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.

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About minimalrho
Unemployed guy with a PhD in math.

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