Math 202B: Lecture 26

We begin by recalling our general algebraic perspective on the discrete Fourier transform (DFT). A Fourier basis of an algebra \mathcal{A} is a basis \{F^\lambda \colon \Lambda\} of pairwise orthogonal selfadjoint idempotents. If such a basis exists, then \mathcal{A} is necessarily commutative. Note also that since we have not mentioned a scalar product on the algebra \mathcal{A}, the orthogonality of the Fourier basis vectors can only refer to the fact that the vector product of two distinct Fourier basis vectors is the zero vector.

There is no guarantee that a given algebra \mathcal{A} admits a Fourier basis, but if it does then every element A \in \mathcal{A} can be expanded on the Fourier basis,

A = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)F^\lambda,

and the coefficients in this expansion define a function \widehat{A} \in \mathcal{F}(\Lambda). One of our most basic but important theorems from early on in the course is that the map A \mapsto \widehat{A} is an algebra isomorphism \mathcal{A} \to \mathcal{F}(\Lambda). This algebra isomorphism is called the Fourier transform on \mathcal{A}.

The above construction cannot work for noncommutative algebras, in the sense that a noncommutative algebra cannot be isomorphic to a function algebra. We do however have an alternative path to pursue in this case. Instead of looking for a Fourier basis in \mathcal{A}, we look for a Frobenius scalar product on \mathcal{A}. When such a scalar product exists, then \mathcal{A} becomes a Hilbert space in a way which is compatible with its algebra structure. In this case, we have shown that \mathcal{A} is isomorphic not to an algebra of functions on a finite set, but rather to an algebra of operators on a finite-dimensional Hilbert space. More precisely, we have an algebra homomorphism \Phi \colon \mathcal{A} \to \mathrm{End} \mathcal{A} which sends each A \in \mathcal{A} to the corresponding left multiplication operator \Phi(A) \in \mathrm{End}\mathcal{A}. A good name for the homomorphism \Phi would be the Wedderburn transform; the pair (\mathcal{A},\Phi) together is the left-regular representation of \mathcal{A}. The Wedderburn transform is an injective algebra homomorphism, and it replaces the Fourier transform in the sense that it sets up an isomorphism between the abstract algebra \mathcal{A} and the subalgebra of \mathrm{End}\mathcal{A} consisting of left multiplication operators \Phi(A), A \in \mathcal{A}.

Although the above describes the Wedderburn transform as a replacement for the Fourier transform in the noncommutative setting, there is no reason that it cannot be applied in the commutative setting. Namely, the best case scenario is that we have a commutative algebra \mathcal{A} in which we have successfully identified a Fourier basis \{F^\lambda \colon \lambda \in \Lambda\}, and we also are able to identify a Frobenius scalar product on the commutative algebra \mathcal{A}. In this case, both the Fourier transform \mathcal{A} \to \mathcal{F}(\Lambda) and the Wedderburn transform \mathcal{A} \to \mathrm{End}\mathcal{A} are in play. They are moreover related in a very useful way: the Fourier basis \{F^\lambda \colon \lambda \in \Lambda\} is an eigenbasis for the Wedderburn transform \Phi(A) of every A \in \mathcal{A}, and the corresponding eigenvalues are given by the Fourier transform of A,

\Phi(A)F^\lambda = \widehat{A}(\lambda)F^\lambda, \quad \lambda \in \Lambda.

We have concretely implemented this compelling abstract picture in the case where \mathcal{A}=\mathcal{C}(G) is the convolution algebra of a finite abelian group, G. In order to construct a Fourier basis of \mathcal{C}(G), we started from the observation that the set \{\chi^\lambda \colon \lambda \in \Lambda\} of multiplicative characters of G, which is parameterized by an explicit set \Lambda corresponding to the decomposition of G as a product of cyclical groups, forms an orthogonal basis of \mathcal{C}(G). This allows us to define a Fourier basis in \mathcal{C}(G) consisting of the normalized multiplicative characters

F^\lambda = \frac{1}{|G|}\sum\limits_{g \in G} \chi^\lambda(g)E_g.

In order to see how the Fourier transform and the Wedderburn transform work together in the setting of convolution algebras of finite abelian groups, I asked you to work out the case where G is a cyclic group in detail, where the synchrony between the two transforms gives a method to diagonalize circulant matrices. I hope this was instructive, and it is definitely something worth reviewing more than once.

Now suppose G is a nonabelian group. Then, \mathcal{C}(G) is a noncommutative algebra which cannot be isomorphic to a function algebra. However, the center \mathcal{K}(G) of \mathcal{C}(G), which consists of functions on G which are constant on conjugacy classes of G, is a commutative algebra associated to G. In this lecture we will see that the Fourier transform can indeed be concretely implemented for the class algebra of any finite group, which generalizes our prior implementation for convolution algebras of abelian groups. To achieve this extension, we will rely on our extension of multiplicative characters of finite abelian groups to irreducible characters of arbitrary finite groups. I hope that at this point you are close to internalizing the fact that, for abelian groups, multiplicative characters and irreducible characters are the same thing.

Let \Lambda be a set parameterizing irreducible unitary representations of G (or equivalently irreducible linear representations of \mathcal{C}(G)) up to isomorphism. For each \lambda \in \Lambda, let (V^\lambda,\varphi^\lambda) be a representative of the corresponding isomorphism class, and let

\chi^\lambda(g) = \mathrm{Tr} \varphi^\lambda(g), \quad g \in G.

We know that the set \{\chi^\lambda \colon \lambda \in \Lambda\} forms an orthogonal basis of \mathcal{K}(G). Let \{C_\alpha \colon \alpha \in \Lambda\} be the set of conjugacy classes in G. For each \alpha \in \Lambda, let K_\alpha be the indicator function of C_\alpha. Then, the set \{K_\alpha \colon \alpha \in \Lambda\} is an orthogonal basis of \mathcal{K}(G),

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

The connection coefficients of the class basis,

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

count solutions to certain equations in the underlying group, G. Namely, c_{\alpha\beta\gamma} is the number of pairs (h,k) \in C_\alpha \times C_\beta such that g=hk, where g \in C_\gamma. Let \chi^\lambda_\alpha denote \chi^\lambda(g) for any g \in C_\alpha. In Lecture 25, we posed the following problem which expresses the connection coefficients of \mathcal{K}(G) in terms of the irreducible characters of G. You should reflect on what this says in the case where G is abelian.

Theorem 26.1. 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}.

Using this result, we construct a Fourier basis of \mathcal{K}(G). Here is the basic claim: the central functions

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

form a Fourier basis of \mathcal{K}(G). Let us first check that these candidate functions are nonzero.

Problem 26.1. Compute the L^2-norm of F^\lambda, and show that \sum_{\lambda \in \Lambda} \|F^\lambda\|^2=1.

Now, on to the main event.

Theorem 26.1. The |\Lambda| nonzero functions

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

are pairwise orthogonal selfadjoint idempotents in \mathcal{K}(G).

Proof: 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 26.1, 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.

This completes the orthogonal idempotent part of the proof. It remains to check that the proposed Fourier basis consists of selfadjoint elements, and we leave this as an exercise. \square

Published by Jonathan Novak

Mathematician at UC San Diego.

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