## Unappreciated Exposition

During my time in China, I often had a conversation with various people about the completion of the Elliott classification program for nuclear C*-algebras. Often, the main question was: Who’s going to write the book on the topic? The general answer seemed to be no one. Despite the usefulness of a good exposition on the subject, it seems that there is no incentive for writing such a book. And there is no incentive because no math department will consider such a book to be a worthwhile research accomplishment. And my chain of whys comes to a sudden halt here. I don’t know why expositionary work wouldn’t be rewarded (or even if this is always true). The effort to read the proofs and to write a text for a slightly more general audience (by which I mean, more than the handful of people responsible for the proof) is not worth the opportunity cost of conducting original research.

It may be the case that this is a particular problem in my field. I recall in my undergraduate days there was an attempt to write an exposition of the classification of finite simple groups (a much more difficult project, I presume). This may have more to do with where C*-algebras are right now, rather than being a permanent feature. Honestly, I don’t know, but it is disappointing to me that exposition of “known” results is less significant than even marginal original research.

## Functors Between Metric-Enriched Categories: Is This A Stupid Idea?

Let Met denote the category whose objects consist of metric spaces (for convenience, we will allow metrics to take the value of infinity) and whose morphisms are (weakly) contractive maps (aka short maps, nonexpansive maps, 1-Lipschitz maps), or more precisely functions satisfying $d(f(x),f(y)) \leq d(x,y)$Met is a monoidal category where $X\otimes Y$ is the product space with metric given by $d((x_1,y_1),(x_2,y_2) = d_X(x_1, x_2) + d(y_1, y_2)$ (the identity object is the singleton).

It is my (perhaps mistaken) opinion that a lot of functional analysis can be done using categories enriched in Met. (Actually, I think that for more generality, one ought to replace metric spaces with semi-metric spaces (aka quasi-metric spaces) where distinct elements can have zero distance. It may also be possible that Lawvere metric spaces are the appropriate choice here, but I’m not yet convinced of this.) Of course, one “trivial” yet important example is using the discrete metric. So for any two objects $A, B$ in a locally small category, we can define a metric on $\operatorname{Hom}(A,B)$ by $d(f,g) = 1$ if $f\neq g$ and $d(f,g) = 0$ if $f = g$.

A category enriched in Met enables one to talk about approximately commuting diagrams. This has been explored in approximate Fraïssé limits, though I don’t know enough logic to understand it.

Given two categories $\mathcal{C}, \mathcal{D}$ enriched in Met, we can define a type of continuity (except that word’s already taken) where a functor $F: \mathcal{C} \to \mathcal{D}$ is “continuous” if for any $\epsilon > 0$, there exists $\delta > 0$ such that for any objects $A,B\in \mathcal{C}$ and any morphisms $f,g\in \operatorname{Hom}(A,B)$, if $d(f,g) < \delta$, then $d(F(f),F(g)) < \epsilon$.

Apologies for the underdeveloped ideas, I have been thinking about this for quite a while (throughout my time as a graduate student) and I have some trouble formulating what I want to say. Talking to a few people about this did not generate a terrible amount of interest, but I was curious if people had any insight as to:

1. Has there been any work done along these lines?
2. Do you think this might be a potentially interesting idea?

My category theory is limited, and what I’ve learned does not seem to have this type of idea in mind, but I’d love to hear some thoughts!

## A Constructive Proof that Integer-Valued Continuous Functions form a Free Abelian Group

Disclaimer: I am not a constructivist, nor do I entirely know what strictly speaking entails a constructive proof. The adjective “constructive” used in the title of this post is meant to be illustrative and not technical.

Let $X$ be a compact Hausdorff space. The abelian group $C(X,\mathbb{Z})$ of continuous integer-valued functions on $X$ is free (as a $\mathbb{Z}$-module). One proof of this is that the group of all integer-valued functions is free (prove using Zorn’s lemma, I suppose?) and that a subgroup of a free abelian group is also free (more generally, the submodule of a free module over a PID is free; this is sometimes proved during the classification of finitely generated modules over a PID, though not always).

