Math 202B: Lecture 11

Let \mathcal{A} be an abelian subalgebra of the endomorphism algebra \mathrm{End}(V) of a finite dimensional Hilbert space V. We are going to use Math 202A linear algebra to show that \mathcal{A} is isomorphic to a function algebra. Combining this with the work we did last week, this result allows us to conclude the following: if \mathcal{A} is a commutative algebra which supports a scalar product satisfying the left-Frobenius identity (or, equivalently, admits a faithful state), then \mathcal{A} is isomorphic to a function algebra.

Problem 11.1. Prove that a one-dimensional subalgebra of \mathrm{End}(V) is isomorphic to the function algebra of a point.

Now let \mathcal{A} be an arbitrary m-dimensional abelian subalgebra of \mathrm{End}(V), and let \{A_1,\dots,A_m\} be a basis of \mathcal{A}. Since \mathcal{A} is commutative, all its elements are normal, as we proved in Week 1. Thus, A_1,\dots,A_m are commuting normal operators on V and we can apply the spectral theorem.

Theorem 11.1. There exists an orthonormal basis X \subset V such that

A_1x = \widehat{A}_1(x),\dots,A_mx=\widehat{A}_m(x)x, \quad x \in X,

where \widehat{A}_1(x), \dots, \widehat{A}_m(x) are scalars.

In possibly more familiar terms, Theorem 11.1 says that a finite family of commuting normal operators is simultaneously diagonalizable. For our purposes, we want to interpret this result as defining a function \mathcal{A} \to \mathcal{F}(X). Since \{A_1,\dots,A_m\} is a basis of \mathcal{A}, every A \in \mathcal{A} can be uniquely represented as a linear combination

A=\alpha_1A_1,\dots,\alpha_mA_m,

so we have a well-defined function

\mathcal{A} \longrightarrow \mathcal{F}(X)

which sends A \in \mathcal{A} to the function \widehat{A} \in \mathcal{F}(X) defined by

\widehat{A}(x) = \alpha_1\widehat{A}_1(x) + \dots + \alpha_m\widehat{A}_m(x), \quad x \in X.

We call this mapping \mathcal{A} \to \mathcal{F}(X) the spectral transform on \mathcal{A} relative to the orthonormal basis X \subset V.

Theorem 11.2. The spectral transform is an injective algebra homomorphism.

Proof: We carefully checked this in class, and if you were not there you should do the same. \square

Theorem 11.2 proves that every abelian subalgebra \mathcal{A} of \mathrm{End}(V) is isomorphic to a subalgebra of \mathcal{F}(X) for a finite set X. Since we have already shown in Math 202B that every subalgebra of a function algebra is isomorphic to a function algebra, this completes the proof that every commutative subalgebra of an endomorphism algebra is isomorphic to a function algebra.

However, we can be more precise than this: we can say which subalgebra of \mathcal{F}(X) the abelian subalgebra \mathcal{A} \leq \mathrm{End}(V) is transformed into. Namely, each of the operators A_i in our chosen basis \{A_1,\dots,A_m\} of \mathcal{A} induces a partition \mathfrak{p}_i of X obtained by partition the points of X into distinct eigenspaces. That is, two points x,y \in X are in the same block of \mathfrak{p}_i if and only if \widehat{A}_i(x)=\widehat{A}_i(y).

Problem 11.2. Prove that the image of \mathcal{A} under the spectral transform is the subalgebra of \mathcal{F}(X) consisting of all functions constant on the blocks of \mathfrak{q}, where \mathfrak{q} is the coarsest partition of X finer than each of the partitions \mathfrak{p}_1,\dots,\mathfrak{p}_m. (Hint: Show that \mathcal{A} maps into the stated subalgebra is straightforward, and Stone-Weierstrass finishes the job).

Published by Jonathan Novak

Mathematician at UC San Diego.

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