Typo fixes.

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}
+

src/op/system/two.tex

@@ -261,7 +261,7 @@
     Thus $\dpn{P^{-1/2}\xi, P^{-1/2}\xi}{H} \ge \dpn{(P+\eps)^{-1/2}\xi, (P+\eps)^{-1/2}\xi}{H}$ for all $\xi \in P^{1/2}(H)$. In particular, since $A(H) \subset P^{1/2}(H)$, 
     \begin{align*}
     \dpn{P^{-1/2}A\xi, P^{-1/2}A\xi}{H} &\ge \dpn{(P+\eps)^{-1/2}A\xi, (P+\eps)^{-1/2}A\xi}{H} \\
-    \dpn{(P^{-1/2}A)^*(P^{1/2}A) \xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H}
+    \dpn{(P^{-1/2}A)^*(P^{-1/2}A) \xi, \xi}{H} &\ge \dpn{A^*(P+\eps)^{-1}A\xi,\xi}{H}
     \end{align*}
     
     so
@@ -310,7 +310,7 @@
 
     and
     \begin{align*}
-        \dpn{(A + A^*)\xi, \xi}{H} &= 2\text{Re}\dpn{A + A^*\xi, \xi}{H} \\
+        \dpn{(A + A^*)\xi, \xi}{H} &= 2\text{Re}\dpn{(A + A^*)\xi, \xi}{H} \\
         |\dpn{(A + A^*)\xi, \xi}{H}| &\le 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} \\
         \dpn{(P + A + A^* + Q)\xi, \xi}{H} &\le \dpn{P\xi, \xi}{H} + 2 \dpn{P\xi, \xi}{H}^{1/2}\dpn{Q\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H} \\
         &= (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\