Various fixes.

document.tex

@@ -17,6 +17,7 @@ Occasionally, I make up some definitions to play with. These definition blocks w
 \input{./src/harmonic/index}
 \input{./src/dg/index}
 \input{./src/op/index}
+\input{./src/vna/index}
 \input{./src/dyn/index}
 %\input{./src/process/index}
 

refs.bib

@@ -320,3 +320,10 @@
   year={1958},
   publisher={Duke University Press}
 }
+
+@misc{AnantharamanPopa,
+  author       = {Anantharaman, Claire and Popa, Sorin},
+  title        = {An introduction to {II$_1$} factors},
+  howpublished = {\url{https://www.math.ucla.edu/~popa/Books/IIunV15.pdf}},
+  note         = {Book draft}
+}

src/cat/cat/cat-func.tex

@@ -39,3 +39,32 @@
         \item[(FN2')] For any $A, B, C \in \obj{\mathfrak{A}}$, $f \in \mor{A, B}$, and $g \in \mor{B, C}$, $\lambda(g \circ f) = \lambda(f) \circ \lambda(g)$.
     \end{enumerate}
 \end{definition}
+
+
+\begin{definition}[Injective]
+\label{definition:injective-cat}
+    Let $\catc$ be a category with a notion of embedding and $I \in \obj(\catc)$, then $I$ is \textbf{injective} if for any objects $X, Y \in \obj(\catc)$, embedding $f: X \to Y$, and morphism $\phi: X \to I$, there exists an extension $\Phi: Y \to I$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    & I \\
+    X \ar@{->}[ru]^{\phi} \ar@{^{(}->}[r]_{f} & Y \ar@{->}[u]_{\Phi}
+    }
+    \]
+\end{definition}
+
+\begin{definition}[Sufficiently Many Injectives]
+\label{definition:suff-many-injective}
+    Let $\catc$ be a category with a notion of embedding, then $\catc$ has \textbf{sufficiently many injectives} if every object embeds into an injective object. 
+\end{definition}
+
+\begin{definition}[Extension]
+\label{definition:category-extension-object}
+    Let $\catc$ be a category with a notion of embedding and $X \in \obj(\catc)$, then an \textbf{extension} of $X$ is a pair $(Y, \iota)$ where $Y \in \obj(\catc)$ and $\iota: X \to Y$ is an embedding. 
+    
+    The extension $(Y, \iota)$ is \textbf{essential} if for any $Z \in \obj(\catc)$ and $\phi: Y \To Z$, $\phi \circ \iota: X \to Z$ being an embedding implies that $\phi$ is also an embedding. 
+\end{definition}
+
+
+
+
+

src/dyn/meas/definitions.tex

@@ -0,0 +1,74 @@
+\section{Measurable Dynamical Systems}
+\label{section:measurable-dynamics-intro}
+
+\begin{definition}[Measure Preserving]
+\label{definition:measure-preserving}
+    Let $(X, \cm, \mu)$ be a measure space and $T: X \to X$ be measurable, then $T$ is \textbf{measure preserving} if $\mu(T^{-1}(E)) = \mu(E)$ for all $E \in \cm$. 
+\end{definition}
+
+\begin{definition}[Automorphism of Measure Space]
+\label{definition:measure-preserving-automorphism}
+    Let $(X, \cm, \mu)$ be a measure space and $T: X \to X$ be measurable, then $T$ is an \textbf{automorphism} if $T$ is bijective and measure preserving. The set $\text{Aut}(X, \cm, \mu)$ is the \textbf{automorphism group} of $(X, \cm, \mu)$.
+\end{definition}
+
+\begin{definition}[Measure Preserving Action]
+\label{definition:measure-preserving-action}
+    Let $\Gamma$ be a group and $(X, \cm, \mu)$ be a measure space, then a \textbf{measure preserving action} of $\Gamma$ on $(X, \cm, \mu)$ is a group homomorphism $\alpha: \Gamma \to \text{Aut}(X, \cm, \mu)$. The pair $(\Gamma, (X, \cm, \mu)) = \Gamma \curvearrowright (X, \cm, \mu)$ is a \textbf{measurable dynamical system}.
+\end{definition}
+
+\begin{definition}[Invariant]
+\label{definition:measurable-invariant-set}
+    Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$ and $E \in \cm$, then $E$ is \textbf{$\Gamma$-invariant} if $gE = E$ for all $g \in \Gamma$. For any separable metric space $Y$ and measurable function $f: X \to Y$, $f$ is \textbf{$\Gamma$-invariant} if $f \circ g = f$ a.e. for all $g \in \Gamma$. 
+\end{definition}
+
+\begin{definition}[Ergodic]
+\label{definition:ergodic}
+    Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$, then the action is \textbf{ergodic} if every $\Gamma$-invariant set is either null or conull. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:measurable-group-composition}
+    Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$ and $Y$ be a separable metric space, then $\Gamma$ induces an action\footnote{W.r.t. the topology of convergence in measure.}
+    \[
+    \Gamma \to \text{Aut}(L^0(X; Y)) \quad g \cdot f = f \circ g
+    \]
+
+    For any Banach space $E$\footnote{Separability not required since $L^p$ here refers to strongly measurable functions.}, this induces a representation
+    \[
+    \Gamma \to L^p(X; E) \quad g \cdot f = f \circ g
+    \]
+
+    for all $p \in [1, \infty]$. Moreover, the following are equivalent:
+    \begin{enumerate}
+        \item The action of $\Gamma$ on $(X, \cm, \mu)$ is ergodic. 
+        \item For every $f \in L^0(X; \real)$, $f$ is $\Gamma$-invariant if and only if $f$ is constant almost everywhere.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $t \in \real$, let $A_t = \bracsn{f \le t}$, then since the action of $\Gamma$ is ergodic, $A_t$ is either null or conull. Let
+    \[
+    \alpha = \sup\bracsn{t \in \real| A_t \text{ null}} \in \real
+    \]
+
+    then $A_\alpha \supset \bigcup_{n \in \natp}A_{\alpha + 1/n}$ is conull. Therefore $\bracsn{f = \alpha} = A_\alpha \setminus \bigcup_{n \in \natp} A_{\alpha - 1/n}$ is conull. 
+\end{proof}
+
+
+\begin{example}
+    \label{example:ergodic-fourier}
+    Let $\mathbb{T} = \mathbb{R}/\mathbb{Z} = \partial B_\complex(0, 1)$, equipped with the normalised Lebesgue measure. For each $\alpha \in \real$, let $R_\alpha: \mathbb{T} \to \mathbb{T}$ be defined by $x \mapsto x + \alpha \mod 1$, then $R_\alpha$ induces an $\integer$-action on $\mathbb{T}$, defined by $n \cdot x = x + n\alpha$, which is ergodic if and only if $\alpha \in \real \setminus \mathbb Q$. 
+\end{example}
+\begin{proof}
+    Let $f \in L^2(\mathbb{T}; \complex)$, then
+    \[
+    f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n x} \quad R_\alpha f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n (x + \alpha)}
+    \]
+
+    where the series converges in $L^2(\mathbb{T}; \complex)$. If $f$ is $R_\alpha$-invariant, then
+    \[
+    f(x) = \sum_{n \in \integer} \hat f(n) e^{2\pi i n (x + \alpha)} = \sum_{n \in \integer} \hat f(n) e^{2\pi i n \alpha} e^{2\pi i n x}
+    \]
+
+    As such, $\hat f(n)e^{2\pi i n \alpha} = \hat f(n)$ for all $n \in \integer$. As $\alpha \not\in \mathbb Q$, $e^{2\pi i n \alpha} \ne 1$ for all $n \in \integer \setminus \bracsn{0}$, so $\hat f(n) = 0$ for all $n \in \integer \setminus \bracsn{0}$. Therefore $f = \hat f(0)$ almost everywhere. By \autoref{proposition:measurable-group-composition}, $R_\alpha$ is ergodic. 
+    
+\end{proof}

src/dyn/meas/index.tex

@@ -0,0 +1,5 @@
+\chapter{Measurable Dynamics}
+\label{chap:measurable-dynamics}
+
+\input{./definitions.tex}
+

src/dyn/meas/rec.tex

@@ -0,0 +1,67 @@
+\section{Measurable Recurrence}
+\label{section:measure-recurrence}
+
+\begin{theorem}[Poincaré Recurrence]
+\label{theorem:pointcare-recurrence}
+    Let $(\Omega, \cf, \bp)$ be a probability space, $T: \Omega \to \Omega$ be an automorphism, and $A \in \cf$, then $\bracsn{n \in \natp|T^n\omega \in A}$ is infinite for almost every $\omega \in A$. 
+\end{theorem}
+\begin{proof}
+    Let $W = \bracsn{\omega \in A| T^n\omega \not\in A \forall n \in \natp}$. Since $W \subset A$, $T^nW \cap W = \emptyset$ for all $\in \natp$. Thus $\bracsn{T^nW}_0^\infty \subset \cf$ is pairwise disjoint. As $T$ is probability measure preserving,
+    \[
+    1 \ge \bp\paren{\bigsqcup_{n \in \natz}T^nW} = \sum_{n \in \natz}\bp(T^nW) = \sum_{n \in \natz}\bp(W)
+    \]
+
+    so $\bp(W) = 0$. 
+
+    Let $E = \bracsn{\omega \in A|T^n\omega \in A \text{ finitely many times}}$, then for each $\omega \in E$, there exists $n_\omega \in \natp$ such that $T^{n_\omega}\omega \in W$. In other words, $E \subset \bigcup_{k \in \natp} T^{-k}W$, and $E$ is also null. Since $\bracsn{n \in \natp|T^n\omega \in A}$ is infinite for all $\omega \in A \setminus E$, $\bracsn{n \in \natp|T^n\omega \in A}$ is infinite for almost every $\omega \in A$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:measure-set-recurrence}
+    Let $(\Omega, \cf, \bp)$ be a probability space, $T: \Omega \to \Omega$ be an automorphism, and $A \in \cf$ with $\bp(A) > 0$, then there exists $n \in \natp$ such that $\bp(A \cap T^{-n}A) > 0$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[Poincaré Recurrence Theorem]{theorem:pointcare-recurrence}, $A \subset \bigcup_{n \in \natp}A \cap T^{-n}A$. Since $\bp(A) > 0$, \hyperref[subadditivity]{proposition:measure-properties}, there exists $n \in \natp$ with $\bp(A \cap T^{-n}A) > 0$. 
+\end{proof}
+
+\begin{definition}[Koopman Representation]
+\label{definition:koopman-representation}
+    Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$, then the unitary representation
+    \[
+    \pi: \Gamma \to U(B(L^2(X, \cm, \mu; \complex))) \quad \pi(\gamma)(f) = f \circ \gamma
+    \]
+
+    is the \textbf{Koopman representation} of $\Gamma$. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:koopman-ergodic}
+    Let $\Gamma$ be a group acting on a measure space $(X, \cm, \mu)$ and $\pi: \Gamma \to U(B(L^2(X, \cm, \mu; \complex)))$ be its Koopman representation, then the subspace
+    \[
+    \text{Fix}(\Gamma) = \bracsn{f \in L^2(X, \cm, \mu; \complex)| \pi(\gamma) f = f \forall \gamma \in \Gamma}
+    \]
+
+    is equal to $\complex \one$ if and only if the action of $\Gamma$ is ergodic. 
+\end{proposition}
+
+\begin{definition}[Averaging Operator]
+\label{definition:ergodic-averaging-operator}
+    Let $(X, \cm, \mu)$ be a measure space, $T: X \to X$ be an automorphism, $E$ be a Banach space, $p \in [1, \infty]$, and $N \in \natp$, then the \textbf{$N$-th time averaging operator} is the mapping
+    \[
+    A_N: L^p(X; E) \to L^p(X; E) \quad A_Nf = \frac{1}{N}\sum_{n = 0}^{N-1} f \circ T^n
+    \]
+
+    which is a contraction on $L^p(X; E)$. 
+\end{definition}
+
+\begin{theorem}[von Neumann's Mean Ergodic Theorem]
+\label{theorem:vn-mean-ergodic}
+    Let $H$ be a Hilbert space, $U: H \to H$ be a unitary operator, $P \in B(H)$ be the orthogonal projection onto the fixed points of $U$. For each $N \in \natp$, let
+    \[
+    A_N: H \to H \quad \xi \mapsto \frac{1}{N}\sum_{n = 0}^{N-1} U^n\xi
+    \]
+
+    be the $N$-th time averaging operator. Then, for each $\xi \in H$, $P\xi = \slim_{N \to \infty}A_N\xi$. 
+\end{theorem}
+

src/dyn/top/def.tex

@@ -4,6 +4,16 @@
 \begin{definition}[Topological Dynamical System]
 \label{definition:topological-dynamical-system}
     Let $G$ be a group and $X$ be a (non-empty) compact Hausdorff space, then a \textbf{$G$-flow/topological dynamical system} is a homomorphism $\alpha: G \to \text{Aut}(X)$. In which case, the system is denoted by $G \curvearrowright X$. 
+
+    Let $G \curvearrowright X$ and $G \curvearrowright$ be $G$-flows, then a \textbf{morphism of $G$-flows} is a continuous map $\pi: X \to Y$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    X \ar@{->}[r]^{\pi} \ar@{->}[d]_{g} & Y \ar@{->}[d]^{g} \\
+    X \ar@{->}[r]_{\pi} & Y
+    }
+    \]
+
+    for all $g \in G$. 
 \end{definition}
 
 \begin{definition}[Invariant]
@@ -59,6 +69,20 @@
 \end{example}
 % Proof omitted. 
 
+
+\begin{definition}[Pointed Flow]
+\label{definition:tds-pointed}
+    Let $G$ be a group, then a \textbf{pointed $G$-flow} is a pair $(X, x_0)$, where $X$ is a $G$-flow, and $x_0 \in X$ is a distinguished point. The mapping $\rho_x: G \to X$ defined by $g \mapsto gx_0$ is the \textbf{orbit map} of $(X, x_0)$. 
+
+    Let $(G \curvearrowleft X, x_0)$ and $(G \curvearrowright Y, y_0)$ be pointed $G$-flows, then a \textbf{morphism of pointed $G$-flows} is a morphism of $G$-flows $\pi: X \to Y$ such that $\pi(x_0) = y_0$. 
+\end{definition}
+
+\begin{definition}[Ambit]
+\label{definition:tds-ambit}
+    Let $G$ be a group and $(G \curvearrowright X, x_0)$ be a pointed $G$-flow, then $(G \curvearrowright X, x_0)$ is an \textbf{ambit} if $x_0$ is a topologically transitive point. 
+\end{definition}
+
+
 \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/hales-jewett.tex

@@ -0,0 +1,114 @@
+\section{The Hales-Jewett Theorem}
+\label{section:hales-jewett}
+
+
+\begin{definition}[Variable Words]
+\label{definition:variable-words}
+    Let $A$ be a finite alphabet, $S = A^+$ be the free semigroup of all non-empty words in $A$, $T = (A \sqcup \bracsn{*})^+$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$, then the set of \textbf{variable words} in $A$ is the two-sided ideal $V := T \setminus S$ of all words in $A \sqcup \bracsn{*}$ containing $*$. 
+
+    For each $a \in A$, the semigroup homomorphism
+    \[
+    \pi_a: T \to S \quad b \mapsto \begin{cases}
+        b &b \in A \\
+        a &b = *
+    \end{cases}
+    \]
+
+    is the \textbf{substitution map} from $*$ to $a$, with $\pi_a|_S = \text{Id}_S$. 
+\end{definition}
+
+\begin{definition}[Combinatorial Line]
+\label{definition:combinatorial-line}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $V$ be the set of variable words in $A$, and $v \in V$, then
+    \[
+    L(v) = \bracsn{\pi_a(v)|a \in A} \subset S
+    \]
+
+    is the \textbf{combinatorial line}\footnote{Line in the sense that this is a straight line in the hypercube $A^{|v|}$. } generated by $v$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:hales-jewett-minimal-idempotent}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $T = (A \sqcup \bracsn{*})^*$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$, and $V$ be the set of variable words in $A$, then:
+    \begin{enumerate}
+        \item The inclusions $S \to T$ and $V \to T$ admit extensions into embeddings $\beta S \to T$ and $\beta V \to T$, which allows identifying $\beta T = \beta S \bigsqcup \beta V$. Moreover, $\beta S$ may be identified as a subgroup of $T$. In particular, every idempotent in $\beta S$ is idempotent in $\beta T$. 
+        \item For any minimal idempotent $p \in \beta S \setminus S$, $p$ is not minimal in $\beta T$. 
+        \item For any minimal idempotent $p \in \beta S \setminus S$, there exists a minimal idempotent $q \in \beta S \setminus \beta T$ such that $pq = qp = q$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By (U) of the \hyperref[Stone-Čech compactification]{definition:stone-cech}.
+
+    (2): Suppose for contradiction that $p$ is a minimal idempotent in $\beta T$, then $L = (\beta T)p$ is a minimal left ideal in $\beta T$. Since $(\beta T)*p \subset \beta V \cap (\beta T)p$, $(\beta T)p \cap \beta V =  (\beta T)p$ by minimality, and $(\beta T)p \subset \beta V$. As such, $(\beta S)p \subset \beta S \cap (\beta T)p \cap \beta V = \emptyset$, which is impossible. 
+
+    (3): By \autoref{theorem:ellis-semigroup}, $S \setminus \beta S$ admits a minimal idempotent $p$, which is idempotent in $\beta T$. By the \hyperref[idempotent refinement lemma]{lemma:idempotent-refinement}, there exists a minimial idempotent $q \in \beta T$ such that $pq = qp = q$. 
+
+    Suppose again for contradiction that $q \in \beta S$, then $pq = q \in (\beta S)p$, which is minimal in $\beta S$. As such, $(\beta S)q \subset (\beta S)p$, and $(\beta S)q = (\beta S)p \subset \beta S$, so $q$ is a minimal idempotent in $\beta S$. By (2), $q$ is not minimal in $\beta T$, which is impossible. 
+\end{proof}
+
+\begin{theorem}[Hales-Jewett]
+\label{theorem:hales-jewett}
+    Let $A$ be a finite alphabet, $S$ be the free semigroup of non-empty words in $A$, $V$ be the set of variable words in $A$, then:
+    \begin{enumerate}
+        \item[(IF)] Every finite colouring of $S$ contains a monochromatic combinatorial line. 
+        \item[(F)] For every $r \ge 1$, then there exists $N = N(r) \ge 1$ such that every $r$-colouring of $A^N$ contains a monochromatic combinatorial line. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (IF): Let $T = (A \sqcup \bracsn{*})^+$ be the free semigroup of all non-empty words in $A \sqcup \bracsn{*}$.
+    
+    For each $a \in A$, let $\pi_a: T \to S$ be the corresponding substitution map, and $\beta \pi_a: \beta T \to \beta S$ be its continuous extension. Let $p \in \beta S \setminus S$ be a minimal idempotent. By \autoref{lemma:hales-jewett-minimal-idempotent}, there exists a minimal idempotent $q \in \beta T$ be a minimal idempotent with $pq = qp = q$ and $q \in \beta T \setminus \beta S = \beta V$.
+
+    As $\pi_a|_S = \text{Id}_S$, $\beta \pi_a|_{\beta S} = \text{Id}_{\beta S}$, so $\beta \pi_a(p) = p$. Since $q$ is idempotent and $\beta \pi_a$ is a semigroup homomorphism, $\beta \pi_a(q) \in \beta S$ is also idempotent. Further, 
+    \[
+    \beta \pi_a(q)p = \beta \pi_a(q)\beta \pi_a(p) = \beta\pi_a(qp) = \beta \pi_a(q)
+    \]
+
+    and
+    \[
+    p\beta \pi_a(q) = \beta \pi_a(p)\beta \pi_a(q) = \beta\pi_a(qp) = \beta \pi_a(q)
+    \]
+
+    so $\beta \pi_a(q)p = p\beta \pi_a(q) = \beta \pi_a(q)$. In particular, $\beta \pi_a(q) = \beta \pi_a(q)p \in (\beta S)p$, where $(\beta S)p$ is a minimal left ideal in $\beta S$. By minimality, $(\beta S)\beta \pi_a(q) = (\beta S)p$, and there exists $u \in \beta S$ with $p = u\beta \pi_a(q)$. As such, 
+    \[
+    \beta \pi_a(q) = p\beta \pi_a(q) = u\beta \pi_a(q)^2 = u\beta \pi_a(q) = p
+    \]
+
+    Let $S = \bigsqcup_{j = 1}^r C_r$ be a finite colouring of $S$, then there exists a unique $1 \le i \le r$ such that $C_i \in p$. Since $\beta \pi_a(q) = p$ for all $a \in A$, $\bigcap_{a \in A}\pi_a^{-1}(C_i) \in q$. As $q \in \beta T \setminus \beta S = \beta V$, $V \in q$, and $V \cap \bigcap_{a \in A}\pi_a^{-1}(C_i) \ne \emptyset$. Therefore any element of $V \cap \bigcap_{a \in A}\pi_a^{-1}(C_i)$ generates a combinatorial line, contained in $C_i$. 
+
+    (F): Let $r \ge 1$ and suppose for contradiction that for every $N \in \natp$, there exists an $r$-colouring $c_N: A^N \to \bracsn{1, \cdots, r}$ such that $c_N$ contains no monochromatic combinatorial line. Define
+    \[
+    c: S \to \bracsn{1, \cdots, r} \quad w \mapsto c_{|w|}(w)
+    \]
+
+    then $c$ is a $r$-colouring of $S$ containing no monochromatic combinatorial lines, which contradicts (IF).
+\end{proof}
+
+
+\begin{theorem}[van der Waerden]
+\label{theorem:van-der-waerden}
+    For every $k \in \natp$ and finite colouring $c: \integer \to \bracsn{1, \cdots, r}$, there exists $a \in \integer$ and $n \ge 1$ such that $\bracsn{a + jn|0 \le j \le k}$ is monochromatic. In other words, there exists monochromatic arithmetic progressions of arbitrary length. 
+\end{theorem}
+\begin{proof}
+    Let $A = \bracsn{1, \cdots, k}$ and
+    \[
+    \tilde c: A^N \to \bracsn{1, \cdots, r} \quad x_1 \cdots x_N \mapsto c(x_1 + \cdots + x_N)
+    \]
+
+    By the \hyperref[Hales-Jewett Theorem]{theorem:hales-jewett}, there exists $N = N(r)$ such that $\tilde c$ contains a monochromatic combinatorial line. In other words, there exists a variable word $v \in (A \sqcup \bracsn{*})^N \setminus A^N$ such that the set
+    \[
+    L(v) = \bracsn{\pi_j(v)|1 \le j \le k}
+    \]
+
+    is monochromatic. Let $J = \bracsn{1 \le j \le N|v_j = *}$, then as $v \not\in A^N$, $J$ is non-empty. Take $n = |J|$ and $a = \sum_{j \not\in J}v_j$, then for each $1 \le i \le k$, 
+    \[
+    \tilde c(\pi_i(v)) = c\paren{i\sum_{j \in J}j + \sum_{j \not\in J}v_j} = c\paren{in +a}
+    \]
+
+    In particular, for each $1 \le i, i' \le k$, 
+    \[
+    c\paren{in +a} = \tilde c(\pi_i(v)) = \tilde c(\pi_{i'}(v)) = c(i'n +a)
+    \]
+
+    Therefore the arithmetic progression $\bracsn{a + in|1 \le i \le k}$ is monochromatic.
+\end{proof}

src/dyn/top/index.tex

@@ -4,3 +4,5 @@
 \input{./def.tex}
 \input{./minimal.tex}
 \input{./recurrence.tex}
+\input{./stone-cech.tex}
+\input{./hales-jewett.tex}

src/dyn/top/recurrence.tex

@@ -14,6 +14,7 @@
     \]
 \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.

src/dyn/top/stone-cech.tex

@@ -0,0 +1,241 @@
+\section{The Universal Flow}
+\label{section:stone-cech-flow}
+
+\begin{definition}[Greatest Ambit]
+\label{definition:stone-cech-flow}
+    Let $G$ be a discrete group, and
+    \[
+    \beta L: G \to \text{Aut}(\beta G) \quad \beta L_g(\fU) = g\fU
+    \]
+
+    then
+    \begin{enumerate}
+        \item $(\beta L: G \curvearrowright \beta G, 1_G)$ is an ambit.  
+        \item[(U)] For any ambit $(G \curvearrowright X, x_0)$, there exists a unique morphism of pointed $G$-flows $\pi: (\beta G, 1_G) \to (X, x_0)$. Moreover, $\pi$ is surjective. 
+    \end{enumerate}
+    
+    Any ambit $(Y, y_0)$ satisfying (U) is the \textbf{greatest $G$-ambit}. 
+\end{definition}
+\begin{proof}
+    (1): Let
+    \[
+    L: G \to \text{Perm}(G) \quad L_g(h) = gh
+    \]
+
+    be the left translation action on itself. For each $g \in G$, the functoriality of the \hyperref[Stone-Čech compactification]{definition:stone-cech} implies that $L_g$ extends into a continuous map $\beta L_g: \beta G \to \beta G$. By uniqueness of continuous extensions, $\beta L_g(\fU) = g\fU$ for all $\fU \in \beta G$. In particular, $\text{Id}_{\beta G} = \beta L_{e}$. 
+    
+    Moreover, for any $g, h \in G$, $\beta L_{gh} = \beta L_{g} \circ \beta L_{h}$, so $g \mapsto L_g$ is a homomorphism. Finally, for any $g \in G$, 
+    \[
+    \beta L_{g^{-1}} \circ \beta L_{g} = \beta L_{1_G} = \beta L_{g} \circ \beta L_{g^{-1}}
+    \]
+    
+    and $\beta L_g \in \text{Aut}(\beta G)$. 
+
+    Since $\bracsn{L_g1_G|g \in G} = G$ is dense in $\beta G$, $(G \curvearrowright \beta G, 1_G)$ is an ambit. 
+
+    (2): For each $g \in G$, let $\pi_0(g) = gx_0$, then $\pi_0 \in C(G; X)$, $\pi_0(1_G) = x_0$, and the following diagram commutes
+    \[
+    \xymatrix{
+    G \ar@{->}[r]^{\pi_0} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
+    G \ar@{->}[r]_{\pi_0} & X
+    }
+    \]
+
+    for all $g \in G$. By universal property of $\beta G$, $\pi_0$ admits a unique continuous extension $\pi \in C(\beta G; X)$. By uniqueness of continuous extensions, the following diagram commutes
+    \[
+    \xymatrix{
+    \beta G \ar@{->}[r]^{\pi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
+    \beta G \ar@{->}[r]_{\pi} & X
+    }
+    \]
+
+    for all $g \in G$. Finally, since $\beta G$ is compact and $X$ is Hausdorff, $\pi(\beta G)$ is \hyperref[also compact]{proposition:compact-extensions} and \hyperref[hence closed]{proposition:compact-closed}. As $(X, x_0)$ is an ambit and $\pi(\beta G) \supset Gx_0$, $\pi(\beta G) \supset \ol{Gx_0} = X$. 
+\end{proof}
+
+\begin{definition}
+\label{definition:stone-cech-semigroup}
+    Let $G$ be a discrete group. For each $\fU, \fV \in \beta G$, let $\fU \cdot \fV = \lim_{g \to \fU}g\fV$, then
+    \begin{enumerate}
+        \item For each $A \subset G$, $A \in \fU \cdot \fV$ if and only if $\bracsn{g \in G|g^{-1}A \in \fV} \in \fU$. 
+        \item $\beta G$ equipped with the composition law $(\fU, \fV) \mapsto \fU \cdot \fV$ is a semigroup with identity $1_G$. 
+        \item (\textbf{Right Topological}) For each $\fV \in \beta G$, the mapping $\beta G \to \beta G$ defined by $\fU \mapsto \fU \cdot \fV$ is continuous. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (3): By (U2) of the \hyperref[Stone-Čech compactification]{definition:stone-cech}. 
+\end{proof}
+
+\begin{definition}[Induced Action]
+\label{definition:stone-cech-semigroup-induced-action}
+    Let $G$ be a group, $(X, x_0)$ be a pointed $G$-flow, and $p_x: G \to X$ be the orbit map, then the mapping
+    \[
+    \pi: \beta G \to C(X; X) \quad \pi_\fU x  = \fU x = \lim_{g \to \fU}gx
+    \]
+
+    is a semigroup action of $\beta G$ on $X$, known as the \textbf{induced action} of $\beta G$ on $X$. 
+\end{definition}
+
+
+
+\begin{proposition}
+\label{proposition:tds-return-set-stone-cech}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$. For each $y \in X$ and $\fU \in \beta G$, then $y = \fU x$ if and only if $N(x, U) \in \fU$.
+\end{proposition}
+\begin{proof}
+    By definition of filter limits, $y = \fU x$ if and only if $p_x^{-1}(U) \in \fU$ for all $U \in \cn_X(y)$. Since $p_x^{-1}(U) = N(x, U)$, $y = \fU x$ if and only if $N(x, U) \in \fU$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:z-stone-cech-fixed-point}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $U \in \cn_X(x)$, $N(x, U)$ is infinite. 
+        \item There exists $\fU \in \beta G \setminus G$ such that $\fU x = x$. 
+    \end{enumerate}
+\end{proposition}
+% "obvious"
+
+
+\begin{definition}[Left Ideal]
+\label{definition:left-ideal-semigroup}
+    Let $G$ be a semigroup and $\emptyset \ne I \subset G$, then $I$ is a \textbf{left ideal} if for any $g \in G$ and $h \in I$, $gh \in I$. The left ideal $I$ is \textbf{minimal} if there exists no left ideal $J \subsetneq I$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:stone-cech-ideal-subflow}
+    Let $G$ be a group and $\emptyset \ne I \subset \beta G$. If $I$ is closed, then the following are equivalent:
+    \begin{enumerate}
+        \item $L$ is a $G$-subflow of $\beta G$. 
+        \item $L$ is a closed left ideal of $\beta G$. 
+    \end{enumerate}
+
+    Moreover, for arbitrary $\emptyset \ne I \subset \beta G$, the following are equivalent
+    \begin{enumerate}[label=(M\arabic*)]
+        \item $L$ is a minimal $G$-subflow of $\beta G$.
+        \item $L$ is a minimal left ideal of $\beta G$. 
+    \end{enumerate}
+    
+    In particular, minimal left ideals of $\beta G$ are closed. 
+\end{proposition}
+
+\begin{corollary}
+\label{corollary:stone-cech-minimal-ideal-existence}
+    Let $G$ be a group, then $\beta G$ admits a minimal ideal. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:tds-minimal-existence} and \autoref{proposition:stone-cech-ideal-subflow}. 
+\end{proof}
+
+\begin{definition}[Universal Minimal Flow]
+\label{definition:universal-minimal-flow}
+    Let $G$ be a group, then there exists a flow $L$ such that:
+    \begin{enumerate}
+        \item $G \curvearrowright L$ is a minimal $G$-flow. 
+        \item[(U)] For any minimal $G$-flow $X$, there exists a continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
+        \[
+        \xymatrix{
+        L \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
+        L \ar@{->}[r]_{\Phi} & X
+        }
+        \]
+
+        for all $g \in G$.
+    \end{enumerate}
+
+    The flow $(M, \alpha)$ is the \textbf{universal minimal $G$-flow}.
+\end{definition}
+\begin{proof}
+    (1): Using \autoref{corollary:stone-cech-minimal-ideal-existence}, let $L \subset \beta G$ be a minimal left ideal, then $G \curvearrowright L$ is a minimal $G$-flow. 
+
+    (U): Let $G \curvearrowright X$ be a minimal $G$-flow, $x_0 \in X$, and $m_0 \in M$. By (U) of the \hyperref[universal flow]{definition:stone-cech-flow}, there exists a unique continuous map $\phi \in C(\beta G; X)$ such that $\phi(1_G) = x_0$ and the following diagram commutes
+    \[
+    \xymatrix{
+    \beta G \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
+    \beta G \ar@{->}[r]_{\Phi} & X
+    }
+    \]
+
+    for all $g \in G$. 
+
+    Let $\Phi = \phi|_{L} \in C(L; X)$, then as $L$ is $G$-invariant, $\Phi(L) \supset G\phi(x_0)$. By \autoref{proposition:compact-extensions}, $\Phi(L) = \ol{G\phi(x_0)} = X$.  
+\end{proof}
+
+
+\begin{definition}[Minimal Idempotent]
+\label{definition:tds-minimal-idempotent}
+    Let $S$ be a compact Hausdorff right topological semigroup and $e \in S$ be an idempotent, then $e$ is \textbf{minimal} if there exists a minimal left ideal $I \subset G$ such that $e \in I$. 
+\end{definition}
+
+\begin{theorem}[Ellis]
+\label{theorem:ellis-semigroup}
+    Let $S$ be a non-empty compact Hausdorff right topological semigroup and $I \subset S$ be a closed left ideal, then there exists a minimal left ideal $I \subset S$ and a minimal idempotent $e \in I$. 
+\end{theorem}
+\begin{proof}
+    By Zorn's Lemma, $I$ admits a minimal non-empty closed right topological subsemigroup $K \subset S$. For each $a \in K$, $Ka$ is closed by continuity of right multiplication and \autoref{proposition:compact-extensions}. By minimality of $K$, $K = Ka$, so there exists $e \in K$ such that $a = ea$. Let $E = \bracsn{b \in K|ba = a}$, then $E$ is also a closed and right topological subsemigroup. By minimality of $K$, $E = K$, and in particular $a \in E$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:stone-cech-minimal-idempotent}
+    Let $G$ be an infinite group, then $\beta G$ admits a minimal idempotent in $\beta G \setminus G$. 
+\end{corollary}
+\begin{proof}
+    Let $L \subset (\beta G \setminus G)$ be a minimal left ideal, then $L$ is a non-empty compact Hausdorff right topological semigroup. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, there exists $e \in L$ with $e^2 = e$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:stone-cech-flow-minimal-idempotent}
+    Let $G$ be an infinite group, $G \curvearrowright X$ be a minimal $G$-flow, then there exists a minimal idempotent $\fU \in \beta G \setminus G$ such that $\fU x = x$. 
+\end{proposition}
+\begin{proof}
+    Let $L \subset \beta G$ be a minimal left ideal, then the mapping $\fU \mapsto \fU x$ is a continuous surjection. As $x \in X = \bracsn{\fU x|x \in L}$, there exists $\fU \in L$ with $\fU x = x$. Let $F = \bracsn{\fU \in L|\fU x = x}$, then $F$ is a non-empty closed right topological subsemigroup of $L$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $F$ contains an idempotent. 
+\end{proof}
+
+
+\begin{theorem}[Infinite Ramsey Theorem for Pairs]
+\label{theorem:ramsey-pairs}
+    Let
+    \[
+    [\natp]^2 = \bracsn{\bracsn{m, n} \in \nat^2|m, n \in \natp, m \ne n}
+    \]
+    
+    be the set of unordered pairs of distinct elements of $\natp$ and $c: [\natp]^2 \to \bracsn{1, \cdots, r}$, then there exists $A \subset \natp$ infinite such that for any $m, n, m', n' \in A$, $c(\bracsn{m, n}) = c(\bracsn{m', n'})$. 
+\end{theorem}
+\begin{proof}
+    Let $\fU \subset 2^{\natp}$ be a free ultrafilter. For each $n \in \natp$ and $1 \le i \le r$, let $A(n, i) = \bracsn{m > n| c(\bracsn{m,n}) = i}$, then $\natp = \bracsn{1, \cdots, n} \bigsqcup_{i = 1}^r A(n, i)$. As $\fU$ is a free ultrafilter, there exists a unique $1 \le i(n) \le r$ such that $A(n, i(n)) \in \fU$. 
+
+    For each $1 \le i \le r$, let $B_i = \bracsn{n \in \natp|i(n) = i}$, then $\natp = \bigsqcup_{i = 1}^r B_i$. As $\fU$ is an ultrafilter, there exists a unique $1 \le i \le r$ with $B_i \in \fU$. 
+    
+    Let $n_1 \in B_i$, so $i(n_1) = i$. Assume inductively that $\bracsn{n_j}_1^N \subset \natp$ has been defined such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $1 \le j \le N - 1$, $n_j < n_{j+1}$. 
+        \item For each $1 \le j \le N$, $i(n_j) = i$. 
+        \item For each $1 \le j < k \le N$, $c(\bracsn{n_j, n_k}) = i$. 
+    \end{enumerate}
+
+    Let $n_{N+1} \in B_i \cap \bigcap_{j = 1}^N A(n_j, i)$, then:
+    \begin{enumerate}
+        \item As $n_{N+1} \in A(n_N, i)$, $n_{N+1} > n_N$. 
+        \item Since $n_{N+1} \in B_i$, $i(n_{N+1}) = i$. 
+        \item For each $1 \le j \le N$, $n_{N+1} \in A(n_j, i) = A(n_j, i(n))$, so $c(\bracsn{n_j, n_{N+1}}) = i$. 
+    \end{enumerate}
+
+    By induction, there exists $\seq{n_j} \subset \natp$ such that $c(\bracsn{n_j, n_k}) = i$ for all $j, k \in \natp$ with $j \ne k$. 
+\end{proof}
+
+
+\begin{lemma}[Idempotent Refinement]
+\label{lemma:idempotent-refinement}
+    Let $S$ be a semigroup and $f \in \beta S$ be idempotent, then there exists a minimal idempotent $e \in \beta S$ such that $ef = fe = e$. 
+\end{lemma}
+\begin{proof}
+    By \autoref{proposition:compact-extensions}, $(\beta S)f$ is a closed left ideal in $S$. By \hyperref[Ellis' Theorem]{theorem:ellis-semigroup}, $(\beta S)f$ contains a minimal left ideal $L$, and a minimal idempotent $r \in L$. As $L \subset (\beta S)f$, there exists $t \in \beta S$ such that $r = tf$. 
+
+    Let $e = fr \in L$, then $e^2 = frfr = fr^2 = fr = e$. Moreover, 
+    \[
+    ef = frf = ftf^2 = ftf = fr = e = fr = ffr = fe
+    \]
+\end{proof}
+
+
+
+

src/fa/convex/legendre.tex

@@ -269,7 +269,7 @@
 
 \begin{corollary}
 \label{corollary:separable-legendre}
-    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ be convex and lower semicontinuous, then there exists $\seq{(\phi_n, \alpha_n)} \subset F \times \real$ such that for each $x \in E$, 
+    Let $\dpn{E, F}{\lambda}$ be a duality over $\real$ with $E$ being separable, and $f: E \to (-\infty, \infty]$ with $f \ne \infty$ be convex and lower semicontinuous, then there exists $\seq{(\phi_n, \alpha_n)} \subset F \times \real$ such that for each $x \in E$, 
     \[
     f(x) = \sup_{n \in \natp} \dpn{x, \phi_n}{\lambda} - \alpha_n
     \]

src/fa/lc/compact.tex

@@ -29,7 +29,7 @@
 
 \begin{definition}[Extreme Subset]
 \label{definition:extreme-subset}
-    Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$, then $A$ is \textbf{extreme set} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.  
+    Let $E$ be a vector space over $\real$, $K \subset E$ be convex, and $A \subset K$ be convex, then $A$ is \textbf{extreme} if for any $x \in A$ and $y, z \in K$ such that $x \in (y, z)$, $y, z \in A$ as well.  
 \end{definition}
 
 \begin{lemma}
@@ -40,6 +40,14 @@
     Let $y, z \in K$ such that $x \in (y, z)$. Since $A$ is extreme, $y, z \in A$. As $x$ is extremal in $A$, $x = y = z$, so $x$ is extremal in $K$. 
 \end{proof}
 
+\begin{lemma}
+\label{lemma:extremal-face-map}
+    Let $E, F$ be locally convex spaces over $\real$, $K \subset E$ be convex, $T: K \to F$ be affine, and $A \subset T(K)$ be extreme, then $T^{-1}(A)$ is also extreme. 
+\end{lemma}
+\begin{proof}
+    Let $x \in T^{-1}(A)$ and $y, z \in K$ with $x \in (y, z)$, then $Tx \in (Ty, Tz)$. As $A$ is extreme, $Ty, Tz \in A$, so $y, z \in T^{-1}(A)$. 
+\end{proof}
+
 \begin{lemma}
 \label{lemma:extremal-face}
     Let $E$ be a locally convex space over $\real$, $K \subset E$ be non-empty and compact, and $\phi \in E^*$. Let $\alpha = \sup\bracs{\dpn{x, \phi}{E}|x \in K}$, then $A = \bracs{\phi = \alpha} \cap K$ is a non-empty extreme subset of $K$. 
@@ -50,18 +58,19 @@
 
 \begin{theorem}[Krein-Milman]
 \label{theorem:krein-milman}
-    Let $E$ be a separated locally convex space over $\real$ and $K \subset E$ be a compact convex set, then $K$ is the closed convex hull of its extreme points. 
+    Let $E$ be a separated locally convex space over $\real$ and $\emptyset \ne 
+    K \subset E$ be a compact convex set, then 
+    \begin{enumerate}
+        \item Every non-empty, closed, extreme subset of $K$ contains an extreme point of $K$. 
+        \item $K$ is the closed convex hull of its extreme points. 
+    \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 1.12.5]{Bogachev}}}. ]
-    Assume without loss of generality that $K \ne \emptyset$. 
-    
-    Let $K_0 \subset K$ be a closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. 
+    (1): Let $K_0 \subset K$ be a non-empty, closed, extreme subset of $K$, and $\mathcal{E}(K_0) \subset 2^K$ be the collection of all non-empty closed extreme subsets of $K$ contained in $K_0$. Since $K_0 \in \mathcal{E}(K_0)$, $\mathcal{E}(K_0) \ne \emptyset$. Since $K$ is compact, for any chain $\mathcal{C} \subset \mathcal{E}$, $\bigcap_{A \in \mathcal{C}}A$ is also non-empty, closed, and extreme. 
     
