Fixed typo.

preamble.sty

@@ -232,3 +232,10 @@
 \newcommand{\sotlim}{\operatorname*{\text{\small SOT}\text{-}\!\lim}}
 \newcommand{\wotlim}{\operatorname*{\text{\small WOT}\text{-}\!\lim}}
 
+% VNA Types
+\newcommand{\vnI}{\mathrm{I}}
+\newcommand{\vnII}{\mathrm{II}}
+\newcommand{\vnIIo}{\mathrm{II}_{1}}
+\newcommand{\vnIIi}{\mathrm{II}_{\infty}}
+\newcommand{\vnIII}{\mathrm{III}}
+

refs.bib

@@ -268,4 +268,28 @@
   year      = {2002},
   isbn      = {978-1-85233-437-6},
   doi       = {10.1007/978-1-4471-3903-4}
+}
+
+@article{ArensBilinear,
+ ISSN = {00029939, 10886826},
+ URL = {http://www.jstor.org/stable/2031695},
+ author = {Richard Arens},
+ journal = {Proceedings of the American Mathematical Society},
+ number = {6},
+ pages = {839--848},
+ publisher = {American Mathematical Society},
+ title = {The Adjoint of a Bilinear Operation},
+ urldate = {2026-08-13},
+ volume = {2},
+ year = {1951}
+}
+
+
+@MISC {TownesType1,
+    title = {Classification of Type 1 factors},
+    author = {leslie townes},
+    howpublished = {Mathematics Stack Exchange},
+    note = {URL:https://math.stackexchange.com/q/150258 (version: 2012-05-27)},
+    eprint = {https://math.stackexchange.com/q/150258},
+    url = {https://math.stackexchange.com/q/150258}
 }

src/cat/cat/cat-func.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Category]
 \label{definition:category}
-    A \textbf{category} $\catc$ is a collection of objects $\obj{\catc}$, such that for any $A, B, C \in \obj{\catc}$, there exists sets $\mor{A, B}$, $\mor{B, C}$, and a composition law
+    A \textbf{category} $\catc$ is a collection of objects $\obj{\catc}$, such that for any $A, B, C \in \obj{\catc}$, there exist sets $\mor{A, B}$, $\mor{B, C}$, and a composition law
     \[
     \mor{A, B} \times \mor{B, C} \to \mor{A, C}
     \]

src/fa/lc/compact.tex

@@ -33,7 +33,6 @@
 \end{definition}
 
 
-
 \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 $K$. 

src/fa/lp/definition.tex

@@ -1,6 +1,12 @@
 \section{Basic Properties}
 \label{section:lp-basic}
 
+\begin{definition}[$B^\infty$ Spaces]
+\label{definition:bounded-borel-function}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a normed vector space, then the set $B^\infty(X; E)$ is the \textbf{space of bounded $E$-valued strongly measurable functions} on $X$, and the set $B^\infty(X)$ is the space of bounded complex-valued Borel measurable functions on $X$. 
+\end{definition}
+
+
 \begin{definition}[$\mathcal{L}^p$ Spaces]
 \label{definition:lp-unequivalence}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $f: X \to E$ be strongly measurable, and $p \in [1, \infty)$, then $f$ is \textbf{$p$-integrable} if
@@ -8,17 +14,19 @@
     \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
     \]
 
-    The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.
+    The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $E$-valued $p$-integrable functions on $X$.
 \end{definition}
 
 \begin{definition}[Essential Supremum]
 \label{definition:esssup}
     Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if
     \[
-    \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
+    \norm{f}_{\mathcal{L}^\infty(X; E)} = \norm{f}_{\mathcal{L}^\infty(\mu; E)} = \norm{f}_{\mathcal{L}^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
     \]
 
-    In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
+    In which case, $\norm{f}_{\mathcal{L}^\infty(X; E)}$ is the \textbf{essential supremum} of $f$. 
+    
+    The set $\mathcal{L}^\infty(X; E) = \mathcal{L}^\infty(\mu; E) = \mathcal{L}^\infty(X, \cm, \mu; E)$ is the space of all $E$-valued essentially bounded functions on $X$. 
 \end{definition}
 
 \begin{definition}[Hölder conjugates]

src/fa/lp/duality.tex

@@ -166,8 +166,8 @@ After the duality of $L^p$ and $L^q$ is established for Hölder conjugate expone
     \[
     \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi_g}_{L^1(X; H)^*}
     \]
-    
-    The above argument shows that the truncation argument was technically not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
+
+    A posteriori, the truncation argument was not required. By applying the truncated case again, $\norm{g}_{L^q(X; F)} = \norm{\phi_g}_{L^p(X; E)^*}$.
 \end{proof}
 
 
@@ -178,7 +178,7 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
     Let $(X, \cm, \mu)$ be a measure space, $K \in \RC$, $H$ be a Hilbert space over $K$, $p, q \in [1, \infty]$ be Hölder conjugates such that one of the following holds:
     \begin{enumerate}[label=(\alph*)]
         \item $p \in (1, \infty)$ and $q \in (1, \infty)$.  
-        \item $p = 1$, $q = \infty$, and $\mu$ is $\sigma$-finite\footnote{This should become localisable. }. 
+        \item $p = 1$, $q = \infty$, $H$ is separable, and $\mu$ is localisable. 
     \end{enumerate}
 
     For each $g \in L^q(X, \cm, \mu; H)$, let
@@ -203,25 +203,27 @@ The typical argument for $L^p$ duality requires using the Radon-Nikodym theorem
 
     By \autoref{theorem:lp-dual-function}, $g \in L^q(X; H)$. 
 
-    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a $\sigma$-finite set satisfying the same restriction condition. 
+    (Arbitrary): In the case of (a), by \autoref{lemma:lp-functional-support}, there exists a $\sigma$-finite set $A \in \cm$ such that for each $f \in L^p(X; H)$, $\dpn{f, \phi}{L^p(X; H)} = \dpn{\one_A \cdot f, \phi}{L^p(X; H)}$. In the case of (b), $A = X$ is a localisable set satisfying the same restriction condition. 
 
-    Let $\seq{A_n} \subset \cm$ such that $\mu(A_n) < \infty$ for all $n \in \natp$, and $A = \bigsqcup_{n \in \natp}A_n$. By the finite case, there exists $\seq{g_n} \subset L^q(X; H)$ such that for each $n \in \natp$ and $f \in L^p(X; H)$, 
+    Let $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. By the finite case, there exists $g_F \in L^q(F; H)$ such that for every $f \in L^p(X; H)$, 
     \[
-    \int \dpn{f, g_n}{H} d\mu = \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)}
-    \] 
-
-    Let $g = \sum_{n = 1}^\infty g_n$. If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
-    \[
-    \norm{g}_{L^\infty(X; H)} \le \sup_{n \in \natp}\norm{g_n}_{L^\infty(X; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \int_F \dpn{f, g_F}{H} d\mu = \dpn{\one_{F} \cdot f, \phi}{L^p(X; H)}
     \]
+
+    In the case of (a), there exists a countable exhaustion $\seq{F_n} \subset \cm$ of $A$ with sets of finite measure. For each $n \in \natp$, a representative of $g_{F_n}$ may be taken to have separable range. In the case of (b), such a representative may be chosen for every $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. Thus by the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g: X \to H$ such that $g|_{F} = g_F$ almost everywhere for all $F \in \cm$ with $F \subset A$ and $\mu(F) < \infty$. 
+
+    If $q < \infty$, then $g \in L^q(X; H)$ by the \hyperref[Monotone Convergence Theorem]{theorem:mct}. Otherwise,
+    \[
+    \norm{g}_{L^\infty(X; H)} \le \sup_{\substack{F \in \cm \\ F \subset A \\ \mu(F) < \infty}}\norm{g_F}_{L^\infty(F; H)} \le \norm{\phi}_{L^1(X; H)^*}
+    \]
+
+    Hence $g \in L^q(X; H)$ with $\norm{g}_{L^q(X; H)} \le \norm{\phi}_{L^1(X; H)^*}$. 
     
-    For every $f \in L^p(X; H)$, 
+    Finally, let $f \in L^p(X; H)$, then there exists $\seq{F_n} \subset \cm$ such that $F_n \upto \bracsn{f \ne 0} \cap A$ and $\mu(F_n) < \infty$ for all $n \in \natp$. In which case, by the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \begin{align*}
-    \int \dpn{f, g}{H} d\mu &= \sum_{n = 1}^\infty \int \dpn{f, g_n}{H} d\mu = \sum_{n = 1}^\infty \dpn{\one_{A_n} \cdot f, \phi}{L^p(X; H)} \\
-    &= \dpn{f, \phi}{L^p(X; H)}
+        \int \dpn{f, g}{H} d\mu &= \limv{n}\int_{F_n} \dpn{f, g}{H} d\mu = \limv{n}\int_{F_n} \dpn{f, g_{F_n}}{H} d\mu \\
+        &= \limv{n}\dpn{\one_{F_n} \cdot f, \phi}{L^p(X; H)} = \limv{n}\dpn{f, \phi}{L^p(X; H)}
     \end{align*}
-    
-    by the \hyperref[Dominated Convergence Theorem]{theorem:dct}.
 
     Therefore the mapping is surjective, and hence an isomorphism. 
 \end{proof}

src/fa/norm/absolute.tex

@@ -33,7 +33,7 @@
 
 \begin{theorem}[Riemann's Rearrangement Theorem]
 \label{theorem:riemann-rearrangement}
-    Let $\seq{x_n} \subset \real$ and $N = P \sqcup N$ such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then
+    Let $\seq{x_n} \subset \real$ and $\natp = P \sqcup N$ be a partition such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then
     \begin{enumerate}
         \item If $\sum_{n \in P}x_n = \infty$ and $\sum_{n \in N}x_n = -\infty$, then there exists bijections $\sigma, \tau: \natp \to \natp$ such that
         

