41.9 Completely Positive Maps into $M_{n}(\complex)$

Lemma 41.9.1.label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, and $\phi \in L(S; M_{n}(\complex))$. Define

\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]

then:

  1. (1)

    Let $\xi = \sum_{j = 1}^{n} e_{j} \otimes e_{j} \in \complex^{n} \otimes \complex^{n}$, then

    \[\dpn{a, \tilde \phi}{M_n(S)}= \dpn{\phi_n(a)\xi, \xi}{\complex^n \otimes \complex^n}\]

    for all $a \in M_{n}(S)$.

  2. (2)

    The mapping $\phi \mapsto \tilde \phi$ is a linear isomorphism between $M_{n}(S)^{*}$ and $L(S; M_{n}(\complex))$.

Proof. (1): For any $a \in A$ and $1 \le k, l \le n$,

\begin{align*}\dpn{\phi_n(E_{kl} \otimes a)\xi, \xi}{\complex^n \otimes \complex^n}&= \sum_{i, j = 1}^{n} \dpn{\phi_n(E_{kl} \otimes a)(e_j \otimes e_j), (e_i \otimes e_i)}{\complex^n \otimes \complex^n}\\&= \sum_{i, j = 1}^{n} \dpn{E_{kl} e_j, e_i}{\complex^n}\dpn{\phi(a)e_j, e_i}{\complex^n}= \phi(a)_{kl}\end{align*}

(2): Let $\phi_{0}: M_{n}(S) \to \complex$ be a linear map, and let

\[\phi: S \to M_{n}(\complex) \quad a \mapsto \sum_{i, j = 1}^{n} \dpn{E_{ij} \otimes a, \phi_0}{M_n(S)}\cdot E_{ij}\]

then for any $1 \le k, l \le n$ and $a \in A$,

\[\dpn{E_{kl} \otimes a, \tilde \phi}{M_n(S)}= \phi(a)_{kl}= \dpn{E_{kl} \otimes a, \phi_0}{M_n(S)}\]

Therefore the mapping $\phi \to \tilde \phi$ is an isomorphism.$\square$

Theorem 41.9.2.label Let $S$ be an operator system, $\phi \in L(S; M_{n}(\complex))$, and

\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]

then the following are equivalent:

  1. (1)

    $\phi$ is completely positive.

  2. (2)

    $\phi$ is $n$-positive.

  3. (3)

    $\tilde \phi \in M_{n}(S)^{*}$ is a positive linear functional.

Proof. See Proposition 41.11.6.$\square$

Theorem 41.9.3 (Krein’s Extension Theorem).label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, and $\phi: S \to M_{n}(\complex)$ be completely positive, then there exists a completely positive map $\Phi: A \to M_{n}(\complex)$ such that $\Phi|_{S} = \phi$.

Proof. Let

\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]

then by Theorem 41.9.2, $\tilde \phi$ is a positive linear functional. As $S$ is an operator system, $\normn{\tilde \phi}_{S^*}= \tilde \phi(1_{A})$. By the Hahn-Banach Theorem and Lemma 41.9.1, there exists $\Phi \in L(A; M_{n}(\complex))$ such that $\tilde \Phi|_{M_n(S)}= \tilde \phi$ and $\Phi|_{S} = \phi$. By Theorem 39.9.2, $\tilde \Phi$ is positive, so $\Phi$ is completely positive by Theorem 41.9.2.$\square$

Definition 41.9.4 (Non-Entangled Cone).label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, then

\[M_{n}(\complex)_{+} \otimes S_{+} := \bracs{\sum_{j = 1}^na_j \otimes b_j \bigg | \seqf{a_j} \subset M_n(\complex)_+, \seqf{b_j} \subset S_+, n \in \natp}\]

is the non-entangled positive cone of $M_{n}(S)$.

Lemma 41.9.5.label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be an operator system, $\phi \in L(S; M_{n}(\complex))$, and

\[\tilde \phi: M_{n}(S) \to \complex \quad E_{ij}\otimes a \mapsto \phi(a)_{ij}\]

then $\phi$ is positive if and only if $\tilde \phi|_{M_n(\complex)_+ \otimes S_+}\ge 0$.

Proof. ($\Rightarrow$): Let $a \in M_{n}(\complex)_{+}$ and $b \in S_{+}$. Using the Spectral Theorem, assume without loss of generality that $a$ is a rank one projection. That is, there exists $\alpha = (\alpha_{1}, \cdots, \alpha_{n}) \subset \complex^{n}$ such that $a = \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} E_{ij}$. As such,

\[\dpn{a \otimes b, \tilde \phi}{M_n(S)}= \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} \phi(b)_{ij}= \dpn{\phi(b)\alpha, \alpha}{\complex^n}\ge 0\]

($\Leftarrow$): Let $b \in S_{+}$, $\alpha = (\alpha_{1}, \cdots, \alpha_{n}) \in \complex^{n}$, and $a = \alpha\alpha^{*}$, then

\[\dpn{\phi(b)\alpha, \alpha}{\complex^n}= \sum_{i, j = 1}^{n} \ol \alpha_{i} \alpha_{j} \phi(b)_{ij}= \dpn{a \otimes b, \tilde \phi}{M_n(S)}\ge 0\]

$\square$

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