Unitarily invariant norms on finite von Neumann algebras

Abstract

We give a characterization of all the unitarily invariant norms on a finite von Neumann algebra acting on a separable Hilbert space. The characterization is analogous to von Neumann’s characterization for the $n\times n$ complex matrices and the characterization in Fang et al. (J Funct Anal 255(1):142–183, 2008) for $II_{1}$ factors.

1 Introduction

Since John von Neumann’s beautiful characterization of the unitarily invariant norms for the $n\times n$ complex matrices $\mathbb {M}_{n}\left( \mathbb {C}\right)$, there have been over four hundred papers related to this subject. In the 1930’s von Neumann [24] showed that there is a natural one-to-one correspondence between the unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$ and the normalized symmetric gauge norms on $\mathbb {C}^{n}$. As pointed out by the referee, these norms have been generalized and utilized in several contexts (see [20] or [21]). More recently, J. Fang, D. Hadwin, E. A. Nordgren and J. Shen [10] showed that there is an analogous correspondence between the unitarily invariant norms on a $II_{1}$ factor von Neumann algebra $\mathcal {M}$ and the normalized symmetric gauge norms on $L^{\infty }\left[ 0,1\right]$. Although the proofs of both results relied on s-numbers, the proof of the latter result was different from von Neumann’s proof. We provide a new proof of the $II_{1}$ factor result that more closely parallels the proof for $\mathbb {M}_{n}\left( \mathbb {C}\right)$. The key ingredient is an “approximate” version of the Ky Fan Lemma that is used in the finite-dimensional case.

It is our goal to find a similar characterization of all the unitarily invariant norms on a finite von Neumann algebra $\mathcal {R}$ acting on a separable Hilbert space H. A von Neumann algebra on H is a unital subalgebra of the algebra $B\left( H\right)$ of all operators on H that is closed under the adjoint operation and is closed in the weak operator topology. A von Neumann algebra is a factor if it cannot be written as the direct sum of two von Neumann algebras; equivalently, if its center contains no projection P with $0\ne P\ne 1$. A von Neumann algebra on a separable Hilbert space can be written as a direct integral (i.e., continuous direct sum) of factor von Neumann algebras, which is called the central decomposition. A finite factor is a factor that has a faithful normal tracial state. The finite-dimensional finite factors are all isomorphic to $\mathbb {M}_{n}\left( \mathbb {C}\right)$ for some positive integer n. An infinite-dimensional finite factor is called a $II_{1}$ factor. A finite von Neumann algebra on a separable Hilbert space is a direct integral of finite factors. General references for von Neumann algebras are [16] and [23].

Thus the results in [24] and [10] characterize the unitarily invariant norms on finite factors. To make these two examples look the same, we want to view $\mathbb {C}^{n}$ as $L^{\infty }\left( J_{n},\delta _{n}\right) ,$ where $\left( J_{n},\delta _{n}\right)$ is a probability space. We also want to have $J_{n}\subset \left[ 0,1\right]$. Our choice is $J_{n}=\left\{ \frac{1}{n},\ldots ,\frac{n}{n}\right\}$ and $\delta _{n}$ is normalized counting measure, i.e.,

$$\begin{aligned} \delta _{n}\left( E\right) =\frac{1}{n}\textrm{Card}\left( E\right) . \end{aligned}$$

We define $J_{\infty }=\left[ 0,1\right]$ and $\delta _{\infty }$ to be Lebesgue measure.

A finite von Neumann algebra $\mathcal {R}$ on a separable Hilbert space can be decomposed into a direct integral of factors that are either isomorphic to $\mathbb {M}_{n}\left( \mathbb {C}\right)$ or are $II_{1}$ factors. Each finite factor von Neumann algebra has a unique tracial state. From the central decomposition we can define a tracial state $\tau$ on $\mathcal {R}$. The problem is to identify the corresponding measure space $\left( \Lambda ,\lambda \right)$. A key observation is that every maximal abelian selfadjoint subalgebra (masa) of $\mathbb {M}_{n}\left( \mathbb {C} \right)$ is isomorphic to $\mathbb {C}^{n}=L^{\infty }\left( J_{n},\delta _{n}\right)$ and with the normalized trace corresponding to integration with respect to $\delta _{n}$, and each masa in a $II_{1}$ factor is isomorphic to $L^{\infty }\left[ 0,1\right] =L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$ with the unique tracial state $\tau$ corresponding to integration with respect to $\delta _{\infty }$. It is known [22] that any two masas on a finite von Neumann algebra are isomorphic. If $\mathcal {A}$ is a masa in $\mathcal {R}$, then the central decomposition of $\mathcal {R}$ decomposes $\mathcal {A}$ to a direct integral of algebras that are masas in the corresponding factors. We must analyze this decomposition carefully to see that the masas are all isomorphic, in a very special way, to $L^{\infty }\left( \Lambda ,\lambda \right)$ for some measure space $\left( \Lambda ,\lambda \right)$. Once we find the measure space, we find a certain group $\mathbb {G}\left( \mathcal {R}\right)$ of invertible measure-preserving transformations. We then have to show how the unitarily invariant norms on $\mathcal {R}$ correspond to the normalized $\mathbb {G} \left( \mathcal {R}\right)$-symmetric gauge norms on $L^{\infty }\left( \Lambda ,\lambda \right)$. We will see that $\mathcal {R}$ is a factor if and only if $\mathbb {G}\left( \mathcal {R}\right)$ is the group of all invertible measure-preserving transformations. This involves defining the analogue of the “s-numbers” and proving a general approximate Ky Fan Lemma. To show that things are independent of the choices of the masas used, we need a result on approximate unitary equivalence [4].

In Sect. 2 we discuss the basic properties of unitarily invariant norms, give a brief description of von Neumann’s characterization for $\mathbb {M} _{n}\left( \mathbb {C}\right)$ and give our new proof of the characterization of [10] for $II_{1}$ factors.

In Sect. 3, we prove a reformulation of a result of Ding and Hadwin [4] for approximate unitary equivalence of representations of a (not necessarily separable) abelian C*-algebra into a finite von Neumann algebra $\mathcal {R}$ in terms of the center-valued trace on $\mathcal {R}$.

In Sect. 4 we discuss the definitions and techniques of direct integrals a a canonical way to represent the center valued trace of a finite von Neumann in terms of its central decomposition.

In Sect. 5 we describe, for a given finite von Neumann algebra $\mathcal {R}$, two measure spaces with measures $\mu$ and $\lambda$ so that $L^{\infty }\left( \mu \right)$ is isomorphic to the center $\mathcal {Z} \left( \mathcal {R}\right)$ of $\mathcal {R}$ and, such that every masa $\mathcal {A}$ of $\mathcal {R}$ is isomorphic to $L^{\infty }\left( \lambda \right)$. We prove (Theorem 6) the isomorphism from $L^{\infty }\left( \lambda \right)$ to a masa $\mathcal {A}$ of $\mathcal {R}$ can be chosen in a special way so that a certain commutative diagram holds.

In Sect. 4 we discuss invertible measure-preserving transformations, nonincreasing rearrangements, a function version of s-numbers, and define our group $\mathbb {G}\left( \mathcal {R}\right)$ of measure-preserving transformations on the measure space $\left( \Lambda ,\lambda \right)$. We prove (Theorem 8) a general “approximate” version of the Ky Fan Lemma.

In Sect. 5 we put things together and prove our main theorem, which characterizes the unitarily invariant norms on $\mathcal {R}$ in terms of $\mathbb {G}\left( \mathcal {R}\right)$-symmetric normalized gauge norms on $L^{\infty }\left( \lambda \right)$.

2 Preliminaries

2.1 Unitarily invariant norms

If $\mathcal {A}$ is a unital C*-algebra, $\mathcal {U}\left( \mathcal {A} \right)$ denotes the set of all unitary elements of $\mathcal {A}$. If $T\in \mathcal {A}$ we define $\left| T\right| =\left( T^{*}T\right) ^{1/2}$.

Lemma 1

Suppose $\mathcal {A}$ is a unital C*-algebra and $\alpha$ is a norm on $\mathcal {A}$ such that $\alpha \left( 1\right) =1$. The following are equivalent.

• (1) For every $T\in \mathcal {A}$ and for every $U\in \mathcal {U}\left( \mathcal {A}\right)$,

  $$\begin{aligned} \alpha \left( T\right) =\alpha \left( \left| T\right| \right) =\alpha \left( U^{*}TU\right) . \end{aligned}$$

• (2) For all U, V in $\mathcal {U}\left( \mathcal {A}\right)$,

  $$\begin{aligned} \alpha \left( T\right) =\alpha \left( UTV\right) . \end{aligned}$$

Proof

Suppose $T\in \mathcal {A}$ and for every $U\in \mathcal {U}\left( \mathcal {A} \right)$, we have $\alpha \left( T\right) =\alpha \left( \left| T\right| \right) =\alpha \left( U^{*}TU\right) .$ Then

$$\begin{aligned}{} & {} \alpha \left( UTV\right) =\alpha \left( \left| UTV\right| \right) =\alpha \left( [(UTV)^{*}(UTV)]^{1/2}\right) \\{} & {} \quad =\alpha \left( V^{*}(T^{*}T)^{1/2}V\right) =\alpha \left( |T|\right) =\alpha \left( T\right) \text {.} \end{aligned}$$

Suppose $T\in \mathcal {A}$ and $\alpha \left( T\right) =\alpha \left( UTV\right)$ for every $U,V\in \mathcal {U}\left( \mathcal {A}\right) .$ It is clear that $\alpha \left( T\right) =\alpha \left( U^{*}TU\right)$. To prove $\alpha \left( T\right) =\alpha \left( \left| T\right| \right)$, we can assume, using the universal representation [23, Chapter 3: Sect. 2], that $\mathcal {A}\subset B\left( \mathcal {H}\right)$ for some Hilbert space $\mathcal {H}$ such that the second dual $\mathcal {A} ^{\# \#}$ of $\mathcal {A}$ is isomorphic to $\mathcal {A}^{\prime \prime }$, the weak operator closure of $\mathcal {A}$. Furthermore we have that the weak operator topology on $\mathcal {A}^{\# \#}$ coincides with the weak*-topology, in fact, if $\varphi$ is a continuous linear functional on $\mathcal {A}$, then there are vectors $e,f\in \mathcal {H}$ such that $\varphi =e\otimes f$, i.e., for every $A\in \mathcal {A}$,

$$\begin{aligned} \varphi \left( A\right) =\left( e\otimes f\right) \left( A\right) =\left\langle Ae,f\right\rangle \text {.} \end{aligned}$$

Now suppose $A,B\in \mathcal {A}$, $W\in \mathcal {A}^{\prime \prime }$, $\left\| W\right\| \le 1$ and $A=WB$. We claim

$$\begin{aligned} A\in \overline{{\text {co}}}^{\left\| {}\right\| }\left( \left\{ UB:U\in \mathcal {U}\left( \mathcal {A}\right) \right\} \right) , \end{aligned}$$

where $\overline{{\text {co}}}^{\left\| {}\right\| }$ represents the norm-closed convex hull. Since $\left\{ e^{it}U:t\in \mathbb {R},U\in \mathcal {U}\right.$$\left. \left( \mathcal {A}\right) \right\} =\mathcal {U}\left( \mathcal {A}\right)$, we see that $\overline{{\text {co}}}^{\left\| {}\right\| } \left( \left\{ UB:U\in \mathcal {U}\left( \mathcal {A}\right) \right\} \right)$ is absolutely convex and norm closed. It follows from the Hahn Banach theorem that co$^{-\left\| {}\right\| }( \{ UB:U\in$ $\mathcal {U}\left( \mathcal {A}\right) \} )$ closed in the weak topology on $\mathcal {A}$. If we assume, via contradiction, that the claim is false, there are $e,f\in \mathcal {H}$ and a number $r\in \mathbb {R}$ such that, for all $C\in$ co$^{\left\| {}\right\| }\left( \mathcal {U}\left( \mathcal {A}\right) \right)$,

$$\begin{aligned} \left| \left\langle CBe,f\right\rangle \right| \le r<{\text {Re}}\left( \left\langle Ae,f\right\rangle \right) \le \left| \left\langle Ae,f\right\rangle \right| \text {.} \end{aligned}$$

We know from the Russo-Dye Theorem [7] that co$^{\left\| {}\right\| }\left( \mathcal {U}\left( \mathcal {A}\right) \right)$ is $\{ C\in \mathcal {A}:\left\| C\right\| \le 1 \}$. It follows from the Kaplansky density theorem [16] that there is a net $\left\{ C_{j}\right\}$ in the closed unit ball of $\mathcal {A}$ such that $C_{\lambda }\rightarrow W$ in the weak operator topology. Thus,

$$\begin{aligned} r<\left| \left\langle Ae,f\right\rangle \right| =\left| \left\langle WBe,f\right\rangle \right| =\lim _{\lambda }\left| \left\langle C_{j}Be,f\right\rangle \right| \le r, \end{aligned}$$

which is the desired contradiction. We know from the Russo-Dye Theorem [7] that if $\left\| S\right\| <1$ then $S\in$ co $\left( \mathcal {U}\left( \mathcal {A}\right) \right)$, which implies $\alpha \left( S\right) \le 1$. Hence $\alpha \left( S\right) \le \left\| S\right\|$ for every $S\in \mathcal {A}$. Since the claim is true, and it follows that $\alpha \left( A\right) \le \alpha \left( B\right)$. Since $\mathcal {A}^{\prime \prime }$ is a von Neumann algebra and $T\in \mathcal {A}$, there is a partial isometry $W\in \mathcal {A}^{\prime \prime }$ such that $T=W\left| T\right|$ and $\left| T\right| =W^{*}T$, from which we can conclude $\alpha \left( T\right) =\alpha \left( \left| T\right| \right)$. $\square$

Definition 1

If $\mathcal {A}$ is a unital $C^{*}$-algebra and $\mathcal {\alpha }$ is a norm on $\mathcal {A}$ satisfying $\alpha \left( 1\right) =1$ and either of the two conditions in Lemma 1, we say that $\alpha$ is a unitarily invariant norm on $\mathcal {A}$. It is clear that when A is commutative, a unitarily invariant norm $\alpha$ need only satisfy $\alpha (1)=1$ and $\alpha (T)=\alpha (|T|)$.

Below are some properties about unitarily invariant norms.

Proposition 1

If $\mathcal {A}$ is a unital $C^{*}$-algebra and $\mathcal {\alpha }$ is a unitarily invariant norm on $\mathcal {A}$, and $T,A,B\in \mathcal {A},$ we have the following:

• (1) $\alpha \left( T\right) \le \left\| T\right\| ,$

• (2) $\alpha \left( T\right) =\alpha \left( T^{*}\right)$,

• (3) $\alpha \left( ATB\right) \le \left\| A\right\| \alpha \left( T\right) \left\| B\right\| ,$

• (4) $0\le A\le B$ implies $\alpha \left( A\right) \le \alpha \left( B\right) .$

Note: Whenever we discuss a measure space $\left( \Omega ,\Sigma ,\mu \right)$ we always assume that the space is complete in the sense that, whenever $F\in \Sigma$, $E\subset F$ and $\mu \left( F\right) =0$, we have $E\in \Sigma$. The following lemma is an immediate consequence of Proposition 1.

Lemma 2

If $\alpha$ is a unitarily invariant norm on a unital C*-algebra $\mathcal {R}$, $S,T\in \mathcal {R}$, and $\left\{ U_{j}\right\}$ is a net of unitary operators in $\mathcal {R}$ such that

$$\begin{aligned} \lim _{j}\left\| S-U_{j}^{*}TU_{j}\right\| =0, \end{aligned}$$

then

$$\begin{aligned} \alpha \left( S\right) =\alpha \left( T\right) \text {.} \end{aligned}$$

Definition 2

If $\left( \Omega ,\mu \right)$ is a probability space, then $L^{\infty }\left( \mu \right)$ is a von Neumann algebra, and a unitarily invariant norm $\alpha$ on $L^{\infty }\left( \mu \right)$ is called a normalized gauge norm on $L^{\infty }\left( \mu \right)$. In this case all we need require of $\alpha$ is that $\alpha \left( 1\right) =1$ and $\alpha \left( f\right) =\alpha \left( \left| f\right| \right)$ for every $f\in L^{\infty }\left( \mu \right)$. We let $\mathbb{M}\mathbb{P}\left( \Omega ,\mu \right)$ denote the group (under composition) of all invertible measure-preserving transformations from $\Omega$ to $\Omega$. We say that a gauge norm $\alpha$ on $L^{\infty }\left( \mu \right)$ is symmetric if, for every $\gamma \in \mathbb{M}\mathbb{P}\left( \Omega ,\mu \right)$ and every $f\in L^{\infty }\left( \mu \right)$, we have

$$\begin{aligned} \alpha \left( f\circ \gamma \right) =\alpha \left( f\right) \text {.} \end{aligned}$$

In [24], J. von Neumann characterized all of the unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$, which is the $n\times n$ full matrix algebra with entries in $\mathbb {C}$. Also [10] characterizes the unitarily invariant norms on a $II_{1}$ factor von Neumann algebra. The goal of this paper is to give a characterization of all unitarily invariant norms of a finite von Neumann algebra acting on a separable Hilbert space. Along the way we give a new proof of the characterization of unitarily invariant norms on a $II_{1}$ factor in [10].

2.1.1 Unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$

In this section we give a brief description of von Neumann’s characterization [24] of unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$. Let $\tau _{n}$ be the normalized trace on $\mathbb {M} _{n}\left( \mathbb {C}\right) ,$ i.e., $\tau _{n}=\frac{1}{n}Trace.$

Lemma 3

Suppose $T\in \mathbb {M}_{n}\left( \mathbb {C}\right) ,$ then there exists a unitary matrix $U\in \mathcal {U}\left( \mathbb {M}_{n}\left( \mathbb {C} \right) \right)$ and numbers $s_{T}\left( \frac{1}{n}\right) \geqslant s_{T}\left( \frac{2}{n}\right) \geqslant \cdots$ $s_{T}\left( \frac{n}{n}\right) \ge 0$ such that

$$\begin{aligned} U^{*}|T|U=\left( \begin{array}{cccc} s_{T}\left( \frac{1}{n}\right) &{} 0 &{} \cdots &{} 0\\ 0 &{} s_{T}\left( \frac{2}{n}\right) &{} &{} \vdots \\ \vdots &{} &{} \ddots &{} 0\\ 0 &{} \cdots &{} 0 &{} s_{T}\left( \frac{n}{n}\right) \end{array} \right) . \end{aligned}$$

The numbers $s_{T}\left( \frac{1}{n}\right) ,s_{T}\left( \frac{1}{n}\right) ,...,s_{T}\left( \frac{n}{n}\right)$ are unique and are called the s-numbers of the matrix T. Define

$$\begin{aligned} s\left( T\right) =\left( s_{T}\left( \frac{1}{n}\right) ,s_{T}\left( \frac{2}{n}\right) ,\cdots ,s_{T}\left( \frac{n}{n}\right) \right) \text {.} \end{aligned}$$

If $\alpha$ is a unitarily invariant norm on $\mathbb {M}_{n}\left( \mathbb {C}\right)$, then

$$\begin{aligned} \alpha \left( T\right) =\alpha \left( \left| T\right| \right) =\alpha \left( U^{*}|T|U\right) =\alpha \left( \begin{array}{cccc} s_{T}\left( \frac{1}{n}\right) &{} 0 &{} \cdots &{} 0\\ 0 &{} s_{T}\left( \frac{2}{n}\right) &{} &{} \vdots \\ \vdots &{} &{} \ddots &{} 0\\ 0 &{} \cdots &{} 0 &{} s_{T}\left( \frac{n}{n}\right) \end{array} \right) , \end{aligned}$$

and thus $\alpha \left( T\right)$ depends only on the s-numbers of T.

Note that $s\left( T\right) \in \mathbb {C}^{n}$, and in classical matrix theory [2] the standard notation is $s_{k}\left( T\right)$ instead of our $s_{T}\left( \frac{k}{n}\right)$ for $1\le k\le n$. We know that $\mathbb {C}^{n}$ is isomorphic to $L^{\infty }\left( \delta _{n}\right)$, where $\delta _{n}$ is normalized counting measure on $\left\{ \frac{1}{n},\ldots ,\frac{n}{n}\right\}$. Let $\mathbb {S}_{n}$ be the permutation group (i.e., all the bijective functions on $J_{n}=$ $\left\{ \frac{1}{n},\ldots ,\frac{n}{n}\right\}$). It is clear that $\mathbb {S} _{n}=\mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right)$.

In this case a normalized gauge norm $\beta$ on $\mathbb {C} ^{n}=L^{\infty }\left( \delta _{n}\right)$ is symmetric if, for every $f\in L^{\infty }\left( \delta _{n}\right)$ and every $\sigma \in \mathbb {S}_{n}$,

$$\begin{aligned} \beta \left( f\right) =\beta \left( f\circ \sigma \right) , \end{aligned}$$

that is

$$\begin{aligned} \beta \left( \left( a_{1},\ldots ,a_{n}\right) \right) =\beta \left( \left( a_{\sigma \left( 1\right) },\ldots ,a_{\sigma \left( n\right) }\right) \right) \text {.} \end{aligned}$$

We know that, for each $x=\left( x_{1},\ldots ,x_{n}\right) \$in $\mathbb {C}^{n}$ with $\left| x\right| =\left( \left| x_{1}\right| ,\ldots ,\left| x_{n}\right| \right)$, there is a $\sigma \in \mathbb {S}_{n}$ such that

$$\begin{aligned} \sigma \left( \left| x\right| \right) =\left( \left| x_{\sigma \left( 1\right) }\right| ,\cdots ,\left| x_{\sigma \left( n\right) }\right| \right) \underset{\textrm{def}}{=}\left( s_{\left| x\right| }\left( \frac{1}{n}\right) ,s_{\left| x\right| }\left( \frac{2}{n}\right) ,\cdots ,s_{\left| x\right| }\left( \frac{n}{n}\right) \right) =s_{\left| x\right| }\text {,} \end{aligned}$$

where $s_{x}\left( \frac{1}{n}\right) \ge s_{x}\left( \frac{2}{n}\right) \ge \cdots \ge s_{x}\left( \frac{n}{n}\right) \ge 0.$ We call $s_{\left| x\right| }$ the nonincreasing rearrangement of $\left| x\right|$. Note that, although $\sigma$ may not be unique, $s_{\left| x\right| }$ is unique.

Given a unitarily invariant norm $\alpha$ on $\mathbb {M}_{n}\left( \mathbb {C}\right) ,$ define $\beta _{\alpha }$ on $\mathbb {C}^{n}$ by

$$\begin{aligned} \beta _{\alpha }\left( x\right) =\beta _{\alpha }(x_{1},...,x_{n})=\alpha \left( \begin{array}{ccc} x_{1} &{} &{} \\ &{} \ddots &{} \\ &{} &{} x_{n} \end{array} \right) =\alpha \left( \begin{array}{ccc} s_{\left| x\right| }\left( \frac{1}{n}\right) &{} &{} \\ &{} \ddots &{} \\ &{} &{} s_{\left| x\right| }\left( \frac{n}{n}\right) \end{array} \right) . \end{aligned}$$

Clearly, permutations on $J_{n}$ corresponds to unitary conjugations by permutation matrices in $\mathbb {M}_{n}\left( \mathbb {C}\right)$. Hence $\beta _{\alpha }$ is a normalized gauge norm on $L^{\infty }\left( \delta _{n}\right) =$ $\mathbb {C}^{n}.$

Conversely, given a symmetric normalized gauge norm $\beta$ on $\mathbb {C} ^{n}$, we would like to define $\alpha _{\beta }$ on $\mathbb {M}_{n}\left( \mathbb {C}\right)$ by

$$\begin{aligned} \alpha _{\beta }\left( T\right) =\beta \left( s_{T}\left( \frac{1}{n}\right) ,s_{T}\left( \frac{2}{n}\right) ,\cdots ,s_{T}\left( \frac{n}{n}\right) \right) . \end{aligned}$$

We need to check that $\alpha _{\beta }$ is a norm. Clearly, $s_{\lambda T}\left( \frac{1}{n}\right) =\left| \lambda \right| s_{T}\left( \frac{1}{n}\right)$, so

$$\begin{aligned} \alpha _{\beta }\left( \lambda T\right) =\beta \left( s_{\lambda T}\left( \frac{1}{n}\right) ,s_{\lambda T}\left( \frac{2}{n}\right) ,\cdots ,s_{\lambda T}\left( \frac{n}{n}\right) \right) =\left| \lambda \right| \alpha _{\beta }\left( T\right) . \end{aligned}$$

Also, $\alpha _{\beta }\left( T\right) \ge 0$ and $\alpha _{\beta }\left( T\right) =0$ implies $T=0.$ The big problem is the triangle inequality: $s_{A+B}\left( \frac{k}{n}\right) \le s_{A}\left( \frac{k}{n}\right) +s_{B}\left( \frac{k}{n}\right)$ can fail if $k>1.$ When $k=1,$ $s_{T}\left( \frac{k}{n}\right) =\left\| T\right\| .$

Example 1

$A=\left( \begin{array}{cc} \frac{1}{2} &{} 0\\ 0 &{} 1 \end{array} \right) ,$ $B=\left( \begin{array}{cc} 1 &{} 0\\ 0 &{} \frac{3}{4} \end{array} \right) .$

In this example, $s_{A+B}\left( \frac{2}{n}\right) =\frac{3}{2},s_{A}\left( \frac{2}{n}\right) +s_{B}\left( \frac{2}{n}\right) =\frac{5}{4}.$

In order to prove the triangle inequality of $\alpha _{\beta }$, the Ky Fan Norms play a central role. For $1\le k\le n$ we define $KF_{\frac{k}{n}}:\mathbb {M}_{n}\left( \mathbb {C}\right) \rightarrow [0,\infty )$ and $KF_{\frac{k}{n}}:\mathbb {C}^{n}\rightarrow [0,\infty ),$ by

$$\begin{aligned} { }KF_{\frac{k}{n}}\left( T\right) =\frac{s_{T}\left( \frac{1}{n}\right) +\cdots +s_{T}\left( \frac{k}{n}\right) }{k}\text { and } KF_{\frac{k}{n}}\left( x\right) =\frac{s_{\left| x\right| }\left( \frac{1}{n}\right) +\cdots +s_{\left| x\right| }\left( \frac{k}{n}\right) }{k}. \end{aligned}$$

To prove $KF_{\frac{k}{n}}$ is a norm on $\mathbb {M}_{n}\left( \mathbb {C} \right)$ and on $\mathbb {C}^{n}$, we use the following Lemma whose proof can be found in [7]. Once we know $\alpha =KF_{\frac{k}{n}}$ is a norm on $\mathbb {M}_{n}\left( \mathbb {C}\right)$, it easily follows that $KF_{\frac{k}{n}}=\beta _{\alpha }$ is a symmetric gauge norm on $\mathbb {C} ^{n}$.

Lemma 4

For $T\in \mathbb {M}_{n}\left( \mathbb {C}\right) ,$ $KF_{\frac{k}{n}}\left( T\right) =\sup \{ \left| \textrm{Tr}\left( UTP\right) \right| :U$ is unitary, P is a projection of rank $k\}.$

We easily obtain the following corollary.

Corollary 1

${{\sum _{m=1}^{k}}}s_{A+B}(\frac{m}{n} )\le$ ${{{\sum _{m=1}^{k}}}}\left[ s_{A}(\frac{m}{n} )+s_{B}(\frac{m}{n})\right]$ for $A,B\in \mathbb {M}_{n}\left( \mathbb {C} \right)$ and $1\le k\le n.$

The key result relates the Ky Fan norms to arbitrary unitarily invariant norms. The proof can be found in [9].

Lemma 5

Suppose $n\in \mathbb {N}$, $a=(a_{1},...,a_{n} ),b=(b_{1},...,b_{n})\in \mathbb {C} ^{n},$ $a_{1}\ge a_{2}\ge \cdots \ge a_{n}\ge 0$, and $b_{1}\ge b_{2} \ge \cdots b_{n}\ge 0$. If $KF_{\frac{k}{n}}\left( a\right) \le$ $KF_{\frac{k}{n}}\left( b\right)$ for $1\le k\le n,$ then there exists $N\in \mathbb {N}$, $\sigma _{1},\cdots ,\sigma _{N}\in \mathbb {S}_{n}$, $0\le t_{j}\le 1$, with ${{\sum _{j=1}^{N}}}t_{j}=1$ such that $a\le {{\sum _{j=1}^{N}}}t_{j}\left( \sigma _{j}\left( b\right) \right) ,$ i.e.,

$$\begin{aligned} \left( a_{1},\ldots ,a_{n}\right) \le \sum _{j=1}^{N}t_{j}\left( b_{\sigma _{j}\left( 1\right) },\ldots ,b_{\sigma _{j}\left( n\right) }\right) \text {.} \end{aligned}$$

Corollary 2

Suppose $a,b\in \mathbb {C} ^{n}$ with $KF_{\frac{k}{n}}\left( a\right) \le$ $KF_{\frac{k}{n}}\left( b\right)$ for $1\le k\le n,$ then, for every symmetric gauge norm $\beta$ on $\mathbb {C}^{n}$, $\beta \left( a\right) \le \beta \left( b\right) .$

Proof

$\beta \left( a\right) \le \beta \left( \underset{j=1}{\overset{N}{\sum } }t_{j}\sigma _{j}\left( b\right) \right) \le \underset{j=1}{{ } \overset{N}{\sum }}t_{j}\beta \left( \sigma _{j}\left( b\right) \right) =\left( \underset{j=1}{\overset{N}{\sum }}t_{j}\right) \beta \left( b\right) =\beta \left( b\right) .$ $\square$

Lemma 6

If $\beta$ is a symmetric normalized gauge norm on $\mathbb {C}^{n}$, then $\alpha _{\beta }$ is a unitarily invariant norm on $\mathbb {M}_{n}\left( \mathbb {C}\right)$.

