Example 41.4.9 (Choi).label Let $A$ be a $C^{*}$-algebra, $\lambda \in \complex$ with $|\lambda| = 1$, and $U_{\lambda} = \text{diag}(\lambda^{1}, \cdots, \lambda^{n})$, then:
- (1)
For each $X = \sum_{i, j = 1}^{n} E_{ij}\otimes x_{ij}\in M_{n}(A)$,
\[U_{\lambda}^{*} X U_{\lambda}= \sum_{i, j = 1}^{n} E_{ij}\otimes (\lambda^{j - i}x_{ij})\] - (2)
The mapping $\phi: M_{n}(A) \to M_{n}(A)$ defined by $\phi(X) = n\text{diag}(X) - X$ is completely positive.
- (3)
The mapping $\psi: M_{n}(A) \to M_{n}(A)$ defined by $\psi(X) = (n-1)\text{diag}(X) - X$ is not completely positive.
Proof, [Exercise 3.6, Po02]. Throughout this proof, if $A$ is not unital, view $A$ as an operator system in its unitisation, and $M_{n}(A)$ as an operator system in $M_{n}$ of its unitisation.
(1): Let $X = \sum_{i, j = 1}^{n} E_{ij}\otimes x_{ij}\in M_{n}(A)$, then
(2): Let $\omega = e^{2\pi i /n}$, then the $n$-th roots of unity are $\bracsn{\omega^k}_{k = 1}^{n}$. By (1), for any $X = \sum_{i, j = 1}^{n} E_{ij}\otimes x_{ij}$,
where
so
By Proposition 41.5.3, $\phi$ is completely positive.
(3): Let $x \in A$ with $x \ge 0$ and $x \ne 0$, and $X = \sum_{i, j = 1}^{n} E_{ij}\otimes x$, then
However, if $V = \sum_{i = 1}^{n} E_{i1}\otimes 1_{A}$, then
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.$\square$
Post a Comment