Various fixes.

src/op/system/complete-upgrade.tex

@@ -31,7 +31,7 @@ The following is a slight generalisation of one form of the above example, but s
 
 \begin{proposition}
 \label{proposition:multiply-completely-bounded}
-    Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = x\phi(s)x^*$ for all $s \in S$, then $\phi$ is completely positive. 
+    Let $A$ be a $C^*$-algebra, $S, T \subset A$ be operator systems, and $\phi: S \to T$. If there exists $x \in A$ such that $\phi(s) = xsx^*$ for all $s \in S$, then $\phi$ is completely positive. 
 \end{proposition}
 
 \begin{proposition}
@@ -144,7 +144,7 @@ The following is a slight generalisation of one form of the above example, but s
     &= \sum_{i, j, k, l = 1}^n (E_{ij} \otimes a_{i, j})^*(E_{kl} \otimes a_{k, l}) = \sum_{i, j, k, l = 1}^n (E_{ji}E_{kl}) \otimes (a_{i, j}^* a_{k, l})
     \end{align*}
     
-    For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{il}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, 
+    For each $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{jl}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, 
     \[
         P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) =  \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})
     \]
@@ -193,7 +193,7 @@ The following is a slight generalisation of one form of the above example, but s
     \sum_{k = 1}^n E_{kk} \otimes \phi(T) = \sum_{k = 1}^n S_k(T \otimes 1_B)S_k^*
     \]
 
-    What the above shows, is that, under the identification that
+    What the above shows is that, under the identification that
     \[
     M_n(\complex) \iso M_n(\complex) \otimes 1_B \subset M_n(B)
     \]
@@ -233,7 +233,7 @@ The following is a slight generalisation of one form of the above example, but s
 
     (Not Completely Positive): For each $1 \le i, j \le n$, $\text{Tr}(E_{ij}) = 1$ if and only if $i = j$. So the Choi matrix is given by
     \begin{align*}
-    C_{ij} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\
+    C_{\phi} &= \phi_n\paren{\sum_{i, j = 1}^n E_{ij} \otimes E_{ij}} = \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) \\
     &= \sum_{i, j = 1}^n E_{ij} \otimes (\one_{\bracsn{i = j}} \otimes 1_{M_n(\complex)} - E_{ij}) \\
     &= \sum_{i = 1}^n E_{ii} \otimes 1_{M_n(\complex)} - \sum_{i, j = 1}^n E_{ij} \otimes E_{ij} \\
     &= 1_{M_n(\complex)} \otimes 1_{M_n(\complex)}- \sum_{i, j = 1}^n E_{ij} \otimes E_{ij}
@@ -246,20 +246,20 @@ The following is a slight generalisation of one form of the above example, but s
 \label{proposition:choi-kraus}
     Let $\phi: M_n(\complex) \to M_m(\complex)$ be a completely positive map, then there exists $\bracsn{V_j}_1^{mn} \subset L(\complex^m; \complex^n)$ such that
     \[
-    \phi(a) = \sum_{j = 1}^{nk}V_j^* a V_j
+    \phi(a) = \sum_{j = 1}^{mn}V_j^* a V_j
     \]
 
     for all $a \in M_n(\complex)$. 
 \end{proposition}
 \begin{proof}
-    Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\phi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \phi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$. 
+    Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\pi(M_n(\complex))V\complex^m$ is dense in $K$, $K = \pi(M_n(\complex))V\complex^m$, and $\dim(K) \le n^2m$. 
 
     As $M_n(\complex)$ is simple, the only (non-zero) irreducible representation of $M_n(\complex)$ is the identity. Since $(K, \pi)$ is a direct sum of irreducible representations, after appending additional factors, assume without loss of generality that $K = \bigoplus_{k = 1}^{nm}\complex^n$ and that $\pi(a) = I_{M_{mn}(\complex)} \otimes a$ for all $a \in M_n(\complex)$.
 