When I was giving a talk last June, a senior mathematician found the provided proof unsatisfactory. He and a colleague of mine gave a more constructive proof for the case where the space is the Cantor set, basically illustrating the general case. I provide a sketch of the proof below.

First, we can obtain a quotient space by identifying the points within a connected component. This space, which we call $Y$, is totally disconnected, and since integer-valued continuous functions must be constant on connected components, we see that $C(X,\mathbb{Z}) \cong C(Y,\mathbb{Z})$. So it suffices to prove the theorem in the case where $Y$ is totally disconnected. If $Y$ contains finitely many points, then $C(Y,\mathbb{Z})$ is finitely generated and we are done (since the group has no torsion elements). Otherwise, as noted yesterday, $Y$ is the inverse limit of finite spaces, and so $C(Y,\mathbb{Z})$ is the inductive limit of finitely generated free abelian groups. Of course, not every inductive limit of finitely generated free abelian groups is free. But in this case, since the connecting morphisms are induced from continuous maps of spaces, we know something about the connecting morphisms; characteristic functions (which help form bases) must map to characteristic functions.

Let $C(Y,\mathbb{Z}) = \lim_{k\to\infty} \mathbb{Z}^{r(k)}$. In terms of the original finite spaces, $r(k)$ denotes the number of points in each finite space and characteristic functions of singletons form a basis for the group. Let $e^k_i$ denote this basis for $k=1,2,...$ and $i=1,2,...,r(k)$. Let $\sigma_k: \mathbb{Z}^{r(k)} \to \mathbb{Z}^{r(k+1)}$ denote the $k$th connecting morphism. Since characteristic functions map to characteristic functions, for each $k$ and $i$, there exists a set $S^k_i \subseteq \{1,2,\dotsc,r(k+1)\}$ such that $\sigma_k(e^k_i) = \sum_{j\in S^k_i} e_j$. Furthermore, these sets are disjoint (since otherwise $\sigma_k(e_{i_1} + e_{i_2})$ would not be a characteristic function) and the union of the sets is the entire set $\{1,...,r(k+1)\}$ (if not, it can be arranged to do so). For the C*-algebraists, this is basically saying that the Bratteli diagram of every commutative AF-algebra is a rooted tree.

Now, here I get hand-wavy. The basis $B$ is constructed so that for each $i, k$, $B$ contains all but one element of $S^k_i$.

To see that $B$ generates the group, it suffices to see that $e^k_i$ is generated by $B$ for all $k$ and $i$. If $e^k_i \not\in B$, then by construction $e^{k-1}_i\in B$ and all other elements of $S^k_i$ are in $B$. So $e^k_i$ can be written as the difference between $e^{k-1}_i$ and the sum of the elements in $S^k_i$.

To prove independence, we first prove by induction on $t$ that for every $i,k,t$, if the equation $me^k_i = \sum_{i'=i+1}^{i+t}\sum_{k'} m_{i',j'}e^{k'}_{i'}$ holds where $e^{k'}_{i'}\in B$, then $m = 0$. From this, we can establish the general result by multiplying through by $e^k_i$ in a general linear combination, where $i$ is the smallest value of the index in a linear combination and $k$ is chosen arbitrarily.

Unfortunately, this is a very visual concept and so my verbal description might not be so great. If you know about Bratteli diagrams, I suggest trying to draw a picture with rooted trees.

Correction (2016/12/12): The previous version of this proof was incorrect. In particular, I included too many elements in my basis and my proof of independence was also incorrect.

Correction (2017/01/28): A brief sketch of the proof of independence was added.