     By Zorn's lemma, there exists a minimal element $A$ of $\mathcal{E}(K_0)$. Let $x, y \in A$, $\phi \in E^*$, and $\alpha = \sup_{z \in A}\dpn{z, \phi}{E}$, then $\bracs{\phi = \alpha} \cap K$ is a non-empty, closed, and extreme subset of $K$ by \autoref{lemma:extremal-face}, so $A \cap \bracs{\phi = \alpha}$ is also extreme. By minimality of $A$, $A \subset \bracs{\phi = \alpha}$. Thus by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $A$ consists of exactly one point. In which case, for any $y, z \in E$ with $A \subset (y, z) \subset K$, $y = z \in A$. Therefore there exists an extreme point of $K$ in $K_0$. 
     
-    Since $K$ itself is an extreme subset, a minimal element of $\mathcal{E}(K)$ represents an extreme point, so $K$ admits at least one extreme point. 
-    
-    Now, let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By the preceding discussion, there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
+    (2): Let $C$ be the collection of all extreme points in $K$. Assume for contradiction that $\ol{\conv}(C) \subsetneq K$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $x \in K$ and $\phi \in E^*$ such that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \dpn{x, \phi}{E}$. Let $\alpha = \sup_{z \in K}\dpn{z, \phi}{E}$, then by \autoref{lemma:extremal-face}, $K \cap \bracs{\phi = \alpha}$ is a non-empty, closed, and extreme subset of $K$. By (1), there exists an extreme point of $K$ in $K \cap \bracs{\phi = \alpha}$, which contradicts the fact that $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
 \end{proof}
 
 \begin{theorem}[Markov-Kakutani Fixed Point Theorem]

src/fa/norm/hilbert.tex

@@ -369,4 +369,160 @@ A significant property of Hilbert spaces is that every closed subspace is comple
 \end{proof}
 
 
+\begin{definition}[Quadratic Function]
+\label{definition:hilbert-quadratic-function}
+    Let $H$ be a complex Hilbert space and $q: H \to \complex$, then $q$ is \textbf{quadratic} if:
+    \begin{enumerate}
+        \item For each $\lambda \in \complex$ and $\xi \in H$, $q(\lambda \xi) = |\lambda|^2 q(\xi)$. 
+        \item (\textbf{Parallelogram Law}) For each $\xi, \eta \in H$, 
+        \[
+        q(\xi + \eta) + q(\xi - \eta) = 2q(\xi) + 2q(\eta)
+        \]
+    \end{enumerate}
+
+    and \textbf{bounded} if
+    \begin{enumerate}[start=2]
+        \item There exists $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{proposition}
+\label{proposition:quadratic-function-sesquilinear}
+    Let $H$ be a complex Hilbert space, $q: H \to \complex$ be a bounded \hyperref[quadratic function]{definition:hilbert-quadratic-function}, and
+    \[
+        H \times H \to \complex \quad (\xi, \eta) \mapsto \dpn{\xi, \eta}{q} := \frac{1}{4}\sum_{k = 0}^3i^kq(\xi + i^k\eta)
+    \]
+
+    then
+
+
+    \begin{enumerate}
+        \item $\dpn{\cdot, \cdot}{q}$ is the unique bounded sesquilinear form on $H$ such that $\dpn{\xi, \xi}{q} = q(\xi)$ for all $\xi \in H$. 
+        \item There exists a unique $T \in B(H)$ such that $q(\xi) = \dpn{T\xi, \xi}{H}$ for all $\xi \in H$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Uniqueness is given by \hyperref[polarisation]{proposition:polarisation-complex}. Existence requires a large number of checkpoints, which are detailed below. Let $M \ge 0$ such that $|q(\xi)| \le M\norm{\xi}_H^2$ for all $\xi \in H$. 
+    \begin{enumerate}[label=(1.\roman*)]
+        \item For each $\xi, \eta, \theta \in H$, 
+        \[
+        \dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q} = 2 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
+        \]
+
+        and
+        \[
+        \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2 \angles{\xi, \frac{\eta + \theta}{2}}_{q}
+        \]
+        \item For each dyadic rational number $\lambda \in \mathbb{D}$ with $\lambda \in [0, 1]$, 
+        \[
+        \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+        \]
+
+        \item For each $\xi, \eta, \theta \in H$, 
+        \[
+        \dpn{\xi + \theta, \eta}{q} = \dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q}
+        \]
+
+        and
+        \[
+        \dpn{\xi, \eta + \theta}{q} = \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q}
+        \]
+
+        \item For each dyadic rational number $\lambda \in \mathbb{D}$, 
+        \[
+        \dpn{\lambda\xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+        \]
+
+        \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le M(\norm{\xi}_H^2 +\norm{\eta}_H^2)$.
+        \item For each $\xi, \eta \in H$, $|\dpn{\xi, \eta}{q}| \le 3M \norm{\xi}_H\norm{\eta}_H$. 
+
+        \item $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form on $H$.
+        \item For all $\xi \in H$, $\dpn{\xi, \xi}{q} = q(\xi)$. 
+    \end{enumerate}
+    
+
+    (1.i): Let $\xi, \eta, \theta \in H$, then
+    \[
+        4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q}  = \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)}
+    \]
+
+    For each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $((\xi + \theta)/2 + i^k\eta, (\xi - \theta)/2)$, 
+    \[
+    q(\xi + i^k\eta) + q(\theta + i^k\eta) = 2\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}}
+    \]
+
+    so
+    \begin{align*}
+        4\dpn{\xi, \eta}{q} + 4 \dpn{\theta, \eta}{q}  &= \sum_{k = 0}^3 i^k\braks{q(\xi + i^k\eta) + q(\theta + i^k\eta)} \\ 
+        &= 2\sum_{k = 0}^3 i^k\braks{q\paren{\frac{\xi + \theta}{2} + i^k\eta} + q\paren{\frac{\xi - \theta}{2}}} \\
+        &= 2\sum_{k = 0}^3 i^k{q\paren{\frac{\xi + \theta}{2} + i^k\eta}} = 8 \angles{\frac{\xi + \theta}{2}, \eta}_{q}
+    \end{align*}
+
+    Similarly, for each $0 \le k \le 3$, by the Parallelogram Law applied to the pair $(\xi + i^k(\eta + \theta)/2, i^k(\eta - \theta)/2)$,
+    \[
+    q(\xi + i^k\eta) + q(\xi + i^k\theta) = 2\braks{q\paren{\xi + i^k\frac{\eta + \theta}{2}} + q\paren{i^k\frac{\eta - \theta}{2}}}
+    \]
+
+    and
+    \[
+    \dpn{\xi, \eta}{q} + \dpn{\xi, \theta}{q} = 2\angles{\xi, \frac{\eta + \theta}{2}}_q
+    \]
+    
+
+    
+
+    (1.ii): By definition of $\dpn{\cdot, \cdot}{q}$, the claim holds for $0$ and $1$. Let $n \in \natz$ and suppose inductively that the claim holds for all $\lambda \in \mathbb{D}_n \cap [0, 1]$. Take $\lambda \in \mathbb{D}_{n+1} \cap [0, 1]$, then there exists $\mu, \nu \in \mathbb{D}_n \cap [0, 1]$ such that $\lambda = (\mu + \nu)/2$. In which case, by (1.i),
+    \begin{align*}
+        \dpn{\lambda \xi, \eta}{q} &= \angles{\frac{\mu + \nu}{2}\xi, \eta}_{q} = \frac{1}{2}\dpn{\mu\xi, \eta}{q} + \frac{1}{2}\dpn{\nu\xi, \eta}{q} \\
+        &= \frac{\mu + \nu}{2}\dpn{\xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}
+    \end{align*}
+
+    By induction, $\dpn{\lambda \xi, \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. 
+
+    Similarly, $\dpn{\xi, \lambda \eta}{q} = \lambda \dpn{\xi, \eta}{q}$ for all $\lambda \in \mathbb{D} \cap [0, 1]$. 
+
+    (1.iii): Let $\xi, \eta, \theta \in H$, then by (1.i) and (1.ii), 
+    \[
+        \frac{1}{2}(\dpn{\xi, \eta}{q} + \dpn{\theta, \eta}{q}) = \angles{\frac{\xi + \theta}{2}, \eta}_q = \frac{1}{2}\dpn{\xi + \theta, \eta}{q}
+    \]
+    
+    (1.iv): By (1.ii) and repeated application of (1.iii), 
+    \[
+    \dpn{\lambda \xi, \eta}{q} = \lambda\dpn{\xi, \eta}{q} = \dpn{\xi, \lambda \eta}{q}
+    \]
+
+    for all $\lambda \in \mathbb{D}$. 
+    
+    (1.v): Let $\xi, \eta \in H$, then
+    \begin{align*}
+        |\dpn{\xi, \eta}{q}| &\le \frac{1}{4}\sum_{k = 0}^3 |q(\xi + i^k\eta)| \le \frac{M}{4}\sum_{k = 0}^3\normn{\xi + i^k\eta}_H^2 \\
+        &\le \frac{M}{4}\sum_{k = 0}^3 [\norm{\xi}_H^2 + \norm{\eta}_H^2 + 2\text{Re}\dpn{i^{-k}\xi, \eta}{H}] = M(\norm{\xi}_H^2 + \norm{\eta}_H^2)
+    \end{align*}
+    
+    
+    
+    (1.vi): Let $\xi, \eta \in H$ and assume without loss of generality that $\xi, \eta \ne 0$. Since $\norm{\eta}_H/\norm{\xi}_H > 0$, there exists $n \in \integer$ such that $\norm{\eta}_H/\norm{\xi}_H \in [2^n, 2^{n+1}]$. In which case, by (1.iv), 
+    \begin{align*}
+    |\dpn{\xi, \eta}{q}| &= 2^{-n}|\dpn{2^n\xi, \eta}{q}| \le 2^{-n}M(4^n \norm{\xi}_H^2 + \norm{\eta}_H^2) = 2^nM\norm{\xi}_H^2 + 2^{-n}M\norm{\eta}_H^2 \\
+    &\le M\norm{\xi}_H\norm{\eta}_H + 2M\norm{\xi}_H\norm{\eta}_H = 3M\norm{\xi}_H\norm{\eta}_H
+    \end{align*}
+    
+    (1.vii): By (1.iv) and (1.vi), $\dpn{\cdot, \cdot}{q}$ is a $\real$-bilinear form on $H$. For any $\xi, \eta \in H$, by absolute homogeneity of $q$, 
+    \begin{align*}
+        \dpn{i\xi, \eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(i\xi + i^k\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k-1}\eta) \\
+        &= \frac{i}{4}\sum_{k = 0}^3 i^{k-1} q(\xi + i^{k-1}\eta) = i\dpn{\xi, \eta}{q} \\
+        \dpn{\xi, i\eta}{q} &= \frac{1}{4}\sum_{k = 0}^3 i^k q(\xi + i^{k+1}\eta) = \frac{1}{4}\sum_{k = 0}^3 i^k q(i^{-1}\xi + i^{k}\eta) \\
+        &= \dpn{-i\xi, \eta}{q} =  -i \dpn{\xi, \eta}{q}
+    \end{align*}
+
+    Therefore $\dpn{\cdot, \cdot}{q}$ is a sesquilinear form. 
+    
+    (1.viii): Let $\xi \in H$, then by absolute homogeneity of $q$, 
+    \begin{align*}
+        \dpn{\xi, \xi}{q} &= \frac{1}{4}\braks{q(\xi + \xi) - q(\xi - \xi) + iq(\xi + i\xi) - iq(\xi - i\xi)} \\
+        &= \frac{1}{4}\braks{4q(\xi) + iq(\xi + i\xi) - iq(i\xi + \xi)} = q(\xi)
+    \end{align*}
+    
+    (2): By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}.
+\end{proof}
 

src/fa/norm/normed.tex

@@ -161,3 +161,45 @@
     then the identity $(K^d, \norm{\cdot}_1) \to (K^d, [\cdot])$ is continuous. Let $S = \bracsn{x \in K^d|\ \norm{x}_1 = 1}$, then $S$ is compact with respect to $\norm{\cdot}_1$ and hence compact with respect to $[\cdot]$. By continuity of $[\cdot]$, $\min_{x \in S}[x]$ exists and is non-zero. 
 \end{proof}
 
+
+\begin{theorem}
+\label{theorem:reflexive-banach-spaces}
+    Let $E$ be a Banach space over $K \in \RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item $E$ is reflexive. That is, $E = E^{**}$. 
+        \item $\ol{B_E(0, 1)}^E$ is weakly compact. 
+        \item The weak and weak* topologies on $E^*$ coincide.
+        \item $E^*$ is reflexive. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. 
+
+    (2) $\Rightarrow$ (1): By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak} and \autoref{proposition:compact-extensions},
+    \[
+    \ol{B_{E^{**}}(0, 1)}^{E^{**}} = \ol{B_E(0, 1)}^{\sigma(E^{**}, E^*)} = \ol{B_E(0, 1)}^E
+    \]
+
+    (1) $\Leftrightarrow$ (3): By \autoref{lemma:duality-dual}. 
+
+    (4) $\Rightarrow$ (1): By (1) $\Rightarrow$ (4), $E^{**}$ is reflexive. Since $\ol{B_{E}(0, 1)}^{E}$ is norm closed in $E^{**}$, it is weakly closed in $E^{**}$, and hence weak*-closed in $E^{**}$. By \hyperref[Goldstine's Theorem]{theorem:goldstine-weak}, $\ol{B_E(0, 1)}^E = \ol{B_{E^{**}}(0, 1)}^{E^{**}}$. 
+\end{proof}
+
+% This should be a three space property. 
+\begin{corollary}
+\label{corollary:reflexive-banach-three-spaces}
+    Let $E$ be a Banach space over $K \in \RC$ and $F \subset E$ be a closed subspace, then $F$ and $E/F$ are reflexive if and only if $E$ is reflexive. 
+\end{corollary}
+\begin{proof}
+    ($F$): Assume that $E$ is reflexive. As $F$ is closed, it is weakly closed. Thus $\ol{B_F(0, 1)} = \ol{B_E(0, 1)} \cap F$ is weakly compact. By (2) of \autoref{theorem:reflexive-banach-spaces}, $E$ is reflexive. 
+
+    ($E/F$): Assume that $E$ is reflexive. Let $\pi: E \to E/F$ be the canonical map, then $\pi$ is also weakly continuous. Hence $\pi(\ol{B_E(0, 1)})$ is weakly compact in $E/F$. By (2) of \autoref{theorem:reflexive-banach-spaces}, $E/F$ is reflexive. 
+
+    ($E$): Assume that $F$ and $E/F$ are reflexive. Let $x_0 \in E^{**}$, and $T: F^\perp \to (E/F)^*$ be the canonical isomorphism, then $x_0 \circ T^{-1} \in (E/F)^{**}$. Given that $E/F$ is reflexive, there exists $x \in E$ such that $x_0 \circ T^{-1} = x + F$. In particular, for any $\phi \in F^\perp$, 
+    \[
+    \dpn{\phi, x_0}{E^*} = \dpn{T\phi, x_0 \circ T^{-1}}{(E/F)^*} = \dpn{x + F, T\phi}{E/F} = \dpn{x, \phi}{E} = \dpn{\phi, x}{E^{**}}
+    \]
+    
+    so $x_0 - x \in F^{\perp\perp}$. By the \hyperref[Bipolar Theorem]{theorem:bipolar}, $F^{\perp\perp} = F^{**} = F$. Therefore $x_0 - x \in F$, and $x_0 \in E$. 
+\end{proof}
+

src/fa/norm/separable.tex

@@ -12,7 +12,7 @@
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 
+    (1), (2): Let $D \subset E$ be a countable dense subset. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the topology of pointwise convergence on $E$ and the topology of pointwise convergence on $E$ coincide on $S$, so $S$ is embedded as a subspace of $K^D$. By \autoref{theorem:uniform-metrisable}, $K^D$ is metrisable. By \autoref{proposition:separable-product}, $K^D$ is separable. Thus $S$ is also metrisable and separable by \autoref{proposition:separable-metric-space}. 
 
     (3): For any $A \subset E$, $A = \bigcup_{n \in \natp}A \cap nS$. By \autoref{proposition:separable-metric-space}, $A \cap nS$ is separable for each $n \in \natp$. Therefore $A$ is also separable. 
 \end{proof}
@@ -63,8 +63,8 @@
     Finally, \autoref{lemma:compact-embed} provides yet another linear isometric embedding $C(\mathcal{C}; K) \to C([0, 1]; K)$. Composing the above maps as follows
     \[
     \xymatrix{
-E \ar@{->}[r] & C(B; K) \ar@{->}[r] & C(2^{{\mathbb N}^+}; K) \ar@{->}[r] & C(\mathcal{C}; K) \ar@{->}[r] & C([0, 1]; K)
-}
+    E \ar@{->}[r] & C(B; K) \ar@{->}[r] & C(2^{{\mathbb N}^+}; K) \ar@{->}[r] & C(\mathcal{C}; K) \ar@{->}[r] & C([0, 1]; K)
+    }
     \]
 
     yields the desired embedding. 

src/fa/order/order.tex

@@ -12,6 +12,30 @@
     The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$. 
 \end{definition}
 
+\begin{definition}[Order Unit]
+\label{definition:order-unit}
+    Let $E$ be an ordered vector space and $e \in E$, then $e$ is an \textbf{order unit} if $[-e, e] \subset E$ is radial. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:order-unit-span}
+    Let $E$ be an ordered vector space with order unit $e$, then for any $x \in E$, there exists $x^+, x^- \ge 0$ such that $x = x^+ - x^-$. 
+\end{lemma}
+\begin{proof}
+    Let $\lambda > 0$ such that $\lambda e + s, \lambda e - s \ge 0$, then
+    \[
+    s = \frac{\lambda e + s}{2} - \frac{\lambda e - s}{2}
+    \]
+\end{proof}
+
+
+\begin{definition}[Archimedean]
+\label{definition:order-archimedean}
+    Let $E$ be an ordered vector space, then $E$ is \textbf{Archimedean} if for each $x \in E$, if there exists $y \in E$ such that $nx \le y$ for all $n \in \natp$, then $x \le 0$. 
+\end{definition}
+
+
+
 \begin{definition}[Ordered Topological Vector Space]
 \label{definition:ordered-tvs}
     Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed. 

src/fa/tvs/complete-metric.tex

@@ -150,7 +150,6 @@
 
 \end{proof}
 
-
 \begin{example}
     \label{example:open-mapping-counters}
 
@@ -160,3 +159,31 @@
 \end{example}
 
 
+\begin{theorem}[Closed Complementation Theorem]
+\label{theorem:closed-complement}
+    Let $E$ be a complete metric TVS, $\bracsn{0} \ne F \subsetneq E$ be a closed subspace, then the following are equivalent:
+    \begin{enumerate}
+        \item $F$ admits a closed complementary subspace. 
+        \item There exists $P \in L(E; E)$ such that $P^2 = P$ and $P(E) = F$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (2) $\Rightarrow$ (1): Let $G = \ker(P)$, then the mapping
+    \[
+    E \to F \times G \quad x \mapsto (Px, (1 - P)x)
+    \]
+
+    is a toplinaer isomorphism. 
+
+    (1) $\Rightarrow$ (2): Let $G$ be a closed complementary subspace of $E$, so that the mapping
+    \[
+    T: F \times G \to E \quad (x, y) \mapsto x + y
+    \]
+
+    is a continuous, bijective linear map. As $F, G \subset E$ are both closed, they are complete by \autoref{proposition:complete-closed}, and $E \times F$ is a complete metric TVS by \autoref{proposition:product-complete}. By the \hyperref[Open Mapping Theorem]{theorem:open-mapping}, the $T$ is a toplinear isomorphism. As such, $P$ is given by the following composition
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T^{-1}} & F \times G \ar@{->}[r]^{\pi_1} & F
+    }
+    \]
+\end{proof}

src/harmonic/index.tex

@@ -3,4 +3,5 @@
 
 \input{./lcg/index.tex}
 \input{./rep/index.tex}
+\input{./commutative/index.tex}
 \input{./notation.tex}

src/measure/measurable-maps/product.tex

@@ -37,21 +37,26 @@
 \end{enumerate}
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}]
+\begin{proposition}
 \label{proposition:product-sigma-algebra-metric}
-    Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then:
+    Let $\seq{X_j}$ be topological spaces, $X = \prod_{n \in \natp} X_n$, then:
     \begin{enumerate}
-        \item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$.
-        \item If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$
+        \item $\bigotimes_{n \in \natp} \cb_{X_n} \subset \cb_{X}$.
+        \item If $\seq{X_n}$ are separable and metrisable, then $\bigotimes_{n \in \natp} \cb_{X_n} = \cb_{X}$. 
     \end{enumerate}
 \end{proposition}
-\begin{proof}
-    (2): By \autoref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
-
-    Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
+\begin{proof}[Proof, {{\cite[Proposition 1.5]{Folland}}} and {{\cite[Proposition 8.1.7]{CohnMeasure}}}. ]
+    (2): For each $J \subset \natp$ finite, $x \in X$, and $\eps > 0$, define
     \[
-    U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
+    B_J(x, \eps) = \bigcap_{j \in J}\pi_j^{-1}(B_{X_j}(x_j, \eps))
     \]
 
-    The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
+    By \autoref{proposition:separable-product}, $X$ is also separable. Let $\seq{x_n} \subset X$ be a countable dense subset, and
+    \[
+    \fB = \bracsn{B_J(x_n, 1/k)|n, k \in \natp, J \subset \natp \text{ finite}}
+    \]
+
+    For each $x \in X$ and $U \in \cn_X(x)$, there exists $J \subset X$ finite and $k \in \natp$ such that $x \in B_J(x, 1/k) \subset U$. By density of $\seq{x_n}$, there exists $n \in \natp$ such that $x_n \in B_J(x, 1/(2k))$. In which case, $x \in B_J(x_n, 1/(2k)) \subset B_J(x, 1/k) \subset U$. Thus $\fB$ is a countable base for the product topology, consisting of cylinder sets in $\bigotimes_{n \in \natp} \cb_{X_n}$. 
+
+    As such, every open set in $X$ is a countable union of elements of $\fB$. Therefore $\bigotimes_{n \in \natp} \cb_{X_n} \supset \cb_{X}$.
 \end{proof}

src/op/c-star/gns.tex

@@ -70,6 +70,8 @@
     \end{enumerate}
     
     The representation $(H_\phi, \pi_\phi)$ is the \textbf{cyclic representation of $A$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
+
+    If $\phi$ is faithful, then $L^2(M, \phi)$ also denotes $H_\phi$. 
 \end{definition}
 \begin{proof}[Proof, {{\cite[Proposition 14.2]{Zhu}}}. ]
     (2): Fix $x \in A$, then for each $y_1, y_2 \in A$ with $y_1 - y_2 \in N_\phi$, $x(y_1 - y_2) \in N_\phi$ by \autoref{lemma:cstar-state-kernel}, so $\pi_\phi^0(x)$ is well-defined on $A/N_\phi$. 

src/op/c-star/order.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Positive]
 \label{definition:positive-c-star-algebra}
-    Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$. 
+    Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$. The set $A_+$ is the cone of positive elements in $A$. 
 \end{definition}
 
 \begin{proposition}

src/op/example/disk.tex

@@ -21,9 +21,134 @@
     is a homeomorphism. 
 \end{proposition}
 \begin{proof}
-    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(x_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
+    Let $\phi \in \Omega(D)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A(D)^*} = 1$. Let $p$ be the identity polynomial, and $z_0 = \phi(p)$, then for every $q \in \complex[x]$, $\phi(q) = q(z_0)$. By density of polynomials in $A(D)$, $\phi(f) = f(z_0)$ for all $f \in A(D)$. Therefore $\ol D$ is in bijection with $\Omega(D)$. 
 
     Since $A(D) \subset C(\ol D)$, $E$ is continuous. As $\ol D$ is compact and $\Omega(D)$ is Hausdorff, $E$ is a homeomorphism by \autoref{proposition:compact-hausdorff-homeomorphism}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:disk-algebra-calculus-upper-triangular}
+    Let $a, b, c \in \complex$ with $a \ne b$, and
+    \[
+    T = \begin{bmatrix}
+        a &c \\
+        0 &b
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For each $p \in \complex[z]$, 
+        \[
+        p(T) = \begin{bmatrix}
+            p(a) &c \frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)
+        \end{bmatrix}
+        \]
+
+        \item If $|a|, |b| \le 1$, then for each $f \in A(D)$, 
+        \[
+        f(T) = \begin{bmatrix}
+            f(a) &c \frac{f(a) - f(b)}{a - b} \\
+            0 & f(b)
+        \end{bmatrix}
+        \]
+
+        \item Suppose that $|a|, |b| < 1$ and let
+        \[
+        \varphi: \complex \setminus \bracsn{b/|b|^2} \to \complex \quad \varphi(z) = \frac{z - b}{1 - \ol b z}
+        \]
+
+        then $\norm{T}_{M_2(\complex)} \le 1$ if and only if
+        \[
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2
+        \]
+        \item If $|a|, |b| \le 1$, then $\norm{T}_{M_2(\complex)} \le 1$ if and only if
+        \[
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2
+        \]
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1): Suppose that (1) holds for $p, q \in \complex[z]$, then 
+    \begin{align*}
+        p(T)q(T) &= \begin{bmatrix}
+            p(a) &c \frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)
+        \end{bmatrix}\begin{bmatrix}
+            q(a) &c \frac{q(a) - q(b)}{a - b} \\
+            0 & q(b)
+        \end{bmatrix} \\
+        &= \begin{bmatrix}
+            p(a)q(a) &c p(a)\frac{q(a) - q(b)}{a - b} + cq(b)\frac{p(a) - p(b)}{a - b} \\
+            0 & p(b)q(b)
+        \end{bmatrix}\\
+        &= \begin{bmatrix}
+            p(a)q(a) &c \frac{p(a)q(a) + p(b)q(b)}{a - b} \\
+            0 & p(b)q(b)
+        \end{bmatrix}
+    \end{align*}
+
+    As (1) already holds for $1, z \in \complex[z]$, by induction, (1) holds for all $p \in \complex[z]$. 
+
+    (3): For each $z \in \partial B_\complex(0, 1)$, 
+    \[
+    |\varphi(z)| = \abs{\frac{z - b}{1 - \ol{b}z}} = \abs{\frac{z - b}{\ol{z} - \ol{b}}} = 1
+    \]
+
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\varphi \in A(D)$ with $\norm{\varphi}_{A(D)} = 1$. By (2), 
+    \[
+    \varphi(T) = \begin{bmatrix}
+        \varphi(a) & c\frac{\varphi(a)}{a - b} \\
+        0 & 0
+    \end{bmatrix} \quad \norm{\varphi(T)}_{M_2(\complex)}^2 = |\varphi(a)|^2  + \abs{c\frac{\varphi(a)}{a - b}}^2
+    \]
+
+    so $\norm{\varphi(T)}_{M_2(\complex)} \le 1$ if and only if
+    \begin{align*}
+        1 &\ge |\varphi(a)|^2 \paren{1 + \frac{|c|^2}{|a - b|^2}} \\
+        |a - b|^2 &\ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)
+    \end{align*}
+    
+
+    (3, $\Rightarrow$): Suppose that $\norm{T}_{M_2(\complex)} \le 1$. By continuity of the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} and \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial}, $\norm{\varphi(T)}_{M_2(\complex)} \le 1$, and $|a - b|^2 \ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)$.
+    
+    (3, $\Leftarrow$): Suppose that $|a - b|^2 \ge |\varphi(a)|^2 (|a - b|^2 + |c|^2)$, or equivalently that $\norm{\varphi(T)}_{M_2(\complex)} \le 1$. 
+    
+    Let $\psi(z) = \frac{z + b}{1 + \ol b z}$, then $\psi \circ \varphi = \text{Id}$. For each $z \in \partial B_\complex(0, 1)$, 
+    \[
+    |\psi(z)| = \abs{\frac{z + b}{1 + \ol b z}} = \abs{\frac{z + b}{\ol z + \ol b}} = 1
+    \]
+
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\psi \in A(D)$ with $\norm{\psi}_{A(D)} = 1$. Since $\norm{\psi(T)}_{M_2(\complex)} \le 1$, the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} and \hyperref[von Neumann's inequality]{theorem:von-neumann-inequality-polynomial} imply that 
+    \[
+    \norm{T}_{M_2(\complex)} = \norm{\psi(\varphi(T))}_{M_2(\complex)} \le \norm{\psi}_{A(D)} = 1
+    \]
+    
+    (4): For each $t \in (0, 1)$, let $\varphi_t(z) = (z - tb)/(1 - t\ol b z)$. 
+    
+    If $\norm{T}_{M_2(\complex)} \le 1$, then $\norm{tT}_{M_2(\complex)} \le 1$ for all $t \in (0, 1)$, so
+    \begin{align*}
+        |\varphi(a)|^2(|a - b|^2 + |c|^2) &= \lim_{t \upto 1}|\varphi_t(ta)|^2(|ta - tb|^2 + |tc|^2) \\
+        &\le \lim_{t \upto 1}|ta - tb|^2 = |a - b|^2
+    \end{align*}
+    
+    On the other hand, if $|\varphi(a)|^2(|a - b|^2 + |c|^2) \le |a - b|^2$, then
+    \begin{align*}
+        |1 - t^2\ol b a|^2 &= 1 - 2t^2 \text{Re}(\ol b a) + t^4|a|^2|b|^2 \\
+        &= t^2(1 - 2\text{Re}(\ol b a) + |a|^2|b|^2) + (1 - t^2)(1 - t^2|a|^2|b|^2) \\
+        &\ge t^2(1 - 2\text{Re}(\ol b a) + |a|^2|b|^2) = t^2|1 - \ol ba|^2
+    \end{align*}
+    so
+
+    \begin{align*}
+        |\varphi_t(ta)|^2(|ta - tb|^2 + |tc|^2) &= t^2\abs{\frac{ta - tb}{1 - t^2\ol b a}}^2(|a - b|^2 + |c|^2) \\
+        &\le t^2 \abs{\frac{a - b}{1 - \ol b a}}^2(|a - b|^2 + |c|^2) \\
+        &\le |ta - tb|^2
+    \end{align*}
+
+    and $\norm{tT}_{M_2(\complex)} \le 1$ for all $t \in (0, 1)$. Therefore $\norm{T}_{M_2(\complex)} \le 1$. 
+\end{proof}
+
 

src/op/example/hardy.tex

@@ -3,9 +3,66 @@
 
 \begin{definition}[Hardy Space]
 \label{definition:hardy-space}
-    Let $D = B_\complex(0, 1)$, then the \textbf{Hardy space} $H^\infty(D)$ is the algebra of all bounded holomorphic functions on $D$, equipped with the uniform norm. 
+    Let $D = B_\complex(0, 1)$ and $p \in [1, \infty]$. For each $f \in H(D; \complex)$, let
+    \[
+    \norm{f}_{H^p(D)} = \sup_{r \in [0, 1)} \braks{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^p d\theta}^{1/p}
+    \]
+
+    if $p < \infty$, and $\norm{f}_{H^\infty(D)} = \norm{f}_u$ if $p = \infty$. Then the \textbf{Hardy space} is the space
+    \[
+    H^p(D) = \bracsn{f \in H(D; \complex)|\ \normn{f}_{H^p(D; \complex)} < \infty}
+    \]
+
+    In particular, $H^\infty(D)$ is a Banach algebra under pointwise multiplication. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:hardy-space-h2}
+    Let $D = B_\complex(0, 1)$ and $H^2(D)$ be the Hardy space, then:
+    \begin{enumerate}
+        \item $H^2(D)$ is a Hilbert space. For each $f, g \in H^2(D)$, 
+        \[
+        \dpn{f, g}{H^2(D)} = \lim_{r \upto 1}\frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})} d\theta
+        \]
+        \item The mapping
+        \[
+        l^2(\natz; \complex) \to H^2(D) \quad x \mapsto \sum_{n = 0}^\infty x_nz^n
+        \]
+
+        is unitary. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $r \in [0, 1)$ and $f \in H^2(D)$, let
+    \[
+    [f]_{r} = \braks{\frac{1}{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2 d\theta}^{1/2}
+    \]
+
+    Then, $[\cdot]_r$ is induced by the pseudo inner product
+    \[
+    \dpn{f, g}{r} = \frac{1}{2\pi}\int_0^{2\pi}f(re^{i\theta})\ol{g(re^{i\theta})} d\theta = \frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_{r}^2
+    \]
+
+    By the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}, $\norm{f}_{H^2(D)} = \lim_{r \upto 1}[f]_r$, so the inner product
+    \begin{align*}
+    \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\dpn{f, g}{r} = \lim_{r \upto 1}\frac{1}{4}\sum_{k = 0}^3 i^k [f + i^kg]_r^2 \\
+    &= \frac{1}{4}\sum_{k = 0}^3 i^k \normn{f + i^kg}_{H^2(D)}^2
+    \end{align*}
+    
+
+    exists, and defines the norm on $H^2(D)$. 
+
+    (2): For any $x \in l^2(\natz; \complex) \subset l^\infty(\natz; \complex)$, $\limsup_{n \to \infty}|x_n|^{1/n} \le 1$, so the given map is well-defined. 
+
+    On the other hand, for any $f = \sum_{n = 0}^\infty x_nz^n, g = \sum_{n = 0}^\infty y_nz^n \in H^2(D)$ and $r \in (0, 1)$, 
+    \begin{align*}
+        \dpn{f, g}{r} &= \sum_{n = 0}^\infty \sum_{m = 0}^\infty x_n\ol{y_m}\dpn{z^n, z^m}{r} = \sum_{n = 0}^\infty x_n\ol{y_n}r^{2n} \\
+        \dpn{f, g}{H^2(D)} &= \lim_{r \upto 1}\sum_{n = 0}^\infty x_n\ol{y_n}r^{2n} = \sum_{n = 0}^\infty x_n\ol{y_n}
+    \end{align*}
+
+    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}. Thus $\bracsn{z^n}_0^\infty$ forms an orthonormal basis for $H^2(D)$ and the given map is unitary. 
+\end{proof}
+
 \begin{proposition}
 \label{proposition:hardy-spectrum}
     Let $f \in H^\infty(D)$, then $\sigma(f) = \ol{f(D)}$. 
@@ -20,7 +77,7 @@
 
     then:
     \begin{enumerate}
-        \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(z_0)$. 
+        \item If $\fU \to x_0 \in \ol{D}$, then $f \in A(D)$, $\phi_{\fU}(f) = f(x_0)$. 
         \item $\phi_{\fU}$ is a multiplicative linear functional on $H^\infty(D)$. 
     \end{enumerate}
 \end{proposition}
@@ -31,4 +88,49 @@
 \end{proof}
 
 
+\begin{lemma}
+\label{lemma:hardy-left-multiplication}
+    Let $f \in H^\infty(D)$ and $M_f \in B(H^2(D))$ be the multiplication map $g \mapsto fg$, then $\norm{f}_{H^\infty(D)} = \norm{M_f}_{B(H^2(D))}$. 
+\end{lemma}
 
+
+
+
+\begin{proposition}
+\label{proposition:toeplitz-triangular-norm}
+    Let $\bracsn{a_j}_0 \subset \complex$ and
+    \[
+    T = \begin{bmatrix}
+        a_0 &0 &\cdots &0 \\
+        a_1 &a_0 &\ddots &\vdots \\
+        \vdots &\ddots &\ddots &0 \\
+        a_n &\cdots &a_1 &a_0
+    \end{bmatrix} \in M_{n+1}(\complex)
+    \]
+
+    then
+    \begin{enumerate}
+        \item Let $p = \sum_{k = 0}^n a_k z^k$, $H^2_n(D) = \bracsn{q \in \complex[z]|\deg(q) \le n}$, $P \in B(H^2(D))$ be the orthogonal projection onto $H^2_n(D)$, and identify $H^2_n(D) = \complex^{n+1}$, then $Tf = Ppf$ for all $f \in H^2_n(D)$. 
+        \item For any $q \in \complex[z]$, $P(p + z^{n+1}q)f = Ppf$ for all $f \in H^2(D)$. 
+        \item For any $q \in \complex[z]$, $\norm{T}_{M_{n+1}(\complex)} \le \normn{p + z^{n+1}q}_{H^\infty(D)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $f = \sum_{k = 0}^n \xi_k z^k$, then
+    \begin{align*}
+        pf &= \braks{\sum_{k = 0}^n a_kz^k}\braks{\sum_{k = 0}^n \xi_k z^k} = \sum_{j, k = 0}^n a_j\xi_k z^{j+k} \\
+        Ppf &= \sum_{k = 0}^n z^k\sum_{j = 0}^k a_{k-j}\xi_{j} = Tf
+    \end{align*}
+    
+    (2): For any $f \in H^2(D)$, $z^{n+1}qf \perp H^2_n(D)$, so $P(p + z^{n+1}q)f = Ppf$.
+
+    (3): For each $g \in H^\infty(D)$, let $M_g \in B(H^2(D))$ be the multiplication operator by $g$. By (1), $\norm{T}_{M_{n+1}(\complex)} = \normn{PM_p}_{B(H^2(D))}$. By (2), 
+    \[
+    \normn{PM_p}_{B(H^2(D))} = \normn{PM_{p + z^{n+1}q}}_{B(H^2(D))} \le \norm{M_{p+z^{n+1}q}}_{B(H^2(D))}
+    \]
+
+    By \autoref{lemma:hardy-left-multiplication}, 
+    \[
+        \norm{T}_{M_{n+1}(\complex)} = \normn{PM_p}_{B(H^2(D))} \le \normn{M_{p+z^{n+1}q}}_{B(H^2(D))} = \normn{p + z^{n+1}q}_{H^\infty(D)}
+    \]
+\end{proof}

src/op/example/index.tex

@@ -8,4 +8,3 @@
 \input{./disk.tex}
 \input{./convolution.tex}
 \input{./bc.tex}
-\input{./bb.tex}

src/op/index.tex

@@ -1,9 +1,8 @@
-\part{Operator Algebras}
+\part{Fundamentals of Operator Algebras}
 \label{part:operator-algebras}
 
 \input{./banach/index.tex}
 \input{./c-star/index.tex}
-\input{./vn/index.tex}
 \input{./example/index.tex}
 \input{./system/index.tex}
 \input{./notation.tex}

src/op/notation.tex

@@ -7,6 +7,7 @@
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
     $Z(A)$ & Centre of a Banach algebra. & \autoref{definition:banach-algebra-centre} \\ 
     $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
+    $U(A)$ & Unitary group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
     $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
     $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\
     $\sigma_A(x) = \sigma(x)$ & The spectrum of $x$ in $A$. & \autoref{definition:spectrum} \\
@@ -20,12 +21,7 @@
     $PS(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
     $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
     $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
-    $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
-    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\ 
-    $\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
-    $Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\ 
-    $P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\ 
-    $P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent} \\ 
+    $L^2(M, \phi)$ & Hilbert space of the GNS triple associated with a faithful $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
     
