Commutative von Neumann algebras

Last Thursday, I gave a talk on commutative von Neumann algebras. I revised my notes slightly and filled in some gaps in my presentation. All the material comes from G. Pedersen’s book C*-algebras and their Automorphism Groups. Thanks go out to Eusebio Gardella and Michael Sun for helping me fill in my ignorance and seeing the argument more fully.

A von Neumann algebra is a weak operator closed, self-adjoint subalgebra of bounded linear operators on a Hilbert space. For the sake of convenience, we will assume that every von Neumann algebra contains the identity operator, thus forming a unital algebra. In addition and for more than convenience, we will be assuming that our Hilbert spaces are separable.

Clearly the space of all bounded linear operators forms a von Neumann algebra and will be denoted $B(H)$. Immediately, we notice in contrast to the $C^{*}$-algebra case that the definition of von Neumann algebra we provided is distinctly an operator algebra. Nonetheless, we have a characterization of commutative von Neumann algebras in terms of function spaces.

Let $X$ be a compact, Hausdorff space and $\mu$ a probability Borel measure. We can embed $L^{\infty}(X,\mu)$ into $B(L^2(X,\mu)$ by considering functions as multiplication operators, i.e. $g\in L^{\infty}(X,\mu)$ is identified with the operator $f\in L^2(X,\mu)\mapsto fg\in L^2(X,\mu)$. This embedding is in fact, an isometric homomorphism and the image is weak-operator closed, thereby forming a von Neumann algebra. From now on, we will be making this identification implicitly and simply say that $L^{\infty}(X,\mu)$ is a von Neumann algebra.

The purpose of this post is to provide a proof of the fact that every commutative von Neumann algebra (on a separable Hilbert space) is isomorphic to $L^{\infty}(X,\mu)$  for appropriate choice of $X$ and $\mu$. But first we need to make two observations about $L^{\infty}(X,\mu)$.

First, $L^{\infty}(X,\mu)$ is maximal in the sense that there is no commutative subalgebra that properly contains $L^{\infty}(X,\mu)$ in $B(L^2(X,\mu)$. And second, the weak closure of $C(X)$ identified as multiplication operators is $L^{\infty}(X,\mu)$. For good reasons that won’t be discussed, we denote the weak operator closure of a subalgebra $A$ to be $A''$, and therefore $C(X)''=L^{\infty}(X,\mu)$.

We will first prove the theorem for maximal commutative von Neumann algebras, and then I will subsequently handwave the general case, since this is what I did during the talk. The benefit of having a maximal commutative algebra is the existence of cyclic vectors. A cyclic vector of an algebra $A$ (or even subset) of $B(H)$ is a vector $v\in H$ such that the set $Av=\{av: a\in A\}$ is dense in $H$. The existence of cyclic vectors guarantee that certain isomorphisms of concrete $C^*$-algebras are in fact unitary equivalences and that therefore isomorphisms of $C^*$-algebras lift to their weak operator closures and extend to isomorphisms of von Neumann algebras.

Theorem. Let $A\subseteq B(H_1)$ and $B\subseteq B(H_2)$ be isomorphic $C^*$-algebras with cyclic vectors $v_1$ and $v_2$, respectively and isomorphism $\phi:A\rightarrow B$. If $(av_1,v_1)=(\phi(a)v_2,v_2)$ for all $a\in A$, then there exists a unitary $U:H_1\rightarrow H_2$ such that $U(v_1)=v_2$ and $UAU^{*}=\phi(A)$.

First we define $U_{0}:Av_{1}\rightarrow Bv_{2}$ by $U_0(av_1)=\phi(a)v_2$. This map is clearly linear (provided that it is well-defined). Note that
$\|U(av_1)\|^2=(U(av_1),U(av_1))=(\phi(a^{*}a)v_2,v_2)=(a^{*}av_1,v_1)=\|av_1\|^2$. So $U_{0}$ is a well-defined unitary, and therefore continuous.

Thus $U_{0}$ extends to a map $U:H_1\rightarrow H_2$, which is unitary. Finally, $U(abv_1)=\phi(ab)v_2=\phi(a)U(bv_1)$. So $Ux=\phi(x)U$.

Corollary. If $A\subseteq B(H_1)$ and $B\subseteq B(H_2)$ be isomorphic $C^*$-algebras with cyclic vectors $v_1$ and $v_2$ and isomorphism $\phi:A\rightarrow B$, then $A''$ is isomorphic to $B''$ provided $(av_1,v_1)=(\phi(a)v_2,v_2)$.

This follows from the fact that conjugation by a unitary is weakly continuous. Now we turn our attention back to maximal commutative von Neumann algebras.

Theorem. If $H$ is separable and $\mathcal{N}\subseteq B(H)$ is a maximal commutative von Neumann algebra, then $\mathcal{N}$ has a cyclic vector.

By Zorn’s lemma, there exists a maximal set of unit vectors $v_i\in H$ whose projections $p_i$ onto the closure of $\mathcal{N}v_i$ are mutually orthogonal. This is countable by separability, since maximal sets of orthonormal vectors are countable. The span of the spaces $\mathcal{N}v_i$ is $H$, since otherwise there would be a unit vector in the orthogonal complement $v_0$ and it can be seen that $\mathcal{N}v_0$ would be in the orthogonal complement, contradicting maximality.

Set $v_0=\sum_{n=1}^{\infty}2^{-n}v_n$. Since $\mathcal{N}v_n=\mathcal{N}p_nv_0\subseteq Nv_0$. And therefore $v_0$ is a cyclic vector of $\mathcal{N}$.

Theorem. Every maximal commutative von Neumann algebra on $B(H)$ is unitarily equivalent to $L^{\infty}(X,\mu)$ for some compact, metrizable space $X$ and probability measure $\mu$.

Since the unit ball of $B(H)$ is weakly compact, metrizable, so is the unit ball of $\mathcal{N}$ and thus separable. Take a $C^*$-subalgebra generated by a countable weakly dense subset of unit vectors and call it $A$.

By Gelfand’s representation theorem, $A\cong C(X)$ for some compact, metrizable space $X$. Furthermore, the map $a\mapsto (av,v)$ defines a linear functional on $C(X)$, where $v\in H$ is a cyclic vector of $\mathcal{N}$. By Riesz representation theorem, there is a positive Borel measure $\mu$ such that $(fv_1,v_1)=\int\!f\,d\mu$ Note that $\mu(X)=(v_1,v_1)=1$. Notice that if $v_1$ is a cyclic vector of $\mathcal{N}$, then $v_1$ is a cyclic vector of $A$ and also the constant function $1$ is a cyclic vector of $C(X)$, since $C(X)$ is dense in $L^2(X,\mu)$. So by our corollary, we see that the isomorphism between $A$ and $C(X)$ lifts to an isomorphism between $\mathcal{N}=A''$ and $L^{\infty}(X,\mu)=C(X)''$.

Corollary. Every commutative von Neumann algebra on $B(H)$ is unitarily equivalent to $L^{\infty}(X,\mu)$ for some compact, metrizable space $X$ and probability measure $\mu$.

Unfortunately, I didn’t get to the proof of this. Nor given the above can I prove it now. The basic idea is take our von Neumann algebra and prove that it is isomorphic to a corner, where it has a new representation where our algebra is maximal and then apply the previous theorem.