Proof

We just need to prove the triangle inequality. Suppose $A,B\in \mathbb {M} _{n}\left( \mathbb {C}\right)$. If

$$\begin{aligned} a&=\left( s_{A+B}\left( \frac{1}{n}\right) ,s_{A+B}\left( \frac{2}{n}\right) ,...,s_{A+B}\left( \frac{n}{n}\right) \right) \text { and,}\\ b&=\left( s_{A}\left( \frac{1}{n}\right) +s_{B}\left( \frac{1}{n}\right) ,s_{A}\left( \frac{2}{n}\right) +s_{B}\left( \frac{2}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) +s_{B}\left( \frac{n}{n}\right) \right) , \end{aligned}$$

then, by Corollary 1, we know that $KF_{\frac{k}{n}}\left( a\right) \le KF_{\frac{k}{n}}\left( b\right)$ for $1\le k\le n$. It follows from Corollary 2 that $\beta \left( a\right) \le \beta \left( b\right) .$ However,

$$\begin{aligned}{} & {} \alpha _{\beta }\left( A+B\right) =\beta \left( a\right) \le \beta \left( b\right) =\beta \left( s_{A}+s_{B}\right) \\{} & {} \quad \le \beta \left( s_{A}\right) +\beta \left( s_{B}\right) =\alpha _{\beta }\left( A\right) +\alpha _{\beta }\left( B\right) . \end{aligned}$$

This completes the proof. $\square$

It is easy to see that $\alpha _{\beta _{\alpha }}=\alpha$ and $\beta _{\alpha _{\beta }}=\beta$ always hold. This gives us von Neumann’s characterization of unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$.

Theorem 1

[24]. There is a one to one correspondence between symmetric gauge norms on $\mathbb {C}^{n}$ and unitarily invariant norms on $\mathbb {M}_{n}\left( \mathbb {C}\right)$.

2.1.2 Approximate discrete Ky-Fan Lemma

We want to prove an approximate version of the Ky-Fan Lemma. Suppose $n\in \mathbb {N}$, $a=(1,0,...,0),b=(\frac{1}{n},\frac{1}{n},...,\frac{1}{n})\in \mathbb {C} ^{n}$. The Ky-Fan Lemma says that a convex combination of permutations of a is greater than or equal to b. It is clear that each permutation of a has only one nonzero entry, so the number of permutations must be at least n. We want, given a positive number $\varepsilon ,$ to find a number k that is independent of n,  so that when $0\le a,b\in \mathbb {C}^{n}$ and $KF_{j}\left( a\right) \ge KF_{j}\left( b\right)$ for $1\le j\le n$, there is an average of k permutations of a that is greater than or equal to $\left( b_{1}-\varepsilon ,b_{2}-\varepsilon ,\ldots ,b_{n} -\varepsilon \right)$.

Suppose $n\in \mathbb {N}$, $f:\left\{ 1,\ldots ,n\right\} \rightarrow \mathbb {C}$ and $\gamma :\left\{ 1,\ldots ,n\right\} \rightarrow \left\{ 1,\ldots ,n\right\}$ is bijective. Let $\mathbb {S}_{n}$ denote the set of all bijective functions $\gamma :\left\{ 1,\ldots ,n\right\} \rightarrow \left\{ 1,\ldots ,n\right\}$ with the identity map $\textrm{id}\in \mathbb {S}_{n}$ defined by

$$\begin{aligned} \textrm{id}\left( k\right) =k\text { for all }1\le k\le n\text {.} \end{aligned}$$

Let $\mathbb {C}^{n}$ denote the set of all functions $f:\left\{ 1,\ldots ,n\right\} \rightarrow \mathbb {C}$. For each $\gamma \in \mathbb {S}_{n}$, define the map $C_{\gamma }:\mathbb {C}^{n}\rightarrow \mathbb {C}^{n}$ by

$$\begin{aligned} C_{\gamma }\left( f\right) =f\circ \gamma \text {.} \end{aligned}$$

If N is a positive integer, we define

$$\begin{aligned} \mathcal {C}_{N}=\left\{ \frac{1}{N}\sum _{k=1}^{N}C_{\gamma _{k}}:\gamma _{1},\ldots ,\gamma _{N}\in \mathbb {S}_{N}\right\} \text {.} \end{aligned}$$

It is easily seen that if $\varphi _{1}\in \mathcal {C}_{N_{1}}$ and $\varphi _{2}\in \mathcal {C}_{N_{2}}$, then $\varphi _{2}\circ \varphi _{1}\in \mathcal {C}_{N_{1}N_{2}}$. Also if $m_{1},\ldots m_{k}\in \mathbb {N}$ and $\gamma _{1},\ldots ,\gamma _{k}\in \mathbb {S}_{n}$, and if $\sum _{j=1}^{k} m_{j}=N$, then

$$\begin{aligned} \sum _{j=1}^{k}\frac{m_{j}}{N}C_{\gamma _{j}}\in \mathcal {C}_{N}\text {.} \end{aligned}$$

Suppose $\left( X_{1},<_{1}\right) ,$ $\left( X_{2},<_{2}\right)$ are strictly linearly ordered sets. We let < $=\left(<_{1},<_{2}\right)$ denote the lexicographical order on $X_{1}\times X_{2},$ i.e.,

$$\begin{aligned} \left( a_{1},b_{1}\right)<\left( a_{2},b_{2}\right) \Leftrightarrow a_{1}<_{1}a_{2}\text {, or }a_{1}=a_{2}\text { and }b_{1}<_{2}b_{2}\text {. } \end{aligned}$$

Then $\left(<_{1},<_{2}\right)$ is a strict linear order on $X_{1}\times X_{2}$.

Suppose $m\in \mathbb {N}$. Let $E_{m}=\left\{ 1,\ldots ,m\right\} \times \left\{ 1,\ldots ,m\right\}$, linearly ordered by $\prec$ $=\left(<,<\right)$ (where < is the usual order). Let $\prec ^{\prime }=\left( \prec ,\prec \right)$ on $E_{m}\times E_{m}$. This makes $E_{m}\times E_{m}$ order-isomorphic to $\left\{ 1,2,\ldots ,m^{4}\right\}$.

Lemma 7

Suppose $m,n\in \mathbb {N}$ and $f:\left\{ 1,\ldots ,n\right\} \rightarrow \left\{ 1,\ldots ,m\right\}$ and $h:\left\{ 1,\ldots ,n\right\} \rightarrow \left\{ 0,1,\ldots ,m\right\}$ are nonincreasing and, for every $k\in \left\{ 1,\ldots ,n\right\}$

$$\begin{aligned} \sum _{j=1}^{k}h\left( j\right) \le \sum _{j=1}^{k}f\left( j\right) \text {.} \end{aligned}$$

If $N=\left( m!\right) ^{m^{5}}$, then there are permutations $\gamma _{1},\ldots ,\gamma _{N}$ of $\left\{ 1,\ldots ,n\right\}$ such that

$$\begin{aligned} \frac{1}{N}\sum _{k=1}^{N}f\circ \gamma _{k}\ge h\text {.} \end{aligned}$$

Proof

Suppose $g:\left\{ 1,\ldots ,n\right\} \rightarrow \left\{ 0,1,\ldots ,m\right\}$ and $g\not \ge h$. We define

$$\begin{aligned}{} & {} q_{g}=\min \left\{ k:g\left( k\right) <h\left( k\right) \right\} \text { and} \\{} & {} q_{g}^{\prime }=\max \left\{ k:\left( g\left( k\right) ,h\left( k\right) \right) =\left( g\left( q_{g}\right) ,h\left( q_{g}\right) \right) \right\} \text {.} \end{aligned}$$

We say g is nice if $g\not \ge h$ and

• (1)
  $$\begin{aligned} \sum _{j=1}^{k}h\left( j\right) \le \sum _{j=1}^{k}g\left( j\right) \end{aligned}$$

  for $q_{g}^{\prime }\le k\le n,$

• (2) g is nonincreasing on $\left\{ k:q_{g}\le k\le n\right\} .$

Suppose now that g is “nice”. It is clear from the definition of $q_{g}$ and the fact that h is nonincreasing that $\left( g\left( q_{g}\right) ,h\left( q_{g}\right) \right)$ is the largest element (with respect to $\prec$) of

$$\begin{aligned} \left\{ \left( g\left( k\right) ,h\left( k\right) \right) :1\le k\le n\text {, }g\left( k\right) <h\left( k\right) \right\} . \end{aligned}$$

It follows from $\left( 2\right)$ that $g\left( k\right) \le g\left( q_{g}\right)$ whenever $q_{g}\le k\le n$. Since g and h are nonincreasing on $\left\{ k:q_{g}\le k\le n\right\}$, we know that

$$\begin{aligned} \left\{ k:\left( g\left( k\right) ,h\left( k\right) \right) =\left( g\left( q_{g}\right) ,h\left( q_{g}\right) \right) \right\} =\left\{ k:q_{g}\le k\le q_{g}^{\prime }\right\} \text {.} \end{aligned}$$

Thus we have

• (3) for $q_{g}\le k\le n$.

  $$\begin{aligned} \sum _{j=1}^{k}h\left( j\right) \le \sum _{j=1}^{k}g\left( j\right) . \end{aligned}$$

  Let $F=\left\{ k:q_{g}\le k\le q_{g}^{\prime }\right\}$, and let

  $$\begin{aligned} b=Card\left( F\right) =q_{g}^{\prime }-q_{g}+1\text {.} \end{aligned}$$

It follows from $\left( 3\right)$, with $k=q_{g}$, that there is an integer $k_{0}$ with $1\le k_{0}<q_{g}$ such that $g\left( k_{0}\right) >h\left( k_{0}\right)$. Let $p_{g}$ be the smallest positive integer such that

$$\begin{aligned} g\left( p_{g}\right) =\max \left\{ g\left( k\right) :1\le k\le n\text {, }g\left( k\right) >h\left( k\right) \right\} \text {.} \end{aligned}$$

If $g\left( k\right) =g\left( p_{g}\right) ,$ then

$$\begin{aligned} g\left( k\right) =g\left( p_{g}\right) \ge g\left( k_{0}\right)>h\left( k_{0}\right) \ge h\left( q_{g}\right) >g\left( q_{g}\right) , \end{aligned}$$

which means, from $\left( 2\right)$, that $k<q_{g}$. In particular, $p_{g}<q_{g}$.

It is now clear that $\left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right)$ is the largest element (with respect to $\prec$) of $\left\{ \left( g\left( k\right) ,h\left( k\right) \right) :g\left( k\right) >h\left( k\right) \right\}$.

Let

$$\begin{aligned} E=\left\{ k:1\le k\le n,\left( g\left( k\right) ,h\left( k\right) \right) =\left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) \right\} \end{aligned}$$

and let $a=Card\left( E\right)$. Since, for each $k\in E$, $g\left( k\right) =g\left( p_{g}\right) >g\left( q_{g}\right) ,$ so $E\subset [1,q_{g}).$

A simple computation shows that if $t_{g}=\frac{g\left( p_{g}\right) -h\left( p_{g}\right) }{g\left( p_{g}\right) -g\left( q_{g}\right) }$ and $1-t_{g}=\frac{h\left( p_{g}\right) -g\left( q_{g}\right) }{g\left( p_{g}\right) -g\left( q_{g}\right) },$ then

$$\begin{aligned}{} & {} \left( 1-t_{g}\right) g\left( p_{g}\right) +t_{g}g\left( q_{g}\right) =h\left( p_{g}\right) ,\text { and} \\{} & {} \left( 1-t_{g}\right) g\left( q_{g}\right) +t_{g}g\left( p_{g}\right) =g\left( q_{g}\right) +\left[ g\left( p_{g}\right) -h\left( p_{g}\right) \right] \text {.} \end{aligned}$$

Note that $1\le g\left( p_{g}\right) -g\left( q_{g}\right) \le m$. Also

$$\begin{aligned}{} & {} g\left( q_{g}\right) +\left[ g\left( p_{g}\right) -h\left( p_{g}\right) \right] =g\left( p_{g}\right) +\left[ g\left( q_{g}\right) -h\left( p_{g}\right) \right] \\{} & {} g\left( p_{g}\right) +\left[ g\left( q_{g}\right) -h\left( q_{g}\right) \right] <g\left( p_{g}\right) . \end{aligned}$$

We now set up some notation. Suppose $D\subset E$ and $d=Card\left( D\right) \le Card\left( F\right) .$ Let $\alpha _{D,g}:D\rightarrow \left\{ k:q_{g}\le k<q_{g}+d\right\}$ be the unique order-preserving bijection. Define $\gamma _{D,g}\in \mathbb {S}_{n}$ by

$$\begin{aligned} \gamma _{D,g}\left( k\right) =\left\{ \begin{array}{ll} \alpha _{D,g}\left( k\right) &{} \quad \text {if }k\in D\\ \alpha _{D,g}^{-1}\left( k\right) &{} \quad \text {if }k\in \alpha _{D,g}\left( D\right) \\ k &{} \quad \text {otherwise} \end{array} \right. {.} \end{aligned}$$

Define $\varphi _{D,g}\in \mathcal {C}_{m!}$ by

$$\begin{aligned} \varphi _{D,g}=\left( 1-t_{g}\right) C_{\gamma _{D,g}}+t_{g} C_{{\text {id}}}\text {.} \end{aligned}$$

Thus

$$\begin{aligned} \varphi _{D,g}\left( g\right) \left( k\right) =\left\{ \begin{array}{ll} h\left( k\right) &{} \text {if }k\in D\\ g\left( k\right) +\left( g\left( p_{g}\right) -h\left( p_{g}\right) \right) =g\left( q_{g}\right) +\left( g\left( p_{g}\right) -h\left( p_{g}\right) \right) &{}\text {if }k\in \alpha _{D,g}\left( D\right) \\ g\left( k\right) &{} \text {otherwise} \end{array} \right. \text {.} \end{aligned}$$

Also

$$\begin{aligned}{} & {} g\left( q_{g}\right)<\varphi _{D,g}\left( g\right) \left( q_{g}\right) =g\left( q_{g}\right) +\left[ g\left( p_{g}\right) -h\left( p_{g}\right) \right] =g\left( p_{g}\right) +\left[ g\left( q_{g}\right) -h\left( p_{g}\right) \right] \\{} & {} \quad \le g\left( p_{g}\right) +\left[ g\left( q_{g}\right) -h\left( q_{g}\right) \right] <g\left( p_{g}\right) . \end{aligned}$$

Since $g\left( p_{g}\right) -h\left( p_{g}\right) >0$, it is clear that $\varphi _{D,g}\left( g\right)$ is nonincreasing on $\left\{ k:q_{g}\le k\le n\right\}$. If $\varphi _{D,g}\left( g\right) \not \ge h,$ then $q_{\varphi _{D,g}\left( g\right) }^{\prime }\ge q_{g}^{\prime }$. Since

$$\begin{aligned} \sum _{j=1}^{q_{g}}\varphi _{D,g}\left( g\right) \left( j\right) =\sum _{j=1}^{q_{g}}g\left( j\right) , \end{aligned}$$

it follows from $\left( 1\right)$ that, whenever $k\ge q_{\varphi _{D,g}\left( g\right) }^{\prime }\ge q_{g}^{\prime }$,

$$\begin{aligned} \sum _{j=1}^{k}h\left( j\right) =\sum _{j=1}^{k}\varphi _{D,g}\left( g\right) \left( j\right) . \end{aligned}$$

Thus if $\varphi _{D,g}\left( g\right) \not \ge h$, then $\varphi _{D,g}\left( g\right)$ is nice.

For any nice g we define $\rho \left( g\right) \in E_{m}\times E_{m}$ by

$$\begin{aligned} \rho \left( g\right) =\left( \left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) ,\left( g\left( q_{q}\right) ,h\left( q_{g}\right) \right) \right) . \end{aligned}$$

CLAIM: There is a $\hat{\varphi }\in \mathcal {C}_{\left( m!\right) ^{m}}$ such that $\hat{g}=\hat{\varphi }\left( g\right)$ satisfies $\hat{g}\ge h$ or $\hat{g}$ is nice and $\rho \left( \hat{g}\right) <\rho \left( g\right)$.

We now consider a few cases.

Case 1: :

$a\le b.$ Let $D=E$. If $\varphi _{\alpha }\left( g\right) \not \ge h$, then $\varphi _{\alpha }\left( g\right)$ is nice and

$$\begin{aligned} \left( \varphi _{\alpha }\left( g\right) \left( p_{\varphi _{\alpha }\left( g\right) }\right) ,h\left( p_{\varphi _{\alpha }\left( g\right) }\right) \right) \prec \left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) \text {.} \end{aligned}$$

It follows that

$$\begin{aligned} \rho \left( \varphi _{\alpha }\left( g\right) \right) \prec ^{\prime } \rho \left( g\right) . \end{aligned}$$

In this case we define $\hat{g}=\varphi _{\alpha }\left( g\right)$ and $\hat{\varphi }=\varphi _{\alpha }$.

We now can assume that $a>b$. There is a smallest positive integer w such that

$$\begin{aligned} g\left( q_{g}\right) +w\left( g\left( p_{g}\right) -h\left( p_{g}\right) \right) \ge h\left( q_{g}\right) , \end{aligned}$$

and there is a positive integer v and a disjoint collection $\left\{ E_{0},E_{1},\ldots E_{v}\right\}$ whose union is E and such that

$$\begin{aligned} Card\left( E_{k}\right) =b\text { for }1\le k\le v\text {.} \end{aligned}$$

Let $u=\min \left( v,w\right)$, and define bijections $\alpha _{k}:E_{k}\rightarrow F$ for $1\le k\le u$. We let

$$\begin{aligned} t_{k}=\frac{g\left( p_{g}\right) -h\left( p_{g}\right) }{g\left( p_{g}\right) -\left[ g\left( q_{g}\right) +k\left( g\left( q_{g}\right) +\left( k-1\right) \left[ g\left( p_{g}-h\left( p_{g}\right) \right) \right] \right) \right] }\text {,} \end{aligned}$$

and

$$\begin{aligned} \varphi _{\alpha _{k}}= \left( 1-t_{k}\right) C_{\gamma _{k}}+t_{k} {\text {id}.} \end{aligned}$$

For $1\le s\le u$, let

$$\begin{aligned} \varphi _{s}=\varphi _{\alpha _{s}}\circ \cdots \circ \varphi _{\alpha _{1}}. \end{aligned}$$

We see that

$$\begin{aligned} \varphi _{s}\left( g\right) \left( k\right) =\left\{ \begin{array}{ll} h\left( k\right) &{} \text {if }k\in \cup _{1\le j\le s}E_{j}\\ g\left( q_{g}\right) +s\left( g\left( p_{q}\right) -h\left( p_{g}\right) \right) &{} \text {if }k\in F\\ g\left( k\right) &{} \text {otherwise} \end{array} \right. \text {.} \end{aligned}$$

We see that $\varphi _{u-1}\left( g\right) \not \ge h$ and $\varphi _{u-1}\left( g\right)$ is nice. Thus $\varphi _{u}\left( g\right) \ge h$ or $\varphi _{u}\left( g\right)$ is nice.

We now consider more cases.

Case 2: :

$v<w$. We know that $\varphi _{v}\left( g\right)$ is nice. We let $D=E_{0}$ and let $\alpha :E_{0}\rightarrow \left\{ k:q_{g}\le k<q_{q}+Card\left( E_{0}\right) \right\}$ be a bijection and let $\hat{g}=\varphi _{\alpha }\left( g_{v}\right) .$ Then $\hat{g}\ge h$ or $\hat{g}$ is nice and

$$\begin{aligned} \left( \hat{g}\left( p_{\hat{g}}\right) ,h\left( p_{\hat{g}}\right) \right) \prec \left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) . \end{aligned}$$

Thus if $\hat{g}\not \ge h$, then

$$\begin{aligned} \rho \left( \hat{g}\right) \prec ^{\prime }\rho \left( g\right) . \end{aligned}$$

$v>w$ or $v=w$ and $E_{0}\ne \varnothing .$ In this case $g_{w}$ is nice and if $g_{w}\not \ge h$, then $g_{\omega }$ is nice,

$$\begin{aligned} \left( g_{\omega }\left( p_{g_{w}}\right) ,h\left( p_{g_{w}}\right) \right) =\left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) \end{aligned}$$

and

$$\begin{aligned} \left( g_{\omega }\left( q_{g_{w}}\right) ,h\left( p_{g_{w}}\right) \right) =\left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) . \end{aligned}$$

Thus if $\hat{g}=g_{w}$ and $\hat{\varphi }=\varphi _{\alpha _{w}}\circ \cdots \circ \varphi _{1}$, then $\hat{g}=\hat{\varphi }\left( g\right)$, either $\hat{g}\ge h$ or $\hat{g}$ is nice, and

$$\begin{aligned} \rho \left( \hat{g}\right) \prec ^{\prime }\rho \left( g\right) . \end{aligned}$$

Case 3::

$v=w$ and $E=\varnothing$. In this case $\varphi _{w}\left( g\right) \ge h$ or $\varphi _{w}\left( g\right)$ is nice. If $\varphi _{w}\left( g\right)$ is nice, then

$$\begin{aligned} \left( g_{\omega }\left( p_{g_{w}}\right) ,h\left( p_{g_{w}}\right) \right) \prec \left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) \end{aligned}$$

and

$$\begin{aligned} \left( g_{\omega }\left( q_{g_{w}}\right) ,h\left( p_{g_{w}}\right) \right) \prec \left( g\left( p_{g}\right) ,h\left( p_{g}\right) \right) . \end{aligned}$$

Thus if $\hat{g}=g_{w}$ and $\hat{\varphi }=\varphi _{\alpha _{w}}\circ \cdots \circ \varphi _{1}$, then $\hat{g}=\hat{\varphi }\left( g\right)$ and either $\hat{g}\ge h$ or $\hat{g}$ is nice, and

$$\begin{aligned} \rho \left( \hat{g}\right) \prec ^{\prime }\rho \left( g\right) . \end{aligned}$$

$\hat{\varphi }_{\hat{g}}\in \mathcal {C}_{\left( m!\right) ^{m}}$ such that $\hat{g}=\hat{\varphi }_{g}\left( g\right)$ is nice and $\rho \left( \hat{g}\right) \prec ^{\prime }\rho \left( g\right) .$

It follows from Cases 1, 2, 3 that the claim is proved.

If we let $f_{0}=f$ and if $f_{k}$ is nice and $f_{k}\not \ge h$, then $f_{k+1}=\hat{f}_{k}$ and $\hat{f}_{k+1}=\varphi _{k+1}\left( f_{k}\right)$ with $\varphi _{k+1}\in \mathcal {C}_{\left( m!\right) ^{m}}.$ It follows from the Claim that there is a smallest $k\le m^{4}$ such that $f_{k}\ge h$. Then

$$\begin{aligned} \varphi =\varphi _{k}\circ \cdots \circ \varphi _{1}\in \mathcal {C}_{\left( \left( m!\right) ^{m}\right) ^{k}}\subset \mathcal {C}_{\left( m!\right) ^{m^{5}}} \end{aligned}$$

and $f_{k}=\varphi \left( f\right)$. Thus the lemma is proved. $\square$

2.1.3 Unitarily invariant norms on a $II_{1}$ factor

In this section we give a new proof of the characterization in [10] of unitarily invariant norms on a $II_{1}$ factor von Neumann algebra $\mathcal {M}$. If $\mathcal {\mathcal {M}}$ is a type $II_{1}$ factor von Neumann algebra. then $\mathcal {M}$ has a unique faithful normal tracial state $\tau$ with the property that if P and Q are projections in $\mathcal {M}$, then P and Q are unitarily equivalent in $\mathcal {M}$ if and only if $\tau \left( P\right) =\tau \left( Q\right)$. In this case the measure space $\left( J_{n},\delta _{n}\right)$ is replaced with the measure space $\left( J_{\infty },\delta _{\infty }\right)$, where $J_{\infty }=\left[ 0,1\right]$ and $\delta _{\infty }$ is Lebesgue measure. A normalized gauge norm $\beta$ on $L^{\infty }\left[ 0,1\right] =L^{\infty }\left( \delta _{\infty }\right)$ is symmetric if, for every $\gamma \in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right)$ and every $f\in L^{\infty }\left( \delta _{\infty }\right) ,$ we have $\beta (f)=\beta (f\circ \gamma ).$

The main result in [10] is that there is a one-to-one correspondence between the unitarily invariant norms on $\mathcal {M}$ and the symmetric normalized gauge norms on $L^{\infty }\left( \delta _{\infty }\right)$. This looks just like von Neumann’s result for $\mathbb {M}_{n}\left( \mathbb {C} \right)$.

The definition of the s-numbers for a function in $L^{\infty }\left[ 0,1\right]$ can be obtained from nonincreasing rearrangements in measure theory. The proof in [10] doesn’t use a version of the Ky Fan Lemma (Lemma 5); we present a new proof here using an “approximate” version of the Ky Fan Lemma (Theorem 2).

The first result we need is nonincreasing rearrangements from [11, Chapter 37].

Lemma 8

Suppose $f:\left[ 0,1\right] \rightarrow \mathbb {C}$ is measurable. Then there is a $\gamma \in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right)$ such that $s_{f}\underset{\textrm{def}}{=}\left| f\right| \circ \gamma$ is nonincreasing on $\left[ 0,1\right]$. The transformation $\gamma$ may not be unique, but $s_{f}$ is unique (a.e.). It therefore follows that if $f_{1}$,$f_{2}:\left[ 0,1\right] \rightarrow \mathbb {C}$ are measurable, then

$$\begin{aligned} s_{f_{1}}=s_{f_{2}}\text { if and only if }\left| f_{1}\right| =\left| f_{2}\right| \circ \gamma \text { for some }\gamma \in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right) . \end{aligned}$$

The function $s_{f}$ is called the nonincreasing rearrangement of $\left| f\right|$.

For $0<t\le 1,$ we define the Ky Fan norm $KF_{t}$ on $L^{\infty }\left[ 0,1\right]$ by

$$\begin{aligned} KF_{t}\left( f\right) =\frac{1}{t}\int _{0}^{t}s_{f}d\delta _{\infty }\text {.} \end{aligned}$$

For an operator $T\in \mathcal {M}$ and $0\le t\le 1$, the $t^{th}$ s-number of T, denoted by $s_{T}\left( t\right)$, was defined by Fack and Kosaki in [8] as

$$\begin{aligned} s_{T}\left( t\right) =\inf \{ \left\| TE\right\| :E\text { is a projection in }\mathcal {M}\text { with }\tau (E^{\perp })\le t\}. \end{aligned}$$

It is clear that the map $t\mapsto s_{T}\left( t\right)$ is nonincreasing on $\left[ 0,1\right]$. The $t^{th}$ Ky Fan norm $KF_{t}\left( T\right)$ is defined as

$$\begin{aligned} KF_{t}\left( T\right) =\left\{ \begin{array}{ll} \left\| T\right\| &{} \text {if }t=0\\ \frac{1}{t}\int _{0}^{t}s_{T}\left( t\right) d\delta _{\infty } &{}\quad \text {if } 0<t\le 1 \end{array} \right. . \end{aligned}$$

In the matrix case $\left| T\right|$ is unitarily equivalent to a diagonal matrix, which naturally corresponds to an element of $\mathbb {C}^{n}$. In the $II_{1}$ factor case we need a more complicated approach.

Definition 3

A normal $*$-isomorphism $\pi :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {M}$ such that, for every $f\in L^{\infty }\left( \delta _{\infty }\right)$,

$$\begin{aligned} \left( \tau \circ \pi \right) \left( f\right) =\int _{J_{\infty }} fd\delta _{\infty }. \end{aligned}$$

is called a tracial embedding.

In the matrix case, the assertion that $\left| T\right|$ is unitarily equivalent to a diagonal matrix can be rephrased as $\left| T\right|$ is contained in a maximal abelian selfadjoint algebra (i.e., masa) of $\mathbb {M}_{n}\left( \mathbb {C}\right)$, and every masa in $\mathbb {M} _{n}\left( \mathbb {C}\right)$ is unitarily equivalent to the algebra of diagonal $n\times n$ matrices. Here is the analogue for a $II_{1}$ factor. This can be found in [22].

Lemma 9

Suppose $\mathcal {A}$ is a masa in a type $II_{1}$ factor $\mathcal {M}$. Then there is a surjective tracial embedding $\pi :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {A}$.

Lemma 10

Suppose $\mathcal {A}$ is a masa in a type $II_{1}$ factor $\mathcal {M}$. If $f\in L^{\infty }\left[ 0,1\right]$ and $\pi \left( f\right) =T$, then, for almost every $t\in \left[ 0,1\right]$,

$$\begin{aligned} s_{f}\left( t\right) =s_{\pi \left( f\right) }\left( t\right) =s_{T}\left( t\right) . \end{aligned}$$

The following lemma is a consequence of Hadwin-Ding in [4].

