Updated notes from course.

src/op/system/aos.tex

@@ -0,0 +1,76 @@
+\section{Abstract Operator Systems}
+\label{section:abstract-operator-system}
+
+\begin{definition}[Matrix Order]
+\label{definition:matrix-order}
+    Let $S$ be a complex vector space equipped with a \hyperref[complex conjugation]{definition:complex-conjugation} $*: S \to S$, then $S$ is \textbf{matrix ordered} if:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $M_n(S)_{sa}$ is equipped with a vector space ordering. 
+        \item For each $m, n \in \natp$, $A \in M_{n \times m}(\complex)$, and $x \in M_n(S)_{sa}$ with $x \ge 0$, $A^*xA \ge 0$ in $M_m(S)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Matrix Order Unit]
+\label{definition:matrix-order-unit}
+    Let $S$ be a matrix ordered space and $e \in S$, then $e$ is a \textbf{matrix order unit} if $I_n \otimes e \in M_n(S)$ is an order unit for all $n \in \natp$. 
+\end{definition}
+
+\begin{definition}[Extended Map]
+\label{definition:abstract-induced-matrix-map}
+    Let $S$, $T$ be complex vector spaces, $\phi: S \to T$, and $n \in \natp$, then $\phi_n := \text{Id} \otimes \phi: M_n(S) \to M_n(T)$ is the \textbf{extended map} of $\phi$ to $M_n(S)$. 
+\end{definition}
+
+\begin{definition}[Abstract Operator System]
+\label{definition:a-operator-system}
+    Let $S$ be a matrix ordered space, then $S$ is an \textbf{abstract operator system} if $M_n(S)$ is Archimedean for all $n \in \nat$, and $S$ admits a matrix order unit. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:matrix-order-linear-functional}
+    Let $S$ be a matrix ordered space, then the mapping
+    \[
+    \text{hom}(S; M_n(\complex)) \to \text{hom}(M_n(S); \complex) \quad \angles{E_{ij} \otimes a, \phi}_{M_n(S)} = \phi(a)_{ij}
+    \]
+
+    induces an isomorphism $\text{hom}(M_n(S); \complex) \iso \text{hom}(S; M_n(\complex))$. 
+\end{proposition}
+
+\begin{proposition}
+\label{proposition:matrix-order-positive-characterisation}
+    Let $S$ be a matrix ordered space, $\phi \in \text{hom}(S; M_n(\complex))$ and $s \in \text{hom}(M_n(S); \complex)$, and
+    \[
+    s: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $s$ is a positive linear functional. 
+        \item $\phi$ is $n$-positive. 
+        \item $\phi$ is completely positive. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 13.2]{Paulsen}}}. ]
+    (1) $\Rightarrow$ (3): Let $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, $\eta \in (\eta_1, \cdots, \eta_n) \in \complex^n$, and $x \in S$, then
+    \[
+        \dpn{\phi(x)\xi, \eta}{\complex^n} = \sum_{i, j = 1}^n \phi(x)_{ij}\xi_j \ol {\eta_i} = s({\eta^* x\xi})
+    \]
+
+    Let $m \in \natp$, $\xi = (\xi_1, \cdots, \xi_m) \in \complex^m \otimes \complex^n$, and $A = \sum_{i, j = 1}^m E_{ij} \otimes a_{ij} \in M_n(S)$ with $a \ge 0$, then
+    \begin{align*}
+        \dpn{\phi_n(A)\xi, \xi}{\complex^m \otimes \complex^n} &= \sum_{i, j = 1}^m \dpn{\phi(a_{ij})\xi_j, \xi_i}{\complex^n} = \sum_{i, j = 1}^m s(\xi_i^*a_{ij}\xi_j) \\
+        &= s\braks{\begin{bmatrix}
+            \xi_1 \\
+            \vdots \\
+            \xi_n
+        \end{bmatrix}^*A\begin{bmatrix}
+            \xi_1 \\
+            \vdots \\
+            \xi_n
+        \end{bmatrix}} \ge 0
+    \end{align*}
+\end{proof}
+
+
+
+
+