     For each $1 \le k \le mn$, let $P_k \in B(K)$ be the orthogonal projection onto the subspace corresponding to the $k$-th summand, and $Q_k \in L(K; \complex^n)$ be the orthogonal projection onto the $k$-th summand itself, then
     \begin{align*}
-    \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_j\pi(a)P_j)V = \sum_{k = 1}^{mn}V^*Q_j^*aQ_jV \\
+    \phi(a) &= V^*\pi(a)V = \sum_{k = 1}^{mn}V^* (P_k\pi(a)P_k)V = \sum_{k = 1}^{mn}V^*Q_k^*aQ_kV \\
-    &= \sum_{k = 1}^{mn} (Q_jV)^* a (Q_jV)
+    &= \sum_{k = 1}^{mn} (Q_kV)^* a (Q_kV)
     \end{align*}
     
 \end{proof}

src/op/system/completely.tex

@@ -84,7 +84,15 @@
     \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 $\sum_{i, j = 1}^n E_{ij} \otimes b$ for $b \in A$. 
+    The sharp case is given by operators of the form $E = \sum_{i, j = 1}^n E_{ij} \otimes E_{ji}$\footnote{A minimum "complexity" of $M_n$ is required on $A$ for the sharp case to hold.}. 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*}

src/op/system/dilation.tex

@@ -116,7 +116,7 @@
         &\le \sum_{i, j = 1}^n \norm{a^*a}_A \dpn{\phi(b_i^*b_j)\xi_j, \xi_i}{H} = \norm{a}_A^2 \cdot \norm{u + N}_{K_0}^2
     \end{align*}
 
-    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(H)$, and $\pi: A \to B(H)$ is a bounded linear map. 
+    so $\norm{\pi(a)}_{B(K_0)} \le \norm{a}_A$. By the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $\pi(a)$ extends uniquely to an operator in $B(K)$, and $\pi: A \to B(K)$ is a bounded linear map. 
 
     By definition, $\pi$ is a unital homomorphism of Banach algebras. Finally, for any $a \in A$ and $b \otimes \xi, c \otimes \eta \in A \otimes H$, 
     \begin{align*}
@@ -148,7 +148,7 @@
 
     (3): If $\phi$ is unital, then for each $\xi \in H$,
     \[
-    \dpn{V\xi, V\xi}{H} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
+    \dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
     \]
 
     so $V$ is an isometry. 
@@ -174,10 +174,10 @@
 
     By minimality, $\pi_1(A)V_1H$ is dense in $K_1$, and $\pi_2(A)V_2H$ is dense in $K_2$. As such, it is sufficient to show that the mapping $\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j \mapsto \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j$ is a well-defined isometry. To this end, note that
     \begin{align*}
-        \norm{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j}_{K_1}^2 &= \sum_{i, j = 1}^n \dpn{\pi_1(a_j)V_1\xi_j, \pi_1(a_i)V_1\xi_j}{K_1} \\
+        \norm{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j}_{K_1}^2 &= \sum_{i, j = 1}^n \dpn{\pi_1(a_j)V_1\xi_j, \pi_1(a_i)V_1\xi_i}{K_1} \\
-        &= \sum_{i, j = 1}^n \dpn{V_1^*\pi_1(a_i^*a_j)V_1\xi_j, \xi_j}{H} = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_j}{H} \\
+        &= \sum_{i, j = 1}^n \dpn{V_1^*\pi_1(a_i^*a_j)V_1\xi_j, \xi_i}{H} = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} \\
-        &= \sum_{i, j = 1}^n \dpn{V_2^*\pi_2(a_i^*a_j)V_2\xi_j, \xi_j}{H} \\
+        &= \sum_{i, j = 1}^n \dpn{V_2^*\pi_2(a_i^*a_j)V_2\xi_j, \xi_i}{H} \\
-        &= \sum_{i, j = 1}^n \dpn{\pi_2(a_j)V_2\xi_j, \pi_2(a_i)V_2\xi_j}{K_2} = \norm{\sum_{j = 1}^n \pi_2(a_j)V_2\xi_j}_{K_2}^2
+        &= \sum_{i, j = 1}^n \dpn{\pi_2(a_j)V_2\xi_j, \pi_2(a_i)V_2\xi_i}{K_2} = \norm{\sum_{j = 1}^n \pi_2(a_j)V_2\xi_j}_{K_2}^2
     \end{align*}
     
