Bokuan Li
+103
-11
@@ -39,6 +39,100 @@
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\end{definition}
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\begin{proposition}[Tomiyama]
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\label{proposition:tomiyama-matrix-norm-bound}
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Let $A, B$ be unital $C^*$-algebras and $S \subset A$ be an operator system, then:
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\begin{enumerate}
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\item For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(A)$,
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\[
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\norm{a} \le \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} \le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
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\]
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where all inequalities are sharp.
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\item For any bounded linear map $\phi: S \to B$, $\norm{\phi_n}_{L(M_n(S); M_n(B))} \le n\norm{\phi}_{L(S; B)}$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Exercise 3.10]{Paulsen}}}. ]
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(1): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $A$ is a $C^*$-subalgebra of $B(H)$.
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For each $\xi = (\xi_1, \cdots, \xi_n), \eta = (\eta_1, \cdots, \eta_n) \in H^n$,
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\begin{align*}
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\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} \\
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|\dpn{a\xi, \eta}{H^n}| &\le \sum_{i, j = 1}^n \norm{a_{ij}}_A\norm{\xi_j}_H\norm{\eta_i}_H \\
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&= \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} \\
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&\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)}
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\end{align*}
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The sharp case is given by the identity.
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For each $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, by the \hyperref[Cauchy-Schwarz Inequality]{proposition:cauchy-schwarz},
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\begin{align*}
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\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 \\
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&\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}\braks{\sum_{j = 1}^n |\xi_j|^2} \\
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\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}
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\end{align*}
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The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$.
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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
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\begin{align*}
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\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 \\
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&\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 \\
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\alpha^{1/2}\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} &\le n\norm{a}_{M_n(A)}
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\end{align*}
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As the above holds for all $\alpha \in (0, 1)$,
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\[
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\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
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\]
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The sharp case is given by operators of the form $\sum_{i, j = 1}^n E_{ij} \otimes b$ for $b \in A$.
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(2): For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(S)$,
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\begin{align*}
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\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 \\
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&\le n^2\norm{\phi}_{L(S; B)}^2\norm{a}_{M_n(A)}^2
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\end{align*}
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\end{proof}
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\begin{proposition}
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\label{proposition:positive-column-test}
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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:
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\begin{enumerate}
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\item $b$ is positive in $M_n(A)$.
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\item For each $\seqf{a_j} \subset A$, $\sum_{i, j = 1}^n a_ib_{ij}a_j^*$ is positive in $A$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1) $\Rightarrow$ (2): For any $\seqf{a_j} \subset A$, let $a = \sum_{i = 1}^n E_{1i} \otimes a_i$, then
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\begin{align*}
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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} \\
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&= \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^*
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\end{align*}
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so $\sum_{i, j = 1}^n a_ib_{ij}a_j^* \ge 0$ in $A$.
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(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 $H = [l^2(S(A)); H_\phi]$,
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\[
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\pi : A \to B(H) \quad \pi(a)(\xi)_\phi = \pi_\phi(a)(\xi_\phi)
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\]
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be the universal representation of $A$, and identify $A$ as a $C^*$-subalgebra of $B(H)$ through $\pi$.
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For any $\seqf{a_j} \subset A$ and $\phi \in S(A)$,
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\begin{align*}
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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} \\
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&= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi}
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\end{align*}
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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
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\[
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0 \le \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} = \dpn{b\xi, \xi}{H^n}
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\]
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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 $M_n(A)$.
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\end{proof}
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\begin{definition}[Completely Positive Map]
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\label{definition:completely-positive-map}
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@@ -115,8 +209,6 @@
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\[
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\phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B \phi(x^*x)
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\]
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xxx
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\end{proposition}
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\begin{proof}
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Since
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@@ -220,7 +312,7 @@
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so
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\begin{align*}
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\sum_{k = 1}^n U_{\omega^k}^*XU_{\omega^k} &= \sum_{i = 1}^n E_{ii} \otimes (nx_{ii}) = n\text{diag}(X) \\
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\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
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\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
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\end{align*}
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By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive.
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@@ -240,7 +332,7 @@
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V^*\psi(X)V &= (n-1)n(E_{11} \otimes x) - n^2(E_{11} \otimes x) = -n(E_{11} \otimes x)
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\end{align*}
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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.
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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.
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\end{proof}
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@@ -292,17 +384,17 @@
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&= \begin{bmatrix}
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1_A & U & UV \\
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U^* & 1_A & U^*UV \\
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V^*U^* & VV^*U^* & _A
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V^*U^* & V^*U^*U & 1_A
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\end{bmatrix}\\
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&= \begin{bmatrix}
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1_A & U & UV \\
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U^* & 1_A & U^*UV \\
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V^*U^* & V^*U^*U & A_A
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V^*U^* & V^*U^*U & 1_A
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\end{bmatrix}\\
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&= \begin{bmatrix}
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1_A & U & UV \\
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U^* & 1_A & V \\
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V^*U^* & V^* & A_A
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V^*U^* & V^* & 1_A
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\end{bmatrix}
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\end{align*}
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@@ -378,7 +470,7 @@
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\begin{bmatrix}
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1_A & U & UV \\
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U^* & 1_A & V \\
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V^*U^* & V^* & A_A
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V^*U^* & V^* & 1_A
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\end{bmatrix} \ge 0
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\]
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@@ -387,14 +479,14 @@
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\begin{bmatrix}
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1_A & U & VU \\
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U^* & 1_A & V \\
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U^*V^* & V^* & A_A
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U^*V^* & V^* & 1_A
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\end{bmatrix} = \begin{bmatrix}
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1_A & U & U \circ V \\
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U^* & 1_A & V \\
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V^* \circ U^* & V^* & A_A
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V^* \circ U^* & V^* & 1_A
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\end{bmatrix} \ge 0
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\]
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in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V$. Therefore $A$ is commutative.
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in $M_3(A^{\text{op}})$. By (4), $UV = U \circ V = VU$. Therefore $A$ is commutative.
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\end{proof}
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Reference in New Issue
Block a user