src/op/system/arveson.tex

@@ -0,0 +1,62 @@
+\section{Arveson's Extension Theorem}
+\label{section:arveson-extension}
+
+\begin{lemma}
+\label{lemma:tensor-trace-approx}
+    Let $E, F$ be locally convex spaces of $K \in \RC$, $B \subset L(E; F^*)$ be equicontinuous, then the following two topologies on $B$ coincide:
+    \begin{enumerate}
+        \item The weak* topology under the identification $L(E; F^*) = (E \wh \otimes_\pi F)^*$. 
+        \item The topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    As $B$ is equicontinuous, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that the weak* topology coincides with the topology of pointwise convergence on any dense subset. By linearity, the topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$ coincides with the topology of pointwise convergence on $E \otimes F$. As $E \otimes F$ is dense $E \wh \otimes_\pi F$, the topology of pointwise convergence on $E \otimes F$ coincides with the topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$. 
+\end{proof}
+
+
+\begin{definition}[Point-Weak* Topology]
+\label{definition:bw-topology}
+    Let $E$ be a Banach space and $H$ be a Hilbert space. Identify
+    \[
+    L(S; B(H)) = L(S; (H \otimes_\pi H)^*) = [S \otimes_\pi (H \otimes_\pi H)]^*
+    \] 
+
+    then the weak* topology on $L(E; B(H))$ induced by $S \otimes_\pi (H \otimes_\pi H)$ is the \textbf{point-weak* topology}/\textbf{BW topology} on $L(E; B(H))$. 
+\end{definition}
+
+
+
+\begin{proposition}
+\label{proposition:bw-topology-gymnastics}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $H$ be a complex Hilbert space. Identify
+    \[
+    L(S; B(H)) = L(S; (H \otimes_\pi H)^*) = [S \otimes_\pi (H \otimes_\pi H)]^*
+    \] 
+\end{proposition}
+
+\begin{theorem}[Arveson's Extension Theorem]
+\label{theorem:arveson-extension}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a completely positive map, then $\phi$ admits a completely positive extension to $A$. 
+\end{theorem}
+\begin{proof}
+    Let $\mathcal{F} \subset 2^H$ be the collection of all finite-dimensional subspaces of $H$, directed under inclusion. 
+    
+    Let $F \in \mathcal{F}$ and $P_F \in B(H)$ be the orthogonal projection onto $H$, then the corner $P_FB(H)P_F$ is isomorphic to $M_{|F|}(\complex)$. Let
+    \[
+    \phi_F: S \to P_FB(H)P_F \quad \phi_F(a) = P_F\phi(a)P_F
+    \]
+
+    be the compression of $\phi$ onto $F$, then $\phi_F$ is also completely positive by \autoref{proposition:multiply-completely-bounded}. By the \hyperref[Krein Extension Theorem]{theorem:krein-extension}, $\phi_F$ admits an extension to a completely positive map $\Phi_F: A \to P_FB(H)P_F$. 
+
+    By \autoref{proposition:completely-positive-norm}, $\sup_{F \in \mathcal{F}} \norm{\Phi_F}_{L(A; B(H))} \le \norm{\phi}_{L(S; H)}$, so the family $\mathcal{E} = \bracsn{\phi_F|F \in \mathcal{F}}$ is equicontinuous. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, it admits a point-weak* convergent subnet $\anglesn{\Phi_{F_\beta}}_{\beta \in B}$. 
+
+    Let $\Phi = \text{\small BW}\text{-}\!\lim_{\beta \in B}\Phi_{F_\beta}$, then $\Phi$ is also completely positive, with
+    \[
+    \Phi|_{A} = \text{\small BW}\text{-}\!\lim_{\beta \in B}\Phi_{F_\beta}|_{S} = \text{\small BW}\text{-}\!\lim_{\beta \in B}P_F\phi P_F = \phi
+    \]
+\end{proof}
+
+
+
+
+

src/op/system/complete-upgrade.tex

@@ -261,6 +261,9 @@ The following is a slight generalisation of one form of the above example, but s
     \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_k\pi(a)P_k)V = \sum_{k = 1}^{mn}V^*Q_k^*aQ_kV \\
     &= \sum_{k = 1}^{mn} (Q_kV)^* a (Q_kV)
     \end{align*}
-    
 \end{proof}
 
+
+
+
+

src/op/system/completely.tex

@@ -550,3 +550,6 @@
     in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V = V \circ U = UV$. Therefore $A$ is commutative.
 \end{proof}
 
+
+
+

src/op/system/dilation.tex

@@ -184,27 +184,6 @@
 \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 $P = VV^* \in B(K)$ 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}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ 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}
-
-
 
 
 \begin{theorem}[Sz.-Nagy's Dilation Theorem]

src/op/system/index.tex

@@ -1,4 +1,4 @@
-\chapter{Operator Systems}
+\chapter{Operator Systems and Completely Positive Maps}
 \label{chap:opertaor-systems}
 
 \input{./os.tex}
@@ -9,3 +9,7 @@
 \input{./numerical-radius.tex}
 \input{./schur.tex}
 \input{./dilation.tex}
+\input{./os-mn.tex}
+\input{./arveson.tex}
+\input{./aos.tex}
+

src/op/system/module.tex

@@ -4,7 +4,7 @@
 
 \begin{definition}[Module Map]
 \label{definition:module-map}
-    Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. 
+    Let $A, B, C$ be unital $C^*$-algebras with $C \subset A \cap B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. 
     
     Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$. 
 \end{definition}
@@ -48,9 +48,44 @@
     and the given matrix is not positive. 
 \end{proof}
 
+\begin{lemma}
+\label{lemma:positive-multiplicative-domain-estimate}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map with $\norm{\phi(1_A)}_{B(H)} \le 1$, then for any $a, b \in A$, 
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+\end{lemma}
+\begin{proof}
+    Let $(\pi, V, K)$ be a \hyperref[Stinespring representation]{theorem:stinespring-dilation} of $\phi$, then $\norm{V}_{L(H; K)}^2 = \norm{\phi(1_A)}_{B(H)} \le 1$. As such, for any $a, b \in A$, 
+    \begin{align*}
+    \phi(ba) &= V^*\pi(ba)V = V^*\pi(b)I_K\pi(a)V \\
+    \phi(b)\phi(a) &= V^*\pi(b)VV^*\pi(a)V\\
+        \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} &= \normn{V^*\pi(b)(I_K - VV^*)\pi(a)V}_{B(H)} \\
+        &\le \norm{b}_A \normn{(I_K - VV^*)\pi(a)V}_{L(H; K)}
+    \end{align*}
+
+    Let $T = (I_K - VV^*)\pi(a)V$, then $\norm{T}_{L(H; K)}^2 = \norm{T^*T}_{B(H)}$. As $\norm{V}_{L(H; K)}^2 \le 1$, $0 \le I_K - VV^* \le I_K$, so
+    \[
+    T^*T = V^*\pi(a^*)(I_K - VV^*)^2\pi(a)V \le V^*\pi(a^*)(I_K - VV^*)\pi(a)V
+    \]
+
+    and
+
+    \begin{align*}
+        \norm{T}_{L(H; K)}^2 &\le  \norm{V^*\pi(a^*)(I_K - VV^*)\pi(a)V}_{B(H)} \\
+        &= \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}
+    \end{align*}
+    
+    Therefore
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+\end{proof}
+
+
 \begin{definition}[Multiplicative Domains]
 \label{definition:positive-multiplicative-domains}
-    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital\footnote{For without this assumption the domains will not be populated.} $4$-positive map, then
+    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital\footnote{For without this assumption the domains will not be populated.} 4-positive map, then
     \begin{enumerate}
         \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space
         \[
@@ -69,7 +104,7 @@
 
     The spaces $R$ and $L$ are the \textbf{right} and \textbf{left} \textbf{multiplicative domains} of $\phi$, respectively. The $C^*$-subalgebra $C$ is the \textbf{multiplicative domain} of $\phi$. 
 \end{definition}
-\begin{proof}[Proof, {{\cite[Theorem 3.18]{Paulsen}}}. ]
+\begin{proof}[Proof of 4-positive case, using Kadison-Schwarz, {{\cite[Theorem 3.18]{Paulsen}}}. ]
     (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$. 
 
     (1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive}, 
@@ -104,6 +139,105 @@
 
