Added multiplicative domains.

src/op/system/dilation.tex

@@ -47,113 +47,6 @@
     Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. 
 \end{lemma}
 
-\begin{theorem}[Stinespring's Dilation Theorem]
-\label{theorem:stinespring-dilation}
-    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(K, V, \pi)$, where:
-    \begin{enumerate}[label=(\alph*)]
-        \item $K$ is a complex Hilbert space. 
-        \item $V: H \to K$ is a continuous linear map.
-        \item $\pi: A \to B(K)$ is a unital *-homomorphism.
-    \end{enumerate}
-
-    such that:
-    \begin{enumerate}
-        \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
-        \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. 
-    \end{enumerate}
-
-    Moreover, if $\phi$ is unital, then:
-    \begin{enumerate}[start=2]
-        \item $V$ is an isometry. 
-        \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
-    \end{enumerate}
-    
-    The triple $(K, V, \pi)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(K, V, \pi)$ is \textbf{minimal}.
-\end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
-    (a): Let $A \otimes H$ and define the Hermitian form
-    \[
-    \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
-    \]
-
-    For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and
-    \[
-    a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
-    \]
-
-    then since $\phi$ is completely positive, 
-    \[
-        \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
-    \]
-
-    and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. 
-    
-    Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
-    \[
-    \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
-    \]
-
-    so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. 
-
-    Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. 
-
-    (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, 
-    \[
-    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
-    \]
-
-    so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. 
-
-    (c): For each $a \in A$, let
-    \[
-    \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
-    \]
-
-    then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, 
-    \begin{align*}
-        \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, b_i \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\
-        &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2
-    \end{align*}
-
-    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. 
-
-    By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, 
-    \begin{align*}
-        \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
-        &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
-        &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
-    \end{align*}
-
-    By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. 
-
-    (1): For each $\xi \in H$, 
-    \[
-    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
-    \]
-
-    so
-    \[
-    \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
-    \]
-
-    by \autoref{proposition:bh-numerical-radius-gymnastics}. 
-
-    (2): For each $a \in A$ and $\xi, \eta \in H$, 
-    \[
-        \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
-    \]
-
-    so $V^*\pi(a)V = \phi(a)$. 
-
-    (3): If $\phi$ is unital, then for each $\xi \in H$,
-    \[
-    \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
-    \]
-
-    so $V$ is an isometry. 
-    
-\end{proof}
-
 
 \begin{theorem}[Sz.-Nagy's Dilation Theorem]
 \label{theorem:sznagy-dilation}
@@ -393,3 +286,140 @@
     by (3).
 \end{proof}
 
+
+
+\begin{theorem}[Stinespring's Dilation Theorem]
+\label{theorem:stinespring-dilation}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, where:
+    \begin{enumerate}[label=(\alph*)]
+        \item $K$ is a complex Hilbert space. 
+        \item $V: H \to K$ is a continuous linear map.
+        \item $\pi: A \to B(K)$ is a unital *-homomorphism.
+    \end{enumerate}
+
+    such that:
+    \begin{enumerate}
+        \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
+        \item For all $a \in A$, $\phi(a) = V^*\phi(a)V$. 
+    \end{enumerate}
+
+    Moreover, if $\phi$ is unital, then:
+    \begin{enumerate}[start=2]
+        \item $V$ is an isometry. 
+        \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
+    \end{enumerate}
+    
+    The triple $(\pi, V, K)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(\pi, V, K)$ is \textbf{minimal}.
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
+    (a): Let $A \otimes H$ and define the Hermitian form
+    \[
+    \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
+    \]
+
+    For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and
+    \[
+    a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
+    \]
+
+    then since $\phi$ is completely positive, 
+    \[
+        \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
+    \]
+
+    and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. 
+    
+    Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \[
+    \dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
+    \]
+
+    so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in K_0}$. 
+
+    Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. 
+
+    (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
+    \]
+
+    so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. 
+
+    (c): For each $a \in A$, let
+    \[
+    \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
+    \]
+
+    then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, 
+    \begin{align*}
+        \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, b_i \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\
+        &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2
+    \end{align*}
+
+    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. 
+
+    By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, 
+    \begin{align*}
+        \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
+        &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
+        &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
+    \end{align*}
+
+    By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. 
+
+    (1): For each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
+    \]
+
+    so
+    \[
+    \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
+    \]
+
+    by \autoref{proposition:bh-numerical-radius-gymnastics}. 
+
+    (2): For each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
+    \]
+
+    so $V^*\pi(a)V = \phi(a)$. 
+
+    (3): If $\phi$ is unital, then for each $\xi \in H$,
+    \[
+    \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
+    \]
+
+    so $V$ is an isometry. 
+
+    (4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
+    \]
+
+    Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$. 
+    
+\end{proof}
+
+\begin{proposition}[Multiplicative Domain of Stinespring Representations]
+\label{proposition:stinespring-multiplicative-domain}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
+    \begin{enumerate}
+        \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$. 
+        \item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = V^* \in B(K)$ be the orthogonal projection onto $H$. 
+    
+    Let $a \in A$ and $\xi, \eta \in H$, then
+    \begin{align*} 
+    \dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
+    \dpn{\phi(a^*a)\xi, \eta}{H} &= \dpn{\pi(a^*a)V\xi, V\eta}{K} = \dpn{\pi(a)^*\pi(a)P\xi, P\eta}{K}
+    \end{align*}
+
+    By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*P\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$. 
+    
+\end{proof}
+

src/op/system/index.tex

@@ -1,7 +1,6 @@
 \chapter{Operator Systems}
 \label{chap:opertaor-systems}
 
-\input{./dilation.tex}
 \input{./os.tex}
 \input{./two.tex}
 \input{./completely.tex}
@@ -9,3 +8,4 @@
 \input{./module.tex}
 \input{./numerical-radius.tex}
 \input{./schur.tex}
+\input{./dilation.tex}

src/op/vn/vn.tex

@@ -100,9 +100,19 @@
     \begin{enumerate}
         \item $T(M) \subset M$. 
         \item $PTP = TP$. 
+        \item $T^*PTP = T^*TP$. 
     \end{enumerate}
 \end{lemma}
-% Proof omitted due to obviousness
+\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]