## How I Came to Terms with a Proof in Davidson’s Book

The book C*-Algebras by Example by Ken Davidson is a standard text for C*-algebraists and well-praised. But a proof of one of the theorems bothered me a little. The theorem states that the C*-algebra of continuous functions on the Cantor set is AF (approximately finite). It’s a pretty easy result. The general statement that the C*-algebra of continuous functions on a compact totally disconnected space is AF is not difficult. By considering the nerves of open covers, one can see that a compact Hausdorff space is totally disconnected (which is equivalent to being zero-dimensional in this case) if and only if the space is the inverse limit of spaces with finitely many points. So by taking the contrapositive functor of taking continuous functions, we see that the C*-algebra of continuous functions over our space is the inductive limit of finite-dimensional algebras, and thus AF. A nice clean proof.

But this isn’t the proof that Davidson uses! Instead, he considers the standard “middle-third” picture of the Cantor set and then considers locally constant functions (i.e. constant on each connected component) on each intermediate set. (If my memory serves, this book and most of my books are currently in storage.) The algebras of these locally constant functions are finite-dimensional and the inductive limit of these algebras is the algebra of continuous functions over the Cantor set, and the result is proved. There is nothing technically wrong with the proof, but it didn’t seem to fit into the larger picture. This algebra being AF is exclusively a property of the underlying space being totally disconnected and not of the standard picture. So it was slightly annoying.

But this isn’t as problematic as I first thought. For starters, the middle-thirds picture reflects (refinements of) clopen covers and so this picture does reflect the total disconnectedness of the Cantor set. Also given a finite clopen cover of a compact space, we can identify the (geometric realization of the) nerve of the covering with the quotient space by identifying points within a connected component. As I mentioned yesterday, the quotient corresponds to a subalgebra of the larger algebra, which is the locally constant functions! So in fact, this proof that seemed off was a rephrasing of the proof that I had in mind from the start!

## Application of Approximate Diagonalization of Commutative C*-Algebras to Invariants

The premise of my dissertation is founded on the idea that matrices over C*-algebras are important and that approximate diagonalization would make dealing with such matrices easier. I have yet to find any useful application of my own result, though I find some mildly amusing applications to results of matrices over commutative C*-algebras.

Definition. Let $A$ be a C*-algebra and let $n$ be a positive integer. A normal matrix $a\in M_n(A)$ is approximately diagonalizable if for every $\varepsilon > 0$, there exist elements $a_1, \dotsc, a_n\in A$ and a unitary $u\in M_n(A)$ such that

$\lVert uau^{*} - \mathrm{diag}(a_1,\dotsc,a_n) \rVert < \varepsilon$.

Next, we present the theorem that we will be applying:

Theorem. Let $X$ be a compact metrizable space. Every self-adjoint matrix in $M_n(C(X))$ is approximately diagonalizable if and only if $\dim(X) \leq 2$ and $\check{H}^2(X) = 0$.

The first mildly interesting application is that on the connection between K-theory and approximate diagonalization of matrices over commutative C*-algebras.

Theorem. (Theorem 3.4 and 4.1 of [3]) Let $X$ be a compact metrizable space. If for every positive integer $n$, every projection in $M_n(C(X))$ is approximately diagonalizable, then $K_0(C(X)) \cong C(X,\mathbb{Z})$.

Proof. First, note that the $K_0$ class of any diagonal projection is an element of $C(X,\mathbb{Z})$. This is because projections in $C(X)$ are characteristic (indictator) functions of clopen subsets and so the $K_0$-class of a diagonal projection is the sum of indicator functions of clopen subsets, which is a continuous, (non-negative) integer-valued function.

Given an approximately diagonalizable projection $p$, there exist projections $p_1, p_2, \dotsc, p_n \in C(X)$ and a unitary $u\in M_n(C(X))$ such that

$\lVert upu^{*} - \mathrm{diag}(p_1, p_2, \dotsc, p_n)\rVert < 1$.

