Math 202B: Lecture 12

Let \mathcal{C}(G) be the convolution algebra of a finite group G. Then, the left regular representation

\mathsf{L} \colon \mathcal{C}(G) \longrightarrow \mathcal{L}(V)

is an injective algebra homomorphism from the convolution algebra \mathcal{C}(G) into the the linear algebra of \mathcal{C}(G) viewed as a Hilbert space. That is, the regular representation converts convolution of functions into matrix multiplication.

If G is abelian, then the Fourier transform

\widehat{\cdot} \colon \mathcal{C}(G) \longrightarrow \mathcal{F}(\Lambda)

is an algebra isomorphism between the convolution algebra \mathcal{C}(G) and the function algebra of the dual group \Lambda of G. That is, the Fourier transform converts convolution into pointwise multiplication.

Let us reconcile these two points of view on the convolution algebra of a finite abelian group. Let A \in mathcal{C}(G) be any function on G. Let

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

be the Fourier expansion of A. Taking the left regular representation, we have

\mathsf{L}(A) = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)P^\lambda,

where P^\lambda=\mathsf{L}(F^\lambda) is the linear operator on the Hilbert space V=\mathcal{C}(G) given by the image of the Fourier basis vector F^\lambda under the injective algebra homomorphism \mathsf{L}. In other words,

P^\lambda, \quad \lambda \in \Lambda,

is a basis of orthogonal projections for the image of \mathcal{C}(G) in \mathcal{L}(V) under \mathsf{L}, and

\mathsf{L}(A) = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)P^\lambda

is the spectral decomposition of the operator \mathsf{L}(A). This says that the operators \mathsf{L}(A) making up the image of the commutative algebra \mathcal{C}(G) in \mathcal{L}(V) are simultaneously diagonalizable, which we already knew from our characterization of commutative subalgebras of \mathcal{L}(V), and moreover that calculating the eigenvalues of the operator \mathsf{L}(A) \in \mathcal{L}(V) is the same thing as calculating the Fourier transform of the function A \in \mathcal{C}(G).

Problem 12.3. A circulant matrix is a complex square matrix of the form

C=\begin{bmatrix} c_0 & c_1 & \dots & c_{n-1} \\ c_{n-1} & c_0 & \dots & c_{n-2} \\ \vdots & \vdots & {} & \vdots \\ c_1 & c_2 & \dots & c_0 \end{bmatrix}.

Determine the eigenvalues and eigenvectors of C. Hint: show that every n \times n circulant matrix is the image of a function on the cyclic group of order n in the regular representation.

As an extension to the above problem, you may consider how the uncertainty principle for the Fourier transform relates the sparsity of a circulant matrix to the dimension of its kernel. (Note to self: perhaps do this in Lecture in the next iteration of Math 202B).

Published by Jonathan Novak

Mathematician at UC San Diego.

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