213 lines
8.7 KiB
TeX
213 lines
8.7 KiB
TeX
\section{Abstract Operator Systems}
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\label{section:abstract-operator-system}
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\begin{definition}[Matrix Order]
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\label{definition:matrix-order}
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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:
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\begin{enumerate}
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\item For each $n \in \natp$, $M_n(S)_{sa}$ is equipped with a vector space ordering.
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\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 $M_m(S)$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Matrix Order Unit]
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\label{definition:matrix-order-unit}
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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 $n \in \natp$.
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\end{definition}
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\begin{definition}[Extended Map]
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\label{definition:abstract-induced-matrix-map}
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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 $M_n(S)$.
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\end{definition}
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\begin{definition}[Abstract Operator System]
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\label{definition:a-operator-system}
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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.
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\end{definition}
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\begin{proposition}
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\label{proposition:matrix-order-linear-functional}
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Let $S$ be a matrix ordered space, then the mapping
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\[
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\text{hom}(S; M_n(\complex)) \to \text{hom}(M_n(S); \complex) \quad \angles{E_{ij} \otimes a, \phi}_{M_n(S)} = \phi(a)_{ij}
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\]
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induces an isomorphism $\text{hom}(M_n(S); \complex) \iso \text{hom}(S; M_n(\complex))$.
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\end{proposition}
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\begin{proposition}
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\label{proposition:matrix-order-positive-characterisation}
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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
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\[
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s: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
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\]
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then the following are equivalent:
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\begin{enumerate}
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\item $s$ is a positive linear functional.
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\item $\phi$ is $n$-positive.
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\item $\phi$ is completely positive.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 13.2]{Paulsen}}}. ]
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(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
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\[
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\dpn{\phi(x)\xi, \eta}{\complex^n} = \sum_{i, j = 1}^n \phi(x)_{ij}\xi_j \ol {\eta_i} = s({\eta^* x\xi})
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\]
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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
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\begin{align*}
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\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) \\
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&= s\braks{\begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}^*A\begin{bmatrix}
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\xi_1 \\
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\vdots \\
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\xi_n
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\end{bmatrix}} \ge 0
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:abstract-operator-system-norm}
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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 $M_{2n}(S)$.}
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\[
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\norm{s}_{M_n(S)} = \inf\bracs{r > 0 \bigg | \begin{bmatrix}
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rI_{M_n(S)} \otimes e &s \\
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s^* & rI_{M_n(S)} \otimes e
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\end{bmatrix} \ge 0}
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\]
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then
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\begin{enumerate}
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\item $\norm{\cdot}_{M_n(S)}$ is a norm.
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\item For each $s \in M_n(S)$, $\norm{s}_{M_n(S)} = \normn{s^*}_{M_n(S)}$.
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\item $\bracsn{s \in M_n(S)|s \ge 0}$ is closed with respect to $\norm{\cdot}_{M_n(S)}$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(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 $r > 0$,
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\[
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\begin{bmatrix}
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re &s \\
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s^* & re
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\end{bmatrix} \ge 0 \quad \forall r > 0
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\]
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As such, for each $\lambda \in \complex$,
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\[
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0 \le \begin{bmatrix}
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1 & \ol \lambda
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\end{bmatrix}\begin{bmatrix}
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re &s \\
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s^* & re
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\end{bmatrix}\begin{bmatrix}
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1 \\
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\lambda
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\end{bmatrix} = r(1 + |\lambda|^2)e + \lambda s + (\lambda s)^*
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\]
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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 $s = 0$.
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(2): For any $s \in S$ and $r > 0$,
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\[
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\begin{bmatrix}
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0 & 1 \\
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1 & 0
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\end{bmatrix}\begin{bmatrix}
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re &s \\
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s^* & re
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\end{bmatrix}\begin{bmatrix}
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0 & 1 \\
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1 & 0
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\end{bmatrix} = \begin{bmatrix}
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re &s^* \\
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s & re
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\end{bmatrix}
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\]
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so
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\[
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\begin{bmatrix}
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re &s \\
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s^* & re
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\end{bmatrix} \ge 0 \quad \Leftrightarrow \quad \begin{bmatrix}
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re &s^* \\
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s & re
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\end{bmatrix} \ge 0
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\]
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(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,
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\begin{align*}
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0 &\le \begin{bmatrix}
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re &s - t \\
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s - t & re
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\end{bmatrix} \\
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0 &\le \begin{bmatrix}
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1 & 1
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\end{bmatrix} \begin{bmatrix}
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re &s - t \\
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s - t & re
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\end{bmatrix}\begin{bmatrix}
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1 \\ 1
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\end{bmatrix} = 2re + 2s - 2t
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\end{align*}
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so $re + s \ge 0$. By the Archimedean property, $s \ge 0$.
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\end{proof}
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\begin{theorem}[Choi-Effros]
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\label{theorem:choi-effros}
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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 $\phi(e) = I_H$.
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\end{theorem}
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\begin{proof}
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For each $n \in \natp$, let
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\[
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K_n = \bracsn{\phi: S \to M_n(\complex)| \phi \text{ is unital, completely positive}}
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\]
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and $K = \bigsqcup_{n \in \natp}K_n$\footnote{The matrix state space.}. Define\footnote{The summand on each $\phi \in K_n$ is $M_n(\complex)$.}
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\[
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\Phi: S \to [l^\infty(K); M_n(\complex)] \quad \Phi(s)_\phi = \phi(s)
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\]
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then $[l^\infty(K); M_n(\complex)] \subset B([l^2(K); \complex^n])$, and $\Phi$ is unital and completely positive.
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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,
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\[
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\sum_{i, j = 1}^n \dpn{\phi(s_{ij})e_j, e_i}{\complex^n} = \tilde \phi(s) < 0
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\]
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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.
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\end{proof}
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\begin{definition}[Maximal Operator System Structure]
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\label{definition:maximal-operator-system}
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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 $S$.
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\end{definition}
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\begin{example}
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\label{example:matrix-os-embedding-not-homomorphism}
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Let
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\[
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\phi: M_n(\complex) \to M_{n+1}(\complex) \quad a \mapsto \begin{bmatrix}
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a & 0 \\
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0 & \frac{\tau(a)}{n}
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\end{bmatrix}
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\]
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then $\phi$ is an embedding of operator systems, but $\phi(M_n(\complex))$ is not a $C^*$-subalgebra.
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\end{example}
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\begin{example}
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\label{example:arveson-nc-function}
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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, $C(\mathbb T) = C^*_{\min}(S)$.
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\end{example}
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