Housekeeping: moved von Neumann algebras to its own part.
This commit is contained in:
Bokuan Li
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\section{$L^\infty$}
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\label{section:l-infty-algebra}
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\begin{proposition}
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\label{proposition:measures-dual-algebra}
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Let $X$ be a compact Hausdorff space and $\mathscr{M} \subset M_R(X; \complex)$ be a closed subspace such that:
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\begin{enumerate}
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\item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
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\end{enumerate}
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and
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\[
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J: C(X; \complex) \to \mathscr{M}^* \quad \dpn{\mu, J(f)}{\mathscr{M}} = \int f d\mu
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\]
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then
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\begin{enumerate}
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\item $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$.
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\item There exists a unique weak*-continuous involution on $\mathscr{M}^*$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by
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\[
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\dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
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\]
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\item There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
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\item $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^*$-algebra.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using \autoref{theorem:hilbert-measures-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, identify
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\[
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\mathscr{M} = [l^1(I); L^1(\mu_i; \complex)] \quad \mathscr{M}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]
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\]
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(1): Under the above, $C(X; \complex)$ may be identified as the diagonal
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\[
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\bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I} \subset [l^\infty(I); L^\infty(\mu_i; \complex)]
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\]
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By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. As a result, $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$.
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(2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$,
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\[
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\dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
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\]
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so the conjugation map is weak*-continuous.
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(3): Let $f, g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$ and $\mu \in [l^1(I); L^1(\mu_i; \complex)]$,
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\[
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\dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]}
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\]
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so the composition map is separately weak*-continuous.
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(4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra.
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\end{proof}
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\begin{lemma}
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\label{lemma:linfty-multiply-bound}
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Let $(X, \cm, \mu)$ be a measure space, $f: X \to \complex$ be a measurable function, and $C \ge 0$ such that $\norm{fg}_{L^2(X; \complex)} \le C\norm{g}_{L^2(X; \complex)}$ for all $g \in L^2(X; \complex) \cap \Sigma(X; \complex)$, then $f \in L^\infty(X; \complex)$ with $\norm{f}_{L^\infty(X; \complex)} \le C$.
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\end{lemma}
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\begin{proof}
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For any $h \in L^1(X; \complex) \cap L^\infty(X; \complex)$,
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\begin{align*}
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\normn{f^2h}_{L^1(X; \complex)} &= \int_X |f^2| \cdot |h| = \int_X |f||h|^{1/2} \cdot |f||h|^{1/2} = \normn{f|h|^{1/2}}_{L^2(X; \complex)}^2 \\
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&\le C^2 \normn{|h|^{1/2}}_{L^2(X; \complex)}^2 = C^2 \norm{h}_{L^1(X; \complex)}
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\end{align*}
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By \autoref{theorem:lp-dual-function}, $f^2 \in L^\infty(X; \complex)$ with $\normn{f^2}_{L^\infty(X; \complex)} \le C^2$. Hence $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le C$.
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\end{proof}
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\begin{proposition}
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\label{proposition:linfty-von-neumann-algebra}
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Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
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\begin{enumerate}
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\item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
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\item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
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\item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is maximal abelian.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$.
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(3): Using \autoref{proposition:maximal-commutative-vn}, it is sufficient to show that $L^\infty(X; \complex)' = L^\infty(X; \complex)$. For each $g \in L^\infty(X; \complex)$, denote $M_g \in B(L^2(X; \complex))$ as the multiplication map by $g$.
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First suppose that $\mu(X) < \infty$ and let $f = T\one_X$, then for each $g \in L^\infty(X; \complex)$,
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\[
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M_fg = M_g f= M_g T\one_X = TM_g\one_X = Tg
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\]
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By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex) \subset L^\infty(X; \complex)$ is dense in $L^2(X; \complex)$, so $M_fg = Tg$ for all $g \in L^2(X; \complex)$.
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Now suppose that $\mu$ is an arbitrary localisable measure. For each $A \in \cm$ with $\mu(A) < \infty$, the finite case shows that there exists $f_A \in L^\infty(X; \complex)$ such that
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\begin{enumerate}[label=(\roman*)]
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\item $f_A|_{X \setminus A} = 0$ almost everywhere.
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\item $Tg = M_{f_A}g$ for all $g \in L^2(X; \complex)$ with $f_A|_{X \setminus A} = 0$ almost everywhere.
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\end{enumerate}
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By (ii), $f_A|_{A \cap B} = f_B|_{A \cap B}$ for all $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$. By the \hyperref[gluing lemma for meaasurable functions]{lemma:gluing-measurable}, there exists $f \in L^\infty(X; \complex)$ such that $f|_A = f_A|_A$ for all $A \in \cm$ with $\mu(A) < \infty$.
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For each $g \in L^2(X; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. As such, there exists $\seq{A_n} \subset \cm$ such that $\bracsn{g \ne 0} = \bigsqcup_{n \in \natp}A_n$ and $\mu(A_n) < \infty$ for all $n \in \natp$. In which case,
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\[
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Tg = \sum_{n = 1}^\infty T(\one_{A_n}g) = \sum_{n = 1}^\infty M_{f_{A_n}} \one_{A_n}g = M_fg
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\]
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By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. Therefore $T = M_f$, and $L^\infty(X; \complex)$ is maximal abelian.
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\end{proof}
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\begin{theorem}[Order Structure of $L^\infty$]
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\label{theorem:l-infty-dedekind-complete}
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Let $(X, \cm, \mu)$ be a localisable measure space, then
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\begin{enumerate}
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\item $L^\infty(X; \real)$ is order complete.
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\item For each $\phi \in L^1(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^\infty(X; \real)$,
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\[
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\sup_{f \in S} \dpn{\phi, f}{L^1(X; \real)} = \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}
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\]
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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By \autoref{proposition:linfty-von-neumann-algebra}, $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra.
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(1): Since $L^\infty(X; \real)$ is a lattice, it is order complete by \autoref{theorem:existence-of-projections-vna}.
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(2): By \autoref{theorem:existence-of-projections-vna}, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of \autoref{proposition:linfty-von-neumann-algebra}, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)} = \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.
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\end{proof}
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\begin{corollary}
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\label{corollary:linfinity-extremally-disconnected}
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Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremally disconnected.
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\end{corollary}
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\begin{proof}
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By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremally disconnected.
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\end{proof}
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\begin{lemma}
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\label{lemma:l-infty-order-isotone}
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Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^\infty(X; \complex)\to L^\infty(Y; \complex)$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item $T$ is an isometric isomorphism.
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\item For each $f, g \in L^\infty(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$.
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\end{enumerate}
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then:
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\begin{enumerate}
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\item $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$.
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\item $T$ is weak*-continuous.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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Let $\phi \in L^1(Y; \complex)$ with $\phi \ge 0$, then $T^*\phi \in L^\infty(X; \complex)^*$. For each $\seq{B_n} \subset \cm$ and $B \in \cm$ with $B_n \upto B$, $\one_B = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^\infty(X; \complex)$. Thus (b) and \autoref{theorem:l-infty-dedekind-complete} imply that
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\begin{align*}
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\sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)} &= \sup_{n \in \natp} \dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)} = \dpn{\phi, T\one_B}{L^1(Y; \complex)} \\
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&= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}
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\end{align*}
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Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(X; \complex)$ with $f \ge 0$ such that $\int_B f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$.
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Let $g \in L^\infty(X; [0, 1])$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists simple functions $\seq{g_n} \subset \Sigma(X; [0, 1])$ such that $g_n \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_n$ as an element of $L^\infty(X; \real)$, so (b) and \autoref{theorem:l-infty-dedekind-complete} imply that,
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\begin{align*}
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\dpn{\phi, Tg}{L^1(Y; \complex)} &= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)} = \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)} \\
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&= \dpn{f, g}{L^1(X; \complex)}
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\end{align*}
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By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)} = \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^\infty(X; \complex)$. Therefore $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$, and $T$ is weak*-continuous.
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\end{proof}
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\begin{theorem}[Uniqueness of $L^\infty$]
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\label{theorem:linfty-uniqueness}
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Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_X \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^\infty(\mu; \complex) \to L^\infty(\nu; \complex)$ such that:
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\begin{enumerate}[label=(\alph*)]
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\item $\Phi$ is a *-isomorphism.
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\item $\Phi|_{C(X; \complex)}$ is the identity.
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\end{enumerate}
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then:
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\begin{enumerate}
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\item $\mu$ and $\nu$ are equivalent.
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\item $L^\infty(\mu; \complex) = L^\infty(\nu; \complex)$.
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\item $\Phi$ is the identity map.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 21.4]{Zhu}}}. ]
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As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^\infty(\mu; \complex)$.
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(1): By \autoref{lemma:l-infty-order-isotone}, there exists $f \in L^1(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0} implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent.
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(3): By \autoref{lemma:l-infty-order-isotone}, $\Phi$ is also weak*-continuous. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $L^\infty(\mu; \complex)$ and $L^\infty(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^\infty(\mu; \complex)$.
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\end{proof}
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@@ -0,0 +1,133 @@
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\section{The Hyperfinite $\vnII_1$ Factor}
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\label{section:hyperfinite-ii1-factor}
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\begin{definition}[Inductive Limit of Tracial von Neumann Algebras]
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\label{definition:inductive-limit-tracial-von-neumann-algebra}
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Let $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$ be an upward-directed system of tracial von Neumann algebras, where $\phi^i_j: M_i \to M_j$ is a trace-preserving *-homomorphism for each $i, j \in I$ with $i \le j$, then there exists a pair $((M, \tau), \bracsn{\phi^i}_{i \in I})$ such that:
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\begin{enumerate}
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\item $(M, \tau)$ is a tracial von Neumann algebra.
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\item For each $i \in I$, $\phi^i: M_i \to M$ is a trace-preserving *-homomorphism.
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\item For each $i, j \in I$ with $i \le j$, the following diagram commutes:
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\[
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\xymatrix{
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M_i \ar@{->}[d]_{\phi^i_j} \ar@{->}[rd]^{\phi^i} & \\
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M_j \ar@{->}[r]_{\phi^j} & M
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}
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\]
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\item[(U)] For any pair $((N, \sigma), \bracsn{\psi^i}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique trace-preserving *-homomorphism $\psi: M \to N$ such that the following diagram commutes:
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\[
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\xymatrix{
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M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\
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M \ar@{->}[r]_{\psi} & N
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}
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\]
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for all $i \in I$.
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\item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$.
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\end{enumerate}
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The pair $((M, \tau), \bracsn{\phi^i}_{i \in I}) = \lim_{\longrightarrow}(M_i, \tau_i)$ is the \textbf{inductive limit} of $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$.
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\end{definition}
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\begin{proof}
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Let $(M_0, \bracsn{\phi^i|i \in I})$ be the inductive limit of $(\bracsn{M_i}_{i \in I}, \bracsn{\phi^i_j|i \le j})$ as vector spaces. Since $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are trace-preserving and $\bracsn{\tau_i}_{i \in I}$ are faithful, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are also unital isometric embeddings. As such, $\bracsn{M_i}_{i \in I}$ may be identified as subspaces of $M_0$, the norms on $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ are isometric embeddings for all $i \in I$.
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For each $i, j \in I$ with $i \le j$, $\phi^i_j: M_i \to M_j$ is trace-preserving. By universal property of the inductive limit, there exists a linear functional $\tau: M_0 \to \complex$ such that $\tau \circ \phi^i = \tau_i$ for all $i \in I$. Similarly, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are *-homomorphisms, the products and involutions from $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ is a unital trace-preserving *-homomorphism for all $i \in I$. In particular, $1_{M_0} = \phi^i(1_{M_i})$ for all $i \in I$.
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For each $i \in I$, $\tau_i: M_i \to \complex$ is a faithful tracial state. As such, for any $x \in M_0$, $x = 0$ if and only if $\tau(x^*x) = 0$, so the pairing $\dpn{x, y}{\tau} = \tau(y^*x)$ is an inner product on $M_0$. Let $H$ be the completion of $M_0$ with respect to $\dpn{\cdot, \cdot}{\tau}$. For each $x \in M_0$, let $L_x: M_0 \to M_0$ be defined by $L_x(y) = xy$ for all $y \in M_0$. Then, for any $x, y, z \in M_0$,
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\[
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\dpn{L_xy, L_xy}{\tau} = \tau(y^*x^*xy) \le \norm{x}_{M_0}^2 \tau(y^*y) = \norm{x}_{M_0}^2 \dpn{y, y}{\tau}
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\]
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and
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\[
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\dpn{L_xy, z}{\tau} = \tau(z^*xy) = \tau((x^*z)^*y) = \dpn{y, L_{x^*}z}{\tau}
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\]
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with
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\[
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\dpn{L_x 1_{M_0}, L_x 1_{M_0}}{\tau} = \tau(x^*x) \le \tau(\normn{x}_{M_0}^2) = \norm{x}_{M_0}^2
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\]
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Thus the mapping $M_0 \to B(H)$ with $x \mapsto L_x$ allows identifying $M_0$ as an involutive subalgebra of $B(H)$.
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|
||||
Finally, let $M = M_0'' \subset B(H)$ and extend $\tau$ to $M$ by $\tau(x) = \dpn{x1_{M_0}, 1_{M_0}}{\tau}$, then $\tau$ is a vector state on $M$.
|
||||
|
||||
(1): By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $M$ is a von Neumann algebra. As $\tau(z) = \dpn{z1_{M_0}, 1_{M_0}}{\tau}$ for all $z \in M$, $\tau$ is strong-operator continuous, and the mapping $(x, y) \mapsto \tau(xy)$ is separately strong-operator continuous. Since $\tau$ is the extension of $\bracsn{\tau_i}_{i \in I}$, $\tau|_{M_i}$ is tracial for all $i \in I$, and $\tau|_{M_0}$ is also tracial. By strong-operator density of $M_0$ in $M$, $\tau$ is a tracial state.
|
||||
|
||||
(2), (3): By definition of $\tau$ and universal property of inductive limits for vector spaces.
|
||||
|
||||
(4): Let $(K, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\sigma$, then since $\sigma$ is a faithful state, the representation $\pi: N \to B(K)$ is faithful. Using \autoref{lemma:tracial-von-neumann-gns}, identify $N = \pi(N) \subset B(K)$ as a von Neumann subalgebra of $B(K)$.
|
||||
|
||||
By universal property of inductive limits of vector spaces, there exists a trace-preserving *-homomorphism $\phi_0: M_0 \to N$ such that the following diagram commutes
|
||||
\[
|
||||
\xymatrix{
|
||||
M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} & \\
|
||||
M_0 \ar@{->}[r]_{\phi_0} & N
|
||||
}
|
||||
\]
|
||||
|
||||
for all $i \in I$. Let
|
||||
\[
|
||||
U: H \to K \quad \dpn{UL_x1_{M_0}, \pi(y)\xi}{K} = \sigma(y^*\phi_0(x))
|
||||
\]
|
||||
|
||||
then for any $x \in M_0$, $\norm{UL_x1_{M_0}}_K^2 = \sigma(\phi_0(x^*x)) = \tau(x^*x) = \norm{L_x1_{M_0}}_H^2$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_0$ and $y, z \in N$,
|
||||
\begin{align*}
|
||||
\dpn{(\pi \circ \phi_0)(x)\pi(y)\xi, \pi(z)\xi}{K} &= \sigma(z^*\phi_0(x)y) = \sigma(yz^*\phi_0(x)) = \sigma((zy^*)^*\phi_0(x)) \\
|
||||
&= \dpn{UL_x1_{M_0}, \pi(zy^*)\xi}{K} = \dpn{L_x1_{M_0}, U^*\pi(zy^*)\xi}{H}
|
||||
\end{align*}
|
||||
|
||||
so $x \mapsto \dpn{\phi_0(x)\pi(y)\xi, \pi(z)\xi}{K}$ is a weak-operator continuous functional on $M_0$. Since $\pi(N)\xi$ is dense in $K$, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the weak operator topology on $\ol{B_{\pi(N)}(0, 1)}$ is induced by maps of the form $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{K}$ with $y, z \in N$. As such, $\phi_0: M_0 \to N$ is weak-operator continuous when restricted to bounded sets.
|
||||
|
||||
By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density} and \autoref{theorem:uniform-continuous-extension}, $\phi_0$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\phi$ is also trace-preserving. Finally, if $\Phi: M \to N$ is another such map, then it is also weak-operator continuous on bounded sets by \autoref{lemma:tracial-von-neumann-gns}. Therefore $\Phi = \phi$ by uniqueness of continuous extensions.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Hyperfinite $\vnII_1$ Factor]
|
||||
\label{definition:hyperfinite-ii1}
|
||||
For each $n \in \natp$, equip $M_{2^n}(\complex)$ with the normalised trace $\tau_n: M_{2^n}(\complex) \to \complex$, and
|
||||
\[
|
||||
\phi_n: M_{2^n}(\complex) \to M_{2^{n+1}}(\complex) \quad x \mapsto \begin{bmatrix}
|
||||
x & 0 \\
|
||||
0 & x
|
||||
\end{bmatrix}
|
||||
\]
|
||||
|
||||
Then, $\phi_n$ is a trace-preserving *-homomorphism, and the resulting inductive limit $\lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is a factor.
|
||||
|
||||
The tracial von Neumann algebra $(M_2(\complex)^{\ol \otimes \infty}, \tau) := \lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is the \textbf{hyperfinite $\vnII_1$ factor}.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
For each $n \in \natp$, identify $M_{2^n}(\complex)$ as a subalgebra of $M_2(\complex)^{\ol \otimes \infty}$. Let $M_0 = \bigcup_{n \in \natp}M_{2^n}(\complex)$, then by (4) of \autoref{definition:inductive-limit-tracial-von-neumann-algebra}, $M_2(\complex)^{\ol \otimes \infty} = M_0''$.
|
||||
|
||||
Let $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\tau$. Using \autoref{lemma:tracial-von-neumann-gns}, identify $M_2(\complex)^{\ol \otimes \infty}$ as a von Neumann subalgebra of $B(H)$. Let $x \in Z(M_2(\complex)^{\ol \otimes \infty})$, then by the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, there exists a net $\angles{x_\alpha}_{\alpha \in A} \subset M_0$ such that $\sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}} < \infty$ and $x_\alpha \to x$ in the strong operator topology.
