Bokuan Li
+38
-164
@@ -298,58 +298,6 @@
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Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^* \ge 0$ as well.
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\end{proof}
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\begin{proposition}
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\label{proposition:completely-positive-norm}
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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
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\[
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\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
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\]
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\end{proposition}
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\begin{proof}
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After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital.
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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
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\[
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\begin{bmatrix}
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1_{M_n(A)} &T \\
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T^* &1_{M_n(A)}
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\end{bmatrix} \ge 0
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\]
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in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive,
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\[
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\phi_{2n}\begin{bmatrix}
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1_{M_n(A)} &T \\
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T^* &1_{M_n(A)}
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\end{bmatrix} = \begin{bmatrix}
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\phi_n(1_{M_n(A)}) &\phi_n(T) \\
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\phi_n(T)^* & \phi_n(1_{M_n(A)})
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\end{bmatrix} \ge 0
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\]
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First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then
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\begin{align*}
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\phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\
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1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\
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\normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1}
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\end{align*}
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By the \hyperref[Schur complement condition]{lemma:schur-complement-positive},
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\begin{align*}
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\phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\
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&\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\
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\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
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\end{align*}
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and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$.
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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,
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\[
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\norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
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\]
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\end{proof}
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\begin{example}
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\label{example:transpose-not-completely-positive}
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Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive.
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@@ -378,135 +326,61 @@
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is invertible, with negative determinant, so $T_2(C)$ is not positive.
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\end{proof}
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\begin{example}
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\label{example: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{example}
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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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\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{definition}[Module Map]
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\label{definition:module-map}
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Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication.
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Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$.
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\end{definition}
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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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\label{theorem:completely-positive-bimodule}
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Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital completely positive map, 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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\item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space $R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)}$ is a subalgebra of $A$, and $\phi|_R$ is a homomorphism.
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\item For any $a \in A$, $\phi(a)\phi(a)^* = \phi(aa^*)$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space $L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)}$ is a subalgebra of $A$, and $\phi|_L$ is a homomorphism.
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\item $L \cap R$ is a $C^*$-subalgebra of $A$, and $\phi|_{L \cap R}$ is a *-homomorphism.
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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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\begin{proof}
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(1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$.
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(1):
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(1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive},
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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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\phi_2\begin{bmatrix}
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a & b^* \\
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0 & 0
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\end{bmatrix}^*\phi_2\begin{bmatrix}
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a & b^* \\
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0 & 0
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\end{bmatrix} \le \phi_2\begin{bmatrix}
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a^*a & a^*b^* \\
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ba & bb^*
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\end{bmatrix}
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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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\[
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\dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0
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\begin{bmatrix}
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\phi(a^*)\phi(a) & \phi(a^*)\phi(b) \\
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\phi(b)\phi(a) & \phi(b)\phi(b^*)
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\end{bmatrix} \le
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\begin{bmatrix}
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\phi(a^*a) & \phi(a^*b) \\
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\phi(ba) & \phi(bb^*)
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\end{bmatrix}
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\]
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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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and
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\[
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\norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps
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\begin{bmatrix}
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0 & \phi(a^*b^*) - \phi(a^*)\phi(b) \\
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\phi(b)\phi(a) - \phi(ba) & \phi(b)\phi(b^*) - \phi(bb^*)
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\end{bmatrix} \ge 0
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\]
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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{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 $E_{ij}$ be the standard matrix units for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Chi 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): The matrix $E = (E_{ij})$ is positive in $M_n(\complex)$.
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(3) $\Rightarrow$ (1):
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$\phi, \psi \in S(G)$, $K \subset G$ be compact, and $V \in \cn_G(1_G)$ be relatively compact, then
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\begin{align*}
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\end{align*}
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Since the top left entry is zero, $\phi(b)\phi(a) - \phi(ba) = 0$.
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\end{proof}
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