Course related house keeping.

src/op/system/aos.tex

@@ -70,7 +70,143 @@
     \end{align*}
 \end{proof}
 
+\begin{proposition}
+\label{proposition:abstract-operator-system-norm}
+    Let $S$ be an abstract operator system with order unit $e$. For each $s \in M_n(S)$, let\footnote{Infimum exists because $I_{M_{2n}(S)} \otimes e$ is the order unit in $M_{2n}(S)$.}
+    \[
+    \norm{s}_{M_n(S)} = \inf\bracs{r > 0 \bigg | \begin{bmatrix}
+        rI_{M_n(S)} \otimes e &s \\
+        s^* & rI_{M_n(S)} \otimes e
+    \end{bmatrix} \ge 0}
+    \]
 
+    then 
+    \begin{enumerate}
+        \item $\norm{\cdot}_{M_n(S)}$ is a norm. 
+        \item For each $s \in M_n(S)$, $\norm{s}_{M_n(S)} = \normn{s^*}_{M_n(S)}$. 
+        \item $\bracsn{s \in M_n(S)|s \ge 0}$ is closed with respect to $\norm{\cdot}_{M_n(S)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): By replacing $S$ with $M_n(S)$, assume without loss of generality that $n = 1$. Let $s \in S$ with $\norm{s}_{M_n(S)} = 0$, then for each $r > 0$, 
+    \[
+    \begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix} \ge 0 \quad \forall r > 0
+    \]
 
+    As such, for each $\lambda \in \complex$, 
+    \[
+    0 \le \begin{bmatrix}
+        1 & \ol \lambda
+    \end{bmatrix}\begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix}\begin{bmatrix}
+        1 \\
+        \lambda
+    \end{bmatrix} = r(1 + |\lambda|^2)e + \lambda s + (\lambda s)^*
+    \]
 
+    As the above holds for all $r > 0$, $\lambda s + (\lambda s)^* \ge 0$. In particular, $0 \le s + s^* \le 0$ and $0 \le is + (is)^* \le 0$, so $\text{Re}(s) = \text{Im}(s) = 0$, and $s = 0$. 
+
+    (2): For any $s \in S$ and $r > 0$, 
+    \[
+    \begin{bmatrix}
+        0 & 1 \\
+        1 & 0
+    \end{bmatrix}\begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix}\begin{bmatrix}
+        0 & 1 \\
+        1 & 0
+    \end{bmatrix} = \begin{bmatrix}
+        re &s^* \\
+        s & re
+    \end{bmatrix}
+    \]
+
+    so
+    \[
+    \begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix} \ge 0 \quad \Leftrightarrow \quad \begin{bmatrix}
+        re &s^* \\
+        s & re
+    \end{bmatrix} \ge 0
+    \]
+
+    (3): Let $s \in S^+$, where the closure is taken with respect to $\normn{\cdot}_{M_1(S)}$. By (2), $\ol{S^+} \subset \ol S_{sa} = S$. Let $r > 0$, then there exists $t \in S^+$ and $\normn{s - t}_{S} < r$. As such, 
+    \begin{align*}
+        
+    0 &\le \begin{bmatrix}
+        re &s - t \\
+        s - t & re
+    \end{bmatrix} \\
+    0 &\le \begin{bmatrix}
+    1 & 1
+    \end{bmatrix} \begin{bmatrix}
+        re &s - t \\
+        s - t & re
+    \end{bmatrix}\begin{bmatrix}
+    1 \\ 1
+    \end{bmatrix} = 2re + 2s - 2t
+    \end{align*}
+
+    so $re + s \ge 0$. By the Archimedean property, $s \ge 0$. 
+    
+\end{proof}
+
+\begin{theorem}[Choi-Effros]
+\label{theorem:choi-effros}
+    Let $S$ be an abstract operator system with order unit $e$, then there exists a complex Hilbert space $H$, an operator system $S_1 \subset B(H)$, and a complete order isomorphism $\Phi: S \to S_1$ with $\phi(e) = I_H$. 
+\end{theorem}
+\begin{proof}
+    For each $n \in \natp$, let
+    \[
+    K_n = \bracsn{\phi: S \to M_n(\complex)| \phi \text{ is unital, completely positive}}
+    \]
+
+    and $K = \bigsqcup_{n \in \natp}K_n$\footnote{The matrix state space.}. Define\footnote{The summand on each $\phi \in K_n$ is $M_n(\complex)$.}
+    \[
+    \Phi: S \to [l^\infty(K); M_n(\complex)] \quad \Phi(s)_\phi = \phi(s)
+    \]
+
+    then $[l^\infty(K); M_n(\complex)] \subset B([l^2(K); \complex^n])$, and $\Phi$ is unital and completely positive. 
+
+    Let $n \in \natp$ and $s \in M_n(S)$ with $\Phi_n(s) \ge 0$. Suppose for contradiction that $s < 0$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\tilde \phi \in M_n(S)^*$ such that $\tilde \phi (s) \not\ge 0$ but $\tilde \phi(t) \ge 0$ for all $t \in M_n(S)$ with $t \ge 0$. As such, $\tilde \phi$ is a positive linear functional on $M_n(S)^*$. By \autoref{proposition:matrix-order-positive-characterisation}, the associated map $\phi: S \to M_n(\complex)$, defined by $\phi(t)_{ij} = \tilde \phi(E_{ij} \otimes t)$, is completely positive. However, 
+    \[
+    \sum_{i, j = 1}^n \dpn{\phi(s_{ij})e_j, e_i}{\complex^n} = \tilde \phi(s) < 0
+    \]
+
+    so $\phi_n(s) \not\ge 0$ as well. After composing with the range projection of $\phi(e)$, assume without loss of generality that $\phi(e)$ is invertible. In which case, $\phi' = \phi(e)^{-1/2}\phi \phi(e)^{-1/2}$ is unital and completely positive with $\phi_n'(s) \not\ge 0$, contradiction.
+\end{proof}
+
+\begin{definition}[Maximal Operator System Structure]
+\label{definition:maximal-operator-system}
+    Let $S$ be an operator system with order unit $e$. For each $n \in \natp$, $C_n = \ol{M_n(\complex)_+ \otimes S_+} \subset M_n(S)$, then $\seq{C_n}$ defines a matrix order on $S$, and $(S, \seq{C_n}, e)$ is an operator system. The triple $\text{OMAX}(S) = (S, \seq{C_n}, e)$ is the \textbf{maximal} operator system structure on $S$. 
+\end{definition}
+
+\begin{example}
+    \label{example:matrix-os-embedding-not-homomorphism}
+
+    Let
+    \[
+    \phi: M_n(\complex) \to M_{n+1}(\complex) \quad a \mapsto \begin{bmatrix}
+        a & 0 \\
+        0 & \frac{\tau(a)}{n}
+    \end{bmatrix}
+    \]
+
+    then $\phi$ is an embedding of operator systems, but $\phi(M_n(\complex))$ is not a $C^*$-subalgebra. 
+\end{example}
+
+\begin{example}
+    \label{example:arveson-nc-function}
+
+    Let $S = \text{span}\bracsn{1, z, \ol z} \subset C(\mathbb{T})$, then $C^*(S) = C(\mathbb{T})$, but $S$ also embeds into and generates $C(\mathbb D)$. In this particular case, $C(\mathbb T) = C^*_{\min}(S)$. 
+\end{example}