src/fa/norm/arens.tex

@@ -0,0 +1,117 @@
+\section{The Arens Product}
+\label{section:arens-product}
+
+\begin{definition}[Arens Extension]
+\label{definition:arens-product}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$ and $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then there exists a unique bilinear mapping $\Lambda_1: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_1(x, y) = \lambda(x, y)$. 
+        \item For each $x \in E$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+        \item For each $y \in F^{**}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    Similarly, there exists a unique bilinear mapping $\Lambda_2: E^{**} \times F^{**} \to G^{**}$ such that:
+    \begin{enumerate}
+        \item For each $(x, y) \in E \times F$, $\Lambda_2(x, y) = \lambda(x, y)$. 
+        \item[(2')] For each $x \in E^{**}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+        \item[(3')] For each $y \in F$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+    \end{enumerate}
+
+    The mappings $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ are the \textbf{first} and \textbf{second} \textbf{Arens extensions} of $\lambda$, respectively. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Section 1, Theorem 3.2]{ArensBilinear}}}. ]
+    For each $x \in E$, the mapping $\lambda(x, \cdot) \in L(F; G)$ admits an adjoint $\lambda^*(x, \cdot) \in L(G^*; F^*)$, which induces an adjoint of the bilinear map as follows
+    \[
+    \lambda^*: E \times G^* \to F^* \quad \dpn{y, \lambda^*(x, \phi)}{F} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    Applying the above operation again yields a second adjoint
+    \[
+    \lambda^{**}: F^{**} \times G^* \to E^* \quad \dpn{x, \lambda^{**}(y, \phi)}{E} = \dpn{\lambda^*(x, \phi), y}{F^*}
+    \]
+
+    and finally, applying the adjoint operation a third time gives
+    \[
+    \Lambda_1 = \lambda^{***}: E^{**} \times F^{**} \to G^{**} \quad \dpn{\phi, \lambda^{***}(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}
+    \]
+
+    (1): Let $(x, y) \in E \times F$, then for each $\phi \in G^*$, 
+    \[
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{\lambda^*(x, \phi), y}{F^*} = \dpn{\lambda(x, y), \phi}{G}
+    \]
+
+    By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $\Lambda_1$ is an extension of $\lambda$. 
+
+    (2): Fix $x \in E$ and $\phi \in G^*$, then for each $y \in F^{**}$, 
+    \begin{align*}
+        \dpn{\phi, \Lambda_1(x, y)}{G^*} &= \dpn{\lambda^{**}(y, \phi), x}{E^*} = \dpn{x, \lambda^{**}(y, \phi)}{E} \\
+        &= \dpn{\lambda^*(x, \phi), y}{F^*}
+    \end{align*}
+
+    Since $\lambda^*(x, \phi) \in F^*$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+
+    (3): Fix $y \in F^{**}$ and $\phi \in G^*$, then for each $x \in E^{**}$, $\dpn{\phi, \Lambda_1(x, y)}{G^*} = \dpn{\lambda^{**}(y, \phi), x}{E^*}$. Since $\lambda^{**}(y, \phi) \in E^*$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+
+    (Uniqueness): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $E$ is weak*-dense in $E^{**}$, and $F$ is weak*-dense in $F^{**}$, so the extension is uniquely determined.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:arens-extension-continuous}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, then $\Lambda_1, \Lambda_2 \in L^2(E^{**}, F^{**}; G^{**})$, where
+    \[
+    \norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})} = \norm{\Lambda_2}_{L^2(E^{**}, F^{**}; G^{**})}
+    \]
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $\norm{\lambda}_{L^2(E, F; G)} = 1$. Fix $x \in \ol{B_{E}(0, 1)}$, then
+    \[
+    \Lambda_1(x, \ol{B_F(0, 1)}) = \lambda(x, \ol{B_F(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+    
+    
+    Since for any $z \in G^{**}$, $\norm{z}_{G^{**}} = \sup_{\phi \in G^*, \norm{\phi}_{G^*} \le 1}\dpn{z, \phi}{G^*}$, the norm on $G^{**}$ is weak*-lower semicontinuous, and $\ol{B_{G^{**}}(0, 1)}$ is weak*-closed. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. As such, weak*-continuity of $\Lambda_1(x, \cdot)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(x, \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}$ as well. 
+
+    Now, fix $y \in \ol{B_{F^{**}}(0, 1)}$, then $\Lambda_1(\ol{B_E(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$. Thus the weak*-continuity of $\Lambda_1(\cdot, y)$ and \autoref{proposition:closure-of-image} implies that $\Lambda_1(\ol{B_{E^{**}}(0, 1)}, y) \subset \ol{B_{G^{**}}(0, 1)}$. Therefore
+    \[
+    \Lambda_1(\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}) \subset \ol{B_{G^{**}}(0, 1)}
+    \]
+
+    and $\norm{\lambda}_{L^2(E, F; G)} = \norm{\Lambda_1}_{L^2(E^{**}, F^{**}; G^{**})}$. 
+\end{proof}
+
+
+\begin{definition}[Arens Regularity]
+\label{definition:arens}
+    Let $E, F, G$ be normed vector spaces over $K \in \RC$, $\lambda \in L^2(E, F; G)$ be a continuous bilinear mapping, and $\Lambda_1, \Lambda_2: E^{**} \times F^{**} \to G^{**}$ be its first and second Arens extensions, respectively, then the following are equivalent:  
+    \begin{enumerate}
+        \item $\Lambda_1 = \Lambda_2$. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous. 
+        \item There exists an extension $\Lambda: E^{**} \times F^{**} \to G^{**}$ of $\lambda$ that is separately weak*-continuous when restricted to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$.
+    \end{enumerate}
+    
+    If the above holds, then $\lambda$ is an \textbf{Arens regular} bilinear map, and $\Lambda = \Lambda_1 = \Lambda_2$ is \textit{the} \textbf{Arens extension} of $\lambda$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 3.3]{ArensBilinear}}}. ]
+    (3) $\Rightarrow$ (1): By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $\ol{B_{E}(0, 1)}$ is weak*-dense in $\ol{B_{E^{**}}(0, 1)}$, and $\ol{B_{F}(0, 1)}$ is weak*-dense in $\ol{B_{F^{**}}(0, 1)}$. Thus the restrictions of $\Lambda_1$ and $\Lambda_2$ to $\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}$ are uniquely determined by the value of $\lambda$ on $\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}$, in the following sense:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_E(0, 1)}$, $\Lambda_1(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F^{**}}(0, 1)}$, $\Lambda_1(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+        \item $\Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}$ is the unique extension of $\lambda|_{\ol{B_{E}(0, 1)} \times \ol{B_{F}(0, 1)}}$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item For each $x \in \ol{B_{E^{**}}(0, 1)}$, $\Lambda_2(x, \cdot)$ is weak*-continuous. 
+            \item For each $y \in \ol{B_{F}(0, 1)}$, $\Lambda_2(\cdot, y)$ is weak*-continuous. 
+        \end{enumerate}
+    \end{enumerate}
+
+    Since the given extension $\Lambda$ satisfies (i.a), (i.b), (ii.a), and (ii.b), 
+    \[
+    \Lambda_1|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}} = \Lambda = \Lambda_2|_{\ol{B_{E^{**}}(0, 1)} \times \ol{B_{F^{**}}(0, 1)}}
+    \]
+
+    As $\Lambda_1, \Lambda_2$ are bilinear, the above implies that $\Lambda_1 = \Lambda_2$. 
+\end{proof}
+
+

src/fa/norm/index.tex

@@ -1,11 +1,13 @@
 \chapter{Normed Vector Spaces}
 \label{chap:normed-spaces}
 
+
 \input{./normed.tex}
 \input{./absolute.tex}
 \input{./linear.tex}
 \input{./separable.tex}
 \input{./multilinear.tex}
+\input{./arens.tex}
 \input{./hilbert.tex}
 \input{./compact.tex}
 \input{./ap.tex}

src/fa/norm/separable.tex

@@ -12,8 +12,8 @@
     \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 $\real^D$. By \autoref{theorem:uniform-metrisable}, $\real^D$ is metrisable. By \autoref{proposition:separable-product}, $\real^D$ 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}, $S$ 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}. 
+
     (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}
 

src/fa/tvs/vector-function.tex

@@ -27,7 +27,7 @@
 
     (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
 
-    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+    (B) $\Rightarrow$ (T): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
     \begin{align*}
     \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
     &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))

src/measure/bochner-integral/strongly.tex

@@ -1,6 +1,7 @@
 \section{Strongly Measurable Functions}
 \label{section:strongly-measurable}
 
+
 \begin{definition}[Strongly Measurable Function]
 \label{definition:strongly-measurable}
     Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:

src/measure/measurable-maps/local-in-measure.tex

@@ -37,7 +37,7 @@
 \label{proposition:convergence-in-measure}
     Let $(X, \cm, \cf, \mu)$ be a \hyperref[scaffolded]{definition:measure-scaffold} measure space, $(Y, d)$ be a separable metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is Cauchy in measure if and only if:
     \begin{enumerate}
-        \item[(L)] $\fF$ is \hyperref[definition:locally-in-measure]{definition:locally-in-measure}. 
+        \item[(L)] $\fF$ is Cauchy \hyperref[locally in measure]{definition:locally-in-measure}. 
         \item[(T)] For each $\eps, \delta > 0$, there exists $F \in \fF$ and $A \in \cf$ such that
         \[
         \sup_{f, g \in F}\mu(A^c \cap \bracs{d(f, g) > \delta}) < \eps

src/measure/measure/localisable.tex

@@ -198,43 +198,3 @@
         \item[(U)] For all $A \in \cf$, $f|_A = g|_A$ almost everywhere. Since $\cf$ is a scaffold for $\mu$, $f = g$ almost everywhere. 
     \end{enumerate}
 \end{proof}
-
-\begin{corollary}
-\label{corollary:l-infty-dedekind-complete}
-    Let $(X, \cm, \mu)$ be a localisable measure space, then $L^\infty(X; \real)$ is order complete. 
-\end{corollary}
-\begin{proof}
-    Let $\seqi{f} \subset L^\infty(X; \real)$ and $M \in \real$ such that $f_i \le M$ almost everywhere for all $i \in I$. 
-
-    Fix $A \in \cm$ with $\mu(A) < \infty$, and let
-    \[
-    \mathcal{S}_A = \bracs{g \in L^\infty(A; \real)| f_i|_A \le g \text{ almost everywhere }\forall i \in I}
-    \]
-
-    then since $f_i \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_A \ne \emptyset$, and $m_A = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.
-    
-    Let $\seq{g_{A, n}} \subset \mathcal{S}_A$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_A g_{A, n} d\mu \downto m_A$. Take $g_A = \limv{n}g_{A, n}$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\int g_A d\mu = m_A$. 
-
-    For each $i \in I$, since $g_{A, n} \ge f_i|_A$ almost everywhere for all $n \in \natp$, $g_A \ge f_i|_A$ almost everywhere as well. Thus $g_A \in \mathcal{S}_A$. For any $h \in \mathcal{S}_A$, $g_A \wedge h \in \mathcal{S}_A$ with
-    \[
-    m_A \le \int_A g_A \wedge h d\mu \le \int_A g_A d\mu = m_A 
-    \]
-
-    Thus $g_A \wedge h = g_A$ almost everywhere, so $g_A \le h$ almost everywhere, and $g_A$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$. 
-
-    Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B} g_B + \one_{A \setminus B}M \in \mathcal{S}_A$, and
-    \[
-    m_A \le \int_A g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M)d\mu \le \int_A g_A d\mu = m_A
-    \]
-
-    Thus $g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M) = g_A$ almost everywhere, so $g_A|_{A \cap B} \le g_B|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_A|_{A \cap B} = g_B|_{A \cap B}$ almost everywhere. 
-
-    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g:X \to \real$ such that $g|_A = g_A$ for all $A \in \cm$ with $\mu(A) < \infty$. 
-
-    Let $h \in L^\infty(X; \real)$ with $h \ge f_i$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$, 
-    \[
-    \mu(\bracs{h < g} \cap A) \le \mu(\bracs{h|_A < g_A} \cup \bracs{g|_A \ne g_A}) = 0
-    \]
-    
-    As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_A \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^\infty(X; \real)$ is indeed the essential supremum of $\seqi{f}$. 
-\end{proof}

src/measure/vector/fin.tex

@@ -117,6 +117,7 @@ Despite not covering the full dual space, the bounded Borel functions still form
     (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. 
 \end{proof}
 