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/op/system/aos.tex

@@ -0,0 +1,212 @@
+\section{Abstract Operator Systems}
+\label{section:abstract-operator-system}
+
+\begin{definition}[Matrix Order]
+\label{definition:matrix-order}
+    Let $S$ be a complex vector space equipped with a \hyperref[complex conjugation]{definition:complex-conjugation} $*: S \to S$, then $S$ is \textbf{matrix ordered} if:
+    \begin{enumerate}
+        \item For each $n \in \natp$, $M_n(S)_{sa}$ is equipped with a vector space ordering. 
+        \item For each $m, n \in \natp$, $A \in M_{n \times m}(\complex)$, and $x \in M_n(S)_{sa}$ with $x \ge 0$, $A^*xA \ge 0$ in $M_m(S)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Matrix Order Unit]
+\label{definition:matrix-order-unit}
+    Let $S$ be a matrix ordered space and $e \in S$, then $e$ is a \textbf{matrix order unit} if $I_n \otimes e \in M_n(S)$ is an order unit for all $n \in \natp$. 
+\end{definition}
+
+\begin{definition}[Extended Map]
+\label{definition:abstract-induced-matrix-map}
+    Let $S$, $T$ be complex vector spaces, $\phi: S \to T$, and $n \in \natp$, then $\phi_n := \text{Id} \otimes \phi: M_n(S) \to M_n(T)$ is the \textbf{extended map} of $\phi$ to $M_n(S)$. 
+\end{definition}
+
+\begin{definition}[Abstract Operator System]
+\label{definition:a-operator-system}
+    Let $S$ be a matrix ordered space, then $S$ is an \textbf{abstract operator system} if $M_n(S)$ is Archimedean for all $n \in \nat$, and $S$ admits a matrix order unit. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:matrix-order-linear-functional}
+    Let $S$ be a matrix ordered space, then the mapping
+    \[
+    \text{hom}(S; M_n(\complex)) \to \text{hom}(M_n(S); \complex) \quad \angles{E_{ij} \otimes a, \phi}_{M_n(S)} = \phi(a)_{ij}
+    \]
+
+    induces an isomorphism $\text{hom}(M_n(S); \complex) \iso \text{hom}(S; M_n(\complex))$. 
+\end{proposition}
+
+\begin{proposition}
+\label{proposition:matrix-order-positive-characterisation}
+    Let $S$ be a matrix ordered space, $\phi \in \text{hom}(S; M_n(\complex))$ and $s \in \text{hom}(M_n(S); \complex)$, and
+    \[
+    s: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $s$ is a positive linear functional. 
+        \item $\phi$ is $n$-positive. 
+        \item $\phi$ is completely positive. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 13.2]{Paulsen}}}. ]
+    (1) $\Rightarrow$ (3): Let $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, $\eta \in (\eta_1, \cdots, \eta_n) \in \complex^n$, and $x \in S$, then
+    \[
+        \dpn{\phi(x)\xi, \eta}{\complex^n} = \sum_{i, j = 1}^n \phi(x)_{ij}\xi_j \ol {\eta_i} = s({\eta^* x\xi})
+    \]
+
+    Let $m \in \natp$, $\xi = (\xi_1, \cdots, \xi_m) \in \complex^m \otimes \complex^n$, and $A = \sum_{i, j = 1}^m E_{ij} \otimes a_{ij} \in M_n(S)$ with $a \ge 0$, then
+    \begin{align*}
+        \dpn{\phi_n(A)\xi, \xi}{\complex^m \otimes \complex^n} &= \sum_{i, j = 1}^m \dpn{\phi(a_{ij})\xi_j, \xi_i}{\complex^n} = \sum_{i, j = 1}^m s(\xi_i^*a_{ij}\xi_j) \\
+        &= s\braks{\begin{bmatrix}
+            \xi_1 \\
+            \vdots \\
+            \xi_n
+        \end{bmatrix}^*A\begin{bmatrix}
+            \xi_1 \\
+            \vdots \\
+            \xi_n
+        \end{bmatrix}} \ge 0
+    \end{align*}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:abstract-operator-system-norm}
+    Let $S$ be an abstract operator system with order unit $e$. For each $s \in M_n(S)$, let\footnote{Infimum exists because $I_{M_{2n}(S)} \otimes e$ is the order unit in $M_{2n}(S)$.}
+    \[
+    \norm{s}_{M_n(S)} = \inf\bracs{r > 0 \bigg | \begin{bmatrix}
+        rI_{M_n(S)} \otimes e &s \\
+        s^* & rI_{M_n(S)} \otimes e
+    \end{bmatrix} \ge 0}
+    \]
+
+    then 
+    \begin{enumerate}
+        \item $\norm{\cdot}_{M_n(S)}$ is a norm. 
+        \item For each $s \in M_n(S)$, $\norm{s}_{M_n(S)} = \normn{s^*}_{M_n(S)}$. 
+        \item $\bracsn{s \in M_n(S)|s \ge 0}$ is closed with respect to $\norm{\cdot}_{M_n(S)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): By replacing $S$ with $M_n(S)$, assume without loss of generality that $n = 1$. Let $s \in S$ with $\norm{s}_{M_n(S)} = 0$, then for each $r > 0$, 
+    \[
+    \begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix} \ge 0 \quad \forall r > 0
+    \]
+
+    As such, for each $\lambda \in \complex$, 
+    \[
+    0 \le \begin{bmatrix}
+        1 & \ol \lambda
+    \end{bmatrix}\begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix}\begin{bmatrix}
+        1 \\
+        \lambda
+    \end{bmatrix} = r(1 + |\lambda|^2)e + \lambda s + (\lambda s)^*
+    \]
+
+    As the above holds for all $r > 0$, $\lambda s + (\lambda s)^* \ge 0$. In particular, $0 \le s + s^* \le 0$ and $0 \le is + (is)^* \le 0$, so $\text{Re}(s) = \text{Im}(s) = 0$, and $s = 0$. 
+
+    (2): For any $s \in S$ and $r > 0$, 
+    \[
+    \begin{bmatrix}
+        0 & 1 \\
+        1 & 0
+    \end{bmatrix}\begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix}\begin{bmatrix}
+        0 & 1 \\
+        1 & 0
+    \end{bmatrix} = \begin{bmatrix}
+        re &s^* \\
+        s & re
+    \end{bmatrix}
+    \]
+
+    so
+    \[
+    \begin{bmatrix}
+        re &s \\
+        s^* & re
+    \end{bmatrix} \ge 0 \quad \Leftrightarrow \quad \begin{bmatrix}
+        re &s^* \\
+        s & re
+    \end{bmatrix} \ge 0
+    \]
+
+    (3): Let $s \in S^+$, where the closure is taken with respect to $\normn{\cdot}_{M_1(S)}$. By (2), $\ol{S^+} \subset \ol S_{sa} = S$. Let $r > 0$, then there exists $t \in S^+$ and $\normn{s - t}_{S} < r$. As such, 
+    \begin{align*}
+        
+    0 &\le \begin{bmatrix}
+        re &s - t \\
+        s - t & re
+    \end{bmatrix} \\
+    0 &\le \begin{bmatrix}
+    1 & 1
+    \end{bmatrix} \begin{bmatrix}
+        re &s - t \\
+        s - t & re
+    \end{bmatrix}\begin{bmatrix}
+    1 \\ 1
+    \end{bmatrix} = 2re + 2s - 2t
+    \end{align*}
+
+    so $re + s \ge 0$. By the Archimedean property, $s \ge 0$. 
+    
+\end{proof}
+
+\begin{theorem}[Choi-Effros]
+\label{theorem:choi-effros}
+    Let $S$ be an abstract operator system with order unit $e$, then there exists a complex Hilbert space $H$, an operator system $S_1 \subset B(H)$, and a complete order isomorphism $\Phi: S \to S_1$ with $\phi(e) = I_H$. 
+\end{theorem}
+\begin{proof}
+    For each $n \in \natp$, let
+    \[
+    K_n = \bracsn{\phi: S \to M_n(\complex)| \phi \text{ is unital, completely positive}}
+    \]
+
+    and $K = \bigsqcup_{n \in \natp}K_n$\footnote{The matrix state space.}. Define\footnote{The summand on each $\phi \in K_n$ is $M_n(\complex)$.}
+    \[
+    \Phi: S \to [l^\infty(K); M_n(\complex)] \quad \Phi(s)_\phi = \phi(s)
+    \]
+
+    then $[l^\infty(K); M_n(\complex)] \subset B([l^2(K); \complex^n])$, and $\Phi$ is unital and completely positive. 
+
+    Let $n \in \natp$ and $s \in M_n(S)$ with $\Phi_n(s) \ge 0$. Suppose for contradiction that $s < 0$, then by the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-2}, there exists $\tilde \phi \in M_n(S)^*$ such that $\tilde \phi (s) \not\ge 0$ but $\tilde \phi(t) \ge 0$ for all $t \in M_n(S)$ with $t \ge 0$. As such, $\tilde \phi$ is a positive linear functional on $M_n(S)^*$. By \autoref{proposition:matrix-order-positive-characterisation}, the associated map $\phi: S \to M_n(\complex)$, defined by $\phi(t)_{ij} = \tilde \phi(E_{ij} \otimes t)$, is completely positive. However, 
+    \[
+    \sum_{i, j = 1}^n \dpn{\phi(s_{ij})e_j, e_i}{\complex^n} = \tilde \phi(s) < 0
+    \]
+
+    so $\phi_n(s) \not\ge 0$ as well. After composing with the range projection of $\phi(e)$, assume without loss of generality that $\phi(e)$ is invertible. In which case, $\phi' = \phi(e)^{-1/2}\phi \phi(e)^{-1/2}$ is unital and completely positive with $\phi_n'(s) \not\ge 0$, contradiction.
+\end{proof}
+
+\begin{definition}[Maximal Operator System Structure]
+\label{definition:maximal-operator-system}
+    Let $S$ be an operator system with order unit $e$. For each $n \in \natp$, $C_n = \ol{M_n(\complex)_+ \otimes S_+} \subset M_n(S)$, then $\seq{C_n}$ defines a matrix order on $S$, and $(S, \seq{C_n}, e)$ is an operator system. The triple $\text{OMAX}(S) = (S, \seq{C_n}, e)$ is the \textbf{maximal} operator system structure on $S$. 
+\end{definition}
+
+\begin{example}
+    \label{example:matrix-os-embedding-not-homomorphism}
+
+    Let
+    \[
+    \phi: M_n(\complex) \to M_{n+1}(\complex) \quad a \mapsto \begin{bmatrix}
+        a & 0 \\
+        0 & \frac{\tau(a)}{n}
+    \end{bmatrix}
+    \]
+
+    then $\phi$ is an embedding of operator systems, but $\phi(M_n(\complex))$ is not a $C^*$-subalgebra. 
+\end{example}
+
+\begin{example}
+    \label{example:arveson-nc-function}
+
+    Let $S = \text{span}\bracsn{1, z, \ol z} \subset C(\mathbb{T})$, then $C^*(S) = C(\mathbb{T})$, but $S$ also embeds into and generates $C(\mathbb D)$. In this particular case, $C(\mathbb T) = C^*_{\min}(S)$. 
+\end{example}
+

src/op/system/arveson.tex

@@ -0,0 +1,62 @@
+\section{Arveson's Extension Theorem}
+\label{section:arveson-extension}
+
+\begin{lemma}
+\label{lemma:tensor-trace-approx}
+    Let $E, F$ be locally convex spaces of $K \in \RC$, $B \subset L(E; F^*)$ be equicontinuous, then the following two topologies on $B$ coincide:
+    \begin{enumerate}
+        \item The weak* topology under the identification $L(E; F^*) = (E \wh \otimes_\pi F)^*$. 
+        \item The topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    As $B$ is equicontinuous, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that the weak* topology coincides with the topology of pointwise convergence on any dense subset. By linearity, the topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$ coincides with the topology of pointwise convergence on $E \otimes F$. As $E \otimes F$ is dense $E \wh \otimes_\pi F$, the topology of pointwise convergence on $E \otimes F$ coincides with the topology of pointwise convergence on $\bracsn{e \otimes f|e \in E, f \in F}$. 
+\end{proof}
+
+
+\begin{definition}[Point-Weak* Topology]
+\label{definition:bw-topology}
+    Let $E$ be a Banach space and $H$ be a Hilbert space. Identify
+    \[
+    L(S; B(H)) = L(S; (H \otimes_\pi H)^*) = [S \otimes_\pi (H \otimes_\pi H)]^*
+    \] 
+
+    then the weak* topology on $L(E; B(H))$ induced by $S \otimes_\pi (H \otimes_\pi H)$ is the \textbf{point-weak* topology}/\textbf{BW topology} on $L(E; B(H))$. 
+\end{definition}
+
+
+
+\begin{proposition}
+\label{proposition:bw-topology-gymnastics}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $H$ be a complex Hilbert space. Identify
+    \[
+    L(S; B(H)) = L(S; (H \otimes_\pi H)^*) = [S \otimes_\pi (H \otimes_\pi H)]^*
+    \] 
+\end{proposition}
+
+\begin{theorem}[Arveson's Extension Theorem]
+\label{theorem:arveson-extension}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a completely positive map, then $\phi$ admits a completely positive extension to $A$. 
+\end{theorem}
+\begin{proof}
+    Let $\mathcal{F} \subset 2^H$ be the collection of all finite-dimensional subspaces of $H$, directed under inclusion. 
+    
+    Let $F \in \mathcal{F}$ and $P_F \in B(H)$ be the orthogonal projection onto $H$, then the corner $P_FB(H)P_F$ is isomorphic to $M_{|F|}(\complex)$. Let
+    \[
+    \phi_F: S \to P_FB(H)P_F \quad \phi_F(a) = P_F\phi(a)P_F
+    \]
+
+    be the compression of $\phi$ onto $F$, then $\phi_F$ is also completely positive by \autoref{proposition:multiply-completely-bounded}. By the \hyperref[Krein Extension Theorem]{theorem:krein-extension}, $\phi_F$ admits an extension to a completely positive map $\Phi_F: A \to P_FB(H)P_F$. 
+
+    By \autoref{proposition:completely-positive-norm}, $\sup_{F \in \mathcal{F}} \norm{\Phi_F}_{L(A; B(H))} \le \norm{\phi}_{L(S; H)}$, so the family $\mathcal{E} = \bracsn{\phi_F|F \in \mathcal{F}}$ is equicontinuous. By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, it admits a point-weak* convergent subnet $\anglesn{\Phi_{F_\beta}}_{\beta \in B}$. 
+
+    Let $\Phi = \text{\small BW}\text{-}\!\lim_{\beta \in B}\Phi_{F_\beta}$, then $\Phi$ is also completely positive, with
+    \[
+    \Phi|_{A} = \text{\small BW}\text{-}\!\lim_{\beta \in B}\Phi_{F_\beta}|_{S} = \text{\small BW}\text{-}\!\lim_{\beta \in B}P_F\phi P_F = \phi
+    \]
+\end{proof}
+
+
+
+
+

src/op/system/complete-upgrade.tex

@@ -0,0 +1,269 @@
+\section{Sufficient Conditions for Complete Positivity}
+\label{section:completely-upgrade}
+
+
+\begin{proposition}
+\label{proposition:homomorphism-completely-positive}
+    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a *-homomorphism, then $\phi$ is completely positive. 
+\end{proposition}
+\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}
+
+The following is a slight generalisation of one form of the above example, but stating it helps greatly. 
+
+\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 $\phi(s) = xsx^*$ for all $s \in S$, then $\phi$ is completely positive. 
+\end{proposition}
+
+\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 for any $a \in A$ with $a \ge 0$, 
+    \[
+    \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0
+    \]
+
+    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{definition}[Matrix Unit]
+\label{definition:choi-matrix-unit}
+    Let $n \in \natp$, $\seqf{e_j}$ be the standard orthonormal basis for $\complex^n$, and $1 \le i, j \le n$, then the \textbf{$(i, j)$-th matrix unit} is the operator $E_{ij} = e_j \otimes e_i$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:cstar-positive-matrix-decompose-sum}
+    Let $A$ be a $C^*$-algebra and $P \in M_n(A)$ be positive, then there exists $\bracsn{a_{k, j}}_{j, k = 1}^n$ such that:
+    \begin{enumerate}
+        \item $P = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})$. 
+        \item For each $1 \le k \le n$,
+        \[
+        \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) = \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 3.13]{Paulsen}}}. ]
+    Since $P \ge 0$, there exists $\bracsn{a_{j, k}}_{j, k = 1}^n$ such that
+    \begin{align*}
+    P &= \braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}}^*\braks{\sum_{j, k = 1}^n E_{jk} \otimes a_{j, k}} \\
+    &= \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 $1 \le i, j, k, l \le n$, $E_{ji}E_{kl} = E_{jl}$ if $i = k$, and $E_{ji}E_{kl} = 0$ otherwise. Thus after relabeling, 
+    \[
+        P = \sum_{j, k, l = 1}^nE_{jl} \otimes (a_{k, j}^*a_{k, l}) =  \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j})
+    \]
+    
+    Finally, for each $1 \le k \le n$, undoing the product shows that
+    \begin{align*} 
+    \sum_{i, j = 1}^n E_{ij} \otimes (a_{k, i}^*a_{k, j}) &= \sum_{i, j = 1}^n (E_{ik}E_{kj}) \otimes (a_{k, i}^*a_{k, j}) \\
+    &= \sum_{i, j = 1}^n (E_{ki} \otimes a_{k, i})^* (E_{kj} \otimes a_{k, j}) \\
+    &= \braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}}^*\braks{\sum_{i = 1}^n E_{ki} \otimes a_{k, i}} \ge 0
+    \end{align*}
+    
+\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 $E_{ij}$\footnote{$e_j \otimes e_i \in \complex^n \otimes \complex^n$.} be the standard matrix unit for $M_n(\complex)$, then $C_\phi = (\phi(E_{ij})) \in M_n(B)$ is the \textbf{Choi 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): For each $1 \le i, j \le n$, $E_{ij} \otimes E_{ij}$ is a positive multiple of a rank one projection, and hence positive. Since $C_\phi = \sum_{i, j = 1}^n \phi(E_{ij} \otimes E_{ij})$ is a sum of positive elements, $C_\phi$ is positive. 
+
+    (3) $\Rightarrow$ (1): By \autoref{lemma:cstar-positive-matrix-decompose-sum}, there exist $\bracsn{b_{i, j}}_{i, j = 1}^n \subset B$ such that
+    \[
+    \sum_{i, j = 1}^n E_{ij} \otimes \phi(E_{ij}) = C_\phi = \sum_{k = 1}^n \sum_{i, j = 1}^n E_{ij} \otimes b_{k, i}^* b_{k, j}
+    \]
+
+    So for any $1 \le i, j \le n$,
+    \[
+    \phi(E_{ij}) = \sum_{k = 1}^n b_{k, i}^* b_{k, j}
+    \]
+
+    For each $1 \le k \le n$, let $S_k = \sum_{l = 1}^n E_{kl} \otimes b_{k, l}^*$. Let $1 \le i, j \le n$, then
+    \begin{align*}
+    \sum_{k = 1}^n S_k (E_{ij} \otimes 1_B) S_k^* &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes b_{k, s}^*) (E_{ij} \otimes 1_B) (E_{kt} \otimes b_{k, t}^*)^* \\
+    &= \sum_{k = 1}^n \sum_{s, t = 1}^n (E_{ks} \otimes b_{k, s}^*) (E_{ij} \otimes 1_B) (E_{tk} \otimes b_{k, t}) \\
+    &= \sum_{k = 1}^n E_{kk} \otimes b_{k, i}^* b_{k, j} = \sum_{k = 1}^n E_{kk} \otimes \phi(E_{ij})
+    \end{align*}
+
+    By linearity, for each $T \in M_n(\complex)$, 
+    \[
+    \sum_{k = 1}^n E_{kk} \otimes \phi(T) = \sum_{k = 1}^n S_k(T \otimes 1_B)S_k^*
+    \]
+
+    What the above shows is that, under the identification that
+    \[
+    M_n(\complex) \iso M_n(\complex) \otimes 1_B \subset M_n(B)
+    \]
+    
+    
+    and
+    \[
+    B \iso \bracs{\sum_{k = 1}^n E_{kk} \otimes b \bigg |b \in B} = \bracs{\text{Diag}(b, \cdots, b)|b \in B} \subset M_n(B)
+    \]
+
+    The map $\phi$ takes the form
+    \[
+    \phi(T) = \sum_{k = 1}^n S_kTS_k^* \quad \forall T \in M_n(\complex)
+    \]
+
+    Therefore $\phi$ is completely positive by \autoref{proposition:multiply-completely-bounded}. 
+    
+    
+\end{proof}
+
+\begin{remark}
+\label{remark:choi-positive}
+    The Choi matrix yields a correspondence between linear maps and operator-valued matrices, and provides a way to check complete positivity. However, it is significantly harder to verify positivity through the matrix.
+\end{remark}
+
+\begin{example}[Reduction Map]
+\label{example:matric-reduction-map}
+    Let $n \ge 2$ and $\phi: M_n(\complex) \to M_n(\complex)$ be defined by
+    \[
+    \phi(x) = \text{Tr}(x)1_{M_n(\complex)} - x
+    \]
+
+    then $\phi$ is positive, but not completely positive. 
+\end{example}
+\begin{proof}
+    (Positive): Let $x \in M_n(\complex)$ with $x \ge 0$, then $\norm{x}_{M_n(\complex)}$ is the largest eigenvalue of $x$, which is bounded by the trace of $x$. As such, $\text{Tr}(x)1_{M_n(\complex)} \ge x$, and $\phi(x)$ is positive. 
+
+    (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_{\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}
+    \end{align*}
+
+    which is not positive. By \hyperref[Choi's Theorem]{theorem:choi-matrix-positive}, $\phi$ is not completely positive. 
+\end{proof}
+
+\begin{proposition}[Choi-Kraus]
+\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
+    \[
+    \phi(a) = \sum_{j = 1}^{mn}V_j^* a V_j
+    \]
+
+    for all $a \in M_n(\complex)$. 
+\end{proposition}
+\begin{proof}
+    Let $(\pi, V, K)$ be a minimal Stinespring representation for $\phi$. Since $\pi(M_n(\complex))V\complex^m$ 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_k\pi(a)P_k)V = \sum_{k = 1}^{mn}V^*Q_k^*aQ_kV \\
+    &= \sum_{k = 1}^{mn} (Q_kV)^* a (Q_kV)
+    \end{align*}
+\end{proof}
+
+
+
+
+

src/op/system/completely.tex

@@ -39,6 +39,106 @@
 \end{definition}
 
 
+\begin{proposition}[Tomiyama]
+\label{proposition:tomiyama-matrix-norm-bound}
+    Let $A, B$ be unital $C^*$-algebras and $S \subset A$ be an operator system, then:
+    \begin{enumerate}
+        \item For any $a = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(A)$, 
+        \[
+        \norm{a} \le \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} \le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)} 
+        \]
+
+        where all inequalities are sharp. 
+        \item For any bounded linear map $\phi: S \to B$, $\norm{\phi_n}_{L(M_n(S); M_n(B))} \le n\norm{\phi}_{L(S; B)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.10]{Paulsen}}}. ]
+    (1): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $A$ is a $C^*$-subalgebra of $B(H)$. 
+
+    For each $\xi = (\xi_1, \cdots, \xi_n), \eta = (\eta_1, \cdots, \eta_n) \in H^n$,
+    \begin{align*}
+        \dpn{a\xi, \eta}{H^n} &= \sum_{i, j = 1}^n \dpn{(E_{ij} \otimes a_{ij})\xi, \eta}{H^n} = \sum_{i, j = 1}^n \dpn{a_{ij}\xi_j, \eta_i}{H} \\
+        |\dpn{a\xi, \eta}{H^n}| &\le \sum_{i, j = 1}^n \norm{a_{ij}}_A\norm{\xi_j}_H\norm{\eta_i}_H \\
+        &= \sum_{i, j = 1}^n\dpn{\norm{a_{ij}}_AE_{ij}(\norm{\xi_1}_H, \cdots, \norm{\xi_n}_H), (\norm{\eta_1}_H, \cdots, \norm{\eta_n}_H)}{\complex^n} \\ 
+        &\le \norm{\xi}_{H^n} \cdot \norm{\eta}_{H^n} \cdot \norm{\sum_{i, j = 1}^n\norm{a_{ij}}_AE_{ij}}_{M_n(\complex)} 
+    \end{align*}
+
+    The sharp case is given by the identity.
+
+    For each $\xi = (\xi_1, \cdots, \xi_n) \in \complex^n$, by the \hyperref[Cauchy-Schwarz Inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+        \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n}^2 &= \sum_{i = 1}^n \abs{\sum_{j = 1}^n \norm{a_{ij}}_{A}\xi_j}^2 \\
+        &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}\braks{\sum_{j = 1}^n |\xi_j|^2} \\
+        \norm{\sum_{i, j = 1}^n \norm{a_{ij}}_AE_{ij}\xi}_{\complex^n} &\le \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_A^2}^{1/2} \cdot \norm{\xi}_{\complex^n}
+    \end{align*}
+
+    The sharp case is given by operators of the form $E_{ij} \otimes b$ for $b \in A$. 
+    
+    Finally, 
+    \[
+    \sum_{i, j = 1}^n \norm{a_{ij}}_A^2 = \sum_{i, j = 1}^n \norm{E_{ii} \cdot a \cdot E_{jj}}_{M_n(A)}^2 \le n^2\norm{a}_{M_n(A)}^2
+    \]
+
+    and
+    \[
+    \braks{\sum_{i, j = 1}^n \norm{a_{ij}}_{A}^2}^{1/2} \le n\norm{a}_{M_n(A)}
+    \]
+
+    The sharp case is given by operators of the form $E = \sum_{i, j = 1}^n E_{ij} \otimes E_{ji}$\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*}
+        \normn{\phi(a)}_{M_n(B)}^2 &\le \sum_{i, j = 1}^n \norm{\phi(a_{ij})}_{B}^2 \le \norm{\phi}_{L(S; B)}^2\sum_{i, j = 1}^n \norm{a_{ij}}_A^2 \\
+        &\le n^2\norm{\phi}_{L(S; B)}^2\norm{a}_{M_n(A)}^2
+    \end{align*}
+\end{proof}
+
+\begin{proposition}
+\label{proposition:positive-column-test}
+    Let $A$ be a unital $C^*$-algebra and $b = \sum_{i, j = 1}^n E_{ij} \otimes b_{ij} \in M_n(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $b$ is positive in $M_n(A)$. 
+        \item For each $\seqf{a_j} \subset A$, $\sum_{i, j = 1}^n a_ib_{ij}a_j^*$ is positive in $A$.
+    \end{enumerate}
+\end{proposition}
+\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_{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*}
+    
+    so $\sum_{i, j = 1}^n a_ib_{ij}a_j^* \ge 0$ in $A$. 
+
+    (2) $\Rightarrow$ (1): For each state $\phi \in S(A)$, let $(H_\phi, \pi_\phi, \xi_\phi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\phi$. Let $H = [l^2(S(A)); H_\phi]$,
+    \[
+    \pi : A \to B(H) \quad \pi(a)(\xi)_\phi = \pi_\phi(a)(\xi_\phi)
+    \]
+
+    be the universal representation of $A$, and identify $A$ as a $C^*$-subalgebra of $B(H)$ through $\pi$. 
+    
+    For any $\seqf{a_j} \subset A$ and $\phi \in S(A)$, 
+    \begin{align*}
+    0 \le \sum_{i, j = 1}^n \dpn{a_i^*b_{ij}a_j , \phi}{A} &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij}a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} \\
+    &= \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi}
+    \end{align*}
+
+    Identify $H_\phi$ as a subspace of $H$ and let $\xi = (\pi_\phi(a_1)\xi_\phi, \cdots, \pi_\phi(a_n)\xi_\phi) \in H^n$, then 
+    \[
+    0 \le \sum_{i, j = 1}^n\dpn{\pi_\phi(b_{ij})\pi_\phi(a_j)\xi_\phi, \pi_\phi(a_i)\xi_\phi}{H_\phi} = \dpn{b\xi, \xi}{H^n}
+    \]
+
+    As $(H_\phi, \pi_\phi)$ is a cyclic representation, the above implies that $\dpn{b\eta, \eta}{H^n} \ge 0$ for all $\eta \in H_\phi^n$. Since this holds for all $\phi \in S(A)$, $b \ge 0$ in $M_n(A)$. 
+\end{proof}
+
 
 \begin{definition}[Completely Positive Map]
 \label{definition:completely-positive-map}
@@ -88,216 +188,6 @@
 \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
@@ -335,7 +225,7 @@
         \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}, 
+    By the \hyperref[Schur complement condition]{proposition: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)^* \\
@@ -350,6 +240,75 @@
     \]
 \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 $2$-positive map, then for each $x \in A$,
+    \[
+    \phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B \phi(x^*x)
+    \]
+\end{proposition}
+\begin{proof}[Proof, general case. ]
+    Since
+    \[
+    \begin{bmatrix}
+            1_A &x \\
+            x^* &x^*x
+        \end{bmatrix} = \begin{bmatrix}
+            1_A &x \\
+            0 &0
+        \end{bmatrix}^*\begin{bmatrix}
+            1_A &x \\
+            0 &0
+        \end{bmatrix} \ge 0
+    \]
+
+    and $\phi$ is $2$-positive, 
+    \[
+    \begin{bmatrix}
+            1_B\norm{\phi(1_A)}_B &\phi(x) \\
+            \phi(x)^* &\phi(x^*x)
+        \end{bmatrix} \ge \begin{bmatrix}
+            \phi(1_A) &\phi(x) \\
+            \phi(x)^* &\phi(x^*x)
+        \end{bmatrix} = \phi_2\begin{bmatrix}
+            1_A &x \\
+            x^* &x^*x
+        \end{bmatrix} \ge 0
+    \]
+
+    By \autoref{lemma:norm-matrix-test}, $\phi(x)^*\phi(x) \le \norm{\phi(1_A)}_B\phi(x^*x)$. 
+\end{proof}
+\begin{proof}[Proof, completely positive case. ]
+    Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $B \subset B(H)$. Assume in addition that $\phi$ is completely positive. Let $(K, V, \pi)$ be a \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$, then for each $a \in A$ and $\xi \in H$, 
+    \begin{align*}
+        \dpn{\phi(a^*)\phi(a)\xi, \xi}{H} &= \dpn{V^*\pi(a^*)VV^*\pi(a)V\xi, \xi}{H} = \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\xi}{K} \\
+        &\le \norm{VV^*}_{B(K)}\dpn{\pi(a^*)\pi(a)V\xi, V\xi}{K} \\
+        &\le \norm{V}_{B(K)}^2 \dpn{\pi(a^*a)V\xi, V\xi}{K} = \norm{\phi(1_A)}_{B(H)}\dpn{\phi(a^*a)\xi, \xi}{K}
+    \end{align*}
+\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. 
@@ -378,137 +337,219 @@
     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}
+\begin{example}[Choi]
+    \label{example:choi-roots-of-unity}
 
-    Let $A$ be a $C^*$-algebra, $y, z \in A$, $\phi: A \to A$ be defined by $x \mapsto yxz$, then
+    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 $\phi$ is completely bounded.
-        \item If $y = z^*$, then $\phi$ is completely positive. 
+        \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}
-    (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{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. 
 
-\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 $\xi, \eta \in \complex^n$, 
+    (1): Let $X = \sum_{i, j = 1}^n E_{ij} \otimes x_{ij} \in M_n(A)$, then
     \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 
+        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
-    \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
     \[
-    \sum_{i, j = 1}^n a_{ij}\eta_j\ol{\eta_i} = (\xi \otimes \one)^* \cdot a \cdot (\xi \otimes \one) \ge 0
+    \sum_{k = 1}^n \omega^{k(j-i)} = \begin{cases}
+        0 &j \ne i \\
+        n &j = i
+    \end{cases}
     \]
 
     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*}
-        
+    \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 $x \ne 0$, $-n(E_{11} \otimes x)$ 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^* & V^*U^*U & 1_A
+        \end{bmatrix}\\
+        &= \begin{bmatrix}
+            1_A & U & UV \\
+            U^* & 1_A & U^*UV \\
+            V^*U^* & V^*U^*U & 1_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), 
+    \[
+    \begin{bmatrix}
+        1_A & U & UV \\
+        U^* & 1_A & V \\
+        V^*U^* & V^* & 1_A
+    \end{bmatrix} \ge 0
+    \]
+
+    If the identity is completely positive, then
+    \[
+    \begin{bmatrix}
+        1_A & U & V \circ U \\
+        U^* & 1_A & V \\
+        U^* \circ V^* & V^* & 1_A
+    \end{bmatrix} = \begin{bmatrix}
+        1_A & U & UV \\
+        U^* & 1_A & V \\
+        V^*U^* & V^* & 1_A
+    \end{bmatrix} \ge 0
+    \]
+
+    in $M_3(A^{\text{op}})$. By (4), $VU = U \circ V = V \circ U = UV$. Therefore $A$ is commutative.
 \end{proof}
 
 
 
+

src/op/system/dilation.tex

@@ -18,19 +18,184 @@
     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-polynomial}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$, $(K, S)$ be a power dilation of $T$, and $P \in B(K)$ be the orthogonal projection onto $H$, then
+    \begin{enumerate}
+        \item $(K, S^*)$ is a power dilation of $T^*$. 
+        \item For any $p, q \in \complex[z]$, 
+        \[
+        p(T) + q(T)^* = P[p(S) + q(S)^*]|_{H}
+        \]
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $\xi, \eta \in H$ and $n \in \natp$, $P\xi = \xi$ and $P\eta = \eta$, so
+    \begin{align*}
+        \dpn{(T^*)^n\xi, \eta}{H} &= \dpn{\xi, T^n\eta}{H} = \dpn{\xi, PS^n\eta}{K} = \dpn{P\xi, S^n\eta}{K} \\
+        &= \dpn{(S^*)^nP\xi, P\eta}{K} = \dpn{P(S^*)^n\xi, \eta}{H}
+    \end{align*}
+\end{proof}
+
+
+
 \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 $R$. 
+    Similarly, if $(K, S)$ is a power dilation of $R$ and $(L, T)$ is a power dilation of $S$, then $(L, T)$ is a power dilation of $R$. 
 \end{lemma}
 
+
+\begin{theorem}[Stinespring's Dilation Theorem]
+\label{theorem:stinespring-dilation}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map, then there exists a triple $(\pi, V, K)$, where:
+    \begin{enumerate}[label=(\alph*)]
+        \item $K$ is a complex Hilbert space. 
+        \item $V: H \to K$ is a continuous linear map.
+        \item $\pi: A \to B(K)$ is a unital *-homomorphism.
+    \end{enumerate}
+
+    such that:
+    \begin{enumerate}
+        \item $\norm{\phi(1_A)}_{B(H)} = \norm{V}_{L(H; K)}^2$. 
+        \item For all $a \in A$, $\phi(a) = V^*\pi(a)V$. 
+    \end{enumerate}
+
+    Moreover, if $\phi$ is unital, then:
+    \begin{enumerate}[start=2]
+        \item $V$ is an isometry. 
+        \item Under the identification that $H = V(H) \subset K$, $(K, \pi(a))$ is a dilation of $\phi(a)$ for all $a \in A$. 
+    \end{enumerate}
+    
+    The triple $(\pi, V, K)$ is a \textbf{Stinespring representation} of $\phi$. If $\bracsn{\pi(a)VH|a \in A}$ is dense in $K$, then $(\pi, V, K)$ is \textbf{minimal}\footnote{$K$ can always be trimmed so that $(\pi, V, K)$ is minimal.}.
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 4.1]{Paulsen}}}. ]
+    (a): Let $A \otimes H$ be the algebraic tensor product and define the Hermitian form
+    \[
+    \dpn{\cdot, \cdot}{\phi}: (A \otimes H) \times (A \otimes H)\to \complex \quad \dpn{a \otimes \xi, b \otimes \eta}{\phi} = \dpn{\phi(b^*a)\xi, \eta}{H}
+    \]
+
+    For each $\sum_{j = 1}^n a_j \otimes \xi_j \in A \otimes H$, let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ and
+    \[
+    a = \sum_{i, j = 1}^n E_{ij} \otimes a_{i}^* a_j = \braks{\sum_{i = 1}^n E_{1i} \otimes a_{i}}^*\braks{\sum_{j = 1}^n E_{1j} \otimes a_j} \ge 0
+    \]
+
+    then since $\phi$ is completely positive, 
+    \[
+        \angles{\sum_{j = 1}^n a_j \otimes \xi_j, \sum_{j = 1}^n a_j \otimes \xi_j}_\phi = \sum_{i, j = 1}^n \dpn{\phi(a_i^*a_j)\xi_j, \xi_i}{H} = \dpn{\phi_n(a)\xi, \xi}{H^n} \ge 0
+    \]
+
+    and $\dpn{\cdot, \cdot}{\phi}$ is a pseudo-inner product. 
+    
+    Let $N = \bracsn{u \in A \otimes H| \dpn{u, u}{\phi} = 0}$. For any $u \in N$ and $v \in A \otimes H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \[
+    |\dpn{u, v}{\phi}| \le \dpn{u, u}{\phi}^{1/2} \dpn{v, v}{\phi}^{1/2} = 0
+    \]
+
+    so $N = \bracsn{u \in A \otimes H|\dpn{u, v}{\phi} = 0 \forall v \in A \otimes H}$. 
+
+    Let $K_0 = (A \otimes H)/N$ and $K$ be the completion of $K_0$ with respect to $\dpn{\cdot, \cdot}{\phi}$, then $K$ is a complex Hilbert space. 
+
+    (b): Let $V: H \to K_0$ be defined by $\xi \mapsto 1_A \otimes \xi + N$, then for each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} \le \norm{\phi(1_A)}_{B(H)}\norm{\xi}_H^2
+    \]
+
+    so $\norm{V}_{L(H; K)}^2 \le \norm{\phi(1_A)}_{B(H)}$. 
+
+    (c): For each $a \in A$, let
+    \[
+    \pi(a): K_0 \to K_0 \quad \pi(a)(b \otimes \xi + N) = (ab) \otimes \xi + N
+    \]
+
+    then for any $u = \sum_{j = 1}^n b_j \otimes \xi_j \in A \otimes H$, 
+    \begin{align*}
+        \norm{\pi(a)(u+N)}_{K_0}^2 &= \sum_{i, j = 1}^n \dpn{(ab_j) \otimes \xi_j, (ab_i) \otimes \xi_i}{\phi} = \sum_{i, j = 1}^n \dpn{\phi(b_i^*a^*ab_j)\xi_j, \xi_i}{H} \\
+        &\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 $B(K)$, 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*}
+        \dpn{\pi(a^*)(b \otimes \xi + N), c \otimes \eta + N}{K_0} &= \dpn{(a^*b) \otimes \xi, c \otimes \eta}{\phi}\\
+        &= \dpn{\phi(c^*a^*b)\xi, \eta}{H} = \dpn{b \otimes \xi, (ac) \otimes \eta}{\phi} \\
+        &= \dpn{b \otimes \xi + N, \pi(a)(c \otimes \eta + N)}{K_0}
+    \end{align*}
+
+    By linearity, $\dpn{\pi(a^*)u, v}{K_0} = \dpn{u, \pi(a)v}{K_0}$ for all $u, v \in K_0$. As $K_0$ is dense in $K$, $\pi(a^*) = \pi(a)^*$, so $\pi$ is a unital *-homomorphism. 
+
+    (1): For each $\xi \in H$, 
+    \[
+    \norm{V\xi}_{K_0}^2 = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H}
+    \]
+
+    so
+    \[
+    \norm{V}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\norm{V\xi}_{K_0}^2 = \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{\phi(1_A)\xi, \xi}{H} = w(\phi(1_A)) = \norm{\phi(1_A)}_{B(H)}
+    \]
+
+    by \autoref{proposition:bh-numerical-radius-gymnastics}. 
+
+    (2): For each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{V^*\pi(a)V\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{a \otimes \xi, 1_A \otimes \eta}{\phi} = \dpn{\phi(a)\xi, \eta}{H}
+    \]
+
+    so $V^*\pi(a)V = \phi(a)$. 
+
+    (3): If $\phi$ is unital, then for each $\xi \in H$,
+    \[
+    \dpn{V\xi, V\xi}{K} = \dpn{1_A \otimes \xi, 1_A \otimes \xi}{\phi} = \dpn{\phi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H}
+    \]
+
+    so $V$ is an isometry. 
+
+    (4): Suppose that $\phi$ is unital, then $V$ is an isometry by (3). Under the identification that $H = V(H) \subset K$, let $P \in B(K)$ be the orthogonal projection onto $H$, then for each $a \in A$ and $\xi, \eta \in H$, 
+    \[
+        \dpn{\phi(a)\xi, \eta}{H} = \dpn{\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)P\xi, P\eta}{K} = \dpn{P\pi(a)\xi, \eta}{K}
+    \]
+
+    Therefore $\phi(a) = P\pi(a)|_H$, and $(K, \pi(a))$ is a dilation of $\phi(a)$. 
+    
+\end{proof}
+
+\begin{proposition}[Uniqueness of Minimal Stinespring Representations]
+\label{proposition:stinespring-minimal-uniqueness}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a completely positive map, and $(\pi_1, V_1, K_1), (\pi_2, V_2, K_2)$ be minimal Stinespring representations for $\phi$, then there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$.
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2]{Paulsen}}}. ]
+    Suppose there exists a unitary equivalence $U: K_1 \to K_2$ such that $UV_1 = V_2$, then for each $\seqf{a_j} \subset A$ and $\seqf{\xi_j} \subset H$, 
+    \[
+    U\braks{\sum_{j = 1}^n \pi_1(a_j)V_1\xi_j} = \sum_{j = 1}^n \pi_2(a_j)UV_1\xi_j = \sum_{j = 1}^n \pi_2(a_j)V_2\xi_j
+    \]
+
+    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_i}{K_1} \\
+        &= \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_i}{H} \\
+        &= \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*}
+    
+    
+\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 $T$.
+    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 $(K, U)$ of $T$ such that:
+    \begin{enumerate}
+        \item $K$ is the smallest closed reducing subspace for $U$ containing $H$ such that $T^n = P_HU^n|_H$\footnote{Where $P_H$ is the orthogonal projection onto $H$.} for all $n \in \natz$. 
+        \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}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
-    Using \autoref{lemma:bh-dilation-transitive} it is sufficient to show the following statements:
+\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 $T$. 
         \item For any isometry $V \in B(H)$, there exists a unitary power dilation of $V$. 
@@ -105,18 +270,178 @@
         \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 $V$. 
+    Under the identification that $H = \bracsn{(x, 0)|x \in H} \subset H^2$, let $Q \in B(H^2)$ be the projection onto the first summand, then $QU^n|_{H} = V^n$ for all $n \in \natp$. Therefore $U$ is a unitary power dilation of $V$. 
+\end{proof}
+\begin{proof}[Proof, using Stinespring's Dilation Theorem, {{\cite[Theorem 4.3]{Paulsen}}}. ]
+    (1): Let $\mathbb{T} = \partial B_\complex(0, 1)$. By \autoref{theorem:polynomial-map-positive}, the mapping
+    \[
+    \phi: \complex[z, \ol z] \to B(H) \quad \phi(p + \ol q) = p(T) + q(T)^*
+    \]
+
+    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 $(\pi, V, K)$ be a minimal \hyperref[Stinespring representation]{theorem:stinespring-dilation} for $\phi$ and $U := \pi(z) \in U(B(K))$, then for each $n \in \natz$, 
+    \[
+    T^n = \phi(z^n) = P_H\pi(z^n)|_H = P_HU^n|_H
+    \]
+
+    so $(K, U)$ is a unitary power dilation of $T$. 
+    
+    (U): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $(\pi, V, K)$ is minimal if and only if the span of $\bigcup_{n \in \integer}U^n H$ is dense in $K$, if and only if $U$ admits no non-trivial reducing subspace. 
+    
+    By \hyperref[uniqueness of minimal Stinespring representations]{proposition:stinespring-minimal-uniqueness}, such a dilation is unique. 
 \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 $p \in \complex[x]$, 
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then for any $p, q \in \complex[x]$, 
     \[
-    \norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|
+    \normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
     \]    
 \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 $H$. 
+    By \hyperref[Sz.-Nagy's Dilation Theorem]{theorem:sznagy-dilation}, $T$ admits a unitary power dilation $(K, U)$. Let $P \in B(K)$ be the orthogonal projection onto $H$, then by \autoref{lemma:bh-dilation-transitive}, 
+    \[
+    p(T) + q(T)^* = P[p(U) + q(U)^*]|_{H}
+    \]
 