|
||||
|
||||
Fix $\alpha \in A$, then there exists $n \in \natp$ such that $x_\alpha \in M_{2^n}(\complex)$. As $M_{2^n}(\complex)$ is finite-dimensional\footnote{It seems true in general that inductive limits of $\vnII_1$ factors are also $\vnII_1$ factors. However, this compactness shortcut is not available in the general case.}, \hyperref[Haar's Theorem]{theorem:haar} implies that there exists a left-translation invariant Borel probability measure $\mu: \cb_{U(M_{2^n}(\complex))} \to [0, 1]$. Let $y_\alpha = \int_{U(M_{2^n}(\complex))} ux_\alpha u^* \mu(du)$, then for each $v \in U(M_{2^n}(\complex))$,
|
||||
\begin{align*}
|
||||
vy_\alpha v^* &= \int_{U(M_{2^n}(\complex))} vux_\alpha u^*v^* \mu(du) = \int_{U(M_{2^n}(\complex))} (vu)x_\alpha (vu)^* \mu(du) \\
|
||||
&= \int_{U(M_{2^n}(\complex))} ux_\alpha u^* \mu(du) = y_\alpha
|
||||
\end{align*}
|
||||
|
||||
so $vy_\alpha = y_\alpha v$. By \autoref{corollary:sum-of-unitaries}, $y_\alpha \in Z(M_{2^n}(\complex)) = \complex I_{M_{2^n}(\complex)}$, so there exists $\lambda_\alpha \in \complex$ with $y_\alpha = \lambda_\alpha I_{M_{2^n}(\complex)}$. Since $x \in Z(M_2(\complex)^{\ol \otimes \infty})$ and $\mu$ is a probability measure\footnote{Even though $x$ is a priori not in $M_{2^n}(\complex)$, the integrand here surely has separable range, so the Bochner integral may be used to save some inner product computations. },
|
||||
\[
|
||||
(x - \lambda_\alpha)\xi = \int_{U(M_{2^n}(\complex))} (x - ux_\alpha u^*)\xi \mu(du) = \int_{U(M_{2^n}(\complex))} u(x - x_\alpha) u^*\xi \mu(du)
|
||||
\]
|
||||
|
||||
where
|
||||
\begin{align*}
|
||||
\norm{u(x - x_\alpha) u^*\xi}_H^2 &= \tau(u(x - x_\alpha)^*u^*u(x - x_\alpha)u^*) \\
|
||||
&= \tau((x - x_\alpha)^*(x - x_\alpha)uu^*uu^*) \\
|
||||
&= \tau((x - x_\alpha)^*(x - x_\alpha)) = \norm{(x - x_\alpha)\xi}_H^2
|
||||
\end{align*}
|
||||
|
||||
By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
|
||||
\begin{align*}
|
||||
\normn{(x - \lambda_\alpha)\xi}_H^2 &= \int_{U(M_{2^n}(\complex))} \dpn{u(x - x_\alpha)u^*\xi, (x - \lambda_\alpha)\xi}{H} \mu(du) \\
|
||||
&\le \norm{(x - \lambda_\alpha)\xi}_H\braks{\int_{U(M_{2^n}(\complex))}\normn{u(x - x_\alpha) u^*\xi}_H^2 \mu(du)}^{1/2} \\
|
||||
&= \normn{(x - \lambda_\alpha)\xi}_H \norm{(x - x_\alpha)\xi}_H
|
||||
\end{align*}
|
||||
|
||||
Thus $\limsup_{\alpha \in A}\normn{(x - \lambda_\alpha)\xi}_H \le \lim_{\alpha \in A}\norm{(x - x_\alpha)\xi}_H = 0$. As $x \in Z(M_2(\complex)^{\ol \otimes \infty})$, the above shows that $\lambda_\alpha \to x$ pointwise on $M_2(\complex)^{\ol \otimes \infty}H$. Since $\sup_{\alpha \in A}|\lambda_\alpha| \le \sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}} < \infty$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that $\lambda_\alpha \to x$ in the strong operator topology. Therefore $x \in \complex I_{M_2(\complex)^{\ol \otimes \infty}}$, and $M_2(\complex)^{\ol \otimes \infty}$ is a factor.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
@@ -0,0 +1,8 @@
|
||||
\chapter{Constructions of von Neumann Algebras}
|
||||
\label{chap:vna-example}
|
||||
|
||||
|
||||
\input{./bb.tex}
|
||||
\input{./lg.tex}
|
||||
\input{./hii1.tex}
|
||||
|
||||
@@ -0,0 +1,207 @@
|
||||
\section{Group von Neumann Algebras}
|
||||
\label{section:gvna}
|
||||
|
||||
\begin{definition}[Group von Neumann Algebra]
|
||||
\label{definition:group-von-neumann-algebra}
|
||||
Let $G$ be a locally compact group,
|
||||
\[
|
||||
\lambda: G \to U(L^2(G; \complex)) \quad x \mapsto L_x
|
||||
\]
|
||||
|
||||
be the left regular representation of $G$, and
|
||||
|
||||
\[
|
||||
\lambda: L^1(G; \complex) \to B(L^2(G; \complex)) \quad \lambda(f)(g) = (f * g)
|
||||
\]
|
||||
|
||||
be its extension to $L^1(G; \complex)$, then
|
||||
\[
|
||||
L(G) := \lambda(G)'' = \lambda(L^1(G;\complex))''
|
||||
\]
|
||||
|
||||
is the \textbf{group von Neumann algebra} of $G$. Similarly, if $\rho: G \to U(L^2(G; \complex))$ is the right regular representation of $G$, then $R(G) := \rho(G)'' = \rho(L^1(G;\complex))''$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
Firstly, for each $f \in L^1(G; \complex)$ and $g, h \in L^2(G; \complex)$, by the \hyperref[Fubini-Tonelli Theorem]{theorem:fubini-tonelli},
|
||||
\begin{align*}
|
||||
\dpn{\lambda(f)g, h}{L^2(G; \complex)} &= \int_G f(x)\dpn{L_xg, h}{L^2(G; \complex)} dx = \iint_{G \times G}f(x) g(x^{-1}y) \ol{h}(y) dy dx \\
|
||||
&= \iint_{G \times G}f(x) g(x^{-1}y)dx \ol{h}(y) dy \\
|
||||
&= \int_G (f * g)(y) \ol{h}(y)dy = \dpn{(f * g), h}{L^2(G; \complex)}
|
||||
\end{align*}
|
||||
|
||||
By \autoref{theorem:l1-group-von-neumann}, $\ol{\lambda(G)}^{\text{\small SOT}} = \ol{\lambda(L^1(G;\complex))}^{\text{\small SOT}}$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Left Convolver]
|
||||
\label{definition:left-convolver}
|
||||
Let $G$ be a locally compact group and $f: G \to \complex$ be Borel measurable, then $f$ is a \textbf{left convolver} if for each $g \in L^2(G; \complex)$, $\int_G |f(y)||g(y^{-1}x)|dy < \infty$ for almost every $x \in G$ and $f * g \in L^2(G; \complex)$. The mapping $L_f: L^2(G; \complex) \to L^2(G; \complex)$ defined by $g \mapsto f * g$ is the \textbf{left convolution map} by $f$.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:left-convolver-cc}
|
||||
Let $G$ be a locally compact group and $f \in C_c(G; \complex)$, $f$ is a left convolver.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
By \hyperref[Young's Inequality]{theorem:young-fourier}, for each $g \in L^2(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:left-convolver}
|
||||
Let $G$ be a locally compact group and $f: G \to \complex$ be a left convolver, then:
|
||||
\begin{enumerate}
|
||||
\item $L_f \in B(L^2(G; \complex))$.
|
||||
\item $f \in L_{\text{loc}}^1(G; \complex)$.
|
||||
\item If $f * g = 0$ for all $g \in L^2(G; \complex)$, then $f = 0$ locally almost everywhere.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
(1): Let $\seq{g_n} \subset L^2(G; \complex)$, $\seq{f * g_n} \subset L^2(G; \complex)$, and $g, h \in L^2(G; \complex)$ such that $g_n \to g$ and $f * g_n \to h$ in $L^2(G; \complex)$ as $n \to \infty$. After taking a subsequence, assume without loss of generality that:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $f * g_n \to h$ almost everywhere.
|
||||
\item $\sum_{n \in \natp}\norm{g_n - g}_{L^2(G; \complex)} < \infty$.
|
||||
\end{enumerate}
|
||||
|
||||
In which case, for almost every $x \in G$,
|
||||
\begin{align*}
|
||||
f * (g_N - g)(x) &= \int_G f(y)(g_N - g)(y^{-1}x)dy \\
|
||||
|f * (g_N - g)(x)| &\le \int_G|f(y)|\sum_{n \ge N}|(g_n - g)(y^{-1}x)| dy
|
||||
\end{align*}
|
||||
|
||||
where $\int_G|f(y)|\sum_{n \in \natp}|(g_n - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $f * g_N \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the \hyperref[Closed Graph Theorem]{theorem:closed-graph} implies that $L_f$ is continuous.
|
||||
|
||||
(2): Let $U \in \cn_G(1_G)$ be relatively compact with $U = U^{-1}$, then $\one_U \in L^2(G; \complex)$, and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for almost every $x \in G$. For each $x_0 \in G$, by \autoref{proposition:haar-measure-charge}, there exists $x \in G$ such that $x_0 \in xU$ and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$. As such, for any $K \subset G$ compact, there exists $F \subset G$ finite such that $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for each $x \in F$ and $K \subset \bigcup_{x \in F}xU$. In which case,
|
||||
\[
|
||||
\int_K |f(y)|dy \le \sum_{x \in F}\int_G |f(y)|\one_U(y^{-1}x)dy < \infty
|
||||
\]
|
||||
|
||||
(3): By \autoref{lemma:left-convolver}, $f$ is locally integrable. Let $K \subset G$ be relatively compact. Using \autoref{lemma:lch-compact-neighbour}, assume without loss of generality that $K$ is also open.
|
||||
|
||||
Let $g \in C_c(G; \complex)$ with $\supp{g} \subset K^{-1}$, then there exists $V \in \cn_G(1_G)$ such that $\supp{g}^{-1}V \subset K^{-1}$. In which case, for each $x \in V$,
|
||||
\[
|
||||
(f * g)(x) = \int_G f(y)g(y^{-1}x) dy = \int_K f(y)g(y^{-1}x) dy = (\one_Kf * g)(x)
|
||||
\]
|
||||
|
||||
so $f * g$ is continuous on $V$ by \hyperref[Young's Inequality]{theorem:young-fourier}. In particular, $f * g|_{V} = 0$ almost everywhere implies that $\int_G f(y)g(y^{-1}) dy = 0$. As such, $\int_G fg = 0$ for all $g \in C_c(G; \complex)$ with $\supp{g} \subset K$. By uniqueness from the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $f|_K = 0$ almost everywhere. Therefore $f = 0$ locally almost everywhere.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Space of Left Convolvers]
|
||||
\label{definition:left-convolver-algebra}
|
||||
Let $G$ be a group, $LC(G) := \bracsn{L_f| f: G \to \complex\text{ is a left convolver}} \subset B(l^2(G; \complex))$, and $f, g: G \to \complex$ be left convolvers, then:
|
||||
\begin{enumerate}
|
||||
\item $f \in l^2(G; \complex)$.
|
||||
\item $L_f^* = L_{f^*}$, where $f^*(x) = \ol{f(x^{-1})}$ for all $x \in G$.
|
||||
\item $L_{f * g} = L_fL_g$.
|
||||
\item $LC(G) \subset B(l^2(G; \complex))$ is a von Neumann algebra.
|
||||
\end{enumerate}
|
||||
|
||||
The set $LC(G)$ is the \textbf{algebra of left convolvers} on $G$.
|
||||
\end{definition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 1.3.5]{AnantharamanPopa}}}. ]
|
||||
(1): $f * \delta_e = f \in l^2(G; \complex)$.
|
||||
|
||||
(2): Let $x, y \in G$, then
|
||||
\begin{align*}
|
||||
\dpn{L_f\delta_x, \delta_y}{l^2(G; \complex)} &= (f * \delta_x)(y) = f(yx^{-1}) \\
|
||||
\dpn{\delta_x, L_{f^*}\delta_y}{l^2(G;\complex)} &= \ol{(f^* * \delta_y)(x)} = \ol{f^*(xy^{-1})} = f(yx^{-1})
|
||||
\end{align*}
|
||||
|
||||
|
||||
(3): By (1), $f * g \in l^2(G; \complex)$ is defined. For any $h \in l^2(G; \complex)$, by associativity of convolution,
|
||||
\[
|
||||
L_{f * g}h = (f * g) * h = f * (g * h) = L_fL_gh
|
||||
\]
|
||||
|
||||
(4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_e \in l^2(G; \complex)$, then
|
||||
\[
|
||||
f = T\delta_e = \paren{\sotlim_{\alpha \in A}L_{f_\alpha}}\delta_e = l^2\text{-}\lim_{\alpha \in A}f_\alpha * \delta_e
|
||||
\]
|
||||
|
||||
For any $g \in l^2(G; \complex)$,
|
||||
\[
|
||||
\norm{Tg - f * g}_{l^\infty(G; \complex)} \le \limsup_{\alpha \in A}\norm{g}_{l^2(G; \complex)}\norm{f - f_\alpha}_{l^2(G; \complex)} = 0
|
||||
\]
|
||||
|
||||
by \hyperref[Young's Inequality]{theorem:young-fourier}. Therefore $Tg = f * g$, $T = L_f$, and $LC(G)$ is strong-operator closed. By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $LC(G)$ is a von Neumann algebra.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:group-algebra-identified}
|
||||
Let $G$ be a group, then $L(G) = LC(G) = R(G)'$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 1.3.6]{AnantharamanPopa}}}. ]
|
||||
($\subset$): For any $x, y, z \in G$ and $f \in l^2(G; \complex)$,
|
||||
\[
|
||||
(\lambda(x)\rho(y)f)(z) = (\rho(y)f)(x^{-1}z) = f(x^{-1}zy) = (\lambda(x)f)(zy) = (\rho(y)\lambda(x)f)(z)
|
||||
\]
|
||||
|
||||
so $\lambda(G) \subset R(G)'$. By \autoref{definition:left-convolver-algebra}, $L(G) \subset LC(G) \subset R(G)'$.
|
||||
|
||||
($\supset$): Let $T \in R(G)'$ and $f = T\delta_e \in l^2(G; \complex)$, then for each $x \in G$,
|
||||
\[
|
||||
T\delta_x = T\rho(x^{-1})\delta_e = \rho(x^{-1})T\delta_e = f * \delta_x
|
||||
\]
|
||||
|
||||
As such, $Tg = f * g$ for all $g \in l^2(G; \complex)$, and $T = L_f$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}
|
||||
\label{definition:group-algebra-fourier-series}
|
||||
Let $G$ be a group. For each $x \in G$, let $u_x = \lambda(x)$ and identify it with $\delta_x$, then $\lambda(G) = \bracsn{u_x|x \in G}$ are the \textbf{canonical unitaries} of $L(G)$.
|
||||
|
||||
For each $f \in l^2(G)$, $f = \sum_{x \in G}f(x)u_x$. In particular, for each $T \in L(G) = LC(G) \subset l^2(G; \complex)$, there exists $\bracsn{T_x}_{x \in G} \subset \complex$ such that $T$ may be written as $T = \sum_{x \in G}T_x u_x$\footnote{This sum does not converge in the strong-operator or the weak-operator topology. }. In which case, the scalars $\bracsn{T_x}_{x \in G}$ are the \textbf{Fourier coefficients} of $T$.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[ICC]
|
||||
\label{definition:icc}
|
||||
Let $G$ be a group, then $G$ has \textbf{infinite conjugacy classes (ICC)} if every non-trivial conjugacy class of $G$ is infinite.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:icc-factor}
|
||||
Let $G$ be a group, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $L(G)$ is a factor.
|
||||
\item $G$ is ICC.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 1.3.9]{AnantharamanPopa}}}. ]
|
||||
(2) $\Rightarrow$ (1): Let $T \in Z(L(G))$, then for each $x \in G$,
|
||||
\[
|
||||
T\delta_e = \lambda(x)T\lambda(x^{-1})\delta_e = \lambda(x)T\rho(x)\delta_e = \lambda(x)\rho(x)(T\delta_e)
|
||||
\]
|
||||
|
||||
so $T\delta_e$ is invariant under conjugation, and hence constant on conjugacy classes of $G$.
|
||||
|
||||
Since $G$ is ICC and $T\delta_e \in l^2(G; \complex)$, there exists $\lambda \in \complex$ with $T\delta_e = \lambda \delta_e$. As such, $T = \lambda I$, and $Z(L(G)) = \complex I$.
|
||||
|
||||
$\neg$ (2) $\Rightarrow$ $\neg$ (1): Let $C \subset G$ be a non-trivial finite conjugacy class, then for each $x \in G$, $\delta_x * \one_C = \one_{xC} = \one_{Cx} = \one_C * \delta_x$, so $L_{\one_C}$ is a non-trivial element of $Z(L(G))$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Wreath Product]
|
||||
\label{definition:wreath-product}
|
||||
Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, $H^{(I)} = \bigoplus_{i \in I}H$, and $\Gamma$ act on $H^{(I)}$ by left translation ($(\gamma \xi)_i = \xi_{\gamma^{-1}i}$), then the \textbf{generalised wreath product} $H \wr_I \Gamma$ is the semidirect product $H^{(I)} \rtimes \Gamma$. The \textbf{wreath product} $H \wr \Gamma$ is the special case where $I = \Gamma$, and $\Gamma$ acts by left translation.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:wreath-product-icc}
|
||||
Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, and $\Gamma_f \subset \Gamma$ be the subgroup of elements whose conjugacy class is finite. If
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item The orbits of $\Gamma \curvearrowleft I$ are infinite.
|
||||
\item The restricted action $\Gamma_f \curvearrowright I$ is faithful.
|
||||
\end{enumerate}
|
||||
|
||||
then the wreath product $H \wr_I \Gamma$ is ICC.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 1.3.10]{AnantharamanPopa}}}. ]
|
||||
For each $g \in H \wr_I \Gamma$ and subgroup $K \subset H \wr_I \Gamma$, let $g^K = \bracsn{kgk^{-1}|k \in K}$.
|
||||
|
||||
Let $g = (\xi, \gamma) \in H \wr_I \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^G \supset g^\Gamma$ is infinite.
|
||||
|
||||
Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_f$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_f$. By (b), there exists $i_0 \in I$ with $\gamma i_0 \ne i_0$. As $H$ is non-trivial, there exists $\xi_0 \in H^{(I)} \setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_0 = (\xi_0, e_{\Gamma})$, then
|
||||
\[
|
||||
g_0^{-1}gg_0 = (\xi_0^{-1}\gamma\xi_0, \gamma)
|
||||
\]
|
||||
|
||||
and hence $g$ has infinite conjugacy class because $\xi_0^{-1}\gamma\xi_0 \ne e_{H^{(I)}}$. Therefore $g^G$ is also infinite.
|
||||
\end{proof}
|
||||
@@ -0,0 +1,99 @@
|
||||
\section{The Fundamental Group}
|
||||
\label{section:vna-fundamental-group}
|
||||
|
||||
\begin{definition}[Induced von Neumann Algebra]
|
||||
\label{definition:induced-vna-commutant}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in M'$, then $Mq$ is a von Neumann algebra on $qH$, known as the \textbf{induced von Neumann algebra} of $M$ with respect to $q$.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:induced-vna-commutant-gymnastics}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in M$, then:
|
||||
\begin{enumerate}
|
||||
\item For any central projection $e \in Z(pMp)$, $e = Z(e)p$.
|
||||
\item $Z(pMp) = Z(M)p$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.2.1]{AnantharamanPopa}}}. ]
|
||||
(1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^* = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by \autoref{proposition:central-support-vna}, and $e = z(e)p$.
|
||||
|
||||
(2): By \autoref{theorem:vn-projection-norm-dense}.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:vna-induced-commutants}
|
||||
Let $H$ be a complex Hilbert space and $M \subset B(H)$ be a von Neumann algebra, then
|
||||
\begin{enumerate}
|
||||
\item For any $p \in \text{Proj}(M)$, the commutant $(pMp)'$ in $B(pH)$ is $pM'$.
|
||||
\item For any $q \in \text{Proj}(M')$, the commutant $(Mq)'$ in $B(qH)$ is $qM'q$.
|
||||
\end{enumerate}
|
||||
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.2.2]{AnantharamanPopa}}}. ]
|
||||
(1): For any $x \in M$ and $y \in M'$, $pxp \cdot yp = ypxp = yp \cdot pxp$, so $pMp \subset (M'p)'$.
|
||||
|
||||
For any $x \in (M'p)' \subset B(pH)$, let $\tilde x = xp = pxp \in B(H)$, then for any $y \in M'$,
|
||||
\[
|
||||
\tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x
|
||||
\]
|
||||
|
||||
so $\tilde x \in M'$, $x = p\tilde x p \in pMp$, and $pMp \subset (M'p)'$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:amplified-ii1-factor}
|
||||
Let $(M, \tau)$ be a $\vnII_1$ factor, then $M_n(M)$ is a $\vnII_1$ factor, with normalised trace $\frac{1}{n}\text{Tr}_n \otimes \tau$.
|
||||
\end{proposition}
|
||||
|
||||
|
||||
\begin{definition}[Amplification]
|
||||
\label{definition:ii1-amplification}
|
||||
Let $(M, \tau)$ be a $\vnII_1$ factor, $t > 0$, $n \in \natp$ with $n \ge t$, and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$, then the von Neumann algebra $M^t := pM_n(M)p \subset M_n(M)$ is an \textbf{amplification} of $M$.
|
||||
|
||||
The spatial isomorphism class of $M^t$ is independent of the choice of $n \ge t$ and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
By \autoref{proposition:ii1-trace-projection-equivalence-classes}.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:ii1-amplification-homomorphism}
|
||||
Let $(M, \tau)$ be a $\vnII_1$ factor and $s, t > 0$, then $(M^s)^t \iso M^{st}$.
