Taken from C*-algebras and Operator Theory by Gerald Murphy.

A useful way of thinking of the theory of C*-algebras is as “non-commutative topology”. This is justified by the correspondence between abelian C*-algebras and locally compact Hausdorff spaces given by the Gelfand representation. The algebras studied in this chapter, von Neumann algebras, are a class of C*-algebras whose study can be thought of as “non-commutative measure theory”. The reason for the analogy in this case is that the abelian von Neumann algebras are (up to isomorphism) of the form $L^\infty(\Omega,\mu)$, where $(\Omega,\mu)$ is a measure space.

The theory of von Neumann algebras is a vast and very well-developed area of the theory of operator algebras. We shall be able only to cover some of the basics. The main results of this chapter are the von Neumann double commutant theorem and the Kaplansky density theorem.

## 4.1 The Double Commutant Theorem

There are a number of topologies on $B(H)$ ($H$ a Hilbert space), apart from the norm topology, that play a crucial role, and each has valuable properties that the others lack. The two most important are the strong (operator) topology and the weak (operator) topology. This section is concerned with the former (we shall introduce the weak topology in the next section). One of the reasons for the usefulness of the strong topology is the “order completeness” property asserted in Vigier’s theorem which is analogous to the order completeness property of the order completeness property of the real numbers $\mathbb{R}$.

Henceforth, we shall be using results concerning locally convex spaces.

Let $H$ be a Hilbert space, and $x\in H$. Then the function

$p_x:B(H)\rightarrow\mathbb{R}^+,\,T\mapsto \|Tx\|$,

is a semi-norm on $B(H)$. The locally convex topology on $B(H)$ generated by the separating family $\{p_x\}_{x\in H}$ is called the strong topology on $B(H)$. Thus, a net $\{u_\lambda\}$ converges strongly to an operator $T$ on $H$ iff $Tx=\lim_\lambda T_\lambda(x)$ for all $x\in H$. It follows that the strong topology is weaker than the norm topology on $B(H)$ (This is a theorem in topology. Suppose that $\tau_1,\,\tau_2$ are topologies on a set $X$. Suppose further that whenever $\{x_\lambda\}\subset X$ is a net convergent with respect to the topology $\tau_1$, then $\{x_\lambda\}$ is also convergent with respect to the topology $\tau_2$. Then $\tau_2\subset \tau_1$ — $\tau_2$ is weaker than $\tau_1$.).

With respect to the strong topology, $B(H)$ is a topological vector space, so the operations of addition and scalar multiplication are strongly continuous. This is not the case in general for the multiplication and involution operations.

### Example 4.1.1

Let $H$ be an infinite-dimensional Hilbert space with an orthonormal basis $\{e_n\}_{n\geq 1}$. Set $U_n=|e_1\rangle\langle e_n|$. If $x\in H$, then $U_n(x)=\langle e_n|x\rangle |e_1\rangle$, so $\lim_{n\rightarrow\infty}\|U_n(x)\|=\lim_{n\rightarrow \infty}|\langle x,e_n\rangle|=0$. Thus, the sequence $\{U_n\}$ is strongly convergent to to zero in $B(H)$. Now $U^\star_n=|e_n\rangle\langle e_1$, so $\|U^\star_n(x)\|=|\langle x,e_1\rangle|$. Therefore $\lim_{n\rightarrow \infty}\|U_n^\star(x)\|=1$ for $x=e_1$, so the sequence $\{U_n^\star\}$ does not converge strongly to zero. This shows that the operation $T\mapsto T^*$ on $B(H)$ is not strongly continuous, and therefore the strong and norm topologies on $B(H)$ do not coincide.

Observe also that $\|U_n\|=\|e_1\|\|e_n\|=1$, so the sequence $\{U_n\}$ is not convergent to zero, and therefore the norm $\|\cdot\|:B(H)\rightarrow \mathbb{R}^+$ is not strongly continuous.

The operation of multiplication $B(H)\times B(H)\rightarrow B(H)$$(T,S)\mapsto TS$ is not strongly continuous either (Cf. Exercise 4.3).

The preceding example shows that the strong topology behaves badly in some respects, but it also has some very good qualities, as we shall prove in the next theorem.

Let $H$ be an arbitrary Hilbert space and suppose that $\{T_\lambda\}_{\lambda\in\Lambda}$ is an increasing net in $B(H)_{\text{sa}}$ that converges strongly to $T$ (so $T$ also belongs to $B(H)_{\text{sa}}$). Then $T=\sup_\lambda T_\lambda$ and $\langle Tx,x\rangle=\sup_\lambda\langle T_\lambda x,x\rangle$ (for $x\in H$). The corresponding statement for decreasing nets in $B(H)_{\text{sa}}$ is also true. Both of these observations follow from the fact that if a net $\{T_\lambda\}_{\lambda\in\Lambda}$ converges strongly to an operator $T$, then $\langle Tx,y\rangle=\lim_\lambda\langle T_\lambda x,y\rangle$ ($x,y\in H$).