Note that $p_1, p_2, \dotsc, p_n$ can be chosen to be projections by the stability of the projection relations (i.e. being self-adjoint and idempotent, see Lemma 2.5.4 of [1] for details). Since close projections (norm distance strictly less than 1) are unitary equivalent (see Lemma 2.5.1 of [1]), their $K_0$-classes are the same. So every projection has the same $K_0$-class as of a diagonal projection, and thus belongs to $C(X, \mathbb{Z})$. $\Box$

The observant reader can see that we actually proved every approximately diagonalizable projection is in fact diagonalizable. Combining this with Xue’s theorem, we get the corollary:

Corollary. Let $X$ be a compact metrizable space such that $\dim(X) \leq 2$ and $\check{H}^2(X) = 0$. Then $K_0(C(X)) \cong C(X,\mathbb{Z})$.

Perhaps of more slightly more interest is the fact that we can use this method to prove the following:

Theorem. Let $X$ be a compact metrizable space. If for every positive integer $n$, every positive matrix in $M_n(C(X))$ is approximately diagonalizable, then $W(C(X)) \cong \mathrm{lsc}(X,\mathbb{N}\cup \{0\})$.

Here $W(A) = (A\otimes M_{\infty})/\sim$ denotes the Cuntz semigroup, where $\sim$ denotes Cuntz equivalence. The proof of this theorem falls out in the same way as the previous theorem: the Cuntz class of positive elements correspond to characteristic functions of open sets, which sum to non-negative integer-valued lower semicontinuous functions and it is clear that every approximately diagonalizable matrix is Cuntz equivalent to a diagonal matrix.

When we combine this theorem with Xue’s theorem, we have the following corollary:

Collorary. Let $X$ be a compact metrizable space such that $\dim(X) \leq 2$ and $\check{H}^2(X) = 0$. Then $W(C(X)) \cong \mathrm{lsc}(X,\mathbb{N})$.

This is finite and unital version of Theorem 1.3 of [2]. Perhaps the only real upside to this, is that this proof directly deals with positive elements in contrast to Robert’s proof using Hilbert modules.

I’m still looking for some applications of approximate diagonalization, especially my dissertation results. Any help in this respect would be appreciated.

[1] Lin, Huaxin. An Introduction to the Classification of Amenable C*-Algebras. World Scientific, 2001.

[2] Robert, Leonel. “The Cuntz semigroup of some spaces of dimension at most two” C. R. Math. Acad. Sci. Soc. R. Can. 35:1 (2013) pp. 22-32.

[3] Xue, Yifeng. “Approximate Diagonalization of Self-Adjoint Matrices over $C(M)$Funct. Anal. Approx. Comput. 2:1 (2010) pp. 53-65. [pdf]

## A Sledgehammer Proof of the Spectral Theorem for Normal Matrices

My research is about the structure of homomorphisms between certain C*-algebras. To illustrate what I mean and its possible importance, consider an elementary case given below.

Proposition. Let $X$ be a compact, Hausdorff space, $N$ be a positive integer, and $\phi\colon C(X) \to M_N$ be a homomorphism. There exist mutually orthogonal rank one projections $p_n$  and points $\xi_n\in X$ for $n=1,2,\dotsc,N$ such that

$\phi(f) = \sum_{n=1}^{N} f(\xi_n) p_n$

for all $f \in C(X)$.

Proof: Since $C(X)/\ker\,\phi$ is a commutative C*-algebra, by Gelfand’s representation theorem, there exists a compact Hausdorff space $Y$ such that $C(X)/\ker\,\phi \cong C(Y)$. Furthermore, since there is a surjective homomorphism from $C(X)$ onto $C(Y)$, there is a continuous embedding from $Y$ into $X$. We assume, without loss of generality, that $Y\subseteq X$ and that the canonical projection homomorphism is given by restriction.