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 4.2.3]{AnantharamanPopa}}}. ]
|
||||
Let $m \in \natp$ and $p \in \text{Proj}(M_m(M))$ with $(\text{Tr}_m \otimes \tau)(p) = s$, so that $M^s = pM_m(M)p$, and $\tau_s = s^{-1}(\text{Tr}_m \otimes \tau)|_{M^s}$ denotes the normalised tracial state on $M^s$.
|
||||
|
||||
Let $n \in \natp$ and $q \in \text{Proj}(M_n(M^s))$ with $(\text{Tr}_n \otimes \tau_s)(p) = t$, so that $(M^s)^t = qM_n(M^s)q$. Identify $M_n(M^s) \subset M_n(M_m(M))$, so that $q \in M_n(M_m(M))$, then
|
||||
\begin{align*}
|
||||
(M^s)^t &= qM_n(M^s)q = qM_n(pM_m(M)p)q \\
|
||||
&= q(1_{M_n(\complex)} \otimes p)M_n(M_m(M))(1_{M_n(\complex)} \otimes p)q
|
||||
\end{align*}
|
||||
|
||||
|
||||
Since $q \in M_n(M^s)$, $p$ is the identity on entries of $q$, so
|
||||
\[
|
||||
q(1_{M_n(\complex)} \otimes p) = q = (1_{M_n(\complex)} \otimes p)q
|
||||
\]
|
||||
|
||||
and $(M^s)^t = qM_n(M_m(M))q = qM_{mn}(M)q$. Finally,
|
||||
\[
|
||||
(\text{Tr}_{mn} \otimes \tau)(q) = (\text{Tr}_n \otimes \text{Tr}_{m} \otimes \tau)(q) = s(\text{Tr}_n \otimes \tau_s)(q) = st
|
||||
\]
|
||||
|
||||
Therefore $(M^s)^t = M^{st}$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{definition}[Fundamental Group]
|
||||
\label{definition:ii1-fundamental-group}
|
||||
Let $M$ be a $\vnII_1$ factor, then the \textbf{fundamental group} of $M$ is the family
|
||||
\[
|
||||
\fF(M) = \bracsn{t > 0| M^t \iso M} \subset \real_{>0}^\times
|
||||
\]
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
By \autoref{lemma:ii1-amplification-homomorphism}, $\fF(M)$ is a group.
|
||||
\end{proof}
|
||||
|
||||
|
||||
@@ -0,0 +1,5 @@
|
||||
\chapter{Finite and $\vnII_1$ Factors}
|
||||
\label{chap:ii1-factors}
|
||||
|
||||
\input{./trace.tex}
|
||||
\input{./fg.tex}
|
||||
@@ -0,0 +1,190 @@
|
||||
\section{Traces}
|
||||
\label{section:tracial-von-neumann-algebra}
|
||||
|
||||
\begin{definition}[Trace]
|
||||
\label{definition:trace}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\tau: M_+ \to [0, \infty]$, then $\tau$ is a \textbf{trace} on $M$ if:
|
||||
\begin{enumerate}
|
||||
\item For each $x, y \in M_+$ and $\lambda \ge 0$, $\tau(\lambda x + y) = \lambda \tau(x) + \tau(y)$.
|
||||
\item For each $x \in M_+$ and $u \in U(M)$, $\tau(uxu^*) = \tau(x)$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Tracial von Neumann Algebra]
|
||||
\label{definition:tracial-von-neumann-algebra}
|
||||
Let $M$ be a von Neumann algebra and $\tau: M \to \complex$ be a faithful tracial state, then $(M, \tau)$ is a \textbf{tracial von Neumann algebra} if $\tau$ is normal.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Standard Representation]
|
||||
\label{definition:tracial-vna-standard}
|
||||
Let $(M, \tau)$ be a tracial von Neumann algebra, then its \hyperref[GNS]{definition:gns-triple} representation on $L^2(M, \tau)$ is the \textbf{standard representation} of $(M, \tau)$.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:cstar-trace-vna}
|
||||
Let $M$ be a unital $C^*$-algebra, $\tau: M \to \complex$ be a faithful tracial state, and $(L^2(M, \tau), \pi, 1_M)$ be its \hyperref[GNS]{definition:gns-triple} triple, then:
|
||||
\begin{enumerate}
|
||||
\item The uniform structure of pointwise convergence\footnote{Strong operator topology.} on $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ coincides with the uniform structure of $\ol{B_M(0, 1)} \subset L^2(M, \tau)$ induced by the $L^2(M, \tau)$ norm.
|
||||
\item $\pi(M) \subset B(L^2(M, \tau))$ is a von Neumann algebra if and only if $\ol{B_M(0, 1)}$ is complete in $L^2(M, \tau)$. In which case, $\tau$ is normal.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 2.6.4]{AnantharamanPopa}}}. ]
|
||||
(1): For any $x, y \in \ol{B_M(0, 1)}$,
|
||||
\begin{align*}
|
||||
\norm{x - y}_{L^2(M, \tau)}^2 &= \tau((x - y)^*(x - y)) = \dpn{\pi(x - y)^*\pi(x - y)1_M, 1_M}{L^2(M, \tau)} \\
|
||||
&= \dpn{\pi(x - y)1_M, \pi(x - y)1_M}{L^2(M, \tau)} = \norm{\pi(x - y)1_M}_{L^2(M, \tau)}^2
|
||||
\end{align*}
|
||||
|
||||
so the uniform structure of pointwise convergence is stronger than the uniformity induced by the $L^2(M, \tau)$ norm.
|
||||
|
||||
On the other hand, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the uniform structure of pointwise convergence on $\ol{B_M(0, 1)}$ coincides with the uniform structure of pointwise convergence on any dense subset of $L^2(M, \tau)$. In particular, this includes $M \subset L^2(M, \tau)$. For each $x, y \in \ol{B_M(0, 1)}$ and $t \in M$, since $\tau$ is tracial,
|
||||
\begin{align*}
|
||||
\norm{(x - y)t}_{L^2(M, \tau)}^2 &= \tau(t^*(x - y)^*(x - y)t) \le \tau(tt^*(x - y)^*(x - y)) \\
|
||||
&\le \norm{t}_M^2 \tau((x - y)^*(x - y)) = \norm{t}_M^2 \norm{x - y}_{L^2(M, \tau)}^2
|
||||
\end{align*}
|
||||
|
||||
so the uniformity induced by the $L^2(M, \tau)$ norm is stronger than the uniform structure of pointwise convergence.
|
||||
|
||||
(2): Using (1), it is sufficient to show that $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ is complete if and only if it is strong-operator closed. By \autoref{proposition:complete-closed}, $\ol{B_M(0, 1)}$ is complete if and only if it is closed in $L^2(M, \tau)^{L^2(M, \tau)}$ with respect to the product topology. Finally, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that $\ol{B_M(0, 1)}$ is closed in $L^2(M, \tau)^{L^2(M, \tau)}$ if and only if it is strong-operator closed in $B(L^2(M, \tau))$.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:tracial-separable-condition}
|
||||
Let $(M, \tau)$ be a tracial von Neumann algebra, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $M$ is separable.
|
||||
\item $\ol{B_M(0, 1)}$ is strong-operator separable.
|
||||
\item $\ol{B_M(0, 1)}$ is separable in $L^2(M, \tau)$.
|
||||
\item $L^2(M, \tau)$ is separable.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
(1) $\Rightarrow$ (2): Let $H$ be a separable Hilbert space with $M \subset B(H)$ and $D \subset H$ be countable and dense. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the strong operator topology on $\ol{B_M(0, 1)}$ coincides with the topology of pointwise convergence on $D$. By \autoref{theorem:uniform-metrisable} and \autoref{proposition:separable-product}, $H^D$ is metrisable and separable. Therefore $\ol{B_M(0, 1)} \subset H^D$ is also metrisable and separable.
|
||||
|
||||
(2) $\Rightarrow$ (3): By \autoref{proposition:cstar-trace-vna}.
|
||||
|
||||
(3) $\Rightarrow$ (4): By \autoref{lemma:tracial-von-neumann-gns}, the standard representation $(L^2(M, \tau), \pi, 1_M)$ is faithful and normal. As $M$ is dense in $L^2(M, \tau)$, $\ol{B_M(0, 1)}$ being separable in $L^2(M, \tau)$ implies that $L^2(M, \tau)$ is separable.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:tracial-von-neumann-factor}
|
||||
Let $(M, \tau)$ be a tracial von Neumann factor, then exactly one of the following holds:
|
||||
\begin{enumerate}
|
||||
\item There exists $n \in \natp$ such that $M \iso M_n(\complex)$.
|
||||
\item $M$ is of type $\vnII_1$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
(1): Suppose that $M$ has a minimal projection. By \autoref{theorem:type1-bh}, there exists a complex Hilbert space $H$ such that $M \iso B(H)$. Let $\seqi{e} \subset H$ be an orthonormal basis, then $e_i \otimes e_i \in \text{Proj}(M)$ is minimal for all $i \in I$, and all such rank one projections are equivalent. As such, for any $j \in I$,
|
||||
\[
|
||||
\tau(1_M) = \sum_{i \in I}\tau(e_i \otimes e_i) = |I| \cdot \tau(e_j \otimes e_j)
|
||||
\]
|
||||
|
||||
so $I$ must be finite, and $M \iso M_{|I|}(\complex)$.
|
||||
|
||||
(2): Suppose that $M$ has no minimal projections. By \autoref{lemma:abelian-minimal-factor}, $M$ has no abelian projections. As $\tau$ is a faithful tracial state, $1_M$ is finite. Therefore $M$ is a $\vnII_1$ factor.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:diffuse-factor-subsplit}
|
||||
Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 \le p$.
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
|
||||
Since $p$ is not minimal, there exists a projection $q \le p$ with $0 \ne q \ne p$. As $M$ is a factor, $Z(q)Z(p - q) = 1_M$. By \autoref{proposition:central-support-mvn}, there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$, $p_1 \le q$, and $p_2 \le p - q$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:diffuse-factor-split}
|
||||
Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 = p$.
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
|
||||
Let
|
||||
\[
|
||||
\mathcal{F} = \bracsn{(q, r) \in \text{Proj}(M)^2|q \sim r, q \perp r, q + r \le p}
|
||||
\]
|
||||
|
||||
and $\mathcal{M} \subset \mathcal{F}$ be a maximal family such that $\bracsn{q, r|(q,r) \in \mathcal{M}}$ are all mutually orthogonal. Let $q = \sum_{(q, r) \in \mathcal{M}}q$ and $r = \sum_{(q, r) \in \mathcal{M}}r$, then by \autoref{lemma:mvn-equivalent-direct-sum}, $q \sim r$, $q \perp r$, and $q + r \le p$.
|
||||
|
||||
If $q + r \ne p$, then by \autoref{lemma:diffuse-factor-subsplit} applied to $p - (q + r)$, there exist non-zero projections $q', r' \in \text{Proj}(M)$ with $q' \sim r'$, $q' \perp r'$, and $q' + r' \le p - (q + r)$, which contradicts the maximality of $\mathcal{M}$.
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{corollary:diffuse-factor-dyadic}
|
||||
Let $M$ be a diffuse factor, then for each $n \in \natp$, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. If $\tau: M \to \complex$ is a tracial state, then $\tau(p_j) = 2^{-n}$ for all $1 \le j \le 2^n$.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
By \autoref{lemma:diffuse-factor-split}.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:tracial-factor-simple}
|
||||
Let $M$ be a factor, then $M$ admits at most one tracial state.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.1.3]{AnantharamanPopa}}}. ]
|
||||
Using \autoref{proposition:tracial-von-neumann-factor}, assume without loss of generality that $M$ is diffuse.
|
||||
|
||||
Let $p \in \text{Proj}(M)$ with $0 \le p < 1_M$. Let $n \in \natp$, then by \autoref{corollary:diffuse-factor-dyadic}, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. By the \hyperref[corollary:vna-factor-comparability]{theorem:vna-comparability}, there exists a unique $0 \le k \le 2^n - 1$ such that
|
||||
\[
|
||||
\sum_{j = 1}^k p_j \preceq p \prec \sum_{j = 1}^{k+1}p_j
|
||||
\]
|
||||
|
||||
As such, for any tracial state $\tau: M \to \complex$,
|
||||
\[
|
||||
\frac{k}{2^n} \le \tau(p) \le \frac{k+1}{2^n}
|
||||
\]
|
||||
|
||||
so $\tau(p)$ is independent of the choice of $\tau$. By \autoref{theorem:vn-projection-norm-dense}, $\tau$ is unique.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:tracial-factor-unique-trace}
|
||||
Let $(M, \tau)$ be a tracial von Neumann algebra, then $M$ is a factor if and only if $\tau$ is the unique normal faithful tracial state on $M$.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.1.4]{AnantharamanPopa}}}. ]
|
||||
If $M$ is a factor, then $\tau$ is unique by \autoref{proposition:tracial-factor-simple}. Suppose that $M$ is not a factor, then $M$ admits a non-trivial central projection $z \in Z(M)$. In which case, for any $\alpha \in (0, 1)$,
|
||||
\[
|
||||
\tilde \tau(x) = \frac{\alpha}{\tau(z)} \tau(xz) + \frac{1 - \alpha}{1 - \tau(z)}\tau(x(1 - z))
|
||||
\]
|
||||
|
||||
is also a normal faithful tracial state. In particular,
|
||||
\[
|
||||
\tilde \tau(z) = \frac{\alpha}{\tau(z)} \tau(z) + \frac{1 - \alpha}{1 - \tau(z)}\tau(z(1 - z)) = \alpha
|
||||
\]
|
||||
|
||||
so if $\alpha \ne \tau(z)$, then $\tilde \tau \ne \tau$.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:tracial-factor-calkin}
|
||||
Let $(M, \tau)$ be a tracial factor, then $M$ admits no non-trivial two-sided ideal.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.1.5]{AnantharamanPopa}}}. ]
|
||||
Let $I \subset M$ be a non-zero two-sided ideal. By \autoref{proposition:vna-two-sided-ideal}, $I$ admits a non-zero projection $e \in I$. Let $\mathcal{P} \subset \text{Proj}(M)$ be a maximal family of mutually orthogonal projections equivalent to $e$. Since $\tau$ is faithful, $\tau(p) > 0$ for all $p \in \mathcal{P}$, so $\mathcal{P}$ is finite.
|
||||
|
||||
Let $q = 1_M - \sum_{p \in \mathcal{P}}p$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $q \preceq e$ or $e \prec q$. If $e \prec q$, then there exists a subprojection $p \le q$ with $e \sim p$, which contradicts the maximality of $\mathcal{P}$. Thus $q \preceq e$.
|
||||
|
||||
Let $\mathcal{Q} = \mathcal{P} \sqcup \bracsn{q}$, then $p \preceq e$ for all $p \in \mathcal{Q}$ and $1_M = \sum_{p \in \mathcal{Q}}p$. For each $p \in \mathcal{Q}$, let $v_p \in M$ be a partial isometry such that $v_pv_p^* = p$ and $v_p^*v_p \le e$. As such, $p = v_pev_p^* \in I$ for all $p \in \mathcal{Q}$. Therefore $1_M = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:ii1-trace-projection-equivalence-classes}
|
||||
Let $M$ be a $\vnII_1$ factor with tracial normal state $\tau: M \to \complex$, then the mapping $p \mapsto \tau(p)$ induces a bijection from the equivalence classes of $\text{Proj}(M)$ to $[0, 1]$.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 4.1.6]{AnantharamanPopa}}}. ]
|
||||
(Injective): Let $p, q \in \text{Proj}(M)$ with $\tau(p) = \tau(q)$. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $p \prec q$, $p \sim q$, or $q \prec p$. Since $\tau$ is faithful, $p \prec q$ implies that $\tau(p) < \tau(q)$, and $q \prec p$ implies that $\tau(q) < \tau(p)$. Therefore the only possibility is $p \sim q$.
|
||||
|
||||
(Surjective): By \autoref{lemma:diffuse-factor-split}, there exists a family $\bracsn{p_r|r \in \mathbb D \cap [0, 1]} \subset \text{Proj}(M)$ such that:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item For each dyadic rational number $r \in \mathbb D \cap [0, 1]$, $\tau(p_r) = r$.
|
||||
\item For each $r, s \in \mathbb D \cap [0, 1]$ with $r \le s$, $p_r \le p_s$.
|
||||
\end{enumerate}
|
||||
|
||||
For each $t \in [0, 1]$, let $p_t = \sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}$, then since $\tau$ is normal,
|
||||
\begin{align*}
|
||||
\tau(p_t) &= \tau(\sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}) \\
|
||||
&= \sup\bracsn{\tau(p_r)|r \in \mathbb D \cap [0, 1], r \le t} = t
|
||||
\end{align*}
|
||||
|
||||
|
||||
\end{proof}
|
||||
|
||||
@@ -0,0 +1,9 @@
|
||||
\part{Von Neumann Algebras}
|
||||
\label{part:vna}
|
||||
|
||||
\input{./vn/index.tex}
|
||||
\input{./example/index.tex}
|
||||
\input{./ii1/index.tex}
|
||||
\input{./notation.tex}
|
||||
|
||||
|
||||
@@ -0,0 +1,13 @@
|
||||
\chapter{Notations}
|
||||
\label{chap:vna-notations}
|
||||
|
||||
\begin{tabular}{lll}
|
||||
\textbf{Notation} & \textbf{Description} & \textbf{Source} \\
|
||||
\hline
|
||||
|
||||
$E_{x, y}$ & Section of a spectral measure $E$, where $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\
|
||||
$\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
|
||||
$Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\
|
||||
$P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\
|
||||
$P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent}
|
||||
\end{tabular}
|
||||
@@ -0,0 +1,53 @@
|
||||
\section{The Cayley Transform}
|
||||
\label{section:cayley-transform}
|
||||
|
||||
\begin{definition}[Cayley Transform]
|
||||
\label{definition:cayley-transform-bounded}
|
||||
Let $H$ be a complex Hilbert space and $T \in B(H)$ with $-i \not\in \sigma_A(H)$, then $U(T) = (T - i)(T + i)^{-1}$ is the \textbf{Cayley transform} of $T$.
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:cayley-sa-uni}
|
||||
Let $H$ be a complex Hilbert space and $U_1$ be the set of unitary operators on $H$ with $1$ not in their spectrum, then the Cayley transform $T \mapsto (T - i)(T + i)^{-1}$ is a strong-operator continuous bijection between $B(H)_{sa}$ and $U_1$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 19.3]{Zhu}}}. ]
|
||||
Let $T \in B(H)$ be self-adjoint. By \autoref{proposition:self-adjoint-spectrum}, $\sigma_{B(H)}(T) \subset \real$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous},
|
||||
\[
|
||||
\sigma_{B(H)}[(T-i)(T+i)^{-1}] \subset \bracsn{(t - i)/(t + i)|t \in \real} \subset \partial B_\complex(0, 1) \setminus \bracsn{1}
|
||||
\]
|
||||
|
||||
Hence $(T - i)(T+i)^{-1}$ is a well-defined unitary element of $B(H)$ whose spectrum does not contain $1$.
|
||||
|
||||
Since the mapping $t \mapsto -i(t + 1)/(t - 1)$ is the inverse of $t \mapsto (t - i)/(t + i)$ on $\partial B_\complex(0, 1)$, the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous} implies that $T \mapsto -i(T + I)(T - I)^{-1}$ is the inverse of the Cayley transform on $U_1$.
|
||||
|
||||
For any self-adjoint elements $S, T \in B(H)$,
|
||||
\begin{align*}
|
||||
U(S) - U(T) &= (S + i)^{-1}(S - i) - (T - i)(T+i)^{-1} \\
|
||||
&= (S + i)^{-1}[(S - i)(T + i) - (S + i)(T - i)](T + i)^{-1} \\
|
||||
&= 2i(S + i)^{-1}(S - T)(T + i)^{-1}
|
||||
\end{align*}
|
||||
|
||||
so for any $x \in H$,
|
||||
\[
|
||||
\normn{[U(S) - U(T)]x}_H \le 2\normn{(S + i)^{-1}}_{B(H)} \cdot \normn{(S - T)(T+i)^{-1}x}_H
|
||||
\]
|
||||
|
||||
By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\normn{(S+i)^{-1}}_{B(H)} \le 1$. Therefore the Cayley transform is strong-operator continuous.