Lemma 11

If $\pi$ and $\rho$ are tracial embeddings into a $II_{1}$ factor $\mathcal {M}$, then $\pi$ and $\rho$ are approximately unitarily equivalent in $\mathcal {M}$, i.e., there is a net $\left\{ U_{j}\right\}$ of unitary operators in $\mathcal {M}$ such that, for every $f\in L^{\infty }\left( \delta _{\infty }\right)$,

$$\begin{aligned} \left\| U_{j}^{*}\pi \left( f\right) U_{j}-\rho \left( f\right) \right\| \rightarrow 0. \end{aligned}$$

Corollary 3

If $\pi :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {M}$ is a tracial embedding and $\gamma \in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right)$, then $\rho :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {M}$ defined by $\rho \left( f\right) =\pi \left( f\circ \gamma \right)$ is also a tracial embedding. Hence, there is a net $\left\{ U_{j}\right\}$ of unitary operators in $\mathcal {M}$ such that, for every $f\in L^{\infty }\left( \delta _{\infty }\right)$,

$$\begin{aligned} \left\| U_{j}^{*}\pi \left( f\right) U_{j}-\pi \left( f\circ \gamma \right) \right\| \rightarrow 0. \end{aligned}$$

As in the matrix case we need to prove $KF_{t}$ is a norm on $\mathcal {M}$ by giving an alternate characterization given in [10, Lemma 5.1]

Lemma 12

If $T\in \mathcal {M}$ and $0<t\le 1$, then

$$\begin{aligned} KF_{t}\left( T\right) =\sup \left\{ \left| \tau \left( UTP\right) \right| :U\in \mathcal {U}\left( \mathcal {M}\right) ,P\text { is a projection in }\mathcal {M}\text {, }\tau \left( P\right) =t\right\} \text {.} \end{aligned}$$

Suppose $\alpha$ is a unitarily invariant norm on $\mathcal {M}$. We can choose a tracial embedding $\pi :L^{\infty }\left( J_{\infty },\delta _{\infty }\right) \rightarrow \mathcal {M}$ and define a norm $\beta _{\alpha }$ on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$ by

$$\begin{aligned} \beta _{\alpha }\left( f\right) =\alpha \left( \pi \left( f\right) \right) . \end{aligned}$$

We need to show that the definition does not depend on the embedding $\pi .$ If $\rho :$ $L^{\infty }\left( J_{\infty },\delta _{\infty }\right) \rightarrow \mathcal {M}$ is another tracial embedding, then by Lemma 11, there is a net $\left\{ U_{j}\right\}$ of unitary operators in $\mathcal {M}$ such that, for every $f\in L^{\infty }\left( J_{\infty },\delta \right)$

$$\begin{aligned} \left\| U_{j}^{*}\pi \left( f\right) U_{j}-\rho \left( f\right) \right\| \rightarrow 0. \end{aligned}$$

Since

$$\begin{aligned} \left| \beta \left( \pi \left( f\right) \right) -\beta \left( \rho \left( f\right) \right) \right|&=\left| \beta \left( U_{j}^{*}\pi \left( f\right) U_{j}\right) -\beta \left( \rho \left( f\right) \right) \right| \\&\le \beta \left( U_{j}^{*}\pi \left( f\right) U_{j}-\rho \left( f\right) \right) \le \left\| U_{j}^{*}\pi \left( f\right) U_{j} -\rho \left( f\right) \right\| \rightarrow 0, \end{aligned}$$

we see that $\beta \left( \pi \left( f\right) \right) =\beta \left( \rho \left( f\right) \right)$. Moreover, it follows from Corollary 3 that, the gauge norm $\beta _{\alpha }$ is symmetric. A simple consequence is that $KF_{t}=\beta _{KF_{t}}$ is a symmetric gauge norm on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$.

Next suppose $\beta$ is a symmetric gauge norm on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$. We want to define $\alpha _{\beta }$ on $\mathcal {M}$. If $T\in \mathcal {M}$, we can choose a masa $\mathcal {A}$ in $\mathcal {M}$ such that $\left| T\right| \in \mathcal {A}$. We then choose a surjective tracial embedding $\pi :L^{\infty }\left( J_{\infty },\delta _{\infty }\right) \rightarrow \mathcal {A}$ and choose $f\in L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$ such that $\pi \left( f\right) =\left| T\right|$ and then define

$$\begin{aligned} \alpha _{\beta }\left( T\right) =\beta \left( f\right) =\beta \left( s_{f}\right) \text {.} \end{aligned}$$

Since

$$\begin{aligned} s_{f}\left( t\right) =s_{\pi \left( f\right) }\left( t\right) =s_{\left| T\right| }\left( t\right) , \end{aligned}$$

we see that the definition is independent of $\mathcal {A}$ and $\pi$. As in the matrix case, the main difficulty is proving that $\alpha _{\beta }$ satisfies the triangle inequality. In [10] this was done using an approach that avoids proving an analogue of the matrix Ky Fan Lemma (Lemma 5). Here we prove a general “continuous” version of the approximate Ky Fan Lemma that we will need later in our paper.

Lemma 13

Suppose $f,h\in L^{\infty }\left[ 0,1\right] ,$ and $0\le f,h\le 1,\left\| f\right\| _{\infty }=1.$ Suppose f, h are non-increasing, then there exist step functions $s_{f}^{[m]}\ge f$ and $s_{h}^{[m]}\le h$ with ranges contained in $\left\{ \frac{k}{m}:0\le k\le m\right\}$ such that $\frac{1}{m}\le s_{f}^{[m]}\le 1$ and $0\le$ $s_{h}^{[m]}\le \frac{m-1}{m}\$and $f\le s_{f}^{[m]}\le f+\frac{1}{m}$ and $\max \left( h-\frac{1}{m},0\right) \le s_{h}^{[m]}\le h$. It follows that $KF_{t}\left( s_{h}^{[m]}\right) \le KF_{t}\left( h\right)$ and $KF_{t}\left( f\right) \le KF_{t}\left( s_{f}^{[m]}\right)$ for every $t\in (0,1].$

Proof

For every $m\in \mathbb {N},$ let $p_{i}=\sup f^{-1}\left( (1-\frac{i}{m},1-\frac{i-1}{m}]\right)$,

$q_{i}=\inf h^{-1}\left( (1-\frac{i}{m},1-\frac{i-1}{m}]\right) ,$ $i=1,...,m.$ Let $p_{0}=q_{0}=0$. Then define

$$\begin{aligned} s_{f}^{[m]}\left( x\right)&=\sum _{i=0}^{m-1}\left( 1-\frac{i}{m}\right) \chi _{[p_{i},p_{i+1})}\left( x\right) \text { for }i=0,...,m-1.\\ s_{h}^{[m]}\left( x\right)&=\sum _{i=0}^{m-1}\left( 1-\frac{i+1}{m}\right) \chi _{[q_{i},q_{i+1})}\left( x\right) \text { for }i=0,...,m-1. \end{aligned}$$

It is easy to see that $f\le s_{f}^{[m]}\le f+\frac{1}{m}$; thus $\left\| f-s_{f}^{[m]}\right\| _{\infty }\le \frac{1}{m}.$ Also

$\max \left( h-\frac{1}{m},0\right) \le s_{h}^{[m]}\le h$; so $\left\| h-s_{h} ^{[m]}\right\| _{\infty }\le \frac{1}{m}.$

Therefore, $KF_{t}\left( s_{h}^{[m]}\right) \le KF_{t}\left( h\right)$ and $KF_{t}\left( f\right) \le KF_{t}\left( s_{f}^{[m]}\right)$ for every $t\in (0,1]$ $\square$

Lemma 14

Suppose f is a step function on $\left[ a,b\right]$ and $k\in \mathbb {N}$, then there exists an invertible measure preserving map $\varphi _{k}:\left[ a,b\right] \rightarrow \left[ a,b\right]$ such that

$$\begin{aligned} \left\| \frac{1}{k}\underset{j=1}{\overset{k}{\sum }}f\circ \varphi _{k}^{\left( j\right) }-\frac{1}{b-a}\int _{a}^{b}f\left( x\right) d\delta _{\infty }\right\| _{\infty }\le \eta \left\| f\right\| _{\infty }\frac{4}{k}, \end{aligned}$$

where $\eta =card$ $f\left( \left[ a,b\right] \right)$, $\varphi _{k}^{\left( j\right) }$ is the composition of j of the $\varphi _{k}$’s,  i.e., $\varphi _{k}\circ \varphi _{k}\circ \cdots \circ \varphi _{k}.$

Proof

Define $\varphi _{k}:\left[ a,b\right] \rightarrow \left[ a,b\right]$ by

$$\begin{aligned} \varphi _{k}\left( x\right) =\left\{ \begin{array}{ll} x+\frac{b-a}{k} &{} \text {if }a\le x\le b-\frac{b-a}{k}\\ x+\frac{b-a}{k}-b+a &{} \text {if }b-\frac{b-a}{k}<x\le b. \end{array} \right. \end{aligned}$$

Then $\varphi _{k}^{\left( k\right) }$ is the identity map.

If we define $\rho _{k}\left( f\right) =\frac{1}{k}{\sum _{j=1}^{k} }f\circ \varphi _{k}^{\left( j\right) }-\frac{1}{b-a}\int _{a}^{b} fd\delta _{\infty }$, then $\rho _{k}$ is linear and $\left\| \rho _{k}\right\| \le 2$ (with $\rho _{k}$ acting as an operator on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$). Suppose $0\le j<k.$ Then $\rho _{k}\left( \chi _{[a+j\frac{b-a}{k},a+\left( j+1\right) )\frac{b-a}{k})}\right) =0$ a.e. $\left( \delta _{\infty }\right)$. However, $\rho _{k}$ is linear; therefore

$\rho _{k}\left( \chi _{[a+j_{1}\frac{b-a}{k},a+\left( j_{2}\right) )\frac{b-a}{k})}\right) =0$ whenever $0\le j_{1}<j_{2}\le k$. Suppose $a\le \alpha <\beta \le b$. We choose $j_{1}$ and $j_{2}$ such that $j_{1}$ is the largest j, $1\le j\le k$ for which $a+j_{1}\frac{b-a}{k}\le \alpha$ and choose $j_{2}$ to be the smallest j, $1\le j\le k$ for which $\beta \le a+j_{2}\frac{b-a}{k}$. Then

$$\begin{aligned} \chi _{[a+j_{1}\frac{b-a}{k},a+\left( j_{2}\right) )\frac{b-a}{k} )}-\chi _{[\alpha ,\beta )}=\chi _{[a+j_{1}\frac{b-a}{k},\alpha )}-\chi _{[\beta ,a+\left( j_{2}\right) )\frac{b-a}{k})}. \end{aligned}$$

Hence

$$\begin{aligned} \rho _{k}\left( \chi _{[\alpha ,\beta )}\right) =\rho _{k}\left( \chi _{[a+j_{1}\frac{b-a}{k},\alpha )}\right) -\rho _{k}\left( \chi _{[\beta ,a+\left( j_{2}\right) )\frac{b-a}{k})}\right) . \end{aligned}$$

However, if $E\in \left\{ [a+j_{1}\frac{b-a}{k},\alpha ),[\beta ,a+\left( j_{2}\right) )\frac{b-a}{k})\right\}$ and $f=\chi _{E}$ then, since $f\circ \varphi _{k}^{\left( j\right) }=\chi _{\left( \varphi _{k}^{\left( j\right) }\right) ^{-1}\left( E\right) }$ and the collection $\left\{ \left( \varphi _{k}^{\left( j\right) }\right) ^{-1}\left( E\right) :1\le j\le k\right\}$ is disjoint, we have

$$\begin{aligned} \left\| \frac{1}{k}\underset{j=1}{\overset{k}{\sum }}f\circ \varphi _{k}^{\left( j\right) }\right\| _{\infty }\le \frac{1}{k} \end{aligned}$$

and

$$\begin{aligned} \frac{1}{b-a}\int _{a}^{b}\chi _{E}d\delta _{\infty }\le \frac{1}{b-a}\frac{b-a}{k}=\frac{1}{k}, \end{aligned}$$

we have $\left\| \rho _{k}\left( E\right) \right\| _{\infty }\le \frac{2}{k}$. Hence

$$\begin{aligned} \rho _{k}\left( \chi _{[\alpha ,\beta )}\right) \le \frac{4}{k}. \end{aligned}$$

Suppose f is a step function, then $f={\sum _{j=1}^{n}} a_{j}\chi _{[\alpha _{j},\alpha _{j+1})}$ for some $n\in \mathbb {N}.$ Denote $f_{j}=\chi _{[\alpha _{j},\alpha _{j+1})}.$ Then

$$\begin{aligned} f=\underset{j=1}{\overset{n}{\sum }}a_{j}f_{j}\int _{a}^{b}f\left( x\right) d\delta _{\infty }=\underset{j=1}{\overset{n}{\sum }}a_{j}\int _{a}^{b} \chi _{[\alpha _{j},\alpha _{j+1})}d\delta _{\infty }. \end{aligned}$$

Thus

$$\begin{aligned} \left\| \rho _{k}\left( f\right) \right\| _{\infty }&\le \underset{j=1}{\overset{n}{\sum }}\left| a_{j}\right| \left\| \rho _{k}\left( f_{j}\right) \right\| _{\infty }\\&\le \left( \overset{n}{\underset{j=1}{\sum }}\left| a_{j}\right| \right) \frac{2}{k}\le \eta \left\| f\right\| _{\infty }\left( \frac{4}{k}\right) . \end{aligned}$$

$\square$

We call the following the approximate Ky Fan Lemma for $L^{\infty }\left( \delta _{\infty }\right) .$

Theorem 2

Suppose m is a positive integer. Then whenever $0\le f,h\le 1$ in $L^{\infty }\left( \delta _{\infty }\right)$ satisfy

$$\begin{aligned} KF_{t}\left( h\right) \le KF_{t}\left( f\right) \text { for all rational numbers }0<t\le 1, \end{aligned}$$

there are, $\gamma _{1},\ldots ,\gamma _{m^{m^{2}}}\in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right) ,$ such that

$$\begin{aligned} s_{h}\le \frac{1}{m^{m^{2}}}\overset{m^{m^{2}}}{\underset{i=1}{\sum }} s_{f}\circ \gamma _{i}+\frac{2}{m}. \end{aligned}$$

Hence $\beta \left( h\right) \le \beta \left( f\right)$ for every symmetric gauge norm $\beta$ on $L^{\infty }\left( \delta _{\infty }\right) .$

Proof

If $f\in L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$, then the map $t\mapsto KF_{t}\left( f\right)$ is continuous on (0, 1]. Hence we have $KF_{t}\left( h\right) \le KF_{t}\left( f\right)$ for all $0<t\le 1.$ We know that $KF_{t}\left( f\right) =KF_{t}\left( s_{f}\right)$ and $\beta \left( f\right) =\beta \left( s_{f}\right)$ for every $f\in L^{\infty }\left( \delta _{\infty }\right) .$ We may assume that f, h are nonincreasing, and we let u, w be the step functions defined in the proof of Lemma 13. Then u, w satisfy $f\le u\le f+\frac{1}{m}$and $\max \left( h-\frac{1}{m},0\right) \le w\le h$. Recall that

$$\begin{aligned} u&=\left( 1-\frac{i}{m}\right) \chi _{[p_{i},p_{i+1})}\left( x\right) \text { for }i=0,...,m-1.\\ w&=\left( 1-\frac{i+1}{m}\right) \chi _{[q_{i},q_{i+1})}\left( x\right) \text { for }i=0,...,m-1. \end{aligned}$$

and it is easy to see that

$$\begin{aligned} \int _{0}^{t}fd\delta _{\infty }+\frac{t}{m}\ge \int _{0}^{t}ud\delta _{\infty } \ge \int _{0}^{t}wd\delta _{\infty }\ge \int _{0}^{t}hd\delta _{\infty }-\frac{t}{m}\text {,} \end{aligned}$$

for all $0\le t\le 1.$

By Lemma 14, for each $m\in \mathbb {N},$ there exists a measure preserving map $\varphi _{m}:\left[ 0,1\right] \rightarrow \left[ 0,1\right]$ such that

$$\begin{aligned} \left\| \frac{1}{m}\underset{j=1}{\overset{m}{\sum }}u\circ \varphi _{m}^{\left( j\right) }-\int _{0}^{1}ud\delta _{\infty }\right\| _{\infty }\le \eta \left\| u\right\| _{\infty }\frac{4}{m}. \end{aligned}$$

where $\eta =card\left( Ran\left( u\right) \right)$

Let $l(t)=\frac{1}{t}\int _{0}^{t}ud\delta _{\infty },$ then $l:[0,1]\longrightarrow [0,\infty )$ is a continuous function. There are 2 cases to consider:

• Case 1: If $l(1)=\int _{0}^{1}ud\delta _{\infty }\ge b_{1}=\max \left\{ w\left( t\right) :0<t\le 1\right\}$, then by Lemma 14, for $\mathcal {8}k=m^{2}\in \mathbb {N},$ there exists $\varphi _{k}\in \mathbb {MP[}0,1\mathbb {]}$ such that

  $$\begin{aligned} \left\| \frac{u\circ \varphi _{k}^{\left( 1\right) }+\cdots +u\circ \varphi _{k}^{\left( m^{2}\right) }}{m^{2}}-\int _{0}^{1}ud\delta _{\infty }\right\| _{\infty }\le \frac{4\eta \left\| u\right\| _{\infty }}{m^{2} }\le \frac{4}{m}, \end{aligned}$$

  where $\eta =\textrm{card}\left( u\right) \le m.$Denote $\varphi _{k}^{\left( i\right) }$ by $\gamma _{j}$ Then we have

  $$\begin{aligned} \frac{1}{m^{2}}\underset{j=1}{\overset{m^{2}}{\sum }}u\circ \gamma _{j}\ge w-\frac{6}{m}. \end{aligned}$$

  Therefore $\frac{1}{m^{2}}\underset{j=1}{\overset{m^{2}}{\sum }}f\circ \varphi _{\left( j\right) }+\frac{1}{m}\ge h-\frac{4}{m}$ follows from Lemma 13. That is

  $$\begin{aligned} \frac{1}{m^{2}}\underset{j=1}{\overset{m^{2}}{\sum }}f\circ \varphi _{\left( j\right) }\ge h-\frac{3}{m}. \end{aligned}$$

  We can view it as

  $$\begin{aligned} \frac{1}{m^{2m}}\underset{j=1}{\overset{m^{2m}}{\sum }}f\circ \varphi _{\left( j\right) }\ge h-\frac{3}{m}, \end{aligned}$$

  where $\varphi _{\left( i+m^{2}t\right) }=\varphi _{\left( i\right) }$ for $1\le i\le m^{2}$ and $0\le t\le m^{2\,m-2}-1.$

• Case 2: $l(1)=\int _{0}^{1}ud\delta _{\infty }<b_{1}.$ Then there must exist $p_{1}^{^{\prime }}\in \left( 0,1\right) ,$ so that

  $$\begin{aligned} l(p_{1}^{^{\prime }})=\frac{1}{p_{1}^{^{\prime }}}\int _{0}^{p_{1}^{^{\prime }} }ud\delta _{\infty }=b_{1}. \end{aligned}$$

Define $u^{\left( 1\right) }$ in the following way

$$\begin{aligned} u^{\left( 1\right) }\left( x\right) =\left\{ \begin{array}{l} b_{1}\\ u(x) \end{array} \right. \begin{array}{l} 0\le x\le p_{1}^{\prime }\\ p_{1}^{\prime }<x\le 1. \end{array} \end{aligned}$$

Then for every $t>p_{1}^{^{\prime }},$

$$\begin{aligned}{} & {} l\left( t\right) =\int _{0}^{t}ud\delta _{\infty }\geqslant \int _{0}^{t} wd\delta _{\infty }\implies \int _{0}^{p_{1}^{\prime }}ud\delta _{\infty } +\int _{p_{1}^{\prime }}^{t}ud\delta _{\infty }\geqslant \int _{0}^{p_{1}^{\prime } }wd\delta _{\infty }\\{} & {} \quad +\int _{p_{1}^{\prime }}^{t}wd\delta _{\infty }. \end{aligned}$$

Thus we have $b_{1}p_{1}^{\prime }+\int _{p_{1}^{\prime }}^{t}u^{\left( 1\right) }d\delta _{\infty }\geqslant b_{1}p_{1}^{\prime }+\int _{p_{1}^{\prime } }^{t}wd\delta _{\infty }$; therefore

$$\begin{aligned} \int _{q_{1}}^{t}u^{\left( 1\right) }d\delta _{\infty }\geqslant \int _{q_{1} }^{t}wd\delta _{\infty }. \end{aligned}$$

Therefore, for every $0<t\le 1,$ we have

$$\begin{aligned}{} & {} u-\frac{1}{m}\le u^{\left( 1\right) }\le u, \\{} & {} KF_{t}\left( u^{\left( 1\right) }\right) \ge KF_{t}\left( w\right) , \end{aligned}$$

and for every $t\le t_{1},\left\| u^{\left( 1\right) }\right\| _{t}=b_{1}=\left\| h\right\| _{t}$

By Lemma 14 again, for $k=m^{2}\in \mathbb {N},$ there exist $\varphi _{\left( 1\right) },\dots ,\varphi _{\left( m^{2}\right) }:[0,1]\longrightarrow [0,1]$ such that

$$\begin{aligned} \left\| \frac{1}{m^{2}}\overset{m^{2}}{\underset{i=1}{\Sigma }}u\circ \varphi _{\left( i\right) }-\int _{0}^{1}ud\delta _{\infty }\right\| _{\infty }\le \eta \left\| u\right\| _{\infty }\frac{4}{m^{2}}\le \frac{4}{m}. \end{aligned}$$

Let $\varphi _{\left( r\right) }^{\left( 1\right) }(t)=\left\{ \begin{array}{l} \varphi _{\left( r\right) }(t)\\ t \end{array} \right. \begin{array}{l} t\le q_{1}\\ t>q_{1} \end{array}$, $r=1,\dots ,m^{2}.$ Then $\varphi _{\left( r\right) }^{\left( 1\right) }\in \mathbb{M}\mathbb{P}[0,1]$ for all $1\le r\le m^{2}$ and

$$\begin{aligned} \left\| \frac{1}{m^{2}}\overset{m^{2}}{\underset{r=1}{\Sigma }}u\circ \varphi _{\left( r\right) }^{\left( 1\right) }-u^{\left( 1\right) }\right\| _{\infty }\le \frac{2}{m}. \end{aligned}$$

That is $u^{\left( 1\right) }\approx \frac{1}{m^{2}}\overset{m^{2}}{\underset{r=1}{\Sigma }}u\circ \varphi _{\left( r\right) }^{\left( 1\right) }$ and Ran$(u^{\left( 1\right) })\subseteq \{b_{1},a_{2},...,a_{m}\}.$

If $\frac{1}{q_{1}}\int _{q_{1}}^{1}u^{\left( 1\right) }d\delta _{\infty }\geqslant b_{2},$ go to case 1.

If $\frac{1}{q_{1}}\int _{q_{1}}^{1}u^{\left( 1\right) }d\delta _{\infty }<b_{2},$ do the similar process as case 2 above, we have $u^{\left( 2\right) }$ and

$$\begin{aligned} \left\| \frac{1}{m^{2}}\overset{m^{2}}{\underset{i=1}{\Sigma }}u^{\left( 1\right) }\circ \varphi _{i}^{\left( 2\right) }-u^{\left( 2\right) }\right\| _{\infty }\le \frac{2}{m}\text {.} \end{aligned}$$

That is

$$\begin{aligned} u^{\left( 2\right) }&\approx \frac{1}{m^{2}}\overset{m^{2}}{\underset{i=1}{\Sigma }}u^{\left( 1\right) }\circ \varphi _{i}^{\left( 2\right) }\\&=\frac{1}{m^{2}}\overset{m^{2}}{\underset{i_{1}=1}{\Sigma }}\left( \overset{m^{2}}{\underset{i_{2}=1}{\Sigma }}\frac{1}{m^{2}}\left( u\circ \varphi _{i_{1}}^{\left( 1\right) }\right) \right) \circ \varphi _{i_{2}}^{\left( 2\right) }\\&=\frac{1}{m^{4}}\overset{m^{2}}{\underset{i=1}{\Sigma }}\overset{m^{2} }{\underset{j=1}{\Sigma }}(u\circ \varphi _{i_{1}}^{\left( 1\right) } \circ \varphi _{i_{2}}^{\left( 2\right) }). \end{aligned}$$

and Ran$\left( \left( u\right) ^{\left( 2\right) }\right) \subseteq \{b_{1},b_{2},a_{3},...,a_{m}\}.$

Finally, after r steps (at most m), we will have

$$\begin{aligned} u^{\left( r\right) }\approx \frac{1}{m^{2r}}\overset{m^{2}}{\underset{i_{1}=1}{\Sigma }}\cdots \overset{m^{2}}{\underset{i_{r}=1}{\Sigma }} (u\circ \varphi _{i_{1}}^{\left( 1\right) }\circ \varphi _{i_{2}}^{\left( 2\right) }\cdots \circ \varphi _{i_{r}}^{(r)}), \end{aligned}$$

and thus $u^{\left( r\right) }\geqslant w.$

Since $m^{2r}|m^{m^{2}},$ as in case 1, we can view this as

$$\begin{aligned} \frac{1}{m^{m^{2}}}\underset{j=1}{\overset{m^{m^{2}}}{\sum }}u\circ \varphi _{\left( j\right) }\ge w-\frac{2}{m}. \end{aligned}$$

In conclusion, for every m,  there is an integer $N=m^{m^{2}}$, and there are $\gamma _{1},\dots \gamma _{N}\in \mathbb{M}\mathbb{P}\left( J_{\infty },\delta _{\infty }\right)$ such that

$$\begin{aligned} \frac{1}{N}\overset{N}{\underset{i=1}{\sum }}u\circ \gamma _{i}\ge w-\frac{2m}{N}. \end{aligned}$$

By Lemma 13, we know that $f\ge u-\frac{1}{m}$ and $h\le w+\frac{1}{m}$

Thus, $\frac{1}{N}\overset{N}{\underset{i=1}{\sum }}s_{f}\circ \gamma _{i} +\frac{2\,m}{N}+\frac{1}{m}\ge s_{h}$.

Therefore, $\beta \left( f\right) \ge \beta \left( h\right)$ as $m\rightarrow \infty .$ $\square$

Corollary 4

If $\beta$ is a symmetric gauge norm on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right)$, then $\alpha _{\beta }$ is a norm on $\mathcal {M}$.

Proof

We need only prove the triangle inequality. If $A,B\in \mathcal {M}$, we define $h\left( t\right) =s_{A+B}\left( t\right)$ and $f\left( t\right) =s_{A}\left( t\right) +s_{B}\left( t\right)$. Then $KF_{t}\left( h\right) =KF_{t}\left( A+B\right)$ and $KF_{t}\left( f\right) =KF_{t}\left( A\right) +KF_{t}\left( B\right)$, so Lemma 2 applies, and we get

$$\begin{aligned}{} & {} \alpha _{\beta }\left( A+B\right) =\beta \left( h\right) \le \beta \left( f\right) =\beta \left( s_{A}\left( t\right) +s_{B}\left( t\right) \right) \\{} & {} \quad \le \beta \left( s_{A}\left( t\right) \right) +\beta \left( s_{B}\left( t\right) \right) =\alpha _{\beta }\left( A\right) +\alpha _{\beta }\left( B\right) \text {.} \end{aligned}$$

$\square$

Since it is easily seen that $\alpha =\alpha _{\beta _{\alpha }}$ and $\beta =\beta _{\alpha _{\beta }}$, we obtain the characterization [10] of the unitarily invariant norms on a $II_{1}$ factor von Neumann algebra.

Theorem 3

Let $\mathcal {M}$ be a type $II_{1}$ factor von Neumann algebra, then there is a one-to-one correspondence between unitarily invariant norms on $\mathcal {M}$ and symmetric gauge norms on $L^{\infty }\left( J_{\infty },\delta _{\infty }\right) .$

2.2 Approximate unitary equivalence

The following is a consequence of a result of Hadwin and Ding [4]. Suppose $\mathcal {R}$ is a von Neumann algebra and $T\in \mathcal {R}$. $\mathcal {Z}\left( \mathcal {R}\right) =\mathcal {R\cap R}^{\prime }$ is the center. In [12] the $\mathcal {R}$-rank of T was defined to be the Murray-von Neumann equivalence class of the projection $P_{T}$ onto the closure of the range of T. We let $\left( SOT\right)$ and $\left( WOT\right)$ denote, respectively, the strong and weak operator topologies. Note that

$$\begin{aligned} P_{T}=\lim _{n\rightarrow \infty }\left( TT^{*}\right) ^{1/n}{ }\left( SOT\right) , \end{aligned}$$

so $P_{T}\in \mathcal {M}$.

The following theorem can be found in [13]. The center-valued trace $\Phi$ is described in Sect. 2.3.6.

Theorem 4

Suppose $\mathcal {R}$ is a finite von Neumann algebra acting on a separable Hilbert space H. Let $\Phi :\mathcal {R}\rightarrow \mathcal {Z}\left( \mathcal {R}\right)$ be the unique center-valued trace on $\mathcal {R}$. Suppose $\mathcal {A}$ is a unital commutative C* algebra and $\pi ,\rho :\mathcal {A}\rightarrow \mathcal {R}$ are unital $*$-homomorphisms. The following are equivalent:

(1):

There is a net $\left\{ U_{j}\right\}$ of unitary operators in $\mathcal {R}$ such that, for every $a\in \mathcal {A}$,

$$\begin{aligned} \left\| U_{j}^{*}\pi \left( a\right) U_{j}-\rho \left( a\right) \right\| \rightarrow 0. \end{aligned}$$

(2):

$\Phi \circ \pi =\Phi \circ \rho$.

2.3 The central decomposition

We refer the reader to [16] for the theory of direct integrals and the central decomposition of a von Neumann algebra acting on a separable Hilbert space. Since we are only interested in the von Neumann algebra $\mathcal {R}$ and not how it acts on a Hilbert space, we can ignore multiplicities when using the central decomposition [16]. Suppose $\mathcal {R}$ is a finite von Neumann algebra acting on a separable Hilbert space. Then we can write

$$\begin{aligned} \mathcal {R=}{ }\left[ \mathcal {R}_{1}\oplus \mathcal {R}_{2}\oplus \cdots \right] \oplus \mathcal {R}_{\infty }, \end{aligned}$$

where $\mathcal {R}_{k}$ is of type $I_{k}$ for $1\le k<\infty$ and $\mathcal {R}_{\infty }$ is a type $II_{1}$ von Neumann algebra.

