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/Part 7: Operator Algebras/Chapter 41: Operator Systems

41.4 Sufficient Conditions for Complete Positivity

Proposition 41.4.1.label Let $A, B$ be $C^{*}$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive.

Proof. For each $n \in \natp$, the induced map $\phi_{n}: M_{n}(A) \to M_{n}(B)$ is a *-homomorphism.$\square$

Example 41.4.2.label Let $A$ be a $C^{*}$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then

  1. (1)

    $\phi$ is completely bounded.

  2. (2)

    If $y = z^{*}$, then $\phi$ is completely positive.

Proof. (1): For each $n \in \natp$ and $T \in M_{n}(A)$,

\[\phi_{n}(T) = (1_{M_n(\complex)}\otimes y)T(1_{M_n(\complex)}\otimes z)\]

$\square$

The following is a slight generalisation of one form of the above example, but stating it helps greatly.

Proposition 41.4.3.label Let $A$ be a $C^{*}$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = x\phi(s)x^{*}$ for all $s \in S$, then $\phi$ is completely positive.

Proposition 41.4.4.label Let $A, B$ be $C^{*}$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then

\[\norm{\phi}_{\text{cb}}= \norm{\phi}_{L(S; B)}= \norm{\phi(1_A)}_{B}\]

Proof. After taking a unitisation, assume without loss of generality that $B$ is unital.

It is sufficient to show that $\norm{\phi}_{\text{cb}}\le \norm{\phi(1_A)}_{B}$. To this end, let $T \in M_{n}(S)$ with $\norm{T}_{M_n(S)}\le 1$, then

\[\begin{bmatrix}1_{M_n(A)}&T \\ T^{*}&1_{M_n(A)}\end{bmatrix} \ge 0\]

in $M_{2}(M_{n}(S)) = M_{2n}(S)$ by Lemma 41.3.6. Since $\phi$ is completely positive,

\[\phi_{2n}\begin{bmatrix}1_{M_n(A)}&T \\ T^{*}&1_{M_n(A)}\end{bmatrix} = \begin{bmatrix}\phi_{n}(1_{M_n(A)})&\phi_{n}(T) \\ \phi_{n}(T)^{*}&\phi_{n}(1_{M_n(A)})\end{bmatrix} \ge 0\]

First suppose that $\phi_{n}(1_{M_n(A)})$ is invertible, then

\begin{align*}\phi_{n}(1_{M_n(A)})&\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}\cdot 1_{M_n(B)}\\ 1_{M_n(B)}&\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}\cdot \phi_{n}(1_{M_n(A)})^{-1}\\ \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}&\le \phi_{n}(1_{M_n(A)})^{-1}\end{align*}

By the Schur complement condition,

\begin{align*}\phi_{n}(1_{M_n(A)})&\ge \phi_{n}(T)\phi_{n}(1_{M_n(A)})^{-1}\phi_{n}(T)^{*} \\&\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_{n}(T)\phi_{n}(T)^{*} \\ \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{2}&\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)}= \normn{\phi_n(T)}_{M_n(B)}^{2}\end{align*}

and $\normn{\phi_n(T)}_{M_n(B)}\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}}\le \normn{\phi(1_A)}_{B}$.

Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_{B}\psi$ is also a completely positive map. In which case,

\[\norm{\phi}_{\text{cb}}\le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}}\le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_{B} = \norm{\phi(1_A)}_{B}\]

$\square$

Proposition 41.4.5.label Let $X$ be a compact Hausdorff space, then:

  1. (1)

    $C(X; M_{n}(\complex))$ equipped with pointwise product, pointwise involution, and the norm

    \[\norm{f}_{C(X; M_n(\complex))}= \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}\]

    is a $C^{*}$-algebra.

  2. (2)

    The mapping

    \[M_{n}(C(X; \complex)) \to C(X; M_{n}(\complex)) \quad M \otimes f \mapsto M \cdot f\]

    is a *-isomorphism.

  3. (3)

    For any $f \in C(X; M_{n}(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$.

Proof. (1): For any $f \in C(X; M_{n}(\complex))$,

\[\sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^{2} = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)}= \norm{f^*f}_{C(X; M_n(\complex))}\]

(3): If $f \ge 0$, then there exists $g \in C(X; M_{n}(\complex))$ such that $f(x) = g^{*}(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$.

On the other hand, if $f(x) \ge 0$ in $M_{n}(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the continuous functional calculus, $g$ is continuous from $X$ to $M_{n}(\complex)$. Therefore there exists $g \in C(X; M_{n}(\complex))$ such that $f = g^{*}g$.$\square$

Theorem 41.4.6.label Let $A$ be a unital $C^{*}$-algebra, $S \subset A$ be a subspace, $X$ be a compact Hausdorff space, and $\phi \in L(S; C(X; \complex))$, then:

  1. (1)

    $\norm{\phi}_{\text{cb}}= \norm{\phi}_{L(S; C(X; \complex))}$.

  2. (2)

    If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive.

