Proposition 41.3.3: Tomiyama | Jerry's Digital Garden

Proposition 41.3.3 (Tomiyama).label Let $A, B$ be unital $C^{*}$-algebras and $S \subset A$ be an operator system, then:

(1)
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.
(2)
For any bounded linear map $\phi: S \to B$, $\norm{\phi_n}_{L(M_n(S); M_n(B))}\le n\norm{\phi}_{L(S; B)}$.

Proof, [Exercise 3.10, Po02]. (1): Using the Gelfand-Naimark Theorem, 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 Cauchy-Schwarz Inequality,

\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,

\[\sum_{i, j = 1}^{n} \norm{a_{ij}}_{A}^{2} = \sum_{i, j = 1}^{n} \norm{E_{ii} \cdot a \cdot E_{jj}}_{M_n(A)}^{2} \le n^{2}\norm{a}_{M_n(A)}^{2}\]

and

\[\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2}\le n\norm{a}_{M_n(A)}\]

The sharp case is given by operators of the form $E = \sum_{i, j = 1}^{n} E_{ij}\otimes E_{ji}$^[1]. As

\begin{align*}\braks{\sum_{i, j = 1}^n E_{ij} \otimes E_{ji}}^{*}\braks{\sum_{i, j = 1}^n E_{ij} \otimes E_{ji}}&= \sum_{i, j, k, l = 1}^{n} E_{ij}E_{kl}\otimes E_{ji}E_{lk}\\&= \sum_{i, j, l = 1}^{n} E_{il}\otimes E_{ji}E_{lj}= \sum_{i, j = 1}^{n} E_{ii}\otimes E_{jj}\\&= I_{M_n(\complex)}\otimes I_{M_n(\complex)}= I_{M_{n \times n}(\complex)}\end{align*}

the operator $E$ is of norm $1$, but $\sum_{i, j = 1}^{n} \norm{E_{ji}}_{M_n(\complex)}^{2} = n^{2}$.

(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*}

$\square$

A minimum ”complexity” of $M_{n}$ is required on $A$ for the sharp case to hold.keyboard_return

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