252 lines
14 KiB
TeX
252 lines
14 KiB
TeX
\section{Von Neumann Algebras}
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\label{section:vna}
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\begin{theorem}[Existence of Projections]
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\label{theorem:existence-of-projections-vna}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a strong-operator closed $C^*$-subalgebra, then:
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\begin{enumerate}
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\item For any bounded directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$.
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\item For any family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$.
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\end{enumerate}
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and
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\begin{enumerate}[start=2]
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\item Let $T \in A_{sa}$ with $0 \le T \le I$ and $P \in B(H)$ be the orthogonal projection onto $\ol{T(H)}$, then $P = \sotlim_{n \to \infty}T^{1/n} \in A$.
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\item For each $T \in A$, the orthogonal projection onto $\ol{T(H)}$ is in $A$.
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\end{enumerate}
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Finally,
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\begin{enumerate}[start=4]
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\item There exists a maximum projection $P \in A$ such that $PT = TP = T$ for all $T \in A$, which is the multiplicative unit of $A$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Section 17]{Zhu}}}. ]
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(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 \autoref{theorem:c-star-normal-spectral-radius}, where the spectral radius is taken with respect to $B(H)$.
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Thus $\cf \subset \ol{B_{A}(0, 1)}$, and is relatively compact in the weak operator topology by the \hyperref[Banach-Alaoglu Theorem]{proposition:bh-ultraweak-bounded}. 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 \autoref{proposition:bh-operator-topologies-facts}. 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 $T \in \cf$.
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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 $\cf$.
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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 $S \ge T$,
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\[
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\normn{(R - S)^{1/2}x}_H^2 = \dpn{(R - S)x, x}{H} \le \dpn{(R - T)x, x}{H} \le \eps
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\]
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By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} and \autoref{theorem:c-star-normal-spectral-radius},
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\[
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\normn{(R - S)^{1/2}}_{B(H)} = \norm{R - S}_{B(H)}^{1/2} \le \sqrt{2}
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\]
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so
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\[
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\norm{(R - S)x}_H \le \normn{(R - S)^{1/2}}_{B(H)} \cdot \normn{(R - S)^{1/2}x}_H \le \sqrt{2 \eps}
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\]
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for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$.
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(2): Assume without loss of generality that $\mathcal{P}$ is directed. 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,
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\[
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\dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
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\]
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Thus $x \in \sup(\mathcal{P})(H)$, and $\sup(\mathcal{P})(H) \supset {\ol{\bigcup_{P \in \mathcal{P}}P(H)}}$.
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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
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\[
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\ol{\bigcup_{P \in \mathcal{P}}P(H)} = Q(H) \supset \sup(\mathcal{P})(H)
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\]
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(3): As $0 \le T \le I$, $\sigma_{B(H)}(T) \subset [0, 1]$ by the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}. 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 \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\bracsn{f_n(T)}_1^\infty = \bracsn{T^{1/n}}_1^\infty$ is an increasing sequence that lies between $0$ and $I$.
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Let $n \in \natp$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, there exist polynomials $\seq{p_{n, k}} \subset \real[x]$ such that:
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\begin{enumerate}[label=(\roman*)]
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\item For each $k \in \natp$, $p_{n, k}(0) = 0$.
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\item $p_{n, k} \to f_n$ uniformly on $\sigma_{B(H)}(T)$.
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\end{enumerate}
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As $A$ is a uniformly closed subalgebra of $B(H)$, $f_n(T) \in A$ for all $n \in \natp$.
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By (1), the supremum $Q = \sup_{n \in \natp}T^{1/n} = \sotlim_{n \to \infty}T^{1/n}$ exists in $A$. Since
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\[
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Q^2 = \sotlim_{n \to \infty}T^{2/n} = \sotlim_{n \to \infty}T^{1/n} = Q
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\]
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and $Q$ is self-adjoint, $Q$ is a projection.
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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 $P = Q$.
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(4): Assume without loss of generality that $T \ne 0$. Let $x \in H$, then
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\[
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\norm{T^*x}_H^2 = \dpn{T^*x, T^*x}{H} = \dpn{TT^*x, x}{H}
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\]
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Thus $\ker(T^*) = \ker(TT^*)$, and
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\[
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\ol{T(H)} = \ker(T^*)^\perp = \ker(TT^*)^\perp = \ol{TT^*(H)}
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\]
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By (3) applied to $TT^*/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$.
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(5): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (2), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$.
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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 $A$.
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\end{proof}
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\begin{lemma}
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\label{lemma:invariant-projection-test}
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Let $H$ be a complex Hilbert space, $M \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $M$, and $T \in B(H)$, then the following are equivalent:
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\begin{enumerate}
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\item $T(M) \subset M$.
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\item $PTP = TP$.
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\end{enumerate}
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\end{lemma}
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% Proof omitted due to obviousness
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\begin{definition}[Reducing Subspace]
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\label{definition:reducing-subspace}
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Let $H$ be a complex Hilbert space, $M \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $P$, and $T \in B(H)$, then the following are equivalent:
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\begin{enumerate}
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\item $T(M) \subset M$ and $T^*(M) \subset M$.
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\item $TP = PT$.
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\end{enumerate}
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If the above holds, then $M$ is a \textbf{reducing subspace} of $T$.
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\end{definition}
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\begin{proof}[Proof, {{\cite[Corollary 18.3]{Zhu}}}. ]
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(1) $\Rightarrow$ (2): By \autoref{lemma:invariant-projection-test}, $PTP = TP$ and $PT^*P = T^*P$. Thus $TP = PTP = PT$.
