267 lines
13 KiB
TeX
267 lines
13 KiB
TeX
\section{Sufficient Conditions for Complete Positivity}
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\label{section:completely-upgrade}
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\begin{proposition}
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\label{proposition:homomorphism-completely-positive}
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Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive.
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\end{proposition}
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\begin{proof}
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For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism.
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\end{proof}
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\begin{example}
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\label{example:multiply-completely-bounded}
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Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then
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\begin{enumerate}
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\item $\phi$ is completely bounded.
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\item If $y = z^*$, then $\phi$ is completely positive.
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\end{enumerate}
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\end{example}
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\begin{proof}
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(1): For each $n \in \natp$ and $T \in M_n(A)$,
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\[
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\phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z)
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\]
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\end{proof}
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The following is a slight generalisation of one form of the above example, but stating it helps greatly.
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\begin{proposition}
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\label{proposition:multiply-completely-bounded}
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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 $\phi(s) = xsx^*$ for all $s \in S$, then $\phi$ is completely positive.
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\end{proposition}
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\begin{proposition}
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\label{proposition:commutative-mn-facts}
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Let $X$ be a compact Hausdorff space, then:
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\begin{enumerate}
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\item $C(X; M_n(\complex))$ equipped with pointwise product, pointwise involution, and the norm
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\[
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\norm{f}_{C(X; M_n(\complex))} = \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}
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\]
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is a $C^*$-algebra.
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\item The mapping
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\[
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M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f
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\]
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is a *-isomorphism.
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\item 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 $x \in X$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): For any $f \in C(X; M_n(\complex))$,
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\[
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\sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^2 = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)} = \norm{f^*f}_{C(X; M_n(\complex))}
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\]
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(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 $x \in X$.
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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 \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $g$ is continuous from $X$ to $M_n(\complex)$. Therefore there exists $g \in C(X; M_n(\complex))$ such that $f = g^*g$.
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\end{proof}
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\begin{theorem}
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\label{theorem:bounded-functional-completely-bounded}
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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:
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\begin{enumerate}
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\item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$.
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\item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive.
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\end{enumerate}
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ]
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By \autoref{proposition:commutative-mn-facts}, 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 $C(X; \complex) = \complex$.
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Let $a = (a_{ij}) \in M_n(S)$. For each $\xi, \eta \in \complex^n$,
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\begin{align*}
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\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
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\end{align*}
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where
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\begin{align*}
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\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i} &= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one)
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\end{align*}
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(1):
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\begin{align*}
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|\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} \\
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\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)} \\
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\norm{\phi_n(a)}_{M_n(\complex)} &\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)} \\
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\norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*}
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\end{align*}
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(2): If $\xi = \eta$, then
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\[
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\sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0
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\]
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so
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\[
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\dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0
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\]
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\end{proof}
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\begin{theorem}[Stinespring]
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\label{theorem:stinespring-commutative-completely-positive}
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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.
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\end{theorem}
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\begin{proof}
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Let $P \in C(X; M_n(\complex))$ with $P \ge 0$ and $\eps > 0$. By \autoref{proposition:uniform-continuous-compact} and \autoref{theorem:lch-paracompact}, 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
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\[
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\norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps
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\]
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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, $\phi_n(P) \ge 0$.
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\end{proof}
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\begin{definition}[Matrix Unit]
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\label{definition:choi-matrix-unit}
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Let $n \in \natp$, $\seqf{e_j}$ be the standard orthonormal basis for $\complex^n$, and $1 \le i, j \le n$, then the \textbf{$(i, j)$-th matrix unit} is the operator $E_{ij} = e_j \otimes e_i$.
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\end{definition}
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\begin{lemma}
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\label{lemma:cstar-positive-matrix-decompose-sum}
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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:
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\begin{enumerate}
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\item $P = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})$.
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\item For each $1 \le k \le n$,
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\[
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\sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0
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\]
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\end{enumerate}
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\end{lemma}
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\begin{proof}[Proof, {{\cite[Lemma 3.13]{Paulsen}}}. ]
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Since $P \ge 0$, there exists $\bracsn{a_{j, k}}_{j, k = 1}^n$ such that
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\begin{align*}
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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}} \\
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&= \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})
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\end{align*}
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For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{jl}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling,
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\[
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P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})
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\]
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Finally, for each $1 \le k \le n$, undoing the product shows that
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\begin{align*}
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\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}) \\
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&= \sum_{i, j = 1}^n (E_{ki} \otimes a_{k, i})^* (E_{kj} \otimes a_{k, j}) \\
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&= \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0
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\end{align*}
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\end{proof}
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\begin{theorem}[Choi]
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\label{theorem:choi-matrix-positive}
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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 $E_{ij}$\footnote{$e_j \otimes e_i \in \complex^n \otimes \complex^n$.} be the standard matrix unit for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{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:
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\begin{enumerate}
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\item $\phi$ is completely positive.