## Theorem 4.1.1 (Vigier)

Let $\{T_\lambda\}_{\lambda\in\Lambda}$ be a net of hermitian operators on a Hilbert space $H$. Then $\{T_\lambda\}$ is strongly convergent if it is increasing and bounded above, or if it is decreasing and bounded below.

### Proof

We prove only the case where $\{T_\lambda\}$ is increasing, since the decreasing case can be got from multiplying by minus 1.

Suppose then that $\{T_\lambda\}$ is increasing and bounded above. By truncating the net (that is, by choosing an index $\lambda_0\in\Lambda$ and considering the truncated net $\{T_\lambda\}_{\lambda\geq\lambda_0}$) we may suppose that $\{T_\lambda\}$ is also bounded below, by $S$ say. We may further assume that all $T_\lambda$ are positive (by considering the net $\{T_\lambda-S\}$ if necessary). There is a positive number $M$ such that $\|T_\lambda\|\leq M$. It follows that the increasing net $\{\langle T_\lambda x,x\rangle\}$ is bounded above by $M\|x\|^2$, so this net is convergent (by monotone convergence). Using the Polarisation Identity

$\langle T_\lambda x,y\rangle-\frac{1}{4}\sum_{k=0}^3i^k\langle T_\lambda(x+i^ky),x+i^ky\rangle,$

we see that $\{\langle T_\lambda x,t\rangle\}$ is a convergent net for all $x,y\in H$. Letting $\sigma(x,y)$ denote its limit, it is easy to check that the function

$\sigma:H^2\rightarrow\mathbb{C}, \,(x,y)\mapsto \sigma(x,y)$,

is a sesquilinear form on $H$. Moreover,

$|\sigma(x,y)|=\lim_\lambda |\langle T_\lambda x,y\rangle|\leq M\|x\|\|y\|$,

so $\sigma$ is bounded. Hence, there is an operator $T$ on $H$ such that $\langle Tx,y\rangle=\sigma(x,y)$ for all $x,y$ (Theorem 2.3.6).  Clearly $\|T\|\leq M$ (same theorem), $T$ is hermitian, and $T_\lambda\leq T$ for all $\lambda\in\Lambda$. Also

$\|Tx-T_\lambda x\|^2=\|(T-T_\lambda)^{1/2}(T-T_\lambda)^{1/2}x\|^2$

$\leq \|T-T_\lambda\|\|(T-T_\lambda)^{1/2}x\|^2$

$\leq 2M\langle(T-T_\lambda)x,x\rangle$,

(the $2M$ term comes from a triangle inequality) and $\langle (T-T_\lambda)x,x\rangle=0$ so $Tx=\lim_\lambda x$. Thus, $\{T_\lambda\}$ converges strongly to $T$ $\bullet$

### Remark 4.1.1

If $\{P_\lambda\}$ is a net of projections on a Hilbert space strongly convergent to an operator $P$, then $P$ is a projection. For $P$ is self-adjoint and $\langle Px,y\rangle=\lim_\lambda\langle P_\lambda x,y\rangle=\lim_\lambda\langle P_\lambda x,P_\lambda y\rangle=\langle Px,Py\rangle=\langle P^2x,y\rangle$, so $P=P^2$.

## Theorem 4.1.2

Suppose that $\{P_\lambda\}_{\lambda\in\Lambda}$ is a net of projections on a Hilbert space $H$.

1. If $\{P_\lambda\}$ is increasing, then it is strongly convergent to the projection of $H$ onto the closed vector subspace $\overline{\bigcup_\lambda P_\lambda H}$.
2. If $\{P_\lambda\}$ is decreasing, then it is strongly convergent to the projection of $H$ onto $\bigcap_\lambda P_\lambda H$.

### Proof

Left as an exercise $\bullet$

Just as for normed vector spaces, we say a family $\{x_\lambda\}_{\lambda\in\Lambda}$ of elements of a locally convex space is summable to a point $x$ is the net $\{\sum_{\lambda\in F}x_\lambda\}_F$ (where $F$ runs over all non-empty finite subsets of $\Lambda$) is convergent to $x$, and in this case we write $x=\sum_{\lambda\in\Lambda} x_\lambda$.

## Theorem 4.1.3

Let $\{P_\lambda\}_{\lambda\in\Lambda}$ be a family of projections on a Hilbert $H$ space that are pairwise orthogonal (that is $P_{\lambda_1} P_{\lambda_2}=0$ if $\lambda_1\neq\lambda_2$.) Then $\{P_\lambda\}$ is summable in the strong topology on $B(H)$ to a projection $P$, such that

$\|Px\|=\left(\sum_{\lambda\in\Lambda}\|P_\lambda x\|^2\right)^{1/2}$     $(x\in H)$.

If $P=I$, then the map

$H\rightarrow\bigoplus_\lambda P_\lambda(H)$$x\mapsto \{P_\lambda x\}$,

is a unitary.