2.3.1 Measurable families

Suppose $\mathcal {M}$ is a type $II_{1}$ von Neumann algebra with a faithful tracial state acting on a separable Hilbert space $H\mathcal {=}l_{\infty }^{2}$. We will associate with $\mathcal {M}$ a probability space $(\Omega ,\mu )$ and a unitary operator $U:H\longrightarrow L^{2}(\mu ,H)$ that transforms $\mathcal {M}$ into a certain von Neumann algebra of operators on $L^{2}\left( \mu ,H\right)$ that will be described next.

We let $\mathcal {M}^{\prime }$ denote the commutant of $\mathcal {M}$, i.e., the set of all operators that commute with every operator in $\mathcal {M}$.

For each $\omega \in \Omega$, there is a type $II_{1}$ von Neumann algebra $\mathcal {M}_{\omega }$ in $B\left( H\right)$ that is determined by two sequences of SOT (strong operator topology) measurable operator-valued functions $f_{n}$ and $g_{n}$ from $\Omega$ into the unit ball of $B\left( H\right)$ so that $\mathcal {M}_{\omega }$ is generated by the set $\left\{ f_{n}\left( \omega \right) :n\in \mathbb {N}\right\} ,$ $\mathcal {M}_{\omega }^{^{\prime }}$ is generated by the set $\left\{ g_{n}\left( \omega \right) :n\in \mathbb {N}\right\}$, and each of those sets is SOT dense in the unit ball of the von Neumann algebra it generates. Suppose $\varphi :\Omega \rightarrow B\left( H\right)$ is an SOT-measurable function, and define $\left| \varphi \right| =\left\| \cdot \right\| \circ \varphi$, that is $\left| \varphi \right| \left( \omega \right) =\left\| \varphi \left( \omega \right) \right\|$ for $\omega \in \Omega$. If $\left| \varphi \right| \in L^{\infty }\left( \mu \right) ,$ then let $\left\| \varphi \right\| _{\infty }=\left\| \left| \varphi \right| \right\| _{\infty }$. We will assume that $\left( \Omega ,\mu \right)$, U, and the $f_{n},g_{n},\mathcal {M}_{\omega }$ have been chosen so that

$$\begin{aligned}{} & {} U^{*}\mathcal {M}U= \\{} & {} \quad \left\{ \varphi :\Omega \longrightarrow B(H)|{ }\varphi \text { is SOT-measurable, }\varphi \left( \omega \right) \in \mathcal {M}_{\omega }\text { a.e.}\left( \mu \right) ,{ }\left| \varphi \right| \in L^{\infty }\left( \mu \right) \right\} \text {.} \end{aligned}$$

As usual, $\varphi _{1}=\varphi _{2}$ will mean $\varphi _{1}\left( \omega \right) =\varphi _{2}\left( \omega \right)$ a.e. $(\mu )$, and each $\varphi$ in $U^{*}\mathcal {M}U$ is viewed as the operator on $L^{2}\left( \mu ,H\right)$ defined for $f\in L^{2}\left( \mu ,H\right)$ by

$$\begin{aligned} \left( \varphi f\right) \left( \omega \right) =\varphi \left( \omega \right) f\left( \omega \right) . \end{aligned}$$

2.3.2 Measurable cross-sections

Definition 4

Suppose (X, d) is a metric space and $\mu :\textrm{Bor}(X)\longrightarrow [0,\infty )$ is a finite measure. A subset B of X is called $\mu$-measurable if there are $A,F\in \textrm{Bor}(X)$ such that $B\backslash A\subset F$ and $\mu \left( F\right) =0$. The $\sigma$-algebra of all $\mu$-measurable sets is denoted by $\mathcal {M}_{\mu }$. A subset D of X is absolutely measurable if D is $\mu$-measurable for every finite measure $\mu$ on Bor(X). The $\sigma$-algebra of all absolutely measurable subsets of X is denoted by $\mathbb{A}\mathbb{M}\left( X\right)$. Clearly we have

$$\begin{aligned} \mathbb{A}\mathbb{M}\left( X\right) =\bigcap \left\{ \mathcal {M}_{\mu }:\mu \text { is a finite Borel measure on }X\right\} . \end{aligned}$$

It is obvious that each $\mathcal {M}_{\mu }$ contains $\textrm{Bor}(X)$, so $\textrm{Bor}(X)\subset \mathbb{A}\mathbb{M}\left( X\right)$. However, it is often the case that $\textrm{Bor}(X)\ne \mathbb{A}\mathbb{M}\left( X\right)$. If Y is another metric space, we say that a function $f:X\rightarrow Y$ is absolutely measurable if f is $\mathbb{A}\mathbb{M}\left( X\right)$-$\textrm{Bor}(Y)$ measurable, i.e., for every Borel set $E\subseteq Y,$ $f^{-1}(E)\in \mathbb{A}\mathbb{M}\left( X\right)$. Recall that a finite measure space $\left( \Lambda ,\Sigma ,\lambda \right)$ is complete if, $E\in \Sigma$ whenever $E\subset F,F\in \Sigma$ and $\lambda \left( F\right) =0$, i.e., all subsets of sets of measure 0 are in $\Sigma$. Note that statement $\left( 4\right)$ in Lemma 15 shows how, in the presence of a complete measure space, absolute measurability turns into measurability.

Lemma 15

Suppose X, Y and Z are metric spaces and $f:X\longrightarrow Y$, and $g:Y\rightarrow Z$. Then

(1):

f is absolutely measurable if and only if f is $\mathbb{A}\mathbb{M} \mathcal {(}X\mathcal {)}$-$\mathbb{A}\mathbb{M}\left( Y\right)$ measurable.

(2):

If f and g are absolutely measurable, then $g\circ f:X\rightarrow Z$ is absolutely measurable.

(3):

For every Borel set $E\subseteq Y,$ $f^{-1}(E)$ is absolutely measurable.

(4):

If $\left( \Lambda ,\Sigma ,\lambda \right)$ is a complete finite measure space and $\varphi :\Lambda \rightarrow X$ is Borel measurable, then

(a):

$\varphi$ is $\Sigma$-$\mathbb{A}\mathbb{M}\left( X\right)$ measurable, and,

(b):

If f is absolutely measurable, then $f\circ \varphi :X\rightarrow Y$ is measurable.

Definition 5

If $f:X\rightarrow Y$ and $g:f(X)\longrightarrow X$ satisfy, for every $y\in f(X),$

$$\begin{aligned} f\left( g(y)\right) =y, \end{aligned}$$

then g is called a cross-section for f.

The following Theorem is from Theorem 3.4.3 in [1] and is the key to dealing with direct integrals.

Theorem 5

Suppose X is a Borel subset of a complete separable metric space, and Y is a separable metric space. If $f:X\longrightarrow Y$ is a continuous function, then

(1):

f(X) is an absolutely measurable subset of Y,  and

(2):

f has an absolutely measurable cross-section $g:f(X)\longrightarrow X$.

Here is a simple result proved using measurable cross-section.

Lemma 16

Suppose n is a positive integer and $\mathbb {M}_{n}\left( \mathbb {C} \right) ^{+}$ is the set of $n\times n$ matrices A such that $A\ge 0$. Let $\mathcal {U}_{n}$ be the set of unitary $n\times n$ matrices and let $\mathcal {D}_{n}$ be the set of all diagonal $n\times n$ matrices in $\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$ of the form $\textrm{diag} \left( s_{\frac{1}{n}},\ldots s_{1}\right)$ with $s_{\frac{1}{n}}\ge s_{\frac{2}{n}}\ge \cdots \ge s_{1}\ge 0.$ Then there is an absolutely measurable function $u:\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+} \rightarrow \mathcal {U}_{n}$ such that, for every $A\in \mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$,

$$\begin{aligned} u\left( A\right) ^{*}Au\left( A\right) \in \mathcal {D}_{n}, \end{aligned}$$

i.e.,

$$\begin{aligned} u\left( A\right) ^{*}Au\left( A\right) =\left( \begin{array}{cccc} s_{A}\left( \frac{1}{n}\right) &{} &{} &{} \\ &{} s_{A}\left( \frac{2}{n}\right) &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{} s_{A}\left( \frac{n}{n}\right) \end{array} \right) \text {.} \end{aligned}$$

Hence, for every $T\in \mathbb {M}_{n}\left( \mathbb {C}\right)$,

$$\begin{aligned} u\left( \left| T\right| \right) ^{*}\left| T\right| u\left( \left| T\right| \right) =\left( \begin{array}{cccc} s_{T}\left( \frac{1}{n}\right) &{} &{} &{} \\ &{} s_{T}\left( \frac{2}{n}\right) &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{} s_{T}\left( \frac{n}{n}\right) \end{array} \right) \text {.} \end{aligned}$$

Proof

Let $X=$

$\left\{ (A,U_{A}):A\in \mathbb {M}_{n}\left( \mathbb {C} \right) ^{+},U_{A}\in \mathcal {U}_{n},U_{A}^{*}AU_{A}=\textrm{diag}\left( s_{A}\left( \frac{1}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) \right) \right\}$, which is a subset of $\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}\times \mathcal {U}_{n}.$ For every $\mathcal {(} A_{\lambda },U_{A_{\lambda }}\mathcal {)}\in X,$ and $\mathcal {(}A_{\lambda },U_{A_{\lambda }}\mathcal {)}\longrightarrow (A,U_{A})$, we have $A_{\lambda }\longrightarrow A,U_{A_{\lambda }}\longrightarrow U_{A},$ Thus

$$\begin{aligned} \left\| U_{A}^{*}AU_{A}-U_{A_{\lambda }}A_{\lambda }U_{A_{\lambda } }\right\| \rightarrow 0, \end{aligned}$$

We also know that $\frac{1}{i}\underset{j=1}{\overset{i}{\sum }}s_{A}\left( \frac{j}{n}\right) =KF_{i}\left( A\right)$ for all $1\le i\le n$ and $s_{A}\left( \frac{1}{n}\right) =KF_{1}\left( A\right) \le \left\| A\right\| .$ We can get $s_{A_{\lambda }}\left( \frac{i}{n}\right) \overset{\left\| \cdot \right\| }{\rightarrow }s_{A}\left( \frac{i}{n}\right)$ for all $1\le i\le n$. Thus

$$\begin{aligned} U_{A_{\lambda }}^{*}A_{\lambda }U_{A_{\lambda }}=\textrm{diag}\left( s_{A_{\lambda }}\left( \frac{1}{n}\right) ,\dots ,s_{A_{\lambda }}\left( \frac{n}{n}\right) \right) \overset{\left\| \cdot \right\| }{\longrightarrow }\textrm{diag}\left( s_{A}\left( \frac{1}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) \right) \end{aligned}$$

Therefore $U_{A}^{*}AU_{A}=\textrm{diag}\left( s_{A}\left( \frac{1}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) \right) ,$ and X is a closed subset of a $\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+} \times \mathcal {U}_{n}$, which is a complete separable metric space.

Define $\pi _{1}:$ $X\longrightarrow \mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$ and $\pi _{2}:X\longrightarrow \mathcal {U}_{n}$ by

$$\begin{aligned} \pi _{1}(A,U)=A,\text { }\pi _{2}(A,U)=U. \end{aligned}$$

It is easy to see that $\pi _{1}(X)=\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$.

Since we know for every $A\in \mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$, there exists a unitary $U_{A}$ such that

$$\begin{aligned} U_{A}^{*}AU_{A}=\textrm{diag}\left( s_{A}\left( \frac{1}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) \right) . \end{aligned}$$

Thus by Theorem 5, there exists an absolutely measurable function $g:\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}\longrightarrow X$ such that $\pi _{1}\circ g=id$ on $\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$, for every $A\in \mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$, $g(A)=(A,U_{A}).$ Then we define $u=\pi _{2}\circ g:\mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}\longrightarrow \mathcal {U}_{n},$ it is absolutely measurable.

Therefore, for every $A\in \mathbb {M}_{n}\left( \mathbb {C}\right) ^{+}$,

$$\begin{aligned} u\left( A\right) =U_{A}\text { and }u\left( A\right) ^{*}Au\left( A\right) =\textrm{diag}\left( s_{A}\left( \frac{1}{n}\right) ,\dots ,s_{A}\left( \frac{n}{n}\right) \right) \in \mathcal {D}_{n}. \end{aligned}$$

Hence, for every $T\in \mathbb {M}_{n}\left( \mathbb {C}\right)$,

$$\begin{aligned} u\left( \left| T\right| \right) ^{*}\left| T\right| u\left( \left| T\right| \right) =\left( \begin{array}{cccc} s_{T}\left( \frac{1}{n}\right) &{} &{} &{} \\ &{} s_{T}\left( \frac{2}{n}\right) &{} &{} \\ &{} &{} \ddots &{} \\ &{} &{} &{} s_{T}\left( \frac{n}{n}\right) \end{array} \right) . \end{aligned}$$

$\square$

2.3.3 Direct integrals

Suppose $\Omega \subseteq \mathbb {R}$ is compact, $\mu$ is a probability Borel measure, H is a separable Hilbert space. Define $\int _{\Omega }^{\oplus } Hd\mu =L^{2}\left( \mu ,H\right)$ to be the set of all measurable functions $f:\Omega \rightarrow H$ such that

$$\begin{aligned} \left\| f\right\| _{2}^{2}\underset{\textrm{def}}{=}\int _{\Omega }\left\| f\left( \omega \right) \right\| ^{2}d\mu \left( \omega \right) <\infty \text {. } \end{aligned}$$

We define an inner product $\left\langle \cdot ,\cdot \right\rangle$ on $L^{2}\left( \mu ,H\right)$ by

$$\begin{aligned} \left\langle f,h\right\rangle =\int _{\Omega }\left\langle f\left( \omega \right) ,h\left( \omega \right) \right\rangle d\mu \left( \omega \right) . \end{aligned}$$

In this way $L^{2}\left( \mu ,H\right)$ is a Hilbert space.

We define $L^{\infty }\left( \mu ,B\left( H\right) \right)$ to be the set of all bounded functions $\varphi :\Omega \rightarrow B\left( H\right)$ that are measurable with respect to the weak operator topology (WOT) on $B\left( H\right)$. Although the weak operator topology, strong operator topology (SOT) and $*$-strong operator topology ($*$-SOT) on $B\left( H\right)$ are different, the Borel sets with respect to these topologies are all the same. Suppose the map $\omega \mapsto T_{\omega }$ is in $L^{\infty }\left( \mu ,B\left( H\right) \right)$. We define an operator $T=\int _{\Omega }^{\oplus }T_{\omega }d\mu \left( \omega \right)$ by

$$\begin{aligned} \left( Tf\right) \left( \omega \right) =T_{\omega }\left( f\left( \omega \right) \right) \text {.} \end{aligned}$$

If $\varphi \in L^{\infty }\left( \mu ,B\left( H\right) \right)$ and $T_{\omega }=\varphi \left( \omega \right)$ for $\omega \in \Omega$, we also use the notation $M_{\varphi }$ to denote $\int _{\Omega }^{\oplus }T_{\omega } d\mu \left( \omega \right)$. In this way we can view $L^{\infty }\left( \mu ,B\left( H\right) \right) \subseteq B\left( L^{2}\left( \mu ,H\right) \right) ,$ and we can write $L^{\infty }\left( \mu ,B\left( H\right) \right) =\int _{\Omega }^{\oplus }B\left( H\right) d\mu \left( \omega \right) .$

We have that $L^{\infty }\left( \mu \right)$ can be viewed as the subalgebra $\mathcal {D}$ of $L^{\infty }\left( \mu ,B\left( H\right) \right)$ of all functions $\varphi$ such that $\varphi \left( \omega \right) \in \mathbb {C}\cdot 1$ a.e. $\left( \mu \right)$, that is, by identifying $h\in L^{\infty }\left( \mu \right)$ with the function $\omega \mapsto h\left( \omega \right) 1$. We denote $\mathcal {D}$ by

$$\begin{aligned} \mathcal {D=}\int _{\Omega }^{\oplus } \mathbb {C} \cdot 1d\mu \left( \omega \right) \text {.} \end{aligned}$$

We have $\mathcal {D}^{\prime }=L^{\infty }\left( \mu ,B\left( H\right) \right)$ and $L^{\infty }\left( \mu ,B\left( H\right) \right) ^{\prime }=\mathcal {D},$ therefore $\mathcal {D=}$

$\mathcal {Z}\left( L^{\infty }\left( \mu ,B\left( H\right) \right) \right)$.

Suppose, for each $\omega \in \Omega$, $\mathcal {R}_{\omega }\subset B\left( H\right)$ is a von Neumann algebra. We say that the family $\left\{ \mathcal {R}_{\omega }\right\} _{\omega \in \Omega }$ is a measurable family if there is a countable set $\left\{ \varphi _{1},\varphi _{2},\ldots \right\} \subset L^{\infty }\left( \mu ,B\left( H\right) \right)$ such that

$$\begin{aligned} {\text {ball}}\left( \mathcal {R}_{\omega }\right) =\left\{ \varphi _{1}\left( \omega \right) ,\varphi _{2}\left( \omega \right) ,\ldots \right\} ^{-SOT}\text { a.e. }\left( \mu \right) . \end{aligned}$$

It is known that if $\left\{ \mathcal {R}_{\omega }\right\} _{\omega \in \Omega }$ is a measurable family, then so is $\left\{ \mathcal {R}_{\omega }^{\prime }\right\} _{\omega \in \Omega }$. Moreover, if $\left\{ \mathcal {R}_{\omega }^{\prime }\right\} _{\omega \in \Omega }$ is a measurable family, then there is a sequence $\left\{ \psi _{1},\psi _{2},\ldots \right\} \subset L^{\infty }\left( \mu ,B\left( H\right) \right)$ such that

$$\begin{aligned} {\text {ball}}\left( \mathcal {R}_{\omega }^{\prime }\right) =\left\{ \psi _{1}\left( \omega \right) ,\psi _{2}\left( \omega \right) ,\ldots \right\} ^{-SOT}\text { a.e. }\left( \mu \right) . \end{aligned}$$

If $\left\{ \mathcal {R}_{\omega }\right\} _{\omega \in \Omega }$ is a measurable family, then we define the direct integral $\int _{\Omega }^{\oplus }\mathcal {R}_{\omega }d\mu \left( \omega \right)$ to be the set of all $T=\int _{\Omega }^{\oplus }T_{\omega }d\mu \left( \omega \right) \in L^{\infty }\left( \mu ,B\left( H\right) \right)$ such that

$$\begin{aligned} T_{\omega }\in \mathcal {R}_{\omega }\text { a.e. }\left( \mu \right) \text {.} \end{aligned}$$

It is known [16] that a von Neumann algebra $\mathcal {R}\subset B\left( L^{2}\left( \mu ,H\right) \right)$ can be written as

$$\begin{aligned} \mathcal {R}=\int _{\Omega }^{\oplus }\mathcal {R}_{\omega }d\mu \left( \omega \right) \end{aligned}$$

for a measurable family $\left\{ \mathcal {R}_{\omega }\right\} _{\omega \in \Omega }$ if and only if

$$\begin{aligned} \mathcal {D}=\int _{\Omega }^{\oplus }\mathbb {C}\cdot 1d\mu \left( \omega \right) \subset \mathcal {R}\subset \int _{\Omega }^{\oplus }B\left( H\right) d\mu \left( \omega \right) =\mathcal {D}^{\prime }, \end{aligned}$$

equivalently,

$$\begin{aligned} \mathcal {D\subset Z}\left( \mathcal {R}\right) \text {.} \end{aligned}$$

In particular, since $\mathcal {Z}\left( \mathcal {R}\right) =\mathcal {Z} \left( \mathcal {R}^{\prime }\right) =\mathcal {R}\cap \mathcal {R}^{\prime }$ for every von Neumann algebra $\mathcal {R}$, we see that $\mathcal {R}$ can be decomposed as a direct integral if and only if $\mathcal {R}^{\prime }$ can be decomposed as a direct integral.

Suppose $1\le n\le \infty =\aleph _{0}$. We define $\ell _{n}^{2}$ to be the space of square summable sequences with the inner product $\left\langle x,y\right\rangle ={{\sum _{i=1}^{n}}}x_{i}y_{i}$, where $x,y\in H$ and H is a Hilbert space with dimension n.

Lemma 17

Suppose $\mathcal {A}$ is an abelian von Neumann algebra on a separable Hilbert space H. Then there are compact subsets $\Omega _{n} \subset \mathbb {R}$ for $1\le n\le \infty$ and a Borel measure $\mu _{n}$ on $\Omega _{n}$ such that $\mu _{n}\left( \Omega _{n}\right) \in \left\{ 0,1\right\}$ and $\mathcal {A}$ is unitarily equivalent to $\sum _{1\le n\le \infty }^{\oplus }L^{\infty }\left( \mu _{n},\mathbb {C}\cdot 1\right)$ acting on $\sum _{1\le n\le \infty }^{\oplus }L^{2}\left( \mu _{n},\ell _{n} ^{2}\right)$.

Suppose $\mathcal {R}$ is a von Neumann algebra acting on a separable Hilbert space H. Then the center $\mathcal {Z}\left( \mathcal {R}\right)$ of $\mathcal {R}$ is an abelian von Neumann algebra on H. From Lemma 17 we can write

$$\begin{aligned} H=\sum _{1\le n\le \infty }^{\oplus }L^{2}\left( \mu _{n},\ell _{n}^{2}\right) \end{aligned}$$

and

$$\begin{aligned} \mathcal {Z}\left( \mathcal {R}\right) =\sum _{1\le n\le \infty }^{\oplus }L^{\infty }\left( \mu _{n},\mathbb {C}\cdot 1\right) . \end{aligned}$$

Since $\mathcal {R}$ commutes with $\mathcal {Z}\left( \mathcal {R}\right)$, we can write

$$\begin{aligned} \mathcal {R}=\sum _{1\le n\le \infty }^{\oplus }\mathcal {R}_{n}, \end{aligned}$$

where $\mathcal {R}_{n}\subset B\left( L^{2}\left( \mu _{n},\ell _{n} ^{2}\right) \right)$. It is clear, for $1\le n\le \infty$, that

$$\begin{aligned} \mathcal {Z}\left( \mathcal {R}_{n}\right) =L^{\infty }\left( \mu _{n},\mathbb {C}\cdot 1\right) , \end{aligned}$$

which implies

$$\begin{aligned} \mathcal {R}_{n}\subset \mathcal {Z}\left( \mathcal {R}_{n}\right) ^{\prime }=L^{\infty }\left( \mu _{n},\mathbb {C}\cdot 1\right) ^{\prime }=L^{\infty }\left( \mu _{n},B\left( \ell _{n}^{2}\right) \right) . \end{aligned}$$

Hence, for each n,  $1\le n\le \infty$, there is a measurable family $\left\{ \mathcal {R}_{n}\left( \omega \right) \right\} _{\omega \in \Omega _{n}}$ such that

$$\begin{aligned} \mathcal {R}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {R}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) . \end{aligned}$$

We therefore have

$$\begin{aligned} \mathcal {R}=\sum \nolimits _{1\le n\le \infty }^{\oplus }\int _{\Omega _{n} }^{\oplus }\mathcal {R}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) . \end{aligned}$$

This is called the central decomposition of $\mathcal {R}$.

The following Lemma is a well-known result [16].

Lemma 18

In the central decomposition of $\mathcal {R}$, almost every $\mathcal {R} _{n}\left( \omega \right)$ is a factor von Neumann algebra.

The following lemma can be obtained from [5, Chapter 3 of Part II]; we include a short proof.

Lemma 19

Suppose $\mathcal {A}_{n}$ is a masa of a $\mathcal {R} _{n}$ for $1\le n\le \infty ,$ then there is a measurable family $\left\{ \mathcal {A}_{n}\left( \omega \right) \right\} _{\omega \in \Omega _{n}}$ such that

$$\begin{aligned} \mathcal {A}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) , \end{aligned}$$

where $\mathcal {A}_{n}\left( \omega \right)$ is a masa in $\mathcal {R} _{n}\left( \omega \right) .$

Proof

Suppose

$$\begin{aligned} \mathcal {W}=\left( B\left( l_{n}^{2}\right) \right) \times A\times B\times C\times E\times \mathbb {N\times \mathbb {N}}, \end{aligned}$$

where $A=B=C=\underset{i=1}{\overset{\infty }{\Pi }}\textrm{ball}\left( B\left( l_{n}^{2}\right) \right)$ and $E=\left\{ x\in l_{n} ^{2}:\left\| x\right\| =1\right\}$. Then $\mathcal {W}$ is a complete separable metric space with product topology.

Define $\mathcal {X}_{m,k}$ to be the set of elements $\left( T,\left\{ A_{i}\right\} _{i=1}^{\infty },\left\{ B_{i}\right\} _{i=1}^{\infty },\left\{ C_{i}\right\} _{i=1}^{\infty },\right.$e, m, k) in $\mathcal {W}$ satisfying

$$\begin{aligned} TA_{i}=A_{i}T\text {, }TB_{i}=B_{i}T\text {, }\left\| \left( TC_{m} -C_{m}T\right) e\right\| \ge \frac{1}{k}\text {, for every }i\in \mathbb {N}\text {. } \end{aligned}$$

Then $\mathcal {X}_{m,k}$ is a closed subset of $\mathcal {W}$. We define $\mathcal {X}={{\cup _{m,k=1}^{\infty }}}\mathcal {X}_{m,k}$, then $\mathcal {X}$ is a Borel subset of $\mathcal {W}$.

Let $\pi _{2,3,4}:\mathcal {X\rightarrow }A\times B\times C$ be the projection map. Then $\pi _{2,3,4}\left( \mathcal {X}\right)$ consists of elements $\left( \left\{ A_{i}\right\} _{i=1}^{\infty },\left\{ B_{i}\right\} _{i=1}^{\infty },\left\{ C_{i}\right\} _{i=1}^{\infty }\right)$ so that there exists $T\in \textrm{ball}\left( B\left( l_{n}^{2}\right) \right)$ such that

$$\begin{aligned} T\in \left\{ A_{1},A_{2},\dots \right\} ^{\prime }\cap \left\{ B_{1},B_{2},\dots \right\} ^{\prime }\text { and }T\notin \left\{ C_{1},C_{2},\dots \right\} ^{\prime }. \end{aligned}$$

Suppose there are sequences $\left\{ f_{1},f_{2},\ldots \right\}$, $\left\{ \psi _{1},\psi _{2},\ldots \right\}$ and $\left\{ g_{1},g_{2},\ldots \right\}$ contained in $L^{\infty }\left( \mu _{n},B\left( l_{n}^{2}\right) \right)$ such that

$$\begin{aligned} \begin{array}{l} \textrm{ball}\mathcal {A}_{n}\left( \omega \right) =\left\{ f_{1}\left( \omega \right) ,{ }f_{2}\left( \omega \right) ,\dots \right\} ^{-SOT},\\ \textrm{ball}\mathcal {R}_{n}\left( \omega \right) ^{\prime }=\left\{ \psi _{1}\left( \omega \right) ,\psi _{2}\left( \omega \right) ,\dots \right\} ^{-SOT},\\ \textrm{ball}\mathcal {A}_{n}\left( \omega \right) ^{\prime }=\left\{ g_{1}\left( \omega \right) ,g_{2}\left( \omega \right) ,\dots \right\} ^{-SOT}. \end{array} \end{aligned}$$

By Theorem 5, we know there exists an absolutely measurable function

$\Upsilon :\pi _{2,3,4}\left( \mathcal {X}\right) \longrightarrow \mathcal {X}$ such that $\pi _{2,3,4}\mathcal {\circ }\Upsilon$ is the identity function on $\pi _{2,3,4}\left( \mathcal {X}\right) .$

Define $F:\Omega _{n}\rightarrow A\times B\times C$ by

$$\begin{aligned} F\left( \omega \right) =\left\{ f_{i}\left( \omega \right) \right\} _{i=1}^{\infty }\times \left\{ \psi _{i}\left( \omega \right) \right\} _{i=1}^{\infty }\times \left\{ g_{i}\left( \omega \right) \right\} _{i=1}^{\infty }. \end{aligned}$$

Let

$$\begin{aligned} \begin{array}{c} G=F^{-1}\left( \pi _{2,3,4}\left( \mathcal {X}\right) \right) =\\ \left\{ \omega :\text {there exists }T\in B\left( l_{n}^{2}\right) \text { with }T\left( \omega \right) \in \mathcal {A}_{n}\left( \omega \right) ^{\prime }\cap \mathcal {R}_{n}\left( \omega \right) ,{ }T\left( \omega \right) \notin \mathcal {A}_{n}\left( \omega \right) \right\} . \end{array} \end{aligned}$$

We know from Lemma 15 and the completeness of $\left( \Omega _{n},\mu _{n}\right)$ that G is measurable. We need to prove $\mu _{n}\left( G^{c}\right) =0.$ Suppose not, and let $\pi _{1}:\mathcal {X\rightarrow }B\left( l_{n}^{2}\right)$ be the projection map (into the first coordinate). Then, by Lemma 15, $\pi _{1}\circ \Upsilon \circ F|_{G}$ is a measurable function from G to $B\left( l_{n}^{2}\right)$. We define T by

$$\begin{aligned} T\left( \omega \right) =\left\{ \begin{array}{ll} \left( \pi _{1}\circ \Upsilon \circ F|_{G}\right) \left( \omega \right) &{} \text {if }\omega \in G\\ 0 &{} \text {if }\omega \notin G \end{array} \right. . \end{aligned}$$

Thus

$$\begin{aligned} T=\int _{G}^{\oplus }T\left( \omega \right) d\mu _{n}\left( \omega \right) \oplus \int _{\Omega _{n}\backslash G}^{\oplus }0d\mu _{n}\left( \omega \right) , \end{aligned}$$

then $T\in \mathcal {A}_{n}^{\prime }\cap \mathcal {R}_{n}$ and $T\notin \mathcal {A}_{n}$, which contradicts to the assumption that $\mathcal {A}_{n}$ is a masa. Therefore $\mu _{n}\left( G\right) =0$ and

$$\begin{aligned} \mathcal {A}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) , \end{aligned}$$

$\mathcal {A}_{n}\left( \omega \right)$ is a masa a.e.$\left( \mu _{n}\right)$. This completes the proof. $\square$

2.3.4 Multiplicities for type I$_{n}$ factors

A type I factor von Neumann algebra is isomorphic to $B\left( H\right)$ for some Hilbert space H. However, if m is a cardinal, we can let $H^{\left( m\right) }$ denote a direct sum of m copies of H and, for each $T\in B\left( H\right)$ write $T^{\left( m\right) }$ be a direct sum of m copies of T acting on $H^{\left( m\right) }$, and let $B\left( H\right) ^{\left( m\right) }=\left\{ T^{\left( m\right) }:T\in B\left( H\right) \right\}$. Clearly, $B\left( H\right) ^{\left( m\right) }$ is isomorphic to $B\left( H\right)$. The number m is called the multiplicity of the factor $B\left( H\right) ^{\left( m\right) }$ and it is the minimal rank of a nonzero projection in $B\left( H\right) ^{\left( m\right) }$. If we consider a type I von Neumann algebra acting on a separable Hilbert space as a direct integral of factors, we can change the factors so that they all have multiplicity 1. This gives another von Neumann algebra that is isomorphic to the original one. Since we are interested in finite von Neumann algebras, the type $I_{n}$ algebras, with $1\le n<\infty$, can be written as direct integrals of copies of $\mathbb {M}_{n}\left( \mathbb {C}\right)$, i.e., $\int _{\Omega _{n}}^{\oplus }\mathbb {M}_{n}\left( \mathbb {C}\right) d\mu _{n}\left( \omega \right)$ acting on $L^{2}\left( \mu _{n},\ell _{n}^{2}\right)$ for some probability space $\left( \Omega _{n},\mu _{n}\right)$ where $\mu _{n}$ is a Borel measure on a compact subset $\Omega _{n}$ of $\mathbb {R}$. In this case, $\int _{\Omega _{n}}^{\oplus }\mathbb {M}_{n}\left( \mathbb {C}\right) d\mu _{n}\left( \omega \right)$ is naturally isomorphic to $\mathbb {M}_{n}\left( L^{\infty }\left( \mu _{n}\right) \right)$ acting on $L^{2}\left( \mu _{n}\right) ^{\left( n\right) }$. When we write the type $I_{n}$ part of a von Neumann algebra this way, we have an isomorphic copy, but maybe not a unitarily equivalent copy of the algebra, since we changed all of the multiplicities to be 1. Note that the center $\mathcal {Z}\left( \int _{\Omega _{n}}^{\oplus }\mathbb {M}_{n}\left( \mathbb {C}\right) d\mu _{n}\left( \omega \right) \right) =\int _{\Omega _{n}}^{\oplus }\mathbb {C}\cdot 1\mu _{n}\left( \omega \right)$ acts on $L^{2}\left( \mu _{n},\ell _{n}^{2}\right)$.

