Math 202B: Lecture 3

Let $\mathcal{A}$ be an algebra.

Definition 3.1. A basis of orthogonal projections in $\mathcal{A}$ is said to be a Fourier basis of $\mathcal{A}.$

Let $X$ be a finite nonempty set.

Definition 3.2. The function algebra of $X$ is the vector space

$\mathcal{F}(X) = \{A \colon X \to \mathbb{C}\}$

of $\mathbb{C}$ -valued functions on $X$ with multiplication defined by

$[AB](x) = A(x)B(x), \quad x \in X,$

and conjugation defined by

$A^*(x) = \overline{A(x)}, \quad x \in X.$

One says that the operations in $\mathcal{F}(X)$ are defined pointwise.

Problem 3.1. Prove that $\mathcal{F}(X)$ is indeed an algebra.

There is a natural Fourier basis in $\mathcal{F}(X)$ indexed by the points of $X.$ For each $x \in X,$ define the corresponding elementary function $E_x \in \mathcal{F}(X)$ by

$E_x(y) = \begin{cases} 1, \text{ if }x=y \\ 0,\text{ if }x \neq y\end{cases}.$

That is, $E_x(y) = \delta_{xy}$ where $\delta_{xy}$ is the Kronecker delta. The elementary functions are selfadjoint elements of $\mathcal{F}(X)$ because they are real-valued, and they are orthogonal projections because

$[E_xE_y](z)=E_x(z)E_y(z)=\delta_{xz}\delta_{yz}=\delta_{xy}E_x(z).$

Thus, $\{E_x \colon x \in X\}$ is a linearly independent set in $\mathcal{F}(X)$ by Theorem 1.1 in Lecture 1. To see that $\{E_x \colon x \in X\}$ spans $\mathcal{F}(X)$ , observe that any function $A \in \mathcal{F}(X)$ can be written as

$A = \sum\limits_{x \in X} A(x)E_x.$

Using the elementary basis $\{E_x \colon x \in X\}$ of $\mathcal{F}(X)$ , we can forget that the elements of this algebra are functions on $X$ and view them as linear combinations

$A = \sum\limits_{x\in X} \alpha_xE_x, \quad \alpha_x \in \mathbb{C}.$

Combining the fact that the elementary functions are orthogonal projections with the axioms of bilinearity and antilinearity, we recover multiplication and conjugation in $\mathcal{F}(X)$ in the form

$AB = \left(\sum\limits_{x \in X} \alpha_xE_x\right)\left(\sum\limits_{x \in X} \beta_xE_x\right)=\sum\limits_{x,y\in X}\alpha_x\beta_yE_xE_y = \sum\limits_{x \in X} \alpha_x\beta_xE_x$

and

$A^*= \left(\sum\limits_{x \in X} \alpha_xE_x\right)^*=\sum\limits_{x \in X} \bar{\alpha}_xE_x^* = \sum\limits_{x \in X} \bar{\alpha}_xE_x.$

Two sets $X$ and $Y$ are said to be isomorphic if there exists a bijection between them.

Problem 3.1. Prove that two finite nonempty sets $X$ and $Y$ are isomorphic if and only if their function algebras $\mathcal{F}(X)$ and $\mathcal{F}(Y)$ are isomorphic.

Theorem 3.1. An algebra $\mathcal{A}$ admits a Fourier basis if and only if it is isomorphic to a function algebra.

Proof: We have already shown that a function algebra has a Fourier basis. Conversely, let $\mathcal{A}$ be an algebra, and let $\{F^\lambda \colon \lambda \in \Lambda\}$ be a vector space basis of $\mathcal{A}$ indexed by the points of a finite set $\Lambda$ whose cardinality is the dimension of $\mathcal{A}.$ Let $\mathcal{F}(\Lambda)$ be the function algebra of this set, and define a linear transformation

$\mathsf{T} \colon \mathcal{A} \longrightarrow \mathcal{F}(\Lambda)$

by $\mathsf{T}(F^\lambda)=E_\lambda$ , where $E_\lambda \in \mathcal{F}(\Lambda)$ is the elementary function corresponding to $\lambda \in \Lambda$ . This is a vector space isomorphism, since it maps a basis of $\mathcal{A}$ onto a basis of $\mathcal{F}(\Lambda)$ . If $\{F^\lambda \colon \lambda \in \Lambda\}$ is a basis of orthogonal projections in $\mathcal{A}$ , then $\mathsf{T}$ is also an algebra homomorphism. Indeed, by Theorem 1.2 and linearity of $\mathsf{T}$ we have

$\mathsf{T}(I_\mathcal{A}) = \sum\limits_{\lambda \in \Lambda} \mathsf{T}(F^\lambda) = \sum\limits_{\lambda \in \Lambda} E_\lambda = I_{\mathcal{F}(\Lambda)},$

so $\mathsf{T}$ maps the multiplicative unit of $\mathcal{A}$ to that of $\mathcal{F}(\Lambda).$ Next,

$\mathsf{T}(F^\lambda F^\mu) = \mathsf{T}(\delta_{\lambda\mu}F^\lambda)=\delta_{\lambda\mu}E_\lambda = E_\lambda E_\mu = \mathsf{T}(F^\lambda)\mathsf{T}(F^\mu),$

so $\mathsf{T}$ respects multiplication. Finally,

$\mathsf{T}((F^\lambda)^*)=\mathsf{T}(F^\lambda)=E_\lambda=E_\lambda^*=\mathsf{T}(F^\lambda)^*.$

-QED

Theorem 3.1 is simple but important and you should go over its proof carefully. The argument shows that if $\{F^\lambda \colon \lambda \in \Lambda\}$ is a Fourier basis of $\mathcal{A},$ then the linear transformation

$\mathsf{T} \colon \mathcal{A} \longrightarrow \mathcal{F}(\Lambda)$

defined by

$\mathsf{T}(F^\lambda) = E_\lambda, \quad \lambda \in \Lambda,$

is an algebra isomorphism. This isomorphism is called the Fourier transform on $\mathcal{A}$ , and it is generally denoted as $A \mapsto \widehat{A}$ rather than $A \mapsto \mathsf{T}(A).$ Thus,

$\widehat{F^\lambda} = E_\lambda,\quad \lambda \in \Lambda,$

and the argument above shows that the linear isomorphism $\mathcal{A} \to \mathcal{F}(\Lambda)$ so defined is an algebra isomorphism. Note that the Fourier transform on $\mathcal{A}$ is not canonical – it is defined in terms of a specified basis $\{F^\lambda \colon \lambda \in \Lambda\}$ of orthogonal projections in $\mathcal{A}.$ For any element $A \in \mathcal{A}$ , its expansion in the Fourier basis is written

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

and the coefficients in this expansion are called the Fourier coefficients of $A.$ We thus have

$\widehat{A} = \sum\limits_{\lambda \in \Lambda} \widehat{A}(\lambda)E_\lambda,$

which is the elementary expansion of a function $\widehat{A} \in \mathcal{F}(\Lambda).$ The function $\widehat{A} \in \mathcal{F}(\Lambda)$ is called the Fourier transform of the algebra element $A \in \mathcal{A}.$ The practical value of the Fourier transform lies in the fact that

$\widehat{A * B} = \widehat{A}\widehat{B}$

where $A*B$ is the product of $A,B \in \mathcal{A}$ , which might be convoluted and difficult to calculate. On the other hand, the pointwise product of the corresponding functions $\widehat{A},\widehat{B}\in \mathcal{F}(\Lambda)$ is very simple, both conceptually and computationally.

Math 202B: Lecture 3

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