### Proof

If $F$ is a finite non-empty subset of $latex\Lambda$, then $P_F=\sum_{\lambda\in F}P_\lambda$ is a projection. Therefore, $\{P_F\}_F$ is an increasing net of projections, hence strongly convergent to a projection $P$; that is, the family $\{P_\lambda\}$ is strongly summable to $P$. Moreover

$\|Px\|^2=\lim_F\|P_Fx\|^2=\lim_F\sum_{\lambda\in F}\|P_\lambda x\|^2=\sum_{\lambda\in\Lambda}\|P_\lambda x\|^2$.

The observation concerning the case when $P=I$ is clear $\bullet$

If $C$ is a subset of an algebra $A$, we define its commutant $C^\prime$ to be the set of all elements of $A$ that commute with all the elements of $C$. Observe that $C^\prime$ is a subalgebra of $A$ (straightforward). The double commutant $C^{\prime\prime}$ of $C$ is $(C^\prime)^\prime$. Similarly, $C^{\prime\prime\prime}=(C^{\prime\prime})^\prime$. Always $C\subset C^{\prime\prime}$ and $C^{\prime}\subset C^{\prime\prime\prime}$ (let $c\in C$. If $a\in C^\prime$ then $ac=ca$. That is $c\in C^\prime$. Now let $a\in C^\prime$ so that for all $c\in C$, we have $ac=ca$. Now for any $b\in C^{\prime\prime}$$ab=ba$ so $a$ commutes with everything in $C^{\prime\prime}$. That is $a\in C^{\prime\prime\prime}$. The text claims that in fact $C^\prime=C^{\prime\prime\prime}$. Now let $a\in C'''$. Now $ab=ba$, for all $b\in C''$. However $C\subset C''$ so therefore $ab=ba$ for all $b\in C$. That is $a\in C'$ also.). If $A$ is a normed algebra, then $C^\prime$ is closed (with more structure we could show this as follows. Define $f_c:A\rightarrow A$ by $a\mapsto ac-ca$. Now $f_c^{-1}(\{0\})$ is closed as is $\bigcap_{c\in C}f_c^{-1}(\{0\})=C^\prime$.). If $A$ is a *-algebra and $C$ is self-adjoint, then $C^\prime$ is a *-subalgebra of $A$ (straightforward).

## Lemma 4.1.4

Let $H$ be a Hilbert space and $A$ a *-subalgebra of $B(H)$ containing $I_H$. Then $A$ is strongly dense in $A^{\prime\prime}$.

### Proof

Let $T\in A^{\prime\prime}$, $x\in H$, and $K=\overline{Sx|S\in A}$. Then $K$ is a closed vector subspace of $H$ which is invariant, and therefore reducing ($K$ and $K^\perp$ invariant), for all $S\in A$, since $A$ is self-adjoint. Thus, if $P$ is the projection of $H$ onto $K$, then $P\in A^\prime$, so $PT=TP$. Hence $Tx\in K$, and therefore there is a sequence $\{S_n\}\subset A$ such that $Tx=\lim_{n\rightarrow\infty} S_nx$.

For each positive integer $n$ the map

$\varphi:B(H)\rightarrow B(H^{(n)})$$S\mapsto (\delta_{ij}S)$,

is a unital *-homomorphism, so $\varphi(A)$ is a subalgebra of $B(H^{(n)})$ containing $I_{H^{(n)}}$. Moreover, $\varphi(T)\in(\varphi(A))^{\prime\prime}$, for if $R\in (\varphi(A))^\prime$ and $S\in A$ then $\varphi(S)R=R\varphi(S) \Rightarrow R_{ij}V=VR_{ij}$. Hence $R_{ij}\in A^\prime$, so $TR_{ij}=R_{ij}T$. Therefore, $\varphi(T)R=R\varphi(T)$. Suppose now that $x=(x_1,\dots,x_n)\in H^{(n)}$. Then by the first paragraph of this proof there is a sequence $\{S_m\}_{m\geq 0}$ in $A$ such that $\varphi(T)x=\lim_{n\rightarrow\infty}\varphi(S_m)x$. Hence, $Tx_j=\lim_{n\rightarrow }$ for $j=1,\dots,n$.

We show that this implies that $T$ is in the strong closure of $A$. If $W$ is a strong neighbourhood of $T$, we must show that $W\cap A$ is non-empty, and to do this we may suppose that $W-T$ is a basic neighbourhood of $0$. Therefore, there are elements $x_1,\dots,x_n\in H$ and a positive number $\varepsilon$ such that.

$W-T=\{S\in B(H):\|Tx_j\|<\varepsilon;\,j=1,\dots,n\}$.

Hence, there is a sequence $\{S_m\}_{m\geq0}$ in $A$ such that

$Tx_j=\lim_{m\rightarrow\infty}S_mx_j$   $j=1,\dots,n$.