For example, if a von Neumann algebra is $\int _{E_{1}}^{\oplus }\mathbb {M} _{2}\left( \mathbb {C}\right) d\eta _{1}\left( \omega \right) \oplus \int _{E_{2}}^{\oplus }\mathbb {M}_{2}\left( \mathbb {C}\right) ^{\left( 3\right) }d\eta _{2}\left( \omega \right)$, then it is isomorphic to $\int _{\Omega }^{\oplus }\mathbb {M}_{2}\left( \mathbb {C}\right) d\mu \left( \omega \right)$ where $\Omega$ is the disjoint union of $E_{1}$ and $E_{2}$ and $\mu \left( A\right) =\eta _{1}\left( A\cap E_{1}\right) +\eta _{2}\left( A\cap E_{2}\right)$.

Thus in the central decomposition, we can assume, for each positive integer n (i.e., $1\le n<\infty$), that

$$\begin{aligned} \mathcal {R}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {R}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) =\int _{\Omega _{n}}^{\oplus }\mathbb {M}_{n}\left( \mathbb {C}\right) d\mu _{n}\left( \omega \right) , \end{aligned}$$

and

$$\begin{aligned} \mathcal {Z}\left( \mathcal {R}_{n}\right) =\int _{\Omega _{n}}^{\oplus }\mathbb {C}\cdot 1d\mu _{n}\text {.} \end{aligned}$$

For $1\le n<\infty$ we have that the map is a normal faithful tracial state on $\mathcal {R}_{n}$.

2.3.5 $II_{1}$ von Neumann algebras

Once we have changed the multiplicities of the type $I_{n}$ parts of $\mathcal {R}$, in the decomposition

$$\begin{aligned} \mathcal {R}_{\infty }=\int _{\Omega _{\infty }}^{\oplus }\mathcal {R}_{\infty }\left( \omega \right) d\mu _{\infty }\left( \omega \right) , \end{aligned}$$

we have

$$\begin{aligned} \mathcal {Z}\left( \mathcal {R}_{\infty }\right) =\int _{\Omega _{\infty } }^{\oplus }\mathbb {C}\cdot 1d\mu _{\infty }\left( \omega \right) \text {.} \end{aligned}$$

where each $\mathcal {R}_{\infty }\left( \omega \right)$ must be an infinite dimensional finite factor; this means it must be a type $II_{1}$ factor, and we can assume it acts on $\ell ^{2}$. In this case making the multiplicity infinite can make things more convenient.

We let $\mathcal {R}_{\infty }^{\left( \infty \right) }=\left\{ T^{\left( \infty \right) }=T\oplus T\oplus \cdots :T\in \mathcal {R}_{\infty }\right\}$. Clearly, $\mathcal {R}_{\infty }^{\left( \infty \right) }$ is isomorphic to $\mathcal {R}_{\infty }$, and we have

$$\begin{aligned} \mathcal {R}_{\infty }^{\left( \infty \right) }=\int _{\Omega _{\infty }}^{\oplus }\mathcal {R}_{\infty }^{\left( \infty \right) }\left( \omega \right) d\mu _{\infty }\left( \omega \right) \end{aligned}$$

acting on $L^{2}\left( \mu _{\infty },\left( \ell ^{2}\right) ^{\left( \infty \right) }\right)$. The nice thing about $\mathcal {R}_{\infty }^{\left( \infty \right) }\left( \omega \right)$ is that every normal state $\varphi$ on $\mathcal {R}_{\infty }^{\left( \infty \right) }\left( \omega \right)$ can be written as

$$\begin{aligned} \varphi \left( T^{\left( \infty \right) }\right) =\left\langle T^{\left( \infty \right) }e,e\right\rangle \end{aligned}$$

for some unit vector $e\in \left( \ell ^{2}\right) ^{\left( \infty \right) }$. Since $\left( \ell ^{2}\right) ^{\left( \infty \right) }$ is isomorphic to $\ell ^{2}=\ell _{\infty }^{2}$, we can, by replacing $\mathcal {R}_{\infty }$ with $\mathcal {R}_{\infty }^{\left( \infty \right) }$, assume that every normal state $\varphi$ on $\mathcal {R}_{\infty }\left( \omega \right)$ can be written as

$$\begin{aligned} \varphi \left( T\right) =\left\langle Te,e\right\rangle \end{aligned}$$

for some unit vector e. In particular, since $\mathcal {R}_{\infty }\left( \omega \right)$ is a $II_{1}$ factor, there is a unique normal tracial state $\tau _{\infty ,\omega }$ on $\mathcal {R}_{\infty }\left( \omega \right)$. Hence there is a unit vector $e\left( \omega \right) \in \ell _{\infty }^{2}$ such that, for every $T\in \mathcal {R}_{\infty }\left( \omega \right)$,

$$\begin{aligned} \tau _{\infty ,\omega }\left( T\right) =\left\langle Te\left( \omega \right) ,e\left( \omega \right) \right\rangle . \end{aligned}$$

Using the measurable cross-section theorem, Theorem 5, we can choose $e\left( \omega \right)$ so that the map $e:\Omega _{\infty }\rightarrow \ell _{\infty }^{2}$ is absolutely measurable.

Lemma 20

Suppose $\mathcal {R}_{\infty }$ is a type $II_{1}$ von Neumann algebra with

$$\begin{aligned} \mathcal {R}_{\infty }=\int _{\Omega _{\infty }}^{\oplus }\mathcal {R}_{\infty }\left( \omega \right) d\mu _{\infty }\left( \omega \right) . \end{aligned}$$

Then there exists a map $e\in L^{2}\left( \mu _{\infty },\ell _{\infty } ^{2}\right)$ with $\left\| e\left( \omega \right) \right\| =1$ a.e., such that for every $T=\int _{\Omega _{\infty }}^{\oplus }T_{\omega }d\mu _{\infty }\left( \omega \right) \in \mathcal {R}_{\infty },$ $\left\langle T_{\infty ,\omega }e\left( \omega \right) ,e\left( \omega \right) \right\rangle =\tau _{\infty ,\omega }\left( T_{\omega }\right)$, where $\tau _{\infty ,\omega }$ is the unique normal tracial state on $\mathcal {R}_{\infty }\left( \omega \right) .$

Proof

Suppose

$$\begin{aligned} \mathcal {W}=\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times \underset{i=1}{\overset{\infty }{\Pi }}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times E, \end{aligned}$$

where $E=\left\{ x\in l_{\infty }^{2}:\left\| x\right\| =1\right\}$. Then $\mathcal {W}$ is a complete separable metric space with product topology.

Let $\mathcal {X}$ be the set of elements $\left( T,\left\{ A_{i}\right\} _{i=1}^{\infty },e\right)$ in $\mathcal {W}$ satisfying

$$\begin{aligned} TA_{i}=A_{i}T,\left\langle A_{i}A_{j}e,e\right\rangle =\left\langle A_{j} A_{i}e,e\right\rangle \text { for every }i,j\in \mathbb {N}. \end{aligned}$$

It is easy to verify that $\mathcal {X}$ is closed.

Let $\pi _{2}:\mathcal {X\rightarrow }\underset{i=1}{\overset{\infty }{\Pi } }\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right)$, $\pi _{3}:\mathcal {X\rightarrow }E$ be the projection maps. Then $\pi _{2}\left( \mathcal {X}\right)$ is the set of elements $\left\{ A_{i}\right\} _{i=1}^{\infty }$ so that there exists $T\in \textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right)$ such that

$$\begin{aligned} T\in \left\{ A_{1},A_{2},\dots \right\} ^{\prime }\cap \left\{ B_{1},B_{2},\dots \right\} ^{\prime }\text { and }\left\langle A_{i}A_{j} e,e\right\rangle =\left\langle A_{j}A_{i}e,e\right\rangle \text { for all }i,j\in \mathbb {N}. \end{aligned}$$

There is sequence $\left\{ \psi _{1},\psi _{2},\ldots \right\}$ contained in $L^{\infty }\left( \mu _{\infty },B\left( l_{\infty }^{2}\right) \right)$ such that

$$\begin{aligned} \textrm{ball}\mathcal {R}_{\infty }\left( \omega \right) ^{\prime }=\left\{ \psi _{1}\left( \omega \right) ,\psi _{2}\left( \omega \right) ,\dots \right\} ^{-SOT} \end{aligned}$$

By Theorem 5, we know there exists an absolutely measurable function $\Upsilon :\pi _{2}\left( \mathcal {X}\right) \longrightarrow \mathcal {X}$ such that $\pi _{2}\circ \Upsilon$ is the identity function on $\pi _{2}\left( \mathcal {X}\right) .$

Define $F:\Omega _{\infty }\rightarrow {\Pi }_{i=1}^{{\infty }}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right)$ by

$$\begin{aligned} F\left( \omega \right) =\left\{ \psi _{i}\left( \omega \right) \right\} _{i=1}^{\infty }, \end{aligned}$$

which is measurable, thus, by Lemma 15, $\pi _{3}\circ \Upsilon \circ F$ is a measurable function from $\Omega _{\infty }$ to $l_{\infty }^{2}$. We define e by

$$\begin{aligned} e\left( \omega \right) =\left( \pi _{3}\circ \Upsilon \circ F\right) \left( \omega \right) . \end{aligned}$$

Thus e is a measurable function with $e=\int _{\Omega _{\infty }}^{\oplus }e\left( \omega \right) d\mu _{\infty }\left( \omega \right)$, that is $e\in L^{2}\left( \mu _{\infty },\ell _{\infty }^{2}\right)$ and

$$\begin{aligned} \left\| e\right\| _{2}^{2}=\int _{\Omega _{\infty }}\left\| e\left( \omega \right) \right\| ^{2}d\mu _{\infty }\left( \omega \right) =\int _{\Omega _{\infty }}1d\mu _{\infty }\left( \omega \right) =\mu _{\infty }\left( \Omega _{\infty }\right) =1\text {.} \end{aligned}$$

$\square$

The map

$$\begin{aligned} \tau _{\infty }:\mathcal {R}_{\infty }\rightarrow \mathbb {C} \end{aligned}$$

defined by

$$\begin{aligned} \tau _{\infty }\left( T\right) =\left\langle Te,e\right\rangle =\int _{\Omega _{\infty }}\left\langle T_{\omega }e\left( \omega \right) ,e\left( \omega \right) \right\rangle d\mu _{\infty }\left( \omega \right) \underset{\textrm{def}}{=}\int _{\Omega _{\infty }}\tau _{\infty ,\omega }\left( T_{\omega }\right) d\mu _{\infty }\left( \omega \right) \end{aligned}$$

is a faithful normal trace on $\mathcal {R}_{\infty }$. Since $\tau _{\infty ,\omega }$ is a faithful normal trace on $\mathcal {R}_{\infty }\left( \omega \right)$ and the trace on a type $II_{1}$ factor is unique, it follows that $\tau _{\infty ,\omega }$ is the usual trace.

2.3.6 The center-valued trace

Suppose $\mathcal {R}$ is an arbitrary finite von Neumann algebra, possibly not acting on a separable Hilbert space. There is (see [23]) a unique map $\Phi _{\mathcal {R}}:\mathcal {R}\rightarrow \mathcal {Z}\left( \mathcal {R}\right)$ satisfying

• (1) $\Phi _{\mathcal {R}}$ is linear and completely positive,

• (2) $\Phi _{\mathcal {R}}\left( 1\right) =1,$

• (3) $\Phi _{\mathcal {R}}\left( AB\right) =\Phi _{\mathcal {R}}\left( BA\right)$ for all $A,B\in \mathcal {R}$,

• (4) $\Phi _{\mathcal {R}}$ is weak*-weak* continuous, and

• (5) $\Phi _{\mathcal {R}}\left( ATB\right) =A\Phi _{\mathcal {R}}\left( T\right) B$ for all $T\in \mathcal {R}$ and all $A,B\in \mathcal {Z}\left( \mathcal {R}\right)$.

The map $\Phi _{\mathcal {R}}$ is called the center-valued trace on $\mathcal {R}$.

When $\mathcal {R}$ acts on a separable Hilbert space, we have

$$\begin{aligned} \mathcal {R}=\sum \nolimits _{1\le n\le \infty }^{\oplus }\mathcal {R}_{n}, \end{aligned}$$

we have

$$\begin{aligned} \mathcal {Z}\left( \mathcal {R}\right) =\sum \nolimits _{1\le n\le \infty }^{\oplus }\mathcal {Z}\left( \mathcal {R}_{n}\right) , \end{aligned}$$

and we have

$$\begin{aligned} \Phi _{\mathcal {R}}=\sum \nolimits _{1\le n\le \infty }^{\oplus }\Phi _{\mathcal {R}_{n}}. \end{aligned}$$

We can write each $\Phi _{\mathcal {R}_{n}}$ explicitly in terms of the central decomposition, i.e.,

$$\begin{aligned} \Phi _{\mathcal {R}_{n}}\left( T\right) =\int _{\Omega _{n}}^{\oplus }\tau _{n}\left( T_{\omega }\right) \cdot 1d\mu _{n}\left( \omega \right) \end{aligned}$$

when $1\le n<\infty$, and

$$\begin{aligned} \Phi _{\mathcal {R}_{\infty }}\left( T\right) =\int _{\Omega _{\infty }}^{\oplus }\tau _{\omega }\left( T_{\omega }\right) \cdot 1d\mu _{\infty }\left( \omega \right) . \end{aligned}$$

It is clear that these maps satisfy the defining properties $\left( 1\right)$-$\left( 5\right)$ and the uniqueness tells us that these formulas are correct.

2.3.7 Two simple relations

Suppose $1\le n\le \infty$. There is a normal $*$-isomorphism $\gamma _{n}:L^{\infty }\left( \mu _{n}\right) \rightarrow \mathcal {Z}\left( \mathcal {R}_{n}\right)$ defined by

$$\begin{aligned} \gamma _{n}\left( f\right) =\int _{\Omega _{n}}^{\oplus }f\left( \omega \right) \cdot 1d\mu _{n}\left( \omega \right) . \end{aligned}$$

Recall $\rho _{n}:\mathcal {R}_{n}\rightarrow \mathcal {\mathbb {C}}$ is defined by

$$\begin{aligned} \rho _{n}\left( T\right) =\int _{\Omega _{n}}\tau _{n,\omega }\left( T_{\omega }\right) d\mu _{n}\left( \omega \right) \text {.} \end{aligned}$$

The map $f\mapsto \int {}_{\Omega _{n}}fd\mu _{n}$ is a state on $L^{\infty }\left( \mu _{n}\right)$. The simple relation between this state and the $*$-isomorphism $\gamma _{n}$ and the state $\rho _{n}$ is given by

$$\begin{aligned} \left( \rho _{n}\circ \gamma _{n}\right) \left( f\right) =\int _{\Omega _{n} }fd\mu _{n} \end{aligned}$$

for every $f\in L^{\infty }\left( \mu _{n}\right)$.

Another simple relationship between $\rho _{n}$ and $\Phi _{\mathcal {R}_{n}}$ is

$$\begin{aligned} \rho _{n}=\rho _{n}\circ \Phi _{\mathcal {R}_{n}}\text {.} \end{aligned}$$

2.3.8 Putting things together

We let $\Omega$ be the disjoint union of $\{ \Omega _{n}:1\le n$$\le \infty \}$, which can be represented as a Borel subset of $\mathbb {R}$. We define a probability Borel measure $\mu$ on $\Omega$ by

$$\begin{aligned} \mu \left( E\right) =\frac{1}{2}\mu _{\infty }\left( E\cap \Omega _{\infty }\right) +\sum _{n=1}^{\infty }\frac{1}{2^{n+1}}\mu _{n}\left( E\cap \Omega _{n}\right) \text {.} \end{aligned}$$

Then the von Neumann algebra $L^{\infty }\left( \mu \right)$ can be written as

$$\begin{aligned} L^{\infty }\left( \mu \right) =L^{\infty }\left( \mu _{\infty }\right) \oplus \sum \nolimits _{1\le n\le \infty }^{\oplus }L^{\infty }\left( \mu _{n}\right) \end{aligned}$$

We define an isomorphism

$$\begin{aligned} \gamma :L^{\infty }\left( \mu \right) \rightarrow \mathcal {Z}\left( \mathcal {R}\right) , \end{aligned}$$

by

$$\begin{aligned} \gamma \left( f_{\infty }\oplus f_{1}\oplus f_{2}\oplus \cdots \right) =\gamma _{\infty }\left( f_{\infty }\right) \oplus \gamma _{1}\left( f_{1}\right) \oplus \gamma _{2}\left( f_{2}\right) \cdots \text {.} \end{aligned}$$

We can define a faithful normal tracial state $\rho :\mathcal {R}\rightarrow \mathbb {C}$ by

$$\begin{aligned} \rho \left( \sum \nolimits _{1\le n\le \infty }^{\oplus }T_{n}\right) =\frac{1}{2}\rho _{\infty }\left( T_{\infty }\right) +\underset{1\le n<\infty }{\sum }\frac{1}{2^{n+1}}\rho _{n}\left( T_{n}\right) . \end{aligned}$$

We have

$$\begin{aligned} \begin{array}{l} \text {1. }\rho =\rho \circ \Phi _{\mathcal {R}},\\ \text {2. }\left( \rho \circ \gamma \right) \left( f\right) =\int _{\Omega }fd\mu \text { for every }f\in L^{\infty }\left( \mu \right) \text {, and, as we stated above,}\\ \text {3. }\Phi _{\mathcal {R}}\left( T\right) =\sum \nolimits _{1\le n\le \infty }^{\oplus }\Phi _{\mathcal {R}_{n}}\left( T_{n}\right) \\ { }=\left[ \sum \nolimits _{1\le n<\infty }^{\oplus }\int _{\Omega _{n} }^{\oplus }\tau _{n}\left( T_{n}\left( \omega \right) \right) \cdot 1d\mu _{n}\left( \omega \right) \right] \oplus \int _{\Omega _{\infty }}^{\oplus } \tau _{\omega }\left( T_{\infty }\left( \omega \right) \right) \cdot 1d\mu _{\infty }\left( \omega \right) . \end{array} \end{aligned}$$

3 MASAS in finite von Neumann algebras

A masa in a C*-algebra is a maximal abelian selfadjoint subalgebra. In $B\left( H\right)$ where H is a separable infinite-dimensional Hilbert space there are many different masas. For example, the set of all diagonal operators with respect to some fixed orthonormal basis is a discrete masa. On the other hand $L^{\infty }\left[ 0,1\right] =L^{\infty }\left( \delta _{\infty }\right)$ acting as multiplications on $L^{2}\left[ 0,1\right]$ with Lebesgue measure is also a masa that is not isomorphic to the diagonal masa, since it is diffuse (i.e., has no minimal nonzero projections). It was show by A. Sinclair and R. Smith [22] that in a finite von Neumann algebra acting on a separable Hilbert space all masas are isomorphic. We will prove that all masas are isomorphic in a very special way.

Theorem 6

Suppose $\mathcal {A}$ is a masa in a finite von Neumann algebra $\mathcal {R}$. Then there is an tracial embedding $\pi _{\mathcal {A}}:L^{\infty }\left( \lambda \right) \rightarrow \mathcal {A}$ such that the following diagram commutes

$$\begin{aligned} \begin{array}{ccc} L^{\infty }\left( \lambda \right) &{} \overset{\pi _{\mathcal {A}}}{\rightarrow } &{} \mathcal {A}\\ \downarrow \eta &{} &{} \downarrow \Phi _{\mathcal {R}}\\ L^{\infty }\left( \mu \right) &{} \overset{\gamma }{\rightarrow }\ &{} Z\left( \mathcal {R}\right) \end{array} \end{aligned}$$

Moreover, if $\mathcal {B}$ is another masa in $\mathcal {R}$, then $\mathcal {B}$ is isomorphic to $\mathcal {A}$. In fact, $\pi _{\mathcal {A}}$ and $\pi _{\mathcal {B}}$ are approximately equivalent in $\mathcal {R}$.

We first need to prove this theorem when $\mathcal {R}$ is a finite factor. When $\mathcal {R}$ is a type $I_{n}$ factor, i.e., $\mathcal {R}=\mathbb {M} _{n}\left( \mathbb {C}\right)$, the result is obvious.

Lemma 21

Suppose $\mathcal {A}\subset \mathbb {M}_{n}\left( \mathbb {C}\right)$ is a masa. Then there exists a unitary $U\in \mathcal {U}\left( \mathbb {M}_{n}\left( \mathbb {C}\right) \right)$ such that $U\mathcal {A}U^{*}=\mathcal {D}_{n},$ the $n\times n$ complex diagonal matrices. Hence there is a *-isomorphism $\pi _{\mathcal {A}}:L^{\infty }\left( \delta _{n}\right) \rightarrow \mathcal {A}$ such that, for every $f\in L^{\infty }\left( \delta _{n}\right)$, which is isometrically isomorphic to $\mathbb {C} ^{n}$.

$$\begin{aligned} \tau _{n}\left( \pi _{\mathcal {A}}\left( f\right) \right) =\int _{J_{n} }fd\delta _{n}\text {.} \end{aligned}$$

When $\mathcal {R}$ is a type $II_{1}$ factor the result is well-known [22].

Lemma 22

Suppose $\mathcal {M}$ is a type $II_{1}$ factor von Neumann algebra acting on a separable Hilbert space with a (unique) faithful normal tracial state $\tau$, and suppose $\mathcal {A}$ is a masa in $\mathcal {M}$. Then there is an isomorphism $\pi _{\mathcal {A}}:L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {A}$ such that, for every $f\in L^{\infty }\left( \delta _{\infty }\right)$,

$$\begin{aligned} \tau \left( \pi _{\mathcal {A}}\left( f\right) \right) =\int _{0}^{1}f\left( t\right) d\delta _{\infty }\left( t\right) \text {.} \end{aligned}$$

Corollary 5

Suppose $\mathcal {A}$ is an abelian von Neumann algebra on a separable Hilbert space with a faithful (tracial) state $\tau$. The following are equivalent:

• (1) There is a tracial embedding $\pi :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {A}$ such that, for every $f\in L^{\infty }\left( \delta _{\infty }\right)$,

  $$\begin{aligned} \tau \left( \pi \left( f\right) \right) =\int _{0}^{1}f\left( t\right) d\delta _{\infty }\left( t\right) . \end{aligned}$$

• (2) There is a $T\in \mathcal {A}$ such that

  • (a) $W^{*}\left( T\right) =\mathcal {A}$

  • (b) $T=T^{*}$

  • (c) $\tau \left( T^{n}\right) =\frac{1}{n+1}$ for $n\in \mathbb {N}$

  Moreover, if $\left( 2\right)$ holds, then $0\le T\le 1$, then the map

  $$\begin{aligned} \pi \left( f\right) =f\left( T\right) \end{aligned}$$

  is the required map in $\left( 1\right) .$

Proof

$\left( 1\right) \Rightarrow \left( 2\right)$. Suppose $\pi$ exists as in $\left( 1\right)$. Define $f\left( t\right) =t$ in $L^{\infty }\left( \delta _{\infty }\right)$ and let $T=\pi \left( f\right) .$ Then $0\le T\le 1,$

$$\begin{aligned} \mathcal {A}=\pi \left( L^{\infty }\left( \delta _{\infty }\right) \right) =\pi \left( W^{*}\left( f\right) \right) =W^{*}\left( \pi \left( f\right) \right) =W^{*}\left( T\right) , \end{aligned}$$

and, for each $n\in \mathbb {N}$,

$$\begin{aligned} \tau \left( T^{n}\right) =\tau \left( \pi \left( f^{n}\right) \right) =\int _{0}^{1}t^{n}dt=\frac{1}{n+1}\text {.} \end{aligned}$$

$\left( 2\right) \Rightarrow \left( 1\right)$. Define the state $\rho :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathbb {C}$ by

$$\begin{aligned} \rho \left( f\right) =\int _{0}^{1}f\left( t\right) d\delta _{\infty }\left( t\right) . \end{aligned}$$

Letting $f\in L^{\infty }\left( \delta _{\infty }\right)$ be as above, we have $\tau \left( T^{n}\right) =\rho \left( f^{n}\right) =\frac{1}{n+1}$ for each $n\in \mathbb {N}$. It follows from Lemma 1 in [6] that there is a normal (i.e., weak*-weak* continuous) $*$-isomorphism $\pi :L^{\infty }\left( \delta _{\infty }\right) \rightarrow \mathcal {A}$ such that $\pi \left( f\right) =T$ and such that $\tau \circ \pi =\rho$. It is clear that, for any polynomial $p\left( t\right)$, $\pi \left( p\right) =p\left( T\right)$. Suppose $f\in L^{\infty }\left( \delta _{\infty }\right)$. By changing f on a set of measure 0, we can assume that f is Borel measurable. Then there is a sequence $\left\{ p_{n}\right\}$ of polynomials such that $p_{n}\rightarrow f$ weak*. Thus

$$\begin{aligned} f\left( T\right) =\left( \text {weak*}\right) \lim _{n\rightarrow \infty }p_{n}\left( T\right) =\left( \text {weak*}\right) \lim _{n\rightarrow \infty }\pi \left( p_{n}\right) =\pi \left( f\right) \text {.} \end{aligned}$$

$\square$

From this Lemma, we can see that $\pi \left( f\right) =f\left( T\right)$ and $\tau \left( T^{n}\right) {=}\tau \left( \pi _{A}\left( x^{n}\right) \right) =\int _{0}^{1}x^{n}d\delta _{\infty }=\frac{1}{n+1}$ for $n=1,2,\cdots .$ Let $\mathbb {C}_{\mathbb {Q}}=\mathfrak {\mathbb {Q}}+i\mathbb {Q}$ denote the set of all complex numbers z such that ${\text {Re}}\left( z\right)$ and ${\text {Im}}\left( z\right)$ are rational. Since the set $\mathbb {C}_{\mathbb {Q}}[z]$ of all polynomials with coefficients in $\mathbb {C}_{\mathbb {Q}}$ is countable, we can write

$$\begin{aligned} \mathbb {C}_{\mathbb {Q}}[z]=\left\{ p_{1},p_{2},\ldots \right\} \text {.} \end{aligned}$$

