Added the start to operator systems.

src/dyn/top/def.tex

@@ -48,41 +48,6 @@
     Let $x \in G$ be a topologically transitive point and $U, V \subset X$ be non-empty and open. Since $Gx$ is dense in $G$, $Gx \cap U \ne \emptyset$ and $Gx \cap V \ne \emptyset$. After translating $x$, assume without loss of generality that $x \in U$, then $GU \supset Gx \cap V \ne \emptyset$. Therefore there exists $g \in G$ such that $gU \cap V \ne \emptyset$. 
 \end{proof}
 
-\begin{definition}[Periodic]
-\label{definition:tds-periodic}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{periodic} with respect to $G$ if $Gx$ is finite. 
-\end{definition}
-
-\begin{definition}[Return Set]
-\label{definition:tds-return-set}
-    Let $G \curvearrowright X$ be a $G$-flow, $x \in X$, and $\emptyset \ne U \subset X$, then the \textbf{return set} of $x$ to $U$ is
-    \[
-    N(x, U) = \bracsn{g \in G|gx \in U}
-    \]
-\end{definition}
-
-\begin{definition}[Recurrence]
-\label{definition:tds-recurrence}
-    Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
-\end{definition}
-
-\begin{definition}[Uniformly Recurrent]
-\label{definition:tds-uniformly-recurrent}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
-\end{definition}
-
-
-\begin{definition}[Syndetic]
-\label{definition:group-syndetic}
-    Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
-\end{definition}
-
-\begin{lemma}
-\label{lemma:subgroup-syndetic}
-    Let $G$ be a group and $H \subset G$ be a subgroup, then $H$ is syndetic if and only if the index of $H$ in $G$ is finite. 
-\end{lemma}
-% Proof omitted. 
-
 \begin{example}
     \label{example:z-syndetic}
     Let $S \subset \integer$, then $S$ has \textbf{bounded gaps} if
@@ -94,69 +59,6 @@
 \end{example}
 % Proof omitted. 
 
-\begin{definition}[Minimality]
-\label{definition:tds-minimal}
-    Let $G \curvearrowright X$ be a $G$-flow, then the following are equivalent:
-    \begin{enumerate}
-        \item $G \curvearrowright X$ admits no proper subflows.
-        \item Every point in $X$ is topologically transitive. 
-        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
-    \end{enumerate}
-    
-
-    If the above holds, then $G \curvearrowright X$ is \textbf{minimal}.
-\end{definition}
-\begin{proof}
-    (1) $\Rightarrow$ (2): Let $x \in X$, then $\ol{Gx}$ with the restricted action is a $G$-subflow of $X$. As $\emptyset \ne \ol{Gx}$ and $G \curvearrowright X$ admits no proper subflows, $\ol{Gx} = x$. 
-
-    (2) $\Rightarrow$ (1): Let $G \curvearrowright Y$ be a $G$-subflow of $G \curvearrowright X$. For any $x \in Y$, $Y \supset \ol{Gx} = X$, so $Y = X$. 
-
-    (2) $\Rightarrow$ (3): Let $\emptyset \ne U \subset X$ be open. Since $G \curvearrowright X$ is minimal, $Gx \cap U \ne \emptyset$ for all $x \in X$. Thus $X \subset G^{-1}U = GU$. By compactness of $X$, there exists $F \subset G$ finite such that $FU = X$. 
-
-    Let $x \in X$ be arbitrary, then for each $g \in G$, $gx \in X = FU$. Thus there exists $h \in F$ with $gx \in hU$ and $h^{-1}gx \in U$. Hence $h^{-1}g \in N(x, U)$, and $g \in hN(x, U) \subset FN(x, U)$.  
-\end{proof}
-
-\begin{lemma}
-\label{lemma:tds-minimal-recurrent}
-    Let $G$ be an infinite group and $G \curvearrowright X$ be minimal $G$-flow, then every point in $X$ is recurrent. 
-\end{lemma}
-\begin{proof}
-    Let $x \in X$ and $U \in \cn_X(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$. 
-\end{proof}
-
-
-\begin{definition}[Minimal Point]
-\label{definition:tds-minimal-point}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
-    \begin{enumerate}
-        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
-        \item $x$ is uniformly recurrent. 
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
-
-    (2) $\Rightarrow$ (1): Let $Y = \ol{Gx}$ and $y \in Y$. For any $U \in \cn_X(x)$, since $X$ is a compact Hausdorff space, there exists $V \in \cn_X(x)$ with $\ol{V} \subset U$. As $N(x, V)$ is syndetic in $G$, there exists $F \subset G$ finite such that $G = F^{-1}N(x, V)$. 
-
-    Since $\ol{Gx} = Y$, there exists an ultrafilter $\fF \subset 2^G$ such that $\fF x \to y$. As $G = \bigcup_{f \in F}N(x, V)$, there exists $f \in F$ such that $f^{-1}N(x, V) \in \fF$. Thus $y = \lim \fF x \subset \ol{f^{-1}N(x, V)} \subset \ol{f^{-1}V}$. Equivalently, $fy \in \ol{V}$, and $Gy \cap U \supset Gy \cap f^{-1}\ol{V} \ne \emptyset$. As this holds for all $U \in \cn_X(x)$, $x \in \ol{Gy}$, and $\ol{Gy} = \ol{Gx} = Y$. Therefore every point in $Y$ is transitive, and $G \curvearrowright Y$ is a minimal subflow. 
-\end{proof}
-
-\begin{theorem}[Existence of Minimal Subflows]
-\label{theorem:tds-minimal-existence}
-    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
-\end{theorem}
-% Proof omitted. 
-
-\begin{corollary}
-\label{corollary:tds-recurrent-existence}
-    Let $G$ be an infinite group and $G \curvearrowright X$ be a $G$-flow, then $X$ admits a recurrent point. 
-\end{corollary}
-\begin{proof}
-    By \autoref{theorem:tds-minimal-existence}, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, \autoref{lemma:tds-minimal-recurrent} implies that every point in $Y$ is recurrent. 
-\end{proof}
-
-
-
 \begin{definition}[Bernoulli Shift]
 \label{definition:bernoulli-shift}
     Let $G$ be a group, $X = 2^G$ equipped with the product topology, and