Consequently, for some $N$ the operator $S_N\in W$, so $W\cap A\neq\emptyset$ $\bullet$

Let $H$ be a Hilbert space. If $A$ is a strongly closed *-subalgebra of $B(H)$, we call $A$ a von Neumann algebra on $H$. Since the strong topology is weaker than the norm topology, a strongly closed set is also norm closed. Hence, a von Neumann algebra is a C*-algebra.

Obviously, $B(H)$ is a von Neumann algebra on $H$, as is $\mathbb{C}I_H$. If $(H_\lambda)_{\lambda\in\Lambda}$ is a family of Hilbert spaces and $A_\lambda$ is a von Neumann algebra on $H_\lambda$ for each index $\lambda$, then it is an easy exercise  to show that the direct sum $\oplus_{\lambda} A_\lambda$ is a von Neumann algebra on $\oplus_\lambda H_\lambda$.

If $A$ is a *-algebra on a Hilbert space $H$, then its commutant $A^\prime$ is a von Neumann algebra on $H$ (Let $\{T_\lambda\}$ be a net in $A^\prime$, strongly convergent to $T$, then $T\in A^\prime$. This is a consequence of the fact that $A'$ is strongly closed.)

If $A$ is a von Neumann algebra on $H$ and $P$ is a projection in $A$, then $PAP$ is a von Neumann algebra on $H$ ($PAP=\{PTP:T\in A\}$. We want to show that this set is strongly closed. Let $\{PT_\lambda P\}_{\lambda\in\Lambda}$ be a net in $A$ strongly convergent to $S$. We want to show that $S=PTP$ for some $T$ in $A$. Strong operator convergence implies that for any $x\in H$ we have that $\lim_\lambda \|PT_\lambda Px\|=\|Sx\|$. We know that $S\in A$, as $A$ is strongly closed. The rest follows through as multiplication is continuous). Also, $M_n(A)$ is a von Neumann algebra on $H^{(n)}$.

If $H$ is infinite-dimensional, then $K(H)$ is not a von Neumann algebra. To see this, let $E$ be an orthonormal basis for $H$, and for each finite non-empty subset of $E$ let $P_F=\sum_{e\in F}|e\rangle\langle e|$. Then $P_F$ is a finite rank projection and the net $\{P_F\}_{F}$ converges strongly to $I_H$ on $H$. If $K(H)$ were a von Neumann algebra, this would imply that it contains $I$, and so $\text{dim }H<|infty$, contradicting our assumption on $H$.

A fundamental result concerning von Neumann algebras is the following, known as the Double Commutant Theorem.

## Theorem 4.1.5 (von Neumann’s Double Commutant Theorem)

Let $A$ be a *-algebra on a Hilbert space $H$ and suppose that $I_H\in A$. Then $A$ is a von Neumann algebra on $H$ iff $A=A^{\prime\prime}$.

### Proof

Immediate from Lemma 4.1.4  $\bullet$

The intersection of a family of von Neumann algebras on a Hilbert space is also a von Neumann algebra. Thus, for any set $C\subset B(H)$ there is a smallest von Neumann algebra $A$ containing $C$. We call $A$ the von Neumann algebra generated by $C$. If $C$ is self-adjoint and contains $I_H$, then $A=C^{\prime\prime}$. If in addition $C$ consists of commuting elements, then $A$ is abelian (for in this case $C\subset C^\prime\Rightarrow A=C^{\prime\prime}\subset C^\prime\Rightarrow A\subset A^\prime$ $\checkmark$). This implies that there are non-trivial examples of abelian von Neumann algebras. We give an explicit example.

### Example 4.1.2

Let $X$ be a compact Hausdorff space, and suppose that $\mu$ is a finite positive regular Borel measure on $X$. We saw in Example 2.5.1 (https://jpmccarthymaths.wordpress.com/2011/01/19/c-algebras-and-operator-theory-2-5-the-spectral-theorem/) that the map

$L^\infty(X,\mu)\rightarrow B(L^2(X,\mu))$$\varphi\mapsto M_\varphi$,

is an isometric *-homomorphism. It’s range $A$ is a C*-subalgebra of $B(H)$. Denote by $B$ the C*-subalgebra of all multiplication operators on $L^2(X,\mu)$ with continuous symbol. The commutant of $B$ is $A$. Hence, $A$ is a von Neumann algebra on the Hilbert space $L^2(X,\mu)$. Since $A\subset A^\prime$ (because $A$ is abelian) and $A^\prime\subset B^\prime=A$, therefore $A=A^\prime$. Consequently, $A=B^{\prime\prime}$, so $B$ is strongly dense in $A$ be Lemma 4.1.4.

Let $K$ be a closed vector subspace of a Hilbert space $H$ and let $P$ be the projection of $H$ onto $K$. If $T\in B(H)$, let $T_P=T_K$ be compression (restriction) of $T$ to $K$. It is easy to verify that the map

$PB(H)P\rightarrow B(K)$$T\mapsto T_K$

is a *-isomorphism ($\checkmark$).

If $A$ is a *-algebra on $H$ and $P$ is in $A^\prime$, set $A_P=\{T_P:T\in A\}$.