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{corollary:functional-calculus-c0-self-adjoint}
|
||||
Let $H$ be a complex Hilbert space and $f \in C_0(\real; \complex)$, then the mapping $T \mapsto f(T)$ is strong-operator continuous on $B(H)_{sa}$.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Let
|
||||
\[
|
||||
g: \partial B_\complex(0, 1) \to \complex \quad z \mapsto \begin{cases}
|
||||
f(-i(z+1)/(z-1)) &z \ne 1 \\
|
||||
0 &z = 1
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
then since $f \in C_0(\real; \complex)$, $g \in C(\partial B_\complex(0, 1); \complex)$. For each $T \in B(H)_{sa}$, $f(T) = g(U(T))$. By \autoref{proposition:bh-adjoint-strong-continuous}, the mapping $U \mapsto g(U)$ is strong-operator continuous on the set of unitary operators on $H$. By \autoref{theorem:cayley-sa-uni}, $T \mapsto f(T)$ is the composition of two strong-operator continuous mappings.
|
||||
\end{proof}
|
||||
|
||||
@@ -0,0 +1,56 @@
|
||||
\section{Commutative von Neumann Algebras}
|
||||
\label{section:vn-commutative}
|
||||
|
||||
\begin{definition}[Separating Vector]
|
||||
\label{definition:separating-vector}
|
||||
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.
|
||||
\end{definition}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:maximal-commutative-vn}
|
||||
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 $A = A'$.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
($\Leftarrow$): Let $B \supset A$ be a abelian von Neumann algebra, then $B \subset A' = A$.
|
||||
|
||||
($\Rightarrow$): 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 $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:cyclic-separating-commutant}
|
||||
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 $A'$.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 22.1]{Zhu}}}. ]
|
||||
($\Rightarrow$): 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}.
|
||||
|
||||
($\Leftarrow$): 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 $\ol{Ax} = H$.
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{corollary:cyclic-is-separating-commutative}
|
||||
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 $A$.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by \autoref{proposition:cyclic-separating-commutant}, it is also separating for $A$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:commutative-has-separating}
|
||||
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.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
|
||||
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.
|
||||
|
||||
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.
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{corollary:maximal-abelian}
|
||||
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.
|
||||
\end{corollary}
|
||||
\begin{proof}[Proof, {{\cite[Corollary 22.4]{Zhu}}}. ]
|
||||
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 $A' = A$.
|
||||
\end{proof}
|
||||
|
||||
@@ -0,0 +1,115 @@
|
||||
\section{The $L^\infty$ Functional Calculus}
|
||||
\label{section:linfty-functional-calculus}
|
||||
|
||||
|
||||
|
||||
\begin{definition}[$L^\infty$ Functional Calculus]
|
||||
\label{definition:linfty-functional-calculus}
|
||||
Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
|
||||
\begin{enumerate}
|
||||
\item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that
|
||||
\[
|
||||
T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)
|
||||
\]
|
||||
\item Let $\mathscr{E} \subset M_R(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
|
||||
\[
|
||||
I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE
|
||||
\]
|
||||
|
||||
is a *-isomorphism.
|
||||
\item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$.
|
||||
\end{enumerate}
|
||||
|
||||
The spectral measure $E$ is the \textbf{resolution of the identity} for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
(1), (2): By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
|
||||
\[
|
||||
I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
|
||||
\]
|
||||
|
||||
is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.
|
||||
|
||||
For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}.
|
||||
|
||||
(3): By uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:vn-projection-norm-dense}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
|
||||
\begin{enumerate}
|
||||
\item For each normal operator $T \in A$ and $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$.
|
||||
\item For each $T \in A$ with $0 \le T \le 1$, there exists projections $\seq{P_n} \subset A$ such that
|
||||
\[
|
||||
T = \sum_{n = 1}^\infty \frac{P_n}{2^n}
|
||||
\]
|
||||
\item The linear span of projections in $A$ is norm dense in $A$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}
|
||||
(1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By \hyperref[Urysohn's Lemma]{lemma:urysohn}, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B} = 1$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, there exists a net $\angles{g_\gamma}_{\gamma \in C} \subset C(\sigma_{B(H)}(T); \complex)$ such that $g_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_B = \one_B$, $fg_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well.
|
||||
|
||||
By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_\gamma(T) \in A$ for all $\gamma \in C$. Thus the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus} implies that $\one_B(T) \in A$ as well.
|
||||
|
||||
If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)} \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_B(T) \in A$ as well.
|
||||
|
||||
By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By \autoref{lemma:separable-metric-space-approx-identity}, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$.
|
||||
|
||||
|
||||
|
||||
(2): Let $f = \text{Id}: [0, 1] \to [0, 1]$, then
|
||||
\[
|
||||
f = \sum_{n = 1}^\infty \frac{\one_{\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2}}}{2^n}
|
||||
\]
|
||||
|
||||
Let $E: \cb_{\sigma_{A}(T)} \to \text{Proj}(A)$ be the resolution of the identity for $f$, then by the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus},
|
||||
\begin{align*}
|
||||
T &= f(T) = \sum_{n = 1}^\infty\frac{\one_{\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2}}(T)}{2^n} \\
|
||||
&= \sum_{n = 1}^\infty 2^{-n}E(\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2})
|
||||
\end{align*}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}
|
||||
\label{theorem:von-neumann-group-connected}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
|
||||
\begin{enumerate}
|
||||
\item $G(A)$ is path-connected in the norm topology.
|
||||
\item The unitary group of $A$ is path-connected in the norm topology.
|
||||
\item $I(A)$ is trivial.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}
|
||||
Using \autoref{theorem:existence-of-projections-vna} and after possibly shrinking $H$, assume without loss of generality that $I \in A$.
|
||||
|
||||
After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $e^{\phi(z)} = z$ for all $z \in \complex \setminus \bracs{0}$.
|
||||
\item For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)} \setminus B(0, r)$.
|
||||
\end{enumerate}
|
||||
|
||||
(1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_A(T) \subset \ol{B(0, R)} \setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_A(T)$. By the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$.
|
||||
|
||||
(2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^\infty$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by \autoref{corollary:spectrum-characterisation-iff}.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:vna-two-sided-ideal}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $I \subset M$ be a two-sided ideal, then:
|
||||
\begin{enumerate}
|
||||
\item $I$ is self-adjoint.
|
||||
\item Let $x \in I$ with $x > 0$, $E: \cb_{[0, \infty)} \to \text{Proj}(M)$ be its resolution of the identity, and $t > 0$, then $E([t, \infty)) \in I$.
|
||||
\item If $I$ is weak-operator closed, then there exists a unique projection $z \in Z(M)$ with $I = Mz$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 2.4.15]{AnantharamanPopa}}}. ]
|
||||
(1): Let $x = u|x|$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $x$, then $|x| = u^*x \in I$, and $x^* = |x|u^* \in I$.
|
||||
|
||||
(2): Let $f: [0, \infty) \to \real$ be defined by $f(s) = s^{-1}\one_{\bracsn{s \ge t}}(s)$, then by the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $E([t, \infty)) = xf(x) \in I$.
|
||||
|
||||
(3): By (5) of \autoref{theorem:existence-of-projections-vna}.
|
||||
\end{proof}
|
||||
|
||||
|
||||
@@ -0,0 +1,12 @@
|
||||
\chapter{Fundamentals of von Neumann Algebras}
|
||||
\label{chap:von-neumann-algebras}
|
||||
|
||||
\input{./topologies.tex}
|
||||
\input{./cayley.tex}
|
||||
\input{./vn.tex}
|
||||
\input{./weak.tex}
|
||||
\input{./commutative.tex}
|
||||
\input{./spec.tex}
|
||||
\input{./fc.tex}
|
||||
\input{./projection.tex}
|
||||
\input{./type-decomp.tex}
|
||||
@@ -0,0 +1,296 @@
|
||||
\section{The Projection Lattice}
|
||||
\label{section:vn-projection-lattice}
|
||||
|
||||
\begin{definition}[Projection Lattice]
|
||||
\label{definition:vn-projection-lattice}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\text{Proj}(A)$ be the set of all projections in $A$, then:
|
||||
\begin{enumerate}
|
||||
\item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto the closed subspace generated by ${\bigcup_{Q \in S}Q(H)}$, then $P = \sup(S) \in A$.
|
||||
\item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto $\bigcap_{Q \in S}Q(H)$, then $P = \inf(S) \in A$.
|
||||
\item $\text{Proj}(A)$ is order complete.
|
||||
\end{enumerate}
|
||||
|
||||
The set $\text{Proj}(A)$ is the \textbf{projection lattice} of $A$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
(1): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, the closed subspace generated by $\bigcup_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}.
|
||||
|
||||
(2): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, $\bigcap_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}.
|
||||
\end{proof}
|
||||
|
||||
\subsection{Central Support of Projections}
|
||||
\label{subsection:projection-central-support}
|
||||
|
||||
|
||||
|
||||
\begin{definition}[Central Support]
|
||||
\label{definition:central-support-vna}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $Z(P) = \inf_{Q \in \text{Proj}(Z(A)), Q \ge P}Q$ is the \textbf{central support} of $P$.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Centrally Orthogonal]
|
||||
\label{definition:centrally-orthogonal}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P} \subset \text{Proj}(A)$, then $\seqi{P}$ is \textbf{centrally orthogonal} if $\bracsn{Z(P_i)}_{i \in I}$ is mutually orthogonal.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:central-support-vna}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$. For each $T \in A$, let $R(TP)$ be the orthogonal projection onto $\ol{TP(H)}$, then
|
||||
\[
|
||||
Z(P) = \sup_{T \in A}R(TP)
|
||||
\]
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 24.6]{Zhu}}}. ]
|
||||
Since $Z(P) \in Z(A)$, $Z(P)(H)$ is a reducing subspace of every operator in $A$. As $P \le Z(P)$, $TP(H) \subset T(Z(P)(H)) \subset Z(P)(H)$, so $Z(P) \ge R(TP)$ for all $T \in A$, and $Z(P) \ge \sup_{T \in A}R(TP)$.
|
||||
|
||||
On the other hand, for each $S, T \in A$, $S(TP(H)) \subset \bigcup_{R \in A}RP(H)$. As $A$ is a von Neumann algebra, the range of $\sup_{T \in A}R(TP)$ is a reducing subspace for every operator in $A$. Therefore $\sup_{T \in A}R(TP) \in Z(A)$, and $Z(P) \le \sup_{T \in A}R(TP)$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:central-support-mvn}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $Z(P)Z(Q) \ne 0$.
|
||||
\item $PAQ \ne \bracsn{0}$.
|
||||
\item There exists non-zero projections $P_0 \le P$ and $Q_0 \le Q$ such that $P_0 \sim Q_0$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 24.7]{Zhu}}}. ]
|
||||
(1) $\Rightarrow$ (2): For each $T \in A$, let $R(T)$ be the orthogonal projection onto $\ol{T(H)}$. By \autoref{proposition:central-support-vna},
|
||||
\[
|
||||
Z(P) = \sup_{T \in A}R(TP) \quad Z(Q) = \sup_{T \in A}R(TQ)
|
||||
\]
|
||||
|
||||
Given that $Z(P)Z(Q) \ne 0$, $Z(P)(H) \not\perp Z(Q)(H)$. Since $Z(P)(H) = \ol{\bigcup_{S \in A}SP(H)}$ and $Z(Q)(H) = \ol{\bigcup_{T \in A}TQ(H)}$, there exists $S, T \in A$ such that $SP(H) \not\perp TQ(H)$. As such, there exists $x, y \in H$ with
|
||||
\[
|
||||
0 \ne \dpn{TQx, SPy}{H} = \dpn{PS^*TQx, y}{H}
|
||||
\]
|
||||
|
||||
so $PAQ \ne \bracsn{0}$.
|
||||
|
||||
(2) $\Rightarrow$ (3): Let $T \in A$ with $PTQ \ne 0$. Let $P_0 = R(PTQ)$ and $Q_0 = R(QT^*P)$, then $0 \ne P_0 \le P$, $0 \ne Q_0 \le Q$, and $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-adjoint}.
|
||||
|
||||
(3) $\Rightarrow$ (1): By \autoref{lemma:central-support-mvn-eq}, $Z(P_0) = Z(Q_0)$, so
|
||||
\[
|
||||
Z(P)Z(Q) = Z(P) \wedge Z(Q) \ge Z(P_0) \vee Z(Q_0) \ne 0
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:central-support-mvn-eq}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$ with $P \sim Q$, then:
|
||||
\begin{enumerate}
|
||||
\item $Z(P) = Z(Q)$.
|
||||
\item For any central projection $R \in \text{Proj}(A)$, $PR \sim QR$.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $V \in A$ with $P = V^*V$ and $Q = VV^*$, then $V$ is a partial isometry with initial space $P(H)$ and final space $Q(H)$.
|
||||
|
||||
(1): Since $Z(P) \ge P$ and $Z(P) \in Z(A)$,
|
||||
\[
|
||||
Z(P)Q = Z(P)VV^* = VZ(P)V^* = VV^* = Q
|
||||
\]
|
||||
|
||||
and $Z(P) \ge Q$, and $Z(P) \ge Z(Q)$. By symmetry, $Z(Q) \ge Z(P)$, so $Z(P) = Z(Q)$.
|
||||
|
||||
(2): Let $R$ be a central projection, then
|
||||
\[
|
||||
PR = V^*VR = V^*RV = (VR)^*(VR) \sim (VR)(VR)^* = VRV^* = VV^*R = QR
|
||||
\]
|
||||
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection{Murray-von Neumann Equivalence}
|
||||
\label{subsection:mvn-equivalence}
|
||||
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:projection-mental-gymnastics}
|
||||
Let $H$ be a complex Hilbert space and $P, Q \in B(H)$ be projections, then:
|
||||
\begin{enumerate}
|
||||
\item $\ker(PQ) = \ker(Q) + \ker(P) \cap Q(H)$.
|
||||
\item If $PQ = QP$, then $PQ$ is a projection with $PQ(H) = P(H) \cap Q(H)$.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
(1): Let $x \in \ker(PQ)$, then $Q(x) \in \ker(P)$, so $x = Q(x) + (1 - Q)(x) \in \ker(Q) + \ker(P) \cap Q(H)$.
|
||||
|
||||
(2): Since $PQ = QP$, $(PQ)^2 = P^2Q^2 = PQ$ and $(PQ)^* = Q^*P^* = QP = PQ$, $PQ$ is a projection. As $PQ(H) = P(Q(H)) \subset P(H)$ and $PQ(H) = Q(P(H)) \subset Q(H)$, $PQ(H) \subset P(H) \cap Q(H)$. On the other hand, $PQ$ is the identity on $P(H) \cap Q(H)$, so $PQ(H) = P(H) \cap Q(H)$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\begin{definition}[Murray-von Neumann Equivalent]
|
||||
\label{definition:murray-von-neumann-equivalent}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item There exists $V \in A$ such that $P = V^*V$ and $Q = VV^*$.
|
||||
\item There exists a partial isometry $V \in A$ from $P(H)$ to $Q(H)$.
|
||||
\end{enumerate}
|
||||
|
||||
If the above holds, then $P$ and $Q$ are \textbf{Murray-von Neumann equivalent}, denoted $P \sim Q$. The relation $\sim$ is an equivalence relation on $\text{Proj}(A)$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
(1) $\Rightarrow$ (2): Let $V \in A$ with $P = V^*V$ and $Q = VV^*$. By \autoref{proposition:partial-isometry-characterisation}, $V$ is a partial isometry with initial space $\ker(P)^\perp$, and $V^*$ is a partial isometry with initial space $\ker(Q)^\perp$. Therefore $V$ is a partial isometry from $P(H)$ to $Q(H)$.
|
||||
|
||||
(2) $\Rightarrow$ (1): By \autoref{proposition:partial-isometry-characterisation}, $P = V^*V$ is a projection onto $\ker(V)^\perp$, and $Q = VV^*$ is a projection onto $\ker(V^*)^\perp = V(H)$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Murray-von Neumann Subequivalent]
|
||||
\label{definition:murray-von-neumann-subequivalent}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then $P$ is \textbf{Murray-von Neumann subequivalent} to $Q$, denoted $P \preceq Q$, if there exists $R \in \text{Proj}(A)$ such that $P \sim R$ and $R \le Q$.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:mvn-equivalent-direct-sum}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
|
||||
\begin{enumerate}[label=(\alph*)]
|
||||
\item $\seqi{P}$ is mutually orthogonal.
|
||||
\item $\seqi{Q}$ is mutually orthogonal.
|
||||
\item For each $i \in I$, $P_i \sim Q_i$.
|
||||
\end{enumerate}
|
||||
|
||||
then $\sum_{i \in I}P_i \sim \sum_{i \in I}Q_i$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
For each $i \in I$, let $V_i \in A$ such that $P_i = V_i^*V_i$ and $Q_i = V_iV_i^*$, then $V_i$ is a partial isometry with initial space $P_i(H)$ and final space $Q_i(H)$. As $\seqi{P}$ is mutually orthogonal and $\seqi{Q}$ is mutually orthogonal, the sum $\sum_{i \in I}V_i$ converges in strong operator topology to an operator $V$, where
|
||||
\[
|
||||
V^*V = \sum_{i, j \in I}V_i^*V_j = \sum_{i \in I}V_i^*V_i = \sum_{i \in I}P_i
|
||||
\]
|
||||
|
||||
and
|
||||
\[
|
||||
VV^* = \sum_{i, j \in I}V_iV_j^* = \sum_{i \in I}V_iV_i^* = \sum_{i \in I}Q_i
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:mvn-equivalent-adjoint}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $T \in A$, and $P, Q \in \text{Proj}(A)$ be orthogonal projections onto $\ol{T(H)}$ and $\ol{T^*(H)}$, respectively, then $P \sim Q$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $T = V|T|$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $T$, then $V$ is a partial isometry from $\ol{T^*(H)}$ to $\ol{T(H)}$. Since $V \in A$, $P \sim Q$.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}[Kaplansky's Formula]
|
||||
\label{theorem:kaplansky-formula}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then
|
||||
\[
|
||||
[(P \vee Q) - Q] \sim [(P - P \wedge Q)]
|
||||
\]
|
||||
\end{theorem}
|
||||
\begin{proof}
|
||||
Using \autoref{theorem:existence-of-projections-vna}, assume without loss of generality that $I \in A$. In which case, by (1) of \autoref{lemma:projection-mental-gymnastics},
|
||||
\[
|
||||
|
||||
[(I - Q)P](H)^\perp = \ker(P(I - Q)) = \ker(Q)^\perp + [\ker(Q) \cap \ker(P)]
|
||||
\]
|
||||
|
||||
and since $P \vee Q$ and $Q$ commute,
|
||||
\begin{align*}
|
||||
[(I - Q)P](H) &= [\ker(Q)^\perp + [\ker(Q) \cap \ker(P)]]^\perp \\
|
||||
&= \ker(Q) \cap [\ker(Q) \cap \ker(P)]^\perp \\
|
||||
&= \ker(Q) \cap [\ker(Q)^\perp + \ker(P)^\perp]\\
|
||||
&= (I - Q)(H) \cap [Q(H) + P(H)]\\
|
||||
&= (I - Q)(H) \cap [(Q \vee P)(H)] \\
|
||||
&= (P \vee Q)(I - Q)(H) = [(P \vee Q) - Q](H)
|
||||
\end{align*}
|
||||
|
||||
by (2) of \autoref{lemma:projection-mental-gymnastics}. Similarly,
|
||||
\[
|
||||
[P(I - Q)](H)^\perp = \ker((I - Q)P) = \ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]
|
||||
\]
|
||||
|
||||
so
|
||||
\begin{align*}
|
||||
[P(I - Q)](H) &= [\ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]]^\perp \\
|
||||
&= \ker(P)^\perp \cap [\ker(Q) + \ker(P)] \\
|
||||
&= P(H) \cap [(I - Q)(H) + (I - P)(H)] \\
|
||||
&= P(H) \cap [(I - Q) \vee (I - P)](H) \\
|
||||
&= P(H) \cap [I - (P \wedge Q)](H) \\
|
||||
&= [P - (P \wedge Q)](H)
|
||||
\end{align*}
|
||||
|
||||
Therefore
|
||||
\[
|
||||
[(P \vee Q) - Q](H) = [(I - Q)P](H) \sim [P(I - Q)](H) = [P - (P \wedge Q)](H)
|
||||
\]
|
||||
|
||||
by \autoref{lemma:mvn-equivalent-adjoint}.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{theorem}["Cantor-Bernstein"]
|
||||
\label{theorem:murray-von-neumann-subequivalent-partial-order}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. If $P \preceq Q$ and $Q \preceq P$, then $P \sim Q$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 25.1]{Zhu}}}. ]
|
||||
Let $U, V \in A$ be partial isometries such that $P = U^*U$, $UU^* \le Q$, $Q = V^*V$, and $VV^* \le P$. Denote $Q_0 = Q$ and $P_0 = P$. For each $n \in \natz$, inductively define $P_{n+1} = VQ_nV^*$ and $Q_{n+1} = UP_nU^*$, then:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item For each $n \in \natz$, $P_n, Q_n \in \text{Proj}(A)$.