Proof, [Proposition 3.8, Po02]. By Proposition 41.4.5, the norm and order relation on $C(X; M_{n}(\complex)) \iso M_{n}(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$.

Let $a = (a_{ij}) \in M_{n}(S)$. For each $\xi, \eta \in \complex^{n}$,

\begin{align*}\dpn{\phi_n(a)\xi, \eta}{\complex^n}&= \sum_{i, j = 1}^{n} \xi_{j} \ol{\eta_i}\dpn{a_{ij}, \phi}{S}= \angles{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}, \phi}_{S}\end{align*}

where

\begin{align*}\sum_{i, j = 1}^{n} a_{ij}\xi_{j}\ol{\eta_i}&= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one)\end{align*}

(1):

\begin{align*}|\dpn{\phi_n(a)\xi, \eta}{\complex^n}|&\le \norm{\phi}_{S^*}\cdot \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S}\\ \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S}&\le \norm{\xi}_{\complex^n}\norm{\eta}_{\complex^n}\norm{a}_{M_n(S)}\\ \norm{\phi_n(a)}_{M_n(\complex)}&\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)}\\ \norm{\phi}_{\text{cb}}&\le \norm{\phi}_{S^*}\end{align*}

(2): If $\xi = \eta$, then

\[\sum_{i, j = 1}^{n} a_{ij}\eta_{j}\ol{\eta_i}= (\xi \otimes \one)^{*} \cdot a \cdot (\xi \otimes \one) \ge 0\]

so

\[\dpn{\phi_n(a)\xi, \eta}{\complex^n}= \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S}\ge 0\]

$\square$

Theorem 41.4.7 (Stinespring).label Let $X$ be a compact Hausdorff space, $B$ be a $C^{*}$-algebra, $\phi: C(X; \complex) \to B$ be a positive map, then $\phi$ is completely positive.

Proof. Let $P \in C(X; M_{n}(\complex))$ with $P \ge 0$ and $\eps > 0$. By Proposition 6.4.5 and Theorem 5.20.10, there exists a partition of unity $\seqf{\psi_j}\subset C(X; [0, 1])$ and positive matrices $\seqf{p_j}\subset M_{n}(\complex)$ such that

\[\norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)}< \eps\]

for all $x \in X$. For each $1 \le j \le n$, $\phi_{n}(p_{j}\psi_{j}) = p_{j}\phi(\psi_{j}) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_{n}(P) \ge 0$.$\square$

Definition 41.4.8 (Matrix Unit).label Let $n \in \natp$, $\seqf{e_j}$ be the standard orthonormal basis for $\complex^{n}$, and $1 \le i, j \le n$, then the $(i, j)$-th matrix unit is the operator $E_{ij}= e_{j} \otimes e_{i}$.

Lemma 41.4.9.label Let $A$ be a $C^{*}$-algebra and $P \in M_{n}(A)$ be positive, then there exists $\bracsn{a_{k, j}}_{j, k = 1}^{n}$ such that:

  1. (1)

    $P = \sum_{k = 1}^{n} \sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})$.

  2. (2)

    For each $1 \le k \le n$,

    \[\sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^{*}\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}\ge 0\]

Proof, [Lemma 3.13, Po02]. Since $P \ge 0$, there exists $\bracsn{a_{j, k}}_{j, k = 1}^{n}$ such that

\begin{align*}P&= \braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}^{*}\braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}\\&= \sum_{i, j, k, l = 1}^{n} (E_{ij}\otimes a_{i, j})^{*}(E_{kl}\otimes a_{k, l}) = \sum_{i, j, k, l = 1}^{n} (E_{ji}E_{kl}) \otimes (a_{i, j}^{*} a_{k, l})\end{align*}

For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl}= E_{il}$ if $i = k$, and $E_{ji}E_{kl}= 0$ otherwise. Thus after relabeling,

\[P = \sum_{j, k, l = 1}^{n}E_{jl}\otimes (a_{k, j}^{*}a_{k, l}) = \sum_{k = 1}^{n} \sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})\]

Finally, for each $1 \le k \le n$, undoing the product shows that

\begin{align*}\sum_{i, j = 1}^{n} E_{ij}\otimes (a_{k, i}^{*}a_{k, j})&= \sum_{i, j = 1}^{n} (E_{ik}E_{kj}) \otimes (a_{k, i}^{*}a_{k, j}) \\&= \sum_{i, j = 1}^{n} (E_{ki}\otimes a_{k, i})^{*} (E_{kj}\otimes a_{k, j}) \\&= \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^{*}\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}\ge 0\end{align*}

$\square$

Theorem 41.4.10 (Choi).label Let $n \in \natp$, $B$ be a $C^{*}$-algebra, and $\phi: M_{n}(\complex) \to B$ be a linear map. For each $1 \le i, j \le n$, let $E_{ij}$[1] be the standard matrix unit for $M_{n}(\complex)$, then $C_{\phi} = (\phi(E_{ij})) \in M_{n}(B)$ is the Choi matrix of $\phi$. The mapping $\phi \mapsto C_{\phi}$ is an isomorphism from $L(M_{n}(\complex); B)$ to $M_{n}(B)$, and the following are equivalent:

  1. (1)

    $\phi$ is completely positive.