## Lemma 4.1.6

Let $A$ be a *-algebra on a Hilbert space $H$, and $P$ a projection in $A^\prime$. Then $PAP$ and $A_P$ are *-algebras on $H$ and $P(H)$, respectively, and the map

$PAP\rightarrow A_P$ , $T\mapsto T_P$,

is  a *-isomorphism. Moreover, if also $P\in A^{\prime\prime}$, then $(A^\prime)_P=(A_P)^\prime$.

### Proof

We only show that $P\in A^{\prime\prime}\Rightarrow P\in (A^\prime)_P=(A_P)^\prime$ — the rest is left to the exercises.

Suppose that $T\in (A^\prime)_P$ and $S\in A_p$. Then there exist $\tilde{T}$ and $\tilde{S}\in A^\prime$ and $A$, respectively, such that $T=\tilde{T}_P$ and $S=\tilde{S}_P$. Now we have

$\tilde{S}=\left(\begin{array}{cc}S&S_1\\S_2&S_3\end{array}\right)\in A$, and

$\tilde{T}=\left(\begin{array}{cc}T&T_1\\T_2\\T_3\end{array}\right)\in A'$,

so that $\tilde{S}=\tilde{T}=\tilde{T}\tilde{S}$ which implies that $ST+S_1T_2=TS+T_1S_2$. With respect to this same basis

$P=\left(\begin{array}{cc}I&0\&0\end{array}\right)$,

and we have that $P\tilde{T}=\tilde{T}P$; equivalently

$\left(\begin{array}{cc}T&T_1\&0\end{array}\right)=\left(\begin{array}{cc}T&0\\T_2&0\end{array}\right)$,

which implies that $T_1=0=T_2$ so that $TS=ST$. Therefore $T\in(A_P)'$, so $(A')_P\subset (A_P)'$.

Conversely, suppose now that $T\in (A_P)^\prime$, and write $T=\tilde{T}_P$ for some $\tilde{T}\in PB(H)P$. Let

$\tilde{T}=\left(\begin{array}{cc}T&0\&0\end{array}\right)$.

Then if $\tilde{S}\in A$

$\tilde{S}=\left(\begin{array}{cc}S&S_1\\S_2&S_3\end{array}\right)$.

So we have

$\tilde{T}\tilde{S}=\left(\begin{array}{cc}TS &TS_1\&0\end{array}\right)$, and

$\tilde{S}\tilde{T}=\left(\begin{array}{cc}ST &0\\ S_2T &0\end{array}\right)$.

But $P\in A'$, so

$P=\left(\begin{array}{cc}I&0\&0\end{array}\right)$ commutes with $\tilde{S}$.

This implies that $S_1=0=S_2$ so that $\tilde{T}\tilde{S}=\tilde{S}\tilde{T}$Consequently, $\tilde{T}\in A^\prime$, and therefore $T\in (A^\prime)_P$. This shows that the inclusion $(A_P)^\prime\subset (A^\prime)_P$ holds $\bullet$

The reader should be aware that some authors define a von Neumann algebra on a Hilbert space $H$ to be a *-algebra $A$ on $H$ such that $A=A^{\prime\prime}$. This automatically ensures that $I_H\in A$. However, proofs appear to run more smoothly if von Neumann algebras are defined as we have done. Moreover, we can frequently reduce to the case where $A=A^{\prime\prime}$, by the trick explained in Remark 4.1.2. Using our definition von Neumann algebras are still unitial, but the unit may not be the identity of the underlying Hilbert space.

## Theorem 4.1.7

If $A$ is a non-zero von Neumann algebra, then it is unital.

### Proof

Suppose that $A$ acts on a Hilbert space $H$, and let $\{u_\lambda\}_{\lambda\in\Lambda}$ be an approximate unit for $A$. By Theorem 4.1.1, $\{u_\lambda\}$ converges strongly to a self-adjoint operator, $U$ say, and obviously $U\in A$, since $A$ is strongly closed. If $x\in H$ and $T\in A$, then $UTx=\lim_\lambda u_\lambda Tx=Tx$, so $UT=T$. Hence, $U$ is a unit for $A$ $\bullet$

### Remark 4.1.2

Let $A$ be a von Neumann algebra on a Hilbert space $H$ and let $U$ be the unit of $A$. Of course, $U$ is a projection in $A^\prime$. The map

$A\rightarrow A_U$$T\mapsto T_U$,

is a *-isomorphism (by Lemma 4.1.6), and $A_U$ is a von Neumann algebra on $U(H)$ containing $I_{U(H)}$, so $A_U=(A_U)^{\prime\prime}$. This device will be frequently used to reduce to the case where the von Neumann algebra is it’s own double commutant (we don’t necessarily have $A=A^{\prime\prime}$ if $I_H\not\in A$).