|
||||
\item For each $n \in \natz$, $P_n \le P$ and $Q_n \le Q$.
|
||||
\item For each $n \in \natz$, $P_{n+1} \le P_n$ and $Q_{n+1} \le Q_n$.
|
||||
\end{enumerate}
|
||||
|
||||
As $\seq{P_n}, \seq{Q_n} \subset \text{Proj}(A)$ are non-increasing sequences, by \autoref{theorem:existence-of-projections-vna}, there exists $P_\infty, Q_\infty \in \text{Proj}(A)$ such that $P_n \to P_\infty$ and $Q_n \to Q_\infty$ in the strong operator topology as $n \to \infty$.
|
||||
|
||||
For each $n \in \natz$, $U(P_n - P_{n+1})U^* = Q_{n+1} - Q_{n+2}$, so
|
||||
\begin{align*}
|
||||
[U(P_n - P_{n+1})]^*[U(P_n - P_{n+1})] &= (P_n - P_{n+1})P(P_n - P_{n+1}) = P_n - P_{n+1} \\
|
||||
[U(P_n - P_{n+1})][U(P_n - P_{n+1})]^* &= U(P_n - P_{n+1})^2U^* = Q_{n+1} - Q_{n+2}
|
||||
\end{align*}
|
||||
|
||||
and $P_n - P_{n+1} \sim Q_{n+1} - Q_{n+2}$. Similarly, $Q_n - Q_{n+1} \sim P_{n+1} - P_{n+2}$. As $P_{n+1} = VQ_nV^*$ for all $n \in \natz$, $P_\infty \sim Q_\infty$ after passing through a strong-operator limit.
|
||||
|
||||
For each $N \in \natz$, $\sum_{n = 0}^N (P_n - P_{n+1}) = P - P_{N+1}$, so $P = P_\infty + \sum_{n = 0}^\infty (P_n - P_{n+1})$. Similarly, $Q = Q_\infty + \sum_{n = 0}^\infty (Q_n - Q_{n+1})$. Therefore
|
||||
\begin{align*}
|
||||
P &= P_\infty + \sum_{n = 0}^\infty (P_{2n} - P_{2n+1}) + \sum_{n = 0}^\infty (P_{2n + 1} - P_{2n+2}) \\
|
||||
&\sim Q_\infty + \sum_{n = 0}^\infty (Q_{2n + 1} - Q_{2n+2}) + \sum_{n = 0}^\infty (Q_{2n} - Q_{2n+1}) = Q
|
||||
\end{align*}
|
||||
|
||||
by \autoref{lemma:mvn-equivalent-direct-sum}.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}[The Comparability Theorem]
|
||||
\label{theorem:vna-comparability}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then there exists a central projection $R$ such that $RP \preceq RQ$ and $(I - R)Q \preceq (I - R)P$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 25.4]{Zhu}}}. ]
|
||||
By Zorn's lemma, there exists maximal families $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $\seqi{P}$ is mutually orthogonal.
|
||||
\item $\seqi{Q}$ is mutually orthogonal.
|
||||
\item For each $i \in I$, $P_i \sim Q_i$.
|
||||
\item For each $i \in I$, $P_i \le P$ and $Q_i \le Q$.
|
||||
\end{enumerate}
|
||||
|
||||
Let $P_0 = \sum_{i \in I}P_i$ and $Q_0 = \sum_{i \in I}Q_i$, then $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-direct-sum}. By maximality, there exists no non-zero $P', Q' \in \text{Proj}(A)$ such that $P' \le P - P_0$, $Q' \le Q - Q_0$, and $P' \sim Q'$. By \autoref{proposition:central-support-mvn}, $Z(P - P_0) Z(Q - Q_0) = 0$.
|
||||
|
||||
Let $R = Z(Q - Q_0)$, then $Q - Q_0 \le R$ and $P - P_0 \le (I - R)$, so $(P - P_0)R = 0$ and $(Q - Q_0)R = Q - Q_0$. By \autoref{lemma:central-support-mvn-eq},
|
||||
\[
|
||||
PR = P_0R \sim Q_0R \le QR
|
||||
\]
|
||||
|
||||
and
|
||||
\[
|
||||
Q(I - R) = Q_0(I - R) \sim P_0(I - R) \le P(I - R)
|
||||
\]
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}
|
||||
\label{corollary:vna-factor-comparability}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then for any $P, Q \in \text{Proj}(A)$, either $P \prec Q$, $P \sim Q$, or $Q \prec P$.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
By the \hyperref[comparability theorem]{theorem:vna-comparability}, there exists a central projection $R$ such that $PR \preceq QR$ and $Q(I - R) \preceq P(I - R)$. As $A$ is a factor, either $R = 0$ or $R = I$. In which case, $P \preceq Q$ or $Q \preceq P$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
@@ -0,0 +1,336 @@
|
||||
\section{The Spectral Theorem}
|
||||
\label{section:spectral-theorem}
|
||||
|
||||
\begin{definition}[Spectral Measure]
|
||||
\label{definition:spectral-measure}
|
||||
Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$, then $E$ is a \textbf{spectral measure relative to $H$} if:
|
||||
\begin{enumerate}
|
||||
\item For each $B \in \cb_X$, $E(B)$ is an orthogonal projection.
|
||||
\item $E(\emptyset) = 0$, $E(X) = I_{B(H)}$.
|
||||
\item For each $B, C \in \cb_X$, $E(B \cap C) = E(B)E(C)$.
|
||||
\item For each $x, y \in H$, the mapping
|
||||
\[
|
||||
E_{x, y}: \cb_X \to \complex \quad B \mapsto \dpn{E(B)x, y}{H}
|
||||
\]
|
||||
|
||||
is a complex Radon measure on $X$.
|
||||
\end{enumerate}
|
||||
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:spectral-measure-properties}
|
||||
Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, then:
|
||||
\begin{enumerate}
|
||||
\item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$.
|
||||
\item For each $x \in H$, $E_{x, x}$ is positive.
|
||||
\end{enumerate}
|
||||
|
||||
Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
|
||||
\begin{enumerate}[start=2]
|
||||
\item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well.
|
||||
\item Let
|
||||
\[
|
||||
J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
|
||||
\]
|
||||
|
||||
then $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
(1): Let $x, y \in H$, $\seqf{B_j} \subset \cb_X$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^n B_j$, then for each $1 \le i < j \le n$, $E(B_i)(H) \perp E(B_j)(H)$, so by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and the \hyperref[Pythagorean Theorem]{theorem:pythagoras},
|
||||
\begin{align*}
|
||||
\sum_{j = 1}^n |\dpn{E(B_j)x, y}{H}| &= \sum_{j = 1}^n |\dpn{E(B_j)x, E(B_j)y}{H}| \\
|
||||
&\le \sum_{j = 1}^n \norm{E(B_j)x}_H \norm{E(B_j)y}_H \\
|
||||
&\le \braks{\sum_{j = 1}^n \norm{E(B_j)x}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n \norm{E(B_j)y}_H^2}^{1/2} \\
|
||||
&= \norm{E(B)x}_H \cdot \norm{E(B)y}_H \le \norm{x}_H \cdot \norm{y}_H
|
||||
\end{align*}
|
||||
|
||||
As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$.
|
||||
|
||||
(2): For each $B \in \cb_X$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H} \ge 0$.
|
||||
|
||||
(3): For each $x, y \in H$ and $B, C \in \cb_X$,
|
||||
\[
|
||||
\int_C \one_B dE_{x, y} = \dpn{E(C \cap B)x, y}{H} = \dpn{E(C)E(B)x, y}{H} = E_{E(B)x, y}(C)
|
||||
\]
|
||||
|
||||
By linearity, $fdE_{x, y} \in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex)$ is dense in $L^1(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^1(\mu; \complex)$ and $\mu \in \mathscr{E}$.
|
||||
|
||||
Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$.
|
||||
|
||||
(4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous.
|
||||
\item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
|
||||
\item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
|
||||
\end{enumerate}
|
||||
|
||||
the mapping $J$ is a unital *-homomorphism.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{definition}[Integration Against Spectral Measure]
|
||||
\label{definition:spectral-measure-integral}
|
||||
Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and
|
||||
\[
|
||||
J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
|
||||
\]
|
||||
|
||||
Then, $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism.
|
||||
|
||||
For each $\phi \in \mathscr{E}^*$, let $I_E(\phi) \in B(H)$ be the operator defined by
|
||||
\[
|
||||
\dpn{I_E(\phi) \cdot x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} \quad \forall x, y \in H
|
||||
\]
|
||||
|
||||
then
|
||||
|
||||
\begin{enumerate}
|
||||
\item $I_E$ is a contraction from $\mathscr{E}^*$ to $B(H)$.
|
||||
\item $I_E$ is continuous from the weak*-topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$.
|
||||
\item $I_E$ is an injective unital *-homomorphism.
|
||||
\end{enumerate}
|
||||
|
||||
For any $\phi \in \mathscr{E}^*$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
(1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then by \autoref{lemma:spectral-measure-properties},
|
||||
\begin{align*}
|
||||
|\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{\mathscr{E}}| \le \norm{E_{x, y}}_{\mathscr{E}} \cdot \norm{\phi}_{\mathscr{E}^{*}} \\
|
||||
&\le \norm{\phi}_{\mathscr{E}^{*}} \cdot \norm{x}_H \cdot \norm{y}_H
|
||||
\end{align*}
|
||||
|
||||
Since the above holds for all $x, y \in H$, $I_E(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)} \le \norm{\phi}_{\mathscr{E}^{*}}$.
|
||||
|
||||
(2): For each $x, y \in H$, $E_{x, y} \in \mathscr{E}$. Since $\angles{\int \phi dE \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}}$ for every $\phi \in \mathscr{E}^{*}$, $I_E$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$.
|
||||
|
||||
(3): By \autoref{lemma:separable-metric-space-approx-identity}, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^\infty(X; \complex)$. Since
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $I_E$ restricted to $J(\Sigma(X; \complex))$ is a *-homomorphism.
|
||||
\item Multiplication and conjugation are continuous in the uniform norm on $B^\infty(X; \complex)$
|
||||
\item Composition and adjunction are continuous in the operator norm on $B(H)$
|
||||
\end{enumerate}
|
||||
|
||||
the map $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism by continuity.
|
||||
|
||||
By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex) \subset B^\infty(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$, so $J(C(X; \complex))$ is weak*-dense in $\mathscr{E}^*$. As
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism.
|
||||
\item The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $\mathscr{E}^{*}$.
|
||||
\item The adjunction $T \mapsto T^*$ is weak-operator continuous on $B(H)$.
|
||||
\item The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $\mathscr{E}^{*}$.
|
||||
\item The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.
|
||||
\end{enumerate}
|
||||
|
||||
the map $I_E$ is a *-homomorphism by the weak* to weak-operator continuity established in (2). Since $E(X) = I_{B(H)}$, $I_E$ is a unital *-homomorphism.
|
||||
|
||||
Finally, let $\phi \in \mathscr{E}^*$ with $I_E(\phi) = 0$, then $\dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} = 0$ for all $x, y \in H$. As $\mathscr{E}$ is the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, $\phi = 0$. Therefore $I_E$ is an injective unital *-homomorphism.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}[Spectral Theorem I]
|
||||
\label{theorem:spectral-theorem-vn-1}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then:
|
||||
\begin{enumerate}
|
||||
\item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that\footnote{Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^*$. }
|
||||
\[
|
||||
T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A
|
||||
\]
|
||||
\item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
|
||||
\[
|
||||
I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE
|
||||
\]
|
||||
|
||||
is a *-isomorphism.
|
||||
\end{enumerate}
|
||||
|
||||
The measure $E$ is the \textbf{spectral measure associated with $A$}, and the homomorphism $I_E$ is the \textbf{extended inverse Gelfand transform} of $A$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ]
|
||||
(1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
|
||||
\[
|
||||
E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H}
|
||||
\]
|
||||
|
||||
which, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists.
|
||||
|
||||
Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
|
||||
\[
|
||||
J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
|
||||
\]
|
||||
|
||||
with $J_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$.
|
||||
|
||||
For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
|
||||
\[
|
||||
E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
|
||||
\]
|
||||
|
||||
defines a spectral measure, it is sufficient to show that $J_E$ is a *-homomorphism.
|
||||
|
||||
Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$,
|
||||
\begin{align*}
|
||||
\dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
|
||||
&= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, J_E(f)^*y}}{C(\Omega(A); \complex)}
|
||||
\end{align*}
|
||||
|
||||
As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$,
|
||||
\begin{align*}
|
||||
\dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, J_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{J_E(\phi)x, J_E(f)^*y}{H} \\
|
||||
&= \dpn{J_E(f)J_E(\phi)x, y}{H} = \dpn{f, E_{J_E(\phi)x, y}}{C(\Omega(A); \complex)}
|
||||
\end{align*}
|
||||
|
||||
so $\phi E_{x, y} = E_{J_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$,
|
||||
\begin{align*}
|
||||
\dpn{J_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{J_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
|
||||
&= \dpn{J_E(\phi)J_E(\psi)x, y}{H}
|
||||
\end{align*}
|
||||
|
||||
and $J_E$ is a homomorphism.
|
||||
|
||||
Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $J_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{J_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $J_E(\phi)$ is self-adjoint, and $J_E$ is a *-homomorphism.
|
||||
|
||||
(2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$.
|
||||
|
||||
Let $J: C(\Omega(A); \complex) \to \mathscr{E}^*$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}} = \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^*$. Since $I_E$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$, $I_E(\mathscr{E}^*) \subset B$ by \autoref{proposition:closure-of-image}.
|
||||
|
||||
On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_E(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_E(\mathscr{E}^*) = B$.
|
||||
\end{proof}
|
||||
|
||||
\begin{theorem}[Spectral Theorem II]
|
||||
\label{theorem:spectral-theorem-vn-2}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $Id \in A$, $B \subset B(H)$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $\seqi{\xi} \subset H$ be a maximal family such that the subspaces $\bracsn{A\xi_i|i \in I}$ are mutually orthogonal, then:
|
||||
\begin{enumerate}
|
||||
\item For each $i \in I$, there exists a finite positive Radon measure $\mu_i \in \mathscr{E}$ on $\Omega(A)$ such that for every Borel set $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$.
|
||||
\item For each $i \in I$, let $P_i: H \to \ol{A\xi_i}$ be the orthogonal projection onto $\ol{A\xi_i}$, then for any $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$.
|
||||
\item The natural map $C(\Omega(A); \complex) \to [l^\infty(I); L^\infty(\mu_i; \complex)]$ is injective. Equivalently, $\ol{\bigcup_{i \in I}\supp{\mu_i}} = \Omega(A)$.
|
||||
\item The space $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$ under the mapping
|
||||
\[
|
||||
\mathscr{M}: [l^1(I); L^1(\mu_i; \complex)] \to \mathscr{E} \quad f \mapsto \sum_{i \in I}f_id\mu_i
|
||||
\]
|
||||
|
||||
and $\mathscr{E}^*$ may be identified as a closed subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ through $\mathscr{M}^*$.
|
||||
\item There exists a unitary equivalence $U: H \to [l^2(I); L^2(\mu_i; \complex)]$ between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$, such that for each $i \in I$, $U|_{\ol{A\xi_i}}$ is an isometry onto the $i$-th factor of $[l^2(I); L^2(\mu_i; \complex)]$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 1.47]{FollandHarmonic}}}. ]
|
||||
(1): Fix $ i \in I$ and let $\mu_i = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_i(C) = 0$. By (1) and (2) of \autoref{lemma:spectral-measure-properties}, $\mu_i$ is a finite positive Radon measure.
|
||||
|
||||
By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$,
|
||||
\[
|
||||
\dpn{E(C)S\xi_i, T\xi_i}{H} = \int_C \Gamma_AS \cdot \ol{\Gamma_AT} dE_{\xi_i, \xi_i}
|
||||
\]
|
||||
|
||||
so $\Gamma_AS \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i} = dE_{S\xi_i, T\xi_i} \ll \mu_i$. By (1) of \autoref{lemma:spectral-measure-properties} and completeness of $L^1(\mu_i; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_i$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$.
|
||||
|
||||
(2): Let $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i} \perp \ol{A\xi_j}$,
|
||||
\[
|
||||
\int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0
|
||||
\]
|
||||
|
||||
As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y} = 0$.
|
||||
|
||||
Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$,
|
||||
\[
|
||||
E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} = \sum_{i \in I}E_{P_ix, P_iy} \in \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])
|
||||
\]
|
||||
|
||||
(3): Let $T \in A$ with $\Gamma_A T = 0$ $\mu_i$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$. Thus for any $x, y \in H$,
|
||||
\[
|
||||
\dpn{Tx, y}{H} = \int_{\Omega(A)}\Gamma_A T dE_{x, y} = \sum_{i \in I}\int_{\Omega(A)}\Gamma_A TdE_{P_ix, P_iy} = 0
|
||||
\]
|
||||
|
||||
Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$.
|
||||
|
||||
(4): By (2), for each $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$, so $\mathscr{E} \subset \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$.
|
||||
|
||||
On the other hand, for each $i \in I$, since $\mu_i$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^1(\mu_i; \complex)$ by \autoref{proposition:radon-cc-dense}. As
|
||||
\begin{align*}
|
||||
\mathscr{E} &\supset \bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \supset \bracsn{fdE_{\xi_i, \xi_i}|f \in C(\Omega(A); \complex)} \\
|
||||
&= \bracsn{fd\mu_i|f \in C(\Omega(A); \complex)}
|
||||
\end{align*}
|
||||
|
||||
|
||||
and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ is closed, $\mathscr{E} \supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$.
|
||||
|
||||
Finally, given that the above holds for all $i \in I$, $\mathscr{E} = \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, $[l^\infty(I); L^\infty(\mu_i; \complex)] = [l^1(I); L^1(\mu_i; \complex)]^*$, so $\mathscr{E}^*$ may be identified with its image under $\mathscr{M}^*$.
|
||||
|
||||
(5): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_i = T\xi_i$,
|
||||
\[
|
||||
\Gamma_AS dE_{\xi_i, \xi_i} = E_{S\xi_i, \xi_i} = E_{T\xi_i, \xi_i} = \Gamma_A T dE_{\xi_i, \xi_i}
|
||||
\]
|
||||
|
||||
so $\Gamma_A S = \Gamma_A T$ $\mu_i$-almost everywhere. Thus the mapping
|
||||
\[
|
||||
U_i: \ol{A\xi_i} \to L^2(\mu_i; \complex) \quad T\xi_i \mapsto \Gamma_AT
|
||||
\]
|
||||
|
||||
is well-defined. Moreover, for any $S, T \in A$,
|
||||
\[
|
||||
\dpn{S\xi_i, T\xi_i}{H} = \int \Gamma_AS \cdot \ol{\Gamma_A T} dE_{\xi_i, \xi_i} = \dpn{\Gamma_A S, \Gamma_A T}{L^2(\mu_i; \complex)}
|
||||
\]
|
||||
|
||||
so $U_i$ extends into an isometry between $\ol{A\xi_i}$ and $L^2(\mu_i; \complex)$. Thus the mapping
|
||||
\[
|
||||
U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix)
|
||||
\]
|
||||
|
||||
is an isometry between $H$ and $[l^2(I); L^2(\mu_i; \complex)]$ such that $U(Tx) = \Gamma_AT \cdot Ux$ for all $x \in H$ and $T \in A$.
|
||||
|
||||
Finally, given that
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^*$.
|
||||
\item The weak* topology on $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is equal to the weak operator topology of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$.