src/dyn/top/index.tex

@@ -2,3 +2,5 @@
 \label{chap:topological-dynamics}
 
 \input{./def.tex}
+\input{./minimal.tex}
+\input{./recurrence.tex}

src/dyn/top/minimal.tex

@@ -0,0 +1,64 @@
+\section{Minimal Flows}
+\label{section:minimal-flow}
+
+\begin{definition}[Minimality]
+\label{definition:tds-minimal}
+    Let $G \curvearrowright X$ be a $G$-flow, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright X$ admits no proper subflows.
+        \item Every point in $X$ is topologically transitive. 
+        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
+    \end{enumerate}
+    
+
+    If the above holds, then $G \curvearrowright X$ is \textbf{minimal}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $x \in X$, then $\ol{Gx}$ with the restricted action is a $G$-subflow of $X$. As $\emptyset \ne \ol{Gx}$ and $G \curvearrowright X$ admits no proper subflows, $\ol{Gx} = x$. 
+
+    (2) $\Rightarrow$ (1): Let $G \curvearrowright Y$ be a $G$-subflow of $G \curvearrowright X$. For any $x \in Y$, $Y \supset \ol{Gx} = X$, so $Y = X$. 
+
+    (2) $\Rightarrow$ (3): Let $\emptyset \ne U \subset X$ be open. Since $G \curvearrowright X$ is minimal, $Gx \cap U \ne \emptyset$ for all $x \in X$. Thus $X \subset G^{-1}U = GU$. By compactness of $X$, there exists $F \subset G$ finite such that $FU = X$. 
+
+    Let $x \in X$ be arbitrary, then for each $g \in G$, $gx \in X = FU$. Thus there exists $h \in F$ with $gx \in hU$ and $h^{-1}gx \in U$. Hence $h^{-1}g \in N(x, U)$, and $g \in hN(x, U) \subset FN(x, U)$.  
+\end{proof}
+
+\begin{lemma}
+\label{lemma:tds-minimal-recurrent}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be minimal $G$-flow, then every point in $X$ is recurrent. 
+\end{lemma}
+\begin{proof}
+    Let $x \in X$ and $U \in \cn_X(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$. 
+\end{proof}
+
+
+\begin{definition}[Minimal Point]
+\label{definition:tds-minimal-point}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
+        \item $x$ is uniformly recurrent. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
+
+    (2) $\Rightarrow$ (1): Let $Y = \ol{Gx}$ and $y \in Y$. For any $U \in \cn_X(x)$, since $X$ is a compact Hausdorff space, there exists $V \in \cn_X(x)$ with $\ol{V} \subset U$. As $N(x, V)$ is syndetic in $G$, there exists $F \subset G$ finite such that $G = F^{-1}N(x, V)$. 
+
+    Since $\ol{Gx} = Y$, there exists an ultrafilter $\fF \subset 2^G$ such that $\fF x \to y$. As $G = \bigcup_{f \in F}N(x, V)$, there exists $f \in F$ such that $f^{-1}N(x, V) \in \fF$. Thus $y = \lim \fF x \subset \ol{f^{-1}N(x, V)} \subset \ol{f^{-1}V}$. Equivalently, $fy \in \ol{V}$, and $Gy \cap U \supset Gy \cap f^{-1}\ol{V} \ne \emptyset$. As this holds for all $U \in \cn_X(x)$, $x \in \ol{Gy}$, and $\ol{Gy} = \ol{Gx} = Y$. Therefore every point in $Y$ is transitive, and $G \curvearrowright Y$ is a minimal subflow. 
+\end{proof}
+
+\begin{theorem}[Existence of Minimal Subflows]
+\label{theorem:tds-minimal-existence}
+    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
+\end{theorem}
+% Proof omitted. 
+
+\begin{corollary}
+\label{corollary:tds-recurrent-existence}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be a $G$-flow, then $X$ admits a recurrent point. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:tds-minimal-existence}, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, \autoref{lemma:tds-minimal-recurrent} implies that every point in $Y$ is recurrent. 
+\end{proof}
+

src/dyn/top/recurrence.tex

@@ -0,0 +1,36 @@
+\section{Topological Recurrence}
+\label{section:tds-recurrence}
+
+\begin{definition}[Periodic]
+\label{definition:tds-periodic}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{periodic} with respect to $G$ if $Gx$ is finite. 
+\end{definition}
+
+\begin{definition}[Return Set]
+\label{definition:tds-return-set}
+    Let $G \curvearrowright X$ be a $G$-flow, $x \in X$, and $\emptyset \ne U \subset X$, then the \textbf{return set} of $x$ to $U$ is
+    \[
+    N(x, U) = \bracsn{g \in G|gx \in U}
+    \]
+\end{definition}
+
+\begin{definition}[Recurrence]
+\label{definition:tds-recurrence}
+    Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
+\end{definition}
+
+\begin{definition}[Syndetic]
+\label{definition:group-syndetic}
+    Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:subgroup-syndetic}
+    Let $G$ be a group and $H \subset G$ be a subgroup, then $H$ is syndetic if and only if the index of $H$ in $G$ is finite. 
+\end{lemma}
+% Proof omitted. 
+
+\begin{definition}[Uniformly Recurrent]
+\label{definition:tds-uniformly-recurrent}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
+\end{definition}

src/measure/measure/lebesgue-stieltjes.tex

@@ -46,7 +46,17 @@
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
     \end{enumerate}
 
-    Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
+    Conversely, if $F: \real \to \real$ is a Stieltjes function, then 
+    \begin{enumerate}
+        \item[(1C)] There exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ such that for all $-\infty < a < b < \infty$, 
+        \[
+        \mu_F((a, b]) = F(b) - F(a)
+        \]
+    \end{enumerate}
+    
+    In which case, $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
+    
+    
 \end{definition}
 \begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
     (1): Let
@@ -71,7 +81,7 @@
     (U): If $G: \real \to \real$ satisfies (1), then $F(x) = G(x) - G(0)$. Hence $F - G$ is constant.
 