+
 \begin{proposition}
 \label{proposition:space-of-measures-extreme-points}
     Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:

src/op/banach/definitions.tex

@@ -18,6 +18,17 @@
     For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
 \end{proof}
 
+\begin{definition}[Centre]
+\label{definition:banach-algebra-centre}
+    Let $A$ be a Banach algebra, then
+    \[
+    Z(A) = \bracsn{x \in A|xy = yx \forall y \in A}
+    \]
+
+    is the \textbf{centre} of $A$. 
+\end{definition}
+
+
 \begin{definition}[Homomorphism]
 \label{definition:banach-algebra-homomorphism}
     Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:

src/op/banach/invertible.tex

@@ -9,7 +9,7 @@
 
 \begin{lemma}[Neumann Series]
 \label{lemma:neumann-series}
-    Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
+    Let $A$ be a unital Banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
     \[
     x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
     \]

src/op/c-star/gelfand.tex

@@ -65,13 +65,5 @@
     By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
 \end{proof}
 
-\begin{corollary}
-\label{corollary:linfinity-extremely-disconnected}
-    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
-\end{corollary}
-\begin{proof}
-    By \autoref{corollary:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
-\end{proof}
-
 
 

src/op/c-star/gns.tex

@@ -127,7 +127,7 @@
 
         is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A} \ne 0$. 
 
-        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(P(A)); H_\phi])$. 
+        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(PS(A)); H_\phi])$. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}
@@ -168,7 +168,7 @@
 
     so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 
 
-    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{P(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
+    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in PS(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{PS(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{PS(A)}(A)$ is closed in $B([l^2(PS(A)); H_\phi])$. 
     
 \end{proof}
 

src/op/c-star/state.tex

@@ -31,12 +31,12 @@
 
 \begin{definition}[Pure State]
 \label{definition:pure-state}
-    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $P(A)$ is the collection of all pure states of $A$. 
+    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $PS(A)$ is the collection of all pure states of $A$. 
 \end{definition}
 
 \begin{proposition}
 \label{proposition:state-space-compact-convex}
-    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $PS(A)$. 
 \end{proposition}
 \begin{proof}
     Since the evaluation map is weak* continuous and
@@ -48,15 +48,15 @@
 
     By \autoref{theorem:cstar-positive-algebraic}, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Therefore $S(A)$ is compact by \autoref{proposition:compact-extensions}. 
 
-    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $PS(A)$. 
 \end{proof}
 
 \begin{proposition}
 \label{proposition:multiplicative-pure-state}
     Let $A$ be a unital $C^*$-algebra, then:
     \begin{enumerate}
-        \item $\Omega(A) \subset P(A)$. 
-        \item If $A$ is commutative, then $\Omega(A) = P(A)$. 
+        \item $\Omega(A) \subset PS(A)$. 
+        \item If $A$ is commutative, then $\Omega(A) = PS(A)$. 
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -80,7 +80,7 @@
     Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra with $1_A \in B$, and $\phi \in S(B)$, then
     \begin{enumerate}
         \item There exists $\Phi \in S(A)$ such that $\Phi|_B = \phi$.
-        \item If $\phi \in P(B)$, then there exists $\Phi \in P(A)$ such that $\Phi|_B = \phi$. 
+        \item If $\phi \in PS(B)$, then there exists $\Phi \in PS(A)$ such that $\Phi|_B = \phi$. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}
@@ -88,7 +88,7 @@
 
     (2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}. 
 
-    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in P(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$. 
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in PS(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in PS(A)$. 
 \end{proof}
 
 
@@ -96,15 +96,15 @@
 \label{corollary:cstar-positive-property-probe}
     Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
     \begin{align*}
-    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
+    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in PS(A)} \\
     &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
     \end{align*}
     
 
-    In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
+    In particular, there exists $\phi \in PS(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
 \end{corollary}
 \begin{proof}
-    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
+    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in PS(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in PS(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in PS(A)}$. 
 
     Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
     \[
@@ -113,7 +113,7 @@
 
     Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
     \begin{align*}
-        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
+        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in PS(A)}) \\
         &\subset \ol{\text{Conv}}(\sigma_A(x))
     \end{align*}
 
@@ -139,35 +139,35 @@
 \label{corollary:cstar-positive-weakstar-dense}
     Let $A$ be a unital $C^*$-algebra, then:
     \begin{enumerate}
-        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. 
-        \item The linear span of $P(A)$ is weak*-dense in $A^*$. 
+        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in PS(A)$. 
+        \item The linear span of $PS(A)$ is weak*-dense in $A^*$. 
     \end{enumerate}
 
     Moreover, for any $x \in A$, 
     \begin{enumerate}[start=2]
-        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
-        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
+        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in PS(A)$. 
+        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in PS(A)$. 
     \end{enumerate}
     
 \end{corollary}
 \begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ]
-    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
+    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in PS(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
 
-    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, 
+    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in PS(A)$, 
     \[
     0 = \text{Re}(\dpn{x, \phi}{A}) = \dpn{\text{Re}(x), \phi}{A}
     \]
 
     because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. 
 
-    (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
+    (2): Since the linear span of $PS(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
 
-    (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
+    (3): Let $\phi \in PS(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
 
-    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
+    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in PS(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
 
-    (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
+    (4): Let $\phi \in PS(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
 
-    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in PS(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
 \end{proof}
 

src/op/example/bb.tex

@@ -0,0 +1,153 @@
+\section{$L^\infty$}
+\label{section:l-infty-algebra}
+
+
+\begin{proposition}
+\label{proposition:measures-dual-algebra}
+    Let $X$ be a compact Hausdorff space and $\mathscr{M} \subset M_R(X; \complex)$ be a closed subspace such that:
+    \begin{enumerate}
+        \item[(A)] For each $\mu \in \mathscr{M}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{M}$.
+    \end{enumerate}
+
+    and
+    \[
+    J: C(X; \complex) \to \mathscr{M}^* \quad \dpn{\mu, J(f)}{\mathscr{M}} = \int f d\mu
+    \]
+
+    then
+    \begin{enumerate}
+        \item $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$. 
+        \item There exists a unique weak*-continuous involution on $\mathscr{M}^*$ such that $J(f^*) = J(f)^*$ for all $f \in C(X; \complex)$, given by
+        \[
+            \dpn{\mu, \phi^*}{\mathscr{M}^*} = \ol{\dpn{\mu, \phi}{\mathscr{M}^*}}
+        \]
+        \item There exists a unique separately weak*-continuous bilinear map on $\mathscr{M}^*$ such that $J(fg) = J(f)J(g)$ for all $f, g \in C(X; \complex)$.
+        \item $\mathscr{M}^{*}$ equipped with the above involution and product is a commutative unital $C^*$-algebra. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Let $\seqi{\mu}$ be a maximal mutually singular family of Radon measures on $X$. Using \autoref{theorem:hilbert-measures-dual} and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, identify
+    \[
+    \mathscr{M} = [l^1(I); L^1(\mu_i; \complex)] \quad \mathscr{M}^* = [l^\infty(I); L^\infty(\mu_i; \complex)]
+    \]
+
+    (1): Under the above, $C(X; \complex)$ may be identified as the diagonal
+    \[
+    \bracsn{f \in C(X; \complex)^I|f_i = f_j \forall i, j \in I} \subset [l^\infty(I); L^\infty(\mu_i; \complex)]
+    \]
+
+    By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. As a result, $J(C(X; \complex))$ is weak*-dense in $\mathscr{M}^*$. 
+
+    (2): For each $g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$, let $g^* = \ol g$. For any $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, 
+    \[
+    \dpn{\mu, g^*}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{\mu, \ol g}{[l^1(I); L^1(\mu_i; \complex)]} = \ol{\dpn{\mu, g}{[l^1(I); L^1(\mu_i; \complex)]}}
+    \]
+
+    so the conjugation map is weak*-continuous. 
+
+    (3): Let $f, g \in [l^\infty(I); L^\infty(\mu_i; \complex)]$ and $\mu \in [l^1(I); L^1(\mu_i; \complex)]$, 
+    \[
+        \dpn{\mu, fg}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{f\mu, g}{[l^1(I); L^1(\mu_i; \complex)]} = \dpn{g\mu, f}{[l^1(I); L^1(\mu_i; \complex)]}
+    \]
+
+    so the composition map is separately weak*-continuous. 
+
+    (4): $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is a commutative unital $C^*$-algebra. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:linfty-von-neumann-algebra}
+    Let $(X, \cm, \mu)$ be a localisable measure space, and let $L^\infty(X; \complex)$ act on $L^2(X; \complex)$ by multiplication, then:
+    \begin{enumerate}
+        \item The weak* topology on $L^\infty(X; \complex)$ is equal to the weak operator topology of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$.
+        \item $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Let $A \subset B(L^2(X; \complex))$ be the von Neumann algebra generated by $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$. By \autoref{theorem:lp-duality}, $L^\infty(X; \complex)$ is the dual of $L^1(X; \complex)$, so (1) and the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu} imply that the closed unit ball of $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is weak-operator closed. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_{L^\infty(X; \complex)}(0, 1)} = \ol{B_A(0, 1)}$. Therefore $L^\infty(X; \complex) = A$. 
+\end{proof}
+
+\begin{theorem}[Order Structure of $L^\infty$]
+\label{theorem:l-infty-dedekind-complete}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then
+    \begin{enumerate}
+        \item $L^\infty(X; \real)$ is order complete. 
+        \item For each $\phi \in L^1(X; \real)$ with $\phi \ge 0$ and bounded directed subset $S \subset L^\infty(X; \real)$,
+        \[
+        \sup_{f \in S} \dpn{\phi, f}{L^1(X; \real)} = \bigg\langle\phi, \sup_{f \in S}f\bigg\rangle_{L^1(X; \real)}
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    By \autoref{proposition:linfty-von-neumann-algebra}, $L^\infty(X; \complex)$ acting on $L^2(X; \complex)$ is a von Neumann algebra. 
+
+    (1): Since $L^\infty(X; \real)$ is a lattice, it is order complete by \autoref{theorem:existence-of-projections-vna}.
+
+    (2): By \autoref{theorem:existence-of-projections-vna}, $\sup(S) = \sotlim_{f \in S}f = \wotlim_{f \in S}f$. By (1) of \autoref{proposition:linfty-von-neumann-algebra}, $\sup_{f \in S}\dpn{\phi, f}{L^1(X; \real)} = \dpn{\phi, \sup_{f \in S}f}{L^1(X; \real)}$.
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:linfinity-extremely-disconnected}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:l-infty-order-isotone}
+    Let $(X, \cm, \mu)$ and $(Y, \cn, \nu)$ be localisable measure spaces, and $T: L^\infty(X; \complex)\to L^\infty(Y; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $T$ is an isometric isomorphism.
+        \item For each $f, g \in L^\infty(X; \complex)$, $f \ge g$ if and only if $Tf \ge Tg$. 
+    \end{enumerate}
+    
+    then:
+    \begin{enumerate}
+        \item $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$.
+        \item $T$ is weak*-continuous. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $\phi \in L^1(Y; \complex)$ with $\phi \ge 0$, then $T^*\phi \in L^\infty(X; \complex)^*$. For each $\seq{B_n} \subset \cm$ and $B \in \cm$ with $B_n \upto B$, $\one_B = \sup_{n \in \natp}\one_{B_n}$ as an element of $L^\infty(X; \complex)$. Thus (b) and \autoref{theorem:l-infty-dedekind-complete} imply that
+    \begin{align*}
+    \sup_{n \in \natp}\dpn{\one_{B_n}, T^*\phi}{L^\infty(X; \complex)} &= \sup_{n \in \natp} \dpn{\phi, T\one_{B_n}}{L^1(Y; \complex)} = \dpn{\phi, T\one_B}{L^1(Y; \complex)} \\
+    &= \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}
+    \end{align*}
+
+    Hence the mapping $B \mapsto \dpn{\one_B, T^*\phi}{L^\infty(X; \complex)}$ is a finite positive measure on $(X, \cm)$, which is absolutely continuous with respect to $\mu$. By the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(X; \complex)$ with $f \ge 0$ such that $\int_B f d\mu = \dpn{\one_{B}, T^*\phi}{L^\infty(X; \complex)}$ for all $B \in \cm$. 
+
+    Let $g \in L^\infty(X; [0, 1])$. By \autoref{lemma:separable-metric-space-approx-identity}, there exists simple functions $\seq{g_n} \subset \Sigma(X; [0, 1])$ such that $g_n \upto g$ pointwise. In which case, $g = \sup_{n \in \natp}g_n$ as an element of $L^\infty(X; \real)$, so (b) and \autoref{theorem:l-infty-dedekind-complete} imply that, 
+    \begin{align*}
+    \dpn{\phi, Tg}{L^1(Y; \complex)} &= \sup_{n \in \natp}\dpn{\phi, Tg_n}{L^1(Y; \complex)} = \sup_{n \in \natp}\dpn{f, g_n}{L^1(X; \complex)} \\
+    &= \dpn{f, g}{L^1(X; \complex)}
+    \end{align*}
+
+    By linearity, $\dpn{\phi, Tg}{L^1(Y; \complex)} = \dpn{f, g}{L^1(X; \complex)}$ for all $g \in L^\infty(X; \complex)$. Therefore $T^*(L^1(Y; \complex)) \subset L^1(X; \complex)$, and $T$ is weak*-continuous. 
+\end{proof}
+
+\begin{theorem}[Uniqueness of $L^\infty$]
+\label{theorem:linfty-uniqueness}
+    Let $X$ be a compact Hausdorff space, $\mu, \nu: \cb_X \to [0, \infty)$ be Radon measures on $X$, and $\Phi: L^\infty(\mu; \complex) \to L^\infty(\nu; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\Phi$ is a *-isomorphism. 
+        \item $\Phi|_{C(X; \complex)}$ is the identity.
+    \end{enumerate}
+
+    then:
+    \begin{enumerate}
+        \item $\mu$ and $\nu$ are equivalent. 
+        \item $L^\infty(\mu; \complex) = L^\infty(\nu; \complex)$. 
+        \item $\Phi$ is the identity map. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 21.4]{Zhu}}}. ]
+    As $\Phi$ is a *-isomorphism, $\Phi f \ge \Phi g$ if and only if $f \ge g$ for any $f, g \in L^\infty(\mu; \complex)$. 
+    
+    (1): By \autoref{lemma:l-infty-order-isotone}, there exists $f \in L^1(\mu; \complex)$ such that $\int f g d\mu = \int g d\nu$ for all $g \in C(X; \complex)$. Thus the uniqueness of the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0} implies that $\nu = f d\mu$. As the argument is symmetric, the two measures are equivalent. 
+
+    (3): By \autoref{lemma:l-infty-order-isotone}, $\Phi$ is also weak*-continuous. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex)$ is weak*-dense in $L^\infty(\mu; \complex)$ and $L^\infty(\nu; \complex)$. As $\Phi$ is the identity on $C(X; \complex)$, $\Phi$ is the identity on $L^\infty(\mu; \complex)$. 
+\end{proof}
+
+