     
@@ -215,7 +215,7 @@
         \item[(U)] For any unitary power dilation $(K', U')$ satisfying (1), there exists a unitary $V: K \to K'$ such that $V|_{H} = I_H$ and $VUV^* = U'$. 
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Constuctive proof without uniqueness, {{\cite[Theorem 1.1]{Paulsen}}}. ]
+\begin{proof}[Constructive proof without uniqueness, {{\cite[Theorem 1.1]{Paulsen}}}. ]
     Using \autoref{lemma:bh-dilation-transitive}, it is sufficient to show the following statements:
     \begin{enumerate}[label=(\roman*)]
         \item For any $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, there exists an isometric power dilation of $T$. 
@@ -299,9 +299,9 @@
     \phi: \complex[z, \ol z] \to B(H) \quad \phi(p + \ol q) = p(T) + q(T)^*
     \]
 
-    extends into a unital positive map $C(\mathbb{T}; \complex) \to B(H)$, which is completely positive by \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}. 
+    extends to a unital positive map $C(\mathbb{T}; \complex) \to B(H)$, which is completely positive by \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}. 
 
-    Let $(\pi, W, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$, 
+    Let $(\pi, V, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$, 
     \[
     T^n = \phi(z^n) = P_H\pi(z^n)|_H = P_HU^n|_H
     \]
@@ -390,7 +390,7 @@
     \begin{align*}
         \dpn{[p(T) + q(T)^*]\xi, \eta}{H} &= \frac{1}{2\pi}\int_0^{2\pi} [p(e^{it}) + \ol{q(e^{it})}]\dpn{P(T; t)\xi, \eta}{H}dt \\
         &= \frac{1}{2\pi}\int_0^{2\pi} [p(e^{it}) + \ol{q(e^{it})}]\dpn{\xi, \eta}{t}dt \\ 
-        |\dpn{[p(T) + q(T)^*]\xi, \eta}{H}| &\le \frac{1}{2\pi}\int_0^{2\pi} M\dpn{\xi, \xi}{t}^{1/2}\dpn{\eta, \eta}t^{1/2}dt\\
+        |\dpn{[p(T) + q(T)^*]\xi, \eta}{H}| &\le \frac{1}{2\pi}\int_0^{2\pi} M\dpn{\xi, \xi}{t}^{1/2}\dpn{\eta, \eta}{t}^{1/2}dt\\
         &\le M\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt}^{1/2}\braks{\frac{1}{2\pi}\int_0^{2\pi}\dpn{\eta, \eta}{t} dt}^{1/2}
     \end{align*}
 
@@ -425,8 +425,8 @@
         f_{\xi, \eta}: B \to \complex \quad (z_1, \cdots, z_n) \mapsto \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n}
         \]
 
-        then $f_{\xi, \eta}$ achieves its maximum modulus on a point $(z_1, \cdots, z_n) \in \complex^n$ such that $|z_j| = 1$ for all $1 \le j \le n$. 
+        then $f_{\xi, \eta}$ achieves its maximum modulus at a point $(z_1, \cdots, z_n) \in \complex^n$ such that $|z_j| = 1$ for all $1 \le j \le n$. 
-        \item For any $(z_1, \cdots, z_n) \in \complex$, if
+        \item For any $(z_1, \cdots, z_n) \in \complex^n$, if
         \[
         \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)} 
         \]