If $T$ is an operator on a Hilbert space $H$, then its range projection $[T]$ is the projection of $H$ on $\overline{T(H)}$. We have $[T]=[(TT^*)^{1/2}]$, since $\overline{T(H)}^{\perp}=\text{ker}(T^*)=\text{ker}(TT^*)^{1/2}=\overline{(TT^*)^{1/2}(H)}^\perp$ ( First equality. Suppose $x\in \overline{T(H)}^\perp$. That is for all $y\in H$ we have $\langle x,Ty\rangle=0\Leftrightarrow \langle T^*x,y\rangle=0\Leftrightarrow x\in\text{ker }T^*$.  The rest by the polar decomposition of $T^*$).

## Theorem 4.1.9

If $A$ is a von Neumann algebra, then it contains the range projections of all of its elements.

### Proof

Let $A$ act on $H$ and let $T\in A$. Since $(TT^*)^{1/2}\in A$, to show that the range projection of $T$ is in $A$, we may suppose that $T\leq 0$. Obviously we may also suppose that $T\geq I$. Let $T_n=T^{1/2^n}$ for $n\in\mathbb{N}$. Then $\{T_n\}$ is an increasing sequence of positive elements in the closed unit ball of $A$, so by Theorem 4.1.1, $\{T_n\}$ is strongly convergent to a positive operator, $P$ say. If $x\in H$, then

$\|(P^2-T_n^2)x\|\leq \|(P^2-T_nP)x\|+\|(T_nP-T_n^2)x\|$

$\leq\|(P-T_n)Px\|+\|(P-T_n)x\|.$

Therefore, $\{T_n^2\}$ converges to $P^2$ strongly. But $T_n^2=T_{n-1}$ for all $n>0$ so $P=P^2$.

The sequence $\{T_n\}$ is in $C^*(T)$, so $P(H)\subset \overline{T(H)}$. The continuous functions

$\sigma(T)\rightarrow\mathbb{R}$$t\mapsto t^{1+1/2^n}$,

form an increasing sequence and converge pointwise to the identity function $t\mapsto t$, so by Dini’s Theorem, they converge uniformly. Therefore, by the functional calculus, $T=\lim_{n\rightarrow\infty} T^{1+1/2^n}$; that is $T=\lim_{n\rightarrow\infty} TT_n$. Hence, $T=TP=PT$, so $\overline{T(H)}\subset P(H)$. Therefore $[T]=P\in A$ $\bullet$

## Theorem 4.1.10

Let $A$ be a von Neumann algebra on a Hilbert space $H$ and $T$ and element of $A$ with polar decomposition $T=U|T|$. Then $U\in A$.

### Proof

Let $V$ be a unitary in the unital C*-algebra $A^\prime$. Then $\tilde{V}=V^*UV$ is a partial isometry on $H$ such that $T=\tilde{V}|T|$ and $\text{ker}(\tilde{V})=\text{ker }T$ (The projection onto the initial space of $\tilde{V}$ is

$\tilde{V}^*\tilde{V}=(V^*UV)^*(V^*UV)=V^*U^*\underbrace{VV^*}_{=I}UV$

$=V^*\underbrace{U^*U}_{=\text{projection onto initial sp. of }U}V$

But the initial space of $U$ is $\text{ran }|T|$, and this projection lies in $A$, so

$V^*U^*UV=U^*UV^*V=U^*U$

that is the initial space of $\tilde{V}$ is the initial space of $U$. Therefore $\text{ker }\tilde{V}=\text{ker }U=\text{ker }T$). It follows, therefore, from the uniqueness of the polar decomposition that $\tilde{V}=U$, so $U$ and $V$ commute ($\checkmark$). But $A^\prime$ is the linear span of its isometries (Suppose that a C*-algebra $A$ is unital and let $a\in A_{\text{sa}}$. Then $\{U_a=a+i(I-a^2)^{1/2}:a\in A_\text{sa}\}$ is a set of unitaries that spans that space.), so $U$ must commute with all elements of $A^\prime$, and therefore $U\in A^{\prime\prime}=(A+\mathbb{C}1)^{\prime\prime}$. By Lemma 4.1.4 there is a net $\{U_\lambda\}_{\lambda\in\Lambda}$ and a net $\{\alpha_\lambda\}_{\lambda\in\Lambda}$  in $\mathbb{C}$ such that the net $\{U_\lambda+\alpha_\lambda 1\}_{\lambda\in\Lambda}$ converges strongly to $U$ on $H$. If $P=[|T|]$, then by Theorem 4.1.9 $P\in A$. Since $(1-P)(H)=\overline{|T|(H)}^\perp=\text{ker }|T|=\text{ker }T=\text{ker }U$ ($\text{ker }T=\text{ker U}$ by Theorem 2.3.4), we have $U(1-P)=0$; that is, $U=UP$. Therefore, $U$ is the strong limit of the net $\{U_\lambda P+\alpha_\lambda P 1\}$ which lies in $A$, so $U\in A$  $\bullet$

## Theorem 4.1.11

Suppose that  $A$ is a von Neumann algebra on a Hilbert space $H$.