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\item $\phi$ is $n$-positive.
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\item $C_\phi$ is positive in $M_n(B)$.
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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(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.
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(3) $\Rightarrow$ (1): By \autoref{lemma:cstar-positive-matrix-decompose-sum}, there exist $\bracsn{b_{i, j}}_{i, j = 1}^n \subset B$ such that
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\[
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\sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) = C_\phi = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes b_{k, i}^* b_{k, j}
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\]
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So for any $1 \le i, j \le n$,
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\[
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\phi(E_{ij}) = \sum_{k = 1}^n b_{k, i}^* b_{k, j}
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\]
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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
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\begin{align*}
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\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}^*)^* \\
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&= \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}) \\
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&= \sum_{k = 1}^n E_{kk} \otimes b_{k, i}^* b_{k, j} = \sum_{k = 1}^n E_{kk} \otimes \phi(E_{ij})
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\end{align*}
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By linearity, for each $T \in M_n(\complex)$,
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\[
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\sum_{k = 1}^n E_{kk} \otimes \phi(T) = \sum_{k = 1}^n S_k(T \otimes 1_B)S_k^*
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\]
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What the above shows is that, under the identification that
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\[
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M_n(\complex) \iso M_n(\complex) \otimes 1_B \subset M_n(B)
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\]
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and
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\[
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B \iso \bracs{\sum_{k = 1}^n E_{kk} \otimes b \bigg |b \in B} = \bracs{\text{Diag}(b, \cdots, b)|b \in B} \subset M_n(B)
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\]
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The map $\phi$ takes the form
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\[
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\phi(T) = \sum_{k = 1}^n S_kTS_k^* \quad \forall T \in M_n(\complex)
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\]
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Therefore $\phi$ is completely positive by \autoref{proposition:multiply-completely-bounded}.
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\end{proof}
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\begin{remark}
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\label{remark:choi-positive}
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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.
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\end{remark}
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\begin{example}[Reduction Map]
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\label{example:matric-reduction-map}
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Let $\phi: M_n(\complex) \to M_n(\complex)$ be defined by
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\[
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\phi(x) = \text{Tr}(x)1_{M_n(\complex)} - x
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\]
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then $\phi$ is positive, but not completely positive.
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\end{example}
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\begin{proof}
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(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.
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(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
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\begin{align*}
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C_{\phi} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\
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&= \sum_{i, j = 1}^n E_{ij} \otimes (\one_{\bracsn{i = j}} \otimes 1_{M_n(\complex)} - E_{ij}) \\
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&= \sum_{i = 1}^n E_{ii} \otimes 1_{M_n(\complex)} - \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} \\
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&= 1_{M_n(\complex)} \otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^n E_{ij} \otimes E_{ij}
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\end{align*}
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which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive.
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\end{proof}
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\begin{proposition}[Choi-Kraus]
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\label{proposition:choi-kraus}
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Let $\phi: M_n(\complex) \to M_m(\complex)$ be a completely positive map, then there exists $\bracsn{V_j}_1^{mn} \subset L(\complex^m; \complex^n)$ such that
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\[
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\phi(a) = \sum_{j = 1}^{mn}V_j^* a V_j
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\]
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for all $a \in M_n(\complex)$.
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\end{proposition}
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\begin{proof}
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Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\pi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \pi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$.
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As $M_n(\complex)$ is simple, the only (non-zero) irreducible representation of $M_n(\complex)$ is the identity. Since $(K, \pi)$ is a direct sum of irreducible representations, after appending additional factors, assume without loss of generality that $K = \bigoplus_{k = 1}^{nm}\complex^n$ and that $\pi(a) = I_{M_{mn}(\complex)} \otimes a$ for all $a \in M_n(\complex)$.
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For each $1 \le k \le mn$, let $P_k \in B(K)$ be the orthogonal projection onto the subspace corresponding to the $k$-th summand, and $Q_k \in L(K; \complex^n)$ be the orthogonal projection onto the $k$-th summand itself, then
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\begin{align*}
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\phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_k\pi(a)P_k)V = \sum_{k = 1}^{mn}V^*Q_k^*aQ_kV \\
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&= \sum_{k = 1}^{mn} (Q_kV)^* a (Q_kV)
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\end{align*}
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\end{proof}
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