|
||||
\item $A$ is weak-operator dense in $B$.
|
||||
\item By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B$.
|
||||
\end{enumerate}
|
||||
|
||||
the mapping $U$ is a unitary equivalence between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$.
|
||||
\end{proof}
|
||||
|
||||
\begin{corollary}[Representation of Commutative von Neumann Algebras]
|
||||
\label{corollary:commutative-von-neumann-linfty}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative von Neumann algebra with $I \in A$, then:
|
||||
\begin{enumerate}
|
||||
\item There exists a LCH space $\Omega$ and a decomposable Radon measure $\mu$ on $\Omega$ such that $A$ is *-isomorphic to $L^\infty(\mu; \complex)$.
|
||||
\item If $A$ admits a cyclic vector, then $\Omega$ may be taken to be compact.
|
||||
\item If $H$ is separable, then $\Omega$ may be taken to be compact.
|
||||
\end{enumerate}
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
Let $E$ be the spectral measure on $\Omega(A)$ associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, $A$ is *-isomorphic to $\mathscr{E}^*$.
|
||||
|
||||
(1): By (3) of \autoref{lemma:spectral-measure-properties} and \autoref{theorem:hilbert-measures-dual}, $A$ is *-isomorphic to $[l^\infty(I); L^\infty(\mu_i; \complex)]$, where $\seqi{\mu} \subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_i$ denote the $i$-th copy of $\Omega(A)$, then
|
||||
\[
|
||||
\mu: \cb_\Omega \to [0, \infty] \quad B \mapsto \sum_{i \in I}\mu_i(B \cap \Omega_i)
|
||||
\]
|
||||
|
||||
is the desired decomposable Radon measure.
|
||||
|
||||
(2): By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton.
|
||||
|
||||
(3): If $H$ is separable, then so is $\mathscr{E}$. As such, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{remark}
|
||||
\label{remark:spectral-theorem-vn-2}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$.
|
||||
|
||||
By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a decomposable measure space $\Omega$, corresponding to a number of copies of $\Omega(A)$, such that $\mathscr{E}$ is a quotient of its $L^1$ space, $\mathscr{E}^*$ is a subspace of its $L^\infty$ space, and $H$ is isomorphic to its $L^2$ space. The preceding isomorphisms are all linked by a unitary equivalence between $B$ acting on $H$, and $\mathscr{E}^*$ acting on the $l^2$ direct sum.
|
||||
|
||||
The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors:
|
||||
\begin{enumerate}
|
||||
\item The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^1$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see \autoref{theorem:hilbert-measures-dual}.
|
||||
\item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremally large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$.
|
||||
\end{enumerate}
|
||||
|
||||
More concretely, (1) manifests as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$.
|
||||
|
||||
By limiting these two sources of complexity, it is possible to remove the need of multiple copies of $\Omega(A)$. In particular,
|
||||
\begin{enumerate}
|
||||
\item If $B$ admits a cyclic vector, then only one copy of $\Omega(A)$ is required for the construction in the Spectral Theorem \cite[Theorem 23.1]{Zhu}.
|
||||
\item If $H$ is separable, then at most countably many copies of $\Omega(A)$ are required for the construction in the Spectral Theorem. In which case, the measures can be summed such that $B$ is isomorphic to an $L^\infty$ space on $\Omega(A)$ \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}.
|
||||
\end{enumerate}
|
||||
\end{remark}
|
||||
|
||||
@@ -0,0 +1,122 @@
|
||||
\section{Topologies on $B(H)$}
|
||||
\label{section:topologies-on-bh}
|
||||
|
||||
Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
|
||||
\[
|
||||
B(H) \times (H \otimes \ol H) \to \complex \quad \dpn{T, \phi \otimes x}{B(H)} = \dpn{Tx, \phi}{H}
|
||||
\]
|
||||
|
||||
Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators:
|
||||
|
||||
\begin{definition}[Ultraweak Topology]
|
||||
\label{definition:bh-ultraweak-topology}
|
||||
Let $H$ be a complex Hilbert space, then the dual of $H \wh \otimes_\pi H$ is $B(H)$, and the $\sigma(B(H), H \wh \otimes_\pi H)$-topology is the \textbf{ultraweak}/\textbf{$\sigma$-weak} topology on $B(H)$.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
By \autoref{proposition:projective-tensor-product-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}.
|
||||
\end{proof}
|
||||
|
||||
A natural topology consistent with the ultraweak topology would be the ultrastrong topology.
|
||||
|
||||
\begin{definition}[Ultrastrong Topology]
|
||||
\label{definition:bh-ultrastrong-topology}
|
||||
Let $H$ be a complex Hilbert space. For each $x = \seq{x_n} \in L^2(\natp; H)$, let
|
||||
\[
|
||||
\Phi_x: B(H) \to L^2(\natp; H) \quad (\Phi_xT)_n = Tx_n
|
||||
\]
|
||||
|
||||
then the \textbf{ultrastrong}/\textbf{$\sigma$-strong} topology on $B(H)$ is the topology generated by the maps $\bracsn{\Phi_x|x \in L^2(\natp; H)}$.
|
||||
\end{definition}
|
||||
|
||||
|
||||
Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:bh-ultraweak-bounded}
|
||||
Let $H$ be a complex Hilbert space, then every bounded subset of $B(H)$ is relatively compact with respect to the ultraweak topology.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.
|
||||
\end{proof}
|
||||
|
||||
Now, a few facts about the more familiar operator topologies:
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:bh-operator-topologies-facts}
|
||||
Let $H$ be a complex Hilbert space, then:
|
||||
\begin{enumerate}
|
||||
\item The dual of $B(H)$ with respect to its strong and weak operator topologies is $H \otimes H$.
|
||||
\item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology.
|
||||
\item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies.
|
||||
\item The composition map $(S, T) \mapsto ST$ is left-hypocontinuous with respect to the strong operator topology and strong-operator bounded subsets of $B(H)$.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
(2): By \autoref{proposition:bh-ultraweak-bounded}.
|
||||
|
||||
(4): By the \hyperref[Banach-Steinhaus Theorem]{theorem:banach-steinhaus}, every strong-operator bounded subset of $B(H)$ is equicontinuous.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:bh-adjoint-strong-continuous}
|
||||
Let $H$ be a complex Hilbert space, then:
|
||||
\begin{enumerate}
|
||||
\item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology.
|
||||
\item $T \mapsto T^*$ restricted to the normal operators is continuous in the strong operator topology.
|
||||
\item For any $f \in C(\complex; \complex)$, the mapping $T \mapsto f(T)$ restricted to any bounded set of normal operators is continuous in the strong operator topology.
|
||||
\end{enumerate}
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Section 19.1]{Zhu}}}. ]
|
||||
(2): Let $S, T \in B(H)$, then
|
||||
\begin{align*}
|
||||
\normn{(S^* - T^*)x}_H^2 &= \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, ST^*x}{H} - \dpn{ST^*x, x}{H} \\
|
||||
&\le \normn{S^*x}_H^2 + \normn{T^*x}_H^2 - \dpn{x, TT^*x}{H} - \dpn{TT^*x, x}{H} \\
|
||||
&+ |\dpn{x, (T - S)T^*x}{H}| + |\dpn{(T - S)T^*x, x}{H}| \\
|
||||
&\le |\normn{S^*x}_H^2 - \normn{T^*x}_H^2| + 2\norm{x}_H\normn{(T - S)T^*x}_H
|
||||
\end{align*}
|
||||
|
||||
Now, if $S$ and $T$ are normal, then $\normn{S^*x}_H = \norm{Sx}_H$ and $\norm{T^*x}_H = \norm{Tx}_H$, so
|
||||
\begin{align*}
|
||||
|\norm{S^*x}_H^2 - \norm{T^*x}_H^2| &= |\norm{Sx}_H^2 - \norm{Tx}_H^2| \\
|
||||
&\le \norm{(S - T)x}_H (\norm{Sx}_H + \norm{Tx}_H) \\
|
||||
&\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H)
|
||||
\end{align*}
|
||||
|
||||
Therefore
|
||||
\begin{align*}
|
||||
\normn{(S^* - T^*)x}_H^2 &\le \norm{(S - T)x}_H (\norm{(S - T)x}_H + 2\norm{Tx}_H) \\
|
||||
&+ 2\norm{x}_H\normn{(T - S)T^*x}_H
|
||||
\end{align*}
|
||||
|
||||
and the adjoint map restricted to normal operators is continuous in the strong operator topology.
|
||||
|
||||
(3): Let $S, T \in B_{B(H)}(0, 1)$ and $x \in H$ and $n \in \natp$, then
|
||||
\begin{align*}
|
||||
\normn{(S^n - T^n)x}_H &\le \sum_{k = 0}^{n-1}\normn{S^{n-1-k}(S - T)T^kx}_{H} \\
|
||||
&\le \sum_{k = 0}^{n - 1}\normn{(S - T)T^kx}_H
|
||||
\end{align*}
|
||||
|
||||
|
||||
so the mapping $T \mapsto T^n$ on $B_{B(H)}(0, 1)$ is continuous in the strong operator topology. By (2), the mapping $T \mapsto p(T, T^*)$ is strong-operator continuous for all $p \in \complex[z, \ol z]$.
|
||||
|
||||
By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exist polynomials $p_n \in \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\ol{B_\complex(0, 1)}$. For any $T \in B_{B(H)}(0, 1)$, $x \in H$, and $n \in \natp$,
|
||||
\begin{align*}
|
||||
\norm{[f(T) - p_n(T)]x}_H &\le \norm{f(T) - p_n(T)}_{B(H)} \cdot \norm{x}_H \\
|
||||
&\le \norm{x}_H \cdot \sup_{z \in \ol{B_\complex(0, 1)}}|f(z) - p_n(z)|
|
||||
\end{align*}
|
||||
|
||||
|
||||
by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:spatial-isomorphism-sot-continuous}
|
||||
Let $A$ be a $C^*$-algebra, $H_1, H_2$ be complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be a unitary equivalence, then the mapping
|
||||
\[
|
||||
\pi_1(A) \to \pi_2(A) \quad T \mapsto UTU^{-1}
|
||||
\]
|
||||
|
||||
is strong-operator and weak-operator continuous.
|
||||
\end{proposition}
|
||||
|
||||
|
||||
@@ -0,0 +1,309 @@
|
||||
\section{Type Decomposition}
|
||||
\label{section:vna-type-decomposition}
|
||||
|
||||
|
||||
\begin{definition}[Finite Projection]
|
||||
\label{definition:finite-projection}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{finite} if for any $Q \in \text{Proj}(A)$ with $P \sim Q$ and $Q \le P$, $P = Q$. For any $P \in \text{Proj}(A)$, $P$ is \textbf{infinite} if it is not finite.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Abelian Projection]
|
||||
\label{definition:abelian-projection}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{abelian} if $PAP$ is abelian.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Minimal Projection]
|
||||
\label{definition:minimal-projection}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $PAP = \complex P$.
|
||||
\item There exists no $Q \in \text{Proj}(A)$ with $0 < Q < P$.
|
||||
\end{enumerate}
|
||||
|
||||
If the above holds, then $P$ is \textbf{minimal}.
|
||||
\end{definition}
|
||||
\begin{proof}
|
||||
(1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$.
|
||||
|
||||
(2) $\Rightarrow$ (1): Given that there exist no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Diffuse]
|
||||
\label{definition:diffuse-vna}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, then $M$ is \textbf{diffuse} if it admits no minimal projections.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:projection-types-gymnastics}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$.
|
||||
\begin{enumerate}
|
||||
\item If $P$ is minimal, then $P$ is abelian.
|
||||
\item If $P$ is abelian, then $P$ is finite.
|
||||
\item If $P$ is finite and $P \sim Q$, then $Q$ is finite.
|
||||
\item If $P$ is finite and $Q \le P$, then $Q$ is finite.
|
||||
\item If $P$ is minimal and $P \sim Q$, then $Q$ is minimal.
|
||||
\item If $P, Q$ are minimal with $P \sim Q$, then for any $U, V \in A$ with $P = U^*U = V^*V$ and $Q = UU^* = VV^*$, there exists $\lambda \in \partial B_{\complex}(0, 1)$ such that $V = \lambda U$.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Section 26.1]{Zhu}}}. ]
|
||||
(1): $PAP = \complex P$ is abelian.
|
||||
|
||||
(2): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, then there exists $V \in A$ such that $R = V^*V$ and $P = VV^*$. Since $V$ has initial space $R(H) \subset P(H)$ and final space $P(H)$, $V = PVP$ and $V^* = PV^*P$. As $PAP$ is abelian,
|
||||
\[
|
||||
R = V^*V = PV^*PPVP = PVPPV^*P = VV^* = P
|
||||
\]
|
||||
|
||||
(3): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q$. Let $V \in A$ with $Q = V^*V$ and $P = VV^*$, then $V$ is a partial isometry with initial space $Q(H)$ and final space $P(H)$. In which case, $P = VQV^*$, and $VRV^* \le VQV^* = P$. Let $U = (VRV^*)V$, then
|
||||
\begin{align*}
|
||||
U^*U &= (VRV^*V)^*(VRV^*V) = V^*VRV^* \cdot VRV^*V \\
|
||||
&= V^*VRV^*V = QRQ = R
|
||||
\end{align*}
|
||||
|
||||
|
||||
and as $Q = V^*PV$,
|
||||
\begin{align*}
|
||||
UU^* &= (VRV^*V)(VRV^*V)^* = VRV^*V \cdot V^*VRV^* \\
|
||||
&= VRV^*PVRV^* = VRQRV^* = VRV^*
|
||||
\end{align*}
|
||||
|
||||
so $VRV^* \sim R \sim Q \sim P$. Given that $P$ is finite, $VRV^* = P$. Therefore
|
||||
\[
|
||||
R = QRQ = V^*VRV^*V = V^*PV = Q
|
||||
\]
|
||||
|
||||
(4): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q \le P$, then $P \sim (P - Q) + R \le P$, so $P - Q + R = P$, and $Q = R$.
|
||||
|
||||
(5): Since $P \sim Q$, there exists $V \in A$ with $P = V^*V$ and $Q = VV^*$. Let $R \in \text{Proj}(A)$ with $0 < R \le Q$, then $0 \le V^*RV \le V^*QV = P$. By minimality of $P$, $V^*RV = P$, so
|
||||
\[
|
||||
R = QRQ = VV^*RVV^* = VPV^* = Q
|
||||
\]
|
||||
|
||||
(6): Let $R = U^*V$, then since $U$ and $V$ are partial isometries with initial space $P(H)$ and final space $Q(H)$,
|
||||
\[
|
||||
PRP = U^*U \cdot U^*V \cdot V^*V = U^*QV = U^*V
|
||||
\]
|
||||
|
||||
so $PRP \in PAP = \complex P$. Thus there exists $\lambda \in \complex$ such that $R = \lambda P$. In which case,
|
||||
\[
|
||||
\lambda U = U \cdot \lambda P = UU^*V = QV = V
|
||||
\]
|
||||
|
||||
and
|
||||
\[
|
||||
P = V^*V = \lambda \ol{\lambda} U^*U = |\lambda|^2 P
|
||||
\]
|
||||
|
||||
so $|\lambda| = 1$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:centrally-orthogonal-sum-properties}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $\seqi{P} \subset \text{Proj}(A)$ be centrally orthogonal, and $P = \sum_{i \in I}P_i$, then
|
||||
\begin{enumerate}
|
||||
\item For each $T \in A$, $PTP = \sum_{i \in I}P_iTP_i$.
|
||||
\item If $\seqi{P}$ are abelian, then $P$ is also abelian.
|
||||
\item If $\seqi{P}$ are finite, then $P$ is also finite.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 26.2]{Zhu}}}. ]
|
||||
(1): For each $i \in I$, $Z(P_i) \ge P_i$, so $Z(P_i)P_i = P_i$. For any $i, j \in I$ with $i \ne j$, $Z(P_i)$ and $Z(P_j)$ are orthogonal, so $Z(P_i)P_j = Z(P_i)Z(P_j)P_j = 0$.
|
||||
|
||||
Let $T \in A$, then by \autoref{proposition:central-support-vna}, $Z(P_i)(H) \supset TP_i(H)$ for all $i \in I$. Therefore
|
||||
\begin{align*}
|
||||
PTP &= \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}Z(P_i)P_i \cdot T \cdot Z(P_j)P_j \\
|
||||
&= \sum_{i, j \in I}Z(P_i)P_i \cdot Z(P_j) \cdot T \cdot Z(P_j)P_j \\
|
||||
&= \sum_{i \in I}Z(P_i)P_i \cdot T \cdot Z(P_i)P_i = \sum_{i \in I}P_i TP_i
|
||||
\end{align*}
|
||||
|
||||
(2): Let $S, T \in A$, then by (1),
|
||||
\begin{align*}
|
||||
PSP \cdot PTP &= \sum_{i, j \in I}P_iSP_i \cdot P_jTP_j = \sum_{i \in I}P_iSP_i \cdot P_iTP_i \\
|
||||
&= \sum_{i \in I}P_iTP_i \cdot P_iSP_i = PTP \cdot PSP
|
||||
\end{align*}
|
||||
|
||||
(3): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, and $V \in A$ with $R = V^*V$ and $P = VV^*$, then for each $i \in I$, $Z(P_i)R \sim Z(P_i)P = P_i$, and $Z(P_i)R \le Z(P_i)P = P_i$. As $\seqi{P}$ are finite, $Z(P_i)R = P_i$ for all $i \in I$. Therefore
|
||||
\[
|
||||
R = RP = R\sum_{i \in I}Z(P_i)P_i = \sum_{i \in I}Z(P_i)RP_i = \sum_{i \in I}P_i = P
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Type $\vnI$]
|
||||
\label{definition:vna-t1}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnI$} if for every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero abelian projection $Q \in \text{Proj}(A)$ with $P \ge Q$.
|
||||
\end{definition}
|
||||
% Todo: add a few equivalent characterisations.
|
||||
|
||||
\begin{definition}[Type $\vnII$]
|
||||
\label{definition:vna-t2}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnII$} if:
|
||||
\begin{enumerate}
|
||||
\item $A$ has no non-zero abelian projections.
|
||||
\item For every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero finite projection $Q \in \text{Proj}(A)$ with $P \ge Q$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Type $\vnII_1$]
|
||||
\label{definition:vna-t21}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_1$} if $I$ is a finite projection.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Type $\vnII_\infty$]
|
||||
\label{definition:vna-t2inf}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_\infty$} if $A$ has no non-zero finite central projections.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Type $\vnIII$]
|
||||
\label{definition:vna-t3}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnIII$} if $A$ has no non-zero finite projections.
|
||||
\end{definition}
|
||||
|
||||
\begin{theorem}[Type Decomposition]
|
||||
\label{theorem:vna-type-decomposition}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then there exist unique von Neumann algebras $A_{\vnI}, A_{\vnII_1}, A_{\vnII_{\infty}}, A_{\vnIII} \subset A$\footnote{Not all four types are guaranteed to be present.} of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively, such that
|
||||
\[
|
||||
A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
|
||||
\]
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 26.3]{Zhu}}}. ]
|
||||
($\vnI$): By Zorn's lemma, there exists a maximal family $\seqi{P} \subset \text{Proj}(A)$ of centrally orthogonal abelian projections. Let $P = \sum_{i \in I}P_i$, then $P$ is abelian by (2) of \autoref{lemma:centrally-orthogonal-sum-properties}.
|
||||
|
||||
Let $P_{\vnI} = Z(P)$, then $A_{\vnI} := P_{\vnI}AP_{\vnI}$ is a von Neumann algebra with identity $P_{\vnI}$. Let $R \in \text{Proj}(Z(A_{\vnI})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnI}$, $RP \le R$ is a non-zero abelian projection. Therefore $A_{\vnI}$ is of type $\vnI$.
|
||||
|
||||
($\vnII$): Assume without loss of generality that $I \in A$. Since $\seqi{P}$ is maximal and $(I - P_{\vnI}) \in Z(A)$, $(I - P_{\vnI})A(I - P_{\vnI})$ has no non-zero abelian projections.
|
||||
|
||||
By Zorn's lemma, there exists a maximal family $\seqj{Q} \subset \text{Proj}((I - P_{\vnI})A(I - P_{\vnI}))$ of centrally orthogonal finite projections. Let $Q = \sum_{j \in J}Q_j$, then $Q$ is finite by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}.