src/op/example/bounded.tex

@@ -18,7 +18,7 @@
 
 \begin{proposition}
 \label{proposition:partial-isometry-characterisation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. 
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*T$ is a projection. In which case, $T^*T$ is a projection onto $\ker(T)^\perp$.  
 \end{proposition}
 \begin{proof}
     ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^\perp$, then $\dpn{Tx, Tx}{H} = \norm{x}_H^2$ and $\dpn{T^*Tx, x}{H} = \norm{x}_H^2$. By \hyperref[polarisation]{proposition:polarisation-complex}, for each $x, y \in \ker(T)^\perp$, 
@@ -27,10 +27,18 @@
         &= \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x + i^ky, x + i^ky}{H} = \dpn{x, y}{H} 
     \end{align*}
 
-    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection. 
+    Therefore $T^*T$ is idempotent. As $T^*T$ is self-adjoint, it is a projection onto $\ker(T)^\perp$. 
 
     ($\Leftarrow$): Suppose that $T^*T$ is a projection, then for each $x \in \ker(T)^\perp$, $\dpn{Tx, Tx}{H} = \dpn{T^*Tx, x}{H} = \norm{x}_H^2$. 
 \end{proof}
+\begin{corollary}
+\label{corollary:partial-isometry-adjoint}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^*$ is a partial isometry. 
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 12.7]{Zhu}}}. ]
+    Suppose that $T$ is a partial isometry, then $P = T^*T$ is a projection onto $\ker(T)^\perp$ by \autoref{proposition:partial-isometry-characterisation}. In which case, $T(T^*T) = T$ and $(TT^*)^2 = T(T^*T)T^* = TT^*$, so $TT^*$ is a projection, and $T^*$ is a partial isometry by \autoref{proposition:partial-isometry-characterisation}. 
+\end{proof}
+
 
 \begin{theorem}[Polar Decomposition]
 \label{theorem:hilbert-polar-decomposition}
@@ -42,21 +50,34 @@
         \item $\ker P = \ker V$.
     \end{enumerate}
 
-    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$. 
+    The pair $(P, V)$ is the \textbf{polar decomposition} of $T$, and
+    \begin{enumerate}[start=4]
+        \item $V$ is a partial isometry from $\ker(T)^\perp$ to $\ol{T(H)}$.
+        \item $P$ and $V$ are contained in the von Neumann algebra generated by $T$. 
+    \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem 12.8]{Zhu}}}. ]
-    Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). For each $x \in H$, 
+\begin{proof}[Proof, {{\cite[Theorem 12.8, Theorem 18.9]{Zhu}}}. ]
+    (1): Let $P = |T| = \sqrt{T^*T}$, then $P$ is positive (1). 
+    
+    (2): For each $x \in H$, 
     \[
         \norm{Px}_H^2 = \dpn{Px, Px}{H} = \dpn{P^*Px, x}{H} = \dpn{T^*Tx, x}{H} = \norm{Tx}_H^2
     \]
 
-    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry (2). Moreover, for any $x \in H$, $Tx = V_0Px = VPx$ (3).
-
-    Finally, since the initial space of $V$ is $\ol{P(H)}$, $\ker(V) = P(H)^\perp = \ker(P)$ (4). 
+    Let $V_0: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_0$ extends to a well-defined isometry $\ol{P(H)} \to H$. Further extend $V_0$ to $V$ by setting its value to $0$ on $P(H)^\perp$, then $V$ is a partial isometry. 
     
-    It remains to show uniqueness. Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
+    (3): For any $x \in H$, $Tx = V_0Px = VPx$. 
+    
+    
+    (4), (5): Since the initial space of $V$ is $\ol{P(H)} = \ker(T)^\perp$, $\ker(V) = P(H)^\perp = \ker(P)$. 
+    
+    (Uniqueness): Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By \autoref{proposition:partial-isometry-characterisation}, $W^*W$ is a projection onto $\ker(W)^\perp = \ker(Q)^\perp = \ol{Q(H)}$. Thus $P^2 = T^*T = QW^*WQ = Q^2$, and $P = Q$ by uniqueness of the positive square root. 
 
-    Now, since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+    Since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^\perp$, $V = W$ on $H$, and the polar decomposition is unique. 
+
+    (6): Let $A$ be the von Neumann algebra generated by $T$. By \autoref{theorem:existence-of-projections-vna}, $A$ is a unital $C^*$-algebra, so $P = \sqrt{T^*T} \in A$. To see that $V \in A$, it is sufficient to apply the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^\perp$, $SV|_{\ker(P)} = 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)} = 0$ as well. Therefore $V \in A'' = A$. 
 \end{proof}
 
 

src/op/example/index.tex

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

src/op/notation.tex

@@ -5,6 +5,7 @@
     \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
     \hline
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+    $Z(A)$ & Centre of a Banach algebra. & \autoref{definition:banach-algebra-centre} \\ 
     $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
     $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
     $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\
@@ -16,10 +17,15 @@
     $\Gamma = \Gamma_A$ & The Gelfand transform on $A$. & \autoref{definition:gelfand-transform} \\ 
     $A[S]$ & $C^*$-subalgebra of $A$ generated by $S \subset A$. & \autoref{definition:generated-subalgebra} \\ 
     $S(A)$ & State space of a $C^*$-algebra $A$. & \autoref{definition:cstar-state} \\ 
-    $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
+    $PS(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
     $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
     $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
     $U(T)$ & Cayley transform of $T$. & \autoref{definition:cayley-transform-bounded} \\ 
+    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \autoref{definition:spectral-measure} \\ 
+    $\text{Proj}(A)$ & Projections in $A$. & \autoref{definition:vn-projection-lattice} \\
+    $Z(P)$ & Central support of $P \in \text{Proj}(A)$. & \autoref{definition:central-support-vna} \\ 
+    $P \sim Q$ & $P, Q \in \text{Proj}(A)$ are Murray-von Neumann equivalent & \autoref{definition:murray-von-neumann-equivalent} \\ 
+    $P \preceq Q$ & $P$ is Murray-von Neumann subequivalent to $Q$. & \autoref{definition:murray-von-neumann-subequivalent} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/op/vn/commutative.tex