  2. (2)

    $\phi$ is $n$-positive.

  3. (3)

    $C_{\phi}$ is positive in $M_{n}(B)$.

Proof. (2) $\Rightarrow$ (3): For each $1 \le i, j \le n$, $E_{ij}\otimes E_{ij}$ is a positive multiple of a rank one projection, and hence positive. Since $C_{\phi} = \sum_{i, j = 1}^{n} \phi(E_{ij}\otimes E_{ij})$ is a sum of positive elements, $C_{\phi}$ is positive.

(3) $\Rightarrow$ (1): By Lemma 41.4.9, there exist $\bracsn{b_{i, j}}_{i, j = 1}^{n} \subset B$ such that

\[\sum_{i, j = 1}^{n} E_{ij}\otimes \phi(E_{ij}) = C_{\phi} = \sum_{k = 1}^{n} \sum_{i, j = 1}^{n} E_{ij}\otimes b_{k, i}^{*} b_{k, j}\]

So for any $1 \le i, j \le n$,

\[\phi(E_{ij}) = \sum_{k = 1}^{n} b_{k, i}^{*} b_{k, j}\]

For each $1 \le k \le n$, let $S_{k} = \sum_{l = 1}^{n} E_{kl}\otimes b_{k, l}^{*}$. Let $1 \le i, j \le n$, then

\begin{align*}\sum_{k = 1}^{n} S_{k} (E_{ij}\otimes 1_{B}) S_{k}^{*}&= \sum_{k = 1}^{n} \sum_{s, t = 1}^{n} (E_{ks}\otimes b_{k, s}^{*}) (E_{ij}\otimes 1_{B}) (E_{kt}\otimes b_{k, t}^{*})^{*} \\&= \sum_{k = 1}^{n} \sum_{s, t = 1}^{n} (E_{ks}\otimes b_{k, s}^{*}) (E_{ij}\otimes 1_{B}) (E_{tk}\otimes b_{k, t}) \\&= \sum_{k = 1}^{n} E_{kk}\otimes b_{k, i}^{*} b_{k, j}^{*} = \sum_{k = 1}^{n} E_{kk}\otimes \phi(E_{ij})\end{align*}

By linearity, for each $T \in M_{n}(\complex)$,

\[\sum_{k = 1}^{n} E_{kk}\otimes \phi(T) = \sum_{k = 1}^{n} S_{k}(T \otimes 1_{B})S_{k}^{*}\]

What the above shows, is that, under the identification that

\[M_{n}(\complex) \iso M_{n}(\complex) \otimes 1_{B} \subset M_{n}(B)\]

and

\[B \iso \bracs{\sum_{k = 1}^n E_{kk} \otimes b \bigg |b \in B}= \bracs{\text{Diag}(b, \cdots, b)|b \in B}\subset M_{n}(B)\]

The map $\phi$ takes the form

\[\phi(T) = \sum_{k = 1}^{n} S_{k}TS_{k}^{*} \quad \forall T \in M_{n}(\complex)\]

Therefore $\phi$ is completely positive by Proposition 41.4.3.$\square$

Remark 41.4.1.label The Choi matrix yields a correspondence between linear maps and operator-valued matrices, and provides a way to check complete positivity. However, it is significantly harder to verify positivity through the matrix.

Example 41.4.11 (Reduction Map).label Let $\phi: M_{n}(\complex) \to M_{n}(\complex)$ be defined by

\[\phi(x) = \text{Tr}(x)1_{M_n(\complex)}- x\]

then $\phi$ is positive, but not completely positive.

Proof. (Positive): Let $x \in M_{n}(\complex)$ with $x \ge 0$, then $\norm{x}_{M_n(\complex)}$ is the largest eigenvalue of $x$, which is bounded by the trace of $x$. As such, $\text{Tr}(x)1_{M_n(\complex)}\ge x$, and $\phi(x)$ is positive.

(Not Completely Positive): For each $1 \le i, j \le n$, $\text{Tr}(E_{ij}) = 1$ if and only if $i = j$. So the Choi matrix is given by

\begin{align*}C_{ij}&= \phi_{n}\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}}= \sum_{i, j = 1}^{n} E_{ij}\otimes \phi(E_{ij}) \\&= \sum_{i, j = 1}^{n} E_{ij}\otimes (\one_{\bracsn{i = j}}\otimes 1_{M_n(\complex)}- E_{ij}) \\&= \sum_{i = 1}^{n} E_{ii}\otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^{n} E_{ij}\otimes E_{ij}\\&= 1_{M_n(\complex)}\otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^{n} E_{ij}\otimes E_{ij}\end{align*}

which is not positive. By Choi’s Theorem, $\phi$ is not completely positive.$\square$

  1. $e_{j} \otimes e_{i} \in \complex^{n} \otimes \complex^{n}$.keyboard_return

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