diff --git a/src/op/system/completely.tex b/src/op/system/completely.tex index d9056b2..066bd25 100644 --- a/src/op/system/completely.tex +++ b/src/op/system/completely.tex @@ -180,3 +180,221 @@ is invertible, with negative determinant, so $T_2(C)$ is not positive. \end{proof} + +\begin{example}[Choi] + \label{example:choi-roots-of-unity} + + Let $A$ be a $C^*$-algebra, $\lambda \in \complex$ with $|\lambda| = 1$, and $U_\lambda = \text{diag}(\lambda^1, \cdots, \lambda^n)$, then: + \begin{enumerate} + \item For each $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij} \in M_n(A)$, + \[ + U_\lambda^* X U_{\lambda} = \sum_{i, j = 1}^n E_{ij} \otimes (\lambda^{j - i}x_{ij}) + \] + \item The mapping $\phi: M_n(A) \to M_n(A)$ defined by $\phi(X) = n\text{diag}(X) - X$ is completely positive. + \item The mapping $\psi: M_n(A) \to M_n(A)$ defined by $\psi(X) = (n-1)\text{diag}(X) - X$ is not completely positive. + \end{enumerate} +\end{example} +\begin{proof}[Proof, {{\cite[Exercise 3.6]{Paulsen}}}. ] + 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 + \begin{align*} + U_\lambda^* X U_{\lambda} &= \sum_{i, j, k, l = 1}^n (E_{ii} \otimes \lambda^{-i})(E_{jk} \otimes x_{jk})(E_{ll} \otimes \lambda^l) \\ + &= \sum_{j, k = 1}^n (E_{jj} \cdot E_{jk} \cdot E_{kk}) \otimes (\lambda^{-j} \cdot x_{jk} \cdot \lambda^k) \\ + &= \sum_{j, k = 1}^n E_{jk} \otimes (\lambda^{k-j} x_{jk}) + \end{align*} + + (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 $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij}$, + \[ + \sum_{k = 1}^{n} U_{\omega^k}^*XU_{\omega^k} = \sum_{i, j, k = 1}^n E_{ij} \otimes \omega^{k(j - i)}x_{ij} + \] + + where + \[ + \sum_{k = 1}^n \omega^{k(j-i)} = \begin{cases} + 0 &j \ne i \\ + n &j = i + \end{cases} + \] + + 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 + \end{align*} + + By \autoref{proposition:multiply-completely-bounded}, $\phi$ 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 + \[ + X = \sum_{i, j = 1}^n E_{ij} \otimes x = \braks{\sum_{i = 1}^n E_{1i} \otimes x^{1/2}}^*\braks{\sum_{i = 1}^n E_{1i} \otimes x^{1/2}} \ge 0 + \] + + However, if $V = \sum_{i = 1}^n E_{i1} \otimes 1_A$, then + \begin{align*} + V^*\psi(X)V &= (n-1)V^*\text{diag}(X)V - V^*XV \\ + V^*\text{diag}(X)V &= \sum_{i, j, k = 1}^n (E_{1i} \otimes 1_A)(E_{jj} \otimes x)(E_{k1} \otimes 1_A) \\ + &= \sum_{i = 1}^n (E_{1i} \otimes 1_A)(E_{ii} \otimes x)(E_{i1} \otimes 1_A) = n(E_{11} \otimes x) \\ + V^*XV &= \sum_{i, j, k, l = 1}^n (E_{1i} \otimes 1_A)(E_{jk} \otimes x)(E_{l1} \otimes 1_A) \\ + &= \sum_{i, j = 1}^n (E_{1i} \otimes 1_A)(E_{ij} \otimes x)(E_{j1} \otimes 1_A) = n^2(E_{11} \otimes x) \\ + 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. + +\end{proof} + +\begin{proposition} +\label{proposition:cstar-reverse-positive} + Let $A$ be a $C^*$-algebra, $A^{\text{op}}$ denote $A$ with the same norm and *-operation, but with the product + \[ + A^{\text{op}} \times A^{\text{op}} \to A^{\text{op}} \quad (x, y) \mapsto x \circ y = y \cdot x + \] + + then: + \begin{enumerate} + \item $A^{\text{op}}$ is a $C^*$-algebra, $\text{Id}: A \to A^{\text{op}}$ is positive. + \item The transpose map $M_n(\complex) \to M_n(\complex)^{\text{op}}$ is a *-isomorphism. + \item $\text{Id}: M_2(\complex) \to M_2(\complex)^{\text{op}}$ is not 2-positive. + \item (\textbf{Walter}) For each $U, V \in U(A)$ and $X \in A$, + \[ + \begin{bmatrix} + 1_A & U & X \\ + U^* & 