@@ -0,0 +1,56 @@
+\section{Commutative von Neumann Algebras}
+\label{section:vn-commutative}
+
+\begin{definition}[Separating Vector]
+\label{definition:separating-vector}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$, and $x \in H$, then $x$ is a \textbf{separating vector} for $A$ if the mapping $A \to H$ defined by $T \mapsto Tx$ is injective.
+\end{definition}
+
+\begin{proposition}
+\label{proposition:maximal-commutative-vn}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a maximal commutative von Neumann algebra if and only if $A = A'$.
+\end{proposition}
+\begin{proof}
+    ($\Leftarrow$): Let $B \supset A$ be a commutative von Neumann algebra, then $B \subset A' = A$. 
+
+    ($\Rightarrow$): For each $T \in (A')_{sa}$, the von Neumann algebra generated by $A$ and $T$ is commutative. As such, $T \in A$. As this holds for all $T \in (A')_{sa}$, $A' = (A')_{sa} + i(A')_{sa} \subset A$.  
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:cyclic-separating-commutant}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra with $I \in A$, and $x \in H$, then $x$ is cyclic for $A$ if and only if $x$ is separating for $A'$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 22.1]{Zhu}}}. ]
+    ($\Rightarrow$): Let $T \in A'$ with $Tx = 0$, then $TSx = STx = 0$ for all $S \in A$. In which case, $T(H) \subset \ol{T(Ax)} = \bracs{0}$ by \autoref{proposition:closure-of-image}.
+
+    ($\Leftarrow$): Let $P \in B(H)$ be the orthogonal projection from $H$ onto $\ol{Ax}$, then as $I \in A$, $x \in \ol{Ax}$. Since $\ol{Ax}$ is a reducing subspace for $A$, $P \in A'$. Thus $I, P \in A'$ and $(I - P)x = 0$. Given that $x$ is separating for $A'$, $I = P$, so $\ol{Ax} = H$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:cyclic-is-separating-commutative}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, and $x \in H$ be a cyclic vector for $A$, then $x$ is also a separating vector for $A$. 
+\end{corollary}
+\begin{proof}
+    Since $A$ is commutative, $A \subset A'$. As $x$ is separating for $A'$ by \autoref{proposition:cyclic-separating-commutant}, it is also separating for $A$. 
+\end{proof}
+
+
+\begin{theorem}
+\label{theorem:commutative-has-separating}
+    Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then $A$ admits a separating vector. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 22.3]{Zhu}}}. ]
+    Let $\seqj{x} \subset H$ be a maximal collection of non-zero vectors such that the spaces $\bracsn{Ax_j|j \in I}$ are mutually orthogonal. Such a collection exists by Zorn's lemma, and must be at most countable given that $H$ is separable. 
+
+    Let $\seq{x_n}$ be an enumeration of such a set, padding by zeroes if necessary, and $x = \sum_{n \in \natp}x_n/2^n$. For any $T \in A$, if $Tx = 0$, then as $\bracsn{Ax_n|n \in \natp}$ are mutually orthogonal, $Tx_n = 0$ for all $n \in \natp$. Since $A$ is commutative, $Ax_n \subset \ker(T)$ for all $n \in \natp$. By maximality of $\seq{x_n}$, $H = [l^2(\natp); \ol{Ax_n}]$, $H \subset \ker(T)$, and $T = 0$. Therefore $x$ is a separating vector. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:maximal-abelian}
+    Let $H$ be a separable Hilbert space and $A \subset B(H)$ be a maximal abelian von Neumann algebra, then $A$ admits a cyclic vector. 
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Corollary 22.4]{Zhu}}}. ]
+    By \autoref{proposition:maximal-commutative-vn}, $A = A'$. By \autoref{theorem:commutative-has-separating}, $A$ admits a separating vector. By \autoref{proposition:cyclic-separating-commutant}, this separating vector for $A$ is a cyclic vector for $A' = A$. 
+\end{proof}
+

src/op/vn/fc.tex

@@ -0,0 +1,80 @@
+\section{The $L^\infty$ Functional Calculus}
+\label{section:linfty-functional-calculus}
+
+
+
+\begin{definition}[$L^\infty$ Functional Calculus]
+\label{definition:linfty-functional-calculus}
+    Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
+    \begin{enumerate}
+        \item There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)} \to A$ such that
+        \[
+        T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^* = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)
+        \]
+        \item Let $\mathscr{E} \subset M_R(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
+        \[
+        I_E: \mathscr{E}^* \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE
+        \]
+
+        is a *-isomorphism. 
+        \item $I_E: \mathscr{E}^* \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_E(\text{Id}) = T$. 
+    \end{enumerate}
+    
+    The spectral measure $E$ is the \textbf{resolution of the identity} for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^*$ is the \textbf{$L^\infty$-functional calculus} of $T$. 
+\end{definition}
+\begin{proof}
+    (1), (2): By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1} applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
+    \[
+    I_E: \mathscr{E}^{*} \to A \quad \phi \mapsto \int_{\sigma_{B(H)}(T)} \phi dE
+    \]
+
+    is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$. 
+    
+    For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by \autoref{definition:spectral-measure-integral}.
+
+    (3): By uniqueness of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, and (2), the mapping $\phi \mapsto \phi(T)$ is unique. 
+\end{proof}
+
+
+\begin{theorem}
+\label{theorem:vn-projection-norm-dense}
+    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 The linear span of projections in $A$ is norm dense in $A$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $B \subset \sigma_{B(H)}(T) \setminus \bracs{0}$ be a Borel set. First suppose that $0 \not\in \ol{B}$. By \hyperref[Urysohn's Lemma]{lemma:urysohn}, there exists $f \in C(\sigma_{B(H)}(T); [0, 1])$ with $f(0) = 0$ and $f|_{\ol B} = 1$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, there exists a net $\angles{g_\gamma}_{\gamma \in C} \subset C(\sigma_{B(H)}(T); \complex)$ such that $g_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$. As $f\one_B = \one_B$, $fg_\gamma \to \one_B$ in the weak* topology of $C(\sigma_{B(H)}(T); \complex)^{**}$ as well. 
+
+    By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, $h(T) \in A$ for all $h \in C(\sigma_{B(H)}(T); \complex)$ with $h(0) = 0$. In particular, $fg_\gamma(T) \in A$ for all $\gamma \in C$. Thus the \hyperref[$L^\infty$ functional calculus]{definition:linfty-functional-calculus} implies that $\one_B(T) \in A$ as well. 
+
+    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 $f(0) = 0$. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:von-neumann-group-connected}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then:
+    \begin{enumerate}
+        \item $G(A)$ is path-connected in the norm topology.
+        \item The unitary group of $A$ is path-connected in the norm topology. 
+        \item $I(A)$ is trivial. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Using \autoref{theorem:existence-of-projections-vna} and after possibly shrinking $H$, assume without loss of generality that $I \in A$. 
+
+    After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $e^{\phi(z)} = z$ for all $z \in \complex \setminus \bracs{0}$. 
+        \item For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)} \setminus B(0, r)$. 
+    \end{enumerate}
+
+    (1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_A(T) \subset \ol{B(0, R)} \setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_A(T)$. By the \hyperref[Borel functional calculus]{definition:linfty-functional-calculus}, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$.
+    
+    (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}
+
+

src/op/vn/index.tex

@@ -3,4 +3,9 @@
 
 \input{./topologies.tex}
 \input{./cayley.tex}
-\input{./vn.tex}
+\input{./vn.tex}
+\input{./commutative.tex}
+\input{./spec.tex}
+\input{./fc.tex}
+\input{./projection.tex}
+\input{./type-decomp.tex}

