37.3 Von Neumann Algebras

Proof, [Section 17, Zhu93]. (1): After rescaling, assume without loss of generality that $-I \le T \le I$ for all $T \in \cf$. Since $\cf \subset A_{sa}$, $\norm{T}_{B(H)}= [T]_{sp}\le 1$ by Theorem 36.4.3, where the spectral radius is taken with respect to $B(H)$.

Thus $\cf \subset \ol{B_{A}(0, 1)}$, and is relatively compact in the weak operator topology by the Banach-Alaoglu Theorem. As such, $\bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}\ne \emptyset$. Let $R \in \bigcap_{T \in \cf}\ol{\bracs{S \in \cf|S \ge T}}^{\text{\small WOT}}$. Since $A$ is strong-operator closed, so is $A_{sa}$ by Proposition 37.1.4. Thus for each $T \in A_{sa}$, $\bracs{S \in A_{sa}|S \ge T}$ is closed in the weak operator topology, and $R \in A_{sa}$ with $R \ge T$ for all $T \in \cf$.

Let $S \in B(H)$ be self-adjoint such that $S \ge T$ for all $T \in \cf$, then $S \ge T$ for all $T \in \ol{\cf}^{\text{\small WOT}}$. In particular, $S \ge R$, thus $R$ is indeed the supremum of $\cf$.

Finally, let $x \in H$ and $\eps > 0$, then there exists $T \in \cf$ such that $\dpn{(R - T)x, x}{H}\le \eps$. For any $S \in \cf$ with $S \ge T$,

By the continuous functional calculus and Theorem 36.4.3,

so

for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$.

(2): By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case,

Thus $x \in \sup(\mathcal{P})(H)$, and $\sup(\mathcal{P})(H) \supset{\ol{\bigcup_{P \in \mathcal{P}}P(H)}}$.

On the other hand, let $Q \in B(H)$ be the orthogonal projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$, then $Q$ is also an upper bound of $\mathcal{P}$. Therefore

(3): As $0 \le T \le I$, $\sigma_{B(H)}(T) \subset [0, 1]$ by the Gelfand-Naimark Theorem. For each $n \in \natp$ and $t \in [0, 1]$, let $f_{n}(t) = t^{1/n}$, then $f_{n}$ is an increasing sequence of continuous functions on $[0, 1]$. By the continuous functional calculus, $\bracsn{f_n(T)}_{1}^{\infty} = \bracsn{T^{1/n}}_{1}^{\infty}$ is an increasing sequence that lies between $0$ and $I$.

Let $n \in \natp$. By the Stone-Weierstrass Theorem, there exist polynomials $\seq{p_{n, k}}\subset \real[x]$ such that:

1. (i)

   For each $k \in \natp$, $p_{n, k}(0) = 0$.

2. (ii)

   $p_{n, k}\to f_{n}$ uniformly on $\sigma_{B(H)}(T)$.

As $A$ is a uniformly closed subalgebra of $B(H)$, $f_{n}(T) \in A$ for all $n \in \natp$.

By (1), the supremum $Q = \sup_{n \in \natp}T^{1/n}= \sotlim_{n \to \infty}T^{1/n}$ exists in $A$. Since

and $Q$ is self-adjoint, $Q$ is a projection.

For each $x \in H \setminus \ker(T)$, $\dpn{Qx, x}{H}\ge \dpn{Tx, x}{H}> 0$ because $T$ is positive. As such, $\ker(Q) \subset \ker(T)$. For any $x \in \ker(T)$, $\dpn{Qx, x}{H}= \limv{n}\dpn{T^{1/n}x, x}{H}= 0$, so $\ker(Q) \supset \ker(T)$. Since both operators are self-adjoint, $\ol{Q(H)}= \ol{T(H)}= P(H)$, and $P = Q$.

(4): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then

Thus $\ker(T^{*}) = \ker(TT^{*})$, and

By (3) applied to $TT^{*}/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$.

(5): Let $\mathcal{P}$ be the set of all projections in $A$, and $\cf \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$. For each $F \in \cf$, let $P_{F}$ be the projection onto $\braks{\sum_{P \in F}P}(H)$, then $P_{F} \ge P$ for all $P \in F$ and $P_{F} \in A$ by (4). Since $\bracsn{P_F}_{F \in \cf}$ is a bounded and directed family of projections, $\sup_{F \in \cf}P_{F} \in A$ by (1). As $\sup_{F \in \cf}P_{F} \in \mathcal{P}$, it is the maximum projection in $A$.

Let $T \in A$, then by (4), $P$ is greater than the projection onto $\ol{T(H)}$, so $PT = T$. On the other hand, since $PT^{*} = T^{*}$, $TP = T$ as well. Therefore $P$ is the multiplicative identity in $A$.$\square$

Proof, [Corollary 18.3, Zhu93]. (1) $\Rightarrow$ (2): By Lemma 37.3.2, $PTP = TP$ and $PT^{*}P = T^{*}P$. Thus $TP = PTP = PT$.