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(2) $\Rightarrow$ (1): Since $TP = PT$, $PTP = PT$, and $T(M) \subset M$ by \autoref{lemma:invariant-projection-test}. Similarly, $T^*P = PT^*$ implies that $PT^*P = PT^*$, and $T^*(M) \subset M$ by \autoref{lemma:invariant-projection-test}.
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\end{proof}
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\begin{definition}[Commutant]
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\label{definition:commutant}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$, then
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\[
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A' = \bracs{T \in B(H)| TS = ST \forall S \in A}
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\]
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is the \textbf{commutant} of $A$.
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\end{definition}
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\begin{lemma}
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\label{lemma:bicommutant-pointwise}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, $T \in A''$, and $x \in H$, then $Tx \in \ol{\bracsn{Sx|S \in A}}$.
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma 18.4]{Zhu}}}. ]
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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, $P \in A'$.
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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 $Tx \in M = \ol{\bracsn{Sx|S \in A}}$.
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\end{proof}
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\begin{lemma}[Amplification]
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\label{lemma:bh-amplification}
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Let $H$ be a complex Hilbert space, $n \in \natp$, and
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\[
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\pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n
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\]
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then for each $\seqf{x_j} \subset H$,
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\[
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\max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H
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\]
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\end{lemma}
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\begin{theorem}[Von Neumann's Bicommutant Theorem]
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\label{theorem:bicommutant}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a unital, self-adjoint subalgebra, then $A''$ is the strong-operator closure of $A$.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Section 18.3]{Zhu}}}. ]
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($\overline{A}^{\text{\small SOT}} \subset A''$): By separate continuity of composition, $A''$ is a strong-operator closed subset that contains $A$. Hence $A''$ contains the strong-operator closure of $A$.
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($\overline{A}^{\text{\small SOT}} \supset A''$): 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
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\begin{enumerate}[label=(\roman*)]
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\item $A^{(n)} = \bracsn{S^{(n)}|S \in A}$ is a unital, self-adjoint subalgebra of $B(H^n)$.
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\item $T^{(n)} \in (A^{(n)})''$.
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\end{enumerate}
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By \autoref{lemma:bicommutant-pointwise},
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\[
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(Tx_1, \cdots, Tx_n) \in \ol{\bracsn{(Sx_1, \cdots, Sx_n)|S \in A}}
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\]
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so $T \in \ol{A}^{\text{\small SOT}}$.
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\end{proof}
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\begin{definition}[Von Neumann Algebra]
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\label{definition:von-neumann-algebra}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a \textbf{von Neumann algebra acting on $H$} if $A$ is closed in the strong operator topology.
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\end{definition}
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\begin{theorem}[Kaplansky Density Theorem]
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\label{theorem:kaplansky-density}
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Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
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\begin{enumerate}
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\item $\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.
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\item $\bracsn{T \in \ol{B_{A}(0, 1)}|T \ge 0}$ is strong-operator dense in $\bracsn{T \in \ol{B_{B}(0, 1)}|T \ge 0}$.
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\item $\ol{B_{A}(0, 1)}$ is strong-operator dense in $\ol{B_{B}(0, 1)}$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 19.5]{Zhu}}}. ]
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(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.
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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
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\[
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f: \real \to \real \quad t \mapsto \begin{cases}
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t &t \in [-1, 1] \\
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1/t &t \in \real \setminus [-1, 1]
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\end{cases}
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\]
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then $f \in C_0(\real; \real)$. By \autoref{corollary:functional-calculus-c0-self-adjoint}, 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 \hyperref[continuous functional calculus]{definition: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 $B_{sa}$.
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(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
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\[
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f: \real \to \real \quad t \mapsto \begin{cases}
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0 &t \le 0 \\
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t &t \in [0, 1] \\
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1/t &t \ge 1
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\end{cases}
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\]
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then $f \in C_0(\real; [0, \infty))$. Since $f \ge 0$, the \hyperref[continuous functional calculus]{definition: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 \autoref{corollary:functional-calculus-c0-self-adjoint}, 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 \autoref{proposition:positive-spectrum}. Thus $f(T) = T$, and $f(T_\gamma) \to T$ in the strong operator topology.
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(3): For each $\mathcal{T} \subset B(H)$, let
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\[
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M_2(\mathcal{T}) = \bracs{\begin{bmatrix} Q & R \\ S & T \end{bmatrix} \bigg | Q, R, S, T \in \mathcal{T}}
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\]
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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
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\[
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T' = \begin{bmatrix} 0 & T \\ T^* & 0 \end{bmatrix}
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\]
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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:
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\begin{enumerate}[label=(\roman*)]
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\item For each $\gamma \in C$,
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\[
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\norm{\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix}}_{B(H^2)} \le 1
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\]
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In particular, $\norm{T_\gamma}_{B(H)} \le 1$.
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\item With respect to the strong operator topology on $B(H^2)$,
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\[
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\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix} \to T'
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\]
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\end{enumerate}
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Therefore $\angles{T_\gamma} \subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology.
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\end{proof}
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\begin{remark}
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\label{remark:kaplansky-unitary}
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The Kaplansky Density Theorem should also apply to the unitary case. Unfortunately, it seems like that the Borel functional calculus is required for an easier proof, so it will be postponed for now.
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\end{remark}
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