Review of previous parts.

src/op/system/completely.tex

@@ -74,14 +74,12 @@
 
     The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$. 
     
-    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
-    \begin{align*}
-        \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 \\
-        &\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 \\
-        \alpha^{1/2}\braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} &\le n\norm{a}_{M_n(A)}
-    \end{align*}
+    Finally, 
+    \[
+    \sum_{i, j = 1}^n \norm{a_{ij}}_A^2 = \sum_{i, j = 1}^n \norm{E_{ii} \cdot a \cdot E_{jj}}_{M_n(A)}^2 \le n^2\norm{a}_{M_n(A)}^2
+    \]
 
-    As the above holds for all $\alpha \in (0, 1)$, 
+    and
     \[
     \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
     \]
@@ -106,7 +104,7 @@
 \begin{proof}
     (1) $\Rightarrow$ (2): For any $\seqf{a_j} \subset A$, let $a = \sum_{i = 1}^n E_{1i} \otimes a_i$, then
     \begin{align*}
-        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} \\
+        0 \le aba^* &= \braks{\sum_{k = 1}^n E_{k1} \otimes a_k^*}\braks{\sum_{j, k = 1}^n E_{jk} \otimes b_{jk}}\braks{\sum_{i = 1}^n E_{i1} \otimes a_i} \\
         &= \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^*
     \end{align*}
     
@@ -130,7 +128,7 @@
     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}
     \]
 
-    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 $M_n(A)$. 
+    As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi^n$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. 
 \end{proof}
 
 
@@ -390,11 +388,6 @@
             1_A & U & UV \\
             U^* & 1_A & U^*UV \\
             V^*U^* & V^*U^*U & 1_A
-        \end{bmatrix}\\
-        &= \begin{bmatrix}
-            1_A & U & UV \\
-            U^* & 1_A & V \\
-            V^*U^* & V^* & 1_A
         \end{bmatrix}
     \end{align*}
     
@@ -477,16 +470,16 @@
     If the identity is completely positive, then
     \[
     \begin{bmatrix}
-        1_A & U & VU \\
+        1_A & U & V \circ U \\
         U^* & 1_A & V \\
-        U^*V^* & V^* & 1_A
+        U^* \circ V^* & V^* & 1_A
     \end{bmatrix} = \begin{bmatrix}
-        1_A & U & U \circ V \\
+        1_A & U & UV \\
         U^* & 1_A & V \\
-        V^* \circ U^* & V^* & 1_A
+        V^*U^* & V^* & 1_A
     \end{bmatrix} \ge 0
     \]
 
-    in $M_3(A^{\text{op}})$. By (4), $UV = U \circ V = VU$. Therefore $A$ is commutative.
+    in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V = V \circ U = UV$. Therefore $A$ is commutative.
 \end{proof}