(2) $\Rightarrow$ (1): Since $TP = PT$, $PTP = PT$, and $T(M) \subset M$ by Lemma 37.3.2. Similarly, $T^{*}P = PT^{*}$ implies that $PT^{*}P = PT^{*}$, and $T^{*}(M) \subset M$ by Lemma 37.3.2.$\square$

Proof, [Lemma 18.4, Zhu93]. Let $M = \ol{\bracsn{Sx|S \in A}}$, then since $A$ is self-adjoint, $M$ is a reducing subspace for each element of $A$. Let $P \in B(H)$ be the orthogonal projection onto $M$, then $PS = SP$ for all $S \in A$. As such, $P \in A'$.

Now, since $T \in A''$, $TP = PT$ as well, so $M$ is a reducing subspace for $T$. As $A$ is unital, $x \in M$, so $Tx \in M = \ol{\bracsn{Sx|S \in A}}$.$\square$

Proof, [Section 18.3, Zhu93]. ($\overline{A}^{\text{\small SOT}}\subset A''$): By separate continuity of composition, $A''$ is a strong-operator closed subset that contains $A$. Hence $A''$ contains the strong-operator closure of $A$.

($\overline{A}^{\text{\small SOT}}\supset A''$): Let $T \in A''$ and $x = \seqf{x_j}\in H^{n}$. For each $S \in B(H)$, denote $S^{(n)}= (S, \cdots, S)$ ($n$-copies), then

1. (i)

   $A^{(n)}= \bracsn{S^{(n)}|S \in A}$ is a unital, self-adjoint subalgebra of $B(H^{n})$.

2. (ii)

   $T^{(n)}\in (A^{(n)})''$.

By Lemma 37.3.5,

so $T \in \ol{A}^{\text{\small SOT}}$.$\square$

Proof, [Theorem 19.5, Zhu93]. (1): Let $T \in \ol{B_{B_{sa}}(0, 1)}$ and $\angles{T_\gamma}_{\gamma \in C}\subset A$ be a net such that $T_{\gamma} \to T$ in the weak operator topology. For each $\gamma \in C$, let $T_{\gamma}' = (T_{\gamma} + T_{\gamma}^{*})/2$, then $T_{\gamma}' \to T$ in the weak operator topology by continuity of the adjoint map in the weak operator topology.

As $A_{sa}$ is a subspace of $B(H)$, its strong and weak-operator closures coincide. Thus there exists a net $\angles{S_\gamma}_{\gamma \in C}\subset A_{sa}$ such that $S_{\gamma} \to T$ in the strong operator topology. In which case, let

then $f \in C_{0}(\real; \real)$. By Corollary 37.2.3, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)}\le 1$, $\sigma_{B}(T) \subset [-1, 1]$. Thus $f(T) = T$, and $f(S_{\gamma}) \to T$ in the strong operator topology. By the continuous functional calculus, $f(S_{\gamma})$ is in the closed unit ball of $A_{sa}$ for all $\gamma \in C$. Therefore the closed unit ball of $A_{sa}$ is strong-operator dense in the closed unit ball of $B_{sa}$.

(2): Let $T \in \ol{B_{B}(0, 1)}$ with $T \ge 0$ and $\angles{T_\gamma}_{\gamma \in C}\subset A_{sa}$ be a net such that $T_{\gamma} \to T$ in the strong operator topology. Define

then $f \in C_{0}(\real; [0, \infty))$. Since $f \ge 0$, the continuous functional calculus then implies that $\norm{f(T_\gamma)}_{B(H)}\le 1$ and $f(T_{\gamma}) \ge 0$ for all $\gamma \in C$. By Corollary 37.2.3, the mapping $S \mapsto f(S)$ is strong-operator continuous. As $\norm{T}_{B(H)}\le 1$ and $T \ge 0$, $\sigma_{B}(T) \subset [0, 1]$ by Proposition 36.8.2. Thus $f(T) = T$, and $f(T_{\gamma}) \to T$ in the strong operator topology.

(3): For each $\mathcal{T}\subset B(H)$, let

then $M_{2}(B)$ is the strong-operator closure of $M_{2}(A)$ in $B(H^{2})$. For each $T \in \ol{B_{B}(0, 1)}$, let

then $T' \in \ol{B_{M_2(B)_{sa}}(0, 1)}$. By (1), there exists a net $\angles{(R_\gamma, S_\gamma, T_\gamma)}_{\gamma \in C}\subset A^{3}$ such that:

1. (i)

   For each $\gamma \in C$,

   \[\norm{\begin{bmatrix}R_{\gamma}&T_{\gamma} \\ T^{*}_{\gamma}&S_{\gamma}\end{bmatrix}}_{B(H^2)}\le 1\]

   
   

   In particular, $\norm{T_\gamma}_{B(H)}\le 1$.

2. (ii)

   With respect to the strong operator topology on $B(H^{2})$,

   \[\begin{bmatrix}R_{\gamma}&T_{\gamma} \\ T^{*}_{\gamma}&S_{\gamma}\end{bmatrix} \to T'\]

   
   

Therefore $\angles{T_\gamma}\subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology.$\square$