diff --git a/src/op/c-star/gelfand.tex b/src/op/c-star/gelfand.tex index 25386f7..c38c48e 100644 --- a/src/op/c-star/gelfand.tex +++ b/src/op/c-star/gelfand.tex @@ -59,10 +59,10 @@ \begin{corollary} \label{corollary:stonean-commutative-algebra} - Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected. + Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremally disconnected. \end{corollary} \begin{proof} - By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected. + By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremally-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremally disconnected. \end{proof} diff --git a/src/op/example/bb.tex b/src/op/example/bb.tex index 1c47f51..159fb60 100644 --- a/src/op/example/bb.tex +++ b/src/op/example/bb.tex @@ -88,11 +88,11 @@ \begin{corollary} -\label{corollary:linfinity-extremely-disconnected} - Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. +\label{corollary:linfinity-extremally-disconnected} + Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremally disconnected. \end{corollary} \begin{proof} - 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 extremely disconnected. + 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. \end{proof} \begin{lemma} diff --git a/src/op/example/bounded.tex b/src/op/example/bounded.tex index 7733455..8ea3824 100644 --- a/src/op/example/bounded.tex +++ b/src/op/example/bounded.tex @@ -158,6 +158,18 @@ so $QU^n|_{H} = V^n$. \end{proof} +\begin{definition}[Compression] +\label{definition:bh-compression} + Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. +\end{definition} + +\begin{definition}[Dilation] +\label{definition:bh-dilation} + Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a Hilbert space $K \supset H$ and an operator $S \in B(K)$ such that $T = PS|_H$, where $P \in B(K)$ is the orthogonal projection onto $H$. If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. +\end{definition} + + + \begin{definition}[Isometric Dilation] \label{definition:isometric-dilation} Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, and @@ -193,7 +205,7 @@ \begin{theorem}[Sz.-Nagy's Dilation Theorem] \label{theorem:sznagy-dilation} - Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. + Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$. \end{theorem} \begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ] Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, diff --git a/src/op/system/definitions.tex b/src/op/system/definitions.tex index 089aebe..c904064 100644 --- a/src/op/system/definitions.tex +++ b/src/op/system/definitions.tex @@ -38,12 +38,13 @@ \begin{proposition} \label{proposition:positive-automatic-continuity} Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then - \[ - \norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B - \] + \begin{enumerate} + \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norom{x}_A$. + \item $\norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B$. + \end{enumerate} \end{proposition} \begin{proof} - Let $x \in T_+$, then $0 \le x \le \norm{x}_A$, and $0 \le \phi(x) \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$, so $\norm{\phi(x)}_B \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$. + (1): Let $x \in T_+$, then $0 \le x \le \norm{x}_A$, and $0 \le \phi(x) \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$, so $\norm{\phi(x)}_B \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$. For any $x, y \in T_+$, $x - y \le x$ and $y - x \le y$. As such, \[ @@ -57,7 +58,7 @@ &\le \norm{s}_A \norm{\phi(1_A)}_B \end{align*} - Finally, for any general $s \in S$, + (2): For any general $s \in S$, \begin{align*} \phi(s) &= \phi(\text{Re}(s)) + i\phi(\text{Im}(s)) \\ \norm{\phi(s)}_B &\le \norm{\phi(\text{Re}(s))}_B + \norm{\phi(\text{Im}(s))}_{B} \\ @@ -65,6 +66,35 @@ \end{align*} \end{proof} +\begin{theorem} +\label{theorem:positive-commutative-norm} + Let $X$ be a compact Hausdorff space, $A$ be a unital $C^*$-algebra and $\phi: C(X; \complex) \to A$ be a positive map, then $\norm{\phi}_{L(C(X; \complex); B)} = \norm{\phi(\one)}_{B}$. +\end{theorem} +\begin{proof} + Assume without loss of generality that $\norm{\phi(\one)}_{B} \le 1$. + + Let $\seqf{h_j} \subset C(X; [0, 1])$ be a partition of unity on $X$, and $\seqf{\lambda_j} \subset \ol{B_\complex(0, 1)}$. + + Since $A$ is a unital $C^*$-algebra, there exists a complex Hilbert space $H$ such that $A$ may be identified with a $C^*$-subalgebra of $B(H)$, with $1_A = 1_{B(H)}$. In which case, for any $\xi, \eta \in H$, + \begin{align*} + \abs{\sum_{j = 1}^n \dpn{\lambda_j \phi(h_j)\xi, \eta}{H}} &\le \sum_{j = 1}^n \abs{\dpn{\phi(h_j)\xi, \eta}{H}} \\ + &= \sum_{j = 1}^n\abs{\dpn{\phi(h_j)^{1/2}\xi, \phi(h_j)^{1/2}\eta}{H}} \\ + &\le \braks{\sum_{j = 1}^n \normn{\phi(h_j)^{1/2}\xi}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n\normn{\phi(h_j)^{1/2}\eta}_H^2}^{1/2}\\ + &\le \braks{\sum_{j = 1}^n \dpn{\phi(h_j)\xi, \xi}{H}}^{1/2} \cdot \braks{\sum_{j = 1}^n\dpn{\phi(h_j)\xi, \xi}{H}}^{1/2} \\ + &\le \braks{\angles{\sum_{j = 1}^n \phi(h_j)\xi, \xi}{H}}^{1/2} \cdot \braks{\angles{\sum_{j = 1}^n \phi(h_j)\eta, \eta}{H}}^{1/2} \\ + &\le (\dpn{\xi, \xi}{H})^{1/2}(\dpn{\eta, \eta}{H})^{1/2} = \norm{\xi}_H \norm{\eta}_H + \end{align*} + + so $\normn{\sum_{j = 1}^n \lambda_j\phi(h_j)}_A \le 1$. + + Let $f \in C(X; \complex)$ with $\norm{f}_u \le 1$ and $\eps > 0$, then there exists a partition of unity $\seqf{h_j} \subset C(X; [0, 1])$ such that for each $1 \le j \le n$, $\sup_{x, y \in \supp{h_j}}|f(x) - f(y)| < \eps$. For each $1 \le j \le n$, choose $x_j \in \supp{h_j}$, then $\norm{f - \sum_{j = 1}^n f(x_j)h_j}_u \le \eps$. + + By the previous case and \autoref{proposition:positive-automatic-continuity}, + \[ + \norm{\phi(f)}_A \le \norm{\sum_{j = 1}^n f(x_j)\phi(h_j)}_A + 2\norm{f - \sum_{j = 1}^n f(x_j)h_j}_u \le 1 + 2\eps + \] +\end{proof} + \begin{example}[Arveson] \label{example:positive-automatic-continuity} @@ -81,7 +111,10 @@ \begin{enumerate} \item $\phi$ is a positive map. \item $\norm{\phi}_{L(C(\mathbb{T} ; \complex); M_2(\complex))} = 2$. + \item $\phi$ admits no extension to a positive map on $C(\mathbb{T}; \complex)$. \end{enumerate} + + As such, $M_2(\complex)$ is not injective with respect to operator systems and positive maps. \end{example} \begin{proof}[Proof, {{\cite[Example 2.2]{Paulsen}}}. ] (1): Let $a + bz + c \ol z \in S$, then the following are equivalent: @@ -95,4 +128,7 @@ (2): $\norm{\phi(z)}_{M_2(\complex)} = 2$. + (3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$. \end{proof} + + diff --git a/src/op/vn/spec.tex b/src/op/vn/spec.tex index 2d49746..52e46f2 100644 --- a/src/op/vn/spec.tex +++ b/src/op/vn/spec.tex @@ -322,7 +322,7 @@ 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 extremely 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 $H$. + \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 $L^1(\Omega) \to \mathscr{E}$. diff --git a/src/topology/main/stonean.tex b/src/topology/main/stonean.tex index 0871b1d..1321bcd 100644 --- a/src/topology/main/stonean.tex +++ b/src/topology/main/stonean.tex @@ -1,18 +1,18 @@ \section{Extremely Disconnected Spaces} -\label{section:extremely-disconnected-space} +\label{section:extremally-disconnected-space} \begin{definition}[Extremely Disconnected] -\label{definition:extremely-disconnected} - Let $X$ be a topological space, then $X$ is \textbf{extremely disconnected} if for every $U \subset X$ open, $\ol U$ is also open. +\label{definition:extremally-disconnected} + Let $X$ be a topological space, then $X$ is \textbf{extremally disconnected} if for every $U \subset X$ open, $\ol U$ is also open. \end{definition} \begin{theorem}[Stone-Nakano] -\label{theorem:stone-nakano-extremely-disconnected} - Let $X$ be a topological space. If $X$ is extremely disconnected, then $C(X; \real)$ is order complete. Conversely, if $X$ is completely regular and $C(X; \real)$ is order complete, then $X$ is extremely disconnected. +\label{theorem:stone-nakano-extremally-disconnected} + Let $X$ be a topological space. If $X$ is extremally disconnected, then $C(X; \real)$ is order complete. Conversely, if $X$ is completely regular and $C(X; \real)$ is order complete, then $X$ is extremally disconnected. \end{theorem} \begin{proof} - ($\Rightarrow$): Suppose that $X$ is extremely disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let + ($\Rightarrow$): Suppose that $X$ is extremally disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let \[ U(q) = \overline{\bigcup_{f \in \cf}\bracs{f > q}} \subset \overline{\bracs{F > q}} \]