Dilation situation.

refs.bib

@@ -293,3 +293,11 @@
     eprint = {https://math.stackexchange.com/q/150258},
     url = {https://math.stackexchange.com/q/150258}
 }
+
+@book{Paulsen,
+  title={Completely bounded maps and operator algebras},
+  author={Paulsen, Vern I and others},
+  volume={78},
+  year={2002},
+  publisher={Cambridge University Press}
+}

src/dyn/top/def.tex

@@ -34,14 +34,65 @@
 \begin{definition}[Transitive]
 \label{definition:tds-transitive}
     Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{topologically transitive} with respect to $G$ if $Gx$ is dense in $X$. 
+    
+    The flow $G \curvearrowright X$ is \textbf{point transitive} if it admits a topologically transitive point. 
+
+    Finally, $G \curvearrowright X$ is \textbf{topologically transitive} if for any $U, V \subset X$ non-empty and open, there exists $g \in G$ such that $gU \cap V \ne \emptyset$\footnote{Approximations of points may be translated into approximations of other points. }. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:tds-point-transitive}
+    Let $G \curvearrowright X$ be a point transitive $G$-flow, then $G \curvearrowright X$ is also topologically transitive. 
+\end{proposition}
+\begin{proof}
+    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}[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. 
+    \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 $Y = X$. 
+\end{proof}
 
 
 
+\begin{definition}[Bernoulli Shift]
+\label{definition:bernoulli-shift}
+    Let $G$ be a group, $X = 2^G$ equipped with the product topology, and
+    \[
+    \alpha: X \to X \quad \alpha_g(x)(h) = x(g^{-1}h)
+    \]
+
+    Viewing $2^G$ as the power set of $G$, $\alpha_g(A) = gA$ for any $A \subset G$. The $G$-flow $G \curvearrowright X$ is the \textbf{Bernoulli shift} of $G$, and $G$-subflows of $X$ are \textbf{subshifts} of $G$. 
+\end{definition}
+
+\begin{example}[The Shift]
+    \label{example:bernoulli-shift-z}
+
+    Let $\integer \curvearrowright 2^\integer$ be the Bernoulli shift of $G$, then:
+    \begin{enumerate}
+        \item For any $x, y \in 2^\integer$, $y \in \ol{\integer x}$ if and only if for each $A \subset \integer$, there exists $n \in \integer$ such that $y|_A = nx|_A$. In other words, $x$ contains a copy of any segment of $y$\footnote{Topological conditions in Bernoulli shifts translate into combinatorial information. }. 
+        \item $\integer \curvearrowright 2^\integer$ is point transitive, and hence topologically transitive\footnote{It is easier to show topological transitivity: move the trees until they are disjoint}. 
+        \item 
+    \end{enumerate}
+    
+\end{example}
+
+
 

src/logic/index.tex

@@ -2,3 +2,6 @@
 \label{part:logic}
 
 This section is plagiarised from 
+
+\input{./tp/index.tex}
+\input{./notation.tex}

src/logic/notation.tex

@@ -0,0 +1,11 @@
+\chapter{Notations}
+\label{chap:logic-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    % ---- Topological Vector Spaces ----
+    $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ & Language $\mathcal{L}$ with function symbols $\mathcal{F}$, relation symbols $\mathcal{R}$, and constant symbols $\mathcal{C}$. & \autoref{definition:language} \\
+    $\mathcal{M} = (M, f^\cm, R^\cm, c^\cm)$ & $\mathcal{L}$-structure with functions $\bracsn{f^\cm|f \in \mathcal{F}}$, relations $\bracsn{R^\cm|R \in \mathcal{R}}$, and constants $\bracsn{c^\cm|c \in \mathcal{C}}$. & \autoref{definition:language-structure}
+\end{tabular}
+

src/logic/tp/lang.tex

@@ -11,5 +11,46 @@
     \end{enumerate}
 \end{definition}
 
+\begin{definition}[$\mathcal{L}$-Structure]
+\label{definition:language-structure}
+    Let $L = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language, then a \textbf{\mathcal{L}-structure} $\mathcal{M}$ is a family
+    \[
+    \mathcal{M} = (M, f^\cm, R^\cm, c^\cm| f \in \mathcal{F}, R \in \mathcal{R}, c \in \mathcal{C})
+    \]
+
+    where
+
+    \begin{enumerate}
+        \item $M$ is a non-empty set, known as the \textbf{universe}, \textbf{domain}, or \textbf{underlying set} of $\mathcal{M}$. 
+        \item For each $f \in \mathcal{F}$, $f^\cm: M^{n_f} \to M$ is a function in $n$-variables in $M$. 
+        \item For each $R \in \mathcal{R}$, $R^\cm \subset M^{n_R}$ is a $n$-ary relation.
+        \item For each $c \in \mathcal{C}$, $c^\cm \in M$ is a constant.  
+    \end{enumerate}
+    
+    The assignments $f^\cm$, $R^\cm$, and $c^\cm$ are the \textbf{interpretations} of $(\mathcal{F}, \mathcal{R}, \mathcal{C})$. 
+\end{definition}
+
+\begin{definition}[Terms]
+\label{definition:logic-terms}
+    Let $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language, then the set of \textbf{$\mathcal{L}$-terms} is the smallest set of expressions $\mathcal{T}(\mathcal{L})$ such that:
+    \begin{enumerate}
+        \item $\mathcal{C} \subset \mathcal{T}(\mathcal{L})$. 
+        \item For each $n \in \natp$, the variable symbol $v_n \in \mathcal{T}(\mathcal{L})$. 
+        \item For each $f \in \mathcal{F}$ and $\bracsn{t_j}_1^{n_f} \subset \mathcal{T}(\mathcal{L})$, the expression $f(t_1, \cdots, t_{n_f}) \in \mathcal{T}(\mathcal{L})$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:logic-terms-induction}
+    Let $\mathcal{L} = (\mathcal{F}, \mathcal{R}, \mathcal{C})$ be a language and $\mathcal{T}_0(\mathcal{L}) = \mathcal{C} \cup \seq{v_n}$. For each $n \in \natp$, inductively define
+    \[
+    \mathcal{T}_{n}(\mathcal{L}) = \mathcal{T}_{n-1}(\mathcal{L}) \cup \bracsn{f(t_1, \cdots, t_{n_f})|f \in \mathcal{F}, \bracsn{t_j}_1^{n_f} \in \mathcal{T}_{n-1}(\mathcal{L})}
+    \]
+
+    then $\mathcal{T} = \bigcup_{n \in \natz}\mathcal{T}_n$. For each $n \in \natz$, $\mathcal{T}_n$ is the set of \textbf{$\mathcal{L}$-terms of complexity at most $n$}.
+\end{lemma}
+
+
+
 
 

src/op/example/bounded.tex

@@ -81,3 +81,123 @@
 \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}[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 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 $H$. 
+\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/example/matrix.tex

@@ -82,5 +82,3 @@
     \]
 \end{example}
 
-
-