 
-    Converse: Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
+    (1C): Let $\ci = \bracs{(a, b]| -\infty < a \le b < \infty}$ be the elementary family of h-intervals and
     \[
     \alg = \bracs{\bigsqcup_{j = 1}^n (a_j, b_j] \bigg | \seqf{(a_j, b_j]} \subset \ci, n \in \natp}
     \]

src/op/c-star/involution.tex

@@ -42,19 +42,7 @@
 \end{proof}
 
 
-\begin{definition}[$M_n(A)$]
-\label{definition:matrix-cstar}
-    Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, equipped with the product
-    \[
-    M_n(A) \times M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab
-    \]
-    and the involution
-    \[
-    *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^*
-    \]
 
-    Let $H$ be a complex Hilbert space and $\pi: A \to B(H)$ be a faithful representation of $A$, then $\pi$ induces a faithful representation $M_n(A) \otimes A \to B(H^n)$. Under the identification that $M_n(A) \subset B(H^n)$, the inherited norm\footnote{Which is equal to the projective norm because $M_n(\complex)$ is finite-dimensional and hence nuclear.} makes $M_n(A)$ a $C^*$-algebra. 
-\end{definition}
 
 \begin{definition}[Completely]
 \label{definition:cstar-completely}

src/op/example/bounded.tex

@@ -95,137 +95,3 @@
 
     To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^\perp$, $SV|_{\ker(P)} = 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)} = 0$ as well. Therefore $V \in A'' = A$. 
 \end{proof}
-
-
-\begin{definition}[Unitary Dilation]
-\label{definition:unitary-dilation}
-    Let $H$ be a complex Hilbert space, $V \in B(H)$ be isometric, $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and
-    \[
-    U = \begin{bmatrix}
-        V & P \\
-        0 & V^*
-    \end{bmatrix}
-    \]
-
-    then:
-    \begin{enumerate}
-        \item $U^*U = UU^* = I_{H^2}$. In other words, $U$ is an isometry on $H^2$. 
-        \item Under the identification that $H = H \times \bracsn{0} \subset H^2$, let $Q \in B(H^2)$ be the projection onto $H$, then $V^n = QU^n|_H$ for all $n \ge 0$. 
-    \end{enumerate}
-    
-    The operator $U \in B(H^2)$ is a \textbf{unitary dilation} of $V$. 
-\end{definition}
-\begin{proof}
-    (1): By direct computation, 
-    \begin{align*}
-        UU^* &= \begin{bmatrix}
-            V & P \\
-            0 & V^*
-        \end{bmatrix} \cdot 
-        \begin{bmatrix}
-            V^* & 0 \\
-            P & V
-        \end{bmatrix} = \begin{bmatrix}
-            VV^* + P & PV \\
-            V^*P & V^*V
-        \end{bmatrix} = I_{H^2} \\
-        U^*U &= \begin{bmatrix}
-            V^* & 0 \\
-            P & V
-        \end{bmatrix} \cdot 
-        \begin{bmatrix}
-            V & P \\
-            0 & V^*
-        \end{bmatrix}
-         = \begin{bmatrix}
-            V^*V & V^*P \\
-            PV & VV^* + P
-        \end{bmatrix} = I_{H^2} \\
-    \end{align*}
-    
-
-    (2): For each $n \in \natp$, 
-    \[
-    U^n = \begin{bmatrix}
-            V & P \\
-            0 & V^*
-        \end{bmatrix}^n = \begin{bmatrix}
-            V^n & \cdots \\
-            0 & (V^*)^n
-        \end{bmatrix}
-    \]
-
-    so $QU^n|_{H} = V^n$.
-\end{proof}
-
-\begin{definition}[Compression]
-\label{definition:bh-compression}
-    Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. 
-\end{definition}
-
-\begin{definition}[Dilation]
-\label{definition:bh-dilation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a Hilbert space $K \supset H$ and an operator $S \in B(K)$ such that $T = PS|_H$, where $P \in B(K)$ is the orthogonal projection onto $H$. If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. 
-\end{definition}
-
-
-
-\begin{definition}[Isometric Dilation]
-\label{definition:isometric-dilation}
-    Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, and
-    \[
-    V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases}
-        Tx_1 &n = 1 \\
-        D_Tx_1 &n = 2 \\
-        x_{n-1} &n \ge 3
-    \end{cases}
-    \]
-
-    then:
-    \begin{enumerate}
-        \item $V$ is isometric on $l^2(\natp; H)$. 
-        \item Under the identification that $H = H \oplus 0 \oplus \cdots$, let $Q$ be the orthogonal projection onto $H$, then $T^n = QV^n|_H$ for all $n \ge 0$. 
-    \end{enumerate}
-
-    The operator $V$ is an \textbf{isometric dilation} of $T$. 
-\end{definition}
-\begin{proof}
-    (1): Since $\norm{T}_H \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. For each $x \in H$, 
-    \begin{align*}
-        \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\
-        &= \dpn{x, x}{H} = \norm{x}_H^2
-    \end{align*}
-
-    Thus for each $x = \seq{x_n} \in l^2(H)$, 
-    \begin{align*}
-        \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\
-        &= \norm{x}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
-    \end{align*}
-\end{proof}
-
-\begin{theorem}[Sz.-Nagy's Dilation Theorem]
-\label{theorem:sznagy-dilation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
-\end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
-    Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, 
-    \begin{align*}
-        V^n &= P_{l^2(\natp; H)}U^n|_{l^2(\natp; H)} \\
-        T^n = P_{H}V^n|_{H} &= P_HU^n|_{H}
-    \end{align*}
-\end{proof}
-
-\begin{corollary}[Von Neumann's Inequality]
-\label{corollary:von-neumann-inequality-polynomial}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, 
-    \[
-    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
-    \]
-\end{corollary}
-\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ]
-    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
-
-    Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
-\end{proof}
-
-