src/op/vn/projection.tex

@@ -0,0 +1,296 @@
+\section{The Projection Lattice}
+\label{section:vn-projection-lattice}
+
+\begin{definition}[Projection Lattice]
+\label{definition:vn-projection-lattice}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\text{Proj}(A)$ be the set of all projections in $A$, then:
+    \begin{enumerate}
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto the closed subspace generated by ${\bigcup_{Q \in S}Q(H)}$, then $P = \sup(S) \in A$. 
+        \item For any $S \subset \text{Proj}(A)$, let $P$ be the orthogonal projection onto $\bigcap_{Q \in S}Q(H)$, then $P = \inf(S) \in A$. 
+        \item $\text{Proj}(A)$ is order complete. 
+    \end{enumerate}
+
+    The set $\text{Proj}(A)$ is the \textbf{projection lattice} of $A$. 
+\end{definition}
+\begin{proof}
+    (1): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, the closed subspace generated by $\bigcup_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+
+    (2): For each $T \in A'$ and $Q \in S$, $TQ = QT$, so $Q(H)$ is a reducing subspace for $T$. As this holds for all $Q \in S$, $\bigcap_{Q \in S}Q(H)$ is a reducing subspace for $T$. Therefore $PT = TP$, and $P \in A$ by the \hyperref[Bicommutant Theorem]{theorem:bicommutant}. 
+\end{proof}
+
+\subsection{Central Support of Projections}
+\label{subsection:projection-central-support}
+
+
+
+\begin{definition}[Central Support]
+\label{definition:central-support-vna}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $Z(P) = \inf_{Q \in \text{Proj}(Z(A)), Q \ge P}Q$ is the \textbf{central support} of $P$. 
+\end{definition}
+
+\begin{definition}[Centrally Orthogonal]
+\label{definition:centrally-orthogonal}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P} \subset \text{Proj}(A)$, then $\seqi{P}$ is \textbf{centrally orthogonal} if $\bracsn{Z(P_i)}_{i \in I}$ is mutually orthogonal. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:central-support-vna}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$. For each $T \in A$, let $R(TP)$ be the orthogonal projection onto $\ol{TP(H)}$, then
+    \[
+    Z(P) = \sup_{T \in A}R(TP)
+    \]
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 24.6]{Zhu}}}. ]
+    Since $Z(P) \in Z(A)$, $Z(P)(H)$ is a reducing subspace of every operator in $A$. As $P \le Z(P)$, $TP(H) \subset T(Z(P)(H)) \subset Z(P)(H)$, so $Z(P) \ge R(TP)$ for all $T \in A$, and $Z(P) \ge \sup_{T \in A}R(TP)$. 
+    
+    On the other hand, for each $S, T \in A$, $S(TP(H)) \subset \bigcup_{R \in A}RP(H)$. As $A$ is a von Neumann algebra, the range of $\sup_{T \in A}R(TP)$ is a reducing subspace for every operator in $A$. Therefore $\sup_{T \in A}R(TP) \in Z(A)$, and $Z(P) \le \sup_{T \in A}R(TP)$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:central-support-mvn}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $Z(P)Z(Q) \ne 0$. 
+        \item $PAQ \ne \bracsn{0}$. 
+        \item There exists non-zero projections $P_0 \le P$ and $Q_0 \le Q$ such that $P_0 \sim Q_0$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 24.7]{Zhu}}}. ]
+    (1) $\Rightarrow$ (2): For each $T \in A$, let $R(T)$ be the orthogonal projection onto $\ol{T(H)}$. By \autoref{proposition:central-support-vna}, 
+    \[
+    Z(P) = \sup_{T \in A}R(TP) \quad Z(Q) = \sup_{T \in A}R(TQ)
+    \]
+
+    Given that $Z(P)Z(Q) \ne 0$, $Z(P)(H) \not\perp Z(Q)(H)$. Since $Z(P)(H) = \ol{\bigcup_{S \in A}SP(H)}$ and $Z(Q)(H) = \ol{\bigcup_{T \in A}TQ(H)}$, there exists $S, T \in A$ such that $SP(H) \not\perp TQ(H)$. As such, there exists $x, y \in H$ with
+    \[
+    0 \ne \dpn{TQx, SPy}{H} = \dpn{PS^*TQx, y}{H}
+    \]
+
+    so $PAQ \ne \bracsn{0}$. 
+
+    (2) $\Rightarrow$ (3): Let $T \in A$ with $PTQ \ne 0$. Let $P_0 = R(PTQ)$ and $Q_0 = R(QT^*P)$, then $0 \ne P_0 \le P$, $0 \ne Q_0 \le Q$, and $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-adjoint}. 
+
+    (3) $\Rightarrow$ (1): By \autoref{lemma:central-support-mvn-eq}, $Z(P_0) = Z(Q_0)$, so
+    \[
+    Z(P)Z(Q) = Z(P) \wedge Z(Q) \ge Z(P_0) \vee Z(Q_0) \ne 0
+    \] 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:central-support-mvn-eq}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$ with $P \sim Q$, then:
+    \begin{enumerate}
+        \item $Z(P) = Z(Q)$. 
+        \item For any central projection $R \in \text{Proj}(A)$, $PR \sim QR$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $V \in A$ with $P = V^*V$ and $Q = VV^*$, then $V$ is a partial isometry with initial space $P(H)$ and final space $Q(H)$. 
+    
+    (1): Since $Z(P) \ge P$ and $Z(P) \in Z(A)$, 
+    \[
+    Z(P)Q = Z(P)VV^* = VZ(P)V^* = VV^* = Q
+    \]
+
+    and $Z(P) \ge Q$, and $Z(P) \ge Z(Q)$. By symmetry, $Z(Q) \ge Z(P)$, so $Z(P) = Z(Q)$. 
+
+    (2): Let $R$ be a central projection, then
+    \[
+    PR = V^*VR = V^*RV = (VR)^*(VR) \sim (VR)(VR)^* = VRV^* = VV^*R = QR
+    \]
+
+
+\end{proof}
+
+\subsection{Murray-von Neumann Equivalence}
+\label{subsection:mvn-equivalence}
+
+
+\begin{lemma}
+\label{lemma:projection-mental-gymnastics}
+    Let $H$ be a complex Hilbert space and $P, Q \in B(H)$ be projections, then:
+    \begin{enumerate}
+        \item $\ker(PQ) = \ker(Q) + \ker(P) \cap Q(H)$. 
+        \item If $PQ = QP$, then $PQ$ is a projection with $PQ(H) = P(H) \cap Q(H)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x \in \ker(PQ)$, then $Q(x) \in \ker(P)$, so $x = Q(x) + (1 - Q)(x) \in  \ker(Q) + \ker(P) \cap Q(H)$.
+    
+    (2): Since $PQ = QP$, $(PQ)^2 = P^2Q^2 = PQ$ and $(PQ)^* = Q^*P^* = QP = PQ$, $PQ$ is a projection. As $PQ(H) = P(Q(H)) \subset P(H)$ and $PQ(H) = Q(P(H)) \subset Q(H)$, $PQ(H) \subset P(H) \cap Q(H)$. On the other hand, $PQ$ is the identity on $P(H) \cap Q(H)$, so $PQ(H) = P(H) \cap Q(H)$. 
+\end{proof}
+
+
+
+
+
+\begin{definition}[Murray-von Neumann Equivalent]
+\label{definition:murray-von-neumann-equivalent}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists $V \in A$ such that $P = V^*V$ and $Q = VV^*$. 
+        \item There exists a partial isometry $V \in A$ from $P(H)$ to $Q(H)$. 
+    \end{enumerate}
+    
+    If the above holds, then $P$ and $Q$ are \textbf{Murray-von Neumann equivalent}, denoted $P \sim Q$. The relation $\sim$ is an equivalence relation on $\text{Proj}(A)$. 
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $V \in A$ with $P = V^*V$ and $Q = VV^*$. By \autoref{proposition:partial-isometry-characterisation}, $V$ is a partial isometry with initial space $\ker(P)^\perp$, and $V^*$ is a partial isometry with initial space $\ker(Q)^\perp$. Therefore $V$ is a partial isometry from $P(H)$ to $Q(H)$.  
+
+    (2) $\Rightarrow$ (1): By \autoref{proposition:partial-isometry-characterisation}, $P = V^*V$ is a projection onto $\ker(V)^\perp$, and $Q = VV^*$ is a projection onto $\ker(V^*)^\perp = V(H)$.
+\end{proof}
+
+\begin{definition}[Murray-von Neumann Subequivalent]
+\label{definition:murray-von-neumann-subequivalent}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then $P$ is \textbf{Murray-von Neumann subequivalent} to $Q$, denoted $P \preceq Q$, if there exists $R \in \text{Proj}(A)$ such that $P \sim R$ and $R \le Q$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:mvn-equivalent-direct-sum}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $\seqi{P}$ is mutually orthogonal.
