Math 202B: Lecture 23

Let G be a finite group and let \Lambda be a set parameterizing isomorphism classes of irreducible unitary representations of G. For each \lambda \in \Lambda, let (V^\lambda,\varphi^\lambda) be a representative of the corresponding isomorphism class, and let \chi^\lambda \colon G \to \mathbb{C} be the corresponding irreducible character, i.e. function G \to \mathbb{C} defined by

\chi^\lambda(g), \quad g \in G.

The character basis \{\chi^\lambda \colon \lambda \in \Lambda\} of the class algebra \mathcal{Z}(G) satisfies the orthogonality relations

\langle \chi^\lambda,\chi^\mu\rangle = \sum\limits_{g \in G} \overline{\chi^\lambda(g)}\chi^\mu(g)=\delta_{\lambda\mu}|G|.

In particular, the cardinality of \Lambda is equal to the dimension of \mathcal{Z}(G), which is the number of conjugacy classes in G.

Problem 23.1. Show that that orthogonality of irreducible characters can equivalently be written

\sum\limits_{g \in G} \chi^\lambda(g^{-1})\chi^\mu(g)=\delta_{\lambda\mu}|G|.

Let \{C_\alpha \colon \alpha \in \Lambda\} be the set of conjugacy classes in G. For each \alpha \in \Lambda, let K_\alpha \colon G \to \mathbb{C} be the indicator function of the class C_\alpha. The class basis \{K_\alpha \colon \alpha \in \Lambda\} satisfies the orthogonality relations

\langle K_\alpha,K_\beta\rangle = \delta_{\alpha\beta}|C_\alpha|, \quad \alpha,\beta \in \Lambda.

Writing

\chi^\lambda = \sum\limits_{\alpha \in \Lambda} \chi^\lambda_\alpha K_\alpha

for the expansion of irreducible characters in the class basis, where \chi^\lambda_\alpha denotes \chi^\lambda(g) for any g \in C_\alpha, character orthogonality takes the form

\langle \chi^\lambda,\chi^\mu \rangle = \sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\mu_\alpha|C_\alpha|=\delta_{\lambda\mu}|G|.

Problem 23.2. We say that G is ambivalent if g and g^{-1} are conjugate for every g \in G. If G is ambivalent, prove that

\langle \chi^\lambda,\chi^\mu \rangle = \sum\limits_{\alpha \in \Lambda}\chi^\lambda_\alpha\chi^\mu_\alpha|C_\alpha|=\delta_{\lambda\mu}|G|.

Theorem 23.1. (Dual character orthogonality) For any \alpha,\beta \in \Lambda, we have

\sum\limits_{\lambda \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\lambda_\beta=\delta_{\alpha\beta}\frac{|G|}{|C_\alpha|}.

Proof: The character table of G is the square matrix

\mathbf{X} = \begin{bmatrix} \chi^\lambda_\alpha \end{bmatrix}_{\alpha,\lambda \in \Lambda},

where we think of \alpha as the column index and \lambda as the row index. The modified character table

\tilde{\mathbf{X}}_G = \begin{bmatrix} \sqrt{\frac{|C_\alpha|}{|G|}}\chi^\lambda_\alpha\end{bmatrix}

is a unitary matrix: the orthonormality of its rows is equivalent to character orthogonality. The transpose of a unitary matrix is again a unitary matrix, and dual character orthogonality is this statement applied to the modified character table of G.

-QED

We will now apply Dual Character Orthogonality to express the multiplication tensor of the class algebra \mathcal{Z}(G), with respect to the class basis, in terms of the irreducible characters of G. This means that we will find a formula for the entries of the three-dimensional array [c_{\alpha\beta\gamma}]_{\alpha,\beta,\gamma \in \Lambda} such that

K_\alpha K_\beta = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma} K_\gamma.

Theorem 23.1. For any \alpha,\beta,\gamma \in \Lambda, we have

c_{\alpha\beta\gamma} = \frac{|C_\alpha||C_\beta|}{|G|}\sum\limits_{\lambda \in \Lambda} \frac{\chi^\lambda_\alpha\chi^\lambda_\beta\overline{\chi^\lambda_\gamma}}{\dim V^\lambda}.

Proof: Let (V^\lambda,\Phi^\lambda) be the irreducible linear representation of \mathcal{C}(G) corresponding to \lambda \in \Lambda. On one hand, we have

\Phi^\lambda(K_\alpha K_\beta)=\Phi^\lambda(K_\alpha)\Phi^\lambda(K_\beta)=\omega^\lambda_\alpha\omega^\lambda_\beta I,

where I \in \mathcal{L}(V^\lambda) is the identity operator. On the other hand,

\Phi^\lambda(K_\alpha K_\beta) = \sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma} \Phi^\lambda(K_\gamma)=\left(\sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma} \omega^\lambda_\gamma\right) I.

