Updated notes from course.

src/op/system/aos.tex

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+\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}
+
+
+
+
+