src/op/index.tex

@@ -5,4 +5,5 @@
 \input{./c-star/index.tex}
 \input{./vn/index.tex}
 \input{./example/index.tex}
+\input{./system/index.tex}
 \input{./notation.tex}

src/op/system/completely.tex

@@ -0,0 +1,514 @@
+\section{Completely Positive and Completely Bounded Maps}
+\label{section:completely-positive-bounded}
+
+\begin{proposition}
+\label{proposition:matrix-cstar}
+    Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, 
+    \[
+    M_n(A) \times M_n(A) \to M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab
+    \]
+    and the involution
+
+    \[
+    *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^*
+    \]
+
+    then there exists a unique norm on $M_n(A)$ such that $M_n(A)$ is a $C^*$-algebra.
+\end{proposition}
+\begin{proof}
+    After applying a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $A$ is unital. 
+    
+    By the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, there exists a faithful representation $(H, \pi)$ of $A$. Under the identification that $H^n = \complex^n \otimes H$, $\pi$ induces a faithful unital *-representation
+    \[
+    \pi_n: M_n(A) \to B(H^n) \quad \pi_n(M \otimes a)(\lambda \otimes \xi) = (M\lambda) \otimes (\pi(a)\xi)
+    \]
+
+    whose image is a uniformly closed subalgebra of $B(H^n)$. As such, $B(H^n)$ induces a norm on $M_n(A)$, making it a $C^*$-algebra. 
+
+    By \autoref{corollary:c-star-unique-norm}, this norm is unique.
+\end{proof}
+
+\begin{definition}[Extended Map]
+\label{definition:matrix-extended-map}
+    Let $A$ be a unital $C^*$-algebra, $B$ be a $C^*$-algebra, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then the mapping
+    \[
+    \phi_n = \text{Id}_{M_n(\complex)} \otimes \phi: M_n(S) \to M_n(B)
+    \]
+
+    is the \textbf{extended map}/\textbf{amplification} of $\phi$ to $M_n(S)$. 
+\end{definition}
+
+
+
+\begin{definition}[Completely Positive Map]
+\label{definition:completely-positive-map}
+    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$, then $\phi$ is \textbf{$n$-positive} if $\phi_n: M_n(S) \to M_n(B)$ is positive. The map $\phi$ is \textbf{completely positive} if $\phi$ is $n$-positive for all $n \in \natp$. 
+\end{definition}
+
+\begin{definition}[Completely Bounded Map]
+\label{definition:completely-bounded-map}
+    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a linear map, then
+    \[
+    \norm{\phi}_{\text{cb}} = \sup_{n \in \natp}\norm{\phi_n}_{L(M_n(S); M_n(B))}
+    \]
+
+    is the \textbf{completely bounded norm} of $\phi$. 
+    
+    The map $\phi$ is \textbf{completely bounded} if $\norm{\phi}_{\text{cb}} < \infty$, and \textbf{completely contractive} if $\norm{\phi}_{\text{cb}} \le 1$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:completely-mental-gymnastics}
+    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, $\phi: S \to B$ be a linear map, and $n \in \natp$. 
+    \begin{enumerate}
+        \item If $\phi$ is $n$-positive, then $\phi$ is $k$-positive for all $1 \le k \le n$. 
+        \item For each $1 \le k \le n$, $\norm{\phi_k}_{L(M_k(S); M_k(B))} \le \norm{\phi_n}_{L(M_n(S); M_n(B))}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    For any $n \in \natp$ and $1 \le k < n$, there is a natural *-homomorphism
+    \[
+    e: M_k(\complex) \to M_n(\complex) \quad M \mapsto \begin{bmatrix}
+        M & 0 \\
+        0 & 0
+    \end{bmatrix}
+    \]
+
+    such that the following diagram commutes:
+    \[
+    \xymatrix{
+    M_k(S) \ar@{->}[d]_{e \otimes \text{Id}_S} \ar@{->}[r]^{\phi_k} & M_k(B) \ar@{->}[d]^{e \otimes \text{Id}_B} \\
+    M_n(S) \ar@{->}[r]_{\phi_n} & M_n(B)
+    }
+    \]
+
+    (1): Since $e \otimes \text{Id}_S$ is the restriction of a *-homomorphism, it is positive. On the other hand, for any $x \in M_k(B)$, $x \ge 0$ if and only if $(e \otimes \text{Id}_B)(x) \ge 0$. Given that $\phi_n$ is positive, $\phi_k$ must also be positive. 
+
+    (2): The maps $e \otimes \text{Id}_S$ and $e \otimes \text{Id}_B$ are both isometric. As such, the norm of $\phi_k$ is bounded above by the norm of $\phi_n$. 
+\end{proof}
+
+
+% Omitted
+
+\begin{lemma}
+\label{lemma:norm-matrix-test}
+    Let $A$ be a unital $C^*$-algebra, then for any $x \in A$ and $y \in A_{sa}$, $x^*x \le y$ if and only if
+    \[
+    \begin{bmatrix}
+        1 &x \\
+        x^* &y
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_2(A)$. In particular, $\norm{x}_A \le 1$ if and only if
+    \[
+    \begin{bmatrix}
+        1 &x \\
+        x^* &1
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_2(A)$.
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 3.1]{Paulsen}}}. ]
+    Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$. For any $\xi, \eta \in H$, 
