Math 202B: Lecture 3

Definition 3.1. A linear transformation $\Phi \colon \mathcal{A} \to \mathcal{B}$ from one algebra to another is said to be an algebra homomorphism if it respects conjugation,

$\Phi(A^*)=\Phi(A)^*, \quad A \in \mathcal{A},$

respects multiplication,

$\Phi(A_1A_1) = \Phi(A_1)\Phi(A_2), \quad A_1,A_2 \in \mathcal{A},$

and is unital,

$\Phi(I_\mathcal{A})=I_\mathcal{B}.$

We can now define the category of algebras.

Definition 3.2. The category $\mathbf{Alg}$ has algebras as its objects and algebra homomorphisms as its morphisms.

Going forward, we will almost exclusively work in the full subcategory $\mathbf{FAlg}$ of $\mathbf{Alg}$ whose objects are finite-dimensional algebras. In order to lighten the terminology, when we say “algebra” we will mean a finite-dimensional algebra, and when dealing with infinite-dimensional objects we will explicitly say “infinite-dimensional algebra.”

Now we come to a basic class of algebras attached to finite sets: function algebras. These will be our model examples of commutative algebras.

Definition 3.3. The function algebra $\mathcal{F}(X)$ of a finite set $X$ is the vector space of functions $A \colon X \to \mathbb{C}$ with conjugation and multiplication defined by

$A^*(x)=\overline{A(x)} \quad\text{and}\quad [AB](x)=A(x)B(x).$

We are already quite familiar with $\mathcal{F}(X)$ as a vector space, since when equipped with the scalar product

$\langle A,B \rangle = \sum\limits_{x\in X} \overline{A(x)}B(x)$

it is the model example of a Hilbert space from Math 202A. In particular, we already know that the set $\{E_x \colon x \in X\}$ consisting of the elementary functions $E_x(y) = \delta_{xy}$ forms an orthonormal basis of the Hilbert space $\mathcal{F}(X).$ Now we are taking the next step of equipping $\mathcal{F}(X)$ with a vector product as well as a scalar product, hence promoting our Math 202A quantization functor to a functor

$\mathcal{F} \colon \mathbf{FSet} \longrightarrow \mathbf{FAlg}.$

When viewing $\mathcal{F}(X)$ as an algebra rather than a Hilbert space, we can ask for the classification of selfadjoint elements and unitary elements (since $\mathcal{F}(X)$ is commutative, all elements are normal). It is straightforward to see that $A \in \mathcal{F}(X)$ is selfjadjoint if and only if it is real-valued, and unitary if and only if it is circle-valued (meaning $|A(x)|=1$ for all $x \in X$ ). It is also clear that the group $I(\mathcal{F}(X))$ of invertible elements in $\mathcal{F}(X)$ consists of non-vanishing functions on $X.$

From the algebra point of view, the elementary basis of $\mathcal{F}(X)$ is orthogonal with respect to the vector product (rather than the scalar product) in the sense that

$E_xE_y = \delta_{xy}E_x.$

In particular, the elementary functions are idempotent in $\mathcal{F}(X),$

$E_x^2=E_x.$

Being real-valued, the elementary functions are also selfadjoint. For abstract algebras, a basis with these properties has a special name.

Definition 3.4. A Fourier basis in an algebra $\mathcal{A}$ is a basis $\{F^\lambda \colon \lambda \in \Lambda\}$ of selfadjoint orthogonal idempotents,

$(F^\lambda)^*=F^\lambda \quad\text{and}\quad F^\lambda F^\mu = \delta_{\lambda\mu}F^\lambda.$

Thanks once again to Lani for paying close attention to definitions in real time and pointing out that selfadjointness should be built into this definition.

Clearly, a necessary condition for an algebra $\mathcal{A}$ to admit a Fourier basis is that it is commutative. However, this condition is not sufficient, and algebras which do admit a Fourier basis are characterized by the following simple but important theorem.

Theorem 3.5. An algebra $\mathcal{A}$ admits a Fourier basis if and only if it is isomorphic to a function algebra.

Proof: We have already seen that a function algebra admits a Fourier basis, namely its elementary basis. Thus, if $\mathcal{A}$ is isomorphic to $\mathcal{F}(X)$ for some set $X$ via an isomorphism $\Phi,$ then $\{\Phi^{-1}(E_x) \colon x \in X\}$ gives a Fourier basis of $\mathcal{A}.$

Conversely, suppose $\mathcal{A}$ admits a Fourier basis $\{F^\lambda \colon \lambda \in \Lambda\},$ and consider the vector space isomorphism

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

defined by

$\Phi(F^\lambda) = E_\lambda, \quad \lambda \in \Lambda.$

We need to check that this vector space isomorphism is an algebra homomorphism. First,

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

so $\Phi$ respects multiplication. Second,

$\Phi((F^\lambda)^*)=\Phi(F^\lambda)=E_\lambda=E_\lambda^*=\Phi(F^\lambda)^*,$

so $\Phi$ respects conjugation. Third, note that in the function algebra $\mathcal{F}(\Lambda)$ the multiplicative identity is

$I_{\mathcal{F}(\Lambda)} = \sum\limits_{\lambda \in \Lambda} E_\lambda.$

We leave it as an exercise to show that any Fourier basis is necessarily a partition of unity, meaning that

$I_\mathcal{A}=\sum\limits_{\lambda \in \Lambda} F^\lambda.$

$\square.$

The algebra isomorphism $\Phi$ constructed in the proof of the preceding theorem is called the Fourier transform on $\mathcal{A}.$ It exists precisely when $\mathcal{A}$ admits a basis of selfadjoint orthogonal idempotents. Thus, if we are given a commutative algebra $\mathcal{A}$ whose multiplication is defined in some horribly convoluted way, if we can construct a Fourier basis in $\mathcal{A}$ then we are able to recognize that this convoluted structure is actually no more complicated than pointwise multiplication of functions. We will soon do this for convolutions algebras of finite abelian groups.

Math 202B: Lecture 3

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