and and restricts to and on them, because each does. If is an isomorphism of representations, then , because for each .
Theorem 13.6.2 (Irreducible characters form an orthonormal basis). The characters of the irreducible representations of form an orthonormal basis of with respect to . Every class function on can be written as In particular, the number of irreducible representations of , up to isomorphism, equals the number of conjugacy classes of .
Proof. Orthonormality is Theorem 13.5.4. We first show that a class function with for all is zero.
The complex conjugate is also a class function. By Proposition 13.6.1, is a homomorphism of representations , so by Schur’s Lemma it equals for some . Taking traces, so for every . Every representation is a direct sum of subrepresentations, each isomorphic to some (Corollary 13.2.3), so by the two facts above, for every representation . Take to be the regular representation , and apply to the basis vector indexed by the identity element: Since the are linearly independent, for every .
Now let be any class function, and let . For each , orthonormality gives , hence . By what we have shown, , which is the stated formula for . So span and form an orthonormal basis of it, and is the number of conjugacy classes. ∎
Corollary 13.6.3 (Square character tables). The character table of a finite group is a square matrix.
Note. By Theorem 3.4.6, two permutations in are conjugate exactly when they have the same cycle type, and cycle types correspond to partitions of . So the number of irreducible representations of is , the number of partitions of . For instance , in agreement with Proposition 13.3.5, while and .
13.6.2 Projection onto Isotypic Components
Let be a representation with a decomposition into irreducible subrepresentations. The isotypic component of of type is the sum of those that are isomorphic to ; it is isomorphic to , where is the multiplicity of in . Weighted averages give an explicit projection onto it.
For irreducible and , take in Proposition 13.6.1. As in the proof of Theorem 13.6.2, is a scalar with , so
Proposition 13.6.4 (Projection onto an isotypic component). Let be a representation of , and let Then is the identity on every subrepresentation of isomorphic to , and zero on every irreducible subrepresentation not isomorphic to . Hence, for any decomposition of into irreducible subrepresentations, is the projection onto the isotypic component of type along the sum of the other isotypic components. In particular, this isotypic component is , so it does not depend on the decomposition.
Proof. We have . On an irreducible subrepresentation , acts as , which is conjugate to . By the computation above, this is the identity if and zero otherwise. ∎
Note. When is the trivial representation, and , and is the averaging projection onto of Proposition 13.5.1.
13.6.3 Column Orthogonality
Orthonormality of the irreducible characters says that the rows of the character table are orthonormal, once the columns are weighted by the sizes of the classes. Because the table is square, its columns are orthogonal as well.
Proposition 13.6.5 (Column orthogonality). Let be the conjugacy classes of , with representatives . Then
Proof. Let be the matrix with entries , which is square by Theorem 13.6.2, and let . Orthonormality of the characters, written class by class, says So the square matrix has inverse , and therefore also , that is, . The entry of this identity is . ∎
Note. By Proposition 3.2.13, , so is the order of the centralizer . Column orthogonality gives an independent check on a computed character table. It also shows that characters separate conjugacy classes: if for every , then , so and lie in the same conjugacy class.
Exercise 13.6.6 (Representations and Characters).
Decompose the permutation representation of into irreducibles using the inner product .
Let be the standard representation of . Use characters and the inner product to decompose , and into irreducibles, and compare with Example 13.3.6.
Compute the character table of the Klein four-group . How many irreducible representations does it have, and what are their dimensions?
Let act on a finite set , and let be the permutation representation.
- Show that is the number of orbits of on , and deduce Burnside’s lemma (Theorem 11.1.3) from Proposition 13.5.1.
- Suppose that and that acts doubly transitively: for any two pairs and of distinct elements of , there is with and . Show that , where is the trivial representation and is irreducible. Conclude that the standard representation of is irreducible for every .
Show solution
(1) The permutation character counts fixed points: , , . With the class sizes , and . So .
(2) The character of is , and and , so . For the symmetric and exterior squares we need : the squares of , , are , , , so . By Proposition 13.4.3(vi), The first is and the second is , so and , as in Example 13.3.6.
(3) The group is abelian of order , so each element is its own conjugacy class, and there are four irreducible representations, all -dimensional (Proposition 13.1.7); indeed . They are the elements of the dual group: for , the character .
(4) (a) A vector is invariant if and only if for all and , that is, if and only if is constant on each orbit. So the sums , one for each orbit , form a basis of . By Proposition 13.5.1 and Proposition 13.4.3(v), the number of orbits is therefore , which is Burnside’s lemma.
(b) The line spanned by is a copy of the trivial representation, is a subrepresentation complementary to it, and . The action is transitive (given , choose and , and use double transitivity), so by (a), . The diagonal and its complement are the orbits of on , and ; so by (a) applied to , . Since is real, and is irreducible by Corollary 13.5.7. The group acts doubly transitively on for , and is then the standard representation.