+    \[
+    \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}}_{H^2} = \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{y\eta, \eta}{H}
+    \]
+
+    ($\Rightarrow$): If $y \ge x^*x$, then by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+    \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            \xi \\
+            \eta
+        \end{bmatrix}}_{H^2} &\ge \norm{\xi}_H^2 + \dpn{x\eta, \xi}{H} + \dpn{\xi, x\eta}{H} + \dpn{x\eta, x\eta}{H} \\
+        &\ge \norm{\xi}_H^2 - 2\norm{\xi}_H\norm{x\eta}_H + \norm{x\eta}_H^2 \ge 0
+    \end{align*}
+
+    ($\Leftarrow$): If the matrix is positive, then for each $\eta \in H$, 
+    \begin{align*}
+        0 &\le \angles{ \begin{bmatrix}
+            1 &x \\
+            x^* &y
+        \end{bmatrix}\begin{bmatrix}
+            -x\eta \\
+            \eta
+        \end{bmatrix}, \begin{bmatrix}
+            -x\eta \\
+            \eta
+        \end{bmatrix}}_{H^2} = \norm{x\eta}_H^2 -2\dpn{x\eta, x\eta}{H} + \dpn{y\eta, \eta}{H} \\
+        &\le \dpn{y\eta, \eta}{H} - \dpn{x\eta, x\eta}{H} = \dpn{(y - x^*x)\eta, \eta}{H}
+    \end{align*}
+
+    so $y \ge x^*x$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:unital-2-positive-contractive}
+    Let $A, B$ be unital $C^*$-algebras, $S \subset A$ be an operator system, $\phi: S \to B$ be a unital $2$-positive map, then $\norm{\phi}_{L(S; B)} \le 1$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 3.2]{Paulsen}}}. ]
+    Let $x \in S$ with $\norm{x}_A \le 1$, then
+    \[
+    \phi_2\begin{bmatrix}
+            1 &x \\
+            x^* &1
+        \end{bmatrix} = \begin{bmatrix}
+            1 &\phi(x) \\
+            \phi(x)^* &1
+        \end{bmatrix} \ge 0
+    \]
+
+    By \autoref{lemma:norm-matrix-test}, $\norm{\phi(x)}_B \le 1$. 
+\end{proof}
+
+\begin{proposition}[Kadison-Schwarz Inequality]
+\label{proposition:cauchy-schwarz-2-positive}
+    Let $A, B$ be unital $C^*$-algebras, $\phi: A \to B$ be a unital $2$-positive map, then for each $x \in A$,
+    \[
+    \phi(x)^*\phi(x) \le \phi(x^*x)
+    \]
+\end{proposition}
+\begin{proof}
+    Since
+    \[
+    \begin{bmatrix}
+            1 &x \\
+            x^* &x^*x
+        \end{bmatrix} = \begin{bmatrix}
+            1 &x \\
+            0 &0
+        \end{bmatrix}^*\begin{bmatrix}
+            1 &x \\
+            0 &0
+        \end{bmatrix} \ge 0
+    \]
+
+    and $\phi$ is unital and $2$-positive, 
+    \[
+    \begin{bmatrix}
+            1 &\phi(x) \\
+            \phi(x)^* &\phi(x^*x)
+        \end{bmatrix} = \phi_2\begin{bmatrix}
+            1 &x \\
+            x^* &x^*x
+        \end{bmatrix} \ge 0
+    \]
+
+    By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \phi(x^*x)$. 
+\end{proof}
+
+\begin{lemma}[Schur Complement Condition for Positivity]
+\label{lemma:schur-complement-positive}
+    Let $H$ be a complex Hilbert space, $x, y \in B(H)$, and $z \in G(B(H))$, then
+    \[
+    \begin{bmatrix}
+            x &y \\
+            y^* &z
+        \end{bmatrix} \ge 0
+    \]
+    if and only if $z \ge 0$ and $x - yz^{-1}y^* \ge 0$. 
+\end{lemma}
+\begin{proof}
+    ($\Leftarrow$): Suppose that $z \ge 0$ and $x - yz^{-1}y^* \ge 0$, then since $z$ is invertible, 
+    \begin{align*}
+    \begin{bmatrix}
+            x &y \\
+            y^* &z
+    \end{bmatrix} &= \begin{bmatrix}
+        I &yz^{-1} \\
+        0 &I
+    \end{bmatrix}\begin{bmatrix}
+        x - yz^{-1}y^* &0 \\
+        0 &z
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        z^{-1}y^* &I
+    \end{bmatrix} \\
+    &= \begin{bmatrix}
+        I &0 \\
+        z^{-1}y^* &I
+    \end{bmatrix}^*\begin{bmatrix}
+        x - yz^{-1}y^* &0 \\
+        0 &z
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        z^{-1}y^* &I
+    \end{bmatrix}
+    \end{align*}
+
+    Since the middle factor is positive, the given matrix is positive as well.
+
+    ($\Rightarrow$): Now suppose that the given matrix is positive, then for any $\xi \in H$, 
+    \[
+    \dpn{z\xi, \xi}{H} =  \angles{\begin{bmatrix}
+            x &y \\
+            y^* &z
+        \end{bmatrix}\begin{bmatrix}
+            0 \\
+            \xi
+        \end{bmatrix}, \begin{bmatrix}
+            0 \\
+            \xi
+        \end{bmatrix}}_{H^2} \ge 0
+    \]
+
+    so $z \ge 0$. As such,
+    \begin{align*}
+    \begin{bmatrix}
+        x - yz^{-1}y^* &0 \\
+        0 &z
+    \end{bmatrix} &= \begin{bmatrix}
+        I &-yz^{-1} \\
+        0 &I
+    \end{bmatrix}\begin{bmatrix}
+            x &y \\
+            y^* &z
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        -z^{-1}y^* &I
+    \end{bmatrix} \\ 
+    &= \begin{bmatrix}
+        I &0 \\
+        -z^{-1}y^* &I