     By \autoref{lemma:matrix-positive-topleft-corner-zero}, $\phi(b)\phi(a) - \phi(ba) = 0$. 
 \end{proof}
+\begin{proof}[Proof of completely positive case, using Stinespring's Dilation Theorem. ]
+    (1, $\Rightarrow$): Suppose that $\phi$ is completely positive, then by \autoref{lemma:positive-multiplicative-domain-estimate}, 
+
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+
+    for all $a, b \in A$. In particular, if $a \in R$, then $\phi(ba) = \phi(b)\phi(a)$. 
+\end{proof}
+
+
+
+
+\begin{proposition}
+\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 $P = VV^* \in B(K)$ 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}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ 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}
+
+\begin{proposition}
+\label{proposition:unital-convergence-multiplicative-domain}
+    Let $A$ be a unital $C^*$-algebra, $K$ be a complex Hilbert space, and $\bracsn{\phi^{(n)}}_{1}^\infty  \subset L(A; B(H))$ be completely positive maps such that $limv{n}\phi^{(n)}(1_A) = I$, then:
+    \begin{enumerate}
+        \item For any $a \in A$, $\limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0$ if and only if $\limv{n}\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a) = 0$ for all $b \in A$. In particular, 
+        \[
+        R = \bracs{a \in A \bigg | \limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0}
+        \]
+
+        is a subalgebra of $A$. 
+        \item For any $a \in A$, $\limv{n}\phi^{(n)}(aa^*) - \phi^{(n)}(a)\phi^{(n)}(a)^* = 0$ if and only if $\limv{n}\phi^{(n)}(ab) - \phi^{(n)}(a)\phi^{(n)}(b) = 0$ for all $b \in A$. In particular, 
+        \[
+        R = \bracs{a \in A \bigg | \limv{n}\phi^{(n)}(aa^*) - \phi^{(n)}(a)\phi^{(n)}(a)^* = 0}
+        \]
+
+        is a subalgebra of $A$. 
+        \item $C = L \cap R$ is a $C^*$-subalgebra of $A$ with $1_C = 1_A$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): It is sufficient to show that if $\limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0$, then $\limv{n}\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a) = 0$ for all $b \in A$. 
+    
+    After rescaling and trimming off an initial segment, assume without loss of generality that $\normn{\phi^{(n)}(1_A)} \le 1$ for all $n \in \natp$. By \autoref{lemma:positive-multiplicative-domain-estimate}, 
+    \[
+    \normn{\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a)}_{B(H)} \le \norm{b}_A \normn{\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a)}_{B(H)}^{1/2}
+    \]
+
+    for all $a, b \in A$. In particular, if $a \in R$, then $\limv{n}{\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a)}_{B(H)} = 0$. 
+    
+    
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:module-map-characterisation}
+    Let $A, B, C$ be $C^*$-algebras with $C \subset A \cap B$ and $1_A = 1_B = 1_C$, and $\phi: A \to B$ be a completely positive map, then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is a $C$-bimodule map. 
+        \item $\phi(c) = c\phi(1_C)$ for all $c \in C$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For any $c \in C$, $\phi(1_C)c = \phi(c^*)^* = \phi(c) = c\phi(1_C)$, so $\phi(1_C)$ commutes with $C$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} and the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, 
+    \[
+    \phi(c) = \phi(1_C)^{1/2} \cdot c \cdot \phi(1_C)^{1/2}
+    \]
+
+    for all $c \in C$. 
+
+
+    Using \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $B \subset B(H)$, and $(\pi, V, K)$ be a Stinespring representation for $\phi$, then for any $c \in C$, 
+    \begin{align*}
+        (\pi(c)V - Vc)^*(\pi(c)V - Vc) &= (V^*\pi(c^*) - c^*V^*)(\pi(c)V - Vc) \\
+        &= V^*\pi(c^*)\pi(c)V - V^*\pi(c^*)Vc \\
+        &- c^*V^*\pi(c)V + c^*V^*Vc \\
+        &= \phi(c^*c) - \phi(c^*)c - c^*\phi(c) + c^*\phi(1_C)c \\
+        &= c^*c\phi(1_C) - c^*\phi(1_C)c + c^*c\phi(1_C) + c^*\phi(1_C)c \\
+        &= 0
+    \end{align*}
+
+    so $\pi(c)V = Vc$. As such, for any $a \in A$, and $c, c' \in C$,
+    \[
+    \phi(cac') = V^*\pi(c)\pi(a)\pi(c')V = cV^*\pi(a)Vc' = c\phi(a)c'
+    \]
+\end{proof}
+
 
 \begin{definition}[Conditional Expectation]
 \label{definition:cstar-conditional-expectation}

