Theorem 41.11.8 (Choi-Effros).label 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 $\phi(e) = I_{H}$.
Proof. For each $n \in \natp$, let
and $K = \bigsqcup_{n \in \natp}K_{n}$[1]. Define[2]
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 Hahn-Banach Theorem, 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 Proposition 41.11.6, the associated map $\phi: S \to M_{n}(\complex)$, defined by $\phi(t)_{ij}= \tilde \phi(E_{ij}\otimes t)$, is completely positive. However,
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.$\square$
- The matrix state space.keyboard_return
- The summand on each $\phi \in K_{n}$ is $M_{n}(\complex)$.keyboard_return
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