+    \end{bmatrix}^*\begin{bmatrix}
+            x &y \\
+            y^* &z
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        -z^{-1}y^* &I
+    \end{bmatrix} \\
+    \end{align*}
+
+    Since the middle factor is positive, the left hand side is also positive. Therefore $x - yz^{-1}y^* \ge 0$ as well. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:completely-positive-norm}
+    Let $A, B$ be $C^*$-algebras with $A$ being unital, $S \subset A$ be an operator system, and $\phi: S \to B$ be a completely positive map, then
+    \[
+    \norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; B)} = \norm{\phi(1_A)}_B
+    \]
+\end{proposition}
+\begin{proof}
+    After taking a \hyperref[unitisation]{proposition:c-star-unitisation}, assume without loss of generality that $B$ is unital. 
+    
+    It is sufficient to show that $\norm{\phi}_{\text{cb}} \le \norm{\phi(1_A)}_B$. To this end, let $T \in M_n(S)$ with $\norm{T}_{M_n(S)} \le 1$, then
+    \[
+    \begin{bmatrix}
+            1_{M_n(A)} &T \\
+            T^* &1_{M_n(A)}
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_{2}(M_n(S)) = M_{2n}(S)$ by \autoref{lemma:norm-matrix-test}. Since $\phi$ is completely positive, 
+    \[
+    \phi_{2n}\begin{bmatrix}
+            1_{M_n(A)} &T \\
+            T^* &1_{M_n(A)}
+    \end{bmatrix} = \begin{bmatrix}
+            \phi_n(1_{M_n(A)}) &\phi_n(T) \\
+            \phi_n(T)^* & \phi_n(1_{M_n(A)})
+    \end{bmatrix} \ge 0
+    \]
+
+    First suppose that $\phi_n(1_{M_n(A)})$ is invertible, then
+    
+    \begin{align*}
+        \phi_n(1_{M_n(A)}) &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot 1_{M_n(B)} \\
+        1_{M_n(B)} &\le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)} \cdot \phi_n(1_{M_n(A)})^{-1} \\
+        \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1} &\le \phi_n(1_{M_n(A)})^{-1} 
+    \end{align*}
+    
+    By the \hyperref[Schur complement condition]{lemma:schur-complement-positive}, 
+    \begin{align*}
+    \phi_n(1_{M_n(A)}) &\ge \phi_n(T)\phi_n(1_{M_n(A)})^{-1}\phi_n(T)^* \\
+    &\ge \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^{-1}\phi_n(T)\phi_n(T)^* \\
+    \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}^2 &\ge \normn{\phi_n(T)\phi_n(T)^*}_{M_n(B)} = \normn{\phi_n(T)}_{M_n(B)}^2 
+    \end{align*}
+
+    and $\normn{\phi_n(T)}_{M_n(B)} \le \normn{\phi_n(1_{M_n(A)})}_{M_n(B)}$. Therefore $\norm{\phi}_{\text{cb}} \le \normn{\phi(1_A)}_B$.
+    
+    Now suppose that $\phi$ is an arbitrary completely positive map. Let $\psi \in S(A)$ be a state and $\eps > 0$, then $\phi + \eps 1_B\psi$ is also a completely positive map. In which case, 
+    \[
+    \norm{\phi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi + \eps1_B\psi}_{\text{cb}} \le \inf_{\eps > 0}\norm{\phi(1_A) + \eps 1_B}_B = \norm{\phi(1_A)}_B
+    \]
+\end{proof}
+
+\begin{example}
+    \label{example:transpose-not-completely-positive}
+    Let $T: M_2(\complex) \to M_2(\complex)$ be the transpose map, then $T$ is positive, but not $2$-positive. 
+\end{example}
+\begin{proof}
+    Let
+    \[
+    C = \begin{bmatrix}
+            1&0&0&1 \\
+            0&0&0&0 \\
+            0&0&0&0 \\ 
+            1&0&0&1
+    \end{bmatrix} = (1, 0, 0, 1) \otimes (1, 0, 0, 1) \ge 0
+    \]
+
+    then
+    \[
+    T_2(C) = \begin{bmatrix}
+            1&0&0&0 \\
+            0&0&1&0 \\
+            0&1&0&0 \\
+            0&0&0&1
+    \end{bmatrix}
+    \]
+
+    is invertible, with negative determinant, so $T_2(C)$ is not positive. 
+\end{proof}
+
+\begin{example}
+    \label{example:homomorphism-completely-positive}
+    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. 
+\end{example}
+\begin{proof}
+    For each $n \in \natp$, the induced map $\phi_n: M_n(A) \to M_n(B)$ is a *-homomorphism. 
+\end{proof}
+
+\begin{example}
+    \label{example:multiply-completely-bounded}
+
+    Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then
+    \begin{enumerate}
+        \item $\phi$ is completely bounded.
+        \item If $y = z^*$, then $\phi$ is completely positive. 
+    \end{enumerate}
+\end{example}
+\begin{proof}
+    (1): For each $n \in \natp$ and $T \in M_n(A)$, 
+    \[
+    \phi_n(T) = (1_{M_n(\complex)} \otimes y)T(1_{M_n(\complex)} \otimes z)
+    \]
+\end{proof}
+
+\begin{proposition}
+\label{proposition:commutative-mn-facts}
+    Let $X$ be a compact Hausdorff space, then:
+    \begin{enumerate}
+        \item $C(X; M_n(\complex))$ equipped with pointwise product, pointwise involution, and the norm
+        \[
+        \norm{f}_{C(X; M_n(\complex))} = \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}
+        \]
+
+        is a $C^*$-algebra. 