Lemma 23

Suppose $A=A^{*}\in B\left( H\right)$. It follows that

$$\begin{aligned} W^{*}\left( A\right) =\left\{ p_{1}\left( A\right) ,p_{2}\left( A\right) ,\cdots \right\} ^{-WOT} \end{aligned}$$

Lemma 24

Suppose $\mathcal {A}_{\infty }$ is a masa of $\mathcal {R}_{\infty }$. Then there exists an operator $T=\int _{\Omega _{\infty }}^{\oplus }T_{\omega }d\mu _{\infty }\left( \omega \right)$ such that $W^{*}\left( T_{\omega }\right) =$ $\mathcal {A}_{\infty }\left( \omega \right)$, and $\tau _{\omega ,\infty }\left( T_{\omega }^{n}\right) =\left\langle T_{\omega }^{n}e\left( \omega \right) ,e\left( \omega \right) \right\rangle =\frac{1}{n+1}$ for $n\ge 1$, $\mathcal {A}_{\infty }=W^{*}\left( T\right) .$

Proof

Let

$$\begin{aligned} \mathcal {Y=}B\left( l_{\infty }^{2}\right) \times \underset{i=1}{\overset{\infty }{\Pi }}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times \underset{i=1}{\overset{\infty }{\Pi }}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times \underset{i=1}{\overset{\infty }{\Pi }}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times E, \end{aligned}$$

where $E=\left\{ x\in l_{\infty }^{2}:\left\| x\right\| =1\right\}$. It is clear that $\mathcal {Y}$ is a complete separable metric space with product topology. Let $\mathcal {X}$ be the set of tuples

$$\begin{aligned} \left( S,\left\{ A_{i}\right\} _{i=1}^{\infty },\left\{ B_{i}\right\} _{i=1}^{\infty },\left\{ C_{i}\right\} _{i=1}^{\infty },x\right) \in \mathcal {Y} \end{aligned}$$

satisfying

$$\begin{aligned} SA_{i}=A_{i}S,SB_{i}=B_{i}S,\left\langle S^{n}x,x\right\rangle =\frac{1}{n+1}\text { for }n\ge 1. \end{aligned}$$

From Lemma 23, we know there exists a sequence $\left\{ p_{n}\right\}$ of polynomials such that $W^{*}\left( S\right) =W^{*}\left( p_{1}\left( T\right) ,p_{2}\left( T\right) ,\ldots \right) .$ Define $\mathcal {W}_{i,k,n}$ to be the subset of $\mathcal {X}$ satisfying

$$\begin{aligned} S=S^{*},{ }d\left( A_{i},p_{n}\left( S\right) \right) \ge \frac{1}{k}\text { for }n\ge 1. \end{aligned}$$

Let $\mathcal {W}_{i,k}={{\bigcap _{n=1}^{\infty }}} \mathcal {W}_{i,k,n}$ and $\mathcal {W}={{\bigcap _{i=1}^{\infty } }}{{\bigcap _{k=1}^{\infty }}}\mathcal {W}_{i,k},$ then $\mathcal {W}$ is a subset of $\mathcal {X}$ satisfying

$$\begin{aligned} A_{i}\notin W^{*}\left( p_{1}\left( T\right) ,p_{2}\left( T\right) ,\cdots \right) ,\text { for }i\ge 1. \end{aligned}$$

Then $\mathcal {X}\backslash \mathcal {W}={ {\bigcap _{i=1}^{\infty }}}{{\bigcap _{i=1}^{\infty }}}\mathcal {X}\backslash \mathcal {W}_{i,k}$ is a subset of $\mathcal {X}$ satisfying

$$\begin{aligned} W^{*}\left( A_{1},A_{2},\cdots \right) \subseteq W^{*}\left( p_{1}\left( T\right) ,p_{2}\left( T\right) ,\cdots \right) , \end{aligned}$$

which is a $G_{\delta }$ set. By [3], there exists an equivalent metric on $\mathcal {X}\backslash \mathcal {W}$ that makes $\mathcal {X} \backslash \mathcal {W}$ a complete separable metric space. If $\pi _{2,3,4}$ is the projection map into the second, third, fourth coordinates, then there exists an absolute measurable function $\Upsilon :\pi _{2,3,4}\left( \mathcal {X}\right) \rightarrow \mathcal {X}$ such that $\pi _{2,3,4} \circ \Upsilon$ is the identity on $\pi _{2,3,4}\left( \mathcal {X}\right) .$

Suppose there are sequences $\left\{ f_{1},f_{2},\cdots \right\} ,$ $\left\{ \psi _{1},\psi _{2},\cdots \right\} ,$ $\left\{ \varphi _{1},\varphi _{2},\cdots \right\}$ contained in $L^{\infty }\left( \mu _{\infty },B\left( l_{\infty }^{2}\right) \right)$ such that, for almost every $\omega$,

$$\begin{aligned} \begin{array}{l} \textrm{ball}\mathcal {A}_{\infty }\left( \omega \right) =\left\{ f_{1}\left( \omega \right) ,{ }f_{2}\left( \omega \right) ,\dots \right\} ^{-SOT},\\ \textrm{ball}\mathcal {R}_{\infty }\left( \omega \right) ^{\prime }=\left\{ \psi _{1}\left( \omega \right) ,\psi _{2}\left( \omega \right) ,\dots \right\} ^{-SOT},\\ \textrm{ball}\mathcal {R}_{\infty }\left( \omega \right) =\left\{ \varphi _{1}\left( \omega \right) ,\varphi _{2}\left( \omega \right) ,\dots \right\} ^{-SOT}. \end{array} \end{aligned}$$

Define $F:\Omega _{\infty }\rightarrow {{\Pi _{i=1}^{\infty }} }\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times {{\Pi _{i=1}^{\infty }}}\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right) \times {\Pi _{i=1}^{\infty } }\textrm{ball}\left( B\left( l_{\infty }^{2}\right) \right)$ by

$$\begin{aligned} F\left( \omega \right) =\left\{ f_{i}\left( \omega \right) \right\} _{i=1}^{\infty }\times \left\{ \psi _{i}\left( \omega \right) \right\} _{i=1}^{\infty }\times \left\{ \varphi _{i}\left( \omega \right) \right\} _{i=1}^{\infty }. \end{aligned}$$

Clearly F is measurable. Thus, by Lemma 15, if $\pi _{1}$ is the projection from $\mathcal {X}\backslash \mathcal {W}$ into its first coordinate, then $T=$ $\pi _{1}\circ \Upsilon \circ F:\omega \longmapsto T_{\omega }$ is the desired measurable function from $\Omega _{\infty }$ to $B\left( l_{\infty } ^{2}\right)$ such that $\textrm{ball}\left( \mathcal {A}_{\infty }\left( \omega \right) \right) =W^{*}\left( T_{\omega }\right)$. $\square$

Lemma 25

Suppose $\mathcal {A}_{n}$ is a masa of $\mathcal {R}_{n}$ for every $1\le n\le \infty$. Then there is an isomorphism $\pi _{\mathcal {A}_{n}}:L^{\infty }\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right) \rightarrow \mathcal {A}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right)$.

Proof

Suppose $1\le n<\infty$. We know that $\mathcal {R}_{n}$ is isomorphic to $\int _{\Omega n}^{\oplus }\mathbb {M}_{n}\left( \mathbb {C}\right) d\mu _{n}\left( \omega \right) ,$ so if $\mathcal {A}_{n}$ is a masa in $\mathcal {R}_{n}$, then $\mathcal {A}_{n}=\int _{\Omega _{n}}^{\oplus } \mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right)$ where each $\mathcal {A}_{n}\left( \omega \right)$ is a masa in $\mathbb {M} _{n}\left( \mathbb {C}\right)$. There is a unitary operator $U_{\omega } \in \mathbb {M}_{n}\left( \mathbb {C}\right)$ such that $\mathcal {A} _{n}\left( \omega \right) =U_{\omega }^{*}\mathcal {D}_{n}\left( \mathbb {C}\right) U_{\omega }$. An easy measurable cross-section proof allows us to choose the $U_{\omega }$’s measurably. However, $\mathcal {D}_{n}$ is isomorphic to $L^{\infty }\left( J_{n},\delta _{n}\right) .$ Define

$$\begin{aligned} \pi _{\mathcal {A}_{n}}:L^{\infty }\left( \Omega _{n}\times J_{n}\right) \rightarrow \int _{\Omega _{n}}^{\oplus }L^{\infty }\left( \delta _{n}\right) d\mu _{n}\left( \omega \right) \end{aligned}$$

by

$$\begin{aligned} \pi _{\mathcal {A}_{n}}\left( f\right) =\int _{\Omega _{n}}^{\oplus }U_{\omega }^{*}\left( \begin{array}{ccc} f\left( \omega ,\frac{1}{n}\right) &{} &{} \\ &{} \ddots &{} \\ &{} &{} f\left( \omega ,\frac{n}{n}\right) \end{array} \right) U_{\omega }d\mu _{n}\left( \omega \right) . \end{aligned}$$

Now suppose $n=\infty$. We choose $\left\{ T_{\omega }\right\}$ as in Lemma 24, and we define

$$\begin{aligned} \pi _{\mathcal {A}_{\infty }}\left( f\right) =\int _{\Omega _{\infty }}^{\oplus }f_{\omega }\left( T_{\omega }\right) d\mu _{\infty }\left( \omega \right) \text {,} \end{aligned}$$

where $f_{\omega }\left( t\right) =f\left( \omega ,t\right)$. $\square$

Suppose now that $\mathcal {R}$ is a finite von Neumann algebra acting on a separable Hilbert space H,

$$\begin{aligned} \mathcal {R=}{ }\left[ \mathcal {R}_{1}\oplus \mathcal {R}_{2}\oplus \cdots \right] \oplus \mathcal {R}_{\infty }\text {.} \end{aligned}$$

For $1\le n<\infty ,$ $\mathcal {R}_{n}$ is a type $I_{n}$ von Neumann algebra acting on $H_{n}$, $\mathcal {R}_{\infty }$ is a type $II_{1}$ von Neumann algebra acting on $H_{\infty },$

$$\begin{aligned} H{ }\mathcal {=}{ }\left[ H_{1}\oplus H_{2}\oplus \cdots \right] \oplus H_{\infty }. \end{aligned}$$

If $\mathcal {A}$ is a masa in $\mathcal {R}$, then, we can write

$$\begin{aligned} \mathcal {A=}{ }\left[ \mathcal {A}_{1}\oplus \mathcal {A}_{2}\oplus \cdots \right] \oplus \mathcal {A}_{\infty }, \end{aligned}$$

where, for $1\le n\le \infty$, $\mathcal {A}_{n}$ is a masa in $\mathcal {R} _{n}$. Clearly, since $\mathcal {A}_{n}$ is a masa in $\mathcal {R}_{n}$, we know that $\mathcal {D}_{n}=\mathcal {Z}\left( \mathcal {R}_{n}\right) \subseteq \mathcal {A}_{n}\subseteq \mathcal {R}_{n}\subseteq L^{\infty }\left( \mu _{n},B\left( H_{n}\right) \right)$. It follows from Lemma 19 that there is a measurable family $\left\{ \mathcal {A} _{n}\left( \omega \right) :\omega \in \Omega _{n}\right\}$ of von Neumann algebras such that

$$\begin{aligned} \mathcal {A}_{n}=\int _{\Omega _{n}}^{\oplus }\mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right) . \end{aligned}$$

If $1\le n<\infty$, then almost every $\mathcal {A}_{n}\left( \omega \right)$ must be a masa in $\mathbb {M}_{n}\left( \mathbb {C}\right)$. If $n=\infty$, then almost every $\mathcal {A}_{n}\left( \omega \right)$ must be a masa in the $II_{1}$ factor $\mathcal {R}_{\infty }\left( \omega \right)$. Since throwing away a set of measure 0 from $\Omega _{n}$ doesn’t change anything, we can assume that, when $1\le n<\infty$ every $\mathcal {A}_{n}\left( \omega \right)$ is a masa in $\mathbb {M}_{n}\left( \mathbb {C}\right)$, and when $n=\infty$, every $\mathcal {A}_{\infty }\left( \omega \right)$ is a masa in $\mathcal {R}_{\infty }\left( \omega \right)$.

If $1\le n\le \infty$, then each $\mathcal {A}_{n}\left( \omega \right)$ is isomorphic to $L^{\infty }\left( \delta _{n}\right)$ (see Lemmas 21 and 22). And $\int _{\Omega _{n} }^{\oplus }\mathcal {A}_{n}\left( \omega \right) d\mu _{n}\left( \omega \right)$ is isomorphic to $\int _{\Omega _{n}}^{\oplus }L^{\infty }\left( \delta _{n}\right) d\mu _{n}\left( \omega \right)$, which is isomorphic to $L^{\infty }\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right)$. The isomorphism sends a function $f\left( \omega ,t\right) \in L^{\infty }\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right)$ to $\int _{\Omega _{n}}^{\oplus }f_{\omega }\left( t\right) d\mu _{n}\left( \omega \right)$, where $f_{\omega }\left( t\right) =f\left( \omega ,t\right)$.

For each n, $1\le n\le \infty$, we define $\Lambda _{n}=\Omega _{n}\times J_{n}$ and we define $\lambda _{n}=\mu _{n}\times \delta _{n}$. We let $\Lambda$ denote the disjoint union of the $\Lambda _{n}$’s for $1\le n\le \infty$, and we can choose $\Lambda$ to be a Borel subset of $\mathbb {R}$, and we define a probability Borel measure $\lambda$ on $\Lambda$ by

$$\begin{aligned} \lambda \left( F\right) =\frac{1}{2}\lambda _{\infty }\left( F\cap \Lambda _{\infty }\right) +\sum _{n=1}^{\infty }\frac{1}{2^{n}}\lambda _{n}\left( F\cap \Lambda _{n}\right) \text {.} \end{aligned}$$

We then have

$$\begin{aligned} \ L^{\infty }\left( \lambda \right) =L^{\infty }\left( \lambda _{\infty }\right) \oplus \underset{1\le n<\infty }{\Pi }L^{\infty }\left( \lambda _{n}\right) \text {.} \end{aligned}$$

For each n, $1\le n\le \infty$, there is a mapping

$$\begin{aligned} \eta _{n}:L^{\infty }\left( \lambda _{n}\right) =L^{\infty }\left( \mu _{n}\times \delta _{n}\right) \rightarrow L^{\infty }\left( \mu _{n}\right) , \end{aligned}$$

defined by

$$\begin{aligned} \eta _{n}\left( f\right) \left( \omega \right) =\int _{J_{n}}^{\oplus }f\left( \omega ,t\right) d\delta _{n}\left( t\right) \text {.} \end{aligned}$$

We define $\eta :L^{\infty }\left( \lambda \right) \rightarrow L^{\infty }\left( \mu \right)$ by

$$\begin{aligned} \eta \left( f\right) =\eta \left( f_{\infty }\oplus f_{1}\oplus f_{2} \oplus \cdots \right) =\eta _{\infty }\left( f_{\infty }\right) \oplus \eta _{1}\left( f_{1}\right) \oplus \eta _{2}\left( f_{2}\right) \oplus \cdots \text {.} \end{aligned}$$

Lemma 26

For $1\le n\le \infty ,$ if $\mathcal {A}_{n}$ is a masa in $\mathcal {R}_{n},$ then there exists a tracial embedding $\pi _{\mathcal {A}_{n}}:L^{\infty }\left( \lambda _{n}\right) =L^{\infty }\left( \mu _{n}\times \delta _{n}\right) \rightarrow \mathcal {A}_{n}$ such that the following diagram commutes

$$\begin{aligned}{} & {} \begin{array}{ccc} L^{\infty }\left( \lambda _{n}\right) &{} \overset{\pi _{\mathcal {A}_{n}} }{\rightarrow } &{} \mathcal {A}_{n}\\ \downarrow \eta _{n} &{} &{} \downarrow \Phi _{n}\\ L^{\infty }\left( \mu _{n}\right) &{} \overset{\gamma _{n}}{\rightarrow } &{} Z\left( \mathcal {R}_{n}\right) \end{array}. \\{} & {} \Phi _{n}\circ \pi _{\mathcal {A}_{n}}=\gamma _{n}\circ \eta _{n}. \end{aligned}$$

where

$$\begin{aligned} \gamma _{n}\left( f\right)&=\int _{\Omega _{n}}^{\oplus }f\left( \omega \right) Id\mu _{n}\left( \omega \right) ,\\ \eta _{n}\left( f\right) \left( \omega ,t\right)&=\int _{J_{n}}f\left( \omega ,t\right) d\delta _{n}\left( t\right) \text { and,}\\ \Phi _{n}\left( \int _{\Omega _{n}}^{\oplus }T\left( \omega \right) d\mu _{n}\left( \omega \right) \right)&=\int _{\Omega _{n}}^{\oplus } \tau _{\omega ,n}\left( T\left( \omega \right) \right) Id\mu _{n}\left( \omega \right) . \end{aligned}$$

Moreover, if $\mathcal {B}_{n}$ is a masa in $\mathcal {R}_{n}$, and there is a tracial embedding $\pi _{\mathcal {B}_{n}}:L^{\infty }\left( \lambda _{n}\right) \rightarrow \mathcal {B}_{n}$ such that $\Phi _{n}\circ \pi _{\mathcal {B}_{n} }=\gamma _{n}\circ \eta _{n}$, then,

if $1\le n<\infty ,$ then there exists a unitary $U\in \mathcal {U}\left( \mathcal {R}_{n}\right)$ such that

$$\begin{aligned} U\pi _{\mathcal {A}_{n}}\left( L^{\infty }\left( \lambda _{n}\right) \right) U^{*}=\pi _{\mathcal {B}_{n}}\left( L^{\infty }\left( \lambda _{n}\right) \right) , \end{aligned}$$

if $n=\infty ,$ then $\pi _{\mathcal {A}_{n}}$ is approximately equivalent to $\pi _{\mathcal {B}_{n}}$ in $\mathcal {R}_{n}.$

Proof

For $1\le n<\infty$, we have

$$\begin{aligned} \gamma _{n}\circ \eta _{n}\left( f\right) \left( \omega \right) =\gamma _{n}\left( \frac{1}{n}\sum _{k=1}^{n}f\left( \omega ,\frac{k}{n}\right) \right) I=\frac{1}{n}\sum _{k=1}^{n}\int _{\Omega _{n}}^{\oplus }f\left( \omega ,\frac{k}{n}\right) Id\mu _{n}\left( \omega \right) , \end{aligned}$$

and

$$\begin{aligned}{} & {} \Phi _{n}\left( \pi _{\mathcal {A}_{n}}\left( f\right) \right) =\Phi _{n}\left( \int _{\Omega _{n}}^{\oplus }U_{\omega }^{*}\left( \begin{array}{ccc} f\left( \omega ,\frac{1}{n}\right) &{} &{} \\ &{} \ddots &{} \\ &{} &{} f\left( \omega ,\frac{n}{n}\right) \end{array} \right) U_{\omega }d\mu _{n}\left( \omega \right) \right) \\{} & {} \quad =\int _{\Omega _{n}}^{\oplus }\tau _{n}\left( U_{\omega }^{*}\left( \begin{array}{ccc} f\left( \omega ,\frac{1}{n}\right) &{} &{} \\ &{} \ddots &{} \\ &{} &{} f\left( \omega ,\frac{n}{n}\right) \end{array} \right) U_{\omega }\right) d\mu _{n}\left( \omega \right) \\{} & {} \quad =\int _{\Omega _{n}}^{\oplus }\frac{1}{n}\sum _{k=1}^{n}f\left( \omega ,\frac{k}{n}\right) Id\mu _{n}\left( \omega \right) . \end{aligned}$$

Thus the diagram commutes. For $n=\infty$, by Lemma 24, we know there exists an operator $T=\int _{\Omega _{\infty }}^{\oplus }T_{\omega } d\mu _{\infty }\left( \omega \right)$ such that $T_{\omega }$ generates $\mathcal {A}_{\infty }\left( \omega \right)$ in weak operator topology with $0\le T_{\omega }\le 1$ and $\tau _{\omega ,\infty }\left( T_{\omega } ^{n}\right) =\frac{1}{n+1}$ for $n\ge 1$. The map $\pi _{\mathcal {A}_{\infty } }:L^{\infty }\left( \delta _{\infty }\right) \rightarrow W^{*}\left( T\right) =\mathcal {A}_{\infty }$ is defined by $\pi _{\mathcal {A}_{\infty } }\left( f\right) =\int _{\Omega _{\infty }}^{\oplus }f_{\omega }\left( T_{\omega }\right) d\mu _{\infty }\left( \omega \right) .$ Thus $\gamma _{\infty }\circ \eta _{\infty }\left( f\right) \left( \omega \right) =\left[ \int _{J_{n}}f\left( \omega ,t\right) d\delta _{n}\left( t\right) \right] I$ and $\Phi _{\infty }\circ \pi _{\mathcal {A}_{\infty }}\left( f\right) \left( \omega \right) =\tau _{\omega ,\infty }\left( f_{\omega }\left( T_{\omega }\right) \right) I=\left[ \int _{J_{n}}f\left( \omega ,t\right) d\delta _{n}\left( t\right) \right] I.$ Therefore the diagram commutes. $\square$

Combining all of these results we obtain Theorem 6.

And we also have the following corollary.

Corollary 6

If $\mathcal {A}$ and $\mathcal {B}$ are masas in $\mathcal {R}$, then the tracial embeddings $\pi _{\mathcal {A}},\pi _{\mathcal {B}}$ are approximately unitarily equivalent in $\mathcal {R}$.

Proof

If $\mathcal {A}$ and $\mathcal {B}$ are masas in $\mathcal {R}$, then there are tracial embeddings $\pi _{\mathcal {A}}$ and $\pi _{\mathcal {B}}$ as in Theorem 6. Thus $\Phi \circ \pi _{\mathcal {A}}=\Phi \circ \pi _{\mathcal {B}}$. By Theorem 4, we have $\pi _{\mathcal {A}}$ and $\pi _{\mathcal {B}}$ are approximately unitarily equivalent in $\mathcal {R}$. $\square$

4 Measure-preserving transformations

4.1 Basic facts

A Borel measurable map $\sigma :\left[ 0,1\right] \rightarrow \left[ 0,1\right]$ is measure-preserving if and only if, for every Borel set $E\subseteq \left[ 0,1\right]$,

$$\begin{aligned} \delta _{\infty }\left( \sigma ^{-1}\left( E\right) \right) =\delta _{\infty }\left( E\right) . \end{aligned}$$

We say that $\sigma :\left[ 0,1\right] \rightarrow \left[ 0,1\right]$ is an invertible measure-preserving map if there are measure-preserving measurable maps $\sigma _{1},\sigma _{2}:\left[ 0,1\right] \rightarrow \left[ 0,1\right]$ such that

$$\begin{aligned} \left( \sigma \circ \sigma _{1}\right) \left( x\right) =x\text { and }\left( \sigma _{2}\circ \sigma \right) \left( x\right) =x,\text { almost everywhere }\left( \delta _{\infty }\right) . \end{aligned}$$

In this case, let $E=\left\{ y\in J_{\infty }:\sigma \circ \sigma _{1}\left( y\right) \ne y\text { or }\sigma _{2}\circ \sigma \left( y\right) \ne y\right\}$ and let S be the semigroup generated by $\sigma ,\sigma _{1},\sigma _{2},id_{\left[ 0,1\right] }$. Then S is countable, thus denoted by $S=\left\{ \widehat{\sigma }_{n}:n\in \mathbb {N}\right\}$. Suppose $F=\left( \underset{n\in \mathbb {N}}{\cup }\widehat{\sigma }_{n}\left( E\right) \right) \cup \left( \underset{n\in \mathbb {N}}{\cup }\widehat{\sigma }_{n}^{-1}\left( E\right) \right) ,$ it follows that $\delta _{\infty }\left( F\right) =0$. and $\sigma \left( F\right) =\sigma _{1}\left( F\right) =\sigma _{2}\left( F\right) =F.$ Therefore, on $J_{\infty }\backslash F,$ the maps $\sigma ,\sigma _{1},\sigma _{2}:J_{\infty }\backslash F\rightarrow J_{\infty }\backslash F$ are bijective, also $\sigma \circ \sigma _{1}=\sigma _{2}\circ \sigma .$ Define $\widetilde{\sigma }$ on $J_{\infty }$ by

$$\begin{aligned} \widetilde{\sigma }\left( y\right) =\left\{ \begin{array}{c} \sigma \left( y\right) \\ y \end{array} \right. \begin{array}{c} y\in J_{\infty }/F\\ y\in F \end{array} \end{aligned}$$

Then $\widetilde{\sigma },$ $\widetilde{\sigma }^{-1}$ are bijective, measurable, and $\widetilde{\sigma }=\sigma$ a.e.$\left( \delta _{\infty }\right) .$ We can change $\sigma$ and $\sigma _{1},\sigma _{2}$ on sets of measure 0 so that $\sigma :J_{\infty }\rightarrow J_{\infty }$ is bijective and $\sigma _{1}=\sigma _{2}=\sigma ^{-1}$ a.e.$\left( \delta _{\infty }\right)$. In the following sections, whenever we talk about an invertible measure-preserving transformation $\sigma$ on $J_{\infty }$, we will mean a bijective map $\sigma :J_{\infty }\rightarrow J_{\infty }$ such that $\sigma$ and $\sigma ^{-1}$ are measurable and measure-preserving.

Let $\mathbb{M}\mathbb{P}\left[ 0,1\right] =$

$$\begin{aligned} \left\{ \sigma |\sigma :\left[ 0,1\right] \rightarrow \left[ 0,1\right] \text { is an invertible measurable preserving transformation}\right\} . \end{aligned}$$

Clearly $\left( \mathbb{M}\mathbb{P}\left[ 0,1\right] ,\circ \right)$ is a group.

Let $\mathcal {V}$ be all unitaries U in $\mathcal {U}\left( B\left( L^{2}\left( \left[ 0,1\right] \right) \right) \right)$ with $U\left( 1\right) =1$, and for all $f,g\in L^{\infty }\left[ 0,1\right]$, $U\left( fg\right) =U\left( f\right) U\left( g\right) .$

Lemma 27

$\mathcal {V}$ is $*$-SOT closed.

Proof

Suppose $\left\{ U_{n}\right\} \subseteq \mathcal {V},$ and $U_{n} \overset{SOT}{\rightarrow }U$, $U_{n}^{*}\overset{SOT}{\rightarrow }U^{*}.$ It is easy to see $U^{*}U=UU^{*}=1$ and $U\left( 1\right) =1.$ And we know that $U_{n}\overset{SOT}{\rightarrow }U$ if and only if

$$\begin{aligned} \overline{sp}\left\{ f\in L^{2}\left[ 0,1\right] :\left\| U_{n} f-Uf\right\| _{2}^{2}\rightarrow 0\right\} =L^{2}\left[ 0,1\right] . \end{aligned}$$

Thus there exists a subsequence $\left\{ U_{n_{k}}\right\}$ such that for all $f,g\in L^{\infty }\left[ 0,1\right] ,$ $Ufg={\lim }_{k\rightarrow \infty }U_{n_{k}}\left( fg\right) ={\lim }_{k\rightarrow \infty }\left( U_{n_{k}}f\right) \left( U_{n_{k}}g\right) =UfUg$, thus $U\in \mathcal {V}$. $\square$

Corollary 7

$\mathcal {V}$ is a complete separable, metric space in the $*$-SOT.

