Jerry's Digital Garden

BibliographyComments
/Part 7: Fundamentals of Operator Algebras/Chapter 41: Operator Systems and Completely Positive Maps

41.11 Abstract Operator Systems

Definition 41.11.1 (Matrix Order).label Let $S$ be a complex vector space equipped with a complex conjugation $*: S \to S$, then $S$ is matrix ordered if:

  1. (1)

    For each $n \in \natp$, $M_{n}(S)_{sa}$ is equipped with a vector space ordering.

  2. (2)

    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 $M_{m}(S)$.

Definition 41.11.2 (Matrix Order Unit).label Let $S$ be a matrix ordered space and $e \in S$, then $e$ is a matrix order unit if $I_{n} \otimes e \in M_{n}(S)$ is an order unit for all $n \in \natp$.

Definition 41.11.3 (Extended Map).label 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 extended map of $\phi$ to $M_{n}(S)$.

Definition 41.11.4 (Abstract Operator System).label Let $S$ be a matrix ordered space, then $S$ is an abstract operator system if $M_{n}(S)$ is Archimedean for all $n \in \nat$, and $S$ admits a matrix order unit.

Proposition 41.11.5.label 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 $\text{hom}(M_{n}(S); \complex) \iso \text{hom}(S; M_{n}(\complex))$.

Proposition 41.11.6.label 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:

  1. (1)

    $s$ is a positive linear functional.

  2. (2)

    $\phi$ is $n$-positive.

  3. (3)

    $\phi$ is completely positive.

Proof, [Proposition 13.2, Po02]. (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*}

$\square$

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

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

\[K_{n} = \bracsn{\phi: S \to M_n(\complex)| \phi \text{ is unital, completely positive}}\]

and $K = \bigsqcup_{n \in \natp}K_{n}$[2]. Define[3]

\[\Phi: S \to [l^{\infty}(K); M_{n}(\complex)] \quad \Phi(s)_{\phi} = \phi(s)\]

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 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,

\[\sum_{i, j = 1}^{n} \dpn{\phi(s_{ij})e_j, e_i}{\complex^n}= \tilde \phi(s) < 0\]

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.$\square$

Definition 41.11.9 (Maximal Operator System Structure).label 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 maximal operator system structure on $S$.

Example 41.11.10.label 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.

Example 41.11.11.label 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, $C(\mathbb{T}) = C^{*}_{\min}(S)$.

  1. Infimum exists because $I_{M_{2n}(S)}\otimes e$ is the order unit in $M_{2n}(S)$.keyboard_return
  2. The matrix state space.keyboard_return
  3. The summand on each $\phi \in K_{n}$ is $M_{n}(\complex)$.keyboard_return

Post a Comment

Name:Email:
Please enter the tag of the current page (1N0) to post the comment.
Tag:

Direct References

Jerry's Digital Garden

BibliographyComments

Direct References