+        \item The mapping
+        \[
+        M_n(C(X; \complex)) \to C(X; M_n(\complex)) \quad M \otimes f \mapsto M \cdot f
+        \]
+
+        is a *-isomorphism. 
+        \item For any $f \in C(X; M_n(\complex))$, $f \ge 0$ if and only if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For any $f \in C(X; M_n(\complex))$, 
+    \[
+    \sup_{x \in X}\norm{f(x)}_{M_n(\complex)}^2 = \sup_{x \in X}\norm{f(x)^*f(x)}_{M_n(\complex)} = \norm{f^*f}_{C(X; M_n(\complex))}
+    \]
+
+    (3): If $f \ge 0$, then there exists $g \in C(X; M_n(\complex))$ such that $f(x) = g^*(x)g(x)$ for all $x \in X$, so $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$. 
+
+    On the other hand, if $f(x) \ge 0$ in $M_n(\complex)$ for all $x \in X$, then $g(x) = f(x)^{1/2}$ is defined for all $x \in X$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $g$ is continuous from $X$ to $M_n(\complex)$. Therefore there exists $g \in C(X; M_n(\complex))$ such that $f = g^*g$. 
+\end{proof}
+
+
+\begin{theorem}
+\label{theorem:bounded-functional-completely-bounded}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be a subspace, $X$ be a compact Hausdorff space, and $\phi \in L(S; C(X; \complex))$, then:
+    \begin{enumerate}
+        \item $\norm{\phi}_{\text{cb}} = \norm{\phi}_{L(S; C(X; \complex))}$. 
+        \item If $S$ is an operator system and $\phi$ is positive, then $\phi$ is completely positive. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Proposition 3.8]{Paulsen}}}. ]
+    By \autoref{proposition:commutative-mn-facts}, the norm and order relation on $C(X; M_n(\complex)) \iso M_n(C(X; \complex))$ are pointwise. As such, assume without loss of generality that $X$ is a single point and $C(X; \complex) = \complex$.
+    
+    Let $a = (a_{ij}) \in M_n(S)$. For each $\xi, \eta \in \complex^n$, 
+    \begin{align*}
+        \dpn{\phi_n(a)\xi, \eta}{\complex^n} &= \sum_{i, j = 1}^n \xi_j \ol{\eta_i} \dpn{a_{ij}, \phi}{S} =  \angles{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}, \phi}_S 
+    \end{align*}
+    
+    where
+    \begin{align*}
+    \sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i} &= (\one \otimes \ol{\eta}) \cdot a \cdot (\xi \otimes \one)
+    \end{align*}
+
+    (1): 
+    \begin{align*}
+        |\dpn{\phi_n(a)\xi, \eta}{\complex^n}| &\le \norm{\phi}_{S^*} \cdot  \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} \\
+    \norm{\sum_{i, j = 1}^n a_{ij}\xi_j\ol{\eta_i}}_{S} &\le \norm{\xi}_{\complex^n}\norm{\eta}_{\complex^n}\norm{a}_{M_n(S)} \\
+    \norm{\phi_n(a)}_{M_n(\complex)} &\le \norm{\phi}_{S^*}\norm{a}_{M_n(S)} \\
+    \norm{\phi}_{\text{cb}} &\le \norm{\phi}_{S^*}
+    \end{align*}
+
+    (2): If $\xi = \eta$, then
+    \[
+    \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0
+    \]
+
+    so
+    \[
+    \dpn{\phi_n(a)\xi, \eta}{\complex^n} = \dpn{(\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one), \phi}{S} \ge 0
+    \]
+\end{proof}
+
+
+\begin{theorem}[Stinespring]
+\label{theorem:stinespring-commutative-completely-positive}
+    Let $X$ be a compact Hausdorff space, $B$ be a $C^*$-algebra, $\phi: C(X; \complex) \to B$ be a positive map, then $\phi$ is completely positive. 
+\end{theorem}
+\begin{proof}
+    Let $P \in C(X; M_n(\complex))$ with $P \ge 0$ and $\eps > 0$. By \autoref{proposition:uniform-continuous-compact} and \autoref{theorem:lch-paracompact}, there exists a partition of unity $\seqf{\psi_j} \subset C(X; [0, 1])$ and positive matrices $\seqf{p_j} \subset M_n(\complex)$ such that
+    \[
+    \norm{P(x) - \sum_{j = 1}^n p_j\psi_j(x)}_{M_n(\complex)} < \eps
+    \]
+
+    for all $x \in X$. For each $1 \le j \le n$, $\phi_n(p_j\psi_j) = p_j\phi(\psi_j) \ge 0$. As $\eps > 0$ is arbitrary, $\phi_n(P) \ge 0$. 
+\end{proof}
+
+\begin{theorem}[Choi]
+\label{theorem:choi-matrix-positive}
+    Let $n \in \natp$, $B$ be a $C^*$-algebra, and $\phi: M_n(\complex) \to B$ be a linear map. For each $1 \le i, j \le n$, let $E_{ij}$ be the standard matrix units for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Chi matrix} of $\phi$. The mapping $\phi \mapsto C_\phi$ is an isomorphism from $L(M_n(\complex); B)$ to $M_n(B)$, and the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is completely positive.
+        \item $\phi$ is $n$-positive. 
+        \item $C_\phi$ is positive in $M_n(B)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (2) $\Rightarrow$ (3): The matrix $E = (E_{ij})$ is positive in $M_n(\complex)$. 
+
+    (3) $\Rightarrow$ (1): 
+    
+    
+    
+    $\phi, \psi \in S(G)$, $K \subset G$ be compact, and $V \in \cn_G(1_G)$ be relatively compact, then
+    \begin{align*}
+        
+    \end{align*}
+    
+\end{proof}
+
+
+