|
||||
|
||||
Let $P_{\vnII} = Z(Q)$ and $A_{\vnII} = P_{\vnII}AP_{\vnII}$, then $A_{\vnII}$ is a von Neumann algebra with identity $P_{\vnII}$. Let $R \in \text{Proj}(Z(A_{\vnII})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnII}$, $RQ \le R$ is a non-zero finite projection by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $A_{\vnII}$ is of type $\vnII$.
|
||||
|
||||
($\vnIII$): Let $P_{\vnIII} = I - P_{\vnI} - P_{\vnII}$ and $A_{\vnIII} = P_{\vnIII}AP_{\vnIII}$. Since $P_{\vnIII} \in Z(A)$ and $\seqi{P}$, $\seqj{Q}$ are maximal, $A_{\vnIII}$ has no non-zero finite projections. Therefore $A_{\vnIII}$ is of type $\vnIII$, and $A = A_{\vnI} \oplus A_{\vnII} \oplus A_{\vnIII}$.
|
||||
|
||||
($\vnII_1$): By Zorn's lemma, there exists a maximal family $\bracsn{R_k}_{k \in K} \subset \text{Proj}(A_{\vnII})$ of orthogonal central finite projections. Let $P_{\vnII_1} = \sum_{k \in K}R_k$, then $P_{\vnII_1}$ is a central finite projection by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}. Hence $A_{\vnII_1} = P_{\vnII_1}AP_{\vnII_1}$ is of type $\vnII_1$.
|
||||
|
||||
($\vnII_\infty$): Let $P_{\vnII_\infty} = P_{\vnII} - P_{\vnII_1}$ and $A_{\vnII_\infty} = P_{\vnII_\infty}AP_{\vnII_\infty}$, then $P_{\vnII_\infty}$ is a central projection. By maximality of $\bracsn{R_k}_{k \in K}$, $A_{\vnII_\infty}$ admits no non-zero finite central projections. Therefore $A_{\vnII_\infty}$ is of type $\vnII_\infty$, $A_{\vnII} = A_{\vnII_1} \oplus A_{\vnII_\infty}$, and
|
||||
\[
|
||||
A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
|
||||
\]
|
||||
|
||||
(Uniqueness): Let $A = A_{\vnI}' \oplus A_{\vnII_1}' \oplus A_{\vnII_{\infty}}' \oplus A_{\vnIII}'$ be a decomposition of $A$ into von Neumann algebras of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively.
|
||||
|
||||
Let $P_{\vnI}'$, $P_{\vnII_1}'$, $P_{\vnII_\infty}'$, and $P_{\vnIII}'$ be the identity elements of $A_{\vnI}'$, $A_{\vnII_1}'$, $A_{\vnII_{\infty}}'$, and $A_{\vnIII}'$, respectively, then
|
||||
\[
|
||||
I = P_{\vnI}' \oplus P_{\vnII_1}' \oplus P_{\vnII_\infty}' \oplus P_{\vnIII}'
|
||||
\]
|
||||
|
||||
is an orthogonal direct sum, and
|
||||
\begin{enumerate}
|
||||
\item[($\vnI$)] Let $P_1 = P'_{\vnI}(I - P_{\vnI})$, then by construction of $P_{\vnI}$, there exists no non-zero abelian projection $R \in \text{Proj}(A)$ with $R \le P_1$. As both $P_{\vnI}'$ and $(I - P_{\vnI})$ are central, $P_1 \in A_{\vnI}'$, so $P_1 = 0$ because $A_{\vnI}'$ is of type $\vnI$. Thus $P_{\vnI}' \le P_{\vnI}$. By symmetry, $P_{\vnI} = P_{\vnI}'$ and $A_{\vnI} = A_{\vnI}'$.
|
||||
\item[($\vnII$, $\vnIII$)] Let $P'_{\vnII} = P_{\vnII_1}' \oplus P_{\vnII_\infty}'$, $A_{\vnII}' = A_{\vnII_1}' \oplus A_{\vnII_\infty}'$, and $P_2 = P'_{\vnII}(I - P_{\vnI} - P_{\vnII})$. By construction of $P_{\vnII}$, there exists no non-zero finite projection $R \in \text{Proj}(A)$ with $R \le P_2$. Since $P_2 \in A_{\vnII}'$ and $A_{\vnII}'$ is of type $\vnII$, $P_2 = 0$ and $P_{\vnII}' \le P_{\vnII}$. By symmetry, $P_{\vnII} = P_{\vnII}'$. Thus $P_{\vnIII} = P_{\vnIII}'$, $A_{\vnII} = A_{\vnII}'$, and $A_{\vnIII} = A_{\vnIII}'$.
|
||||
\item[($\vnII_1$, $\vnII_\infty$)] Let $Q_2 = P'_{\vnII_1}(P_{\vnII} - P_{\vnII_1})$, then there exists no non-zero finite central projection $R \in \text{Proj}(A)$ with $R \le Q_2$. However, since $A_{\vnII_1}'$ is of type $\vnII_1$, $P'_{\vnII_1}$ is itself a finite projection, and every subprojection of $P'_{\vnII_1}$ is finite by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $Q_2 = 0$ and $P_{\vnII_1}' \le P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty}' = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.
|
||||
\end{enumerate}
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:abelian-minimal-factor}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, and $P \in \text{Proj}(A)$ be abelian, then $P$ is minimal.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $Q \in \text{Proj}(A)$ with $0 \le Q \le P$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $Q \preceq P - Q$ or $P - Q \preceq Q$. Assume without loss of generality that $Q \preceq P - Q$.
|
||||
|
||||
Let $V \in A$ such that $Q = V^*V$ and $VV^* \le P - Q$, then $V$ is a partial isometry with initial and final spaces contained in $P(H)$. Since $P$ is abelian, $V \in PAP$, and $Q = V^*V = VV^* \le P - Q$. Therefore $Q = 0$, and $P$ is minimal.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:type1-bh-matrix}
|
||||
Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, $T \in A$, $\seqi{P} \subset \text{Proj}(A)$ be non-zero minimal projections such that $I = \sum_{i \in I}P_i$, and $\bracsn{V_{i, j}}_{i, j \in I} \subset A$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$ for all $i, j \in I$, then there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i, j}V_{i, j}$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Let $i, j \in I$, then since $P_i$ and $P_j$ are minimal,
|
||||
\begin{align*}
|
||||
(P_iTP_j)^*(P_iTP_j) &= P_jT^*P_iTP_j \in \complex P_j \\
|
||||
(P_iTP_j)(P_iTP_j)^* &= P_iTP_jT^*P_i \in \complex P_i
|
||||
\end{align*}
|
||||
|
||||
|
||||
If $P_iTP_j \ne 0$, then $P_jT^*P_iTP_j$ and $P_iTP_jT^*P_i$ are positive, and there exists $\lambda > 0$ such that $\lambda P_iTP_j$ is a partial isometry with initial space $P_j(H)$ and final space $P_i(H)$. By \autoref{lemma:projection-types-gymnastics}, there exists $\mu_{i, j} \in \complex$ such that $P_iTP_j = \mu_{i, j} V_{j, i}$. Therefore
|
||||
\[
|
||||
T = \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}\mu_{i, j}V_{j, i}
|
||||
\]
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{theorem}[Classification of Type $\vnI$ Factors]
|
||||
\label{theorem:type1-bh}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item $A$ is of type $\vnI$.
|
||||
\item There exists a minimal projection $P \in \text{Proj}(A)$.
|
||||
\item For every non-zero $P \in \text{Proj}(A)$, there exists a non-zero minimal projection $Q \in \text{Proj}(A)$ such that $P \ge Q$.
|
||||
\item There exists a complex Hilbert space $K$ and a *-isomorphism $\pi: A \to B(K)$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite{TownesType1}}}. ]
|
||||
(1) $\Rightarrow$ (2): Since $A$ is of type $\vnI$, $A$ admits a non-zero abelian projection $Q \in \text{Proj}(A)$. By \autoref{lemma:abelian-minimal-factor}, $Q$ is minimal.
|
||||
|
||||
(2) $\Rightarrow$ (3): Let $P \in \text{Proj}(A)$ and $Q \in \text{Proj}(A)$ be a minimal projection. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P \preceq Q$ or $Q \preceq P$.
|
||||
|
||||
If $P \preceq Q$, then there exists $R \in \text{Proj}(A)$ with $P \sim R \le Q$. In which case, $R$ is minimal, and $P$ is also minimal by (5) of \autoref{lemma:projection-types-gymnastics}.
|
||||
|
||||
If $Q \preceq P$, then there exists $R \in \text{Proj}(A)$ with $Q \sim R \le P$. By (5) of \autoref{lemma:projection-types-gymnastics}, $R$ is minimal with $R \le P$.
|
||||
|
||||
(3) $\Rightarrow$ (4): By Zorn's lemma, there exists a maximal orthogonal family $\seqi{P} \subset \text{Proj}(A)$ of minimal projections. Since $\seqi{P}$ is maximal and every non-zero projection admits a non-zero minimal subprojection, $I = \sum_{i \in I}P_i$, and $H = \bigoplus_{i \in I}P_iH$.
|
||||
|
||||
For each $i, j \in I$, by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P_i \preceq P_j$ or $P_j \preceq P_i$. In both cases, since both projections are minimal, $P_i \sim P_j$. Thus there exists a partial isometry $V_{i, j} \in A$ with initial space $P_i(H)$ and final space $P_j(H)$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$. By fixing a particular family\footnote{The partial isometries need not be unique. }, assume without loss of generality that $V_{i, j}^* = V_{j, i}$ for all $i, j \in I$.
|
||||
|
||||
Fix $i_0 \in I$, let $H_0 = P_{i_0}H$, and define
|
||||
\[
|
||||
U: H \to l^2(I; H_0) \quad (Ux)_i = V_{i, i_0}P_ix
|
||||
\]
|
||||
|
||||
then since $H = \bigoplus_{i \in I}P_iH$ and each $V_{i, i_0}$ is a partial isometry, $U$ is an isometry with inverse
|
||||
\[
|
||||
U^{-1}: l^2(I; H_0) \to H \quad U^{-1}x = \sum_{i \in I}V_{i_0, i}x_i
|
||||
\]
|
||||
|
||||
For each $i \in I$, denote $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then for every $x \in l^2(I; H_0)$,
|
||||
\[
|
||||
UP_iU^{-1}x = U P_i\sum_{j \in I} V_{i_0, j}x_j = e_{i}V_{i, i_0}P_iV_{i_0, i}x_i = e_iP_{i_0}x_i = e_ix_i
|
||||
\]
|
||||
|
||||
|
||||
so $UP_iU^{-1}$ is the projection onto the $i$-th component of $l^2(I; H_0)$.
|
||||
|
||||
Let $T \in A$, then by \autoref{lemma:type1-bh-matrix}, there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i,j}V_{i, j}$. In which case, for any $x \in l^2(I; \complex)$ and $v \in H_0$,
|
||||
\[
|
||||
UTU^{-1}(xv) = UT \sum_{i \in I}V_{i_0, i}x_i v = U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v
|
||||
\]
|
||||
|
||||
For each $i, j \in I$, there exists $\lambda_{i, j} \in \partial B_\complex(0, 1)$ such that $V_{i, j}V_{i_0, i} = \lambda_{i, j}V_{i_0, j}$ by (6) of \autoref{lemma:projection-types-gymnastics}. For every $i \in I$, let $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then
|
||||
\begin{align*}
|
||||
UTU^{-1}(xv) &=U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v = U\sum_{i, j \in I}\mu_{i, j}x_i \cdot V_{i, j}V_{i_0, i} \cdot v \\
|
||||
&= U\sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot V_{i_0, j} \cdot v = \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot V_{j, i_0}P_jV_{i_0, j} \cdot v \\
|
||||
&= \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot v \in l^2(I; \complex v)
|
||||
\end{align*}
|
||||
|
||||
Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^2(I; \complex)$ implies that $\mu_{i, j} =0 $ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_0$ with $\norm{v}_{H_0} = 1$, the mapping
|
||||
\[
|
||||
\pi: A \to B(l^2(I; \complex v)) \quad T \mapsto UTU^{-1}|_{l^2(I; \complex v)}
|
||||
\]
|
||||
|
||||
is an injective $*$-homomorphism.
|
||||
|
||||
Finally, since $\ol{B_A(0, 1)}$ is weak-operator compact and $U$ is an isometry, $\pi(\ol{B_A(0, 1)})$ is also weak-operator compact, and $\pi(A) \subset B(l^2(I; \complex v))$ is a von Neumann algebra. Let $i, j \in I$, then for each $x \in l^2(I; \complex)$,
|
||||
\begin{align*}
|
||||
\pi(V_{i, j})(xv) &= x_i \cdot e_j \cdot V_{j, i_0}V_{i, j}V_{i_0, i} \cdot v = \lambda_{i, j}x_i \cdot e_j \cdot V_{j, i_0}V_{i_0, j} \cdot v \\
|
||||
&= \lambda_{i, j}x_i \cdot e_j \cdot v
|
||||
\end{align*}
|
||||
|
||||
|
||||
so $\pi(V_{i, j}) = \lambda_{i, j} e_jv \otimes e_iv$. As $\bracsn{e_iv}_{i \in I}$ is an orthonormal basis for $l^2(I; \complex v)$, $\pi(A) = B(l^2(I; \complex v))$.
|
||||
|
||||
(4) $\Rightarrow$ (1): Identify $A = \pi(A) = B(K)$, and let $T \in Z(A)$. For each $v \in K$ with $\norm{v}_K = 1$, $T(v \otimes v) = (v \otimes v)T$, so $Tv = T(v \otimes v)v = (v \otimes v)Tv$, and there exists $\lambda \in \complex$ such that $Tv = \lambda v$.
|
||||
|
||||
For any $w \in K$ linearly independent from $v$, there exists $\mu \in \complex$ with $Tw = \mu w$, and $\rho \in \complex$ with $T(v + w) = \rho(v + w)$. In which case, $\rho v + \rho w = \lambda v + \mu w$, so $\lambda = \rho = \mu$, and $T = \lambda I$. Therefore $A$ is a factor.
|
||||
|
||||
For any $v \in K$ with $\norm{v}_K = 1$, $v \otimes v$ is a minimal, and hence abelian projection by (1) of \autoref{lemma:projection-types-gymnastics}. As $v \otimes v \le I$, $A = B(K)$ is of type $\vnI$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
@@ -0,0 +1,267 @@
|
||||
\section{Von Neumann Algebras}
|
||||
\label{section:vna}
|
||||
|
||||
\begin{theorem}[Existence of Projections]
|
||||
\label{theorem:existence-of-projections-vna}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a strong-operator closed $C^*$-subalgebra, then:
|
||||
\begin{enumerate}
|
||||
\item For any bounded, upward-directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$.
|
||||
\item For any directed family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$.
|
||||
\end{enumerate}
|
||||
|
||||
and
|
||||
\begin{enumerate}[start=2]
|
||||
\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 $P = \sotlim_{n \to \infty}T^{1/n} \in A$.
|
||||
\item For each $T \in A$, the orthogonal projection onto $\ol{T(H)}$ is in $A$.
|
||||
\end{enumerate}
|
||||
|
||||
Finally,
|
||||
\begin{enumerate}[start=4]
|
||||
\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 $A$.
|
||||
\end{enumerate}
|
||||
|
||||
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Section 17]{Zhu}}}. ]
|
||||
(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 $B(H)$.
|
||||
|
||||
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 $T \in \cf$.
|
||||
|
||||
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 $S \ge T$,
|
||||
\[
|
||||
\normn{(R - S)^{1/2}x}_H^2 = \dpn{(R - S)x, x}{H} \le \dpn{(R - T)x, x}{H} \le \eps
|
||||
\]
|
||||
|
||||
By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} and \autoref{theorem:c-star-normal-spectral-radius},
|
||||
\[
|
||||
\normn{(R - S)^{1/2}}_{B(H)} = \norm{R - S}_{B(H)}^{1/2} \le \sqrt{2}
|
||||
\]
|
||||
|
||||
so
|
||||
\[
|
||||
\norm{(R - S)x}_H \le \normn{(R - S)^{1/2}}_{B(H)} \cdot \normn{(R - S)^{1/2}x}_H \le \sqrt{2 \eps}
|
||||
\]
|
||||
|
||||
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,
|
||||
\[
|
||||
\dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
|
||||
\]
|
||||
|
||||
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
|
||||
\[
|
||||
\ol{\bigcup_{P \in \mathcal{P}}P(H)} = Q(H) \supset \sup(\mathcal{P})(H)
|
||||
\]
|
||||
|
||||
(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 $I$.
|
||||
|
||||
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:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item For each $k \in \natp$, $p_{n, k}(0) = 0$.
|
||||
\item $p_{n, k} \to f_n$ uniformly on $\sigma_{B(H)}(T)$.
|
||||
\end{enumerate}
|
||||
|
||||
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
|
||||
\[
|
||||
Q^2 = \sotlim_{n \to \infty}T^{2/n} = \sotlim_{n \to \infty}T^{1/n} = Q
|
||||
\]
|
||||
|
||||
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
|
||||
\[
|
||||
\norm{T^*x}_H^2 = \dpn{T^*x, T^*x}{H} = \dpn{TT^*x, x}{H}
|
||||
\]
|
||||
|
||||
Thus $\ker(T^*) = \ker(TT^*)$, and
|
||||
\[
|
||||
\ol{T(H)} = \ker(T^*)^\perp = \ker(TT^*)^\perp = \ol{TT^*(H)}
|
||||
\]
|
||||
|
||||
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 $A$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:invariant-projection-test}
|
||||
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:
|
||||
\begin{enumerate}
|
||||
\item $T(M) \subset M$.
|
||||
\item $PTP = TP$.
|
||||
\item $T^*PTP = T^*TP$.
|
||||
\end{enumerate}
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
(3) $\Leftarrow$ (1): Let $\xi \in H$, then
|
||||
\begin{align*}
|
||||
\dpn{T^*TP\xi, P\xi}{H} &= \dpn{TP\xi, TP\xi}{H} = \norm{TP\xi}_H^2 \\
|
||||
\dpn{T^*PTP\xi, P\xi}{H} &= \dpn{PTP\xi, TP\xi}{H} = \dpn{PTP\xi, PTP\xi}{H} = \norm{PTP\xi}_H^2
|
||||
\end{align*}
|
||||
|
||||
As $\norm{TP\xi}_H = \norm{PTP\xi}_H$, $TP\xi \in H$, so $T(M) \subset M$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{definition}[Reducing Subspace]
|
||||
\label{definition:reducing-subspace}
|
||||
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:
|
||||
\begin{enumerate}
|
||||
\item $T(M) \subset M$ and $T^*(M) \subset M$.
|
||||
\item $TP = PT$.
|
||||
\end{enumerate}
|
||||
|
||||
If the above holds, then $M$ is a \textbf{reducing subspace} of $T$.
|
||||
\end{definition}
|
||||
\begin{proof}[Proof, {{\cite[Corollary 18.3]{Zhu}}}. ]
|
||||
(1) $\Rightarrow$ (2): By \autoref{lemma:invariant-projection-test}, $PTP = TP$ and $PT^*P = T^*P$. Thus $TP = PTP = PT$.
|
||||
|
||||
(2) $\Rightarrow$ (1): Since $TP = PT$, $PTP = PT$, and $T(M) \subset M$ by \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}.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Commutant]
|
||||
\label{definition:commutant}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$, then
|
||||
\[
|
||||
A' = \bracs{T \in B(H)| TS = ST \forall S \in A}
|
||||
\]
|
||||
|
||||
is the \textbf{commutant} of $A$.
|
||||
\end{definition}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:bicommutant-pointwise}
|
||||
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 $Tx \in \ol{\bracsn{Sx|S \in A}}$.