1_A & V \\ + X^* & V^* & 1_A + \end{bmatrix} \ge 0 + \] + + if and only if $X = UV$. + \item $\text{Id}: A \to A^{\text{op}}$ is completely positive if and only if $A$ is commutative. + \end{enumerate} +\end{proposition} +\begin{proof} + (4, $\Leftarrow$): If $X = UV$, then + \begin{align*} + \begin{bmatrix} + 1_A & U & UV \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix}^*\begin{bmatrix} + 1_A & U & UV \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} &= \begin{bmatrix} + 1_A & 0 & 0 \\ + U^* & 0 & 0 \\ + V^*U^* & 0 & 0 + \end{bmatrix}\begin{bmatrix} + 1_A & U & UV \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} \\ + &= \begin{bmatrix} + 1_A & U & UV \\ + U^* & 1_A & U^*UV \\ + V^*U^* & VV^*U^* & _A + \end{bmatrix}\\ + &= \begin{bmatrix} + 1_A & U & UV \\ + U^* & 1_A & U^*UV \\ + V^*U^* & V^*U^*U & A_A + \end{bmatrix}\\ + &= \begin{bmatrix} + 1_A & U & UV \\ + U^* & 1_A & V \\ + V^*U^* & V^* & A_A + \end{bmatrix} + \end{align*} + + so the given matrix is positive. + + (4, $\Rightarrow$): Suppose that the given matrix is positive. By \autoref{lemma:cstar-positive-matrix-decompose-sum}, there exists $\bracsn{a_{k, j}}_{j, k = 1}^3 \subset A$ such that + + \[ + \begin{bmatrix} + 1_A & U & X \\ + U^* & 1_A & V \\ + X^* & V^* & 1_A + \end{bmatrix} = \sum_{k = 1}^3\begin{bmatrix} + a_{k, 1}^*a_{k,1} & a_{k, 1}^*a_{k,2} & a_{k, 1}^*a_{k,3} \\ + a_{k, 2}^*a_{k,1} & a_{k, 2}^*a_{k,2} & a_{k, 2}^*a_{k,3} \\ + a_{k, 3}^*a_{k,1} & a_{k, 3}^*a_{k,2} & a_{k, 3}^*a_{k,3} \\ + \end{bmatrix} + \] + + Using an elementary column operation, + \[ + \begin{bmatrix} + 1_A & U - U & X \\ + U^* & 1_A - U^*U & V \\ + X^* & V^* - X^*U & 1_A + \end{bmatrix} = \sum_{k = 1}^3\begin{bmatrix} + a_{k, 1}^*a_{k,1} & a_{k, 1}^*(a_{k,2} - a_{k, 1}U) & a_{k, 1}^*a_{k,3} \\ + a_{k, 2}^*a_{k,1} & a_{k, 2}^*(a_{k,2} - a_{k, 1}U) & a_{k, 2}^*a_{k,3} \\ + a_{k, 3}^*a_{k,1} & a_{k, 3}^*(a_{k,2} - a_{k, 1}U) & a_{k, 3}^*a_{k,3} \\ + \end{bmatrix} + \] + + and an elementary row operation, + \[ + \begin{bmatrix} + 1_A & 0 & X \\ + 0 & 0 & V - U^*X \\ + X^* & V^* - X^*U & 1_A + \end{bmatrix} + \] + + is equal to + \[ + \sum_{k = 1}^3\begin{bmatrix} + a_{k, 1}^*a_{k,1} & a_{k, 1}^*(a_{k,2} - a_{k, 1}U) & a_{k, 1}^*a_{k,3} \\ + (a_{k, 2} - a_{k, 1}U)^*a_{k,1} & (a_{k, 2} - a_{k, 1}U)^*(a_{k,2} - a_{k, 1}U) & (a_{k, 2} - a_{k, 1}U)^*a_{k,3} \\ + a_{k, 3}^*a_{k,1} & a_{k, 3}^*(a_{k,2} - a_{k, 1}U) & a_{k, 3}^*a_{k,3} \\ + \end{bmatrix} + \] + + In particular, + \[ + \sum_{k = 1}^3 (a_{k, 2} - a_{k, 1}U)^*(a_{k,2} - a_{k, 1}U) = 0 + \] + + so $a_{k, 2} = a_{k, 1}U$ for $1 \le k \le 3$. Thus + \[ + \begin{bmatrix} + 1_A & 0 & X \\ + 0 & 0 & V - U^*X \\ + X^* & V^* - X^*U & 1_A + \end{bmatrix} = \sum_{k = 1}^3\begin{bmatrix} + a_{k, 1}^*a_{k,1} & 0 & a_{k, 1}^*a_{k,3} \\ + 0 & 0 & 0 \\ + a_{k, 3}^*a_{k,1} &0 & a_{k, 3}^*a_{k,3} \\ + \end{bmatrix} + \] + + and $V = U^*X$, so $X = UV$. + + (5): Let $U, V \in U(A)$, then by (4), + \[ + \begin{bmatrix} + 1_A & U & UV \\ + U^* & 1_A & V \\ + V^*U^* & V^* & A_A + \end{bmatrix} \ge 0 + \] + + If the identity is completely positive, then + \[ + \begin{bmatrix} + 1_A & U & VU \\ + U^* & 1_A & V \\ + U^*V^* & V^* & A_A + \end{bmatrix} = \begin{bmatrix} + 1_A & U & U \circ V \\ + U^* & 1_A & V \\ + V^* \circ U^* & V^* & A_A + \end{bmatrix} \ge 0 + \] + + in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V$. Therefore $A$ is commutative. +\end{proof} +