src/op/system/dilation.tex

@@ -0,0 +1,122 @@
+\section{Dilations}
+\label{section:bh-dilation}
+
+\begin{definition}[Compression]
+\label{definition:bh-compression}
+    Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. 
+\end{definition}
+
+\begin{definition}[Dilation]
+\label{definition:bh-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a pair $(K, S)$ where:
+    \begin{enumerate}
+        \item $K$ is a Hilbert space containing $H$. 
+        \item $S \in B(K)$. 
+        \item If $P \in B(K)$ is the orthogonal projection onto $H$, then $T = PS|_H$. 
+    \end{enumerate}
+
+    If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:bh-dilation-transitive}
+    Let $H$ be a complex Hilbert space, $R \in B(H)$, $(K, S)$ be a dilation of $R$, and $(L, T)$ be a dilation of $S$, then $(L, T)$ is a dilation of $R$. 
+
+    Similarly, if $(K, S)$ is a power dilation of $T$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. 
+\end{lemma}
+
+\begin{theorem}[Sz.-Nagy's Dilation Theorem]
+\label{theorem:sznagy-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
+\end{theorem}
+\begin{proof}[Proof, {{\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 $T$. 
+        \item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$. 
+    \end{enumerate}
+
+    (i): Since $\norm{T}_{B(H)} \le 1$, $1 - T^*T \ge 0$, and $(1 - T^*T)^{1/2}$ is well-defined. Let $D_T = (1 - T^*T)^{1/2}$, and
+    \[
+    V: l^2(\natp; H) \to l^2(\natp; H) \quad (Vx)_n = \begin{cases}
+        Tx_1 &n = 1 \\
+        D_Tx_1 &n = 2 \\
+        x_{n-1} &n \ge 3
+    \end{cases}
+    \]
+    
+    then for each $x \in H$,
+    \begin{align*}
+        \norm{Tx}_H^2 + \norm{D_Tx}_H^2 &= \dpn{T^*Tx, x}{H} + \dpn{(1 - T^*T)x, x}{H} \\
+        &= \dpn{x, x}{H} = \norm{x}_H^2
+    \end{align*}
+
+    Thus for each $x = \seq{x_n} \in l^2(H)$, 
+    \begin{align*}
+        \norm{Vx}_{l^2(\natp; H)}^2 &= \norm{Tx_1}_H^2 + \norm{D_Tx_1}_H^2 + \sum_{n = 3}^\infty \norm{x_{n-1}}_H^2 \\
+        &= \norm{x_1}_H^2 + \sum_{n = 2}^\infty \norm{x_{n}}_H^2 = \norm{x}_{l^2(\natp; H)}^2
+    \end{align*}
+
+    So $V$ is an isometry. Moreover, $[V^n(x, 0, \cdots)]_1 = T^nx$ for all $n \in \natz$. Therefore under the identification that $H = \bracsn{(x, 0, \cdots)|x \in H} \subset l^2(\natp; H)$, $V$ is an isometric power dilation of $T$. 
+
+    (ii): Let $P = I_H - VV^*$ be the projection onto $V(H)^\perp$, and $U \in B(H^2)$ be defined by
+    \[
+    U = \begin{bmatrix}
+        V & P \\
+        0 & V^*
+    \end{bmatrix}
+    \]
+
+    then
+    \begin{align*}
+        UU^* &= \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} = \begin{bmatrix}
+            VV^* + P & PV \\
+            V^*P & V^*V
+        \end{bmatrix} = I_{H^2} \\
+        U^*U &= \begin{bmatrix}
+            V^* & 0 \\
+            P & V
+        \end{bmatrix} \cdot 
+        \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}
+         = \begin{bmatrix}
+            V^*V & V^*P \\
+            PV & VV^* + P
+        \end{bmatrix} = I_{H^2} \\
+    \end{align*}
+
+    so $U$ is unitary. Moreover, for each $n \in \natp$, 
+    \[
+    U^n = \begin{bmatrix}
+            V & P \\
+            0 & V^*
+        \end{bmatrix}^n = \begin{bmatrix}
+            V^n & \cdots \\
+            0 & (V^*)^n
+        \end{bmatrix}
+    \]
+
+    Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. 
+\end{proof}
+
+\begin{theorem}[Von Neumann's Inequality]
+\label{theorem:von-neumann-inequality-polynomial}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p \in \complex[x]$, 
+    \[
+    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
+    \]    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Corollary 1.2]{Paulsen}}}. ]
+    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
+
+    Thus for any $p \in \complex[x]$, $p(T) = P_Hp(U)|_H$. As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U)}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. Therefore $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
+\end{proof}

src/op/system/index.tex

@@ -1,4 +1,6 @@
 \chapter{Operator Systems}
 \label{chap:opertaor-systems}
 
-\input{./definitions.tex}
+\input{./dilation.tex}
+\input{./os.tex}
+\input{./completely.tex}

src/op/system/os.tex

@@ -39,7 +39,7 @@
 \label{proposition:positive-automatic-continuity}
     Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then
     \begin{enumerate}
-        \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norom{x}_A$.
+        \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norm{x}_A$.
         \item $\norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B$. 
     \end{enumerate}
 \end{proposition}