-    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 $\norm{p(T)}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z)|$. 
+    As $U$ is unitary, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} shows that $\norm{p(U) + q(U)^*}_{B(K)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|$. Therefore
+    \[
+    \normn{p(T) + q(T)^*}_{B(H)} \le \sup_{z \in B_\complex(0, 1)}|p(z) + \ol{q(z)}|
+    \] 
 \end{proof}
+\begin{proof}[Proof, {{\cite[Exercise 2.15]{Paulsen}}}. ]
+    Assume without loss of generality that $\norm{T}_{B(H)} < 1$. Let
+    \[
+    P(T; t) = \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1
+    \]
+    
+    
+    Let $p, q \in \complex[z]$ and $\gamma: [0, 2\pi] \to \complex$ be defined by $\gamma(t) = e^{it}$, then by the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus},
+    \[
+        p(T) = \frac{1}{2\pi i}\int_\gamma \frac{p(\lambda)}{\lambda - T} d\lambda
+    \]
+
+    For each $n \in \natz$, since $\gamma([0, 2\pi]) \subset \partial B_\complex(0, 1)$, $\ol{\lambda^n} = \lambda^{-n}$. After passing through a \hyperref[Neumann series]{lemma:neumann-series}, 
+    \begin{align*}
+    \int_\gamma \frac{\ol{\lambda^n}}{\lambda - T} d\lambda &= \int_\gamma \frac{\lambda^{-n}}{\lambda - T} d\lambda = \int_\gamma \frac{\lambda^{-{(n+1)}}}{1 - T/\lambda} d\lambda \\
+    &= \sum_{k = 0}^\infty T^k\int_{\gamma} \lambda^{-(n+k+1)}d\lambda = \begin{cases}
+        2\pi i  & n = 0 \\
+        0 & n > 0
+    \end{cases}
+    \end{align*}
+
+    By linearity and \hyperref[Cauchy's integral formula]{theorem:cauchy-formula}, 
+    \[
+    \frac{1}{2\pi i}\int_\gamma\frac{\ol{q(\lambda)}}{\lambda - T} d\lambda = \ol{q(0)} = \frac{1}{2\pi i}\int_\gamma \frac{\ol{q(\lambda)}}{\lambda} d\lambda
+    \]
+
+    so
+    \begin{align*}
+        p(T) + \overline{q(0)} &= \frac{1}{2\pi i} \int_\gamma \frac{p(\lambda) + \ol{q(\lambda)}}{\lambda - T} d\lambda \\
+        p(T) + \overline{q(0)} &= \frac{1}{2\pi i} \int_0^{2\pi}\frac{ie^{it}[p(e^{it}) + \ol{q(e^{it})}]}{e^{it} - T} dt \\
+        &= \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{-it}T}dt
+    \end{align*}
+    
+    Similarly, 
+    \[
+    q(T)^* + p(0) = \frac{1}{2\pi}\int_0^{2\pi}\frac{p(e^{it}) + \ol{q(e^{it})}}{1 - e^{it}T^*}dt
+    \]
+
+    Therefore
+    \begin{align*}
+        p(T) + q(T)^* + p(0) + \ol{q(0)} &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}](P(T; t) + 1)dt \\
+        p(T) + q(T)^* &= \frac{1}{2\pi}\int_0^{2\pi}[p(e^{it}) + \ol{q(e^{it})}]P(T; t)dt
+    \end{align*}
+    
+    Now, for each $t \in [0, 2\pi]$, 
+    \[
+    (1 - e^{-it}T)(1 - e^{it}T^*) = 1 - e^{-it}T - e^{it}T^* + TT^*
+    \]
+    
+    so by the resolvent identity, 
+    \begin{align*}
+        P(T; t) &= \frac{1}{1 - e^{-it}T} + \frac{1}{1 - e^{it}T^*} - 1 \\
+        &= (1 - e^{-it}T)^{-1}\braks{1 - TT^*}(1 - e^{it}T^*)^{-1} \ge 0
+    \end{align*}
+
+    Let $M = \sup_{z \in \partial B_\complex(0, 1)}|p(z) + \ol{q(z)}|$. For each $\xi, \eta \in H$, let $\dpn{\xi, \eta}{t} = \dpn{P(T; t)\xi, \eta}{H}$, then $\dpn{\cdot, \cdot}{t}$ is a pseudo inner product on $H$. Hence by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \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\\
+        &\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*}
+
+    Since $I_H = \frac{1}{2\pi}\int_0^{2\pi} P(T; t)dt$, 
+    \[
+    \frac{1}{2\pi}\int_0^{2\pi}\dpn{\xi, \xi}{t} dt = \frac{1}{2\pi}\int_0^{2\pi} \dpn{P(T; t)\xi, \xi}{H}dt = \norm{\xi}_H^2
+    \]
+
+    Therefore $\dpn{[p(T) + q(T)^*]\xi, \eta}{H} \le M\norm{\xi}_H\norm{\eta}_H$, and $\norm{[p(T) + q(T)^*]}_{B(H)} \le M$.
+    
+\end{proof}
+
+
+\begin{theorem}[Von Neumann's Inequality for Matrices]
+\label{theorem:von-neumann-matrix-wermer}
+    Let $n \in \natp$, $T \in M_n(\complex)$ with $\norm{T}_{M_n(\complex)} \le 1$, and $U, V, S \in M_n(\complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $U, V$ are unitary.
+        \item $S = \text{diag}(s_1, \cdots, s_n)$, where $\seqf{s_j} \subset [0, 1]$. 
+        \item $T = USV$. 
+    \end{enumerate}
+
+    Let $B = \ol{B_\complex(0, 1)}^n \subset \complex^n$ and 
+    \[
+    T: B \to M_n(\complex) \quad T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V
+    \]
+
+    and $p \in \complex[z]$, then:
+    \begin{enumerate}
+        \item For each $\xi, \eta \in \complex^n$, let
+        \[
+        f_{\xi, \eta}: B \to \complex \quad (z_1, \cdots, z_n) \mapsto \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n}
+        \]
+
+        then $f_{\xi, \eta}$ achieves its maximum modulus 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 $(z_1, \cdots, z_n) \in \complex^n$, if
+        \[
+        \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \max_{w \in \overline{B}}\normn{p(T(w_1, \cdots, w_n))}_{M_n(\complex)} 
+        \]
+
+        then $T(z_1, \cdots, z_n)$ is unitary. 
+        \item $\norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)}$. 
+        \item $\norm{p(T)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Exercise 2.16]{Paulsen}}}, attributed to Wermer. ]
+    (1): By the Maximum Modulus Principle. 
+
+    (2): Let $(z_1, \cdots, z_n) \in \complex^n$ such that $\normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)}$ is maximal. By compactness of the unit sphere in $\complex^n$, there exists $\xi, \eta \in \partial B_{\complex^n}(0, 1)$ such that
+    \[
+    \normn{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} = \dpn{p(T(z_1, \cdots, z_n))\xi, \eta}{\complex^n} = f_{\xi, \eta}(z_1, \cdots, z_n)
+    \]
+
+    By maximality, $f_{\xi, \eta}$ achieves its maximum modulus at $(z_1, \cdots, z_n)$, so (1) implies that $|z_j| = 1$ for all $1 \le j \le n$ and $\text{diag}(z_1, \cdots, z_n)$ is unitary. Therefore $T(z_1, \cdots, z_n) = U\text{diag}(z_1, \cdots, z_n)V$ is also unitary. 
+
+    (3): Since $(s_1, \cdots, s_n) \in B$, (2) implies that  
+    \begin{align*}
+    \norm{p(T)}_{M_n(\complex)} &\le \max_{z \in B}\norm{p(T(z_1, \cdots, z_n))}_{M_n(\complex)} \\
+    &\le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)}
+    \end{align*}
+    
+    (4): For any $W \in U(M_n(\complex))$, the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} implies that
+    \[
+    \norm{p(W)}_{M_n(\complex)} \le \max_{\lambda \in \sigma_{M_n(\complex)}(W)}|p(\lambda)| \le \max_{\lambda \in \partial B_\complex(0, 1)}|p(\lambda)| = \norm{p}_{H^\infty(D)}
+    \]
+
+    As the above holds for all unitary matrices, 
+    \[
+    \norm{p(T)}_{M_n(\complex)} \le \max_{W \in U(M_n(\complex))}\norm{p(W)}_{M_n(\complex)} \le \norm{p}_{H^\infty(D)}
+    \]
+
+    by (3).
+\end{proof}
+
+

src/op/system/index.tex

@@ -1,6 +1,16 @@
-\chapter{Operator Systems}
+\chapter{Operator Systems and Completely Positive Maps}
 \label{chap:opertaor-systems}
 
-\input{./dilation.tex}
 \input{./os.tex}
+\input{./two.tex}
 \input{./completely.tex}
+\input{./complete-upgrade.tex}
+\input{./module.tex}
+\input{./numerical-radius.tex}
+\input{./schur.tex}
+\input{./dilation.tex}
+\input{./os-mn.tex}
+\input{./arveson.tex}
+\input{./aos.tex}
+\input{./tensor.tex}
+

src/op/system/inj.tex

@@ -0,0 +1,103 @@
+\section{Injective Operator Systems}
+\label{section:injective-ops}
+
+
+By \hyperref[Arveson's Extension Theorem]{theorem:arveson-extension}, $B(H)$ is injective for any Hilbert space $H$. Further, the \hyperref[Choi-Effros Theorem]{theorem:choi-effros} implies that every operator system embeds into $B(H)$, for a suitable Hilbert space $H$. Therefore the category of operator systems, with unital complete order isomorphisms, has sufficiently many injectives. 
+
+\begin{theorem}
+\label{theorem:opsys-predual}
+    Let $S, T$ be operator systems. If $T$ admits a predual, then the space
+    \[
+    UCP(S; T) = \bracsn{\phi: S \to T| \phi \text{ unital, completely positive}}
+    \]
+
+    is compact in the point-weak* topology. 
+\end{theorem}
+
+
+\begin{definition}[Injective]
+\label{definition:injective-operator-system}
+    Let $I$ be an operator system, then $I$ is \textbf{injective} if for any pair of operator systems $S, T$ with $S \subset T$ and unital completely positive map $\phi: S \to I$, there exists a unital completely positive map $\Phi: T \to I$ such that the following diagram commutes
+    \[
+    \xymatrix{
+T \ar@{->}[r]^{\Phi} & I \\
+S \ar@{->}[ru]_{\phi} \ar@{^{(}->}[u] & 
+}
+    \]
+\end{definition}
+
+\begin{definition}[Conditional Expectation]
+\label{definition:os-conditional-expectation}
+    Let $S, T$ be operator systems with $S \subset T$ and $\phi: T \to S$ be a unital completely positive map, then $\phi$ is a \textbf{conditional expectation} if $\phi|_{T} = \text{Id}_T$. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:os-injective-ce}
+    Let $I$ be an operator system, then the following are equivalent:
+    \begin{enumerate}
+        \item $I$ is injective. 
+        \item For every complex Hilbert space $H$ and embedding $\iota: I \to B(H)$, there exists a conditional expectation $\phi: B(H) \to \iota(I)$. In other words, $\phi$ is a unital completely positive map such that the following diagram commutes
+        \[
+        \xymatrix{
+        B(H) \ar@{->}[r]^{\phi} & \iota(I) \\
+        I \ar@{=}[ru] \ar@{^{(}->}[u] & 
+        }
+        \]
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): By the \hyperref[Choi-Effros Theorem]{theorem:choi-effros} and \hyperref[Arveson's Extension Theorem]{theorem:arveson-extension}. 
+\end{proof}
+% "obvious"
+
+\begin{example}[Dixmier Algebra]
+    \label{example:sup-not-vna}
+    Let
+    \[
+    A = B^\infty([0, 1])/\bracsn{f \in B^\infty([0, 1])| \bracsn{f \ne 0} \text{ is meagre}}
+    \]
+    
+    then $A$ is an injective, order complete $C^*$-algebra, but $A$ is not a von Neumann algebra. 
+\end{example}
+
+
+\begin{proposition}
+\label{proposition:os-injective-cstar}
+    Let $H$ be a complex Hilbert space, $I \subset B(H)$ be an operator system, and $\phi: B(H) \to I$ be a conditional expectation, then $(x, y) \mapsto \phi(xy)$ is a well-defined product on $I$, making it a $C^*$-algebra. 
+\end{proposition}
+
+
+\begin{proposition}
+\label{proposition:os-injective-cstar}
+    Let $I$ be an injective operator system, then there exists a $C^*$-algebra $A$ such that $I$ is isomorphic to $A$ as operator systems. 
+\end{proposition}
+
+\begin{proposition}
+\label{proposition:os-maximal-essential-extension}
+    Let $S$ be an operator system, then there exists a pair $(\wh S, \iota)$ such that:
+    \begin{enumerate}
+        \item $(\wh S, \iota)$ is an essential extension of $S$. 
+        \item For any proper\footnote{Non-isomorphic.} extension $(T, j)$ of $\wh S$, $(Z, j \circ \iota)$ is not an essential extension of $S$. In other words, every essential extension of $(\wh S, \iota)$ is isomorphic to $(\wh S, \iota)$.
+    \end{enumerate} 
+
+    The pair $(\wh S, \iota)$ is a \textbf{maximal essential extension} of $S$. 
+\end{proposition}
+\begin{proof}
+    Zorn's lemma. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:maximal-essential-injective}
+    
+\end{theorem}
+
+
+
+% I sure wonder how that will happen 
+
+
+
+
+
+

src/op/system/module.tex

@@ -0,0 +1,270 @@
+\section{Multiplicative Domains}
+\label{section:completely-positive-multiplicative-domain}
+
+
+\begin{definition}[Module Map]
+\label{definition:module-map}
+    Let $A, B, C$ be unital $C^*$-algebras with $C \subset A \cap B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. 
+    
+    Let $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{$C$-bimodule map} if $\phi(c_1ac_2) = c_1\phi(a)c_2$ for all $a \in A$ and $c_1, c_2 \in C$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:matrix-positive-topleft-corner-zero}
+    Let $A$ be a $C^*$-algebra and $x, y \in A$, then
+    \[
+    \begin{bmatrix}
+        0 & x \\
+        x^* & y
+    \end{bmatrix} \ge 0
+    \]
+
+    if and only if $x = 0$ and $y \ge 0$. 
+\end{lemma}
+\begin{proof}
+    ($\Rightarrow$): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, assume without loss of generality that there exists a complex Hilbert space $H$ such that $A \subset B(H)$ is a $C^*$-subalgebra. In which case, for any $\xi, \eta \in H$, 
+    \begin{align*}
+        \angles{\begin{bmatrix}
+        0 & x \\
+        x^* & y
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}}_{H^2} &= \angles{\begin{bmatrix}
+    x\eta \\ x^*\xi + y\eta
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}}_{H^2} \\
+    &= \dpn{x^*\xi, \eta}{H} + \dpn{x\eta, \xi}{H} + \dpn{y\eta, \eta}{H} \\
+    &= 2\text{Re}(\dpn{x^*\eta, \xi}{H}) + \dpn{y\eta, \eta}{H}
+    \end{align*}
+
+    If $x \ne 0$, then there exists $\xi_0, \eta_0 \in H$ such that $\dpn{x^*\eta_0, \xi_0}{H}\ne 0$. Therefore
+    \[
+    \inf_{\xi \in H}\text{Re}(\dpn{x^*\eta_0, \xi}{H}) = -\infty
+    \]
+
+    and the given matrix is not positive. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:positive-multiplicative-domain-estimate}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, and $\phi: A \to B(H)$ be a completely positive map with $\norm{\phi(1_A)}_{B(H)} \le 1$, then for any $a, b \in A$, 
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+\end{lemma}
+\begin{proof}
+    Let $(\pi, V, K)$ be a \hyperref[Stinespring representation]{theorem:stinespring-dilation} of $\phi$, then $\norm{V}_{L(H; K)}^2 = \norm{\phi(1_A)}_{B(H)} \le 1$. As such, for any $a, b \in A$, 
+    \begin{align*}
+    \phi(ba) &= V^*\pi(ba)V = V^*\pi(b)I_K\pi(a)V \\
+    \phi(b)\phi(a) &= V^*\pi(b)VV^*\pi(a)V\\
+        \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} &= \normn{V^*\pi(b)(I_K - VV^*)\pi(a)V}_{B(H)} \\
+        &\le \norm{b}_A \normn{(I_K - VV^*)\pi(a)V}_{L(H; K)}
+    \end{align*}
+
+    Let $T = (I_K - VV^*)\pi(a)V$, then $\norm{T}_{L(H; K)}^2 = \norm{T^*T}_{B(H)}$. As $\norm{V}_{L(H; K)}^2 \le 1$, $0 \le I_K - VV^* \le I_K$, so
+    \[
+    T^*T = V^*\pi(a^*)(I_K - VV^*)^2\pi(a)V \le V^*\pi(a^*)(I_K - VV^*)\pi(a)V
+    \]
+
+    and
+
+    \begin{align*}
+        \norm{T}_{L(H; K)}^2 &\le  \norm{V^*\pi(a^*)(I_K - VV^*)\pi(a)V}_{B(H)} \\
+        &= \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}
+    \end{align*}
+    
+    Therefore
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+\end{proof}
+
+
+\begin{definition}[Multiplicative Domains]
+\label{definition:positive-multiplicative-domains}
+    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a unital\footnote{For without this assumption the domains will not be populated.} 4-positive map, then
+    \begin{enumerate}
+        \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$. The space
+        \[
+        R = \bracsn{a \in A|\phi(a)^*\phi(a) = \phi(a^*a)}
+        \]
+        
+        is a subalgebra of $A$, and $\phi|_R$ is a homomorphism. 
+        \item For any $a \in A$, $\phi(a)\phi(a)^* = \phi(aa^*)$ if and only if $\phi(ab) = \phi(a)\phi(b)$ for all $b \in A$. The space
+        \[
+        L = \bracsn{a \in A| \phi(a)\phi(a)^* = \phi(aa^*)}
+        \]
+        
+        is a subalgebra of $A$, and $\phi|_L$ is a homomorphism. 
+        \item $C = L \cap R$ is a $C^*$-subalgebra of $A$ with $1_C = 1_A$, and $\phi|_{C}$ is a unital *-homomorphism. 
+    \end{enumerate}
+
+    The spaces $R$ and $L$ are the \textbf{right} and \textbf{left} \textbf{multiplicative domains} of $\phi$, respectively. The $C^*$-subalgebra $C$ is the \textbf{multiplicative domain} of $\phi$. 
+\end{definition}
+\begin{proof}[Proof of 4-positive case, using Kadison-Schwarz, {{\cite[Theorem 3.18]{Paulsen}}}. ]
+    (1, $\Leftarrow$): Suppose that $\phi(ba) = \phi(b)\phi(a)$ for all $b \in A$, then $\phi(a^*) = \phi(a)^*\phi(a)$ after plugging in $b = a^*$. 
+
+    (1, $\Rightarrow$): Suppose that $\phi(a^*a) = \phi(a)^*\phi(a)$. By the \hyperref[Kadison-Schwarz inequality]{proposition:cauchy-schwarz-2-positive}, 
+    \begin{align*}
+        \phi_2\begin{bmatrix}
+            a & b^* \\
+            0 & 0
+        \end{bmatrix}^*\phi_2\begin{bmatrix}
+            a & b^* \\
+            0 & 0
+        \end{bmatrix} &\le \phi_2\begin{bmatrix}
+            a^*a & a^*b^* \\
+            ba & bb^*
+        \end{bmatrix} \\
+        \begin{bmatrix}
+            \phi(a^*)\phi(a) & \phi(a^*)\phi(b) \\
+            \phi(b)\phi(a) & \phi(b)\phi(b^*)
+        \end{bmatrix} &\le 
+        \begin{bmatrix}
+            \phi(a^*a) & \phi(a^*b) \\
+            \phi(ba) & \phi(bb^*)
+        \end{bmatrix} 
+    \end{align*}
+
+    so
+    \[
+    \begin{bmatrix}
+            0 & \phi(a^*b^*) - \phi(a^*)\phi(b) \\
+            \phi(b)\phi(a) - \phi(ba) & \phi(b)\phi(b^*) - \phi(bb^*)
+        \end{bmatrix} \ge 0
+    \]
+
+    By \autoref{lemma:matrix-positive-topleft-corner-zero}, $\phi(b)\phi(a) - \phi(ba) = 0$. 
+\end{proof}
+\begin{proof}[Proof of completely positive case, using Stinespring's Dilation Theorem. ]
+    (1, $\Rightarrow$): Suppose that $\phi$ is completely positive, then by \autoref{lemma:positive-multiplicative-domain-estimate}, 
+
+    \[
+    \normn{\phi(ba) - \phi(b)\phi(a)}_{B(H)} \le \norm{b}_A \normn{\phi(a^*a) - \phi(a)^*\phi(a)}_{B(H)}^{1/2}
+    \]
+
+    for all $a, b \in A$. In particular, if $a \in R$, then $\phi(ba) = \phi(b)\phi(a)$. 
+\end{proof}
+
+
+
+
+\begin{proposition}
+\label{proposition:stinespring-multiplicative-domain}
+    Let $A$ be a unital $C^*$-algebra, $H$ be a complex Hilbert space, $\phi: A \to B(H)$ be a unital completely positive map, and $(\pi, V, K)$ be a minimal Stinespring representation of $\phi$, then:
+    \begin{enumerate}
+        \item For any $a \in A$, $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$\footnote{This also shows that multiplicative domains of $\phi$ are subalgebras of $A$. }. In other words, $a$ is in the right multiplicative domain of $\phi$ if and only if $V(H)$ is invariant for $\pi(a)$. 
+        \item For any $a \in A$, $a$ is in the multiplicative domain of $\phi$ if and only if $V(H)$ is reducing for $\pi(a)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Since $\phi$ is unital, $V$ is an isometry. Identify $H = V(H) \subset K$ and let $P = VV^* \in B(K)$ be the orthogonal projection onto $H$. 
+    
+    Let $a \in A$ and $\xi, \eta \in H$, then
+    \begin{align*} 
+    \dpn{\phi(a^*)\phi(a)\xi, \eta}{H} &= \dpn{\pi(a^*)VV^*\pi(a)V\xi, V\eta}{K} = \dpn{\pi(a)^*P\pi(a)P\xi, P\eta}{K} \\
+    \dpn{\phi(a^*a)\xi, \eta}{H} &= \dpn{\pi(a^*a)V\xi, V\eta}{K} = \dpn{\pi(a)^*\pi(a)P\xi, P\eta}{K}
+    \end{align*}
+
+    By \autoref{lemma:invariant-projection-test}, $\pi(a)^*P\pi(a)P = \pi(a)^*\pi(a)P$ if and only if $P(K) = V(H)$ is an invariant subspace for $\pi(a)$. Therefore $\phi(a)^*\phi(a) = \phi(a^*a)$ if and only if $V(H)$ is invariant for $\pi(a)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:unital-convergence-multiplicative-domain}
+    Let $A$ be a unital $C^*$-algebra, $K$ be a complex Hilbert space, and $\bracsn{\phi^{(n)}}_{1}^\infty  \subset L(A; B(H))$ be completely positive maps such that $limv{n}\phi^{(n)}(1_A) = I$, then:
+    \begin{enumerate}
+        \item For any $a \in A$, $\limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0$ if and only if $\limv{n}\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a) = 0$ for all $b \in A$. In particular, 
+        \[
+        R = \bracs{a \in A \bigg | \limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0}
+        \]
+
+        is a subalgebra of $A$. 
+        \item For any $a \in A$, $\limv{n}\phi^{(n)}(aa^*) - \phi^{(n)}(a)\phi^{(n)}(a)^* = 0$ if and only if $\limv{n}\phi^{(n)}(ab) - \phi^{(n)}(a)\phi^{(n)}(b) = 0$ for all $b \in A$. In particular, 
+        \[
+        R = \bracs{a \in A \bigg | \limv{n}\phi^{(n)}(aa^*) - \phi^{(n)}(a)\phi^{(n)}(a)^* = 0}
+        \]
+
+        is a subalgebra of $A$. 
+        \item $C = L \cap R$ is a $C^*$-subalgebra of $A$ with $1_C = 1_A$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): It is sufficient to show that if $\limv{n}\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a) = 0$, then $\limv{n}\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a) = 0$ for all $b \in A$. 
+    
+    After rescaling and trimming off an initial segment, assume without loss of generality that $\normn{\phi^{(n)}(1_A)} \le 1$ for all $n \in \natp$. By \autoref{lemma:positive-multiplicative-domain-estimate}, 
+    \[
+    \normn{\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a)}_{B(H)} \le \norm{b}_A \normn{\phi^{(n)}(a^*a) - \phi^{(n)}(a)^*\phi^{(n)}(a)}_{B(H)}^{1/2}
+    \]
+
+    for all $a, b \in A$. In particular, if $a \in R$, then $\limv{n}{\phi^{(n)}(ba) - \phi^{(n)}(b)\phi^{(n)}(a)}_{B(H)} = 0$. 
+    
+    
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:module-map-characterisation}
+    Let $A, B, C$ be $C^*$-algebras with $C \subset A \cap B$ and $1_A = 1_B = 1_C$, and $\phi: A \to B$ be a completely positive map, then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is a $C$-bimodule map. 
+        \item $\phi(c) = c\phi(1_C)$ for all $c \in C$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2) $\Rightarrow$ (1): For any $c \in C$, $\phi(1_C)c = \phi(c^*)^* = \phi(c) = c\phi(1_C)$, so $\phi(1_C)$ commutes with $C$. By the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus} and the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, 
+    \[
+    \phi(c) = \phi(1_C)^{1/2} \cdot c \cdot \phi(1_C)^{1/2}
+    \]
+
+    for all $c \in C$. 
+
+
+    Using \hyperref[Gelfand-Naimark Theorem]{theorem:gns}, let $H$ be a complex Hilbert space such that $B \subset B(H)$, and $(\pi, V, K)$ be a Stinespring representation for $\phi$, then for any $c \in C$, 
+    \begin{align*}
+        (\pi(c)V - Vc)^*(\pi(c)V - Vc) &= (V^*\pi(c^*) - c^*V^*)(\pi(c)V - Vc) \\
+        &= V^*\pi(c^*)\pi(c)V - V^*\pi(c^*)Vc \\
+        &- c^*V^*\pi(c)V + c^*V^*Vc \\
+        &= \phi(c^*c) - \phi(c^*)c - c^*\phi(c) + c^*\phi(1_C)c \\
+        &= c^*c\phi(1_C) - c^*\phi(1_C)c + c^*c\phi(1_C) + c^*\phi(1_C)c \\
+        &= 0
+    \end{align*}
+
+    so $\pi(c)V = Vc$. As such, for any $a \in A$, and $c, c' \in C$,
+    \[
+    \phi(cac') = V^*\pi(c)\pi(a)\pi(c')V = cV^*\pi(a)Vc' = c\phi(a)c'
+    \]
+\end{proof}
+
+
+\begin{definition}[Conditional Expectation]
+\label{definition:cstar-conditional-expectation}
+    Let $A, B$ be unital $C^*$-algebras and $\phi: A \to B$ be a linear map, then $\phi$ is a \textbf{conditional expectation} if:
+    \begin{enumerate}
+        \item $\phi$ is unital and completely positive. 
+        \item $\phi|_B$ is the identity. 
+    \end{enumerate}
+\end{definition}
+
+
+\begin{example}[Conditional Expectation]
+    \label{example:matrix-conditional-expectation}
+    Let $\phi: M_n(\complex) \to \complex^n$ be defined by $\phi(A)_k = a_{kk}$, then:
+    \begin{enumerate}
+        \item $\phi$ is unital and completely positive. 
+        \item The multiplicative domain of $\phi$ is the diagonal matrices on $M_n(\complex)$. 
+    \end{enumerate}
+\end{example}
+\begin{proof}
+    (1): Identify $\complex^n$ as the diagonal matrices in $M_n(\complex)$, then for each $A \in M_n(\complex)$, 
+    \[
+    \phi(A) = \sum_{j = 1}^n E_{jj}AE_{jj}
+    \]
+
+    By \autoref{proposition:multiply-completely-bounded}, $\phi$ is completely positive. 
+
+    (2): ???
+\end{proof}
+

src/op/system/numerical-radius.tex

@@ -0,0 +1,168 @@
+\section{The Numerical Radius}
+\label{section:numerical-radius}
+
+
+\begin{definition}[Numerical Radius]
+\label{definition:bh-numerical-radius}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then
+    \[
+    w(T) = \sup_{\substack{\xi \in H \\ \norm{\xi}_H = 1}}|\dpn{T\xi, \xi}{H}|
+    \]
+
+    is the \textbf{numerical radius} of $T$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:bh-numerical-radius-gymnastics}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The numerical radius $w: B(H) \to [0, \infty)$ is a norm on $B(H)$. 
+        \item For any $T \in B(H)$, $w(T) \le 1$ if and only if $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. 
+        \item For each $T \in B(H)_{sa}$, $w(T) = \norm{T}_{B(H)}$.
+        \item For each $T \in B(H)$, $w(T) \le \norm{T}_{B(H)} \le 2w(T)$. Both inequalities are sharp in general. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For any $S, T \in B(H)$ and $\xi \in H$ with $\norm{\xi}_H = 1$, 
+    \[
+    |\dpn{(S + T)\xi, \xi}{H}| \le |\dpn{S\xi, \xi}{H}| + |\dpn{T\xi, \xi}{H}| \le w(S) + w(T)
+    \]
+
+    As the above holds for all $\xi \in H$, $w(S + T) \le w(S) + w(T)$. Should $w(T) = 0$, then $\dpn{T\xi, \xi}{H} = 0$ for all $\xi \in H$. By \hyperref[polarisation]{proposition:polarisation-complex}, for any $\xi, \eta \in H$, 
+    \[
+    \dpn{T\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} =  0
+    \]
+
+    so $T = 0$ and $w$ is a norm on $B(H)$. 
+
+    (2): For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\lambda \in \partial B_\complex(0, 1)$, 
+    \begin{align*}
+        \dpn{[2 + (\lambda T) + (\lambda T)^*]\xi, \xi}{H} &= 2 + \dpn{\lambda T \xi, \xi}{H} + \dpn{\xi, \lambda T \xi}{H} \\
+        &= 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H}
+    \end{align*}
+
+    If $w(T) \le 1$, then $2\text{Re}(\dpn{\lambda T\xi, \xi}{H}) \ge -2$, and $2 + (\lambda T) + (\lambda T)^* \ge 0$. 
+
+    Suppose that $2 + (\lambda T) + (\lambda T)^* \ge 0$ for all $\lambda \in \partial B_\complex(0, 1)$. For any $\xi \in H$ with $\norm{\xi}_H = 1$ and $\dpn{T\xi, \xi}{H} \ne 0$, let $\lambda = -\ol{\sgn(\dpn{T\xi, \xi}{H})}$, then 
+    \begin{align*}
+        \dpn{\lambda T\xi, \xi}{H} &= \lambda \dpn{T\xi, \xi}{H} = -\ol{\sgn(\dpn{T\xi, \xi}{H})} \dpn{T\xi, \xi}{H} = -|\dpn{T\xi, \xi}{H}| \\
+        0 &\le 2 + 2\text{Re}\dpn{\lambda T \xi, \xi}{H} = 2 - 2|\dpn{T\xi, \xi}{H}| 
+    \end{align*}
+
+    so $|\dpn{T\xi, \xi}{H}| \le 1$, and $w(T) \le 1$. 
+    
+    (3): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_{B(H)}$. On the other hand, if $w(T) \le 1$, then for each $\xi, \eta \in H$ with $\dpn{T\xi, \eta}{H} \in \real$, 
+    \begin{align*}
+        \dpn{T\xi, \eta}{H} &= \frac{1}{4}\sum_{k = 0}^3 i^k\dpn{T(\xi + i^k\eta), \xi + i^k\eta}{H} \\
+        &= \frac{1}{4}(\dpn{T(\xi + \eta), \xi + \eta}{H} - \dpn{T(\xi - \eta), \xi - \eta}{H}) \\
+        &\le \frac{w(T)}{4}(\norm{\xi + \eta}_H^2 + \norm{\xi - \eta}_H^2) \\
+        &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{\eta}_H^2)
+    \end{align*}
+
+    by \hyperref[polarisation]{proposition:polarisation-complex} and the parallelogram law. In particular, 
+    \begin{align*}
+    \norm{T\xi}_{H}^2 &\le \frac{1}{2}(\norm{\xi}_H^2 + \norm{T\xi}_H^2) \\
+    \norm{T\xi}_H^2 &\le \norm{\xi}_H^2
+    \end{align*}
+
+    (4): For any $\xi \in H$ with $\norm{\xi}_H = 1$, $|\dpn{T\xi, \xi}{H}| \le \norm{T}_{B(H)}$, so $w(T) \le \norm{T}_{B(H)}$. The sharp case is given by $I_H$. 
+
+    On the other hand, if $w(T) \le 1$, then $2 + T + T^* = 2 + 2\text{Re}(T) \ge 0$ and $2 - T - T^* = 2 - 2\text{Re}(T) \ge 0$ imply that $\norm{\text{Re}(T)}_{B(H)} \le 1$. Similarly, $\norm{\text{Im}(T)}_{B(H)} \le 1$ as well, so $\norm{T}_{B(H)} \le \norm{\text{Re}(T)}_{B(H)} + \norm{\text{Im}(T)}_{B(H)} \le 2$. Therefore $\norm{T}_{B(H)} \le 2w(T)$. 
+
+    For any $\xi, \eta \in \complex$ with $\norm{(\xi, \eta)}_{\complex^2} = 1$, by \hyperref[Young's inequality]{lemma:young-inequality},
+    \[
+    \abs{\angles{\begin{bmatrix}
+        0 &2 \\
+        0 & 0
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\
+        \eta
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\
+        \eta
+    \end{bmatrix}}_{\complex^2}} = 2|\eta \ol{\xi}| \le |\xi|^2 + |\eta|^2 \le 1
+    \]
+
+    so the above matrix has numerical radius $1$. However, its operator norm is $2$. Thus the second inequality is also sharp. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bh-numerical-radius-positive}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $H$ be a complex Hilbert space, and $\phi: S \to B(H)$ be a unital positive map, then for each $a \in S$, $w(\phi(a)) \le \norm{a}_A$. 
+\end{proposition}
+\begin{proof}
+    Let $a \in S$ with $\norm{a}_A \le 1$, then for each $\lambda \in \ol{B_\complex(0, 1)}$, $2 + (\lambda a) + (\lambda a)^* \ge 0$. Since $\phi$ is positive, $2 + (\lambda \phi(a)) + (\lambda \phi(a))^* \ge 0$ as well. Therefore $w(\phi(a)) \le 1$ by \autoref{proposition:bh-numerical-radius-gymnastics}. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:cyclic-forward-shift-radius}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 1}^{n-1}E_{(j+1)j} \otimes T \in M_n(B(H))$, $\bracsn{e_j}_0^{n-1} \subset \complex^n$ be the canonical basis for $\complex^n$, and
+    \[
+    S_n: \complex^n \to \complex^n \quad S_ne_j = e_{j+1 \mod n}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S_n$ is unitarily equivalent to $\text{diag}(\omega^0, \cdots, \omega^{n-1})$. 
+        \item $w(T) = w(T \otimes S_n)$. 
+        \item $w(R) \le w(T \otimes S_n)$. 
+        \item $\text{Re}\dpn{R\xi, \xi}{H^n} \le 1$ for all $\xi \in H^n$ with $\norm{\xi}_{H^n} = 1$ if and only if $w(R) \le 1$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $0 \le k \le n - 1$, let $f_k = \sum_{j = 0}^{n-1}\omega^{-jk}e_j$, then
+    \begin{align*}
+    S_nf_k &= \sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1 \mod n} = \sum_{j = 0}^{n-1}\omega^{-(j-1)k}e_{j} \\
+    &= \omega^{k}\sum_{j = 0}^{n-1}\omega^{-jk}e_{j} = \omega^kf_k
+    \end{align*}
+
+    For each $0 \le k < l \le n-1$,
+    \[
+        \dpn{f_k, f_l}{\complex^n} = \sum_{j = 0}^{n-1} \omega^{-jk} \cdot \ol{\omega^{-jl}} = \sum_{j = 0}^{n-1} \omega^{-j(k-l)} = 0
+    \]
+
+    so $\bracsn{f_k}_0^{n-1} \subset \complex^n$ is an orthogonal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$. 
+
+    (2): Let $\bracsn{f_j}_{0}^{n-1} \subset \complex^n$ be an orthonormal eigenbasis for $S_n$, with eigenvalues $\bracsn{\omega^k}_0^{n-1}$. For any $\bracsn{\xi_j}_0^{n-1} \subset H$, let $\xi = \sum_{j = 0}^{n-1}\xi_j \otimes f_j$, then 
+    \begin{align*}
+        \dpn{(T \otimes S_n)\xi, \xi}{H^n} &= \sum_{j, k = 0}^{n-1}\dpn{(T \otimes S_n)(\xi_j \otimes f_j), \xi_k \otimes f_k}{H^n} \\
+        &= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H} \dpn{S_nf_j, f_k}{\complex^n} = \sum_{j = 0}^{n-1}\omega^j\dpn{T\xi_j, \xi_j}{H} \\
+        |\dpn{(T \otimes S_n)\xi, \xi}{H^n}| &\le \sum_{j = 0}^{n-1}|\omega^j| \cdot w(T) \cdot \norm{\xi_j}_{H}^2 = w(T)\norm{\xi}_{H^n}^2
+    \end{align*}
+
+    so $w(T \otimes S_n) \le w(T)$. On the other hand, for any $\xi \in H$, 
+    \[
+    |\dpn{T\xi, \xi}{H}| = |\dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H^n}| \le w(T \otimes S_n)\norm{\xi}_{H}^2
+    \]
+
+    so $w(T) \le w(T \otimes S_n)$. 
+
+    (3): Let $\xi = (\xi_1, \cdots, \xi_{n}) \in H^n$, then 
+    \begin{align*}
+        \dpn{R\xi, \xi}{H^n} &= \sum_{j = 1}^{n-1}\dpn{(E_{(j+1)j}\otimes T)\xi, \xi}{H^n} = \sum_{j = 1}^{n-1}  \dpn{T\xi_j, \xi_{j+1}}{H} \\
+        &= \dpn{T0, \xi_1}{H} + \dpn{T\xi_n, 0}{H} + \sum_{j = 1}^{n-1}  \dpn{T\xi_j, \xi_{j+1}}{H} \\
+        &= \dpn{(T \otimes S_{n+1})(\xi_1, \cdots, \xi_n, 0), (\xi_1, \cdots, \xi_n, 0)}{H^{n+1}}
+    \end{align*}
+
+    so $w(R) \le w(T \otimes S_{n+1}) = w(T) = w(T \otimes S_n)$ by (2).
+    
+    (4): Suppose that $\text{Re}\dpn{R\xi, \xi}{H^n} \le 1$ for all $\xi \in H^n$ with $\norm{\xi}_{H^n} \le 1$.
+    
+    Let $\xi = (\xi_1, \cdots, \xi_n) \in H^n$ with $\norm{\xi}_{H^n} = 1$, and assume without loss of generality that $\dpn{R\xi, \xi}{H^n} \ne 0$. Let $\omega = \sgn(\dpn{R\xi, \xi}{H^n})$ and $\xi_\omega = (\omega^1\xi_1, \cdots, \omega^n\xi_n)$, then 
+    \begin{align*}
+        \dpn{R\xi_\omega, \xi_\omega}{H^n} &= \sum_{j = 1}^{n-1}  \dpn{\omega^j T\xi_j, \omega^{j+1}\xi_{j+1}}{H} =  \sum_{j = 1}^{n-1}  \ol{\omega}\dpn{T\xi_j, \xi_{j+1}}{H} \\
+        &= \ol{\sgn(\dpn{R\xi, \xi}{H^n})} \dpn{R\xi, \xi}{H^n} = |\dpn{R\xi, \xi}{H^n}|
+    \end{align*}
+
+    As $\norm{\xi_\omega}_{H^n} = \norm{\xi}_{H^n} = 1$,
+    \[
+    |\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, \xi_\omega}{H^n} = \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n} \le 1
+    \]
+
+    Therefore $w(R) \le 1$ as well. 
+\end{proof}
+
+
+
+