Proof

Since $\mathcal {V}$ is a $*$-SOT closed subset of $\mathcal {U}\left( B\left( L^{2}\left[ 0,1\right] \right) \right)$ and $\mathcal {U}\left( B\left( L^{2}\left[ 0,1\right] \right) \right)$ is a complete separable metric space. It follows that $\mathcal {V}$ is a complete separable metric space. $\square$

Lemma 28

There exists a group isomorphism $\sigma \rightarrow U_{\sigma }$ from $\mathbb{M}\mathbb{P}\left[ 0,1\right]$ onto $\mathcal {V}.$

Proof

If $\sigma \in \mathbb{M}\mathbb{P}\left[ 0,1\right] ,$ define $U_{\sigma }:L^{2}\left[ 0,1\right] \rightarrow L^{2}\left[ 0,1\right]$ by $U_{\sigma } f=f\circ \sigma ^{-1}.$ Since, for every $f\in L^{2}\left[ 0,1\right] ,$

$$\begin{aligned} \left\| U_{\sigma }f\right\| _{2}^{2}=\int _{Y}\left( f\circ \sigma ^{-1}\right) ^{2}d\delta _{\infty }=\int _{Y}\left| f\right| ^{2} \circ \sigma ^{-1}d\delta _{\infty }=\int _{Y}\left| f\right| ^{2} d\delta _{\infty }=\left\| f\right\| _{2}^{2}, \end{aligned}$$

$U_{\sigma }$ is an isometry. Since $U_{\sigma ^{-1}}=U_{\sigma }^{-1}$, $U_{\sigma }$ is unitary. Also $U_{\sigma }\left( fg\right) =\left( fg\right) \circ \sigma =\left( f\circ \sigma \right) \left( g\circ \sigma \right) =\left( U_{\sigma }f\right) \left( U_{\sigma }g\right)$ when $f,g\in L^{\infty }\left[ 0,1\right] .$ Thus $U_{\sigma }\in \mathcal {V}.$

To prove that the map $\sigma \rightarrow U_{\sigma }$ is onto, we suppose $U\in \mathcal {V}.$ Define $x\in L^{2}\left[ 0,1\right]$ by $x\left( t\right) =t,$ and define $\gamma =U\left( x\right)$. We will show that $\gamma \in \mathbb{M}\mathbb{P}\left[ 0,1\right]$. Then $U\left( x^{n}\right) =\gamma ^{n}$ for all $n\ge 1.$ Thus

$$\begin{aligned} \left\| \gamma \right\| _{\infty }&=\lim \left\| \gamma \right\| _{2^{n}}=\lim \left[ \left\| \gamma ^{2^{n-1}}\right\| _{2}\right] ^{1/2^{n-1}}=\lim \left[ \left\| Ux^{2^{n-1}}\right\| _{2}\right] \\&=\left[ \left\| x^{2^{n-1}}\right\| _{2}\right] ^{1/2^{n-1} }=\left\| x\right\| _{\infty }=1. \end{aligned}$$

Also if $\gamma =u+iv$, then

$$\begin{aligned} 4\int v^{2}d\delta _{\infty }&=\int \left\| \gamma -\bar{\gamma }\right\| _{2}^{2}d\delta _{\infty }=\left\| \gamma \right\| _{2}+\left\| \bar{\gamma }\right\| _{2}^{2}-2{\text {Re}}\left\langle \gamma ,\bar{\gamma }\right\rangle \\&=2\left\| \gamma \right\| _{2}^{2}-2\left\langle \gamma ^{2} ,1\right\rangle =2\left\| x\right\| _{2}^{2}-2\int x^{2}d\delta _{\infty }=0. \end{aligned}$$

Thus $\gamma =\bar{\gamma }$. Since

$$\begin{aligned} \int _{0}^{1}\gamma ^{n}d\delta _{\infty }=\int _{0}^{1}x^{n}d\delta _{\infty } =\frac{1}{n+1} \end{aligned}$$

for each $n\ge 1$. It follows from Corollary 5, using $\tau \left( f\right) =\int _{0}^{1}fd\delta _{\infty }$, that $0\le \gamma \le 1$. And the map $\pi \left( f\right) =f\circ \gamma$ is a weak*-continuous automorphism on $L^{\infty }\left( \left[ 0,1\right] \right)$ such that, for every $f\in L^{\infty }\left[ 0,1\right] ,$

$$\begin{aligned} \int _{0}^{1}fd\delta _{\infty }=\tau \left( \pi \left( f\right) \right) =\int _{0}^{1}f\circ \gamma d\delta _{\infty }. \end{aligned}$$

Thus

$$\begin{aligned} \delta _{\infty }\left( \gamma ^{-1}\left( E\right) \right) =\int _{0}^{1} \chi _{E}\circ \gamma d\delta _{\infty }=\delta _{\infty }\left( E\right) . \end{aligned}$$

Hence $\gamma$ is a measure-preserving transformation on $\left[ 0,1\right]$. Furthermore, $U_{\gamma }f=f\circ \gamma$ is an isometry on $L^{2}\left( \left[ 0,1\right] \right)$ and equals U on the dense subset of polynomials. Thus $U=U_{\gamma }$. Since $U_{\gamma }$ is unitary, $\gamma \in \mathbb{M}\mathbb{P}\left[ 0,1\right]$.

Since $\mathcal {V}$ is closed in the $*$-strong operator topology ($*$-SOT), and the closed unit ball of $B\left( L^{2}\left[ 0,1\right] \right)$ is a $*$-SOT complete metric space, we know that $\mathbb{M}\mathbb{P} \left[ 0,1\right]$ is a complete separable metric space with the topology $\gamma _{n}\rightarrow \gamma$ if and only if $U_{\gamma _{n}}\rightarrow U_{\gamma }$ in the $*$-SOT. On $\mathbb{M}\mathbb{P}\left[ 0,1\right]$ this topology is called the weak topology. [14] The metric for the unit ball of $B\left( L^{2}\left[ 0,1\right] \right)$ is rather complicated.

For $\mathbb{M}\mathbb{P}\left[ 0,1\right]$ we have a simpler metric. $\square$

Lemma 29

$\mathbb{M}\mathbb{P}\left[ 0,1\right]$ is a complete separable metric space with the metric d on $\mathbb{M}\mathbb{P}\left[ 0,1\right]$ defined by

$$\begin{aligned} d\left( \gamma _{1},\gamma _{2}\right) =\left\| \gamma _{1}-\gamma _{2}\right\| _{2}+\left\| \gamma _{1}^{-1}-\gamma _{2}^{-1}\right\| _{2} \end{aligned}$$

Proof

Suppose $d\left( \gamma _{n},r\right) \rightarrow 0$, then $\left\| \gamma _{n}-\gamma \right\| _{2}\rightarrow 0$ and $\left\| \gamma _{n} ^{-1}-\gamma ^{-1}\right\| _{2}\rightarrow 0.$ Thus

$\left\| \gamma _{n}^{k}-\gamma ^{k}\right\| _{2}\rightarrow 0$ and $\left\| \left( \gamma _{n}^{-1}\right) ^{k}-\left( \gamma ^{-1}\right) ^{k}\right\| _{2}\rightarrow 0$ for every $k\ge 0.$ Thus

$\left\| U_{\gamma _{n} }x^{k}-U_{\gamma }x^{k}\right\| _{2}\rightarrow 0$ which implies $U_{\gamma _{n}}\rightarrow U$ in SOT and $U_{\gamma _{n}}^{*}=U_{\gamma _{n}^{-1}}\rightarrow U_{\gamma ^{-1}}=U_{\gamma }^{*}$ in SOT. The converse is obvious. To prove completeness, a similar argument to the one above shows that if $\left\{ \gamma _{n}\right\}$ is d-Cauchy, then $\left\{ U_{\gamma _{n}}\right\}$ is $*$-SOT Cauchy, so there is a $\gamma \in \mathbb{M}\mathbb{P}\left[ 0,1\right]$ such that $U_{\gamma _{n}}\rightarrow U_{\gamma }$ in the $*$-SOT. Hence $\gamma _{n}\rightarrow \gamma$ in d. $\square$

We now turn to our measure space $\left( \Lambda ,\lambda \right)$. We want to describe a subgroup $\mathbb {G}\left( \mathcal {R}\right)$ of $\mathbb{M}\mathbb{P}\left( \Lambda ,\lambda \right)$.

Definition 6

Suppose $\sigma \in \mathbb{M}\mathbb{P}\left( \Lambda _{n},\lambda _{n}\right)$. Then $\sigma \in \mathbb {G}_{n}\left( \mathcal {R}\right)$ if and only if, for every measurable $E\subset \Omega _{n}$,

$$\begin{aligned} \sigma \left( E\times J_{n}\right) \subset E\times J_{n}\text {, a.e.,} \end{aligned}$$

i.e.,

$$\begin{aligned} \lambda _{n}\left( \sigma \left( E\times J_{n}\right) \backslash \left( E\times J_{n}\right) \right) =0. \end{aligned}$$

Since it is known that

$$\begin{aligned} \sigma \left( \left( \Omega _{n}\backslash E\right) \times J_{n}\right) \subset \left( \Omega _{n}\backslash E\right) \times J_{n}\text {, a.e.,} \end{aligned}$$

it follows that

$$\begin{aligned} \sigma \left( E\times J_{n}\right) =E\times J_{n}\text {, a.e..} \end{aligned}$$

This implies that $\sigma ^{-1}\in \mathbb {G}_{n}\left( \mathcal {R}\right)$. Clearly, $\mathbb {G}_{n}\left( \mathcal {R}\right)$ is a subgroup of $\mathbb{M}\mathbb{P}\left( \Lambda _{n},\gamma _{n}\right)$.

Definition 7

We define $\mathbb {G}\left( \mathcal {R}\right)$ to be all $\sigma \in \mathbb{M}\mathbb{P}\left( \Lambda ,\lambda \right)$ such that, for $1\le n\le \infty ,$ $\sigma \left( \Lambda _{n}\right) =\Lambda _{n}$ and $\sigma |_{\Lambda _{n}}\in \mathbb {G}_{n}\left( \mathcal {R}\right)$. We see that we can view

$$\begin{aligned} \mathbb {G}\left( \mathcal {R}\right) =\underset{1\le n\le \infty }{\Pi }\mathbb {G}_{n}\left( \mathcal {R}\right) \text {,} \end{aligned}$$

as a product space.

We can express the following lemma as:

$$\begin{aligned} \mathbb {G}\left( \mathcal {R}\right) =\sum _{1\le n\le \infty }^{\oplus } \int _{\Omega _{n}}^{\oplus }\mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right) d\mu _{n}\left( \omega \right) \subset \mathbb{M}\mathbb{P}\left( \Lambda ,\lambda \right) . \end{aligned}$$

In other words, every element of $\mathbb {G}\left( \mathcal {R}\right)$ is a direct integral of invertible measure preserving transformations.

Lemma 30

Suppose $\sigma \in \mathbb {G}_{n}$, $1\le n\le \infty$. Then there is a measurable family $\left\{ \sigma _{\omega }:\omega \in \Omega _{n}\right\}$ in $\mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right)$ such that, for every $f\in L^{\infty }\left( \Lambda _{n},\right)$

$$\begin{aligned} \left( f\circ \sigma \right) \left( \omega ,t\right) =f\left( \omega ,\sigma _{\omega }\left( t\right) \right) . \end{aligned}$$

We write this as

$$\begin{aligned} \sigma =\int _{\Omega _{n}}^{\oplus }\sigma _{\omega }d\mu _{n}\left( \omega \right) \text {.} \end{aligned}$$

Proof

We can view $L^{2}\left( \Lambda _{n},\lambda _{n}\right) =L^{2}\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right)$ as

$$\begin{aligned} \int _{\Omega _{n}}^{\oplus }L^{2}\left( J_{n},\delta _{n}\right) d\mu _{n}\left( \omega \right) \end{aligned}$$

by identifying $f\in$ $L^{2}\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right)$ with

$$\begin{aligned} \int _{\Omega _{n}}^{\oplus }f_{\omega }d\mu _{n}\left( \omega \right) , \end{aligned}$$

where $f_{\omega }\left( t\right) =f\left( \omega ,t\right) .$ Fubini’s theorem shows that this is an isomorphism, i.e.,

$$\begin{aligned} \left\| f\right\| _{2}^{2}=\int _{\Omega _{n}\times J_{n}}\left| f\left( \omega ,t\right) \right| ^{2}d\left( \mu _{n}\times \delta _{n}\right) =\int _{\Omega _{n}}\int _{J_{n}}\left| f_{\omega }\right| ^{2}d\delta _{n}\left( t\right) =\int _{\Omega _{n}}\left\| f_{\omega }\right\| ^{2}d\mu _{n}\left( \omega \right) . \end{aligned}$$

Clearly, $U\left( f\right) =f\circ \sigma$ is a unitary operator on $L^{2}\left( \Lambda _{n},\lambda _{n}\right)$. Suppose $E\subset \Omega _{n}$ is measurable. Then

$$\begin{aligned} P_{E}\underset{{\text {def}}}{=}\int _{\Omega _{n}}^{\oplus }\chi _{E}\left( \omega \right) 1d\mu \left( \omega \right) \in \int _{\Omega _{n} }^{\oplus }B\left( L^{2}\left( J_{n},\delta _{n}\right) \right) d\mu _{n}\left( \omega \right) , \end{aligned}$$

and the definition of $\sigma ^{-1}\in \mathbb {G}_{n}\left( \mathcal {R}\right)$ implies that $P_{E}U=UP_{E}$. Since the linear span of $\left\{ \chi _{E}:E\subset \Omega _{n},{ }E\text { measurable}\right\}$ is dense in $L^{\infty }\left( \Omega _{n},\mu _{n}\right) ,$ we see that U is in the commutant of

$$\begin{aligned} \left\{ \int _{\Omega _{n}}^{\oplus }\varphi \left( \omega \right) 1d\mu _{n}\left( \omega \right) :\varphi \in L^{\infty }\left( \Omega _{n},\mu _{n}\right) \right\} . \end{aligned}$$

Thus there is a measurable family $\left\{ U_{\omega }:\omega \in \Omega _{n}\right\}$ of unitary operators in $B\left( L^{2}\left( J_{n},\delta _{n}\right) \right)$ such that

$$\begin{aligned} U=\int _{\Omega _{n}}^{\oplus }U_{\omega }d\mu \left( \omega \right) . \end{aligned}$$

If $h\in L^{2}\left( J_{n},\delta _{n}\right) ,$ we define $\hat{h}\in L^{2}\left( \Omega _{n}\times J_{n},\mu _{n}\times \delta _{n}\right)$ by

$$\begin{aligned} \hat{h}\left( \omega ,t\right) =h\left( t\right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \hat{h}=\int _{\Omega _{n}}^{\oplus }hd\mu _{n}\left( \omega \right) . \end{aligned}$$

If $h,k\in L^{\infty }\left( J_{n},\delta _{n}\right)$, then $U\left( \hat{h}\hat{k}\right) =U\left( \hat{h}\right) U\left( \hat{k}\right) ,$ so, for almost every $\omega \in \Omega _{n}$,

$$\begin{aligned} U_{\omega }\left( hk\right) =U_{\omega }\left( h\right) U_{\omega }\left( k\right) . \end{aligned}$$

Since $L^{2}\left( J_{n},\delta _{n}\right)$ is separable, there is a countable set $\mathcal {E}$ whose closure in $\left\| \cdot \right\| _{2}$ is

$$\begin{aligned} \left\{ h\in L^{\infty }\left( J_{n},\delta _{n}\right) :\left\| h\right\| _{\infty }\le 1\right\} \end{aligned}$$

(which is $\left\| \cdot \right\| _{2}$-closed). We now have for almost every $\omega \in \Omega _{n}$ and $h,k\in \mathcal {E}$,

$$\begin{aligned} U_{\omega }\left( hk\right) =U_{\omega }\left( h\right) U_{\omega }\left( k\right) . \end{aligned}$$

We can change $U_{\omega }$ on a set of measure 0 and assume that the above relation holds for all $\omega \in \Omega _{n}$. Suppose $h,g\in L^{\infty }\left( J_{n},\delta _{n}\right)$ and $\left\| h\right\| _{\infty },\left\| g\right\| _{\infty }\le 1$ and suppose $\omega \in \Omega _{n}$. We can choose sequences $\left\{ h_{k}\right\}$ and $\left\{ g_{k}\right\}$ in $\mathcal {E}$ such that $\left\| h_{k}-h\right\| _{2}\rightarrow 0$ and $\left\| g_{k}-g\right\| _{2}\rightarrow 0$. By replacing these sequences with appropriate subsequences, we can assume that $h_{k}\left( t\right) \rightarrow h\left( t\right) ,$ $\left( U_{\omega }h_{k}\right) \left( t\right) \rightarrow \left( U_{\omega }h\right) \left( t\right)$, $g_{k}\left( t\right) \rightarrow g\left( t\right) ,$ $\left( U_{\omega }g_{k}\right) \left( t\right) \rightarrow \left( U_{\omega }g\right) \left( t\right)$ a.e. $\left( \delta _{n}\right)$. It follows that $\left\| h_{k}g_{k}-hg\right\| _{2}\rightarrow 0$. Thus

$$\begin{aligned}{} & {} U_{\omega }\left( hg\right) \left( t\right) =\lim _{k\rightarrow \infty }U_{\omega }\left( h_{k}g_{k}\right) \left( t\right) =\lim _{k\rightarrow \infty }\left( U_{\omega }h_{k}\right) \left( t\right) \left( U_{\omega }g_{k}\right) \left( t\right) \\{} & {} \quad =\left( U_{\omega }h\right) \left( t\right) \left( U_{\omega }g\right) \left( t\right) . \end{aligned}$$

It follows from Lemma 28 that, for each $\omega \in \Omega _{n}$, there is a (unique) $\sigma _{\omega }\in \mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right)$ such that, for every $h\in L^{2}\left( J_{n},\delta _{n}\right)$,

$$\begin{aligned} U_{\omega }h=h\circ \sigma _{\omega }\text {.} \end{aligned}$$

Our measurable cross-section theorems can be used to show that there is a measurable choice of the $\sigma _{\omega }$’s, but the uniqueness implies that the map $\omega \mapsto \sigma _{\omega }$ is already measurable on $\Omega _{n}$. $\square$

4.2 Nonincreasing rearrangement functions, s-functions, and Ky Fan functions

Theorem 7

Suppose $f:\Lambda \rightarrow [0,\infty )$ is measurable. Then there is a $\sigma \in \mathbb {G}\left( \mathcal {R}\right)$ such that, for $1\le n\le \infty$, the mapping $t\mapsto \left( f\circ \sigma \right) \left( \omega ,t\right)$ is nonincreasing on $J_{n}$ a.e. $\left( \mu _{n}\right)$.

Proof

Choose $R>\left\| f\right\| _{\infty }$. Suppose $1\le n\le \infty$. Let

$$\begin{aligned} \mathcal {X}=\left\{ \left( h,\sigma \right) \in L^{\infty }\left( \delta _{n}\right) \times \mathbb{M}\mathbb{P}\left( J_{n}\right) :0\le h\le R,h\circ \sigma \text { is nonincreasing on }J_{n}\right\} , \end{aligned}$$

where $\left\{ f:0\le f\le R\right\}$ is given the $\left\| \cdot \right\| _{2,\delta _{n}}$-topology, $\mathbb{M}\mathbb{P}\left( J_{n}\right)$ is given the weak topology, and $L^{\infty }\left( \delta _{n}\right) \times \mathbb{M}\mathbb{P}\left( J_{n}\right)$ is given the product topology. (Note that if $n<\infty$, $\mathbb{M}\mathbb{P}\left( J_{n}\right)$ corresponds to the set of $n\times n$ permutation matrices and has the discrete topology.) Since $\left\| \cdot \right\| _{2}$ convergence implies subsequential convergence almost everywhere, it follows that $\mathcal {X}$ is a complete separable metric space. Since every measurable h has a nonincreasing rearrangement, the map

$$\begin{aligned} \pi _{1}:\mathcal {X}\rightarrow \left\{ h:0\le h\le R\right\} \end{aligned}$$

is onto, so, by Lemma 5, there is an absolutely measurable cross-section $\gamma _{n}:Y\rightarrow \mathcal {X}$ for $\pi _{1}$. Let $\eta _{n}=\pi _{2}\circ \gamma _{n}:Y\rightarrow \mathbb{M}\mathbb{P}\left( J_{n}\right)$.

We now define $s_{n}:\Omega _{n}\rightarrow \mathbb{M}\mathbb{P}\left( J_{n}\right)$ by

$$\begin{aligned} s_{n}\left( \omega \right) =\eta _{n}\left( f_{\omega }\right) \in \mathbb{M}\mathbb{P}\left( J_{n}\right) . \end{aligned}$$

It is clear from the construction that that $f_{\omega }\circ s_{n}\left( \omega \right)$ is a nonincreasing function of t, i.e., $f\left( \omega ,s_{n}\left( \omega \right) \left( t\right) \right)$ is a nonincreasing function of t for each $\omega \in \Omega _{n}$.

We define

$$\begin{aligned} \sigma _{n}\left( \omega ,t\right) =\left( \omega ,s_{n}\left( \omega \right) \left( t\right) \right) . \end{aligned}$$

Then $\sigma =\left\{ \sigma _{n}\right\} _{1\le n\le \infty }\in$ $\mathbb {G}\left( \mathcal {R}\right)$ has the desired properties. $\square$

Note that the function $\sigma$ is not necessarily unique, but the function $f\circ \sigma$ is unique. It is called the nonincreasing rearrangement function for f, and we denote it by $s_{f}$. If f and h are nonnegative measurable functions on $\Lambda$, we say that f and h are $\mathbb {G} \left( \mathcal {R}\right)$-equivalent if and only if $s_{f}=s_{h}$ a.e. $\left( \lambda \right)$. This holds if and only if there is a $\sigma _{1}\in \mathbb {G}\left( \mathcal {R}\right)$ such that $h=f\circ \sigma _{1}$.

For each $\omega \in \Omega _{n}$ and $t\in J_{n}$, $s_{f}\left( \omega ,t\right)$ is call the $t^{th}$  s-number of f  at  $\omega$.

Definition 8

Suppose $T\in \mathcal {R}$. We can write $T=\sum _{1\le n\le \infty } \int _{\Omega _{n}}^{\oplus }T\left( \omega \right) d\mu _{n}\left( \omega \right)$. We define $s_{T}\in L^{\infty }\left( \Lambda ,\lambda \right)$ by

$$\begin{aligned} s_{T}\left( \omega ,t\right) =s_{T\left( \omega \right) }\left( t\right) \end{aligned}$$

when $1\le n\le \infty ,$ $\omega \in \Omega _{n}$ and $t\in J_{n}$.

Definition 9

Suppose $f\in L^{\infty }\left( \Lambda ,\lambda \right)$ and $0\le f$. For each $1\le n\le \infty$, and each $\omega \in \Omega _{n}$, we define $f_{\omega }\in L^{\infty }\left( J_{n},\delta _{n}\right)$ by

$$\begin{aligned} f_{\omega }\left( t\right) =f\left( \omega ,t\right) . \end{aligned}$$

We view

$$\begin{aligned} f=\sum _{1\le n\le \infty }^{\oplus }\int _{\Omega _{n}}^{\oplus }f_{\omega } d\mu _{n}\left( \omega \right) . \end{aligned}$$

We then define $s_{f}\in L^{\infty }\left( \Lambda ,\lambda \right)$ by

$$\begin{aligned} s_{f}\left( \omega ,t\right) =s_{f_{\omega }}\left( t\right) . \end{aligned}$$

Lemma 31

Suppose $0\le f\in L^{\infty }\left( \Lambda ,\lambda \right)$. Then there is a $\sigma \in \mathbb {G}$ such that, $f\circ \sigma =s_{f}$.

Proof

For $1\le n\le \infty$, the map $\omega \mapsto f_{\omega }$ from $\Omega _{n}$ to $L^{\infty }\left( J_{n},\delta _{n}\right)$ is measurable. For each $\omega \in \Omega _{n}$, there is a $\sigma _{\omega }\in \mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right)$ such that $f_{\omega }\circ \sigma _{\omega }=s_{f_{\omega }}$. Using measurable cross-sections, we can choose the $\sigma _{\omega }$’s so that $\left\{ \sigma _{\omega }:\omega \in \Omega \right\}$ is measurable. Thus $\sigma =\sum _{1\le n\le \infty }\int _{\Omega _{n} }^{\oplus }\sigma _{\omega }\in G$ and

$$\begin{aligned} \left( f\circ \sigma \right) \left( \omega ,t\right) =f\left( \omega ,\sigma _{\omega }\left( t\right) \right) =\left( f_{\omega }\circ \sigma _{\omega }\right) \left( t\right) =s_{f_{\omega }}\left( t\right) =s_{f}\left( \omega ,t\right) . \end{aligned}$$

$\square$

Lemma 32

Suppose $T\in \mathcal {R}$, $\mathcal {A}$ is a masa in $\mathcal {R}$, $\left| T\right| \in \mathcal {A}$, $\pi _{\mathcal {A}}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}$ is a tracial embedding as in Theorem 6, and $f\in L^{\infty }\left( \Lambda ,\lambda \right)$ satisfies $\pi _{\mathcal {A} }\left( f\right) =\left| T\right|$. Then $s_{T}=s_{f}$.

Proof

We can write

$$\begin{aligned} \mathcal {A}=\sum _{1\le n\le \infty }\int _{\Omega _{n}}^{\oplus }\mathcal {A} _{\omega }d\mu _{n}\left( \omega \right) , \end{aligned}$$

where, for $1\le n\le \infty$ and $\omega \in \Omega _{n}$, $A_{\omega }$ is a masa in $\mathcal {R}_{\omega }$. We can also write

$$\begin{aligned} \pi _{\mathcal {A}}=\sum _{1\le n\le \infty }\int _{\Omega _{n}}^{\infty } \pi _{\omega }d\mu _{n}\left( \omega \right) , \end{aligned}$$

where, for each $\omega \in \Omega _{n}$, $\pi _{\omega }:L^{\infty }\left( J_{n},\delta _{n}\right) \rightarrow \mathcal {A}_{\omega }$ is a tracial embedding. If $\pi _{\mathcal {A}}\left( f\right) =\left| T\right|$, then, for almost every $\omega$,

$$\begin{aligned} \pi _{\omega }\left( f_{\omega }\right) =\left| T\right| \left( \omega \right) =\left| T_{\omega }\right| \text {.} \end{aligned}$$

Thus, for almost every $\omega \in \Omega$,

$$\begin{aligned} s_{f_{\omega }}=s_{T_{\omega }}\text {.} \end{aligned}$$

Thus $s_{f}=s_{T}$. $\square$

Lemma 33

Suppose $\mathcal {A}_{1},$ $\mathcal {A}_{2}$ are masas in $\mathcal {R}$, $0\le A_{k}\in \mathcal {A}_{k}$, $\pi _{k}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}_{k}$ are the isomorphisms in Theorem 6 and $f_{1},f_{2}\in L^{\infty }\left( \Lambda ,\lambda \right)$ satisfy $\pi _{k}\left( f_{k}\right) =A_{k}$ for $k=1,2$. The following are equivalent:

(1):

$s_{f_{1}}=s_{f_{2}},$

(2):

There is a $\gamma \in \mathbb {G}\left( \mathcal {R}\right)$ such that $f_{2}=f_{1}\circ \gamma ,$

(3):

There is a sequence $\left\{ U_{n}\right\}$ of unitary operators in $\mathcal {R}$ such that

$$\begin{aligned} \left\| U_{n}A_{1}U_{n}^{*}-A_{2}\right\| \rightarrow 0, \end{aligned}$$

(4):

For every unitarily invariant norm $\alpha$ on $\mathcal {R}$

$$\begin{aligned} \alpha \left( A_{1}\right) =\alpha \left( A_{2}\right) , \end{aligned}$$

(5):

For every rational number $t\in (0,1]$ $KF_{t}\left( A_{1}\right) =KF_{t}\left( A_{2}\right) .$

Proof

$\left( 1\right) \Rightarrow \left( 2\right)$. There are $\gamma _{1},\gamma _{2}\in \mathbb {G}\left( \mathcal {R}\right)$ such $s_{f_{k} }=f_{k}\circ \gamma _{k}$ for $k=1,2$. By $\left( 1\right)$ we have $f_{2}=f_{1}\circ \left( \gamma _{1}\circ \gamma _{2}^{-1}\right)$.

$\left( 2\right) \Rightarrow \left( 3\right)$. Define $\pi _{3}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}_{2}$ by

$$\begin{aligned} \pi _{3}\left( f\right) =\pi _{2}\left( f\circ \gamma \right) . \end{aligned}$$

Thus $\pi _{3}\left( f_{1}\right) =A_{2}$. By Theorem 4, $\pi _{1}\sim _{a}\pi _{3}$. Thus there is a net $\left\{ U_{j}\right\}$ of unitary operators in $\mathcal {R}$ such that

$$\begin{aligned} \lim _{j}\left\| U_{j}A_{1}U_{j}^{*}-A_{2}\right\| =\lim _{j}\left\| U_{j}\pi _{1}\left( f_{1}\right) U_{j}^{*}-\pi _{3}\left( f_{1}\right) \right\| =0. \end{aligned}$$

Hence, for every $n\in \mathbb {N}$, there is a unitary $U_{n}$ such that

$$\begin{aligned} \left\| U_{n}A_{1}U_{n}^{*}-A_{2}\right\| <1/n. \end{aligned}$$

$(3)\Rightarrow \left( 4\right)$, $\left( 4\right) \Rightarrow \left( 5\right)$ are trivial.

$\left( 5\right) \Rightarrow \left( 1\right)$. We know that $KF_{t}\left( A_{1}\right) =KF_{t}\left( s_{f_{1}}\right)$ and $KF_{t}\left( s_{f_{2} }\right)$. Let

$$\begin{aligned} E_{t}=\left\{ \omega \in \Omega :KF_{t}\left( s_{f_{1}}\right) \left( \omega \right) \ne KF_{t}\left( s_{f_{2}}\right) \left( \omega \right) \right\} , \end{aligned}$$

and let $E=\cup E_{t}$, then $\lambda \left( E\right) =0$. Therefore $\int _{0}^{t}f_{1}\left( x\right) dx=\int _{0}^{t}f_{2}\left( x\right) dx$ for every $0<t\le 1.$ Thus $f_{1}\left( x\right) =f_{2}\left( x\right)$ except on a countable set. Therefore $f_{1}=f_{2}$ a.e. $\left( \delta _{\infty }\right) .$ $\square$

Corollary 8

Suppose $\mathcal {A}_{1},$ $\mathcal {A}_{2}$ are masas in $\mathcal {R}$, $0\le A\in \mathcal {A}_{k}$, $\pi _{k}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}_{k}$ are the isomorphisms in Theorem 6 and $f_{1},f_{2}\in L^{\infty }\left( \Lambda ,\lambda \right)$ satisfy $\pi _{k}\left( f_{k}\right) =A$ for $k=1,2$. Then $s_{f_{1} }=s_{f_{2}}.$

If $T\in \mathcal {R}$, we define

$$\begin{aligned} KF_{t}\left( T\right) =KF_{t}\left( s\left( f_{T}\right) \right) \end{aligned}$$

We need to define $t^{th}$ Ky Fan function $KF_{t}\left( T\right)$ solely in terms of T and $\mathcal {R}$. (See Lemma 12.)

Note that when $n=\infty ,$ $KF_{t}$ is defined on $L^{\infty }\left( J_{n},\delta _{n}\right)$ for all $0<t\le 1.$ For $1\le n<\infty$, $KF_{t}$ is only defined when $t\in \left\{ \frac{1}{n},\ldots ,\frac{n}{n}\right\}$. The next definition extends this concept.

