\section{Sufficient Conditions for Complete Positivity}
\label{section:completely-upgrade}


\begin{proposition}
\label{proposition:homomorphism-completely-positive}
    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. 
\end{proposition}
\begin{proof}
    For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism. 
\end{proof}


\begin{example}
    \label{example:multiply-completely-bounded}

    Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then
    \begin{enumerate}
        \item $\phi$ is completely bounded.
        \item If $y = z^*$, then $\phi$ is completely positive. 
    \end{enumerate}
\end{example}
\begin{proof}
    (1): For each $n \in \natp$ and $T \in M_n(A)$, 
    \[
    \phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z)
    \]
\end{proof}

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

\begin{proposition}
\label{proposition:multiply-completely-bounded}
    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. 
\end{proposition}



\begin{proposition}
\label{proposition:completely-positive-norm}
    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
    \[
    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
    \]
\end{proposition}
\begin{proof}
    After taking a \hyperref[unitisation]{proposition:c-star-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 \autoref{lemma:norm-matrix-test}. 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 \hyperref[Schur complement condition]{lemma:schur-complement-positive}, 
    \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, 
    \[
    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
    \]
\end{proof}

\begin{proposition}
\label{proposition:commutative-mn-facts}
    Let $X$ be a compact Hausdorff space, then:
    \begin{enumerate}
        \item $C(X; M_n(\complex))$ 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. 
        \item The mapping
        \[
        M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f
        \]

        is a *-isomorphism. 
        \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 $x \in X$. 
    \end{enumerate}
\end{proposition}
\begin{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 \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 $f = g^*g$. 
\end{proof}


\begin{theorem}
\label{theorem:bounded-functional-completely-bounded}
    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:
    \begin{enumerate}
        \item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$. 
        \item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive. 
    \end{enumerate}
\end{theorem}
\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ]
    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 $C(X; \complex) = \complex$.
    
    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
    \[
    \dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0
    \]
\end{proof}


\begin{theorem}[Stinespring]
\label{theorem:stinespring-commutative-completely-positive}
    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. 
\end{theorem}
\begin{proof}
    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
    \[
    \norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps
    \]

    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, $\phi_n(P) \ge 0$. 
\end{proof}



\begin{definition}[Matrix Unit]
\label{definition:choi-matrix-unit}
    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 $E_{ij} = e_j \otimes e_i$. 
\end{definition}

\begin{lemma}
\label{lemma:cstar-positive-matrix-decompose-sum}
    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:
    \begin{enumerate}
        \item $P = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})$. 
        \item For each $1 \le k \le n$,
        \[
        \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0
        \]
    \end{enumerate}
\end{lemma}
\begin{proof}
    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, 
    \[
        P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) =  \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})
    \]
    
    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*}
    
\end{proof}


\begin{theorem}[Choi]
\label{theorem:choi-matrix-positive}
    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:
    \begin{enumerate}
        \item $\phi$ is completely positive.
        \item $\phi$ is $n$-positive. 
        \item $C_\phi$ is positive in $M_n(B)$. 
    \end{enumerate}
\end{theorem}
\begin{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 \autoref{lemma:cstar-positive-matrix-decompose-sum}, 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 $S_k = \sum_{l = 1}^n E_{kl} \otimes a_{k, l}^*$. 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 a_{k, s}^*) (E_{ij} \otimes 1_B) (E_{kt} \otimes a_{k, t}^*)^* \\
    &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes a_{k, s}^*) (E_{ij} \otimes 1_B) (E_{tk} \otimes a_{k, t}) \\
    &= \sum_{k = 1}^n E_{kk} \otimes a_{k, i}^* a_{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
    \[
    \phi(T) = \sum_{k = 1}^n S_kTS_k^* \quad \forall T \in M_n(\complex)
    \]

    Therefore $\phi$ is completely positive by \autoref{proposition:multiply-completely-bounded}. 
    
    
\end{proof}

\begin{remark}
\label{remark:choi-positive}
    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.
\end{remark}

\begin{example}[Reduction Map]
\label{example:matric-reduction-map}
    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. 
\end{example}
\begin{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 \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive. 
\end{proof}