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/Part 7: Fundamentals of Operator Algebras/Chapter 41: Operator Systems and Completely Positive Maps/Section 41.11: Abstract Operator Systems

Proposition 41.11.7.label Let $S$ be an abstract operator system with order unit $e$. For each $s \in M_{n}(S)$, let[1]

\[\norm{s}_{M_n(S)}= \inf\bracs{r > 0 \bigg | \begin{bmatrix}rI_{M_n(S)}\otimes e&s \\ s^{*}&rI_{M_n(S)}\otimes e\end{bmatrix} \ge 0}\]

then

  1. (1)

    $\norm{\cdot}_{M_n(S)}$ is a norm.

  2. (2)

    For each $s \in M_{n}(S)$, $\norm{s}_{M_n(S)}= \normn{s^*}_{M_n(S)}$.

  3. (3)

    $\bracsn{s \in M_n(S)|s \ge 0}$ is closed with respect to $\norm{\cdot}_{M_n(S)}$.

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, $s \ge 0$.$\square$

  1. Infimum exists because $I_{M_{2n}(S)}\otimes e$ is the order unit in $M_{2n}(S)$.keyboard_return

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