src/op/system/schur.tex

@@ -84,14 +84,14 @@
 
     extends into a completely positive map on $C(\mathbb{T}; \complex)$. 
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 2.6]{Paulsen}}}. ]
+\begin{proof}[Proof, using trigonometric moment sequences, {{\cite[Theorem 2.6]{Paulsen}}}. ]
     Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Using \autoref{proposition:trig-moment-sequence-operator}, it is sufficient to show that the Toeplitz matrix
     \[
     A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix}
-        I & T^* &\cdots & (T^*)^n \\
+        I & T^* &\cdots & (T^*)^{n-1} \\
         T &\ddots &\ddots &\vdots \\
         \vdots &\ddots &\ddots & T^* \\
-        T^n &\cdots &T &I
+        T^{n-1} &\cdots &T &I
     \end{bmatrix}
     \]
 
@@ -106,7 +106,7 @@
     \end{bmatrix} = \sum_{j = 1}^{n - 1}E_{(j+1)j} \otimes T
     \]
 
-    then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^n$. In particular, $R^{n+1} = 0$ and
+    then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^k$. In particular, $R^{n+1} = 0$ and
     \[
     A = I + \sum_{k = 1}^n R^k + \sum_{k = 1}^n (R^*)^k = \sum_{k = 0}^\infty R^k + \sum_{k = 0}^\infty (R^*)^k - I
     \]
@@ -115,9 +115,28 @@
     As $\norm{R}_{M_n(B(H))} = \norm{T}_{B(H)} < 1$, by recognising the \hyperref[Neumann series]{lemma:neumann-series} and using the resolvent identity, 
     \begin{align*}
         A &= (I - R)^{-1} + (I - R^*)^{-1} - I \\
-        &= (I - R)^{-1}(I - R^* + I - R + (I - R)(I - R)^*)(1 - R^*)^{-1} \\
+        &= (I - R)^{-1}(I - R^* + I - R - (I - R)(I - R)^*)(I - R^*)^{-1} \\
         &= (I - R)^{-1}(I - RR^*)(1 - R^*)^{-1} \ge 0
     \end{align*}
     
 \end{proof}
+\begin{proof}[Proof, using von Neumann's inequality. ]
+    Let $p = \sum_{j = 0}^n \lambda_j z^j, q = \sum_{j = 0}^n \mu_j z^j \in \complex[z]$ such that $p + \ol q \ge 0$, then
+    \begin{align*}
+    p + \ol q = \text{Re}(p + \ol q) &= \frac{p + \ol p}{2} + \frac{q + \ol q}{2} \\
+    \sum_{j = 0}^n (\lambda_jz^j + \ol{\mu_j} z^{-j}) &= \frac{1}{2}\sum_{j = 0}^n (\lambda_j + \mu_j)z^j + (\ol \lambda_j + \ol \mu_j)z^{-j} 
+    \end{align*}
+
+    so $\lambda_j = (\lambda_j + \mu_j)/2 = \mu_j$ for all $1 \le j \le n$. After rebalancing $\lambda_0$ and $\mu_0$, assume without loss of generality that $q = p$. 
+
+    Let $c = \normn{p + \ol p}_{C(\mathbb{T}; \complex)}/2$, then $\normn{p + \ol p - c}_{C(\mathbb{T}; \complex)} = \norm{p + \ol p}_{C(\mathbb{T}; \complex)}/2$. By \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial}, 
+    \[
+    \normn{p(T) + p(T)^* - c}_{B(H)} \le \normn{p + \ol p - c}_{C(\mathbb{T}; \complex)} = \norm{p + \ol p}_{C(\mathbb{T}; \complex)}/2 = c
+    \]
+
+    As $p(T) + p(T)^* - cI$ is self-adjoint,
+    \[
+    0 \le p(T) + p(T)^* - c + \norm{p(T) + p(T)^* - cI}_{B(H)} \le p(T) + p(T)^*
+    \]
+\end{proof}