Housekeeping: harmonic.

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

@@ -319,4 +319,11 @@
   pages={482--489},
   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 +39,24 @@
         \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}
+
+
+
+

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/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

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

src/dyn/top/stone-cech.tex

@@ -90,7 +90,7 @@
     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 $px = x$. 
+        \item There exists $\fU \in \beta G \setminus G$ such that $\fU x = x$. 
     \end{enumerate}
 \end{proposition}
 % "obvious"
@@ -128,21 +128,114 @@
 
 \begin{definition}[Universal Minimal Flow]
 \label{definition:universal-minimal-flow}
-    Let $G$ be a group, then there exists a flow $(M, \alpha)$ such that:
+    Let $G$ be a group, then there exists a flow $L$ such that:
     \begin{enumerate}
-        \item $(M, \alpha)$ is a minimal $G$-flow. 
+        \item $G \curvearrowright L$ is a minimal $G$-flow. 
-        \item[(U)] For any minimal $G$-flow $(X, \beta)$, $x_0 \in X$, and $m_0 \in M$, there exists a unique continuous map $\Phi \in C(M; X)$ such that $\Phi(m_0) = x_0$ and the following diagram commutes
+        \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
         \[
         \xymatrix{
-        M \ar@{->}[r]^{\Phi} \ar@{->}[d]_{\alpha(g)} & X \ar@{->}[d]^{\beta(g)} \\
+        L \ar@{->}[r]^{\Phi} \ar@{->}[d]_{g} & X \ar@{->}[d]^{g} \\
-        M \ar@{->}[r]_{\Phi} & X
+        L \ar@{->}[r]_{\Phi} & X
         }
         \]
 
         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 $K \ne \emptyset$. 
+    (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. 
-    
-    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. 
     
     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. 
+    (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 $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
-    
-    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 $\sup_{y \in \ol{\conv}(C)}\dpn{y, \phi}{E} < \alpha$. 
 \end{proof}
 
 \begin{theorem}[Markov-Kakutani Fixed Point Theorem]

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/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/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

@@ -20,8 +20,9 @@
     $PS(A)$ & 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} \\ 
+    $L^2(M, \phi)$ & Hilbert space of the GNS triple associated with a faithful $\phi \in S(A)$. & \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} \\ 
+    $E_{x, y}$ & 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} \\ 

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

@@ -261,6 +261,9 @@ The following is a slight generalisation of one form of the above example, but s
     \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

@@ -550,3 +550,6 @@
     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

@@ -184,27 +184,6 @@
 \end{proof}
 
 
-\begin{proposition}[Multiplicative Domain of Stinespring Representations]
-\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{theorem}[Sz.-Nagy's Dilation Theorem]

src/op/system/index.tex

@@ -1,4 +1,4 @@
-\chapter{Operator Systems}
+\chapter{Operator Systems and Completely Positive Maps}
 \label{chap:opertaor-systems}
 
 \input{./os.tex}
@@ -9,3 +9,8 @@
 \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 +1,24 @@
+\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]{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 
+\end{definition}
+
+
+

src/op/system/module.tex

@@ -4,7 +4,7 @@
 
 \begin{definition}[Module Map]
 \label{definition:module-map}
-    Let $A, B, C$ be $C^*$-algebras with $C \subset A, B$ and $1_A = 1_B = 1_C$, then $A$ and $B$ are $C$-bimodules with respect to multiplication. 
+    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}
@@ -48,9 +48,44 @@
     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
+    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
         \[
@@ -69,7 +104,7 @@
 
     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, {{\cite[Theorem 3.18]{Paulsen}}}. ]
+\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}, 
@@ -104,6 +139,105 @@
 
     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}

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/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/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/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/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 $q \in M'$, 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 $p \in M$, 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 $y \in M'$, 
+    \[
+    \tilde x y = pxp \cdot y = pxp \cdot yp = yp \cdot pxp = y \tilde x
+    \]
+
+    so $\tilde x \in M'$, $x = p\tilde x p \in pMp$, and $pMp \subset (M'p)'$.
+\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 $(\text{Tr}_n \otimes \tau_s)(p) = t$, 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/trace.tex

@@ -0,0 +1,190 @@
+\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{definition}[Standard Representation]
+\label{definition:tracial-vna-standard}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then its \hyperref[GNS]{definition:gns-triple} representation on $L^2(M, \tau)$ is the \textbf{standard representation} of $(M, \tau)$.  
+\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 +1,13 @@
+\chapter{Notations}
+\label{chap:vna-notations}
+
+\begin{tabular}{lll}
+    \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
+    \hline
+    
+    $E_{x, y}$ & 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}
+\end{tabular}

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

@@ -1,9 +1,10 @@
-\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}

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/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}
 

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}
+
+