Let $\phi'\colon C(Y) \to M_N$ be the injective homomorphism induced by $\phi$. Since $\phi'$ is an injective linear map and $M_N$ is finite dimensional, $Y$ is finite (i.e. consists of finitely many points). Label these points $\xi_n$ and let $\chi_n \in C(Y)$ denote the indicator functions for $\{\xi_n\}$. Notice that $\phi'(\chi_n)$ are mutually orthogonal projections. Each projection $\phi'(\chi_n)$ can be decomposed into the sum of mutually orthogonal rank one projections $p_n$. By relabeling and repeating $\xi_n$ as necessary, we assume that $p_n$ is associated with the corresponding point $\xi_n$.

Let $\pi_n\colon M_N \to p_n M_N p_n$ be the usual positive linear map. Since $p_n$ has rank one, $p_n M_N p_n \cong \mathbb{C}$. So it is easy to see that $\pi_n \circ \phi'$ is a state of $C(Y)$ and so $\pi_n \circ \phi'(f) = f(\xi_n)$. Therefore,

$\phi(f) = \sum_{n=1}^{N}\pi_n\circ \phi(f) = \sum_{n=1}^{N} f(\xi_n) p_n$.    $\Box$

We can rephrase this proposition to resemble linear algebra more. Since rank one projections correspond to one-dimensional subspaces, we can choose an orthonormal basis $(\delta_n)$, where $\delta_n$ is a unit vector in the range of $p_n$. So there exists a unitary $u \in M_N$ such that $u\delta_n = \epsilon_n$, where $\epsilon_n$ is the standard basis for $\mathbb{C}^n$. So our proposition above can be restated as:

Proposition. Let $X$ be a compact, Hausdorff space, $N$ be a positive integer, and $\phi\colon C(X) \to M_N$ be a homomorphism. There exist points $\xi_n\in X$ for $n=1,2,\dotsc,N$ and a unitary matrix $u\in M_N$ such that

$u\phi(f)u^{*} = \left(\begin{array}{c c c c}f(\xi_1) & 0 & \cdots & 0 \\ 0 & f(\xi_2) & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & f(\xi_N) \end{array}\right)$

for all $f \in C(X)$.

To apply this theorem, let $a \in M_N$ be a normal matrix. Then by continuous functional calculus, there is a homomorphism from $C(X)$ to $M_N$ denoted $f \mapsto f(a)$, where $X = \mathrm{spec}(a)$. In particular, the inclusion function $z \mapsto z$ is sent to the matrix $a$. By applying the proposition to this homomorphism and plugging in the inclusion function, we obtain:

Theorem (Spectral Theorem for Normal Matrices). Let $N$ be a positive integer. For any normal matrix $a \in M_N$, there exist $\xi_n\in \mathbb{C}$ for $n=1,2,\dotsc,N$ and a unitary matrix $u\in M_N$ such that

$u a u^{*} = \left(\begin{array}{c c c c}\xi_1 & 0 & \cdots & 0 \\ 0 & \xi_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \xi_N \end{array}\right)$

As stated, this is the rather significant finite dimensional spectral theorem using rather excessively powerful tools for the proof. But we do obtain a stronger version for “free”. Let $S\subseteq M_N$ be a set of commuting normal matrices. Using the spectral theorem above, $C^{*}(S)$ is a commutative C*-subalgebra of $M_N$ and so by Gelfand’s representation theorem, $C^{*}(S) \cong C(X)$ for some compact Hausdorff space $X$. So we obtain a homomorphism from $C(X)$ to $M_N$ defined by composing the Gelfand transform with inclusion. Since each element of $S$, corresponds to some continuous function of $X$, by applying our proposition and plugging those functions, we obtain a simultaneous diagonalization of the matrices in $S$:

Theorem. Let $N$ be a positive integer. For any set $S \subseteq M_N$ of commuting normal matrices, there exists a unitary $u \in M_N$, such that for any $a\in S$, there exist $\xi_n\in \mathbb{C}$ for $n=1,2,\dotsc,N$ such that

$u a u^{*} = \left(\begin{array}{c c c c}\xi_1 & 0 & \cdots & 0 \\ 0 & \xi_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \xi_N \end{array}\right)$

## 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.