+        \item $\seqi{Q}$ is mutually orthogonal. 
+        \item For each $i \in I$, $P_i \sim Q_i$. 
+    \end{enumerate}
+
+    then $\sum_{i \in I}P_i \sim \sum_{i \in I}Q_i$. 
+\end{lemma}
+\begin{proof}
+    For each $i \in I$, let $V_i \in A$ such that $P_i = V_i^*V_i$ and $Q_i = V_iV_i^*$, then $V_i$ is a partial isometry with initial space $P_i(H)$ and final space $Q_i(H)$. As $\seqi{P}$ is mutually orthogonal and $\seqi{Q}$ is mutually orthogonal, the sum $\sum_{i \in I}V_i$ converges in strong operator topology to an operator $V$, where
+    \[
+    V^*V = \sum_{i, j \in I}V_i^*V_j = \sum_{i \in I}V_i^*V_i = \sum_{i \in I}P_i
+    \]
+
+    and
+    \[
+    VV^* = \sum_{i, j \in I}V_iV_j^* = \sum_{i \in I}V_iV_i^* = \sum_{i \in I}Q_i
+    \]
+\end{proof}
+
+\begin{lemma}
+\label{lemma:mvn-equivalent-adjoint}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $T \in A$, and $P, Q \in \text{Proj}(A)$ be orthogonal projections onto $\ol{T(H)}$ and $\ol{T^*(H)}$, respectively, then $P \sim Q$. 
+\end{lemma}
+\begin{proof}
+    Let $T = V|T|$ be the \hyperref[polar decomposition]{theorem:hilbert-polar-decomposition} of $T$, then $V$ is a partial isometry from $\ol{T^*(H)}$ to $\ol{T(H)}$. Since $V \in A$, $P \sim Q$. 
+\end{proof}
+
+\begin{theorem}[Kaplansky's Formula]
+\label{theorem:kaplansky-formula}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then
+    \[
+    [(P \vee Q) - Q] \sim [(P - P \wedge Q)]
+    \]
+\end{theorem}
+\begin{proof}
+    Using \autoref{theorem:existence-of-projections-vna}, assume without loss of generality that $I \in A$. In which case, by (1) of \autoref{lemma:projection-mental-gymnastics},
+    \[
+    
+        [(I - Q)P](H)^\perp = \ker(P(I - Q)) = \ker(Q)^\perp + [\ker(Q) \cap \ker(P)]
+    \]
+
+    and since $P \vee Q$ and $Q$ commute, 
+    \begin{align*}
+        [(I - Q)P](H) &= [\ker(Q)^\perp + [\ker(Q) \cap \ker(P)]]^\perp \\
+        &= \ker(Q) \cap [\ker(Q) \cap \ker(P)]^\perp \\
+        &= \ker(Q) \cap [\ker(Q)^\perp + \ker(P)^\perp]\\
+        &= (I - Q)(H) \cap [Q(H) + P(H)]\\
+        &= (I - Q)(H) \cap [(Q \vee P)(H)] \\
+        &= (P \vee Q)(I - Q)(H) = [(P \vee Q) - Q](H)
+    \end{align*}
+
+    by (2) of \autoref{lemma:projection-mental-gymnastics}. Similarly, 
+    \[
+    [P(I - Q)](H)^\perp = \ker((I - Q)P) = \ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]
+    \]
+
+    so
+    \begin{align*}
+        [P(I - Q)](H) &= [\ker(P) + [\ker(Q)^\perp \cap \ker(P)^\perp]]^\perp \\
+        &= \ker(P)^\perp \cap  [\ker(Q) + \ker(P)] \\
+        &= P(H) \cap [(I - Q)(H) + (I - P)(H)] \\
+        &= P(H) \cap [(I - Q) \vee (I - P)](H) \\
+        &= P(H) \cap [I - (P \wedge Q)](H) \\
+        &= [P - (P \wedge Q)](H)
+    \end{align*}
+
+    Therefore
+    \[
+    [(P \vee Q) - Q](H) = [(I - Q)P](H)  \sim [P(I - Q)](H) = [P - (P \wedge Q)](H)
+    \]
+    
+    by \autoref{lemma:mvn-equivalent-adjoint}. 
+\end{proof}
+
+
+\begin{theorem}["Cantor-Bernstein"]
+\label{theorem:murray-von-neumann-subequivalent-partial-order}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. If $P \preceq Q$ and $Q \preceq P$, then $P \sim Q$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Lemma 25.1]{Zhu}}}. ]
+    Let $U, V \in A$ be partial isometries such that $P = U^*U$, $UU^* \le Q$, $Q = V^*V$, and $VV^* \le P$. Denote $Q_0 = Q$ and $P_0 = P$. For each $n \in \natz$, inductively define $P_{n+1} = VQ_nV^*$ and $Q_{n+1} = UP_nU^*$, then:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each $n \in \natz$, $P_n, Q_n \in \text{Proj}(A)$.
+        \item For each $n \in \natz$, $P_n \le P$ and $Q_n \le Q$.
+        \item For each $n \in \natz$, $P_{n+1} \le P_n$ and $Q_{n+1} \le Q_n$. 
+    \end{enumerate}
+
+    As $\seq{P_n}, \seq{Q_n} \subset \text{Proj}(A)$ are non-increasing sequences, by \autoref{theorem:existence-of-projections-vna}, there exists $P_\infty, Q_\infty \in \text{Proj}(A)$ such that $P_n \to P_\infty$ and $Q_n \to Q_\infty$ in the strong operator topology as $n \to \infty$. 
+
+    For each $n \in \natz$, $U(P_n - P_{n+1})U^* = Q_{n+1} - Q_{n+2}$, so
+    \begin{align*}
+    [U(P_n - P_{n+1})]^*[U(P_n - P_{n+1})] &= (P_n - P_{n+1})P(P_n - P_{n+1}) = P_n - P_{n+1} \\
+    [U(P_n - P_{n+1})][U(P_n - P_{n+1})]^* &= U(P_n - P_{n+1})^2U^* = Q_{n+1} - Q_{n+2}
+    \end{align*}
+
+    and $P_n - P_{n+1} \sim Q_{n+1} - Q_{n+2}$. Similarly, $Q_n - Q_{n+1} \sim P_{n+1} - P_{n+2}$. As $P_{n+1} = VQ_nV^*$ for all $n \in \natz$, $P_\infty \sim Q_\infty$ after passing through a strong-operator limit. 
+    
+    For each $N \in \natz$, $\sum_{n = 0}^N (P_n - P_{n+1}) = P - P_{N+1}$, so $P = P_\infty + \sum_{n = 0}^\infty (P_n - P_{n+1})$. Similarly, $Q = Q_\infty + \sum_{n = 0}^\infty (Q_n - Q_{n+1})$. Therefore
+    \begin{align*}
+        P &= P_\infty + \sum_{n = 0}^\infty (P_{2n} - P_{2n+1}) + \sum_{n = 0}^\infty (P_{2n + 1} - P_{2n+2}) \\
+        &\sim Q_\infty + \sum_{n = 0}^\infty (Q_{2n + 1} - Q_{2n+2})  + \sum_{n = 0}^\infty (Q_{2n} - Q_{2n+1}) = Q
+    \end{align*}
+
+    by \autoref{lemma:mvn-equivalent-direct-sum}. 
+\end{proof}
+
+\begin{theorem}[The Comparability Theorem]
+\label{theorem:vna-comparability}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$, then there exists a central projection $R$ such that $RP \preceq RQ$ and $(I - R)Q \preceq (I - R)P$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 25.4]{Zhu}}}. ]
+    By Zorn's lemma, there exists maximal families $\seqi{P}, \seqi{Q} \subset \text{Proj}(A)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\seqi{P}$ is mutually orthogonal.
+        \item $\seqi{Q}$ is mutually orthogonal. 
+        \item For each $i \in I$, $P_i \sim Q_i$. 
+        \item For each $i \in I$, $P_i \le P$ and $Q_i \le Q$. 
+    \end{enumerate}
+
+    Let $P_0 = \sum_{i \in I}P_i$ and $Q_0 = \sum_{i \in I}Q_i$, then $P_0 \sim Q_0$ by \autoref{lemma:mvn-equivalent-direct-sum}. By maximality, there exists no non-zero $P', Q' \in \text{Proj}(A)$ such that $P' \le P - P_0$, $Q' \le Q - Q_0$, and $P' \sim Q'$. By \autoref{proposition:central-support-mvn}, $Z(P - P_0) Z(Q - Q_0) = 0$. 
+
+    Let $R = Z(Q - Q_0)$, then $Q - Q_0 \le R$ and $P - P_0 \le (I - R)$, so $(P - P_0)R = 0$ and $(Q - Q_0)R = Q - Q_0$. By \autoref{lemma:central-support-mvn-eq},
+    \[
+    PR = P_0R \sim Q_0R \le QR
+    \]
+
+    and
+    \[
+    Q(I - R) = Q_0(I - R) \sim P_0(I - R) \le P(I - R)
+    \]
+    
+\end{proof}
+
+\begin{corollary}
+\label{corollary:vna-factor-comparability}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then for any $P, Q \in \text{Proj}(A)$, either $P \prec Q$, $P \sim Q$, or $Q \prec P$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[comparability theorem]{theorem:vna-comparability}, there exists a central projection $R$ such that $PR \preceq QR$ and $Q(I - R) \preceq P(I - R)$. As $A$ is a factor, either $R = 0$ or $R = I$. In which case, $P \preceq Q$ or $Q \preceq P$. 
+\end{proof}
+
+
+