src/op/system/os-mn.tex

@@ -0,0 +1,109 @@
+\section{Completely Positive Maps into $M_n(\complex)$}
+\label{section:complete-positive-mn}
+
+
+\begin{lemma}
+\label{lemma:operator-mn-duality-correspondence}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $\phi \in L(S; M_n(\complex))$. Define
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item Let $\xi = \sum_{j = 1}^n e_j \otimes e_j \in \complex^n \otimes \complex^n$, then
+        \[
+        \dpn{a, \tilde \phi}{M_n(S)} = \dpn{\phi_n(a)\xi, \xi}{\complex^n \otimes \complex^n}
+        \]
+
+        for all $a \in M_n(S)$. 
+        \item The mapping $\phi \mapsto \tilde \phi$ is a linear isomorphism between $M_n(S)^*$ and $L(S; M_n(\complex))$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $a \in A$ and $1 \le k, l \le n$, 
+    \begin{align*}
+        \dpn{\phi_n(E_{kl} \otimes a)\xi, \xi}{\complex^n \otimes \complex^n} &= \sum_{i, j = 1}^n \dpn{\phi_n(E_{kl} \otimes a)(e_j \otimes e_j), (e_i \otimes e_i)}{\complex^n \otimes \complex^n} \\
+        &= \sum_{i, j = 1}^n \dpn{E_{kl} e_j, e_i}{\complex^n}\dpn{\phi(a)e_j, e_i}{\complex^n} = \phi(a)_{kl}
+    \end{align*}
+    
+    (2): Let $\phi_0: M_n(S) \to \complex$ be a linear map, and let
+    \[
+    \phi: S \to M_n(\complex) \quad a \mapsto \sum_{i, j = 1}^n \dpn{E_{ij} \otimes a, \phi_0}{M_n(S)} \cdot E_{ij}
+    \]
+
+    then for any $1 \le k, l \le n$ and $a \in A$,
+    \[
+    \dpn{E_{kl} \otimes a, \tilde \phi}{M_n(S)} = \phi(a)_{kl} = \dpn{E_{kl} \otimes a, \phi_0}{M_n(S)}
+    \]
+
+    Therefore the mapping $\phi \to \tilde \phi$ is an isomorphism. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:operator-mn-linear-correspondence}
+    Let $S$ be an operator system, $\phi \in L(S; M_n(\complex))$, and
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is completely positive. 
+        \item $\phi$ is $n$-positive. 
+        \item $\tilde \phi \in M_n(S)^*$ is a positive linear functional. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    See \autoref{proposition:matrix-order-positive-characterisation}. 
+\end{proof}
+
+\begin{theorem}[Krein's Extension Theorem]
+\label{theorem:krein-extension}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, and $\phi: S \to M_n(\complex)$ be completely positive, then there exists a completely positive map $\Phi: A \to M_n(\complex)$ such that $\Phi|_S = \phi$. 
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then by \autoref{theorem:operator-mn-linear-correspondence}, $\tilde \phi$ is a positive linear functional. As $S$ is an operator system, $\normn{\tilde \phi}_{S^*} = \tilde \phi(1_A)$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach} and \autoref{lemma:operator-mn-duality-correspondence}, there exists $\Phi \in L(A; M_n(\complex))$ such that $\tilde \Phi|_{M_n(S)} = \tilde \phi$ and $\Phi|_S = \phi$. By \autoref{theorem:cstar-positive-algebraic}, $\tilde \Phi$ is positive, so $\Phi$ is completely positive by \autoref{theorem:operator-mn-linear-correspondence}. 
+\end{proof}
+
+
+
+\begin{definition}[Non-Entangled Cone]
+\label{definition:non-entangled-positive-cone}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, then
+    \[
+    M_n(\complex)_+ \otimes S_+ := \bracs{\sum_{j = 1}^na_j \otimes b_j \bigg | \seqf{a_j} \subset M_n(\complex)_+, \seqf{b_j} \subset S_+, n \in \natp}
+    \]
+
+    is the \textbf{non-entangled positive cone} of $M_n(S)$. 
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:op-mn-positive}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$ be an operator system, $\phi \in L(S; M_n(\complex))$, and
+    \[
+    \tilde \phi: M_n(S) \to \complex \quad E_{ij} \otimes a \mapsto \phi(a)_{ij}
+    \]
+
+    then $\phi$ is positive if and only if $\tilde \phi|_{M_n(\complex)_+ \otimes S_+} \ge 0$. 
+\end{lemma}
+\begin{proof}
+    ($\Rightarrow$): Let $a \in M_n(\complex)_+$ and $b \in S_+$. Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, assume without loss of generality that $a$ is a rank one projection. That is, there exists $\alpha = (\alpha_1, \cdots, \alpha_n) \subset \complex^n$ such that $a = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j E_{ij}$. As such,
+    \[
+    \dpn{a \otimes b, \tilde \phi}{M_n(S)} = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j \phi(b)_{ij} = \dpn{\phi(b)\alpha, \alpha}{\complex^n} \ge 0
+    \]
+
+    ($\Leftarrow$): Let $b \in S_+$, $\alpha = (\alpha_1, \cdots, \alpha_n) \in \complex^n$, and $a = \alpha\alpha^*$, then
+    \[
+        \dpn{\phi(b)\alpha, \alpha}{\complex^n} = \sum_{i, j = 1}^n \ol \alpha_i \alpha_j \phi(b)_{ij} = \dpn{a \otimes b, \tilde \phi}{M_n(S)} \ge 0
+    \]
+    
+\end{proof}
+
+

src/op/system/os.tex

@@ -21,6 +21,24 @@
     Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{+} = \bracsn{x \in S_{sa}|x \ge 0}$ is the \textbf{positive cone} of $S$. 
 \end{definition}
 
+\begin{lemma}
+\label{lemma:operator-system-positive-approximate}
+    Let $A$ be a $C^*$-algebra, $S \subset A$ be an operator system, then $(\ol{S})_+ = \ol{S_+}$. 
+\end{lemma}
+\begin{proof}
+    Let $x \in (\ol{S})_+$, then there exists $\seq{x_n} \subset S$ such that $x_n \to x$ as $n \to \infty$. In which case, 
+    \[
+    x = \frac{x + x^*}{2} = \limv{n}\frac{x_n + x_n^*}{2} \in \ol{S_{sa}}
+    \]
+
+    Assume without loss of generality that $\seq{x_n} \subset S_{sa}$, so that $\sigma_A(x_n) \subset \real$ for all $n \in \natp$. By \hyperref[upper semicontinuity of the spectrum]{proposition:spectrum-continuous}, $\lim_{n \to \infty}\inf(\sigma_A(x_n)) \ge 0$. Therefore
+    \[
+    x = \limv{n}x_n = \limv{n}[x_n - \inf(\sigma_A(x_n)) \wedge 0] \in \ol{S_+}
+    \]
+
+    by \autoref{corollary:spectrum-characterisation-iff}.
+\end{proof}
+
 
 \begin{proposition}
 \label{proposition:operator-system-positive-span}
@@ -131,4 +149,3 @@
     (3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$. 
 \end{proof}
 
-

src/op/system/schur.tex

@@ -0,0 +1,142 @@
+\section{Schur Products}
+\label{section:schur-product}
+
+
+\begin{lemma}
+\label{lemma:schur-product-positive}
+    Let $H$ be a complex Hilbert space, $A = \sum_{i, j = 1}^n E_{ij} \otimes a_{ij} \in M_n(B(H))$ be positive, and $\Lambda = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \in M_n(\complex)$ be positive, then $S_A(\Lambda) = \sum_{i, j = 1}^n \lambda_{ij}E_{ij} \otimes a_{ij}$ is also positive. 
+\end{lemma}
+\begin{proof}
+    For any unit vector $\xi \in \complex^n$, the projection map $P_\xi$ onto its span is given by $\sum_{i, j = 1}^n \xi_i E_{ij} \ol{\xi_j}$. In which case, 
+    \begin{align*}
+        S_A(P_\xi) &= \sum_{i, j = 1}^n \xi_i \ol{\xi_j} E_{ij} \otimes a_{ij} \\
+        &= \braks{\sum_{i = 1}^n \xi_{i}E_{ii}}\braks{\sum_{i, j = 1}^n E_{ij} \otimes a_{ij}}\braks{\sum_{j = 1}^n {\xi_j}E_{jj}}^* \ge 0
+    \end{align*}
+    
+
+
+    Using the \hyperref[Spectral Theorem]{definition:linfty-functional-calculus}, let $\bracsn{\xi^{(j)}}_1^n \subset \complex^n$ be an orthonormal eigenbasis for $\Lambda$ with eigenvalues $\seqf{\lambda_j} \subset \complex$. For each $1 \le j \le n$, let $P_j \in M_n(\complex)$ be the rank one projection onto the span of $\xi^{(j)}$, then 
+    \[
+        S_A(\Lambda) = \sum_{j = 1}^n \lambda_j S_A(P_{j}) \ge 0
+    \]
+    
+\end{proof}
+
+
+\begin{proposition}[Trigonometric Moment Sequences]
+\label{proposition:trig-moment-sequence-operator}
+    Let $\mathbb{T} = \partial B_\complex(0, 1)$, and $C(\mathbb{T}; \complex)$ act on $L^2(\mathbb{T}; \complex)$ via multiplication. For each $n \in \natp$ and $P = \sum_{-n \le j \le n}\lambda_jz^j \in \complex[z, \ol{z}]$, then the following are equivalent: 
+    \begin{enumerate}
+        \item $P \ge 0$.
+        \item For each $q \in \complex[z, \ol{z}]$, $\dpn{Pq, q}{L^2(\mathbb{T}; \complex)} \ge 0$.
+        \item For each $N \in \natp$\footnote{Padding the coefficients with zeroes if necessary.}, the Toeplitz matrix $\sum_{i, j = 1}^NE_{ij}\lambda_{i - j}$ is positive.
+    \end{enumerate}
+
+    For any complex Hilbert space $H$ and $\bracsn{A_n|n \in \integer} \subset B(H)$, the following are equivalent:
+    \begin{enumerate}[label=(T\arabic*)]
+        \item For each $n \in \natp$, the Toeplitz matrix $\sum_{i, j = 1}^n E_{ij} \otimes A_{i - j}$ is positive. 
+        \item The mapping $\phi: C(\mathbb{T}; \complex) \to B(H)$ defined by extending $\phi(z^n) = A_n$ is completely positive. 
+    \end{enumerate}
+    
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $\complex[z, \ol z]$ is dense in $C(\mathbb{T}; \complex)$. By \autoref{proposition:radon-cc-dense}, $C(\mathbb{T}; \complex)$ is dense in $L^2(\mathbb{T}; \complex)$. 
+
+    (2) $\Leftrightarrow$ (3): Let $q = \sum_{-N \le j \le N}\mu_j z^j$ and $\xi = (\mu_{-N}, \cdots, \mu_{N}) \in \complex^{2N+1}$, then
+    \begin{align*}
+        \dpn{Pq, q}{L^2(\mathbb{T}; \complex)} &= \sum_{-n \le j \le n}\sum_{-N \le k, l \le N}\dpn{\lambda_j z^j \cdot \mu_k z^k, \mu_l z^l}{L^2(\mathbb{T}; \complex)} \\
+        &= \sum_{-N \le k, l \le N}\dpn{\lambda_{l-k}\mu_k z^l, \mu_l z^l}{L^2(\mathbb{T}; \complex)} \\
+        &= \sum_{-N \le k, l \le N} \lambda_{l - k}\mu_k\ol{\mu_l}= \sum_{k, l = 1}^{2N+1} \lambda_{l - k}\xi_{k}\ol{\xi_{l}} \\
+        &= \angles{\paren{\sum_{i, j = 1}^{2N+1} E_{ij}\lambda_{i - j}}\xi, \xi} \\
+    \end{align*}
+
+    Therefore $\dpn{Pq, q}{L^2(\mathbb{T}; \complex)} \ge 0$ for all $q \in \complex[z, \ol z]$ if and only if the Toeplitz matrix $\sum_{i, j = 1}^NE_{ij}\lambda_{i - j}$ is positive for all $N \in \natp$. 
+    
+    (T1) $\Rightarrow$ (T2): Using \hyperref[Stinespring's Theorem]{theorem:stinespring-commutative-completely-positive}, it is sufficient to show that $\phi$ is positive. Using \autoref{lemma:operator-system-positive-approximate} and \autoref{proposition:positive-automatic-continuity}, it is sufficient to show that $\phi(P) \ge 0$ for all $P \in \complex[z, \ol z]$ with $P \ge 0$. 
+    
+    For each $P = \sum_{-n \le j \le n}\lambda_jz^j \in \complex[z, \ol z]$ with $P \ge 0$ and $m \ge n+1$, the matrix 
+    \[
+    P_m = \sum_{i, j = 1}^m \lambda_{i - j}E_{ij} \otimes A_{i - j} \in M_m(B(H))
+    \]
+
+    is a Schur product of positive matrices, and hence positive by \autoref{lemma:schur-product-positive}. For any $\xi \in H$, let $\xi_m = (\xi, \cdots, \xi) \in H^m$, then
+    \begin{align*}
+        0 &\le \dpn{P_m\xi_m, \xi_m}{H^m} = \sum_{i, j = 1}^m \lambda_{i - j}\dpn{A_{i - j}\xi, \xi}{H} =  \sum_{k = -n}^n (m - |k|) \lambda_k \dpn{A_{k}\xi, \xi}{H} \\
+        0 &\le \limv{m}\sum_{k = -n}^n \frac{(m - |k|)}{m} \lambda_k \dpn{A_{k}\xi, \xi}{H} = \sum_{k = -n}^n \dpn{\lambda_k A_k\xi, \xi}{H} = \dpn{\phi(P)\xi, \xi}{H}
+    \end{align*}
+
+    so $\phi(P)$ is positive. 
+
+    (T2) $\Rightarrow$ (T1): For each $n \in \natp$, 
+    \begin{align*}
+    0 &\le \braks{\sum_{j = 1}^n E_{j1} \otimes z^j}\braks{\sum_{j = 1}^n E_{j1} \otimes z^j}^* = \sum_{j, k = 1}^n E_{jk} \otimes z^{j-k} \\
+    0 &\le \phi\braks{\sum_{j, k = 1}^n E_{jk} \otimes z^{j-k}} = \sum_{j, k = 1}^n E_{jk} \otimes A_{j - k}
+    \end{align*}
+\end{proof}
+
+\begin{theorem}
+\label{theorem:polynomial-map-positive}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, and $\mathbb{T} = \partial B_\complex(0, 1)$, then the mapping
+    \[
+    \phi: \complex[z, \ol z] \to B(H) \quad p \mapsto p(T)
+    \]
+
+    extends into a completely positive map on $C(\mathbb{T}; \complex)$. 
+\end{theorem}
+\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
+    \[
+    A = \sum_{i, j = 1}^n E_{ij} \otimes A_{i - j} = \begin{bmatrix}
+        I & T^* &\cdots & (T^*)^{n-1} \\
+        T &\ddots &\ddots &\vdots \\
+        \vdots &\ddots &\ddots & T^* \\
+        T^{n-1} &\cdots &T &I
+    \end{bmatrix}
+    \]
+
+    is positive for all $n \in \natp$. To this end, let
+    \[
+    R = \begin{bmatrix}
+        0 &\cdots &\cdots &\cdots &0 \\
+        T & \ddots & & &\vdots \\
+        0 & \ddots &\ddots & &\vdots \\
+        \vdots &\ddots &\ddots &\ddots &\vdots \\
+        0 &\cdots &0 & T & 0
+    \end{bmatrix} = \sum_{j = 1}^{n - 1}E_{(j+1)j} \otimes T
+    \]
+
+    then for each $k \in \natz$, $R^k = \sum_{j = 1}^{n-k}E_{(j+k)j} \otimes T^k$. In particular, $R^{n+1} = 0$ and
+    \[
+    A = I + \sum_{k = 1}^n R^k + \sum_{k = 1}^n (R^*)^k = \sum_{k = 0}^\infty R^k + \sum_{k = 0}^\infty (R^*)^k - I
+    \]
+    
+    
+    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)^*)(I - R^*)^{-1} \\
+        &= (I - R)^{-1}(I - RR^*)(I - 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}
+

src/op/system/two.tex

@@ -0,0 +1,349 @@
+\section{On 2×2 Matrices}
+\label{section:op-sys-2by2}
+
+\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}[Schur Complement Condition for Positivity, Invertible Case]
+\label{proposition:schur-complement-positive}
+    Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$. Suppose that $P$ is invertible, then
+    \begin{enumerate}[label=(P\arabic*)]
+        \item If $P$ is also positive, then
+        \[
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} = \begin{bmatrix}
+            P^{1/2} &0 \\
+            A^*P^{-1/2} &I
+        \end{bmatrix}\begin{bmatrix}
+            I &0 \\
+            0 & Q - A^*P^{-1}A
+        \end{bmatrix}\begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
+        \end{bmatrix}
+        \]
+        
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $P \ge 0$ and $Q - A^*P^{-1}A \ge 0$.
+    \end{enumerate}
+
+    The operator $Q - A^*P^{-1}A$ is the \textbf{Schur complement} of $P$. Similarly, if $Q$ is invertible, then
+    \begin{enumerate}[label=(Q\arabic*)]
+        \item If $Q$ is also positive, then
+        \[
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} = \begin{bmatrix}
+            I &AQ^{-1/2} \\
+            0 &Q^{1/2}
+        \end{bmatrix}\begin{bmatrix}
+            P - AQ^{-1}A^* &0 \\
+            0 & I
+        \end{bmatrix}\begin{bmatrix}
+            I &0 \\
+            Q^{-1/2}A^* &Q^{1/2}
+        \end{bmatrix}
+        \]
+        
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ if and only if $Q \ge 0$ and $P - AQ^{-1}A^* \ge 0$.
+    \end{enumerate}    
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (P1): Since $P$ is positive and invertible, so is $P^{1/2}$. Thus
+    \begin{align*}
+        \begin{bmatrix}
+            P &A \\
+            A^* &Q
+        \end{bmatrix} &= \begin{bmatrix}
+            P^{1/2}P^{1/2} &P^{1/2}P^{-1/2}A \\
+            A^*P^{-1/2}P^{1/2} &Q
+        \end{bmatrix} \\
+        &= \begin{bmatrix}
+            P^{1/2} &0 \\
+            A^*P^{-1/2} &I
+        \end{bmatrix}\begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} \\
+        \begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} &= \begin{bmatrix}
+            I &0 \\
+            0 &Q - A^*P^{-1}A
+        \end{bmatrix} \begin{bmatrix}
+            P^{1/2} &P^{-1/2}A \\
+            0 &I
+        \end{bmatrix} 
+    \end{align*}
+    
+    and
+    
+    \[
+    \begin{bmatrix}
+        P &A \\
+        A^* &Q
+    \end{bmatrix} = \begin{bmatrix}
+        P^{1/2} &0 \\
+        A^*P^{-1/2} &I
+    \end{bmatrix}\begin{bmatrix}
+        I &0 \\
+        0 & Q - A^*P^{-1}A
+    \end{bmatrix}\begin{bmatrix}
+        P^{1/2} &P^{-1/2}A \\
+        0 &I
+    \end{bmatrix}
+    \]
+
+    (P2): If $P$ is also positive and $Q - A^*P^{-1}A \ge 0$, then by (P1), $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right]$ is of the form $MDM^*$, where $D$ is positive. Thus the given matrix is also positive. 
+
+    On the other hand, if $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then as a diagonal entry, $P$ is also positive. In addition, 
+    \[
+    \begin{bmatrix}
+            I &0 \\
+            0 & Q - A^*P^{-1}A
+        \end{bmatrix} = \begin{bmatrix}
+            P^{-1/2} &0 \\
+            -A^*P^{-1} &I
+        \end{bmatrix}\begin{bmatrix}
+        P &A \\
+        A^* &Q
+    \end{bmatrix} \begin{bmatrix}
+            P^{-1/2} &-P^{-1}A \\
+            0 &I
+        \end{bmatrix} 
+    \]
+
+    is also of the form $MDM^*$, where $D$ is positive, so the diagonal matrix, and hence $Q - A^*P^{-1}A$ are also positive. 
+\end{proof}
+
+
+\begin{proposition}[Schur Complement Condition for Positivity]
+\label{proposition:schur-complement-positive-general}
+    Let $H$ be a complex Hilbert space, $P, Q, A \in B(H)$, $P^{-1/2}: P^{1/2}(H) \to \ker(P)^\perp$ and $Q^{-1/2}: Q^{1/2}(H) \to \ker(Q)^\perp$ be partial inverses of $P^{1/2}$ and $Q^{1/2}$, respectively, then the following are equivalent:
+    \begin{enumerate}
+        \item $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
+        \item For each $\xi, \eta \in H$, $|\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}$.
+        \item $P \ge 0$, $A(H) \subset P^{1/2}(H)$, and $Q - (P^{-1/2}A)^*(P^{-1/2}A) \ge 0$.
+        \item $Q \ge 0$, $A^*(H) \subset Q^{1/2}(H)$, and $P - (Q^{-1/2}A^*)^*(Q^{-1/2}A^*) \ge 0$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (1) $\Rightarrow$ (2): If $\left[\begin{smallmatrix} P & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$, then $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ for all $\eps > 0$. By (P2) of \autoref{proposition:schur-complement-positive}, $P + \eps \ge 0$ and $Q - A^*(P + \eps)^{-1}A \ge 0$. Thus for any $\xi, \eta \in H$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+        |\dpn{A\xi, \eta}{H}| &= |\dpn{(P + \eps)^{-1/2}A\xi, (P + \eps)^{1/2}\eta}{H}| \\
+        &\le \normn{(P + \eps)^{-1/2}A\xi}_{H} \normn{(P + \eps)^{1/2}\eta}_H \\
+        &= \dpn{A^*(P + \eps)^{-1}A\xi, \xi}{H}^{1/2}\normn{(P + \eps)^{1/2}\eta}_H \\
+        &\le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{(P + \eps)\eta, \eta}{H}^{1/2}
+    \end{align*}
+    
+    As the above holds for all $\eps > 0$, 
+    \[
+     |\dpn{A\xi, \eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{P\eta, \eta}{H}^{1/2}
+    \]
+
+    (2) $\Rightarrow$ (3): Let $\xi \in H$, then for all $\eta \in P^{1/2}(H)$, 
+    \[
+     |\dpn{A\xi, P^{-1/2}\eta}{H}| \le \dpn{Q\xi, \xi}{H}^{1/2}\dpn{\eta, \eta}{H}^{1/2}
+    \]
+
+    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\theta \in \ol{P^{1/2}(H)}$ such that 
+    \[
+    \dpn{A\xi, P^{-1/2}\eta}{H} = \dpn{\theta, \eta}{H}
+    \]
+
+    for all $\eta \in P^{1/2}(H)$, so 
+    \[
+    \dpn{A\xi, \eta}{H} = \dpn{\theta, P^{1/2}\eta}{H} = \dpn{P^{1/2}\theta, \eta}{H}
+    \]
+
+    for all $\eta \in H$, and $A\xi \in P^{1/2}(H)$. 
+    
+    Now, for each Cauchy sequence $\seq{\xi_n} \subset H$ such that $\bracsn{P^{-1/2}A\xi_n}_1^\infty \subset \ker(P)^\perp$ is Cauchy, then there exists $\eta \in \ker(P)^\perp$ such that $P^{-1/2}A\xi_n \to \eta$ as $n \to \infty$. By continuity of $P^{1/2}$, $P^{1/2}P^{-1/2}A\xi_n \to P^{1/2}\eta$ as $n \to \infty$. Since $\eta$ is the unique element of $\ker(P)^\perp$ with $P^{1/2}\eta = A\xi = A\limv{n}\xi_n$, $\eta = P^{-1/2}A\limv{n}\xi_n$. By the \hyperref[Closed Graph Theorem]{theorem:closed-graph}, $P^{-1/2}A \in B(H)$, and the second part of the condition is well-defined.
+    
+    
+    Now, for any $\xi \in H$, 
+    \[
+        \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} = \dpn{Q\xi, \xi}{H} - \normn{P^{-1/2}A\xi}_H^2
+    \]
+
+    As $P^{-1/2}A\xi \in \ker(P)^\perp = \ol{P(H)}$, 
+    \begin{align*} 
+    \normn{P^{-1/2}A\xi}_H &= \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{P^{-1/2}A\xi, P^{1/2}\eta}{H}| = \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}|\dpn{A\xi, \eta}{H}| \\
+    &\le \sup_{\substack{\eta \in H \\ \normn{P^{1/2}\eta}_H \le 1}}\dpn{Q\xi, \xi}{H}^{1/2}\dpn{P^{1/2}\eta, P^{1/2}\eta}{H}^{1/2} = \dpn{Q\xi, \xi}{H}^{1/2}
+    \end{align*}
+
+    so
+    \[
+    \dpn{(Q - (P^{-1/2}A)^*(P^{-1/2}A))\xi, \xi}{H} \ge \dpn{Q\xi, \xi}{H} - \dpn{Q\xi, \xi}{H} = 0
+    \]
+    
+
+    (3) $\Rightarrow$ (1): For each $\xi \in H$, 
+    \[
+    \dpn{P^{-1/2}P^{1/2}\xi, P^{-1/2}P^{1/2}\xi}{H} = \dpn{\xi, \xi}{H}
+    \]
+
+    and by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    \dpn{(P+\eps)^{-1}P^{1/2}\xi, P^{1/2}\xi}{H} = \dpn{(P+\eps)^{-1}P\xi, \xi}{H} \le \dpn{\xi, \xi}{H}
+    \]
+
+    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}
+    \end{align*}
+    
+    so
+
+    \[
+    0 \le Q - (P^{-1/2}A)^*(P^{-1/2}A) \le Q - A^*(P+\eps)^{-1}A
+    \]
+
+    and $\left[\begin{smallmatrix} P + \eps & A \\ A^* & Q \end{smallmatrix}\right] \ge 0$ by (P2) of \autoref{proposition:schur-complement-positive}. As this holds for all $\eps > 0$, $\left[\begin{smallmatrix} P& A \\ A^* & Q \end{smallmatrix}\right] \ge 0$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:schur-complement-positive-estimate}
+    Let $H$ be a complex Hilbert space and $P, Q, A \in B(H)$ such that
+    \[
+    \begin{bmatrix}
+        P & A \\
+        A^* & Q
+    \end{bmatrix} \ge 0
+    \]
+
+    then
+    \begin{enumerate}
+        \item For each $\xi \in H$, 
+        \[
+        0 \le \dpn{(P + A + A^* + Q)\xi, \xi}{H} \le (\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2
+        \]
+
+        \item $\normn{P + A + A^* + Q}_{B(H)} \le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2$. 
+        \item If $P = Q$, then $A^*A \le \norm{P}_{B(H)}P$ and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Exercise 3.2]{Paulsen}}}. ]
+    (1): For each $\xi \in H$, 
+    \[
+    0 \le \angles{\begin{bmatrix}
+        P & A \\
+        A^* & Q
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\ \xi
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \xi
+    \end{bmatrix}}_{H^2} = \dpn{(P + A + A^* + Q)\xi, \xi}{H}
+    \]
+
+    and
+    \begin{align*}
+        \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 \\
+    \end{align*}
+
+    (2): By (3) of \autoref{proposition:bh-numerical-radius-gymnastics} and (1),
+    \begin{align*}
+    \norm{P + A + A^* + Q}_{B(H)} &= \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}\dpn{(P + A + A^* + Q)\xi, \xi}{H} \\
+    &\le \sup_{\substack{\xi \in H \\ \norm{\xi}_H \le 1}}(\dpn{P\xi, \xi}{H}^{1/2} + \dpn{Q\xi, \xi}{H}^{1/2})^2 \\
+    &\le (\norm{P}_{B(H)}^{1/2}+\norm{Q}_{B(H)}^{1/2})^2
+    \end{align*}
+    
+    (3): Using (2) of \autoref{proposition:schur-complement-positive}, assume without loss of generality that $P \ne 0$. For each $\xi, \eta \in H$, 
+    \begin{align*}
+        0 &\le \angles{\begin{bmatrix}
+        P & A \\
+        A^* & P
+    \end{bmatrix}\begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}, \begin{bmatrix}
+        \xi \\ \eta
+    \end{bmatrix}}_{H^2} \\
+    &= \dpn{P\xi, \xi}{H} + \dpn{P\eta, \eta}{H} + \dpn{A\eta, \xi}{H} + \dpn{A^*\xi, \eta}{H}
+    \end{align*}
+
+    In particular, for $\xi = -A\eta/\norm{P}_{B(H)}$, 
+    \begin{align*}
+        0 &\le \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}^2} + \dpn{P\eta, \eta}{H} - \frac{\dpn{A\eta, A\eta}{H}}{\norm{P}_{B(H)}} - \frac{\dpn{A^*A\eta, \eta}{H}}{\norm{P}_{B(H)}} \\
+        0 &\le \frac{\dpn{PA\eta, A\eta}{H}}{\norm{P}_{B(H)}} + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2  \\
+        &\le\norm{A\eta}_H^2 + \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - 2\norm{A\eta}_H^2 \\
+        &= \norm{P}_{B(H)}\dpn{P\eta, \eta}{H} - \dpn{A^*A\eta, \eta}{H}
+    \end{align*}
+    
+    so $A^*A \le \norm{P}_{B(H)}P$, and $\norm{A}_{B(H)} \le \norm{P}_{B(H)}$. 
+    
+\end{proof}

src/topology/functions/sw.tex

@@ -1,6 +1,110 @@
 \section{The Stone-Weierstrass Theorem}
 \label{section:stone-weierstrass}
 
+\begin{theorem}[Korovkin]
+\label{theorem:korovkin-convergence}
+    For each $x \in [0, 1]$, let $g_x(t) = (t - x)^2$, then
+    \begin{enumerate}
+        \item For each $f \in C([0, 1]; \complex)$ and $\eps > 0$, there exists $C > 0$ such that
+        \[
+        |f(t) - f(x)| \le \eps + Cg_x(t)
+        \]
+
+        for all $x, t \in [0, 1]$. 
+        \item Let $f \in C([0, 1]; \complex)$ and $\eps > 0$, then there exists $C > 0$ such that for any positive map $\phi: C([0, 1]; \complex) \to C([0, 1]; \complex)$ and $x \in [0, 1]$,
+        \[
+        -\eps\norm{\phi(1)}_u - C\phi(g_x)(x) \le \phi(f)(x) - f(x) \le \eps\norm{\phi(1)}_u + C\phi(g_x)(x)
+        \]
+
+        and
+
+        \[
+        \norm{\phi(f) - f}_u \le \eps \norm{\phi(1)}_u + C\sup_{x \in [0, 1]}|\phi(g_x)(x)|
+        \]
+        \item For each $n \in \natp$, let $\phi_n: C([0, 1]; \complex) \to C([0, 1]; \complex)$ be a positive map. If $\phi_n(1) \to 1$, $\phi_n(t) \to t$, and $\phi_n(t^2) \to t^2$ uniformly as $n \to \infty$, then $\phi_n(f) \to f$ uniformly as $n \to \infty$ for all $f \in C([0, 1]; \complex)$. 
+    \end{enumerate}    
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Exercise 2.20]{Paulsen}}}. ]
+    (1): By \autoref{proposition:uniform-continuous-compact}, there exists $\delta > 0$ such that $|f(t) - f(x)| < \eps$ for all $x, t \in [0, 1]$ with $|x - t| < \delta$. Thus for each $x, t \in [0, 1]$, 
+    \[
+    |f(t) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil
+    \]
+
+    Let $C = \eps/\delta^2$, then for each $s \ge 0$, 
+    \[
+        \eps + Cs^2 = \eps + \eps\paren{\frac{s}{\delta}}^2 \ge \eps + \eps\left\lfloor \abs{\frac{s}{\delta}} \right\rfloor \ge \eps\left\lceil \abs{\frac{s}{\delta}} \right\rceil
+    \]
+
+    so
+    \[
+    |f(t) - f(x)| \le \eps \left\lceil \abs{\frac{x - t}{\delta}} \right\rceil \le \eps + C(x - t)^2 = \eps + Cg_x(t)
+    \]
+    
+    (3): Let $f \in C([0, 1]; \complex)$ and $\eps > 0$. By (2), there exists $C > 0$ such that
+    \[
+    \norm{\phi_n(f) - f}_u \le \eps \norm{\phi_n(1)}_u + C\sup_{x \in [0, 1]}|\phi_n(g_x)(x)|
+    \]
+
+    for all $n \in \natp$, where
+    \begin{align*}
+    \sup_{x \in [0, 1]}|\phi_n(g_x)(x)| &\le \sup_{x \in [0, 1]}|\phi_n(t^2 - 2tx + x^2)(x)| \\
+    &\le \sup_{x \in [0, 1]}|\phi_n(t^2)(x) - 2\phi_n(tx)(x) + x^2| \\
+    &\le \normn{\phi_n(t^2) - t^2}_u + 2\sup_{x \in [0, 1]}\normn{\phi_n(xt) - xt}_u
+    \end{align*}
+
+    which tends to $0$ as $n \to \infty$. Thus
+    \[
+    \limsup_{n \to \infty}\norm{\phi_n(f) - f}_u \le \eps 
+    \]
+
+    As the above holds for all $\eps > 0$, $\phi_n(f) \to f$ uniformly.
+    
+\end{proof}
+
+\begin{theorem}[Weierstrass Approximation Theorem]
+\label{theorem:weierstrass-bernstein-polynomial}
+    For each $n \in \natz$, let
+    \[
+    \phi_n: C([0, 1]; \real) \to C([0, 1]; \real) \quad \phi_n(f)(t) = \sum_{k = 0}^n {n \choose k}f\paren{\frac{k}{n}}t^k(1-t)^{n-k}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For every $n \in \natp$, $\phi_n$ is a positive map.
+        \item For every $n \in \natp$, 
+        \[
+        \phi_n(1) = 1 \quad \phi_n(t) = t \quad \phi_n(t^2) = t^2 + \frac{t - t^2}{n}
+        \]
+        \item For every $f \in C([0, 1]; \real)$, $\phi_n(f) \to f$ uniformly as $n \to \infty$. 
+        \item $\real[t]$ is dense in $C([0, 1]; \real)$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): For each $n \in \natz$ and $0 \le k \le n$, $t^k(1 - t)^{n-k} \ge 0$ on $[0, 1]$. 
+
+    (2): Let $n \in \natp$, then by the Binomial formula, 
+    \[
+    \phi_n(1) = \sum_{k = 0}^n{n \choose k}t^k(1 - t)^{n-k} = (t + (1 - t))^n = 1
+    \]
+
+    and
+    \begin{align*}
+        \phi_n(t) &=  \sum_{k = 1}^n\frac{n!}{k!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\
+        &= \sum_{k = 1}^n\frac{(n-1)!}{(k-1)!(n-k)!}t^k(1 - t)^{n-k} \\
+        &= t\sum_{k = 0}^{n-1}\frac{(n-1)!}{k!(n-k-1)!}t^{k}(1 - t)^{n-k-1} = t
+    \end{align*}
+    
+    Finally, 
+    \begin{align*}
+        \phi_n(t^2) &= \sum_{k = 1}^n \frac{n!}{k!(n-k)!}\frac{k^2}{n^2}t^k(1 - t)^{n-k} \\
+        &= \sum_{k = 1}^n \frac{(n-1)!}{(k-1)!(n-k)!}\frac{k}{n}t^k(1 - t)^{n-k} \\
+        &= t \sum_{k = 0}^{n-1} \frac{(n-1)!}{(k-1)!(n-k-1)!}\frac{k+1}{n}t^k(1 - t)^{n-k} \\
+        &= t \phi_{n-1}\paren{\frac{(n-1)t + 1}{n}} = t^2 + \frac{t - t^2}{n}
+    \end{align*}
+
+    (3): Since $\seq{\phi_n}$ are positive with $\phi_n(1) \to 1$, $\phi_n(t) \to t$, and $\phi_n(t^2) \to t^2$ uniformly as $n \to \infty$, $\phi_n \to \text{Id}$ pointwise by \hyperref[Korovkin's Theorem]{theorem:korovkin-convergence}.
+\end{proof}
+
 \begin{lemma}
 \label{lemma:sqrt-polynomial}
     There exists a sequence of polynomials $\seq{p_n} \subset \real[x]$ such that $p_n(x) \upto \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$. 

src/topology/main/compactify.tex

@@ -125,7 +125,17 @@
     For each $x \in X$, let $U \in \cn_X(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by \autoref{proposition:compact-closed}. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_Y(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V} \supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:stone-cech-not-metrisable}
+    Let $X$ be an infinite discrete topological space, then $\beta X$ is not second countable and hence not metrisable. 
+\end{proposition}
+\begin{proof}
+    Let $\fB \subset 2^X$ be a base for the topology on $\beta X$, then for each $A \subset \beta X$ clopen, there exists $\mathcal{F} \subset \fB$ finite such that $A = \bigcup_{U \in \mathcal{F}}U$. Should $\fB$ be countable, then there are at most countably many clopen subsets of $\beta X$. 
 