Definition 10

Suppose $1\le n<\infty$ and $0<t\le 1$. We choose an integer k,  $1\le k\le n$ such that

$$\begin{aligned} \frac{k-1}{n}<t\le \frac{k}{n}. \end{aligned}$$

We define $KF_{t}$ on $L^{\infty }\left( J_{n},\delta _{n}\right)$ by

$$\begin{aligned} KF_{t}=KF_{\frac{k}{n}}. \end{aligned}$$

For $f\in L^{\infty }\left( \Lambda \right)$ and $1\le n\le \infty$ and $\omega \in \Omega _{n}$ and $t\in J_{n}$, we define

$$\begin{aligned} KF_{t}\left( f\right) \left( \omega ,t\right) =KF_{t}\left( s_{f_{\omega } }\right) , \end{aligned}$$

and we define, for $T\in \mathcal {R}$,

$$\begin{aligned} KF_{t}\left( T\right) =KF_{t}\left( s_{T}\right) \text {.} \end{aligned}$$

We easily have that for $S,T\in \mathcal {R}$

$$\begin{aligned} KF_{t}\left( S+T\right) \le KF_{t}\left( S\right) +KF_{t}\left( T\right) \end{aligned}$$

always holds.

4.3 $\mathbb {G}\left( \mathcal {R}\right)$-symmetric normalized Gauge norms on $L^{\infty }\left( \Lambda ,\lambda \right)$

Suppose $\left( Y,\nu \right)$ is a probability space, and $\mathcal {\mathbb {G}}$ is a subgroup of $\mathbb{M}\mathbb{P}\left( Y,\nu \right)$. A norm $\beta$ on $L^{\infty }\left( Y,\nu \right)$ is called a $\mathbb {G}$-symmetric normalized gauge norm if and only if

• (1) $\beta \left( 1\right) =1$

• (2) $\beta \left( f\right) =\beta \left( \left| f\right| \right)$ for every $f\in L^{\infty }\left( Y,\nu \right)$,

• (3) $\beta \left( f\circ \sigma \right) =\beta \left( f\right)$ for every $f\in L^{\infty }\left( Y,\nu \right)$ and every $\sigma \in \mathcal {\mathbb {G}}$.

The examples that interest us here are for $Y=\Lambda$, $\nu =\lambda$, and $\mathbb {G}=\mathbb {G}\left( \mathcal {R}\right)$, i.e., the $\mathbb {G} \left( \mathcal {R}\right)$-symmetric normalized gauge norms on $L^{\infty }\left( \Lambda ,\lambda \right)$.

Suppose $\beta$ is a $\mathbb {G}\left( \mathcal {R}\right)$-symmetric normalized gauge norm on $L^{\infty }\left( \Lambda ,\lambda \right)$. For every $f\in L^{\infty }\left( \Lambda ,\lambda \right)$, we see that

$$\begin{aligned} \beta \left( f\right) =\beta \left( s_{f}\right) . \end{aligned}$$

4.4 Approximate Ky Fan Lemma

If $T\in \mathcal {R}$, we define

$$\begin{aligned} KF_{t}\left( T\right) =KF_{t}\left( s\left( f_{T}\right) \right) \end{aligned}$$

We can show that $KF_{t}$ satisfies the triangle inequality on $\mathcal {R}$ by describing $KF_{t}\left( T\right)$ directly in terms of T. The Ky Fan Lemma is more complicated. We will apply the Ky Fan Lemmas we have throughout the direct integral. However, this is impossible to do directly as the next examples show.

Example 2

In $\mathbb {C}^{n},$ if $f=\left( 1,0,\ldots ,0\right)$ and $g=\left( \frac{1}{n},\frac{1}{n},\ldots ,\frac{1}{n}\right)$, we have $KF_{\frac{k}{n}}\left( f\right) \ge KF_{\frac{k}{n}}\left( g\right)$ for $1\le k\le n,$ But the number N of permutations $\gamma _{1},\ldots ,\gamma _{N}$ for

$$\begin{aligned} \sum _{j=1}^{N}f\circ \gamma _{j}\ge g \end{aligned}$$

must be at least n since each $f\circ \gamma _{j}$ is nonzero in exactly one coordinate.

Example 3

Suppose $\mathcal {R}=\mathcal {R}_{2}=\mathbb {M}_{2}\left( \mathbb {C}\right) \oplus \mathbb {M}_{2}\left( \mathbb {C}\right)$ and

$$\begin{aligned} A=\sum _{1\le k<\infty }^{\oplus }\left( \begin{array}{cc} 1 &{} \\ &{} 0 \end{array} \right) \end{aligned}$$

and

$$\begin{aligned} B=\sum _{1\le k<\infty }^{\oplus }\left( \begin{array}{cc} \frac{1}{2}+\frac{1}{2^{k}} &{} \\ &{} \frac{1}{2}-\frac{1}{2^{k}} \end{array} \right) \end{aligned}$$

Then there are no $\sigma _{1},\ldots ,\sigma _{N}\in \mathbb {\mathbb {G}}\left( \mathcal {R}\right)$ and $t_{1},\ldots ,t_{N}\in \left[ 0,1\right]$ such that

$$\begin{aligned} \sum _{k=1}^{N}t_{k}\left( s_{A}\circ \sigma _{k}\right) \ge s_{B}\text {.} \end{aligned}$$

This forces us to prove an approximate version of the Ky Fan Lemma that works universally.

Theorem 8

Suppose m is a positive integer. Then, for $1\le n\le \infty$ and for all $0\le f,g\le 1$ in $L^{\infty }\left( J_{n},\delta _{n}\right)$ with

$$\begin{aligned} KF_{t}\left( f\right) \ge KF_{t}\left( g\right) \text { for all }t\in J_{n} \end{aligned}$$

there are $\left\{ \gamma _{j}:1\le j\le m^{2\,m}\right\} \subset \mathbb{M}\mathbb{P}\left( J_{n},\mu _{n}\right)$ such that

$$\begin{aligned} \frac{2}{m}+\frac{1}{m^{2m}}\sum _{j=1}^{m^{2m}}s_{f}\circ \gamma _{j}\ge s_{g}. \end{aligned}$$

Proof

For $1\le n<\infty ,$ it follows from Lemma 2. For $n=\infty ,$ it is proved in Theorem 2. $\square$

Corollary 9

For $1\le n\le \infty ,$ if $KF_{t}\left( f\right) \ge KF_{t}\left( g\right)$ for all $t\in J_{n},$ then $\beta \left( f\right) \ge \beta \left( g\right)$ for all symmetric gauge norms $\beta .$

To prove the approximate Ky Fan Lemma, we need the following Lemmas.

Lemma 34

Suppose m, n are positive integers. $f=\left( f_{1},\cdots ,f_{n}\right) ,h=( h_{1},$$\cdots ,h_{n}),$ where $f_{1},\dots ,f_{n}$ and $h_{1},\dots ,h_{n}$ are integers with $1\le f_{i+1}\le f_{i}\le m,1\le h_{i+1}\le h_{i}\le m.$and $\sum \nolimits _{i=1} ^{k}f_{i}\ge \sum \nolimits _{i=1}^{k}h_{i}$, for $1\le k\le n.$ Then there exists a positive integer $N\le m^{m^{2}}$ and $\gamma _{1},\ldots ,\gamma _{N} \in \mathbb {S}_{n}$ such that

$$\begin{aligned} \frac{1}{N}\sum _{i=1}^{N}f\circ \gamma _{i}\ge h \end{aligned}$$

Proof

Suppose $\mathcal {S=}\left\{ \left( \begin{array}{c} f_{k}\\ h_{k} \end{array} \right) ,1\le k\le n\right\}$, and define an order on $\mathcal {S}$ by

$$\begin{aligned} \left( \begin{array}{c} f_{i}\\ h_{i} \end{array} \right) \ge \left( \begin{array}{c} f_{j}\\ g_{j} \end{array} \right) \text { if }f_{i}>f_{j}\text { or, }f_{i}=f_{j}\text { and }h_{i}\ge h_{j}. \end{aligned}$$

Then $\mathcal {S}$ is a linearly ordered set.

We say $\mathcal {S}$ is trivial if for every $\left( \begin{array}{c} f_{k}\\ h_{k} \end{array} \right) \in \mathcal {S}$, $f_{k}\ge h_{k}$. If $\mathcal {S}$ is trivial, we are done, so we may assume $\mathcal {S}$ is nontrivial. Denote $\mathcal {S} _{0}{=}\mathcal {S}\backslash \left\{ \left( \begin{array}{c} f_{k}\\ f_{k} \end{array} \right) ,f_{k}\in \left\{ 1,\cdots ,m\right\} \right\}$. Define $p\left( \mathcal {S}_{0}\right) =\max \left( f_{k}\right)$, $q\left( \mathcal {S} _{0}\right) =\max \left\{ f_{k}\text {, with }h_{k}>f_{k}\right\}$, where $p\left( \mathcal {S}_{0}\right) ,q\left( \mathcal {S}_{0}\right) \in \left\{ f_{1},\cdots ,f_{n}\right\} ,$ we may assume $p\left( \mathcal {S}_{0}\right) =f_{p},q\left( \mathcal {S}_{0}\right) =f_{q}$. Then denote $l\left( \mathcal {S}_{0}\right) =p\left( \mathcal {S}_{0}\right) -q\left( \mathcal {S}_{0}\right)$. It is not hard to see that $f_{p}>h_{p}\ge h_{q}>f_{q}$, so $f_{p}-f_{q}\ge 2$.

Let $\gamma _{p,q}$ be the permutation that permute $f_{p}$ with $f_{q}$ and leave all other $f_{i}$’s fixed,

define $f^{\left( 1\right) }=\left( f_{1}^{\left( 1\right) },\cdots ,f_{n}^{\left( 1\right) }\right) =\frac{1}{l\left( \mathcal {S}_{0}\right) }\left[ \left( h_{p}-f_{q}\right) f+\left( f_{p}-h_{p}\right) f\circ \gamma _{p,q}\right]$, where $f^{\left( 1\right) }\in \mathbb {N}^{n}$. Then denote $\mathcal {S}^{\left( 1\right) } \mathcal {=}\left\{ \left( \begin{array}{c} f_{k}^{\left( 1\right) }\\ h_{k} \end{array} \right) ,1\le k\le n\right\}$, $\mathcal {S}_{0}^{\left( 1\right) }=\mathcal {S}^{\left( 1\right) }\backslash \left\{ \left( \begin{array}{c} f_{k}^{\left( 1\right) }\\ f_{k}^{\left( 1\right) } \end{array} \right) \right\}$, we form linear convex combination of $f_{i}$’s this way and update f with $f^{\left( 1\right) },\cdots ,f^{\left( r\right) }$ until $l\left( \mathcal {S}_{0}^{\left( r\right) }\right) <l\left( \mathcal {S}_{0}\right)$. We can also see that $l\left( \mathcal {S} _{0}\right) <m$, and $r<m$, so we need at most $m^{m}$ permutations to reduce $l\left( \mathcal {S}_{0}\right)$ for 1. Repeating this process, we need at most $\left( m^{m^{2}}\right)$ permutations to reduce $\mathcal {S}_{0}$ to a trivial set. Note that we can make the number of permutations is exactly $\left( m^{m^{2}}\right) !$, some permutations are duplicate.

Therefore, there exists a positive integer $N=\left( m^{m^{2}}\right) !,\gamma _{1},\cdots ,\gamma _{N}\in \mathbb {S}_{n}$ such that

$$\begin{aligned} \frac{1}{N}\sum _{i=1}^{N}f\circ \gamma _{i}\ge h. \end{aligned}$$

$\square$

Lemma 35

Suppose m, n are positive integers, then there exists a positive integer $N\le m^{m^{2}}$ such that for all $f=\left( f_{1},f_{2},\dots ,f_{n}\right)$ and $h=\left( h_{1},\dots ,h_{n}\right)$ with $1\ge f_{1}\ge \cdots \ge f_{n}\ge 0$, $1\ge h_{1}\ge \cdots \ge h_{n}\ge 0$, and ${\sum }_{j}^{{i=1}}f_{i}\ge {\sum }_{j}^{{i=1}}h_{i}$ for all $1\le j\le n,$ there exist $\gamma _{1},\dots ,\gamma _{N}\in \mathbb {S}_{n}$ such that

$$\begin{aligned} \frac{1}{N}\underset{i=1}{\overset{N}{\sum }}f\circ \gamma _{i}+\frac{2}{m}\ge h. \end{aligned}$$

Proof

For all $1\le i\le n,$ if $\frac{k-1}{m}<f_{i}\le \frac{k}{m}$ for some $k\in \mathbb {N}$, then define $\widetilde{f_{i}}=\frac{k}{m}$ and if $\frac{k-1}{m}\le h_{i}<\frac{k}{m}$ for some $k\in \mathbb {N}$, then define $\tilde{h}_{i}=\frac{k-1}{m}.$ Let $\widetilde{f}=\left( \widetilde{f} _{1},\cdots ,\widetilde{f}_{n}\right)$ and $\widetilde{h}$ $=\left( \widetilde{h}_{1},\cdots ,\widetilde{h}_{n}\right)$. It is easy to check that $f_{i}\le \widetilde{f}_{i}\le f_{i}+\frac{1}{m}$ and $\max \left( h_{i}-\frac{1}{m},0\right) \le \widetilde{h}_{i}\le h_{i}$ for all $1\le i\le n.$ From Lemma 34, we know there exists a positive integer N and $\gamma _{1},\cdots ,\gamma _{N}\in \mathbb {S}_{n}$ such that $\frac{1}{N}\sum _{i=1}^{N}\left( m\widetilde{f}\right) \circ \gamma _{i} \ge \left( m\widetilde{h}\right)$. Therefore, $\frac{1}{N}{{\sum _{j=1}^{N}}}f\circ \gamma _{j}+\frac{2}{m}\ge h.$ $\square$

The following is the Approximate Ky Fan Lemma.

Theorem 9

If $f,g\in L^{\infty }\left( \Lambda ,\lambda \right) ,$ $m\in \mathbb {N}$, $m\ge 2$, and $0\le f,g\le 1$ and $KF_{t}\left( f\right) \ge KF_{t}\left( g\right)$ a.e. $\left( \mu \right)$ for each rational number $t\in (0,1],$ then there are $\sigma _{1},\ldots ,\sigma _{\left( m!\right) ^{m^{5}}m^{m^{2}}}\in \mathbb {G}\left( \mathcal {R}\right)$ such that

$$\begin{aligned} \frac{1}{\left( m!\right) ^{m^{5}}m^{m^{2}}}\sum \nolimits _{k=1}^{\left( m!\right) ^{m^{5}}m^{m^{2}}}f\circ \sigma _{k}+\frac{1}{m}\ge g. \end{aligned}$$

Thus, for every $\mathbb {G}\left( \mathcal {R}\right)$-symmetric normalized gauge norm $\beta$ on $L^{\infty }\left( \Lambda ,\lambda \right) ,$

$$\begin{aligned} \beta \left( f\right) \ge \beta \left( g\right) . \end{aligned}$$

Proof

Suppose $f,g\in L^{\infty }\left( \Lambda ,\lambda \right)$. Since there are $\sigma _{1},\sigma _{2}\in \mathcal {G}\left( \mathcal {R}\right)$ such that $s_{f}=f\circ \sigma _{1}$ and $s_{g}=g\circ \sigma _{2}$, we can assume $f=s_{f}$ and $g=s_{g}$. We know f, g can be viewed as $f=\sum _{1\le n\le \infty }^{\oplus }f_{n}=\sum _{1\le n\le \infty }^{\oplus }\int _{\Omega _{n}}^{\oplus }f_{n,\omega }d\mu _{n}\left( \omega \right)$ and $g=\sum _{1\le n\le \infty }^{\oplus }\int _{\Omega _{n}}^{\oplus }g_{n,\omega }d\mu _{n}\left( \omega \right)$. Suppose $m\in \mathbb {N}$ and $m\ge 2$. For $1\le n\le \infty ,$ let $\mathcal {X}_{n}$ be the set of tuples $\left( F,G,\sigma _{1},\sigma _{2},\cdots ,\sigma _{m^{m^{2}}}\right)$ satisfying $\frac{1}{m^{m^{2}}} \sum \nolimits _{k=1}^{m^{m^{2}}}F\circ \sigma _{k}+\frac{1}{m}\ge G,$ where $0\le f,g\le 1$. Then $\mathcal {X}$ is a closed subset of ball $\left( L^{\infty }\left( J_{n},\delta _{n}\right) \right) \times \textrm{ball}\left( L^{\infty }\left( J_{n},\delta _{n}\right) \right) \times {\Pi }_{i=1}^{{m^{m^{2}}}}\mathbb{M}\mathbb{P}\left( \Lambda ,\lambda \right)$, which is a complete separable metric space with the $\left\| \cdot \right\| _{2}$ on ball$\left( L^{\infty }\left( J_{n},\delta _{n}\right) \right)$. Then by Theorem 5 the projection onto ball$\left( L^{\infty }\left( J_{n},\delta _{n}\right) \right) \times$ball$\left( L^{\infty }\left( J_{n},\delta _{n}\right) \right)$ has an absolutely measurable range $\mathcal {Y}_{n}$ and an absolutely measurable cross-section $\psi$ and we let $\psi _{k}$ be the composition of projection onto the coordinate of $\sigma _{k}$ with $\psi$ for $1\le k\le \left( m^{m^{2}}\right) !$. If $1\le n<\infty ,$ it follows from Lemma 35 and Theorem 2 that

$$\begin{aligned} \left( s_{f_{\omega }},s_{g_{\omega }}\right) \in \mathcal {Y}_{n} \end{aligned}$$

for almost all $\omega \in \Omega _{n}.$ We define, for $1\le k\le \left( m^{m^{2}}\right) !$, $\sigma _{k}\left( \omega \right) \in \mathbb{M}\mathbb{P}\left( J_{n},\delta _{n}\right)$ by

$$\begin{aligned} \sigma _{k}\left( \omega \right) =\psi _{k}\left( s_{f_{\omega }},s_{g_{\omega }}\right) . \end{aligned}$$

This gives $\sigma _{1},\ldots \sigma _{\left( m^{m^{2}}\right) !}\in \mathbb {G}\left( \mathcal {R}\right)$ such that

$$\begin{aligned} \frac{1}{\left( m^{m^{2}}\right) !}\sum \nolimits _{k=1}^{\left( m^{m^{2} }\right) !}s_{f}\circ \sigma _{k}+\frac{1}{m}\ge s_{g}. \end{aligned}$$

If follows that, for any $\mathbb {G}\left( \mathcal {R}\right)$-symmetric normalized gauge norm $\beta$ on $L^{\infty }\left( \Lambda ,\lambda \right)$ that

$$\begin{aligned} \beta \left( g\right)&=\beta \left( s_{g}\right) \le \frac{1}{\left( m^{m^{2}}\right) !}\sum \nolimits _{k=1}^{\left( m^{m^{2}}\right) !} \beta \left( s_{f}\circ \sigma _{k}\right) +\beta \left( \frac{1}{m}\right) \\&=\frac{1}{\left( m^{m^{2}}\right) !}\sum \nolimits _{k=1}^{\left( m^{m^{2}}\right) !}\beta \left( f\right) +\frac{1}{m}=\beta \left( f\right) +\frac{1}{m}. \end{aligned}$$

Since $m\ge 2$ was arbitrary, it follows that $\beta \left( g\right) \le \beta \left( f\right)$. $\square$

5 Main theorem

We are finally ready to prove our main theorem.

Theorem 10

Suppose  $\mathcal {R}$ is a finite von Neumann algebra acting on a separable Hilbert space H. Let the probability space $\left( \Lambda ,\Sigma ,\lambda \right)$ and the group $\mathbb {G}\le \mathbb{M}\mathbb{P} \left( \Lambda ,\Sigma ,\lambda \right)$ be as above. Then there is a natural 1-1 correspondence between the normalized unitarily invariant norms on $\mathcal {R}$ and the normalized $\mathbb {G}$-symmetric gauge norms on $L^{\infty }\left( \Lambda ,\lambda \right)$.

Proof

Suppose $\alpha$ is a normalized unitarily invariant norm on $\mathcal {R}$, choose any masa $\mathcal {A}$ in $\mathcal {R}$, and choose a tracial embedding $\pi _{\mathcal {A}}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}$ as in Theorem 6. Define $\beta _{\alpha }:L^{\infty }\left( \lambda \right) \rightarrow \mathbb {R}$ by

$$\begin{aligned} \beta _{\alpha }\left( f\right) =\alpha \left( \pi _{\mathcal {A}}\left( f\right) \right) , \end{aligned}$$

If $\mathcal {B}$ is another masa in $\mathcal {R}$ and $\pi _{\mathcal {B} }:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {B}$ is as in Theorem 6, we see from Theorem 6 that, if $\Phi :\mathcal {R}\rightarrow \mathcal {Z}\left( \mathcal {R}\right)$ is the center-valued trace on $\mathcal {R}$, then

$$\begin{aligned} \Phi \circ \pi _{\mathcal {A}}=\Phi \circ \pi _{\mathcal {B}}. \end{aligned}$$

Thus, by Theorem 6, $\pi _{\mathcal {A}}$ and $\pi _{\mathcal {B}}$ are approximately equivalent in $\mathcal {R}$. Hence, there is a net $\left\{ U_{j}\right\}$ in $\mathcal {U}\left( \mathcal {R}\right)$ such that, for every $f\in L^{\infty }\left( \Lambda ,\lambda \right) ,$

$$\begin{aligned} \left\| U_{j}^{*}\pi _{\mathcal {A}}\left( f\right) U_{j}-\pi _{\mathcal {B}}\left( f\right) \right\| \rightarrow 0. \end{aligned}$$

It follows from Lemma 2 that, for every $f\in L^{\infty }\left( \Lambda ,\lambda \right)$,

$$\begin{aligned} \alpha \left( \pi _{\mathcal {A}}\left( f\right) \right) =\alpha \left( \pi _{\mathcal {B}}\left( f\right) \right) . \end{aligned}$$

Thus the definition of $\beta _{\alpha }$ is independent of choice of the masa $\mathcal {A}$ and tracial embedding $\pi _{\mathcal {A}}$. It is easy to check that $\beta _{\alpha }$ is norm. To prove $\beta _{\alpha }$ is $\mathbb {G}\left( \mathcal {R}\right)$-symmetric, suppose $\sigma \in \mathbb {G}\left( \mathcal {R}\right) .$ Then, by Lemma 30, there is a measurable family $\left\{ \sigma _{\omega }:\omega \in \Omega \right\}$ with each $\omega \in \Omega _{n}$, such that $\sigma _{\omega }\in \mathbb{M}\mathbb{P}\left( J_{n},\mu _{n}\right)$ and

$$\begin{aligned} \sigma =\int _{\Lambda }^{\oplus }\sigma _{\omega }d\lambda \left( \omega \right) \text {.} \end{aligned}$$

Thus, by Theorem 6,

$$\begin{aligned} \Phi _{n}\left( \pi _{\mathcal {A}}\left( f\circ \sigma \right) \right) =\gamma \circ \eta \left( f\circ \sigma \right) , \end{aligned}$$

but

$$\begin{aligned}{} & {} \eta \left( f\circ \sigma \right) \left( \omega \right) =\int _{J_{n}}\left( f\circ \sigma \right) \left( t,\omega \right) d\delta _{n}\left( t\right) \\{} & {} \quad =\int _{J_{n}}f_{\omega }\left( \sigma _{\omega }\left( t\right) \right) d\delta _{n}\left( t\right) =\int _{J_{n}}f_{\omega }\left( t\right) d\delta _{n}\left( t\right) =\eta \left( f\right) \left( \omega \right) \text {.} \end{aligned}$$

Thus, for every $f\in L^{\infty }\left( \Lambda ,\lambda \right)$,

$$\begin{aligned} \Phi \circ \pi _{\mathcal {A}}\left( f\right) =\Gamma \left( \eta \left( f\right) \right) =\Gamma \left( \eta \left( f\circ \sigma \right) \right) =\Phi \circ \pi _{\mathcal {A}}\left( f\circ \sigma \right) . \end{aligned}$$

Thus, $\rho \left( f\right) =\pi _{\mathcal {A}}\left( f\circ \sigma \right)$ is a tracial embedding as in Theorem 6, which implies $\rho$ is approximately equivalent to $\pi _{\mathcal {A}}$. Hence, by Lemma 2, for every $f\in L^{\infty }\left( \Lambda ,\lambda \right)$, we have

$$\begin{aligned} \beta _{\alpha }\left( f\right) =\alpha \left( \pi _{\mathcal {A}}\left( f\right) \right) =\alpha \left( \pi _{\mathcal {A}}\left( f\circ \sigma \right) \right) =\beta _{\alpha }\left( f\circ \sigma \right) \text {.} \end{aligned}$$

Thus $\beta _{\alpha }$ is a normalized $\mathbb {G}\left( \mathcal {R}\right)$-invariant gauge norm on $L^{\infty }\left( \Lambda ,\lambda \right)$.

Conversely, suppose $\beta$ is a normalized $\mathbb {G}\left( \mathcal {R} \right)$-symmetric gauge norm on $L^{\infty }\left( \Lambda ,\lambda \right)$. If $T\in \mathcal {R}$, then $W^{*}\left( \left| T\right| \right)$ is abelian and is contained in a masa $\mathcal {A}$ of $\mathcal {R}$. By Theorem 6 there is a tracial embedding $\pi _{\mathcal {A}}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {A}$ such that, for every $f\in L^{\infty }\left( \Omega ,\mu \right)$,

$$\begin{aligned} \tau \left( \pi _{\mathcal {A}}\left( f\right) \right) =\int _{\Omega }fd\mu . \end{aligned}$$

Choose $0\le f\in L^{\infty }\left( \Lambda ,\lambda \right)$ with $\pi _{\mathcal {A}}\left( f\right) =\left| T\right| .$ Then we define

$$\begin{aligned} \alpha _{\beta }\left( T\right) =\beta \left( f\right) =\beta \left( \pi _{\mathcal {A}}^{-1}\left( \left| T\right| \right) \right) . \end{aligned}$$

Suppose $\mathcal {B}$ is another masa in $\mathcal {R}$ with $\left| T\right| \in \mathcal {B}$. Then there is a tracial embedding $\pi _{\mathcal {B}}:L^{\infty }\left( \Lambda ,\lambda \right) \rightarrow \mathcal {B}$ and an $0\le h\in L^{\infty }\left( \Lambda ,\lambda \right)$ with $\pi _{\mathcal {B}}\left( h\right) =\left| T\right|$. It follows from Lemma 32 that

$$\begin{aligned} s_{f}=s_{T}=s_{h}. \end{aligned}$$

Hence, by Lemma 33, there is a $\sigma \in \mathbb {G}\left( \mathcal {R}\right)$ such that

$$\begin{aligned} h=f\circ \sigma \text {.} \end{aligned}$$

Thus

$$\begin{aligned} \alpha \left( h\right) =\alpha \left( f\right) =\alpha \left( s_{T}\right) . \end{aligned}$$

Thus the definition of $\alpha _{\beta }\left( T\right) =\beta \left( s_{T}\right)$ is independent of the masa $\mathcal {A}$ or the tracial embedding $\pi _{\mathcal {A}}$. At this point it is easy to see that $\beta _{\alpha _{\beta }}=\beta$ holds for a $\mathbb {G}\left( \mathcal {R} \right)$-symmetric normalized gauge norm on $L^{\infty }\left( \Lambda ,\lambda \right)$.

If U and V are unitaries in $\mathcal {R}$, then, by Lemma 32,

$$\begin{aligned} s_{UTV}=s_{T}\text {.} \end{aligned}$$

Thus $\alpha _{\beta }\left( UTV\right) =\alpha _{\beta }\left( T\right)$ by Lemma 33. Thus $\alpha _{\beta }$ is unitarily invariant.

Clearly, $\alpha _{\beta }\left( 1\right) =1$ and $\alpha _{\beta }\left( zT\right) =\left| z\right| \alpha _{\beta }\left( T\right)$. To show $\alpha _{\beta }$ is a norm, we just need to check the triangle inequality. Suppose $A,B\in \mathcal {R}$. Let $h=s_{A}+s_{B}$. Since, for almost every $\omega \in \Omega$ the functions $s_{A}\left( \omega ,t\right)$ and $s_{B}\left( \omega ,t\right)$ are nonincreasing in t, we see that

$$\begin{aligned} s_{h}=h=s_{A}+s_{B}. \end{aligned}$$

Thus, we have, if $\omega \in \Omega _{n},$ $n\in \mathbb {N}$, and $t=k/n$ with $1\le k\le n$, or if $\omega \in \Omega _{\infty }$ and $0<t\le 1$ is rational, then, for almost every $\omega ,$

$$\begin{aligned}{} & {} KF_{t}\left( s_{h}\right) \left( \omega \right) =KF_{t}\left( s_{A} +s_{B}\right) \left( \omega \right) =KF_{t}\left( s_{A}\right) \left( \omega \right) +KF_{t}\left( s_{B}\right) \left( \omega \right) \\{} & {} \quad =KF_{t}\left( A\right) \left( \omega \right) +KF_{t}\left( B\right) \left( \omega \right) \ge KF_{t}\left( A+B\right) \left( \omega \right) =KF_{t}\left( s_{A+B}\right) \left( \omega \right) . \end{aligned}$$

It follows from the approximate Ky Fan Lemma (Theorem 9) that

$$\begin{aligned} \beta \left( h\right) \ge \beta \left( s_{A+B}\right) , \end{aligned}$$

which means

$$\begin{aligned} \alpha _{\beta }\left( A+B\right) \le \beta \left( h\right) =\beta \left( s_{A}+s_{B}\right) \le \beta \left( s_{A}\right) +\beta \left( s_{B}\right) =\alpha _{\beta }\left( A\right) +\alpha _{\beta }\left( B\right) . \end{aligned}$$

This completes the proof. $\square$