diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index 066bd25..cc7fc05 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -39,6 +39,100 @@ \end{definition} +\begin{proposition}[Tomiyama] +\label{proposition:tomiyama-matrix-norm-bound} + Let $A, B$ be unital $C^*$-algebras and $S \subset A$ be an operator system, then: + \begin{enumerate} + \item For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(A)$, + \[ + \norm{a} \le \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} \le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)} + \] + + where all inequalities are sharp. + \item For any bounded linear map $\phi: S \to B$, $\norm{\phi_n}_{L(M_n(S); M_n(B))} \le n\norm{\phi}_{L(S; B)}$. + \end{enumerate} +\end{proposition} +\begin{proof}[Proof, {{\cite[Exercise 3.10]{Paulsen}}}. ] + (1): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $A$ is a $C^*$-subalgebra of $B(H)$. + + For each $\xi = (\xi_1, \cdots, \xi_n), \eta = (\eta_1, \cdots, \eta_n) \in H^n$, + \begin{align*} + \dpn{a\xi, \eta}{H^n} &= \sum_{i, j = 1}^n \dpn{(E_{ij} \otimes a_{ij})\xi, \eta}{H^n} = \sum_{i, j = 1}^n \dpn{a_{ij}\xi_j, \eta_i}{H} \\ + |\dpn{a\xi, \eta}{H^n}| &\le \sum_{i, j = 1}^n \norm{a_{ij}}_A\norm{\xi_j}_H\norm{\eta_i}_H \\ + &= \sum_{i, j = 1}^n\dpn{\norm{a_{ij}}_AE_{ij}(\norm{\xi_1}_H, \cdots, \norm{\xi_n}_H), (\norm{\eta_1}_H, \cdots, \norm{\eta_n}_H)}{\complex^n} \\ + &\le \norm{\xi}_{H^n} \cdot \norm{\eta}_{H^n} \cdot \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} + \end{align*} + + The sharp case is given by the identity. + + For each $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, by the \hyperref[Cauchy-Schwarz Inequality]{proposition:cauchy-schwarz}, + \begin{align*} + \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n}^2 &= \sum_{i = 1}^n \abs{\sum_{j = 1}^n \norm{a_{ij}}_{A}\xi_j}^2 \\ + &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}\braks{\sum_{j = 1}^n |\xi_j|^2} \\ + \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n} &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}^{1/2} \cdot \norm{\xi}_{\complex^n} + \end{align*} + + The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$. + + Finally, let $\seqf{e_j} \subset \complex^n$ be the standard orthonormal basis for $\complex^n$. Fix $\alpha \in (0, 1)$. For each $1 \le i, j \le n$, let $\xi_{ij} \in H$ such that $\norm{\xi_{ij}}_H = 1$ and $\norm{a_{ij}\xi_{ij}}_H \ge \alpha \norm{a_{ij}}_A$, then + \begin{align*} + \alpha\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2 &\le \sum_{i, j = 1}^n \norm{a_{ij}\xi_{ij}}_H^2 = \sum_{i = 1}^n \norm{a(\xi_{i1}, \cdots, \xi_{in})}_H^2 \\ + &\le \sum_{i = 1}^n \norm{a}_{M_n(A)}^2 \cdot \norm{(\xi_{i1}, \cdots, \xi_{in})}_{H^n}^2 = n^2\norm{a}_{M_n(A)}^2 \\ + \alpha^{1/2}\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} &\le n\norm{a}_{M_n(A)} + \end{align*} + + As the above holds for all $\alpha \in (0, 1)$, + \[ + \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)} + \] + + The sharp case is given by operators of the form $\sum_{i, j = 1}^n E_{ij} \otimes b$ for $b \in A$. + + (2): For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(S)$, + \begin{align*} + \normn{\phi(a)}_{M_n(B)}^2 &\le \sum_{i, j = 1}^n \norm{\phi(a_{ij})}_{B}^2 \le \norm{\phi}_{L(S; B)}^2\sum_{i, j = 1}^n \norm{a_{ij}}_A^2 \\ + &\le n^2\norm{\phi}_{L(S; B)}^2\norm{a}_{M_n(A)}^2 + \end{align*} +\end{proof} + +\begin{proposition} +\label{proposition:positive-column-test} + Let $A$ be a unital $C^*$-algebra and $b = \sum_{i, j = 1}^n E_{ij} \otimes b_{ij} \in M_n(A)$, then the following are equivalent: + \begin{enumerate} + \item $b$ is positive in $M_n(A)$. + \item For each $\seqf{a_j} \subset A$, $\sum_{i, j = 1}^n a_ib_{ij}a_j^*$ is positive in $A$. + \end{enumerate} +\end{proposition} +\begin{proof} + (1) $\Rightarrow$ (2): For any $\seqf{a_j} \subset A$, let $a = \sum_{i = 1}^n E_{1i} \otimes a_i$, then + \begin{align*} + 0 \le aba^* &= \braks{\sum_{i = 1}^n E_{1i} \otimes a_i}\braks{\sum_{j, k = 1}^n E_{jk} \otimes b_{jk}}\braks{\sum_{i = 1}^n E_{i1} \otimes a_i} \\ + &= \sum_{j, k = 1}^n (E_{1j} \cdot E_{jk} \cdot E_{k1}) \otimes (a_jb_{jk}a_k^*) = E_{11} \otimes \sum_{i, j = 1}^n a_ib_{ij}a_j^* + \end{align*} + + so $\sum_{i, j = 1}^n a_ib_{ij}a_j^* \ge 0$ in $A$. + + (2) $\Rightarrow$ (1): For each state $\phi \in S(A)$, let $(H_\phi, \pi_\phi, \xi_\phi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\phi$. Let $H = [l^2(S(A)); H_\phi]$, + \[ + \pi : A \to B(H) \quad \pi(a)(\xi)_\phi = \pi_\phi(a)(\xi_\phi) + \] + + be the universal representation of $A$, and identify $A$ as a $C^*$-subalgebra of $B(H)$ through $\pi$. + + For any $\seqf{a_j} \subset A$ and $\phi \in S(A)$, + \begin{align*} + 0 \le \sum_{i, j = 1}^n \dpn{a_i^*b_{ij}a_j , \phi}{A} &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij}a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} \\ + &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} + \end{align*} + + Identify $H_\phi$ as a subspace of $H$ and let $\xi = (\pi_\phi(a_1)\xi_\phi, \cdots, \pi_\phi(a_n)\xi_\phi) \in H^n$, then + \[ + 0 \le \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} = \dpn{b\xi, \xi}{H^n} + \] + + As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. +\end{proof} + \begin{definition}[Completely Positive Map] \label{definition:completely-positive-map} @@ -115,8 +209,6 @@ \[ \phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B \phi(x^*x) \] - - xxx \end{proposition} \begin{proof} Since @@ -220,7 +312,7 @@ so \begin{align*} \sum_{k = 1}^n U_{\omega^k}^*XU_{\omega^k} &= \sum_{i = 1}^n E_{ii} \otimes (nx_{ii}) = n\text{diag}(X) \\ - \phi(X) = \sum_{k = 1}^{n-1} U_{\omega^k}XU_{\omega^k} &= n\text{diag}(X) - U_1^*XU_1 = n\text{diag}(X) - X + \phi(X) = \sum_{k = 1}^{n-1} U_{\omega^k}^*XU_{\omega^k} &= n\text{diag}(X) - U_1^*XU_1 = n\text{diag}(X) - X \end{align*} By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive. @@ -240,7 +332,7 @@ V^*\psi(X)V &= (n-1)n(E_{11} \otimes x) - n^2(E_{11} \otimes x) = -n(E_{11} \otimes x) \end{align*} - As $x \ne 0$, $-n(E_{ii} \otimes x)$ is not positive. Therefore $\psi(X)$ is not positive, and $\psi$ is not a positive map. + As $x \ne 0$, $-n(E_{11} \otimes x)$ is not positive. Therefore $\psi(X)$ is not positive, and $\psi$ is not a positive map. \end{proof} @@ -292,17 +384,17 @@ &= \begin{bmatrix} 1_A & U & UV \\ U^* & 1_A & U^*UV \\ - V^*U^* & VV^*U^* & _A + V^*U^* & V^*U^*U & 1_A \end{bmatrix}\\ &= \begin{bmatrix} 1_A & U & UV \\ U^* & 1_A & U^*UV \\ - V^*U^* & V^*U^*U & A_A + V^*U^* & V^*U^*U & 1_A \end{bmatrix}\\ &= \begin{bmatrix} 1_A & U & UV \\ U^* & 1_A & V \\ - V^*U^* & V^* & A_A + V^*U^* & V^* & 1_A \end{bmatrix} \end{align*} @@ -378,7 +470,7 @@ \begin{bmatrix} 1_A & U & UV \\ U^* & 1_A & V \\ - V^*U^* & V^* & A_A + V^*U^* & V^* & 1_A \end{bmatrix} \ge 0 \] @@ -387,14 +479,14 @@ \begin{bmatrix} 1_A & U & VU \\ U^* & 1_A & V \\ - U^*V^* & V^* & A_A + U^*V^* & V^* & 1_A \end{bmatrix} = \begin{bmatrix} 1_A & U & U \circ V \\ U^* & 1_A & V \\ - V^* \circ U^* & V^* & A_A + V^* \circ U^* & V^* & 1_A \end{bmatrix} \ge 0 \] - in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V$. Therefore $A$ is commutative. + in $M_3(A^{\text{op}})$. By (4), $UV = U \circ V = VU$. Therefore $A$ is commutative. \end{proof}