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 18.4]{Zhu}}}. ]
|
||||
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 $Tx \in M = \ol{\bracsn{Sx|S \in A}}$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}[Amplification]
|
||||
\label{lemma:bh-amplification}
|
||||
Let $H$ be a complex Hilbert space, $n \in \natp$, and
|
||||
\[
|
||||
\pi: B(H) \to B(H^n) \quad [\pi(T)(x)]_n = Tx_n
|
||||
\]
|
||||
|
||||
then for each $\seqf{x_j} \subset H$,
|
||||
\[
|
||||
\max_{1 \le j \le n}\norm{Tx_j}_H \le \norm{\pi(T)(x)}_{H^n} \le n \max_{1 \le j \le n}\norm{Tx_j}_H
|
||||
\]
|
||||
\end{lemma}
|
||||
|
||||
|
||||
\begin{theorem}[Von Neumann's Bicommutant Theorem]
|
||||
\label{theorem:bicommutant}
|
||||
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 $A$.
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Section 18.3]{Zhu}}}. ]
|
||||
($\overline{A}^{\text{\small SOT}} \subset A''$): 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
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item $A^{(n)} = \bracsn{S^{(n)}|S \in A}$ is a unital, self-adjoint subalgebra of $B(H^n)$.
|
||||
\item $T^{(n)} \in (A^{(n)})''$.
|
||||
\end{enumerate}
|
||||
|
||||
By \autoref{lemma:bicommutant-pointwise},
|
||||
\[
|
||||
(Tx_1, \cdots, Tx_n) \in \ol{\bracsn{(Sx_1, \cdots, Sx_n)|S \in A}}
|
||||
\]
|
||||
|
||||
so $T \in \ol{A}^{\text{\small SOT}}$.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Von Neumann Algebra]
|
||||
\label{definition:von-neumann-algebra}
|
||||
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.
|
||||
\end{definition}
|
||||
|
||||
\begin{definition}[Factor]
|
||||
\label{definition:vna-factor}
|
||||
Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is a \textbf{factor} if $Z(A) = \complex I$.
|
||||
\end{definition}
|
||||
|
||||
|
||||
\begin{theorem}[Kaplansky Density Theorem]
|
||||
\label{theorem:kaplansky-density}
|
||||
Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
|
||||
\begin{enumerate}
|
||||
\item $\ol{B_{A_{sa}}(0, 1)}$ is strong-operator dense in $\ol{B_{B_{sa}}(0, 1)}$.
|
||||
\item $\bracsn{T \in \ol{B_{A}(0, 1)}|T \ge 0}$ is strong-operator dense in $\bracsn{T \in \ol{B_{B}(0, 1)}|T \ge 0}$.
|
||||
\item $\ol{B_{A}(0, 1)}$ is strong-operator dense in $\ol{B_{B}(0, 1)}$.
|
||||
\end{enumerate}
|
||||
\end{theorem}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 19.5]{Zhu}}}. ]
|
||||
(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
|
||||
\[
|
||||
f: \real \to \real \quad t \mapsto \begin{cases}
|
||||
t &t \in [-1, 1] \\
|
||||
1/t &t \in \real \setminus [-1, 1]
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
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 $B_{sa}$.
|
||||
|
||||
(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
|
||||
\[
|
||||
f: \real \to \real \quad t \mapsto \begin{cases}
|
||||
0 &t \le 0 \\
|
||||
t &t \in [0, 1] \\
|
||||
1/t &t \ge 1
|
||||
\end{cases}
|
||||
\]
|
||||
|
||||
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.
|
||||
|
||||
(3): For each $\mathcal{T} \subset B(H)$, let
|
||||
\[
|
||||
M_2(\mathcal{T}) = \bracs{\begin{bmatrix} Q & R \\ S & T \end{bmatrix} \bigg | Q, R, S, T \in \mathcal{T}}
|
||||
\]
|
||||
|
||||
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
|
||||
\[
|
||||
T' = \begin{bmatrix} 0 & T \\ T^* & 0 \end{bmatrix}
|
||||
\]
|
||||
|
||||
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:
|
||||
\begin{enumerate}[label=(\roman*)]
|
||||
\item 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, $\norm{T_\gamma}_{B(H)} \le 1$.
|
||||
\item With respect to the strong operator topology on $B(H^2)$,
|
||||
\[
|
||||
\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix} \to T'
|
||||
\]
|
||||
\end{enumerate}
|
||||
|
||||
Therefore $\angles{T_\gamma} \subset \ol{B_A(0, 1)}$ is a net that converges to $T$ in the strong-operator topology.
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{remark}
|
||||
\label{remark:kaplansky-unitary}
|
||||
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.
|
||||
\end{remark}
|
||||
|
||||
|
||||
@@ -0,0 +1,206 @@
|
||||
\section{Normal Maps}
|
||||
\label{section:ultraweak}
|
||||
|
||||
\begin{lemma}[Jordan Decomposition]
|
||||
\label{lemma:weak-operator-jordan}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a weak-operator continuous linear functional, then there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_0^3$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
Assume without loss of generality that there exist $\xi, \eta \in H$ such that $\dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H}$ for all $x \in M$. In which case, by \hyperref[polarisation]{proposition:polarisation-complex},
|
||||
\[
|
||||
\dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x(\xi + i^k\eta), \xi + i^k\eta}{H}
|
||||
\]
|
||||
|
||||
for all $x \in M$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
\begin{lemma}[Radon-Nikodym]
|
||||
\label{lemma:vna-radon-nikodym}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $\omega: M \to \complex$ be a positive linear functional, and $\xi \in H$ such that $\omega(x) \le \dpn{x\xi, \xi}{H}$ for all $x \in M_+$, then there exists $x' \in M'_+$ such that $\omega(x) = \dpn{x(x'\xi), x'\xi}{H}$ for all $x \in M$.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for each $x, y \in M$,
|
||||
\[
|
||||
|\omega(y^*x)|^2 \le \omega(x^*x)\omega(y^*y) \le \norm{x\xi}_H^2 \norm{y\xi}_H^2
|
||||
\]
|
||||
|
||||
so the mapping $(x, y) \mapsto \omega(y^*x)$ is a bounded sesquilinear form on $M\xi$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $z \in B(\ol{M\xi})$ such that $\omega(y^*x) = \dpn{zx\xi, y\xi}{H}$ for all $x, y \in M$. In particular, $0 \le \omega(x^*x) = \dpn{zx\xi, x\xi}{H}$ for all $x \in M$, so $z \ge 0$. Moreover, for any $x, y, t \in M$,
|
||||
\[
|
||||
\dpn{ztx\xi, y\xi}{H} = \omega(y^*tx) = \omega((t^*y)^*x) = \dpn{zx\xi, t^*y\xi}{H} = \dpn{tzx\xi, y\xi}{H}
|
||||
\]
|
||||
|
||||
so $z \in M'_+$. Finally, let $p: H \to \ol{M\xi}$ be the orthogonal projection onto $\ol{M\xi}$, and $x' = (zp)^{1/2} \in M'$, then for each $x \in M$,
|
||||
\[
|
||||
\omega(x) = \dpn{zx\xi, \xi}{H} = \dpn{zpx\xi, \xi}{H} = \dpn{x(zp)^{1/2}\xi, (zp)^{1/2}\xi}{H}
|
||||
\]
|
||||
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:completely-additive-projection-choose}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\varphi: M \to \complex$ be a positive linear functional. If for any mutually orthogonal projections $\mathcal{P} \subset M$, $\varphi\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\varphi(p)$, then there exist a non-zero projection $p \in M$ and $\xi \in pH$ such that $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$.
|
||||
\end{lemma}
|
||||
\begin{proof}[Proof, {{\cite[Lemma 2.5.6.]{AnantharamanPopa}}}. ]
|
||||
Let $\eta \in H$ with $\varphi(1_M) < \dpn{\eta, \eta}{H}$. Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections with $\varphi(p) \ge \dpn{p\eta, \eta}{H}$ for all $p \in \mathcal{P}$. By assumption,
|
||||
\begin{align*}
|
||||
\varphi\paren{\sum_{p \in \mathcal{P}}p} &\ge \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \\
|
||||
\varphi\paren{1_M - \sum_{p \in \mathcal{P}}p}&= \varphi(1_M) - \varphi\paren{\sum_{p \in \mathcal{P}}p} \\
|
||||
&< \dpn{\eta, \eta}{H} - \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \ge 0
|
||||
\end{align*}
|
||||
|
||||
Let $p := 1_M - \sum_{q \in \mathcal{P}}q > 0$, then by maximality of $\mathcal{P}$, $\varphi(q) < \dpn{q\eta, \eta}{H}$ for all subprojections $q \le p$.
|
||||
|
||||
Let $x \in (pMp)_+$ with $0 \le x \le 1$. By \autoref{theorem:vn-projection-norm-dense}, there exist projections $\seq{q_n} \subset pMp$ such that $x = \sum_{n = 1}^\infty \frac{q_n}{2^n}$. In which case\footnote{Here, only continuity of $\varphi$ is used.},
|
||||
\[
|
||||
\dpn{x\eta, \eta}{H} = \sum_{n = 1}^\infty 2^{-n}\dpn{q_n\eta, \eta}{H} \ge \sum_{n = 1}^\infty \frac{\varphi(q_n)}{2^n} = \varphi(x)
|
||||
\]
|
||||
|
||||
Therefore $\dpn{x\eta, \eta}{H} \ge \varphi(x)$ for all $x \in (pMp)_+$. By the \hyperref[Radon-Nikodym theorem for positive functionals]{lemma:vna-radon-nikodym}, there exists $\xi \in pH$ with $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
\begin{definition}[Normal Linear Functional]
|
||||
\label{definition:vna-normal-functional}
|
||||
Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a positive linear functional, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item For any bounded, upward-directed family $A \subset M$, $\omega(\sup(A)) = \sup(\omega(A))$.
|
||||
\item For any mutually orthogonal projections $\mathcal{P} \subset M$, $\omega\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\omega(p)$.
|
||||
|
||||
\item $\omega$ is the norm limit of a sequence of weak-operator continuous positive linear functionals.
|
||||
\item $\omega$ is weak-operator continuous on bounded subsets of $M$.
|
||||
\item $\omega$ is strong-operator continuous on bounded subsets of $M$.
|
||||
\end{enumerate}
|
||||
\end{definition}
|
||||
\begin{proof}[Proof, {{\cite[Theorem 2.5.5]{AnantharamanPopa}}}. ]
|
||||
(2) $\Rightarrow$ (3): Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections such that for each $p \in \mathcal{P}$, there exists $\xi_p \in pH$ such that $\omega(x) = \dpn{x\xi_p, \xi_p}{H}$ for all $x \in pMp$. Let $q = \sum_{p \in \mathcal{P}}p$, then by maximality of $\mathcal{P}$ and \autoref{lemma:completely-additive-projection-choose} applied to $(1 - q)M(1 - q)$, $\sum_{p \in \mathcal{P}}p = 1_M$.
|
||||
|
||||
For each $x \in M$ and $p \in \mathcal{P}$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
|
||||
\[
|
||||
|\omega(xp)|^2 \le \omega(1) \cdot \omega(px^*xp) = \omega(1) \cdot \norm{x\xi_p}_H^2
|
||||
\]
|
||||
|
||||
so the mapping $y \xi_p \mapsto \omega(yp)$ extends into a bounded linear functional on $\ol{M\xi_p}$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\eta_p \in \ol{M\xi_p}$ such that $\omega(xp) = \dpn{x\xi_p, \eta_p}{H}$ for all $x \in M$.
|
||||
|
||||
Let $\mathcal{F} \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$, directed by inclusion. For each $F \in \mathcal{F}$, let $q_F = \sum_{p \in F}p$ and $\omega_F \in M^*$ be defined by $\omega_F(x) = \omega(q_Fxq_F)$, then for each $x \in M$,
|
||||
\[
|
||||
\omega_F(x) = \sum_{p \in F}\omega(q_Fxp) = \sum_{p \in F}\dpn{q_Fx\xi_p, \eta_p}{H}
|
||||
\]
|
||||
|
||||
so $\omega_F$ is weak-operator continuous. Finally, for any $x \in M$,
|
||||
\begin{align*}
|
||||
|\omega(x) - \omega_F(x)| &\le |\omega(x) - \omega(xq_F)| + |\omega(xq_F) - \omega(q_Fxq_F)| \\
|
||||
&= |\omega(x(1_M - q_F))| + |\omega((1 - q_F)xq_F)| \\
|
||||
&\le 2\norm{x}_M\omega(1)^{1/2} \cdot \omega(1 - q_F)^{1/2}
|
||||
\end{align*}
|
||||
|
||||
By (2), $\lim_{F \in \mathcal{F}}\omega(1 - q_F)^{1/2} = 0$. Therefore $\omega = u\text{-}\lim_{F \in \mathcal{F}}\omega_F$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:normal-functional-bound}
|
||||
Let $M$ be a von Neumann algebra, $\varphi, \omega \in M^*$ be positive linear functionals. If $\omega$ is normal and $\varphi \le \omega$, then $\varphi$ is also normal.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
By (2) of \autoref{definition:vna-normal-functional}.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Normal Positive Map]
|
||||
\label{definition:normal-linear-map}
|
||||
Let $M, N$ be von Neumann algebras and $\phi: M \to N$ be a positive linear map, then the following are equivalent:
|
||||
\begin{enumerate}
|
||||
\item For any bounded, upward-directed family $A \subset M$, $\phi(\sup(A)) = \sup(\phi(A))$.
|
||||
\item For any normal, positive linear functional $\omega \in N^*$, $\omega \circ \phi$ is a normal, positive linear functional.
|
||||
\item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the weak operator topologies.
|
||||
\end{enumerate}
|
||||
|
||||
If $\phi$ is in addition a *-homomorphism, then the above are also equivalent to
|
||||
\begin{enumerate}[start=3]
|
||||
\item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the strong operator topologies.
|
||||
\end{enumerate}
|
||||
|
||||
If the above holds, then $\phi$ is \textbf{normal}.
|
||||
\end{definition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 2.5.8]{AnantharamanPopa}}}. ]
|
||||
(2) $\Rightarrow$ (3): Let $\omega \in N^*$ be a weak-operator continuous functional. By \autoref{lemma:weak-operator-jordan}, there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_0^3$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$. By assumption, $\omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. In which case, $\omega \circ \phi = \sum_{k = 0}^3 i^k \omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. Therefore $\phi$ is weak-operator continuous on bounded subsets of $M$.
|
||||
|
||||
(3) $\Rightarrow$ (1), (4) $\Rightarrow$ (1): By (1) of \autoref{theorem:existence-of-projections-vna}.
|
||||
|
||||
(3) $\Rightarrow$ (4)\footnote{Central here is the observation that $x_\alpha \to 0$ in the strong operator topology i.f.f. $x_\alpha^*x_\alpha \to 0$ in the weak operator topology. }: Suppose that $M$ acts on the Hilbert space $H$, and $N$ acts on the Hilbert space $K$. Let $\angles{x_\alpha}_{\alpha \in A} \subset M$ be a bounded net converging to $x \in M$ in the strong operator topology, then for each $\xi \in H$,
|
||||
\[
|
||||
\lim_{\alpha \in A}\dpn{(x - x_\alpha)^*(x - x_\alpha)\xi, \xi}{H} = \lim_{\alpha \in A}\normn{(x - x_\alpha)\xi}_H^2 = 0
|
||||
\]
|
||||
|
||||
By \hyperref[polarisation]{proposition:polarisation-complex}, $\wotlim_{\alpha \in A}(x - x_\alpha)^*(x - x_\alpha) = 0$. As $\angles{x_\alpha}_{\alpha \in A}$ is bounded,
|
||||
\[
|
||||
\wotlim_{\alpha \in A}\phi((x - x_\alpha)^*(x - x_\alpha)) = 0
|
||||
\]
|
||||
|
||||
as well. In particular, for any $\xi \in K$,
|
||||
\[
|
||||
\lim_{\alpha \in A}\norm{\phi(x - x_\alpha)\xi}_K^2 = \lim_{\alpha \in A}\dpn{\phi(x - x_\alpha)^*\phi(x - x_\alpha)\xi, \xi}{K} = 0
|
||||
\]
|
||||
|
||||
Therefore $\phi$ restricted to bounded subsets of $M$ is strong-operator continuous.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:vna-isomorphism-normal}
|
||||
Let $M, N$ be von Neumann algebras, then every isomorphism $\phi: M \to N$ is normal. In particular, the weak and strong operator topologies on $M$, restricted to bounded subsets, are independent of its representation.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
As $\phi$ is an isomorphism, it is an order isomorphism. By (1) of \autoref{definition:normal-linear-map}, $\phi$ is normal. By (3) and (4) of \autoref{definition:normal-linear-map}, its restriction to bounded subsets of $M$ is both weak-operator and strong-operator continuous.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:vna-normal-bound}
|
||||
Let $M, N$ be von Neumann algebras, $\omega \in M^*$ and $\psi \in N^*$ be normal positive linear functionals with $\psi$ being faithful, and $\phi: M \to N$ be a positive linear map with $\psi \circ \phi \le \omega$, then $\phi$ is normal.
|
||||
\end{proposition}
|
||||
\begin{proof}[Proof, {{\cite[Proposition 2.5.11]{AnantharamanPopa}}}. ]
|
||||
Let $\varphi = \psi \circ \phi \le \omega$, then $\varphi$ is normal by \autoref{lemma:normal-functional-bound}. For any bounded, upward-directed family $A \subset M$,
|
||||
\[
|
||||
\psi(\sup(\phi(A))) =\sup(\psi(\phi(A))) = \sup(\varphi(A)) = \varphi(\sup(A)) = \psi(\phi(\sup(A)))
|
||||
\]
|
||||
|
||||
As $\psi$ is faithful, $\sup(\phi(A)) = \phi(\sup(A))$. Therefore $\phi$ is normal by (1) of \autoref{definition:normal-linear-map}.
|
||||
\end{proof}
|
||||
|
||||
\begin{proposition}
|
||||
\label{proposition:normal-image-vna}
|
||||
Let $M$ be a von Neumann algebra, $H$ be a complex Hilbert space, and $\pi: M \to B(H)$ be a normal unital *-homomorphism, then $\pi(M)$ is a von Neumann algebra.
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\end{proposition}
|
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\begin{proof}
|
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Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_N(0, 1)}^{B(H)} = \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By \autoref{proposition:compact-extensions} and \autoref{proposition:bh-ultraweak-bounded}, $\ol{B_M(0, 1)}^M$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^M)$ is weak-operator compact and contains $\pi(B_M(0, 1))$. Thus
|
||||
\[
|
||||
\ol{B_{\pi(M)}(0, 1)}^{B(H)} \supset \pi(\ol{B_M(0, 1)}^M) \supset \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}} \supset \ol{B_N(0, 1)}^{B(H)}
|
||||
\]
|
||||
|
||||
and $\ol{B_{\pi(M)}(0, 1)}^{B(H)} = \ol{B_{N}(0, 1)}^{B(H)}$, so $\pi(M) = \ol{\pi(M)}^{B(H)} = N$ by \autoref{theorem:cstar-quotient}.
|
||||
\end{proof}
|
||||
|
||||
|
||||
|
||||
|
||||
\begin{lemma}
|
||||
\label{lemma:tracial-von-neumann-gns}
|
||||
Let $M$ be a von Neumann algebra, $\varphi \in S(M)$ be a state, and $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\varphi$, then $\pi$ is normal if and only if $\varphi$ is normal. In which case, $\pi(M) \subset B(H)$ is a von Neumann algebra.
|
||||
\end{lemma}
|
||||
\begin{proof}
|
||||
(1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that the weak operator topology on $\ol{B_{\pi(M)}(0, 1)}$ coincides with the topology of pointwise convergence on $\bracsn{\eta \otimes \theta|\eta, \theta \in \pi(M)\xi}$. For each $x, y, z \in M$,
|
||||
\[
|
||||
\dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H} = \dpn{\pi(z^*)\pi(x)\pi(y)\xi, \xi}{H} = \varphi(z^*xy)
|
||||
\]
|
||||
|
||||
By \hyperref[separate continuity of composition]{proposition:bh-operator-topologies-facts} in the weak operator topology, the functional $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H}$ is weak-operator continuous. As such, $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)} \to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous.
|
||||
|
||||
|
||||
(2): By \autoref{proposition:normal-image-vna}.
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}[Separable]
|
||||
\label{definition:separable-vna}
|
||||
Let $M$ be a von Neumann algebra, then $M$ is \textbf{separable} if there exists a separable Hilbert space $H$ and a faithful normal representation $\pi: M \to B(H)$.
|
||||
\end{definition}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user