+    However, for each $A \subset X$, let $U_A = \bracsn{\fU \in \beta X|A \in \fU}$, then $\bracsn{U_A|A \subset X}$ represents uncountably many clopen sets. Therefore $\fB$ cannot be countable, $\beta X$ is not second countable. 
+
+    If $\beta X$ was metrisable, then it is second countable by compactness and \autoref{proposition:separable-metric-space}. Therefore $\beta X$ is not metrisable. 
+\end{proof}
 
 \begin{definition}[One-Point Compactification]
 \label{definition:alexandroff-compactification}

src/topology/main/definition.tex

@@ -121,6 +121,7 @@
 
 
         is a base for $\topo$.
+        \item For any topological space $Z$ and $F: Z \to X$, $F \in C(Z; X)$ if and only if $f_i \circ F \in C(Z; Y_i)$ for all $i \in I$. 
     \end{enumerate}
 
     The topology $\topo$ is the \textbf{initial/weak topology} generated by the maps $\seqi{f}$.

src/topology/main/filters.tex

@@ -133,7 +133,7 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
     (3) $\Rightarrow$ (2): $E \cup E^c = X \in \fF$.
 \end{proof}
 
-\begin{lemma}
+\begin{lemma}[Ultrafilter Lemma]
 \label{lemma:ultrafilter}
     Let $X$ be a set and $\fF \subset 2^X$, then
     \begin{enumerate}
@@ -206,3 +206,6 @@ The smallest filter $\fF(\fB_0) \subset 2^X$ containing $\fB_0$ is the filter \t
 
     is a \textbf{limit} of $f$ at $y$ with respect to $A$. If $A = X$, then $x \in A$ may be omitted.
 \end{definition}
+
+
+

src/topology/main/interiorclosureboundary.tex

@@ -91,10 +91,16 @@
 
 \begin{proposition}
 \label{proposition:separable-product}
-    Let $\seqf{X_j}$ be separable topological spaces, then $\prod_{j = 1}^n X_j$ is also separable.
+    Let $\seq{X_n}$ be separable topological spaces, then $\prod_{n \in \natp}X_n$ is separable. 
 \end{proposition}
 \begin{proof}
-    Let $\seq{A_j}$ such that $A_j \subset X_j$ is countable and dense for each $1 \le j \le n$. By \autoref{proposition:dense-product}, $\prod_{j = 1}^n A_j$ is countable and dense in $\prod_{j = 1}^n X_j$.
+    For each $n \in \nat$, let $A_n \subset X_n$ be a countable dense subset. For each $N \in \nat$, let $B_N \subset X$ such that:
+    \begin{enumerate}
+        \item $B_N$ is countable.
+        \item For each $\bracsn{a_n}_1^N \subset \prod_{n = 1}^N A_n$, there exists $b \in B_N$ such that $a_n = b_n$ for all $1 \le j \le N$. 
+    \end{enumerate}
+
+    Then, $\bigcup_{N \in \natp}B_N$ is a countable dense subset of $\prod_{n \in \natp}X_n$.
 \end{proof}
 
 

src/topology/main/nets.tex

@@ -6,17 +6,87 @@
     Let $X$ be a set and $A$ be an upward-directed set, then a \textbf{net} in $X$ is a mapping $A \to X$, denoted $\net{x} \subset X$.
 \end{definition}
 
+\begin{definition}[Subnet]
+\label{definition:subnet}
+    Let $X$ be a set, $A, B$ be upward-directed sets, and $\varphi: A \to B$, then $\varphi$ is \textbf{upward-directed} if:
+    \begin{enumerate}
+        \item \textbf{(Isotone)} For any $\alpha, \beta \in A$ with $\alpha \le \beta$, $\varphi(\alpha)  \le \varphi(\beta)$. 
+        \item \textbf{(Cofinality)} For any $\beta \in B$, there exists $\alpha \in A$ such that $\varphi(\alpha) \ge \beta$. 
+    \end{enumerate}
+    
+    For any net $\angles{x_\beta}_{\beta \in B}$ and upward-directed map $\varphi: A \to B$, $\anglesn{x_{\varphi(\alpha)}}_{\alpha \in A}$ is a \textbf{subnet} of $X$. 
+\end{definition}
+
+
 \begin{definition}[Eventually]
 \label{definition:eventually}
-    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{eventually in} $E$ if there exists $\alpha_0 \in A$ such that $x_\alpha \in E$ for all $\alpha \gtrsim \alpha_0$.
+    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{eventually in} $E$ if there exists $\alpha_0 \in A$ such that $x_\alpha \in E$ for all $\alpha \ge \alpha_0$.
+
+    For any filter $\fF \subset 2^X$, $\net{x}$ is \textbf{eventually in} $\fF$ if it is eventually in $E$ for all $E \in \fF$. 
 \end{definition}
 
 \begin{definition}[Frequently]
 \label{definition:frequently}
-    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{frequently in} $E$ if for all $\alpha_0 \in A$, there exists $\alpha \gtrsim \alpha_0$ such that $x_\alpha \in E$.
+    Let $X$ be a set, $\net{x} \subset X$, and $E \subset X$, then $\net{x}$ is \textbf{frequently in} $E$ if for all $\alpha_0 \in A$, there exists $\alpha \ge \alpha_0$ such that $x_\alpha \in E$.
+
+    For any filter $\fF \subset 2^X$, $\net{x}$ is \textbf{frequently in} $\fF$ if it is frequently in $E$ for all $E \in \fF$. 
 \end{definition}
 
 \begin{definition}[Convergence]
 \label{definition:net-convergence}
     Let $X$ be a topological space and $\net{x} \subset X$, then $\net{x}$ \textbf{converges} to $x$ if for every $U \in \cn(x)$, $\net{x}$ is eventually in $U$, denoted $x_\alpha \to x$.
 \end{definition}
+
+
+\begin{definition}[Associated Net]
+\label{definition:filter-associated-net}
+    Let $X$ be a set, $\fF \subset 2^X$ be a filter, and let
+    \[
+    N_{\fF} = \bracsn{(x, F) \in X \times \fF| F \in \fF, x \in F}
+    \]
+
+    under the ordering $(x, E) \le (y, F)$ if $E \supset F$, then $N_{\fF}$ is a directed set. For each $(x, F) \in N_{\fF}$, let $x_{(x, F)} = x$, then $\angles{x_{(x, \fF)}}_{(x, \fF) \in N_{\fF}} \subset X$ is the \textbf{net associated with} $\fF$. 
+\end{definition}
+
+\begin{definition}[Tail Filter]
+\label{definition:tail-filter}
+    Let $X$ be a set and $\angles{x_\alpha}_{\alpha \in A} \subset X$ be a net. For each $\alpha \in A$, let
+    \[
+    T(\alpha) = \bracsn{x_\beta|\beta \in A, \beta \ge \alpha}
+    \]
+
+    then the filter $T_x$ generated by $\bracsn{T(\alpha)|\alpha \in A}$ is the \textbf{tail filter} of $\angles{x_\alpha}_{\alpha \in A}$. 
+\end{definition}
+
+\begin{definition}[Ultranet]
+\label{definition:ultranet}
+    Let $X$ be a set and $\angles{x_\alpha}_{\alpha \in A} \subset X$ be a net, then $\angles{x_\alpha}_{\alpha \in A}$ is an \textbf{ultranet} if for every $E \subset X$, $\angles{x_\alpha}_{\alpha \in A}$ is eventually in $E$ or $E^c$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:filter-subnet-extraction}
+    Let $X$ be a set, $\angles{x_\alpha}_{\alpha \in A} \subset X$ be a net, and $\fF \subset 2^X$ be a filter, then the following are equivalent:
+    \begin{enumerate}
+        \item For each $F \in \fF$, $\angles{x_\alpha}_{\alpha \in A}$ is in $F$ frequently. 
+        \item There exists an subnet $\angles{x_\beta}_{\beta \in B} \subset \angles{x_\alpha}_{\alpha \in A}$ such that for each $F \in \fF$, $\angles{x_\beta}_{\beta \in B}$ is in $F$ eventually. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let
+    \[
+    B = \bracsn{(\alpha, F) \in A \times \fF|x_\alpha \in F, F \in \fF, \alpha \in A}
+    \]
+
+    with the ordering $(\alpha, E) \ge (\beta, F)$ if $\alpha \le \beta$ and $E \supset F$, then $B$ is upward-directed. As $\angles{x_\alpha}_{\alpha \in A}$ is in $F$ frequently for all $F \in \fF$, the mapping $(\alpha, E) \mapsto \alpha$ is upward-directed. 
+    
+    For each $(x, E) \in B$, let $x_{(\alpha, E)} = x_\alpha$, then $\anglesn{x_{(\alpha, E)}}_{(\alpha,E) \in B}$ is a subnet of $\angles{x_\alpha}_{\alpha \in A}$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:ultranet}
+    Let $X$ be a set and $\angles{x_\alpha}_{\alpha \in A} \subset X$ be a net, then $\angles{x_\alpha}_{\alpha \in A}$ admits an ultra subnet. 
+\end{lemma}
+\begin{proof}
+    Let $\fF \subset 2^X$ be the tail filter of $\angles{x_\alpha}_{\alpha \in A}$, then by the \hyperref[ultrafilter lemma]{lemma:ultrafilter}, there exists an ultrafilter $\fU \subset 2^X$ with $\fU \supset \fF$. For each $U \in \fU$, $U \cap F \ne \emptyset$ for all $F \in \fF$. Hence $\angles{x_\alpha}_{\alpha \in A}$ is frequently in $U$ for all $U \in \fU$. By \autoref{lemma:filter-subnet-extraction}, $\angles{x_\alpha}_{\alpha \in A}$ admits a subnet $\angles{x_\beta}_{\beta \in B} \subset \angles{x_\alpha}_{\alpha \in A}$ such that for each $U \in \fU$, $\angles{x_\beta}_{\beta \in B}$ is in $U$ eventually. As $\fU$ is an ultrafilter, this subnet is an ultranet. 
+\end{proof}
+

src/topology/uniform/complete.tex

@@ -16,17 +16,19 @@
 
 \begin{proposition}
 \label{proposition:complete-closed}
-    Let $X$ be a uniform space and $A \subset X$ be closed, then $A$ is complete.
+    Let $X$ be a Hausdorff complete uniform space and $A \subset X$, then $A$ is closed if and only if it is complete.
 \end{proposition}
 \begin{proof}
-    Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \autoref{definition:closure}.
+    ($\Rightarrow$): Suppose that $A$ is closed. Let $\fF$ be a Cauchy filter on $A$, then there exists $x \in X$ such that $\fF$ converges to $x$. Since $A$ is closed, $x \in X$ by (4) of \autoref{definition:closure}.
+
+    ($\Leftarrow$): Suppose that $A$ is complete. Let $x \in \ol A$, then there exists a filter $\fF \subset 2^A$ with $\fF \to x$. By the \hyperref[Cauchy criterion]{proposition:cauchycriterion}, $\fF$ is Cauchy, so there exists $y \in A$ such that $\fF \to y$. Since $X$ is Hausdorff, $x = y \in A$, so $A = \ol A$. 
 \end{proof}
 
 
 
 \begin{lemma}
 \label{lemma:complete-dense}
-    Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is \textbf{complete} if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
+    Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is complete if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
 \end{lemma}
 \begin{proof}
     Let $\fF \subset X$ be a Cauchy filter. Using \autoref{proposition:minimalcauchyexistence}, assume without loss of generality that $\fF$ is minimal. By \autoref{proposition:cauchyinterior}, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_A = \bracs{E \cap A| E \in \fF}$, then $\fF_A$ is a filter base on $A$.

src/vna/example/bb.tex

@@ -55,16 +55,55 @@
     (4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra. 
 \end{proof}
 
+\begin{lemma}
+\label{lemma:linfty-multiply-bound}
+    Let $(X, \cm, \mu)$ be a measure space, $f: X \to \complex$ be a measurable function, and $C \ge 0$ such that $\norm{fg}_{L^2(X; \complex)} \le C\norm{g}_{L^2(X; \complex)}$ for all $g \in L^2(X; \complex) \cap \Sigma(X; \complex)$, then $f \in L^\infty(X; \complex)$ with $\norm{f}_{L^\infty(X; \complex)} \le C$. 
+\end{lemma}
+\begin{proof}
+    For any $h \in L^1(X; \complex) \cap L^\infty(X; \complex)$, 
+    \begin{align*}
+        \normn{f^2h}_{L^1(X; \complex)} &= \int_X |f^2| \cdot |h| = \int_X |f||h|^{1/2} \cdot |f||h|^{1/2} = \normn{f|h|^{1/2}}_{L^2(X; \complex)}^2 \\
+        &\le C^2 \normn{|h|^{1/2}}_{L^2(X; \complex)}^2 = C^2 \norm{h}_{L^1(X; \complex)}
+    \end{align*}
+    
+    By \autoref{theorem:lp-dual-function}, $f^2 \in L^\infty(X; \complex)$ with $\normn{f^2}_{L^\infty(X; \complex)} \le C^2$. Hence $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le C$. 
+\end{proof}
+
 \begin{proposition}
 \label{proposition:linfty-von-neumann-algebra}
     Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
     \begin{enumerate}
         \item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
         \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. 
+        \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is maximal abelian. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
     (2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$. 
+
+    (3): Using \autoref{proposition:maximal-commutative-vn}, it is sufficient to show that $L^\infty(X; \complex)' = L^\infty(X; \complex)$. For each $g \in L^\infty(X; \complex)$, denote $M_g \in B(L^2(X; \complex))$ as the multiplication map by $g$. 
+    
+    First suppose that $\mu(X) < \infty$ and let $f = T\one_X$, then for each $g \in L^\infty(X; \complex)$, 
+    \[
+    M_fg = M_g f= M_g T\one_X = TM_g\one_X = Tg
+    \]
+
+    By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex) \subset L^\infty(X; \complex)$ is dense in $L^2(X; \complex)$, so $M_fg = Tg$ for all $g \in L^2(X; \complex)$. 
+
+    Now suppose that $\mu$ is an arbitrary localisable measure. For each $A \in \cm$ with $\mu(A) < \infty$, the finite case shows that there exists $f_A \in L^\infty(X; \complex)$ such that
+    \begin{enumerate}[label=(\roman*)]
+        \item $f_A|_{X \setminus A} = 0$ almost everywhere. 
+        \item $Tg = M_{f_A}g$ for all $g \in L^2(X; \complex)$ with $f_A|_{X \setminus A} = 0$ almost everywhere. 
+    \end{enumerate}
+
+    By (ii), $f_A|_{A \cap B} = f_B|_{A \cap B}$ for all $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$. By the \hyperref[gluing lemma for meaasurable functions]{lemma:gluing-measurable}, there exists $f \in L^\infty(X; \complex)$ such that $f|_A = f_A|_A$ for all $A \in \cm$ with $\mu(A) < \infty$. 
+
+    For each $g \in L^2(X; \complex)$, $\bracsn{g \ne 0}$ is $\sigma$-finite. As such, there exists $\seq{A_n} \subset \cm$ such that $\bracsn{g \ne 0} = \bigsqcup_{n \in \natp}A_n$ and $\mu(A_n) < \infty$ for all $n \in \natp$. In which case, 
+    \[
+    Tg = \sum_{n = 1}^\infty T(\one_{A_n}g) = \sum_{n = 1}^\infty M_{f_{A_n}} \one_{A_n}g = M_fg
+    \]
+
+    By \autoref{lemma:linfty-multiply-bound}, $f \in L^\infty(X; \complex)$ with $\normn{f}_{B(L^2(X; \complex))} \le \norm{T}_{B(L^2(X; \complex))}$. Therefore $T = M_f$, and $L^\infty(X; \complex)$ is maximal abelian. 
 \end{proof}
 
 \begin{theorem}[Order Structure of $L^\infty$]

src/vna/example/hii1.tex

@@ -0,0 +1,133 @@
+\section{The Hyperfinite $\vnII_1$ Factor}
+\label{section:hyperfinite-ii1-factor}
+
+\begin{definition}[Inductive Limit of Tracial von Neumann Algebras]
+\label{definition:inductive-limit-tracial-von-neumann-algebra}
+    Let $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$ be an upward-directed system of tracial von Neumann algebras, where $\phi^i_j: M_i \to M_j$ is a trace-preserving *-homomorphism for each $i, j \in I$ with $i \le j$, then there exists a pair $((M, \tau), \bracsn{\phi^i}_{i \in I})$ such that:
+    \begin{enumerate}
+        \item $(M, \tau)$ is a tracial von Neumann algebra. 
+        \item For each $i \in I$, $\phi^i: M_i \to M$ is a trace-preserving *-homomorphism. 
+        \item For each $i, j \in I$ with $i \le j$, the following diagram commutes: 
+        \[
+        \xymatrix{
+        M_i \ar@{->}[d]_{\phi^i_j} \ar@{->}[rd]^{\phi^i} &  \\
+        M_j \ar@{->}[r]_{\phi^j} & M
+        }
+        \]
+        \item[(U)] For any pair $((N, \sigma), \bracsn{\psi^i}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique trace-preserving *-homomorphism $\psi: M \to N$ such that the following diagram commutes:
+        \[
+        \xymatrix{
+        M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} &  \\
+        M \ar@{->}[r]_{\psi} & N
+        }
+        \]
+
+        for all $i \in I$. 
+        \item $\braks{\bigcup_{i \in I}\phi^i(M_i)}'' = M$, where the bicommutant is taken in $M$.
+    \end{enumerate}
+
+    The pair $((M, \tau), \bracsn{\phi^i}_{i \in I}) = \lim_{\longrightarrow}(M_i, \tau_i)$ is the \textbf{inductive limit} of $(\bracsn{(M_i, \tau_i)}_{i \in I}, \bracsn{\phi^i_j|i, j \in I, i \le j})$. 
+\end{definition}
+\begin{proof}
+    Let $(M_0, \bracsn{\phi^i|i \in I})$ be the inductive limit of $(\bracsn{M_i}_{i \in I}, \bracsn{\phi^i_j|i \le j})$ as vector spaces. Since $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are trace-preserving and $\bracsn{\tau_i}_{i \in I}$ are faithful, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are also unital isometric embeddings. As such, $\bracsn{M_i}_{i \in I}$ may be identified as subspaces of $M_0$, the norms on $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ are isometric embeddings for all $i \in I$. 
+    
+    For each $i, j \in I$ with $i \le j$, $\phi^i_j: M_i \to M_j$ is trace-preserving. By universal property of the inductive limit, there exists a linear functional $\tau: M_0 \to \complex$ such that $\tau \circ \phi^i = \tau_i$ for all $i \in I$. Similarly, $\bracsn{\phi^i_j|i, j \in I, i \le j}$ are *-homomorphisms, the products and involutions from $\bracsn{M_i}_{i \in I}$ extend to $M_0$, and $\phi^i: M_i \to M_0$ is a unital trace-preserving *-homomorphism for all $i \in I$. In particular, $1_{M_0} = \phi^i(1_{M_i})$ for all $i \in I$. 
+    
+    For each $i \in I$, $\tau_i: M_i \to \complex$ is a faithful tracial state. As such, for any $x \in M_0$, $x = 0$ if and only if $\tau(x^*x) = 0$, so the pairing $\dpn{x, y}{\tau} = \tau(y^*x)$ is an inner product on $M_0$. Let $H$ be the completion of $M_0$ with respect to $\dpn{\cdot, \cdot}{\tau}$. For each $x \in M_0$, let $L_x: M_0 \to M_0$ be defined by $L_x(y) = xy$ for all $y \in M_0$. Then, for any $x, y, z \in M_0$, 
+    \[
+    \dpn{L_xy, L_xy}{\tau} = \tau(y^*x^*xy) \le \norm{x}_{M_0}^2 \tau(y^*y) = \norm{x}_{M_0}^2 \dpn{y, y}{\tau}
+    \]
+    
+    and
+    \[
+    \dpn{L_xy, z}{\tau} = \tau(z^*xy) = \tau((x^*z)^*y) = \dpn{y, L_{x^*}z}{\tau}
+    \]
+    
+    with
+    \[
+    \dpn{L_x 1_{M_0}, L_x 1_{M_0}}{\tau} = \tau(x^*x) \le \tau(\normn{x}_{M_0}^2) = \norm{x}_{M_0}^2
+    \]
+
+    Thus the mapping $M_0 \to B(H)$ with $x \mapsto L_x$ allows identifying $M_0$ as an involutive subalgebra of $B(H)$. 
+    
+    Finally, let $M = M_0'' \subset B(H)$ and extend $\tau$ to $M$ by $\tau(x) = \dpn{x1_{M_0}, 1_{M_0}}{\tau}$, then $\tau$ is a vector state on $M$. 
+
+    (1): By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $M$ is a von Neumann algebra. As $\tau(z) = \dpn{z1_{M_0}, 1_{M_0}}{\tau}$ for all $z \in M$, $\tau$ is strong-operator continuous, and the mapping $(x, y) \mapsto \tau(xy)$ is separately strong-operator continuous. Since $\tau$ is the extension of $\bracsn{\tau_i}_{i \in I}$, $\tau|_{M_i}$ is tracial for all $i \in I$, and $\tau|_{M_0}$ is also tracial. By strong-operator density of $M_0$ in $M$, $\tau$ is a tracial state. 
+
+    (2), (3): By definition of $\tau$ and universal property of inductive limits for vector spaces. 
+
+    (4): Let $(K, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\sigma$, then since $\sigma$ is a faithful state, the representation $\pi: N \to B(K)$ is faithful. Using \autoref{lemma:tracial-von-neumann-gns}, identify $N = \pi(N) \subset B(K)$ as a von Neumann subalgebra of $B(K)$. 
+    
+    By universal property of inductive limits of vector spaces, there exists a trace-preserving *-homomorphism $\phi_0: M_0 \to N$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    M_i \ar@{->}[d]_{\phi^i} \ar@{->}[rd]^{\psi^i} &  \\
+    M_0 \ar@{->}[r]_{\phi_0} & N
+    }
+    \]
+
+    for all $i \in I$. Let
+    \[
+    U: H \to K \quad \dpn{UL_x1_{M_0}, \pi(y)\xi}{K} = \sigma(y^*\phi_0(x))
+    \]
+
+    then for any $x \in M_0$, $\norm{UL_x1_{M_0}}_K^2 = \sigma(\phi_0(x^*x)) = \tau(x^*x) = \norm{L_x1_{M_0}}_H^2$. Thus $U$ extends into an isometry from $H$ to $K$. Moreover, for each $x \in M_0$ and $y, z \in N$, 
+    \begin{align*}
+        \dpn{(\pi \circ \phi_0)(x)\pi(y)\xi, \pi(z)\xi}{K} &= \sigma(z^*\phi_0(x)y) = \sigma(yz^*\phi_0(x)) = \sigma((zy^*)^*\phi_0(x)) \\
+        &= \dpn{UL_x1_{M_0}, \pi(zy^*)\xi}{K} = \dpn{L_x1_{M_0}, U^*\pi(zy^*)\xi}{H}
+    \end{align*}
+
+    so $x \mapsto \dpn{\phi_0(x)\pi(y)\xi, \pi(z)\xi}{K}$ is a weak-operator continuous functional on $M_0$. Since $\pi(N)\xi$ is dense in $K$, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the weak operator topology on $\ol{B_{\pi(N)}(0, 1)}$ is induced by maps of the form $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{K}$ with $y, z \in N$. As such, $\phi_0: M_0 \to N$ is weak-operator continuous when restricted to bounded sets. 
+
+    By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density} and \autoref{theorem:uniform-continuous-extension}, $\phi_0$ extends into a *-homomorphism $\phi: M \to N$. Since $(N, \sigma)$ is a tracial von Neumann algebra, $\phi$ is also trace-preserving. Finally, if $\Phi: M \to N$ is another such map, then it is also weak-operator continuous on bounded sets by \autoref{lemma:tracial-von-neumann-gns}. Therefore $\Phi = \phi$ by uniqueness of continuous extensions. 
+\end{proof}
+
+\begin{definition}[Hyperfinite $\vnII_1$ Factor]
+\label{definition:hyperfinite-ii1}
+    For each $n \in \natp$, equip $M_{2^n}(\complex)$ with the normalised trace $\tau_n: M_{2^n}(\complex) \to \complex$, and
+    \[
+    \phi_n: M_{2^n}(\complex) \to M_{2^{n+1}}(\complex) \quad x \mapsto \begin{bmatrix}
+        x & 0 \\
+        0 & x
+    \end{bmatrix}
+    \]
+
+    Then, $\phi_n$ is a trace-preserving *-homomorphism, and the resulting inductive limit $\lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is a factor. 
+    
+    The tracial von Neumann algebra $(M_2(\complex)^{\ol \otimes \infty}, \tau) := \lim_{\longrightarrow} (M_{2^n}(\complex), \tau_n)$ is the \textbf{hyperfinite $\vnII_1$ factor}. 
+\end{definition}
+\begin{proof}
+    For each $n \in \natp$, identify $M_{2^n}(\complex)$ as a subalgebra of $M_2(\complex)^{\ol \otimes \infty}$. Let $M_0 = \bigcup_{n \in \natp}M_{2^n}(\complex)$, then by (4) of \autoref{definition:inductive-limit-tracial-von-neumann-algebra}, $M_2(\complex)^{\ol \otimes \infty} = M_0''$. 
+
+    Let $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\tau$. Using \autoref{lemma:tracial-von-neumann-gns}, identify $M_2(\complex)^{\ol \otimes \infty}$ as a von Neumann subalgebra of $B(H)$. Let $x \in Z(M_2(\complex)^{\ol \otimes \infty})$, then by the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, there exists a net $\angles{x_\alpha}_{\alpha \in A} \subset M_0$ such that $\sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}} < \infty$ and $x_\alpha \to x$ in the strong operator topology.
+
+    Fix $\alpha \in A$, then there exists $n \in \natp$ such that $x_\alpha \in M_{2^n}(\complex)$. As $M_{2^n}(\complex)$ is finite-dimensional\footnote{It seems true in general that inductive limits of $\vnII_1$ factors are also $\vnII_1$ factors. However, this compactness shortcut is not available in the general case.}, \hyperref[Haar's Theorem]{theorem:haar} implies that there exists a left-translation invariant Borel probability measure $\mu: \cb_{U(M_{2^n}(\complex))} \to [0, 1]$. Let $y_\alpha = \int_{U(M_{2^n}(\complex))} ux_\alpha u^* \mu(du)$, then for each $v \in U(M_{2^n}(\complex))$, 
+    \begin{align*}
+        vy_\alpha v^* &= \int_{U(M_{2^n}(\complex))} vux_\alpha u^*v^* \mu(du) = \int_{U(M_{2^n}(\complex))} (vu)x_\alpha (vu)^* \mu(du) \\
+        &= \int_{U(M_{2^n}(\complex))} ux_\alpha u^* \mu(du) = y_\alpha
+    \end{align*}
+
+    so $vy_\alpha = y_\alpha v$. By \autoref{corollary:sum-of-unitaries}, $y_\alpha \in Z(M_{2^n}(\complex)) = \complex I_{M_{2^n}(\complex)}$, so there exists $\lambda_\alpha \in \complex$ with $y_\alpha = \lambda_\alpha I_{M_{2^n}(\complex)}$. Since $x \in Z(M_2(\complex)^{\ol \otimes \infty})$ and $\mu$ is a probability measure\footnote{Even though $x$ is a priori not in $M_{2^n}(\complex)$, the integrand here surely has separable range, so the Bochner integral may be used to save some inner product computations. },
+    \[
+        (x - \lambda_\alpha)\xi = \int_{U(M_{2^n}(\complex))} (x - ux_\alpha u^*)\xi \mu(du) =  \int_{U(M_{2^n}(\complex))} u(x - x_\alpha) u^*\xi \mu(du)
+    \]
+
+    where
+    \begin{align*}
+    \norm{u(x - x_\alpha) u^*\xi}_H^2 &= \tau(u(x - x_\alpha)^*u^*u(x - x_\alpha)u^*) \\
+    &= \tau((x - x_\alpha)^*(x - x_\alpha)uu^*uu^*) \\
+    &= \tau((x - x_\alpha)^*(x - x_\alpha)) = \norm{(x - x_\alpha)\xi}_H^2
+    \end{align*}
+
+    By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+        \normn{(x - \lambda_\alpha)\xi}_H^2 &= \int_{U(M_{2^n}(\complex))} \dpn{u(x - x_\alpha)u^*\xi, (x - \lambda_\alpha)\xi}{H} \mu(du) \\
+        &\le \norm{(x - \lambda_\alpha)\xi}_H\braks{\int_{U(M_{2^n}(\complex))}\normn{u(x - x_\alpha) u^*\xi}_H^2 \mu(du)}^{1/2} \\
+        &= \normn{(x - \lambda_\alpha)\xi}_H  \norm{(x - x_\alpha)\xi}_H
+    \end{align*}
+
+    Thus $\limsup_{\alpha \in A}\normn{(x - \lambda_\alpha)\xi}_H \le \lim_{\alpha \in A}\norm{(x - x_\alpha)\xi}_H = 0$. As $x \in Z(M_2(\complex)^{\ol \otimes \infty})$, the above shows that $\lambda_\alpha \to x$ pointwise on $M_2(\complex)^{\ol \otimes \infty}H$. Since $\sup_{\alpha \in A}|\lambda_\alpha| \le \sup_{\alpha \in A}\normn{x_\alpha}_{M_2(\complex)^{\ol \otimes \infty}} < \infty$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that $\lambda_\alpha \to x$ in the strong operator topology. Therefore $x \in \complex I_{M_2(\complex)^{\ol \otimes \infty}}$, and $M_2(\complex)^{\ol \otimes \infty}$ is a factor. 
+\end{proof}
+
+
+

src/vna/example/index.tex

@@ -0,0 +1,8 @@
+\chapter{Constructions of von Neumann Algebras}
+\label{chap:vna-example}
+
+
+\input{./bb.tex}
+\input{./lg.tex}
+\input{./hii1.tex}
+

src/vna/example/lg.tex

@@ -0,0 +1,207 @@
+\section{Group von Neumann Algebras}
+\label{section:gvna}
+
+\begin{definition}[Group von Neumann Algebra]
+\label{definition:group-von-neumann-algebra}
+    Let $G$ be a locally compact group, 
+    \[
+    \lambda: G \to U(L^2(G; \complex)) \quad x \mapsto L_x
+    \]
+    
+    be the left regular representation of $G$, and
+
+    \[
+    \lambda: L^1(G; \complex) \to B(L^2(G; \complex)) \quad \lambda(f)(g) = (f * g)
+    \]
+
+    be its extension to $L^1(G; \complex)$, then
+    \[
+    L(G) := \lambda(G)'' = \lambda(L^1(G;\complex))''
+    \]
+
+    is the \textbf{group von Neumann algebra} of $G$. Similarly, if $\rho: G \to U(L^2(G; \complex))$ is the right regular representation of $G$, then $R(G) := \rho(G)'' = \rho(L^1(G;\complex))''$.
+\end{definition}
+\begin{proof}
+    Firstly, for each $f \in L^1(G; \complex)$ and $g, h \in L^2(G; \complex)$, by the \hyperref[Fubini-Tonelli Theorem]{theorem:fubini-tonelli}, 
+    \begin{align*}
+        \dpn{\lambda(f)g, h}{L^2(G; \complex)} &= \int_G f(x)\dpn{L_xg, h}{L^2(G; \complex)} dx = \iint_{G \times G}f(x) g(x^{-1}y) \ol{h}(y) dy dx \\
+        &= \iint_{G \times G}f(x) g(x^{-1}y)dx \ol{h}(y) dy  \\
+        &= \int_G (f * g)(y) \ol{h}(y)dy = \dpn{(f * g), h}{L^2(G; \complex)}
+    \end{align*}
+
+    By \autoref{theorem:l1-group-von-neumann}, $\ol{\lambda(G)}^{\text{\small SOT}} = \ol{\lambda(L^1(G;\complex))}^{\text{\small SOT}}$. 
+\end{proof}
+
+\begin{definition}[Left Convolver]
+\label{definition:left-convolver}
+    Let $G$ be a locally compact group and $f: G \to \complex$ be Borel measurable, then $f$ is a \textbf{left convolver} if for each $g \in L^2(G; \complex)$, $\int_G |f(y)||g(y^{-1}x)|dy < \infty$ for almost every $x \in G$ and $f * g \in L^2(G; \complex)$. The mapping $L_f: L^2(G; \complex) \to L^2(G; \complex)$ defined by $g \mapsto f * g$ is the \textbf{left convolution map} by $f$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:left-convolver-cc}
+    Let $G$ be a locally compact group and $f \in C_c(G; \complex)$, $f$ is a left convolver. 
+\end{lemma}
+\begin{proof}
+    By \hyperref[Young's Inequality]{theorem:young-fourier}, for each $g \in L^2(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:left-convolver}
+    Let $G$ be a locally compact group and $f: G \to \complex$ be a left convolver, then:
+    \begin{enumerate}
+        \item $L_f \in B(L^2(G; \complex))$. 
+        \item $f \in L_{\text{loc}}^1(G; \complex)$. 
+        \item If $f * g = 0$ for all $g \in L^2(G; \complex)$, then $f = 0$ locally almost everywhere. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $\seq{g_n} \subset L^2(G; \complex)$, $\seq{f * g_n} \subset L^2(G; \complex)$, and $g, h \in L^2(G; \complex)$ such that $g_n \to g$ and $f * g_n \to h$ in $L^2(G; \complex)$ as $n \to \infty$. After taking a subsequence, assume without loss of generality that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $f * g_n \to h$ almost everywhere.
+        \item $\sum_{n \in \natp}\norm{g_n - g}_{L^2(G; \complex)} < \infty$.
+    \end{enumerate}
+    
+    In which case, for almost every $x \in G$, 
+    \begin{align*}
+        f * (g_N - g)(x) &= \int_G f(y)(g_N - g)(y^{-1}x)dy \\
+        |f * (g_N - g)(x)| &\le \int_G|f(y)|\sum_{n \ge N}|(g_n - g)(y^{-1}x)| dy
+    \end{align*}
+
+    where $\int_G|f(y)|\sum_{n \in \natp}|(g_n - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $f * g_N \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the \hyperref[Closed Graph Theorem]{theorem:closed-graph} implies that $L_f$ is continuous. 
+
+    (2): Let $U \in \cn_G(1_G)$ be relatively compact with $U = U^{-1}$, then $\one_U \in L^2(G; \complex)$, and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for almost every $x \in G$. For each $x_0 \in G$, by \autoref{proposition:haar-measure-charge}, there exists $x \in G$ such that $x_0 \in xU$ and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$. As such, for any $K \subset G$ compact, there exists $F \subset G$ finite such that $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for each $x \in F$ and $K \subset \bigcup_{x \in F}xU$. In which case, 
+    \[
+    \int_K |f(y)|dy \le \sum_{x \in F}\int_G |f(y)|\one_U(y^{-1}x)dy < \infty
+    \]
+
+    (3): By \autoref{lemma:left-convolver}, $f$ is locally integrable. Let $K \subset G$ be relatively compact. Using \autoref{lemma:lch-compact-neighbour}, assume without loss of generality that $K$ is also open. 
+    
+    Let $g \in C_c(G; \complex)$ with $\supp{g} \subset K^{-1}$, then there exists $V \in \cn_G(1_G)$ such that $\supp{g}^{-1}V \subset K^{-1}$. In which case, for each $x \in V$, 
+    \[
+    (f * g)(x) = \int_G f(y)g(y^{-1}x) dy = \int_K f(y)g(y^{-1}x) dy = (\one_Kf * g)(x)
+    \]
+
+    so $f * g$ is continuous on $V$ by \hyperref[Young's Inequality]{theorem:young-fourier}. In particular, $f * g|_{V} = 0$ almost everywhere implies that $\int_G f(y)g(y^{-1}) dy = 0$. As such, $\int_G fg = 0$ for all $g \in C_c(G; \complex)$ with $\supp{g} \subset K$. By uniqueness from the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $f|_K = 0$ almost everywhere. Therefore $f = 0$ locally almost everywhere. 
+\end{proof}
+
+\begin{definition}[Space of Left Convolvers]
+\label{definition:left-convolver-algebra}
+    Let $G$ be a group, $LC(G) := \bracsn{L_f| f: G \to \complex\text{ is a left convolver}} \subset B(l^2(G; \complex))$, and $f, g: G \to \complex$ be left convolvers, then:
+    \begin{enumerate}
+        \item $f \in l^2(G; \complex)$.
+        \item $L_f^* = L_{f^*}$, where $f^*(x) = \ol{f(x^{-1})}$ for all $x \in G$.
+        \item $L_{f * g} = L_fL_g$. 
+        \item $LC(G) \subset B(l^2(G; \complex))$ is a von Neumann algebra. 
+    \end{enumerate}
+
+    The set $LC(G)$ is the \textbf{algebra of left convolvers} on $G$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 1.3.5]{AnantharamanPopa}}}. ]
+    (1): $f * \delta_e = f \in l^2(G; \complex)$. 
+
+    (2): Let $x, y \in G$, then
+    \begin{align*}
+        \dpn{L_f\delta_x, \delta_y}{l^2(G; \complex)} &= (f * \delta_x)(y) = f(yx^{-1}) \\ 
+    \dpn{\delta_x, L_{f^*}\delta_y}{l^2(G;\complex)} &= \ol{(f^* * \delta_y)(x)} = \ol{f^*(xy^{-1})} = f(yx^{-1})
+    \end{align*}
+    
+
+    (3): By (1), $f * g \in l^2(G; \complex)$ is defined. For any $h \in l^2(G; \complex)$, by associativity of convolution, 
+    \[
+    L_{f * g}h = (f * g) * h = f * (g * h) = L_fL_gh
+    \]
+
+    (4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_e \in l^2(G; \complex)$, then
+    \[
+    f = T\delta_e = \paren{\sotlim_{\alpha \in A}L_{f_\alpha}}\delta_e = l^2\text{-}\lim_{\alpha \in A}f_\alpha * \delta_e
+    \]
+    
+    For any $g \in l^2(G; \complex)$,
+    \[
+    \norm{Tg - f * g}_{l^\infty(G; \complex)} \le \limsup_{\alpha \in A}\norm{g}_{l^2(G; \complex)}\norm{f - f_\alpha}_{l^2(G; \complex)} = 0
+    \]
+
+    by \hyperref[Young's Inequality]{theorem:young-fourier}. Therefore $Tg = f * g$, $T = L_f$, and $LC(G)$ is strong-operator closed. By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $LC(G)$ is a von Neumann algebra. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:group-algebra-identified}
+    Let $G$ be a group, then $L(G) = LC(G) = R(G)'$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.3.6]{AnantharamanPopa}}}. ]
+    ($\subset$): For any $x, y, z \in G$ and $f \in l^2(G; \complex)$, 
+    \[
+    (\lambda(x)\rho(y)f)(z) = (\rho(y)f)(x^{-1}z) = f(x^{-1}zy) = (\lambda(x)f)(zy) = (\rho(y)\lambda(x)f)(z)
+    \]
+
+    so $\lambda(G) \subset R(G)'$. By \autoref{definition:left-convolver-algebra}, $L(G) \subset LC(G) \subset R(G)'$. 
+
+    ($\supset$): Let $T \in R(G)'$ and $f = T\delta_e \in l^2(G; \complex)$, then for each $x \in G$, 
+    \[
+    T\delta_x = T\rho(x^{-1})\delta_e = \rho(x^{-1})T\delta_e = f * \delta_x
+    \]
+
+    As such, $Tg = f * g$ for all $g \in l^2(G; \complex)$, and $T = L_f$. 
+\end{proof}
+
+\begin{definition}
+\label{definition:group-algebra-fourier-series}
+    Let $G$ be a group. For each $x \in G$, let $u_x = \lambda(x)$ and identify it with $\delta_x$, then $\lambda(G) = \bracsn{u_x|x \in G}$ are the \textbf{canonical unitaries} of $L(G)$. 
+
+    For each $f \in l^2(G)$, $f = \sum_{x \in G}f(x)u_x$. In particular, for each $T \in L(G) = LC(G) \subset l^2(G; \complex)$, there exists $\bracsn{T_x}_{x \in G} \subset \complex$ such that $T$ may be written as $T = \sum_{x \in G}T_x u_x$\footnote{This sum does not converge in the strong-operator or the weak-operator topology. }. In which case, the scalars $\bracsn{T_x}_{x \in G}$ are the \textbf{Fourier coefficients} of $T$. 
+\end{definition}
+
+\begin{definition}[ICC]
+\label{definition:icc}
+    Let $G$ be a group, then $G$ has \textbf{infinite conjugacy classes (ICC)} if every non-trivial conjugacy class of $G$ is infinite. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:icc-factor}
+    Let $G$ be a group, then the following are equivalent:
+    \begin{enumerate}
+        \item $L(G)$ is a factor. 
+        \item $G$ is ICC. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 1.3.9]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (1): Let $T \in Z(L(G))$, then for each $x \in G$, 
+    \[
+    T\delta_e = \lambda(x)T\lambda(x^{-1})\delta_e = \lambda(x)T\rho(x)\delta_e = \lambda(x)\rho(x)(T\delta_e)
+    \]
+
+    so $T\delta_e$ is invariant under conjugation, and hence constant on conjugacy classes of $G$. 
+    
+    Since $G$ is ICC and $T\delta_e \in l^2(G; \complex)$, there exists $\lambda \in \complex$ with $T\delta_e = \lambda \delta_e$. As such, $T = \lambda I$, and $Z(L(G)) = \complex I$. 
+
+    $\neg$ (2) $\Rightarrow$ $\neg$ (1): Let $C \subset G$ be a non-trivial finite conjugacy class, then for each $x \in G$, $\delta_x * \one_C = \one_{xC} = \one_{Cx} = \one_C * \delta_x$, so $L_{\one_C}$ is a non-trivial element of $Z(L(G))$. 
+\end{proof}
+
+\begin{definition}[Wreath Product]
+\label{definition:wreath-product}
+    Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, $H^{(I)} = \bigoplus_{i \in I}H$, and $\Gamma$ act on $H^{(I)}$ by left translation ($(\gamma \xi)_i = \xi_{\gamma^{-1}i}$), then the \textbf{generalised wreath product} $H \wr_I \Gamma$ is the semidirect product $H^{(I)} \rtimes \Gamma$. The \textbf{wreath product} $H \wr \Gamma$ is the special case where $I = \Gamma$, and $\Gamma$ acts by left translation. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:wreath-product-icc}
+    Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, and $\Gamma_f \subset \Gamma$ be the subgroup of elements whose conjugacy class is finite. If
+    \begin{enumerate}[label=(\alph*)]
+        \item The orbits of $\Gamma \curvearrowleft I$ are infinite.
+        \item The restricted action $\Gamma_f \curvearrowright I$ is faithful. 
+    \end{enumerate}
+
+    then the wreath product $H \wr_I \Gamma$ is ICC. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 1.3.10]{AnantharamanPopa}}}. ]
+    For each $g \in H \wr_I \Gamma$ and subgroup $K \subset H \wr_I \Gamma$, let $g^K = \bracsn{kgk^{-1}|k \in K}$. 
+
+    Let $g = (\xi, \gamma) \in H \wr_I \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^G \supset g^\Gamma$ is infinite. 
+
+    Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_f$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_f$. By (b), there exists $i_0 \in I$ with $\gamma i_0 \ne i_0$. As $H$ is non-trivial, there exists $\xi_0 \in H^{(I)} \setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_0 = (\xi_0, e_{\Gamma})$, then
+    \[
+    g_0^{-1}gg_0 = (\xi_0^{-1}\gamma\xi_0, \gamma)
+    \]
+
+    and hence $g$ has infinite conjugacy class because $\xi_0^{-1}\gamma\xi_0 \ne e_{H^{(I)}}$. Therefore $g^G$ is also infinite. 
+\end{proof}