This gives the equality

\omega^\lambda_\alpha\omega^\lambda_\beta=\sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma} \omega^\lambda_\gamma,

or equivalently

\frac{|C_\alpha||C_\beta|}{\dim V^\lambda}\chi^\lambda_\alpha \chi^\lambda_\beta=\sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}|C_\gamma|\chi^\lambda_\gamma.

Now we want to extract a particular coefficient on the right hand side. To do so, choose \nu \in \lambda, multiply both sides by \overline{\chi^\lambda_\nu}, and sum over \lambda to obtain

\frac{|C_\alpha||C_\beta|}{|G|}\sum\limits_{\lambda \in \Lambda} \frac{\chi^\lambda_\alpha\chi^\lambda_\beta\overline{\chi^\lambda_\nu}}{\dim V^\lambda} =\sum\limits_{\gamma \in \Lambda} c_{\alpha\beta\gamma}|C_\gamma|\sum\limits_{\lambda \in \Lambda}\chi^\lambda_\gamma\overline{\chi^\lambda_\nu}.

By Dual Character Orthogonality, the sum on the right hand side collapses to a single term, namely

|G|c_{\alpha\beta\nu},

and the result follows.

-QED

We are now in position to generalize the Fourier transform from the convolution algebra \mathcal{C}(G) with G abelian to the class algebra \mathcal{Z}(G) with G arbitrary.

Definition 23.1. The Fourier basis of \mathcal{Z}(G) is defined by

F^\lambda = \frac{\dim V^\lambda}{|G|}\sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}K_\alpha, \quad \lambda \in \Lambda.

The key feature of this putative basis is the following.

Theorem 23.2. The basis \{F^\lambda \colon \lambda \in \Lambda\} consists of orthogonal projections. In particular, the span of \{F^\lambda \colon \lambda \in \Lambda\} is a subalgebra of \mathcal{Z}(G) isomorphic to the pointwise function algebra \mathcal{F}(\Lambda).

Proof: For any \lambda,\mu \in \Lambda, we have

F^\lambda F^\mu = \frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} K_\alpha K_\beta=\frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} c_{\alpha\beta\gamma}K_\gamma.

By Theorem 23.1, the right hand side is

\frac{\dim V^\lambda}{|G|}\frac{\dim V^\mu}{|G|}\sum\limits_{\alpha \in \Lambda}\sum\limits_{\beta \in \Lambda} \sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\alpha} \overline{\chi^\mu_\beta} K_\gamma \frac{|C_\alpha||C_\beta|}{|G|}\sum\limits_{\nu \in \Lambda} \frac{\chi^\nu_\alpha\chi^\nu_\beta\overline{\chi^\nu_\gamma}}{\dim V^\nu}.

This quadruple sum over \Lambda can be reorganized as

\frac{\dim V^\lambda \dim V^\mu}{|G|^3}\sum\limits_{\nu \in \Lambda}\frac{1}{\dim V^\nu} \left(\sum\limits_{\alpha \in \Lambda} \overline{\chi^\lambda_\alpha}\chi^\nu_\alpha|C_\alpha| \right)\left(\sum\limits_{\beta \in \Lambda} \overline{\chi^\mu_\beta}\chi^\nu_\beta|C_\beta|\right)\sum\limits_{\gamma \in \Lambda}\overline{\chi^\nu_\gamma}K_\gamma.

Character orthogonality (original, not dual) collapses the sums over \alpha and \beta, leaving

\frac{\dim V^\lambda \dim V^\mu}{|G|}\sum\limits_{\nu \in \Lambda} \frac{1}{\dim V^\nu}\delta_{\lambda\nu}\delta_{\mu\nu}\sum\limits_{\gamma \in \Lambda}\overline{\chi^\nu_\gamma}K_\gamma =\delta_{\lambda\mu} \frac{\dim V^\lambda}{|G|}\sum\limits_{\gamma \in \Lambda}\overline{\chi^\lambda_\gamma}K_\gamma.

-QED

Now, the set \{F^\lambda \colon \lambda \in \Lambda\} is linearly independent since any set of orthogonal projections is independent. What remains to be shown is that it spans the class algebra \mathcal{Z}(G). In the case of an ambivalent group, the Fourier basis is simply a rescaling of the character basis by nonzero (indeed, positive) factors. In the general case we have the following.

Theorem 23.3. For any \alpha \in \Lambda, we have

K_\alpha = |C_\alpha| \sum\limits_{\lambda \in \Lambda}\frac{\overline{\chi^\lambda_\alpha}}{\dim V^\lambda}F^\lambda.

Problem 23.3. Prove Theorem 23.3.

Published by Jonathan Novak

Mathematician at UC San Diego.

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