src/op/system/os-mn.tex

@@ -0,0 +1,109 @@
+\section{Completely Positive Maps into $M_n(\complex)$}
+\label{section:complete-positive-mn}
+
+
+\begin{lemma}
+\label{lemma:operator-mn-duality-correspondence}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $\phi \in L(S; M_n(\complex))$. Define
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item Let $\xi = \sum_{j = 1}^n e_j \otimes e_j \in \complex^n \otimes \complex^n$, then
+        \[
+        \dpn{a, \tilde \phi}{M_n(S)} = \dpn{\phi_n(a)\xi, \xi}{\complex^n \otimes \complex^n}
+        \]
+
+        for all $a \in M_n(S)$. 
+        \item The mapping $\phi \mapsto \tilde \phi$ is a linear isomorphism between $M_n(S)^*$ and $L(S; M_n(\complex))$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $a \in A$ and $1 \le k, l \le n$, 
+    \begin{align*}
+        \dpn{\phi_n(E_{kl} \otimes a)\xi, \xi}{\complex^n \otimes \complex^n} &= \sum_{i, j = 1}^n \dpn{\phi_n(E_{kl} \otimes a)(e_j \otimes e_j), (e_i \otimes e_i)}{\complex^n \otimes \complex^n} \\
+        &= \sum_{i, j = 1}^n \dpn{E_{kl} e_j, e_i}{\complex^n}\dpn{\phi(a)e_j, e_i}{\complex^n} = \phi(a)_{kl}
+    \end{align*}
+    
+    (2): Let $\phi_0: M_n(S) \to \complex$ be a linear map, and let
+    \[
+    \phi: S \to M_n(\complex) \quad a \mapsto \sum_{i, j = 1}^n \dpn{E_{ij} \otimes a, \phi_0}{M_n(S)} \cdot E_{ij}
+    \]
+
+    then for any $1 \le k, l \le n$ and $a \in A$,
+    \[
+    \dpn{E_{kl} \otimes a, \tilde \phi}{M_n(S)} = \phi(a)_{kl} = \dpn{E_{kl} \otimes a, \phi_0}{M_n(S)}
+    \]
+
+    Therefore the mapping $\phi \to \tilde \phi$ is an isomorphism. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:operator-mn-linear-correspondence}
+    Let $S$ be an operator system, $\phi \in L(S; M_n(\complex))$, and
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is completely positive. 
+        \item $\phi$ is $n$-positive. 
+        \item $\tilde \phi \in M_n(S)^*$ is a positive linear functional. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    See \autoref{proposition:matrix-order-positive-characterisation}. 
+\end{proof}
+
+\begin{theorem}[Krein's Extension Theorem]
+\label{theorem:krein-extension}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $\phi: S \to M_n(\complex)$ be completely positive, then there exists a completely positive map $\Phi: A \to M_n(\complex)$ such that $\Phi|_S = \phi$. 
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then by \autoref{theorem:operator-mn-linear-correspondence}, $\tilde \phi$ is a positive linear functional. As $S$ is an operator system, $\normn{\tilde \phi}_{S^*} = \tilde \phi(1_A)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach} and \autoref{lemma:operator-mn-duality-correspondence}, there exists $\Phi \in L(A; M_n(\complex))$ such that $\tilde \Phi|_{M_n(S)} = \tilde \phi$ and $\Phi|_S = \phi$. By \autoref{theorem:cstar-positive-algebraic}, $\tilde \Phi$ is positive, so $\Phi$ is completely positive by \autoref{theorem:operator-mn-linear-correspondence}. 
+\end{proof}
+
+
+
+\begin{definition}[Non-Entangled Cone]
+\label{definition:non-entangled-positive-cone}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, then
+    \[
+    M_n(\complex)_+ \otimes S_+ := \bracs{\sum_{j = 1}^na_j \otimes b_j \bigg | \seqf{a_j} \subset M_n(\complex)_+, \seqf{b_j} \subset S_+, n \in \natp}
+    \]
+
+    is the \textbf{non-entangled positive cone} of $M_n(S)$. 
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:op-mn-positive}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $\phi \in L(S; M_n(\complex))$, and
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then $\phi$ is positive if and only if $\tilde \phi|_{M_n(\complex)_+ \otimes S_+} \ge 0$. 
+\end{lemma}
+\begin{proof}
+    ($\Rightarrow$): Let $a \in M_n(\complex)_+$ and $b \in S_+$. Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, assume without loss of generality that $a$ is a rank one projection. That is, there exists $\alpha = (\alpha_1, \cdots, \alpha_n) \subset \complex^n$ such that $a = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j E_{ij}$. As such,
+    \[
+    \dpn{a \otimes b, \tilde \phi}{M_n(S)} = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j \phi(b)_{ij} = \dpn{\phi(b)\alpha, \alpha}{\complex^n} \ge 0
+    \]
+
+    ($\Leftarrow$): Let $b \in S_+$, $\alpha = (\alpha_1, \cdots, \alpha_n) \in \complex^n$, and $a = \alpha\alpha^*$, then
+    \[
+        \dpn{\phi(b)\alpha, \alpha}{\complex^n} = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j \phi(b)_{ij} = \dpn{a \otimes b, \tilde \phi}{M_n(S)} \ge 0
+    \]
+    
+\end{proof}
+
+