src/vna/example/tensor.tex

@@ -0,0 +1,16 @@
+\section{Tensor Products of von Neumann Algebras}
+\label{section:vna-tensor}
+
+\begin{definition}[Tensor Product of von Neumann Algebras]
+\label{definition:vna-tensor}
+    Let $H_1, H_2$ be complex Hilbert spaces, $H_1 \wh \otimes_\pi H_2$ be their Hilbert space tensor product, $M_1 \subset B(H_1)$ and $M_2 \subset B(H_2)$ be von Neumann algebras, and
+    \[
+    \iota: M_1 \otimes M_2 \to B(H_1 \wh \otimes_\pi H_2) \quad (x_1 \otimes x_2)(\xi_1 \otimes \xi_2) = (x_1\xi_1) \otimes (x_2\xi_2)
+    \]
+
+    Then, $M_1 \ol{\otimes} M_2 := \iota(M_1 \otimes M_2)'' \subset B(H_1 \wh \otimes_\pi H_2)$ is the \textbf{von Neumann tensor product} of $(M_1, H_1)$ and $(M_2, H_2)$.
+
+    The von Neumann algebra $M_1 \ol{\otimes} M_2$ is unique up to isomorphism, independent of the representations of $M_1$ and $M_2$\footnote{Proof omitted for now.}. 
+\end{definition}
+
+

src/vna/ii1/fg.tex

@@ -0,0 +1,99 @@
+\section{The Fundamental Group}
+\label{section:vna-fundamental-group}
+
+\begin{definition}[Induced von Neumann Algebra]
+\label{definition:induced-vna-commutant}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in \text{Proj}(M')$, then $Mq$ is a von Neumann algebra on $qH$, known as the \textbf{induced von Neumann algebra} of $M$ with respect to $q$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:induced-vna-commutant-gymnastics}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in \text{Proj}(M)$, then:
+    \begin{enumerate}
+        \item For any central projection $e \in Z(pMp)$, $e = Z(e)p$. 
+        \item $Z(pMp) = Z(M)p$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2.1]{AnantharamanPopa}}}. ]
+    (1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^* = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by \autoref{proposition:central-support-vna}, and $e = Z(e)p$. 
+
+    (2): By \autoref{theorem:vn-projection-norm-dense}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:vna-induced-commutants}
+    Let $H$ be a complex Hilbert space and $M \subset B(H)$ be a von Neumann algebra, then
+    \begin{enumerate}
+        \item For any $p \in \text{Proj}(M)$, the commutant $(pMp)'$ in $B(pH)$ is $pM'$. 
+        \item For any $q \in \text{Proj}(M')$, the commutant $(Mq)'$ in $B(qH)$ is $qM'q$. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2.2]{AnantharamanPopa}}}. ]
+    (1): For any $x \in M$ and $y \in M'$, $pxp \cdot yp = ypxp = yp \cdot pxp$, so $pMp \subset (M'p)'$. 
+
+    For any $x \in (M'p)' \subset B(pH)$, let $\tilde x = xp = pxp \in B(H)$, then for any $y \in M'$, 
+    \[
+    \tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x
+    \]
+
+    By the \hyperref[Bicommutant Theorem]{theorem:bicommutant}, $\tilde x \in M$, $x = p\tilde x p \in pMp$, and $pMp \supset (M'p)'$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:amplified-ii1-factor}
+    Let $(M, \tau)$ be a $\vnII_1$ factor, then $M_n(M)$ is a $\vnII_1$ factor, with normalised trace $\frac{1}{n}\text{Tr}_n \otimes \tau$. 
+\end{proposition}
+
+
+\begin{definition}[Amplification]
+\label{definition:ii1-amplification}
+    Let $(M, \tau)$ be a $\vnII_1$ factor, $t > 0$, $n \in \natp$ with $n \ge t$, and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$, then the von Neumann algebra $M^t := pM_n(M)p \subset M_n(M)$ is an \textbf{amplification} of $M$. 
+    
+    The spatial isomorphism class of $M^t$ is independent of the choice of $n \ge t$ and $p \in \text{Proj}(M_n(M))$ with $(\text{Tr}_n \otimes \tau)p = t$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:ii1-trace-projection-equivalence-classes}.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:ii1-amplification-homomorphism}
+    Let $(M, \tau)$ be a $\vnII_1$ factor and $s, t > 0$, then $(M^s)^t \iso M^{st}$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 4.2.3]{AnantharamanPopa}}}. ]
+    Let $m \in \natp$ and $p \in \text{Proj}(M_m(M))$ with $(\text{Tr}_m \otimes \tau)(p) = s$, so that $M^s = pM_m(M)p$, and $\tau_s = s^{-1}(\text{Tr}_m \otimes \tau)|_{M^s}$ denotes the normalised tracial state on $M^s$. 
+    
+    Let $n \in \natp$ and $q \in \text{Proj}(M_n(M^s))$ with $(\text{Tr}_n \otimes \tau_s)(q) = t$, so that $(M^s)^t = qM_n(M^s)q$. Identify $M_n(M^s) \subset M_n(M_m(M))$, so that $q \in M_n(M_m(M))$, then 
+    \begin{align*}
+    (M^s)^t &= qM_n(M^s)q = qM_n(pM_m(M)p)q \\
+    &= q(1_{M_n(\complex)} \otimes p)M_n(M_m(M))(1_{M_n(\complex)} \otimes p)q
+    \end{align*}
+     
+
+    Since $q \in M_n(M^s)$, $p$ is the identity on entries of $q$, so
+    \[
+    q(1_{M_n(\complex)} \otimes p) = q = (1_{M_n(\complex)} \otimes p)q
+    \]
+
+    and $(M^s)^t = qM_n(M_m(M))q = qM_{mn}(M)q$. Finally, 
+    \[
+        (\text{Tr}_{mn} \otimes \tau)(q) = (\text{Tr}_n \otimes \text{Tr}_{m} \otimes \tau)(q) = s(\text{Tr}_n \otimes \tau_s)(q) = st
+    \]
+
+    Therefore $(M^s)^t = M^{st}$. 
+\end{proof}
+
+
+\begin{definition}[Fundamental Group]
+\label{definition:ii1-fundamental-group}
+    Let $M$ be a $\vnII_1$ factor, then the \textbf{fundamental group} of $M$ is the family
+    \[
+    \fF(M) = \bracsn{t > 0| M^t \iso M} \subset \real_{>0}^\times
+    \]
+\end{definition}
+\begin{proof}
+    By \autoref{lemma:ii1-amplification-homomorphism}, $\fF(M)$ is a group. 
+\end{proof}
+
+

src/vna/ii1/index.tex

@@ -0,0 +1,5 @@
+\chapter{Finite and $\vnII_1$ Factors}
+\label{chap:ii1-factors}
+
+\input{./trace.tex}
+\input{./fg.tex}

src/vna/ii1/standard.tex

@@ -0,0 +1,162 @@
+\section{The Standard Representation}
+\label{section:standard-representation}
+
+
+
+\begin{definition}[Standard Representation]
+\label{definition:tracial-vna-standard}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then the \hyperref[GNS representation]{definition:gns-triple}
+    \[
+    (\pi, L^2(M, \tau), \xi_\tau)
+    \]
+
+    is the \textbf{standard representation} of $(M, \tau)$. For each $x \in M$, identify $\wh x = x\xi_\tau \in L^2(M, \tau)$, so that $M = \wh M \subset L^2(M, \tau)$. 
+\end{definition}
+
+\begin{definition}[Canonical Conjugation]
+\label{definition:tracial-vna-standard-conjugate}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then the mapping
+    \[
+    J: L^2(M, \tau) \to L^2(M, \tau) \quad \wh x \mapsto \wh{x^*}
+    \]
+
+    is a conjugate linear isometry on $L^2(M, \tau)$, known as the \textbf{canonical conjugation}. 
+\end{definition}
+
+
+\begin{definition}[Left/Right Regular Representations]
+\label{definition:mvn-standard-regular}
+    Let $(M, \tau)$ be a tracial von Neumann algebra and $x \in M$, then
+    \[
+    L_x: L^2(M, \tau) \to L^2(M, \tau) \quad L_x\wh y = \wh{xy}
+    \]
+    and
+
+    \[
+    R_x: L^2(M, \tau) \to L^2(M, \tau) \quad R_x\wh y = \wh{yx}
+    \]
+
+    are the \textbf{left/right multiplication maps} by $x$, respectively. The mapping
+    \[
+    \lambda: M \to B(L^2(M, \tau)) \quad x \mapsto L_x
+    \]
+
+    is the \textbf{left regular representation} of $M$, with $L(M) = \lambda(M)$ being the \textbf{algebra of left multiplications}. Similarly,
+    \[
+    \rho: M^{op} \to B(L^2(M, \tau)) \quad x \mapsto R_x
+    \]
+
+    is the \textbf{right regular representation} of $M^{op}$, with $R(M) = \rho(M)$ being the \textbf{algebra of right multiplications}. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:vna-standard-conjugation-lr}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then:
+    \begin{enumerate}
+        \item For each $x \in M$, $JL_xJ = R_{x^*}$. 
+        \item $JL(M)J = R(M)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x, y \in M$, 
+    \[
+    JL_xJ \wh y = JL_x \wh{y^*} = J \wh{xy^*} = \wh{yx^*} = R_{x^*}y
+    \]
+\end{proof}
+
+\begin{definition}[Multiplication by Vector]
+\label{definition:vna-standard-vector-multiplication}
+    Let $(M, \tau)$ be a tracial von Neumann algebra and $\xi \in L^2(M, \tau)$, then
+    \[
+    L^0_\xi: \wh M \to L^2(M, \tau) \quad L_\xi^0(\wh y) = R_y(\xi) = \xi y
+    \]
+
+    and
+    \[
+    R^0_\xi: \wh M \to L^2(M, \tau) \quad R_\xi^0(\wh y) = L_y(\xi) = y\xi
+    \]
+
+    are densely defined, closable operators. 
+    
+    Let $L_\xi$ and $R_\xi$ be their respective closures, then $L_\xi$ and $R_\xi$ are the \textbf{left/right multiplication operators by $\xi$}. 
+\end{definition}
+\begin{proof}
+    Let $\seq{x_n} \subset M$ and $\eta \in L^2(M, \tau)$ with $\limv{n}\wh{x_n} = 0$ and $\limv{n}L_\xi^0(\wh{x_n}) = \eta$, then for each $y \in M$, 
+    \begin{align*}
+    \dpn{\eta, \wh y}{L^2(M, \tau)} &= \limv{n} \dpn{L_\xi^0(\wh {x_n}), \wh y}{L^2(M, \tau)} = \limv{n} \dpn{R_{x_n}\xi, \wh y}{L^2(M, \tau)} \\
+    &= \limv{n}\dpn{\xi, R_{x_n}^*\wh y}{L^2(M, \tau)} = \limv{n}\dpn{\xi, R_{x_n^*}\wh y}{L^2(M, \tau)} \\
+    &= \limv{n}\dpn{\xi, \wh {yx_n^*}}{L^2(M, \tau)} \\
+    |\dpn{\eta, \wh y}{L^2(M, \tau)}| &\le \limv{n}\norm{\xi}_{L^2(M, \tau)} \normn{\wh {yx_n^*}}_{L^2(M, \tau)} \\
+    &\le  \limv{n}\norm{\xi}_{L^2(M, \tau)} \normn{y}_{M} \normn{\wh {x_n^*}}_{L^2(M, \tau)} = 0
+    \end{align*}
+    
+\end{proof}
+
+\begin{definition}[Convolvers]
+\label{definition:vna-standard-convolver}
+    Let $(M, \tau)$ be a tracial von Neumann algebra and $\xi \in L^2(M, \tau)$, then $\xi$ is a \textbf{left convolver} if $L_\xi$ is bounded, and $\xi$ is a \textbf{right convolver} if $R_\xi$ is bounded. The sets of left and right convolvers are denoted $LC(M)$ and $RC(M)$, respectively. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:vna-convolver-commute}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then for each $\xi \in LC(M)$ and $\eta \in RC(M)$, $L_\xi R_\eta = R_\eta L_\xi$.
+\end{lemma}
+\begin{proof}
+    First suppose that $\xi, \eta \in M$. In which case, for each $x, y \in M$, 
+    \[
+        \dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{\wh{\xi x \eta}, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}
+    \]
+
+    Now suppose that $\xi \in M$, then for each $\eta \in RC(M)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+    |\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{L_\xi}_{B(L^2(M, \tau))} \norm{x}_M \norm{y}_M \cdot  \norm{\eta}_{L^2(M, \tau)} \\ 
+    |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{x}_M \norm{\xi}_M\norm{y}_M \cdot \norm{\eta}_{L^2(M, \tau)}
+    \end{align*}
+    
+
+    so by density of $\wh M$ in $L^2(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in M$ and $\eta \in RC(M)$. 
+
+    Finally, suppose that $\xi \in LC(M)$ and $\eta \in RC(M)$, then for each $z \in M$, 
+    \begin{align*}
+        \dpn{L_\xi z, \wh y}{L^2(M, \tau)} &= \dpn{\xi z, \wh y}{L^2(M, \tau)} = \dpn{\xi, \wh y z^*}{L^2(M, \tau)} \\
+        &= \dpn{\xi, \wh y J\wh z}{L^2(M, \tau)} = \dpn{y^*\xi, J\wh z}{L^2(M, \tau)} = \dpn{\wh z, Jy^*\xi}{L^2(M, \tau)}
+    \end{align*}
+
+    so
+    \begin{align*}
+        \dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} &= \dpn{L_\xi (x \eta), \wh y}{L^2(M, \tau)} = \dpn{x \eta, J y^* \xi}{L^2(M, \tau)} \\
+        |\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{x \eta}_{L^2(M, \tau)} \norm{y}_M  \cdot \norm{\xi}_{L^2(M, \tau)}\\
+        |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{R_\eta}_{B(L^2(M, \tau))} \norm{x}_M \norm{y}_M \cdot \norm{\xi}_{L^2(M, \tau)}
+    \end{align*}
+    
+    By density of $\wh M$ in $L^2(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in LC(M)$ and $\eta \in RC(M)$. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:vna-convolver-identified}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then:
+    \begin{enumerate}
+        \item $L(M) = LC(M) = R(M)'$. 
+        \item $JL(M)J = R(M) = RC(M) = L(M)'$. 
+        \item In particular, for each $\xi \in L^2(M, \tau)$, $L_\xi \in B(L^2(M, \tau))$ if and only if $\xi \in \wh M$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By definition, $L(M) \subset LC(M)$. By \autoref{lemma:vna-convolver-commute}, $LC(M) \subset RC(M)'$, so
+    \[
+    L(M) \subset LC(M) \subset RC(M)' \subset R(M)'
+    \]
+
+    and
+    \[
+    R(M) \subset RC(M) \subset LC(M)' \subset L(M)'
+    \]
+
+    Using the \hyperref[Bicommutant Theorem]{theorem:bicommutant}, it remains to show that $R(M)' \subset LC(M)$. To this end, let $T \in R(M)'$ and $\xi = T \wh 1$, then for each $x \in M$, 
+    \[
+    T\wh x = TR_x \wh 1 = R_x T \wh 1 = R_x \xi = \xi x = L_\xi^0 \wh x
+    \]
+
+    so $T|_{\wh M} = L_\xi^0$. By density of $\wh M$ in $L^2(M, \tau)$ and continuity of $T$, the closure of $L_\xi^0$ is $T$, hence $T = L_\xi$. 
+\end{proof}
+

src/vna/ii1/trace.tex

@@ -0,0 +1,185 @@
+\section{Traces}
+\label{section:tracial-von-neumann-algebra}
+
+\begin{definition}[Trace]
+\label{definition:trace}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\tau: M_+ \to [0, \infty]$, then $\tau$ is a \textbf{trace} on $M$ if:
+    \begin{enumerate}
+        \item For each $x, y \in M_+$ and $\lambda \ge 0$, $\tau(\lambda x + y) = \lambda \tau(x) + \tau(y)$. 
+        \item For each $x \in M_+$ and $u \in U(M)$, $\tau(uxu^*) = \tau(x)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Tracial von Neumann Algebra]
+\label{definition:tracial-von-neumann-algebra}
+    Let $M$ be a von Neumann algebra and $\tau: M \to \complex$ be a faithful tracial state, then $(M, \tau)$ is a \textbf{tracial von Neumann algebra} if $\tau$ is normal. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:cstar-trace-vna}
+    Let $M$ be a unital $C^*$-algebra, $\tau: M \to \complex$ be a faithful tracial state, and $(L^2(M, \tau), \pi, 1_M)$ be its \hyperref[GNS]{definition:gns-triple} triple, then:
+    \begin{enumerate}
+        \item The uniform structure of pointwise convergence\footnote{Strong operator topology.} on $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ coincides with the uniform structure of $\ol{B_M(0, 1)} \subset L^2(M, \tau)$ induced by the $L^2(M, \tau)$ norm. 
+        \item $\pi(M) \subset B(L^2(M, \tau))$ is a von Neumann algebra if and only if $\ol{B_M(0, 1)}$ is complete in $L^2(M, \tau)$. In which case, $\tau$ is normal. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.6.4]{AnantharamanPopa}}}. ]
+    (1): For any $x, y \in \ol{B_M(0, 1)}$, 
+    \begin{align*}
+    \norm{x - y}_{L^2(M, \tau)}^2 &= \tau((x - y)^*(x - y)) = \dpn{\pi(x - y)^*\pi(x - y)1_M, 1_M}{L^2(M, \tau)} \\
+    &= \dpn{\pi(x - y)1_M, \pi(x - y)1_M}{L^2(M, \tau)} = \norm{\pi(x - y)1_M}_{L^2(M, \tau)}^2
+    \end{align*}
+
+    so the uniform structure of pointwise convergence is stronger than the uniformity induced by the $L^2(M, \tau)$ norm.
+    
+    On the other hand, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the uniform structure of pointwise convergence on $\ol{B_M(0, 1)}$ coincides with the uniform structure of pointwise convergence on any dense subset of $L^2(M, \tau)$. In particular, this includes $M \subset L^2(M, \tau)$. For each $x, y \in \ol{B_M(0, 1)}$ and $t \in M$, since $\tau$ is tracial, 
+    \begin{align*}
+        \norm{(x - y)t}_{L^2(M, \tau)}^2 &= \tau(t^*(x - y)^*(x - y)t) \le \tau(tt^*(x - y)^*(x - y)) \\
+        &\le \norm{t}_M^2 \tau((x - y)^*(x - y)) = \norm{t}_M^2 \norm{x - y}_{L^2(M, \tau)}^2
+    \end{align*}
+
+    so the uniformity induced by the $L^2(M, \tau)$ norm is stronger than the uniform structure of pointwise convergence.
+    
+    (2): Using (1), it is sufficient to show that $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ is complete if and only if it is strong-operator closed. By \autoref{proposition:complete-closed}, $\ol{B_M(0, 1)}$ is complete if and only if it is closed in $L^2(M, \tau)^{L^2(M, \tau)}$ with respect to the product topology. Finally, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that $\ol{B_M(0, 1)}$ is closed in $L^2(M, \tau)^{L^2(M, \tau)}$ if and only if it is strong-operator closed in $B(L^2(M, \tau))$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-separable-condition}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item $M$ is separable. 
+        \item $\ol{B_M(0, 1)}$ is strong-operator separable. 
+        \item $\ol{B_M(0, 1)}$ is separable in $L^2(M, \tau)$. 
+        \item $L^2(M, \tau)$ is separable. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $H$ be a separable Hilbert space with $M \subset B(H)$ and $D \subset H$ be countable and dense. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the strong operator topology on $\ol{B_M(0, 1)}$ coincides with the topology of pointwise convergence on $D$. By \autoref{theorem:uniform-metrisable} and \autoref{proposition:separable-product}, $H^D$ is metrisable and separable. Therefore $\ol{B_M(0, 1)} \subset H^D$ is also metrisable and separable.
+
+    (2) $\Rightarrow$ (3): By \autoref{proposition:cstar-trace-vna}. 
+
+    (3) $\Rightarrow$ (4): By \autoref{lemma:tracial-von-neumann-gns}, the standard representation $(L^2(M, \tau), \pi, 1_M)$ is faithful and normal. As $M$ is dense in $L^2(M, \tau)$, $\ol{B_M(0, 1)}$ being separable in $L^2(M, \tau)$ implies that $L^2(M, \tau)$ is separable. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-von-neumann-factor}
+    Let $(M, \tau)$ be a tracial von Neumann factor, then exactly one of the following holds:
+    \begin{enumerate}
+        \item There exists $n \in \natp$ such that $M \iso M_n(\complex)$. 
+        \item $M$ is of type $\vnII_1$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Suppose that $M$ has a minimal projection. By \autoref{theorem:type1-bh}, there exists a complex Hilbert space $H$ such that $M \iso B(H)$. Let $\seqi{e} \subset H$ be an orthonormal basis, then $e_i \otimes e_i \in \text{Proj}(M)$ is minimal for all $i \in I$, and all such rank one projections are equivalent. As such, for any $j \in I$, 
+    \[
+    \tau(1_M) = \sum_{i \in I}\tau(e_i \otimes e_i) = |I| \cdot \tau(e_j \otimes e_j)
+    \]
+
+    so $I$ must be finite, and $M \iso M_{|I|}(\complex)$. 
+
+    (2): Suppose that $M$ has no minimal projections. By \autoref{lemma:abelian-minimal-factor}, $M$ has no abelian projections. As $\tau$ is a faithful tracial state, $1_M$ is finite. Therefore $M$ is a $\vnII_1$ factor. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:diffuse-factor-subsplit}
+    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 \le p$.  
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
+    Since $p$ is not minimal, there exists a projection $q \le p$ with $0 \ne q \ne p$. As $M$ is a factor, $Z(q)Z(p - q) = 1_M$. By \autoref{proposition:central-support-mvn}, there exist non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$, $p_1 \le q$, and $p_2 \le p - q$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:diffuse-factor-split}
+    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exist $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 = p$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
+    Let
+    \[
+    \mathcal{F} = \bracsn{(q, r) \in \text{Proj}(M)^2|q \sim r, q \perp r, q + r \le p}
+    \] 
+
+    and $\mathcal{M} \subset \mathcal{F}$ be a maximal family such that $\bracsn{q, r|(q,r) \in \mathcal{M}}$ are all mutually orthogonal. Let $q = \sum_{(q, r) \in \mathcal{M}}q$ and $r = \sum_{(q, r) \in \mathcal{M}}r$, then by \autoref{lemma:mvn-equivalent-direct-sum}, $q \sim r$, $q \perp r$, and $q + r \le p$. 
+
+    If $q + r \ne p$, then by \autoref{lemma:diffuse-factor-subsplit} applied to $p - (q + r)$, there exist non-zero projections $q', r' \in \text{Proj}(M)$ with $q' \sim r'$, $q' \perp r'$, and $q' + r' \le  p - (q + r)$, which contradicts the maximality of $\mathcal{M}$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:diffuse-factor-dyadic}
+    Let $M$ be a diffuse factor, then for each $n \in \natp$, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. If $\tau: M \to \complex$ is a tracial state, then $\tau(p_j) = 2^{-n}$ for all $1 \le j \le 2^n$.
+\end{corollary}
+\begin{proof}
+    By \autoref{lemma:diffuse-factor-split}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-simple}
+    Let $M$ be a factor, then $M$ admits at most one tracial state. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.3]{AnantharamanPopa}}}. ]
+    Using \autoref{proposition:tracial-von-neumann-factor}, assume without loss of generality that $M$ is diffuse.
+    
+    Let $p \in \text{Proj}(M)$ with $0 \le p < 1_M$. Let $n \in \natp$, then by \autoref{corollary:diffuse-factor-dyadic}, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. By the \hyperref[corollary:vna-factor-comparability]{theorem:vna-comparability}, there exists a unique $0 \le k \le 2^n - 1$ such that
+    \[
+    \sum_{j = 1}^k p_j \preceq p \prec \sum_{j = 1}^{k+1}p_j
+    \]
+
+    As such, for any tracial state $\tau: M \to \complex$, 
+    \[
+    \frac{k}{2^n} \le \tau(p) \le \frac{k+1}{2^n}
+    \]
+
+    so $\tau(p)$ is independent of the choice of $\tau$. By \autoref{theorem:vn-projection-norm-dense}, $\tau$ is unique.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-unique-trace}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then $M$ is a factor if and only if $\tau$ is the unique normal faithful tracial state on $M$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.4]{AnantharamanPopa}}}. ]
+    If $M$ is a factor, then $\tau$ is unique by \autoref{proposition:tracial-factor-simple}. Suppose that $M$ is not a factor, then $M$ admits a non-trivial central projection $z \in Z(M)$. In which case, for any $\alpha \in (0, 1)$, 
+    \[
+    \tilde \tau(x) = \frac{\alpha}{\tau(z)} \tau(xz) + \frac{1 - \alpha}{1 - \tau(z)}\tau(x(1 - z))
+    \]
+
+    is also a normal faithful tracial state. In particular, 
+    \[
+    \tilde \tau(z) = \frac{\alpha}{\tau(z)} \tau(z) + \frac{1 - \alpha}{1 - \tau(z)}\tau(z(1 - z)) = \alpha
+    \]
+
+    so if $\alpha \ne \tau(z)$, then $\tilde \tau \ne \tau$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-calkin}
+    Let $(M, \tau)$ be a tracial factor, then $M$ admits no non-trivial two-sided ideal. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.5]{AnantharamanPopa}}}. ]
+    Let $I \subset M$ be a non-zero two-sided ideal. By \autoref{proposition:vna-two-sided-ideal}, $I$ admits a non-zero projection $e \in I$. Let $\mathcal{P} \subset \text{Proj}(M)$ be a maximal family of mutually orthogonal projections equivalent to $e$. Since $\tau$ is faithful, $\tau(p) > 0$ for all $p \in \mathcal{P}$, so $\mathcal{P}$ is finite. 
+
+    Let $q = 1_M - \sum_{p \in \mathcal{P}}p$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $q \preceq e$ or $e \prec q$. If $e \prec q$, then there exists a subprojection $p \le q$ with $e \sim p$, which contradicts the maximality of $\mathcal{P}$. Thus $q \preceq e$. 
+
+    Let $\mathcal{Q} = \mathcal{P} \sqcup \bracsn{q}$, then $p \preceq e$ for all $p \in \mathcal{Q}$ and $1_M = \sum_{p \in \mathcal{Q}}p$. For each $p \in \mathcal{Q}$, let $v_p \in M$ be a partial isometry such that $v_pv_p^* = p$ and $v_p^*v_p \le e$. As such, $p = v_pev_p^* \in I$ for all $p \in \mathcal{Q}$. Therefore $1_M = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.  
+\end{proof}
+
+\begin{proposition}
+\label{proposition:ii1-trace-projection-equivalence-classes}
+    Let $M$ be a $\vnII_1$ factor with tracial normal state $\tau: M \to \complex$, then the mapping $p \mapsto \tau(p)$ induces a bijection from the equivalence classes of $\text{Proj}(M)$ to $[0, 1]$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.6]{AnantharamanPopa}}}. ]
+    (Injective): Let $p, q \in \text{Proj}(M)$ with $\tau(p) = \tau(q)$. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $p \prec q$, $p \sim q$, or $q \prec p$. Since $\tau$ is faithful, $p \prec q$ implies that $\tau(p) < \tau(q)$, and $q \prec p$ implies that $\tau(q) < \tau(p)$. Therefore the only possibility is $p \sim q$. 
+
+    (Surjective): By \autoref{lemma:diffuse-factor-split}, there exists a family $\bracsn{p_r|r \in \mathbb D \cap [0, 1]} \subset \text{Proj}(M)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each dyadic rational number $r \in \mathbb D \cap [0, 1]$, $\tau(p_r) = r$. 
+        \item For each $r, s \in \mathbb D \cap [0, 1]$ with $r \le s$, $p_r \le p_s$. 
+    \end{enumerate}
+
+    For each $t \in [0, 1]$, let $p_t = \sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}$, then since $\tau$ is normal, 
+    \begin{align*}
+    \tau(p_t) &= \tau(\sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}) \\
+    &= \sup\bracsn{\tau(p_r)|r \in \mathbb D \cap [0, 1], r \le t} = t
+    \end{align*}
+    
+    
+\end{proof}
+

src/vna/index.tex

@@ -0,0 +1,9 @@
+\part{Von Neumann Algebras}
+\label{part:vna}
+
+\input{./vn/index.tex}
+\input{./example/index.tex}
+\input{./ii1/index.tex}
+\input{./notation.tex}
+
+

src/vna/notation.tex

@@ -0,0 +1,41 @@
+\chapter{Notations}
+\label{chap:vna-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    
+    
+    $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
+    $E_{x, y}$ & Section of a spectral measure $E$, where $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\ 
+    $\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
+    $Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\ 
+    $P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\ 
+    $P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent}
+
+    $A'$ & Commutant of $A \subset B(H)$, the operators in $B(H)$ commuting with every element of $A$. & \autoref{definition:commutant} \\
+    $I_E(\phi) = \int_X \phi dE$ & Integral of $\phi$ with respect to a spectral measure $E$. & \autoref{definition:spectral-measure-integral} \\
+    $\phi(T)$ & $L^\infty$-functional calculus of a normal operator $T$, defined as $\int_{\sigma_{B(H)}(T)}\phi dE$. & \autoref{definition:linfty-functional-calculus} \\
+    $A_{\vnI}, A_{\vnII_1}, A_{\vnII_\infty}, A_{\vnIII}$ & Type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$ summands of a von Neumann algebra $A$. & \autoref{theorem:vna-type-decomposition} \\
+ 
+    $\lambda$ & Left regular representation of $G$, and its extension to $L^1(G; \complex)$. & \autoref{definition:group-von-neumann-algebra} \\
+    $\rho$ & Right regular representation of $G$. & \autoref{definition:group-von-neumann-algebra} \\
+    $L(G)$ & Group von Neumann algebra of $G$, defined as $\lambda(G)''$. & \autoref{definition:group-von-neumann-algebra} \\
+    $R(G)$ & Defined as $\rho(G)''$, the von Neumann algebra generated by the right regular representation. & \autoref{definition:group-von-neumann-algebra} \\
+    $L_f$ & Left convolution map $g \mapsto f * g$ by a left convolver $f$. & \autoref{definition:left-convolver} \\
+    $LC(G)$ & Algebra of left convolvers on $G$. & \autoref{definition:left-convolver-algebra} \\
+    $f^*$ & Defined as $f^*(x) = \ol{f(x^{-1})}$ for a left convolver $f$. & \autoref{definition:left-convolver-algebra} \\
+    $u_x$ & Canonical unitary $\lambda(x)$ of $L(G)$, identified with $\delta_x$. & \autoref{definition:group-algebra-fourier-series} \\
+    $T_x$ & Fourier coefficients of $T \in L(G)$, where $T = \sum_{x \in G}T_xu_x$. & \autoref{definition:group-algebra-fourier-series} \\
+    ICC & Infinite conjugacy classes: every non-trivial conjugacy class is infinite. & \autoref{definition:icc} \\
+    $H^{(I)}$ & Defined as $\bigoplus_{i \in I}H$. & \autoref{definition:wreath-product} \\
+    $H \wr_I \Gamma$ & Generalised wreath product $H^{(I)} \rtimes \Gamma$. & \autoref{definition:wreath-product} \\
+    $H \wr \Gamma$ & Wreath product, the case $I = \Gamma$ with $\Gamma$ acting by left translation. & \autoref{definition:wreath-product} \\
+    $\lim_{\longrightarrow}(M_i, \tau_i)$ & Inductive limit of an upward-directed system of tracial von Neumann algebras. & \autoref{definition:inductive-limit-tracial-von-neumann-algebra} \\
+    $M_2(\complex)^{\ol \otimes \infty}$ & The hyperfinite $\vnII_1$ factor, defined as $\lim_{\longrightarrow}(M_{2^n}(\complex), \tau_n)$. & \autoref{definition:hyperfinite-ii1} \\
+ 
+    $Mq$ & Induced von Neumann algebra of $M$ on $qH$ with respect to $q \in \text{Proj}(M')$. & \autoref{definition:induced-vna-commutant} \\
+    $M^t$ & Amplification of a $\vnII_1$ factor $M$, defined as $pM_n(M)p$ with $(\text{Tr}_n \otimes \tau)p = t$. & \autoref{definition:ii1-amplification} \\
+    $\fF(M)$ & Fundamental group of a $\vnII_1$ factor $M$, defined as $\bracsn{t > 0|M^t \iso M}$. & \autoref{definition:ii1-fundamental-group}
+
+\end{tabular}

src/vna/vn/commutative.tex

@@ -8,12 +8,12 @@
 
 \begin{proposition}
 \label{proposition:maximal-commutative-vn}
-    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal commutative von Neumann algebra if and only if $A = A'$.
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal abelian von Neumann algebra if and only if $A = A'$.
 \end{proposition}
 \begin{proof}
-    ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$. 
+    ($\Leftarrow$): Let $B \supset A$ be a abelian von Neumann algebra, then $B \subset A' = A$. 
 
-    ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.  
+    ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is abelian. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.  
 \end{proof}
 
 

src/vna/vn/fc.tex

@@ -41,6 +41,10 @@
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
     \begin{enumerate}
         \item For each normal operator $T \in A$ and $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$, $f(T) \in A$. 
+        \item For each $T \in A$ with $0 \le T \le 1$, there exists projections $\seq{P_n} \subset A$ such that
+        \[
+        T = \sum_{n = 1}^\infty \frac{P_n}{2^n}
+        \]
         \item The linear span of projections in $A$ is norm dense in $A$. 
     \end{enumerate}
 \end{theorem}
@@ -52,6 +56,20 @@
     If $B$ is arbitrary, then $\one_{B \setminus B_\complex(0, r)} \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as $r \downto 0$. As $\one_{B \setminus B_{\complex}(0, r)}(T) \in A$ for all $r > 0$, $\one_B(T) \in A$ as well. 
 
     By linearity, $g(T) \in A$ for all $g \in \Sigma(\sigma_{B(H)}(T); \complex)$ with $g(0) = 0$. By \autoref{lemma:separable-metric-space-approx-identity}, $\bracsn{g \in \Sigma(\sigma_{B(H)}(T); \complex)|g(0) = 0}$ is uniformly dense in $\bracsn{f \in B^\infty(\sigma_{B(H)(T)}; \complex)|f(0) = 0}$. Therefore $f(T) \in A$ for all $f \in B^\infty(\sigma_{B(H)}(T); \complex)$ with $f(0) = 0$. 
+
+    
+
+    (2): Let $f = \text{Id}: [0, 1] \to [0, 1]$, then
+    \[
+    f = \sum_{n = 1}^\infty \frac{\one_{\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2}}}{2^n}
+    \]
+
+    Let $E: \cb_{\sigma_{A}(T)} \to \text{Proj}(A)$ be the resolution of the identity for $f$, then by the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus}, 
+    \begin{align*}
+    T &= f(T) = \sum_{n = 1}^\infty\frac{\one_{\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2}}(T)}{2^n} \\
+    &= \sum_{n = 1}^\infty 2^{-n}E(\bracsn{x \in [0, 1]| \lceil 2^nx \rceil = 0 \mod 2})
+    \end{align*}
+    
 \end{proof}
 
 \begin{theorem}
@@ -77,4 +95,21 @@
     (2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^\infty$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by \autoref{corollary:spectrum-characterisation-iff}.
 \end{proof}
 
+\begin{proposition}
+\label{proposition:vna-two-sided-ideal}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $I \subset M$ be a two-sided ideal, then:
+    \begin{enumerate}
+        \item $I$ is self-adjoint. 
+        \item Let $x \in I$ with $x > 0$, $E: \cb_{[0, \infty)} \to \text{Proj}(M)$ be its resolution of the identity, and $t > 0$, then $E([t, \infty)) \in I$. 
+        \item If $I$ is weak-operator closed, then there exists a unique projection $z \in Z(M)$ with $I = Mz$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.4.15]{AnantharamanPopa}}}. ]
+    (1): Let $x = u|x|$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $x$, then $|x| = u^*x \in I$, and $x^* = |x|u^* \in I$. 
+
+    (2): Let $f: [0, \infty) \to \real$ be defined by $f(s) = s^{-1}\one_{\bracsn{s \ge t}}(s)$, then by the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $E([t, \infty)) = xf(x) \in I$. 
+
+    (3): By (5) of \autoref{theorem:existence-of-projections-vna}. 
+\end{proof}
+
 

src/vna/vn/index.tex

@@ -1,11 +1,13 @@
-\chapter{Von Neumann Algebras}
+\chapter{Fundamentals of von Neumann Algebras}
 \label{chap:von-neumann-algebras}
 
 \input{./topologies.tex}
 \input{./cayley.tex}
 \input{./vn.tex}
+\input{./weak.tex}
 \input{./commutative.tex}
 \input{./spec.tex}
 \input{./fc.tex}
+\input{./unlimited.tex}
 \input{./projection.tex}
 \input{./type-decomp.tex}

src/vna/vn/spec.tex

@@ -24,10 +24,14 @@
     \begin{enumerate}
         \item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. 
         \item For each $x \in H$, $E_{x, x}$ is positive.
+        \item For each $x, y \in H$ and $B \in \cb_X$, 
+        \[
+        |E_{x, y}(B)| \le E_{x, x}(B)^{1/2} E_{y, y}(B)^{1/2} \le \frac{1}{2}(E_{x, x}(B) + E_{y, y}(B))
+        \]
     \end{enumerate}
 
     Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
-    \begin{enumerate}[start=2]
+    \begin{enumerate}[start=3]
         \item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. 
         \item Let 
         \[
@@ -50,7 +54,17 @@
 
     (2): For each $B \in \cb_X$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H} \ge 0$.
 
-    (3): For each $x, y \in H$ and $B, C \in \cb_X$, 
+    (3): For each $B \in \cb_X$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz},
+    \begin{align*}
+        |E_{x, y}(B)| &= |\dpn{E(B)x, y}{H}| = |\dpn{E(B)x, E(B)y}{H}| \\
+        &\le \dpn{E(B)x, E(B)x}{H}^{1/2}\dpn{E(B)y, E(B)y}{H}^{1/2} \\
+        &= E_{x, x}(B)^{1/2} E_{y, y}(B)^{1/2}
+    \end{align*}
+
+    By \hyperref[Young's Inequality]{lemma:young-inequality}, $E_{x, x}(B)^{1/2} E_{y, y}(B)^{1/2} \le \frac{1}{2}(E_{x, x}(B) + E_{y, y}(B))$. 
+    
+
+    (4): For each $x, y \in H$ and $B, C \in \cb_X$, 
     \[
     \int_C \one_B dE_{x, y} = \dpn{E(C \cap B)x, y}{H} = \dpn{E(C)E(B)x, y}{H} = E_{E(B)x, y}(C)
     \]
@@ -59,7 +73,7 @@
     
     Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$. 
 
-    (4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
+    (5): By (4), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
     \begin{enumerate}[label=(\roman*)]
         \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. 
         \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
@@ -300,7 +314,7 @@
 \begin{proof}
     Let $E$ be the spectral measure on $\Omega(A)$ associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, $A$ is *-isomorphic to $\mathscr{E}^*$.
 