1. $A$ is the closed linear span of its projections.
2. If $I_H\in A$ and $T$ is a normal element of $A$, then $E(X)\in A$ for every Borel set $X$ of $\sigma(T)$, where $E$ is the spectral resolution of the identity for $T$.
3. If $I_H\in A$ and $S\in B(H)$, then $S\in A$ if and only if $S$ commutes with all the projections in $A^\prime$.

### Proof

We may suppose in all cases that $I_H\in A$ (certainly in conclusions 2 and 3. I imagine we will be adapting Remark 4.1.2 to conclusion 1. if $I_H\not\in A$. ). We prove condition 2. first. Let $T$ be a normal element with spectral resolution of the identity for $T$ denoted by $E$. If $S\in A^\prime$, then $ST=TS$ and $ST^*=T^*S$ (as $A^\prime$ is self-adjoint — take adjoints to $S^*T=TS^*$), so $Sf(T)=f(T)S$ for every $f\in B_\infty(\sigma(T))$ (go on the Stone-Weierstrass!). In particular, $S\,E(X)=E(X)\,S$ for every Borel set $X$ of $\sigma(T)$. Therefore, $E(X)\in A^{\prime\prime}=A$.

Condition 1. follows directly from condition 2., using the fact the closed linear span of the characteristic functions $\{\chi_X:X\text{ a Borel set of }\sigma(T)\}$ is $B_\infty (\sigma(T))$ for each normal element $T$ of $A$.

Condition 3. follows immediately from condition 1., since $A^\prime$ is a von Neumann algebra, and therefore it is the closed linear span of its projections $\bullet$

We give an immediate and important application of this result  in the next theorem. First we make some observations.

### Remark 4.1.3

If $H$ is a Hilbert space, then $B(H)^\prime=\mathbb{C}I_H$. For it is obvious that $\mathbb{C}^\prime=B(H)$, and since $\mathbb{C}$ is a von Neumann algebra containing $I_H$, Theorem 4.1.5 implies that $\mathbb{C}=\mathbb{C}^{\prime\prime}$, so $\mathbb{C}=B(H)^\prime$.

### Remark 4.1.4

If $A$ is a C*-algebra acting on a Hilbert space $H$ and $S\subset H$, consider the set $AS:= \langle A(S)\rangle$.

We say $A$ acts non-degerately on $H$ if $\overline{AS}=H$. Equivalently, for each non-zero $x\in H$ there exists $T\in A$ such that $Tx\neq 0$.

If $A$ acts non-degenerately on $H$ and $\{U_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$, then $\{U_\lambda\}$ converges to $I_H$ strongly. To show this we must prove that $\lim_\lambda U_\lambda x=x$, for all $x\in H$. This is clear if $x=Ty$ for some $T\in A$ and $y\in H$ (because $\overline{AH}=H$ — we can fill in the gaps by continuity). Now

$\lim_\lambda U_n x=\lim_\lambda U_nTy=Ty=x$ $\bullet$

If $A$ acts irreducibly on $H$ and $A\neq 0$, then it acts non-degenerately, since $\overline{AH}$ is a non-zero vector subspace  of $H$ invariant for $A$, and therefore equals $A$ ($A$ acting irreducibly on $H$ means that the only closed vector subspaces of $H$ are $\{0\}$ and $H$).

## Theorem 4.1.12

Let $A$ be a non-zero C*-algebra acting on a Hilbert space $H$. The following conditions are equivalent:

1. $A$ acts irreducibly on $H$.
2. $A^\prime=\mathbb{C}I_H$.
3. $A$ is strongly dense in $B(H)$.

### Proof

If $P$ is a projection in $B(H)$, then $P\in A^\prime$ if and only if the closed vector subspace $P(H)$ of $H$ is invariant for $A$. Since $A^\prime$ is a von Neumann algebra, it is the closed linear span of its projections by Theorem 4.1.11, so if $A$ acts irreducibly , then $A^\prime$ has no projections except the trivial ones, and therefore $A^\prime=\mathbb{C}I_H$. Therefore 1. implies 2. The reverse implication is clear.

If we suppose now that $A^\prime=\mathbb{C}I_H$, then $(A+\mathbb{C}I_H)^\prime=\mathbb{C}I_H$ ($\checkmark$), and therefore $(A+\mathbb{C}I_H)^{\prime\prime}=B(H)$. By Lemma 4.1.4 the C*-algebra $A+\mathbb{C}I_H$ is strongly dense in $B(H)$. However, since $A$ acts irreducibly on $H$, it acts non-degenerately, and therefore if $\{U_\lambda\}_{\lambda\in\Lambda}$ is an approximate unit for $A$, it is strongly convergent to $I_H$. Therefore, $I_H\in \overline{A}$, and $\overline{A}=\overline{A+\mathbb{C}I_H}=B(H)$, where the closure is strong closure ($\checkmark$). Hence 2. implies 3.

