57 lines
3.7 KiB
TeX
57 lines
3.7 KiB
TeX
\section{Commutative von Neumann Algebras}
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\label{section:vn-commutative}
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\begin{definition}[Separating Vector]
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\label{definition:separating-vector}
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Let $H$ be a complex Hilbert space, $A \subset B(H)$, and $x \in H$, then $x$ is a \textbf{separating vector} for $A$ if the mapping $A \to H$ defined by $T \mapsto Tx$ is injective.
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\end{definition}
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\begin{proposition}
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\label{proposition:maximal-commutative-vn}
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Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal abelian von Neumann algebra if and only if $A = A'$.
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\end{proposition}
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\begin{proof}
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($\Leftarrow$): Let $B \supset A$ be a abelian von Neumann algebra, then $B \subset A' = A$.
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($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is abelian. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.
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\end{proof}
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\begin{proposition}
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\label{proposition:cyclic-separating-commutant}
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Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra with $I \in A$, and $x \in H$, then $x$ is cyclic for $A$ if and only if $x$ is separating for $A'$.
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 22.1]{Zhu}}}. ]
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($\Rightarrow$): Let $T \in A'$ with $Tx = 0$, then $TSx = STx = 0$ for all $S \in A$. In which case, $T(H) \subset \ol{T(Ax)} = \bracs{0}$ by \autoref{proposition:closure-of-image}.
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($\Leftarrow$): Let $P \in B(H)$ be the orthogonal projection from $H$ onto $\ol{Ax}$, then as $I \in A$, $x \in \ol{Ax}$. Since $\ol{Ax}$ is a reducing subspace for $A$, $P \in A'$. Thus $I, P \in A'$ and $(I - P)x = 0$. Given that $x$ is separating for $A'$, $I = P$, so $\ol{Ax} = H$.
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\end{proof}
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\begin{corollary}
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\label{corollary:cyclic-is-separating-commutative}
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Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, and $x \in H$ be a cyclic vector for $A$, then $x$ is also a separating vector for $A$.
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\end{corollary}
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\begin{proof}
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Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by \autoref{proposition:cyclic-separating-commutant}, it is also separating for $A$.
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\end{proof}
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\begin{theorem}
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\label{theorem:commutative-has-separating}
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Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then $A$ admits a separating vector.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
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Let $\seqj{x} \subset H$ be a maximal collection of unit vectors such that the spaces $\bracsn{Ax_j|j \in J}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable.
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Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_n/2^n$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_n = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_n \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^2(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector.
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\end{proof}
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\begin{corollary}
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\label{corollary:maximal-abelian}
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Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a maximal abelian von Neumann algebra, then $A$ admits a cyclic vector.
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\end{corollary}
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\begin{proof}[Proof, {{\cite[Corollary 22.4]{Zhu}}}. ]
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By \autoref{proposition:maximal-commutative-vn}, $A = A'$. By \autoref{theorem:commutative-has-separating}, $A$ admits a separating vector. By \autoref{proposition:cyclic-separating-commutant}, this separating vector for $A$ is a cyclic vector for $A' = A$.
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\end{proof}
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