-    (1): By (3) of \autoref{lemma:spectral-measure-properties} and \autoref{theorem:hilbert-measures-dual}, $A$ is *-isomorphic to $[l^\infty(I); L^\infty(\mu_i; \complex)]$, where $\seqi{\mu} \subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_i$ denote the $i$-th copy of $\Omega(A)$, then
+    (1): By (4) of \autoref{lemma:spectral-measure-properties} and \autoref{theorem:hilbert-measures-dual}, $A$ is *-isomorphic to $[l^\infty(I); L^\infty(\mu_i; \complex)]$, where $\seqi{\mu} \subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_i$ denote the $i$-th copy of $\Omega(A)$, then
     \[
     \mu: \cb_\Omega \to [0, \infty] \quad B \mapsto \sum_{i \in I}\mu_i(B \cap \Omega_i)
     \]

src/vna/vn/type-decomp.tex

@@ -28,6 +28,12 @@
     (2) $\Rightarrow$ (1): Given that there exist no projections strictly between $0$ and $P$, the only non-zero projection in $PAP$ is $P$ itself. By \autoref{theorem:vn-projection-norm-dense}, the linear span of projections in $PAP$ is norm-dense in $PAP$. Therefore $PAP = \complex P$. 
 \end{proof}
 
+\begin{definition}[Diffuse]
+\label{definition:diffuse-vna}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, then $M$ is \textbf{diffuse} if it admits no minimal projections. 
+\end{definition}
+
+
 \begin{lemma}
 \label{lemma:projection-types-gymnastics}
     Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. 

src/vna/vn/unlimited.tex

@@ -0,0 +1,347 @@
+\section{Unbounded Functional Calculus}
+\label{section:unbounded-functional-calculus}
+
+
+
+\begin{definition}[Closed]
+\label{definition:unbounded-closed}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a linear operator, then $T$ is \textbf{closed} if its graph
+    \[
+    \Gamma(T) = \bracsn{(x, y) \in D(T) \times H| Tx = y}
+    \]
+
+    is a closed subspace of $H \times H$. 
+
+    The operator $T$ is \textbf{closable} if it admits a closed extension $S \supset T$. 
+\end{definition}
+
+
+\begin{definition}[Adjoint]
+\label{definition:unbounded-adjoint}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, 
+    \[
+    D(T^*) = \bracsn{\xi \in H| (\eta \mapsto \dpn{\xi, T\eta}{H}) \text{ is bounded on }D(T)}
+    \]
+
+    and\footnote{$\eta$ sampled from $D(T)$.}
+    \[
+    T^*: D(T^*) \to H \quad \dpn{T^*\xi, \eta}{H} = \dpn{\xi, T\eta}{H}
+    \]
+
+    then $T^*$ is a closed operator, known as the \textbf{adjoint} of $T$. 
+\end{definition}
+\begin{proof}
+    Let $\seq{\xi_n} \subset D(T^*)$ and $\xi, \eta \in H$ such that $\xi_n \to \xi$ and $T^*\xi_n \to \eta$, then for any $\theta \in D(T)$, 
+    \[
+    \dpn{\eta, \theta}{H} = \limv{n}\dpn{T^*\xi_n, \theta}{H} = \limv{n}\dpn{\xi_n, T\theta}{H} = \dpn{\xi, T\theta}{H}
+    \]
+
+    As $D(T)$ is dense, $T \circ \xi \in D(T)^*$, and $\eta = T^*\xi$. Thus the graph of $T^*$ is closed.
+\end{proof}
+
+\begin{definition}[Self-Adjoint]
+\label{definition:unbounded-self-adjoint}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, then $T$ is \textbf{self-adjoint} if $D(T) = D(T^*)$\footnote{Written out to emphasise.} and $T = T^*$. 
+\end{definition}
+
+
+
+\begin{lemma}
+\label{lemma:cayley-transform-unbounded}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined, self-adjoint operator, then:
+    \begin{enumerate}
+        \item For any $\xi \in D(T)$,
+        \[
+        \norm{(T \pm i)\xi}_H^2 = \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \norm{\xi}_H^2
+        \]
+        
+        In particular, $T \pm i$ is injective. 
+        \item $(T \pm i)(H) = H$.
+    \end{enumerate}
+
+    In other words, $(T \pm i)$ is bijective onto $H$. 
+\end{lemma}
+\begin{proof}
+    (1): For any $\xi \in D(T)$, 
+    \begin{align*}
+        \normn{(T + i)\xi}_H^2 &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 + \dpn{T\xi, i\xi}{H} + \dpn{i\xi, T\xi}{H} \\
+        &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 + i\dpn{T\xi, \xi}{H} -i \dpn{\xi, T\xi}{H} \\
+        &= \norm{T\xi}_H^2 + \norm{\xi}_H^2 \ge \normn{\xi}_H^2 
+    \end{align*}
+
+    (2, closed range): For any $\seq{\xi_n} \subset D(T)$ and $\eta \in H$ with $(T + i)\xi_n \to \eta$, (1) implies that $\seq{\xi_n} \subset D(T)$ and $\seq{T\xi_n}$ are both Cauchy. Let $\xi = \limv{n}\xi_n$, then since $T$ is self-adjoint, it is closed, so $T\xi = \limv{n}T\xi_n \in T(H)$, and $\eta = (T + i)\xi \in (T+i)(H)$. 
+
+    (2, dense range): Let $\eta \in H$, then the following are equivalent:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\eta \perp (T + i)(H)$. 
+        \item $\dpn{T\xi, \eta}{H} = \dpn{\xi, i\eta}{H}$ for all $\xi \in D(T)$. 
+        \item $\eta \in D(T^*)$ with $T^*\eta = i\eta$. 
+        \item $\eta \in D(T)$ with $(T - i)\eta = 0$. 
+    \end{enumerate}
+    
+    By (1), $\eta \perp (T+i)(H)$ if and only if $\eta = 0$. 
+\end{proof}
+
+\begin{definition}[Cayley Transform]
+\label{definition:unbounded-cayley-transform}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined, self-adjoint operator, then
+    \[
+    U(T): H \to H \quad U(T) = (T - i)(T + i)^{-1}
+    \]
+
+    is a unitary operator, known as the \textbf{Cayley transform} of $T$.
+\end{definition}
+\begin{proof}
+    By \autoref{lemma:cayley-transform-unbounded}, $(T+i)^{-1}: H \to D(T)$ and $(T - i): D(T) \to H$ are well-defined linear bijections, so $U(T)$ is a well-defined linear isomorphism. For any $\xi \in D(T)$, (1) of \autoref{lemma:cayley-transform-unbounded} implies that $\norm{(T + i)\xi}_H = \norm{(T - i)\xi}_H$. As $(T+i)^{-1}$ and $(T - i)$ are both bijections, $U(T)$ is an isometry. Therefore $U(T)$ is unitary. 
+\end{proof}
+
+\begin{definition}[Unbounded Integration Against Spectral Measure]
+\label{definition:unbounded-spectral-integral}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$. For each Borel measurable function $f: X \to \complex$, let
+    \[
+    D\paren{\int_X f dE} = \bracs{\xi \in H \bigg | \int_X |f|^2 dE_{\xi, \xi} < \infty}
+    \]
+
+    and
+    \[
+    \int_X f dE: D\paren{\int_X f dE} \to H \quad \angles{\int_X fdE \xi, \eta}_H = \int_X f dE_{\xi, \eta}
+    \]
+
+    Then, 
+    \begin{enumerate}
+        \item $\int_X f dE: D(\int_Xf dE) \to H$ is a densely defined linear operator.
+        \item For any Borel measurable functions $f, g: X \to \complex$, 
+        \[
+        D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE}
+        \]
+        
+        and $\int_X f dE \int_X g dE = \int_X fg dE$. 
+        
+        \item For any Borel measurable function $f: X \to \complex$, $(\int_X f dE)^* = \int_X \ol f dE$. In particular, $\int_X f dE$ is always closed. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For each $n \in \natp$, $\one_{\bracsn{f \le n}}f \in B^\infty(X; \complex)$. Thus for each $\xi \in H$, by a monotone class argument, 
+    \begin{align*}
+        \int_X |f|^2 dE_{E(\bracsn{f \le n})\xi, E(\bracsn{f \le n})\xi} &= \int_{\bracsn{f \le n}}|f|^2 dE_{\xi, \xi} < \infty \\
+        \normn{\xi - E(\bracsn{f \le n})\xi}_H^2 &= \normn{E(\bracsn{f > n})\xi}_H^2
+    \end{align*}
+
+    where by \hyperref[continuity from above]{proposition:measure-properties}, $E(\bracsn{f > n})\xi \to 0$ as $n \to \infty$. Therefore $D(\int_X f dE)$ is dense in $H$. 
+
+    It remains to show that $\int_X f dE$ is defined on the claimed domain. Let $\xi \in D(\int_X f dE)$, $\eta \in H$, and $B \in \cb_X$, then by (1) and (3) of \autoref{lemma:spectral-measure-properties}, 
+    \[
+    \int_X \one_B d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2}
+    \]
+
+    For any $\phi \in \Sigma(X; \complex)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \begin{align*}
+    \int_X |\phi| d|E_{\xi, \eta}| &\le \norm{\eta}_H\sum_{\alpha \in \phi(X)} |\alpha| E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2}E_{\eta, \eta}(\bracsn{\phi = \alpha})^{1/2} \\
+    &\le \norm{\eta}_H\braks{\int_X |\phi|^2 dE_{\xi, \xi}}^{1/2}
+    \end{align*}
+    
+
+    By the \hyperref[Monotone Convergence Theorem]{theorem:mct}, 
+    \[
+    \int_X |f| d|E_{\xi, \eta}| \le \norm{\eta}_H \braks{\int_X |f|^2 dE_{\xi, \xi}}^{1/2}
+    \]
+    
+    so $\int_X f dE \xi \in H$ is well-defined. 
+
+    (2): For each $\xi \in D(\int_X g dE)$, by a monotone class argument, 
+    \[
+    \int_X |fg|^2 dE_{\xi, \xi} = \int_X |f|^2 |g|^2 dE_{\xi, \xi} = \int_X |f|^2 dE_{\int_X gdE\xi, \int_X gdE\xi}
+    \]
+
+    so $\int_X g dE \xi \in D(\int_X f dE)$ if and only if $\xi \in D(\int_X fg dE)$, and
+    \[
+    D\paren{\int_X f dE \int_X g dE} = D\paren{\int_X g dE} \cap D\paren{\int_X fg dE}
+    \]
+    
+    
+    Let $\xi \in D(\int_X g dE) \cap D(\int_X fg dE)$ and $\eta \in H$, then
+    \begin{align*}
+        \angles{\int_X f dE \int_X g dE \xi, \eta}_H &= \int_X f dE_{\int_X g dE \xi, \eta} = \int_X fg dE_{\xi, \eta} \\
+        &= \angles{\int_X f  g dE \xi, \eta}_H
+    \end{align*}
+    
+    (3): For any $\xi, \eta \in D(\int_X f dE) = D(\int_X \ol f dE)$, 
+    \begin{align*}
+        \angles{\int_X f dE\xi, \eta}_H &= \int_X f dE_{\xi, \eta} = \ol{\int_X \ol f dE_{\eta, \xi}}  \\
+        &= \ol{\dpn{\int_X \ol f dE\eta, \xi}H} = \dpn{\xi, \int_X \ol f dE \eta}{H}
+    \end{align*}
+    
+    so $\int_X \ol f dE \subset (\int_X f dE)^*$. On the other hand, for each $\xi \in H$ and $\eta \in D[(\int_X f dE)^*]$, by the \hyperref[bounded $L^\infty$ functional calculus]{definition:linfty-functional-calculus} and (2),
+    \begin{align*}
+        \angles{\xi, \int_X \frac{\ol f}{1 + |f|} dE\eta}_H &= \angles{\int_X \frac{f}{1 + |f|} dE\xi, \eta}_H \\
+        &= \angles{\int_X f dE \int_X (1 + |f|)^{-1}dE\xi, \eta}_H \\
+        &=  \angles{ \int_X (1 + |f|)^{-1}dE\xi, \paren{\int_X f dE}^{*}\eta}_H \\
+        &=  \angles{\xi, \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}\eta}_H \\
+    \end{align*}
+
+    so
+    \[
+    \int_X \frac{\ol f}{1 + |f|} dE = \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}
+    \]
+    
+    By (2) and multiplication of $\int_X (1 + |f|) dE$ from the left, $\int_X \ol f dE = \paren{\int_X f dE}^{*}$.     
+\end{proof}
+
+
+\begin{theorem}[Spectral Theorem I (Unbounded)]
+\label{theorem:spectral-theorem-1-unbounded}
+    Let $H$ be a complex Hilbert space and $T: D(T) \to H$ be a densely defined, self-adjoint operator, then there exists a unique spectral measure $E: \cb_\real \to B(H)$ such that $T = \int_{\real} \lambda E(d\lambda)$. 
+\end{theorem}
+\begin{proof}
+    Let $U(T) = (T - i)(T + i)^{-1} \in U(B(H))$ be the Cayley transform of $T$, then $U(T)$ is unitary. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, there exists a unique spectral measure $E: \cb_{\partial B_\complex(0, 1)} \to B(H)$ such that $U(T) = \int_{\partial B_\complex(0, 1)} \lambda E(d\lambda)$. Moreover, 
+    \[
+    (I - U) = (T+i)(T+i)^{-1} - (T - i)(T+i)^{-1} = 2i(T + i)^{-1}
+    \]
+
+    is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let
+    \[
+    \phi: \partial B_\complex(0, 1) \setminus \bracsn{1} \to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda}
+    \]
+
+    then the pushforward $F = \phi_* E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with
+    \begin{align*}
+    \int_\real \lambda F(d\lambda) &= \int_{\partial B_\complex(0, 1) \setminus \bracsn{1}} i\frac{1 + \lambda}{1 - \lambda} E(d\lambda) = i(1 + U)(1 - U)^{-1} \\
+    &= i[(T + i)(T+i)^{-1} + (T-i)(T+i)^{-1}] \cdot (2i(T + i)^{-1})^{-1} \\
+    &= \frac{1}{2}[(T + i)(T+i)^{-1}(T+i) + (T-i)(T+i)^{-1}(T+i)] = T
+    \end{align*}
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:closed-unbonuded-adjoint}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a densely defined linear operator, and
+    \[
+    \Gamma(T) = \bracsn{(\xi, T\xi)|\xi \in D(T)} \subset H \oplus H
+    \]
+
+    be the graph of $T$, then:
+    \begin{enumerate}
+        \item $\Gamma(T)^\perp = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$. 
+        \item If $T$ is closed, then the mapping $I + T^*T: D(T^*T) \to H$ is surjective. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $\Gamma(T) = \bracsn{(\xi, T\xi)| \xi \in D(T)}$ be the graph of $T$, then for any $(\xi, \eta) \in H \oplus H$, the following are equivalent:
+    \begin{enumerate}[label=(\roman*)]
+        \item $(\xi, \eta) \perp \Gamma(T)$. 
+        \item For each $\theta \in D(T)$, $\dpn{\eta, T\theta}{H}  = -\dpn{\xi, \theta}{H}$.
+        \item $\eta \in D(T^*)$, with $\dpn{T^*\eta, \theta}{H} = -\dpn{\xi, \theta}{H}$ for all $\theta \in H$. 
+        \item $\eta \in D(T^*)$ with $\xi = -T^*\eta$. 
+    \end{enumerate}
+
+    so $\Gamma(T)^\perp = \bracsn{(-T^*\eta, \eta)| \eta \in D(T^*)}$.
+
+    (2): Given that $T$ is closed, $H \oplus H = \Gamma(T) \oplus \Gamma(T)^\perp$. In particular, for each $\theta \in H$, there exists $\xi \in D(T)$ and $\eta \in D(T^*)$ such that
+    \[
+    (\theta, 0) = (\xi, T\xi) + (-T^*\eta, \eta)
+    \]
+
+    In which case, $T\xi = -\eta \in D(T^*)$, so $\xi \in D(T^*T)$ with
+    \[
+    \theta = \xi - T^*\eta = \xi + T^*T\xi
+    \]
+
+    so $I + T^*T: D(T^*T) \to H$ is surjective. 
+\end{proof}
+
+\begin{theorem}[von Neumann]
+\label{theorem:von-neumann-closed-operator}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then $T^*T: D(T^*T) \to H$ is a densely defined, self-adjoint, and positive operator. 
+\end{theorem}
+\begin{proof}
+    For each $\xi, \eta \in D(T^*T)$, by definition of the adjoint,
+    \[
+    \dpn{T^*T\xi, \eta}{H} = \dpn{T\xi, T\eta}{H} = \dpn{\xi, T^*T\eta}{H}
+    \] 
+
+    (Densely Defined): Let $\eta \in D(T^*T)^\perp$. By (2) of \autoref{lemma:closed-unbonuded-adjoint}, there exists $\xi \in D(T^*T)$ such that $\eta = \xi + T^*T\xi$. In which case, 
+    \[
+    0 = \dpn{\xi, \eta}{H} = \dpn{\xi, \xi}{H} + \dpn{\xi, T^*T\xi}{H} = \norm{\xi}_H^2 + \norm{T\xi}_H^2
+    \]
+
+    so $T\xi = \xi = 0$, and $\eta = 0$ as well. 
+
+    ($T^*T \subset (T^*T)^*$): For each $\xi, \eta \in D(T^*T)$, $\dpn{T^*T\xi, \eta}{H} = \dpn{\xi, T^*T\eta}{H}$, so $T^*T \subset (T^*T)^*$. 
+
+    ($T^*T \supset (T^*T)^*$): Let $\xi \in D((T^*T)^*)$, then by (2) of \autoref{lemma:closed-unbonuded-adjoint}, there exists $\eta \in D(T^*T)$ such that $\xi + (T^*T)^*\xi = \eta + T^*T\eta$. In which case, for each $\theta \in D(T^*T)$, 
+    \begin{align*}
+        \dpn{(I+T^*T)^*\xi, \theta}{H} &= \dpn{(I+T^*T)\eta, \theta}{H} \\
+        \dpn{\xi, (I + T^*T)\theta}{H} &= \dpn{\eta, (I+T^*T)^*\theta}{H}
+    \end{align*}
+    
+    As the above holds for all $\theta \in D(T^*T)$, (2) of \autoref{lemma:closed-unbonuded-adjoint} implies that $\dpn{\xi, \theta}{H} = \dpn{\eta, \theta}{H}$ for all $\theta \in H$. Therefore $\xi = \eta \in D(T^*T)$, and $T^*T \supset (T^*T)^*$. 
+\end{proof}
+
+\begin{definition}[Absolute Value]
+\label{definition:unbounded-absolute-value}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, and $|T| = (T^*T)^{1/2}: D(|T|) \to H$, then:
+    \begin{enumerate}
+        \item $|T|$ is a densely defined, self-adjoint, and positive operator. 
+        \item Equip $D(T)$ with the inner product $\dpn{\xi, \eta}{D(T)} = \dpn{\xi, \eta}{H} + \dpn{T\xi, T\eta}{H}$, then $D(T)$ is a Hilbert space with $D(T^*T)$ being dense in $D(T)$. 
+        \item Equip $D(|T|)$ with the inner product $\dpn{\xi, \eta}{D(|T|)} = \dpn{\xi, \eta}{H} + \dpn{|T|\xi, |T|\eta}{H}$, then $D(|T|)$ is a Hilbert space with $D(T^*T)$ being dense in $D(|T|)$. 
+        \item $D(T) = D(|T|)$. 
+        \item For each $\xi \in D(T) = D(|T|)$, $\normn{T\xi}_H = \norm{|T|\xi}_H$. In particular, $\ker T = \ker |T|$. 
+    \end{enumerate}
+
+    The operator $|T|$ is the \textbf{absolute value} of $T$. 
+\end{definition}
+\begin{proof}
+    (1): By \hyperref[von Neumann's theorem]{theorem:von-neumann-closed-operator}, $T^*T: D(T^*T) \to H$ is a densely defined, self-adjoint, and positive operator. Thus $|T| = (T^*T)^{1/2}$ is well-defined, self-adjoint, and positive via the \hyperref[spectral theorem]{theorem:spectral-theorem-1-unbounded}. 
+
+    (2): By definition, $D(T)$ is isomorphic to the graph of $T$, which is a Hilbert space given that $T$ is closed. 
+
+    Let $\xi \in D(T)$ with $\xi \perp D(T^*T)$, then for each $\eta \in D(T^*T)$, 
+    \[
+    0 = \dpn{\xi, \eta}{H} + \dpn{T\xi, T\eta}{H} = \dpn{\xi, (I + T^*T)\eta}{H}
+    \]
+
+    By (2) of \autoref{lemma:closed-unbonuded-adjoint}, $(I+T^*T): D(T^*T) \to H$ is surjective, so $\xi = 0$, and $D(T^*T)$ is dense in $D(T)$. 
+
+    (3): By (3) of \autoref{definition:unbounded-spectral-integral}, $|T|$ is self-adjoint, and in particular closed.  Since $D(T)$ is isomorphic to the graph of $|T|$, $D(|T|)$ is a Hilbert space. Similarly, for any $\xi \in D(|T|)$ with $\xi \perp D(T^*T)$ and $\eta \in D(T^*T)$, by the \hyperref[Borel functional calculus]{definition:unbounded-spectral-integral}\footnote{The fact that the spectral measure of a positive element is supported in $[0, \infty)$ is needed here.},
+    \[
+    0 = \dpn{\xi, \eta}{H} + \dpn{|T|\xi, |T|\eta}{H} = \dpn{\xi, (I + T^*T)\eta}{H}
+    \]
+
+    By (2) of \autoref{lemma:closed-unbonuded-adjoint}, $\xi = 0$, and $D(T^*T)$ is dense in $D(|T|)$. 
+
+    (4): For any $\xi \in D(T^*T)$, $\norm{\xi}_{D(T)} = \norm{\xi}_{D(|T|)}$. By (2) and (3), $D(T)$ and $D(|T|)$ are both the completions of $D(T^*T)$ under the same norm, so, $D(T) = D(|T|)$. 
+
+    (5): For each $\xi \in D(T^*T)$, 
+    \[
+    \normn{T\xi}_H^2 = \dpn{T\xi, T\xi}{H} = \dpn{\xi, T^*T\xi}{H} = \dpn{|T|\xi, |T|\xi}{H} = \norm{|T|\xi}_H^2
+    \]
+
+    By the density provided in (2) and (3), $\normn{T\xi}_H^2 = \norm{|T|\xi}_H^2$ for all $\xi \in D(T)$. 
+\end{proof}
+
+
+\begin{theorem}[Polar Decomposition (Unbounded)]
+\label{theorem:unbounded-polar-decomposition}
+    Let $H$ be a complex Hilbert space, $T: D(T) \to H$ be a closed, densely defined linear operator, then there exists a unique partial isometry $U \in B(H)$ with $\ker U = \ker T$ such that $T = U|T|$, where $|T|$ is the \hyperref[absolute value]{definition:unbounded-absolute-value} of $T$. 
+\end{theorem}
+\begin{proof}
+    By \autoref{definition:unbounded-absolute-value}, $|T|: D(T) \to H$ is a densely defined, self-adjoint, and positive operator with $\normn{T\xi}_H = \norm{|T|\xi}_H$ for all $\xi \in D(T)$ and $\ker(T) = \ker(|T|)$. Let
+    \[
+    U: |T|(D(T)) \to T(D(T)) \quad |T|\xi \mapsto T\xi
+    \]
+
+    then $U$ extends into a well-defined isometry $\ol{|T|(D(T))} \to \ol{T(D(T))}$. Further extend $U$ to $H$ by setting $U|_{|T|(D(T))^\perp} = 0$, then $U: H \to \ol{T(D(T))} \subset H$ is a partial isometry with $T = U|T|$. As $|T|$ is self-adjoint, 
+    \[
+    \ker(U) = |T|(D(T))^\perp = \ker(|T|) = \ker(T)
+    \]
+
+    Finally, let $V: H \to H$ be a partial isometry with $T = V|T|$ and $\ker(V) = \ker(T)$, then for each $\xi \in D(T)$, $V|T|\xi = T\xi = U|T|\xi$, so $U = V$ on $|T|(D(T))$. By continuity, $U = V$ on $\ol{|T|(D(T))}$. As
+    \[
+    \ker(V) = \ker(T) = \ker(U) = |T|(D(T))^\perp
+    \]
+
+    and $H = \ker(V) \oplus \ol{|T|(D(T))}$, $U = V$. 
+
+
+\end{proof}
+

src/vna/vn/vn.tex

@@ -5,7 +5,7 @@
 \label{theorem:existence-of-projections-vna}
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a strong-operator closed $C^*$-subalgebra, then:
     \begin{enumerate}
-        \item For any bounded directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$. 
+        \item For any bounded, upward-directed family $\cf \subset A_{sa}$, $\sup(\cf) = \sotlim_{T \in \cf}T \in A_{sa}$. 
         \item For any directed family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
     \end{enumerate}
 
@@ -100,9 +100,19 @@
     \begin{enumerate}
         \item $T(M) \subset M$. 
         \item $PTP = TP$. 
+        \item $T^*PTP = T^*TP$. 
     \end{enumerate}
 \end{lemma}
-% Proof omitted due to obviousness
+\begin{proof}
+    (3) $\Leftarrow$ (1): Let $\xi \in H$, then
+    \begin{align*}
+        \dpn{T^*TP\xi, P\xi}{H} &= \dpn{TP\xi, TP\xi}{H} = \norm{TP\xi}_H^2 \\
+        \dpn{T^*PTP\xi, P\xi}{H} &= \dpn{PTP\xi, TP\xi}{H} = \dpn{PTP\xi, PTP\xi}{H} = \norm{PTP\xi}_H^2
+    \end{align*}
+
+    As $\norm{TP\xi}_H = \norm{PTP\xi}_H$, $TP\xi \in H$, so $T(M) \subset M$. 
+    
+\end{proof}
 
 
 \begin{definition}[Reducing Subspace]

src/vna/vn/weak.tex

@@ -0,0 +1,206 @@
+\section{Normal Maps}
+\label{section:ultraweak}
+
+\begin{lemma}[Jordan Decomposition]
+\label{lemma:weak-operator-jordan}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a weak-operator continuous linear functional, then there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_0^3$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$. 
+\end{lemma}
+\begin{proof}
+    Assume without loss of generality that there exist $\xi, \eta \in H$ such that $\dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H}$ for all $x \in M$. In which case, by \hyperref[polarisation]{proposition:polarisation-complex}, 
+    \[
+    \dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x(\xi + i^k\eta), \xi + i^k\eta}{H}
+    \]
+    
+    for all $x \in M$. 
+\end{proof}
+
+
+\begin{lemma}[Radon-Nikodym]
+\label{lemma:vna-radon-nikodym}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $\omega: M \to \complex$ be a positive linear functional, and $\xi \in H$ such that $\omega(x) \le \dpn{x\xi, \xi}{H}$ for all $x \in M_+$, then there exists $x' \in M'_+$ such that $\omega(x) = \dpn{x(x'\xi), x'\xi}{H}$ for all $x \in M$. 
+\end{lemma}
+\begin{proof}
+    By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for each $x, y \in M$, 
+    \[
+    |\omega(y^*x)|^2 \le \omega(x^*x)\omega(y^*y) \le \norm{x\xi}_H^2 \norm{y\xi}_H^2
+    \]
+
+    so the mapping $(x, y) \mapsto \omega(y^*x)$ is a bounded sesquilinear form on $M\xi$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $z \in B(\ol{M\xi})$ such that $\omega(y^*x) = \dpn{zx\xi, y\xi}{H}$ for all $x, y \in M$. In particular, $0 \le \omega(x^*x) = \dpn{zx\xi, x\xi}{H}$ for all $x \in M$, so $z \ge 0$. Moreover, for any $x, y, t \in M$, 
+    \[
+        \dpn{ztx\xi, y\xi}{H} = \omega(y^*tx) = \omega((t^*y)^*x) = \dpn{zx\xi, t^*y\xi}{H} = \dpn{tzx\xi, y\xi}{H}
+    \]
+
+    so $z \in M'_+$. Finally, let $p: H \to \ol{M\xi}$ be the orthogonal projection onto $\ol{M\xi}$, and $x' = (zp)^{1/2} \in M'$, then for each $x \in M$, 
+    \[
+        \omega(x) = \dpn{zx\xi, \xi}{H} = \dpn{zpx\xi, \xi}{H} = \dpn{x(zp)^{1/2}\xi, (zp)^{1/2}\xi}{H}
+    \]
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:completely-additive-projection-choose}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\varphi: M \to \complex$ be a positive linear functional. If for any mutually orthogonal projections $\mathcal{P} \subset M$, $\varphi\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\varphi(p)$, then there exist a non-zero projection $p \in M$ and $\xi \in pH$ such that $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 2.5.6.]{AnantharamanPopa}}}. ]
+    Let $\eta \in H$ with $\varphi(1_M) < \dpn{\eta, \eta}{H}$. Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections with $\varphi(p) \ge \dpn{p\eta, \eta}{H}$ for all $p \in \mathcal{P}$. By assumption, 
+    \begin{align*}
+    \varphi\paren{\sum_{p \in \mathcal{P}}p} &\ge \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \\
+        \varphi\paren{1_M - \sum_{p \in \mathcal{P}}p}&= \varphi(1_M) - \varphi\paren{\sum_{p \in \mathcal{P}}p} \\
+        &< \dpn{\eta, \eta}{H} - \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H  \ge 0
+    \end{align*}
+
+    Let $p := 1_M - \sum_{q \in \mathcal{P}}q > 0$, then by maximality of $\mathcal{P}$, $\varphi(q) < \dpn{q\eta, \eta}{H}$ for all subprojections $q \le p$. 
+    
+    Let $x \in (pMp)_+$ with $0 \le x \le 1$. By \autoref{theorem:vn-projection-norm-dense}, there exist projections $\seq{q_n} \subset pMp$ such that $x = \sum_{n = 1}^\infty \frac{q_n}{2^n}$. In which case\footnote{Here, only continuity of $\varphi$ is used.}, 
+    \[
+    \dpn{x\eta, \eta}{H} = \sum_{n = 1}^\infty 2^{-n}\dpn{q_n\eta, \eta}{H} \ge \sum_{n = 1}^\infty \frac{\varphi(q_n)}{2^n} = \varphi(x)
+    \]
+
+    Therefore $\dpn{x\eta, \eta}{H} \ge \varphi(x)$ for all $x \in (pMp)_+$. By the \hyperref[Radon-Nikodym theorem for positive functionals]{lemma:vna-radon-nikodym}, there exists $\xi \in pH$ with $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$. 
+\end{proof}
+
+
+
+\begin{definition}[Normal Linear Functional]
+\label{definition:vna-normal-functional}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a positive linear functional, then the following are equivalent:
+    \begin{enumerate}
+        \item For any bounded, upward-directed family $A \subset M$, $\omega(\sup(A)) = \sup(\omega(A))$.
+        \item For any mutually orthogonal projections $\mathcal{P} \subset M$, $\omega\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in \mathcal{P}}\omega(p)$. 
+
+        \item $\omega$ is the norm limit of a sequence of weak-operator continuous positive linear functionals. 
+        \item $\omega$ is weak-operator continuous on bounded subsets of $M$.
+        \item $\omega$ is strong-operator continuous on bounded subsets of $M$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 2.5.5]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (3): Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections such that for each $p \in \mathcal{P}$, there exists $\xi_p \in pH$ such that $\omega(x) = \dpn{x\xi_p, \xi_p}{H}$ for all $x \in pMp$. Let $q = \sum_{p \in \mathcal{P}}p$, then by maximality of $\mathcal{P}$ and \autoref{lemma:completely-additive-projection-choose} applied to $(1 - q)M(1 - q)$, $\sum_{p \in \mathcal{P}}p = 1_M$. 
+
+    For each $x \in M$ and $p \in \mathcal{P}$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \[
+    |\omega(xp)|^2 \le \omega(1) \cdot \omega(px^*xp) = \omega(1) \cdot \norm{x\xi_p}_H^2
+    \]
+
+    so the mapping $y \xi_p \mapsto \omega(yp)$ extends into a bounded linear functional on $\ol{M\xi_p}$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\eta_p \in \ol{M\xi_p}$ such that $\omega(xp) = \dpn{x\xi_p, \eta_p}{H}$ for all $x \in M$. 
+
+    Let $\mathcal{F} \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$, directed by inclusion. For each $F \in \mathcal{F}$, let $q_F = \sum_{p \in F}p$ and $\omega_F \in M^*$ be defined by $\omega_F(x) = \omega(q_Fxq_F)$, then for each $x \in M$, 
+    \[
+    \omega_F(x) = \sum_{p \in F}\omega(q_Fxp) = \sum_{p \in F}\dpn{q_Fx\xi_p, \eta_p}{H}
+    \]
+
+    so $\omega_F$ is weak-operator continuous. Finally, for any $x \in M$, 
+    \begin{align*}
+        |\omega(x) - \omega_F(x)| &\le |\omega(x) - \omega(xq_F)| + |\omega(xq_F) - \omega(q_Fxq_F)| \\
+        &= |\omega(x(1_M - q_F))| + |\omega((1 - q_F)xq_F)| \\
+        &\le 2\norm{x}_M\omega(1)^{1/2} \cdot \omega(1 - q_F)^{1/2}
+    \end{align*}
+    
+    By (2), $\lim_{F \in \mathcal{F}}\omega(1 - q_F)^{1/2} = 0$. Therefore $\omega = u\text{-}\lim_{F \in \mathcal{F}}\omega_F$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:normal-functional-bound}
+    Let $M$ be a von Neumann algebra, $\varphi, \omega \in M^*$ be positive linear functionals. If $\omega$ is normal and $\varphi \le \omega$, then $\varphi$ is also normal. 
+\end{lemma}
+\begin{proof}
+    By (2) of \autoref{definition:vna-normal-functional}. 
+\end{proof}
+
+\begin{definition}[Normal Positive Map]
+\label{definition:normal-linear-map}
+    Let $M, N$ be von Neumann algebras and $\phi: M \to N$ be a positive linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item For any bounded, upward-directed family $A \subset M$, $\phi(\sup(A)) = \sup(\phi(A))$. 
+        \item For any normal, positive linear functional $\omega \in N^*$, $\omega \circ \phi$ is a normal, positive linear functional.
+        \item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the weak operator topologies. 
+    \end{enumerate}
+
+    If $\phi$ is in addition a *-homomorphism, then the above are also equivalent to
+    \begin{enumerate}[start=3]
+        \item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the strong operator topologies. 
+    \end{enumerate}
+    
+    If the above holds, then $\phi$ is \textbf{normal}.
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 2.5.8]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (3): Let $\omega \in N^*$ be a weak-operator continuous functional. By \autoref{lemma:weak-operator-jordan}, there exist positive weak-operator continuous functionals $\bracsn{\omega_k}_0^3$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$. By assumption, $\omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. In which case, $\omega \circ \phi = \sum_{k = 0}^3 i^k \omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. Therefore $\phi$ is weak-operator continuous on bounded subsets of $M$.
+
+    (3) $\Rightarrow$ (1), (4) $\Rightarrow$ (1): By (1) of \autoref{theorem:existence-of-projections-vna}.
+
+    (3) $\Rightarrow$ (4)\footnote{Central here is the observation that $x_\alpha \to 0$ in the strong operator topology i.f.f. $x_\alpha^*x_\alpha \to 0$ in the weak operator topology. }: Suppose that $M$ acts on the Hilbert space $H$, and $N$ acts on the Hilbert space $K$. Let $\angles{x_\alpha}_{\alpha \in A} \subset M$ be a bounded net converging to $x \in M$ in the strong operator topology, then for each $\xi \in H$, 
+    \[
+    \lim_{\alpha \in A}\dpn{(x - x_\alpha)^*(x - x_\alpha)\xi, \xi}{H} = \lim_{\alpha \in A}\normn{(x - x_\alpha)\xi}_H^2 = 0
+    \]
+
+    By \hyperref[polarisation]{proposition:polarisation-complex}, $\wotlim_{\alpha \in A}(x - x_\alpha)^*(x - x_\alpha) = 0$. As $\angles{x_\alpha}_{\alpha \in A}$ is bounded, 
+    \[
+    \wotlim_{\alpha \in A}\phi((x - x_\alpha)^*(x - x_\alpha)) = 0
+    \]
+
+    as well. In particular, for any $\xi \in K$, 
+    \[
+    \lim_{\alpha \in A}\norm{\phi(x - x_\alpha)\xi}_K^2 = \lim_{\alpha \in A}\dpn{\phi(x - x_\alpha)^*\phi(x - x_\alpha)\xi, \xi}{K} = 0
+    \]
+
+    Therefore $\phi$ restricted to bounded subsets of $M$ is strong-operator continuous.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:vna-isomorphism-normal}
+    Let $M, N$ be von Neumann algebras, then every isomorphism $\phi: M \to N$ is normal. In particular, the weak and strong operator topologies on $M$, restricted to bounded subsets, are independent of its representation. 
+\end{proposition}
+\begin{proof}
+    As $\phi$ is an isomorphism, it is an order isomorphism. By (1) of \autoref{definition:normal-linear-map}, $\phi$ is normal. By (3) and (4) of \autoref{definition:normal-linear-map}, its restriction to bounded subsets of $M$ is both weak-operator and strong-operator continuous. 
+\end{proof} 
+
+\begin{proposition}
+\label{proposition:vna-normal-bound}
+    Let $M, N$ be von Neumann algebras, $\omega \in M^*$ and $\psi \in N^*$ be normal positive linear functionals with $\psi$ being faithful, and $\phi: M \to N$ be a positive linear map with $\psi \circ \phi \le \omega$, then $\phi$ is normal. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.5.11]{AnantharamanPopa}}}. ]
+    Let $\varphi = \psi \circ \phi \le \omega$, then $\varphi$ is normal by \autoref{lemma:normal-functional-bound}. For any bounded, upward-directed family $A \subset M$, 
+    \[
+         \psi(\sup(\phi(A))) =\sup(\psi(\phi(A))) = \sup(\varphi(A)) = \varphi(\sup(A)) = \psi(\phi(\sup(A)))
+    \]
+
+    As $\psi$ is faithful, $\sup(\phi(A)) = \phi(\sup(A))$. Therefore $\phi$ is normal by (1) of \autoref{definition:normal-linear-map}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:normal-image-vna}
+    Let $M$ be a von Neumann algebra, $H$ be a complex Hilbert space, and $\pi: M \to B(H)$ be a normal unital *-homomorphism, then $\pi(M)$ is a von Neumann algebra. 
+\end{proposition}
+\begin{proof}
+    Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_N(0, 1)}^{B(H)} = \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By \autoref{proposition:compact-extensions} and \autoref{proposition:bh-ultraweak-bounded}, $\ol{B_M(0, 1)}^M$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^M)$ is weak-operator compact and contains $\pi(B_M(0, 1))$. Thus
+    \[
+    \ol{B_{\pi(M)}(0, 1)}^{B(H)} \supset \pi(\ol{B_M(0, 1)}^M) \supset \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}} \supset \ol{B_N(0, 1)}^{B(H)}
+    \]
+
+    and $\ol{B_{\pi(M)}(0, 1)}^{B(H)} = \ol{B_{N}(0, 1)}^{B(H)}$, so $\pi(M) = \ol{\pi(M)}^{B(H)} = N$ by \autoref{theorem:cstar-quotient}.
+\end{proof}
+
+
+
+
+\begin{lemma}
+\label{lemma:tracial-von-neumann-gns}
+    Let $M$ be a von Neumann algebra, $\varphi \in S(M)$ be a state, and $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\varphi$, then $\pi$ is normal if and only if $\varphi$ is normal. In which case, $\pi(M) \subset B(H)$ is a von Neumann algebra. 
+\end{lemma}
+\begin{proof}
+    (1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that the weak operator topology on $\ol{B_{\pi(M)}(0, 1)}$ coincides with the topology of pointwise convergence on $\bracsn{\eta \otimes \theta|\eta, \theta \in \pi(M)\xi}$. For each $x, y, z \in M$, 
+    \[
+    \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H} = \dpn{\pi(z^*)\pi(x)\pi(y)\xi, \xi}{H} = \varphi(z^*xy)
+    \]
+
+    By \hyperref[separate continuity of composition]{proposition:bh-operator-topologies-facts} in the weak operator topology, the functional $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H}$ is weak-operator continuous. As such, $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)} \to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous. 
+    
+    
+    (2): By \autoref{proposition:normal-image-vna}.
+\end{proof}
+
+\begin{definition}[Separable]
+\label{definition:separable-vna}
+    Let $M$ be a von Neumann algebra, then $M$ is \textbf{separable} if there exists a separable Hilbert space $H$ and a faithful normal representation $\pi: M \to B(H)$. 
+\end{definition}
+
+