Finally, if $A$ is strongly dense in $B(H)$, then $A^\prime=B(H)^\prime=\mathbb{C}$, so 3. implies 2. $\bullet$

If $P,Q$ are projections in a C*-algebra $A$, we say they are Murray-von Neumann equivalent, and we write $P\sim Q$, if there exists $T\in A$ such that $P=T^*T$ and $Q=TT^*$. It is indeed straightforward to show that this is indeed an equivalence relation on the set of projections of $A$ ($P=P^*P=PP^*=P$ so $\sim$ is reflexive. Clearly $\sim$ is symmetric. Now suppose $P=T^*T$ and $Q=TT^*$; and $Q=S^*S$ and $R=SS^*$. I don’t think I need to full result actually…). The relation $\sim$ is of fundamental importance in the classification theory of von Neumann algebras (see the Addenda), and in $K$-theory for C*algebras. However, for the moment we only need one small result concerning $\sim$.

### Remark 4.1.5

If $P,Q$ are infinite-rank projections on a separable Hilbert space $H$, then $P\sim Q$ (I can see how this goes back to the talk on $E-semigroups$.). To see this, choose orthonormal basis $\{e_n\}_{n\geq 1}$ and $\{f_n\}_{n\geq 1}$ for $P(H)$ and $Q(H)$. Let $U:P(H)\rightarrow Q(H)$ be the unitary such that $Ue_n=f_n$ for all $n$. Define $T\in B(H)$ by setting $T=U$ on $P(H)$ and $T=0$ on $(I_H-P)(H)$. It is easily verified that $P=T^*T$ and $Q=TT^*$.

## Theorem 4.1.15

If $H$ is a separable infinite-dimensional Hilbert space, then $K(H)$ is the unique non-trivial closed ideal of $B(H)$.

### Proof

Let $I$ be a non-zero closed ideal of $B(H)$. By Theorems 2.4.5 and 2.4.7, $K(H)\subset I$. Now if $I\not\subset K(H)$, then by Corollary 4.1.14 (not covered as it uses hereditary C*-algebras), $K(H)$ does not contain all of the projections of $I$. Hence, $I$ has an infinite-rank projection, $P$ say. If $Q$ is any other infinite-rank projection on $H$, then as we say in Remark 4.1.5 there is an element $T\in B(H)$ such that $P=T^*T$ and $Q=TT^*$. Therefore, $Q=Q^2=TPT^*$, so $Q$ also belongs to $I$. Hence, $I$ contains all the projections of $B(H)$, whether their rank is finite or infinite, and therefore $I=B(H)$. Thus, we have shown that the only closed ideals of $B(H)$ are $\{0\}$$K(H)$ and $B(H)$ $\bullet$

### Remark 4.1.6

Let $H$ be a separable infinite-dimensional Hilbert space. If $\{x_n\}_{n\geq 1}$ is a dense sequence on $H$, it is easy to check that $K(H)$ is the closed linear span of the operators $|x_n\rangle\langle x_m|$ ($m,n\geq 1$) using the fact that $K(H)$ is the closed linear span of the rank-one projections (Theorems 2.4.5 and 2.4.6). Hence $K(H)$ is separable. However, $B(H)$ is non-separable (and therefore $B(H)/K(H)$ is non-separable). To see this, choose an orthonormal basis $\{e_n\}_{n\geq 1}$ for $H$. For each set $S$ of positive integers, let $P_S$ be the projection onto $\langle e_i:i\in S\rangle$. Clearly, $\|P_S-P_{S^\prime}\|=1$ if $S\neq S^\prime$ ($\checkmark$). Hence, the family of operators $\{P_S\}_{S\subset 2^{\mathbb{N}}}$ cannot be in the closure of the range of any sequence in $B(H)$.

## Theorem 4.1.16

If $H$ is an infinite-dimensional separable Hilbert space, then the Calkin algebra $B(H)/ K(H)$ is a simple C*-algebra.

### Proof

Let $\mathcal{C}$ denote the Calkin algebra and $\pi:B(H)\rightarrow\mathcal{C}$ the quotient map. To see that $\mathcal{C}$ is simple, let $I$ be a closed ideal in $\mathcal{C}$. Then $\pi^{-1}(I)$ is a closed ideal in $B(H)$ (as the quotient map is a homomorphism), and therefore by Theorem 4.1.15, $\pi^{-1}(I)=\{0\},K(H)$ or $B(H)$. Hence, $I=\{0\}$ or $\mathcal{C}$ $\bullet$

### Remark 4.1.7

Let $T,S$ be operators on a Hilbert space $H$ such that $TT^*\leq SS^*$. Then there is an operator $R\in B(H)$ such that $T=SR$. To see this, observe that

$\|T^*x\|^2=\langle TT^*x,x\rangle\leq \langle SS^*x,x\rangle-\|S^*x\|^2$.

Hence, we get a well-defined norm-decreasing linear map $R_0:S^*(H)\rightarrow H$ by setting $R_0S^*x=T^*x$. Clearly, we can extend $R_0$ to a bounded linear $R_1$ on $H$. Setting $R=R_1^*$, we get the required result $T=SR$ $\bullet$