src/op/vn/spec.tex

@@ -0,0 +1,336 @@
+\section{The Spectral Theorem}
+\label{section:spectral-theorem}
+
+\begin{definition}[Spectral Measure]
+\label{definition:spectral-measure}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$, then $E$ is a \textbf{spectral measure relative to $H$} if:
+    \begin{enumerate}
+        \item For each $B \in \cb_X$, $E(B)$ is an orthogonal projection. 
+        \item $E(\emptyset) = 0$, $E(X) = I_{B(H)}$. 
+        \item For each $B, C \in \cb_X$, $E(B \cap C) = E(B)E(C)$. 
+        \item For each $x, y \in H$, the mapping
+        \[
+        E_{x, y}: \cb_X \to \complex \quad B \mapsto \dpn{E(B)x, y}{H}
+        \]
+
+        is a complex Radon measure on $X$. 
+    \end{enumerate}
+    
+\end{definition}
+
+\begin{lemma}
+\label{lemma:spectral-measure-properties}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, and $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, then:
+    \begin{enumerate}
+        \item For each $x, y \in H$, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$. 
+        \item For each $x \in H$, $E_{x, x}$ is positive.
+    \end{enumerate}
+
+    Let $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
+    \begin{enumerate}[start=2]
+        \item For any $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, $\nu \in \mathscr{E}$ as well. 
+        \item Let 
+        \[
+        J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
+        \]
+
+        then $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $x, y \in H$, $\seqf{B_j} \subset \cb_X$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^n B_j$, then for each $1 \le i < j \le n$, $E(B_i)(H) \perp E(B_j)(H)$, so by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz} and the \hyperref[Pythagorean Theorem]{theorem:pythagoras},
+    \begin{align*}
+        \sum_{j = 1}^n |\dpn{E(B_j)x, y}{H}| &= \sum_{j = 1}^n |\dpn{E(B_j)x, E(B_j)y}{H}| \\
+        &\le \sum_{j = 1}^n \norm{E(B_j)x}_H \norm{E(B_j)y}_H \\
+        &\le \braks{\sum_{j = 1}^n \norm{E(B_j)x}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n \norm{E(B_j)y}_H^2}^{1/2} \\
+        &= \norm{E(B)x}_H \cdot \norm{E(B)y}_H \le \norm{x}_H \cdot \norm{y}_H
+    \end{align*}
+
+    As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{\text{var}} \le \norm{x}_H \norm{y}_H$.
+
+    (2): For each $B \in \cb_X$, $E(B)$ is a projection, so $E_{x, x}(B) = \dpn{E(B)x, x}{H} \ge 0$.
+
+    (3): For each $x, y \in H$ and $B, C \in \cb_X$, 
+    \[
+    \int_C \one_B dE_{x, y} = \dpn{E(C \cap B)x, y}{H} = \dpn{E(C)E(B)x, y}{H} = E_{E(B)x, y}(C)
+    \]
+
+    By linearity, $fdE_{x, y} \in \mathscr{E}$ for all $f \in \Sigma(X; \complex)$. For each $f \in \Sigma(X; \complex)$, the mapping $\mu \mapsto f d\mu$ is continuous in the total variation norm, so $fd\mu \in \mathscr{E}$ for all $\mu \in \mathscr{E}$ and $f \in \Sigma(X; \complex)$. By \autoref{proposition:lp-simple-dense}, $\Sigma(X; \complex)$ is dense in $L^1(\mu; \complex)$ for all $\mu \in \mathscr{E}$. Therefore $fd\mu \in \mathscr{E}$ for all $f \in L^1(\mu; \complex)$ and $\mu \in \mathscr{E}$. 
+    
+    Finally, let $\mu \in \mathscr{E}$ and $\nu \in M_R(X; \complex)$ with $\nu \ll \mu$, then by the \hyperref[Radon-Nikodym Theorem]{theorem:lebesgue-radon-nikodym}, there exists $f \in L^1(\mu; \complex)$ such that $d\nu = f d\mu \in \mathscr{E}$. 
+
+    (4): By (3), for any $\mu \in \mathscr{E}$ and $f \in L^1(\mu; \complex)$, $fd\mu \in \mathscr{E}$ as well. By \autoref{proposition:measures-dual-algebra}, there exists a unique weak*-continuous involution and separately weak*-continuous product on $\mathscr{E}^*$ making $\mathscr{E}^*$ a commutative unital $C^*$-algebra, and $J|_{C(X; \complex)}$ a unital *-homomorphism. Since
+    \begin{enumerate}[label=(\roman*)]
+        \item $J$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-$\sigma(\mathscr{E}^*, \mathscr{E})$ continuous. 
+        \item Conjugation on $B^\infty(X; \complex)$ is $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+        \item Multiplication on $B^\infty(X; \complex)$ is separately $\sigma(B^\infty(X; \complex), M_R(X; \complex))$-continuous.
+    \end{enumerate}
+    
+    the mapping $J$ is a unital *-homomorphism.
+\end{proof}
+
+
+\begin{definition}[Integration Against Spectral Measure]
+\label{definition:spectral-measure-integral}
+    Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_X \to B(H)$ be a spectral measure relative to $H$, $\mathscr{E} \subset M_R(X; \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and
+    \[
+    J: B^\infty(X; \complex) \to \mathscr{E}^* \quad \dpn{\mu, J(f)}{\mathscr{E}} = \int_X f d\mu
+    \]
+
+    Then, $\mathscr{E}^*$ admits a unique weak*-continuous involution and a unique separately weak*-continuous product, such that $\mathscr{E}^*$ is a commutative unital $C^*$-algebra, and $J$ is a unital *-homomorphism. 
+    
+    For each $\phi \in \mathscr{E}^*$, let $I_E(\phi) \in B(H)$ be the operator defined by
+    \[
+    \dpn{I_E(\phi) \cdot x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} \quad \forall x, y \in H
+    \]
+
+    then
+
+    \begin{enumerate}
+        \item $I_E$ is a contraction from $\mathscr{E}^*$ to $B(H)$. 
+        \item $I_E$ is continuous from the weak*-topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$. 
+        \item $I_E$ is an injective unital *-homomorphism. 
+    \end{enumerate}
+    
+    For any $\phi \in \mathscr{E}^*$, $I_E(\phi) = \int_X \phi dE$ is the \textbf{integral} of $\phi$ with respect to $E$. 
+\end{definition}
+\begin{proof}
+    (1): Let $\phi \in \mathscr{E}^*$ and $x, y \in H$, then by \autoref{lemma:spectral-measure-properties}, 
+    \begin{align*}
+    |\dpn{I_E(\phi) \cdot x, y}{H}| &= |\dpn{E_{x, y}, \phi}{\mathscr{E}}| \le \norm{E_{x, y}}_{\mathscr{E}} \cdot \norm{\phi}_{\mathscr{E}^{*}} \\
+    &\le \norm{\phi}_{\mathscr{E}^{*}} \cdot \norm{x}_H \cdot \norm{y}_H
+    \end{align*}
+
+    Since the above holds for all $x, y \in H$, $I_E(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)} \le \norm{\phi}_{\mathscr{E}^{*}}$. 
+
+    (2): For each $x, y \in H$, $E_{x, y} \in \mathscr{E}$. Since $\angles{\int \phi dE \cdot x, y}_{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}}$ for every $\phi \in \mathscr{E}^{*}$, $I_E$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$. 
+
+    (3): By \autoref{lemma:separable-metric-space-approx-identity}, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^\infty(X; \complex)$. Since
+    \begin{enumerate}[label=(\roman*)]
+        \item $I_E$ restricted to $J(\Sigma(X; \complex))$ is a *-homomorphism. 
+        \item Multiplication and conjugation are continuous in the uniform norm on $B^\infty(X; \complex)$
+        \item Composition and adjunction are continuous in the operator norm on $B(H)$
+    \end{enumerate}
+
+    the map $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism by continuity. 
+    
+    By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(X; \complex) \subset B^\infty(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$, so $J(C(X; \complex))$ is weak*-dense in $\mathscr{E}^*$. As
+    \begin{enumerate}[label=(\roman*)]
+        \item $I_E$ restricted to $J(B^\infty(X; \complex))$ is a *-homomorphism. 
+        \item The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $\mathscr{E}^{*}$. 
+        \item The adjunction $T \mapsto T^*$ is weak-operator continuous on $B(H)$. 
+        \item The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $\mathscr{E}^{*}$. 
+        \item The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.
+    \end{enumerate}
+
+    the map $I_E$ is a *-homomorphism by the weak* to weak-operator continuity established in (2). Since $E(X) = I_{B(H)}$, $I_E$ is a unital *-homomorphism. 
+
+    Finally, let $\phi \in \mathscr{E}^*$ with $I_E(\phi) = 0$, then $\dpn{I_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{\mathscr{E}} = 0$ for all $x, y \in H$. As $\mathscr{E}$ is the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, $\phi = 0$. Therefore $I_E$ is an injective unital *-homomorphism. 
+\end{proof}
+
+\begin{theorem}[Spectral Theorem I]
+\label{theorem:spectral-theorem-vn-1}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, then:
+    \begin{enumerate}
+        \item There exists a unique spectral measure $E: \cb_{\Omega(A)} \to B(H)$ such that\footnote{Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^*$. }
+        \[
+        T = \int_{\Omega(A)} \Gamma_A T dE \quad \forall T \in A
+        \]
+        \item Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
+        \[
+        I_E: \mathscr{E}^* \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE
+        \]
+
+        is a *-isomorphism. 
+    \end{enumerate}
+    
+    The measure $E$ is the \textbf{spectral measure associated with $A$}, and the homomorphism $I_E$ is the \textbf{extended inverse Gelfand transform} of $A$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 20.2]{Zhu}}}. ]
+    (1): By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, $\Gamma_A: A \to C(\Omega(A); \complex)$ is a *-isomorphism. For each $x, y \in H$, $\Gamma_A^{-1}$ induces a mapping
+    \[
+    E_{x, y}: C(\Omega(A); \complex) \to \complex \quad \dpn{f, E_{x, y}}{C(\Omega(A); \complex)} = \dpn{\Gamma_A^{-1}f \cdot x, y}{H}
+    \]
+
+    which, by the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists. 
+
+    Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*} \le \norm{x}_H\norm{y}_H$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map
+    \[
+    J_E: B^\infty(\Omega(A); \complex) \to B(H) \quad \dpn{J_E(\phi)x, y}{H} = \dpn{E_{x, y}, \phi}{C(\Omega(A); \complex)^*}
+    \]
+
+    with $J_E(f) = \Gamma_A^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$. 
+
+    For any $C \in \cb_{\Omega(A)}$, $\one_C$ is a projection in $B^\infty(\Omega(A); \complex)$. So to see that
+    \[
+    E: \cb_{\Omega(A)} \to B(H) \quad \dpn{E(C)x, y}{H} = E_{x, y}(C)
+    \]
+
+    defines a spectral measure, it is sufficient to show that $J_E$ is a *-homomorphism. 
+
+    Let $x, y \in H$, then as $\Gamma_A$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$, 
+    \begin{align*}
+        \dpn{fg, E_{x, y}}{C(\Omega(A); \complex)} &= \dpn{\Gamma_A^{-1}f \cdot \Gamma_A^{-1}g \cdot x, y}{H} \\
+        &= \dpn{\Gamma_A^{-1}g \cdot x, (\Gamma_A^{-1}f)^* y}{H} = \dpn{g, E_{x, J_E(f)^*y}}{C(\Omega(A); \complex)}
+    \end{align*}
+
+    As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y} = E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^\infty(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$, 
+    \begin{align*}
+        \dpn{E_{x, y}, \phi f}{C(\Omega(A); \complex)^*} &= \dpn{E_{x, J_E(f)^*y}, \phi}{C(\Omega(A); \complex)^*} = \dpn{J_E(\phi)x, J_E(f)^*y}{H} \\
+        &= \dpn{J_E(f)J_E(\phi)x, y}{H} = \dpn{f, E_{J_E(\phi)x, y}}{C(\Omega(A); \complex)}
+    \end{align*}
+
+    so $\phi E_{x, y} = E_{J_E(\phi)x, y}$ for all $\phi \in B^\infty(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^\infty(\Omega(A); \complex)$, 
+    \begin{align*}
+    \dpn{J_E(\phi \psi)x, y}{H} &= \dpn{E_{x, y}, \phi \psi}{C(\Omega(A); \complex)^*} = \dpn{E_{J_E(\psi) x, y}, \phi}{C(\Omega(A); \complex)^*} \\
+    &= \dpn{J_E(\phi)J_E(\psi)x, y}{H}
+    \end{align*}
+
+    and $J_E$ is a homomorphism. 
+
+    Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_A$ is a *-isomorphism, $J_E(f) = \Gamma_A^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)} = \dpn{J_E(f)x, x}{H} \in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^\infty(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H} = \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*} \in \real$ as well. Therefore $J_E(\phi)$ is self-adjoint, and $J_E$ is a *-homomorphism. 
+
+    (2): By \autoref{definition:spectral-measure-integral}, $I_E$ is an injective unital *-homomorphism, so it is sufficient to show that $I_E(\mathscr{E}^*) = B$.
+    
+    Let $J: C(\Omega(A); \complex) \to \mathscr{E}^*$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}} = \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^*$. Since $I_E$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B(H)$, $I_E(\mathscr{E}^*) \subset B$ by \autoref{proposition:closure-of-image}. 
+    
+    On the other hand, by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_E(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_A: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $I_E(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_E(\mathscr{E}^*) = B$.
+\end{proof}
+
+\begin{theorem}[Spectral Theorem II]
+\label{theorem:spectral-theorem-vn-2}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $Id \in A$, $B \subset B(H)$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $\seqi{\xi} \subset H$ be a maximal family such that the subspaces $\bracsn{A\xi_i|i \in I}$ are mutually orthogonal, then:
+    \begin{enumerate}
+        \item For each $i \in I$, there exists a finite positive Radon measure $\mu_i \in \mathscr{E}$ on $\Omega(A)$ such that for every Borel set $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$. 
+        \item For each $i \in I$, let $P_i: H \to \ol{A\xi_i}$ be the orthogonal projection onto $\ol{A\xi_i}$, then for any $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. 
+        \item The natural map $C(\Omega(A); \complex) \to [l^\infty(I); L^\infty(\mu_i; \complex)]$ is injective. Equivalently, $\ol{\bigcup_{i \in I}\supp{\mu_i}} = \Omega(A)$. 
+        \item The space $\mathscr{E}$ is a quotient of $[l^1(I); L^1(\mu_i; \complex)]$ under the mapping
+        \[
+        \mathscr{M}: [l^1(I); L^1(\mu_i; \complex)] \to \mathscr{E} \quad f \mapsto \sum_{i \in I}f_id\mu_i
+        \]
+
+        and $\mathscr{E}^*$ may be identified as a closed subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ through $\mathscr{M}^*$. 
+        \item There exists a unitary equivalence $U: H \to [l^2(I); L^2(\mu_i; \complex)]$ between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$, such that for each $i \in I$, $U|_{\ol{A\xi_i}}$ is an isometry onto the $i$-th factor of $[l^2(I); L^2(\mu_i; \complex)]$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.47]{FollandHarmonic}}}. ]
+    (1): Fix $ i \in I$ and let $\mu_i = E_{\xi_i, \xi_i}$, then for any $C \in \cb_{\Omega(A)}$ with $E_{x, y}(C) = 0$ for all $x, y \in \ol{A\xi_i}$, $\mu_i(C) = 0$. By (1) and (2) of \autoref{lemma:spectral-measure-properties}, $\mu_i$ is a finite positive Radon measure. 
+    
+    By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, for each $S, T \in A$ and $C \in \cb_{\Omega(A)}$, 
+    \[
+    \dpn{E(C)S\xi_i, T\xi_i}{H} = \int_C \Gamma_AS \cdot \ol{\Gamma_AT} dE_{\xi_i, \xi_i}
+    \]
+
+    so $\Gamma_AS \cdot \ol{\Gamma_AT}dE_{\xi_i, \xi_i} = dE_{S\xi_i, T\xi_i} \ll \mu_i$. By (1) of \autoref{lemma:spectral-measure-properties} and completeness of $L^1(\mu_i; \complex)$, $\bracsn{E_{x, y}|x, y \in \ol{A\xi_i}}$ is absolutely continuous with respect to $\mu_i$. Therefore for any $C \in \cb_{\Omega(A)}$, $\mu_i(C) = 0$ if and only if $E_{x, y}(C) = 0$ for all $x, y \in \ol{A \xi_i}$. 
+
+    (2): Let $i, j \in I$ with $i \ne j$, $x \in \ol{A\xi_i}$, $y \in \ol{A\xi_j}$, and $f \in C(\Omega(A); \complex)$, then since $\ol{A\xi_i} \perp \ol{A\xi_j}$,
+    \[
+        \int_{\Omega(A)} f dE_{x, y} = \dpn{\Gamma_A^{-1}(f)x, y}{H} = 0
+    \]
+
+    As the above holds for all $f \in C(\Omega(A); \complex)$, $E_{x, y} = 0$. 
+    
+    Given that $\seqi{\xi}$ is maximal, $x = \sum_{i \in I}P_ix$ for all $x \in H$. Thus for any $x, y \in H$, 
+    \[
+        E_{x, y} = \sum_{i, j \in I}E_{P_ix, P_jy} =  \sum_{i \in I}E_{P_ix, P_iy} \in \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])
+    \]
+
+    (3): Let $T \in A$ with $\Gamma_A T = 0$ $\mu_i$-almost everywhere for all $i \in I$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$. Thus for any $x, y \in H$, 
+    \[
+    \dpn{Tx, y}{H} = \int_{\Omega(A)}\Gamma_A T dE_{x, y} = \sum_{i \in I}\int_{\Omega(A)}\Gamma_A TdE_{P_ix, P_iy} = 0
+    \]
+
+    Therefore $C(\Omega(A); \complex)$ may be identified as a subspace of $[l^\infty(I); L^\infty(\mu_i; \complex)]$. 
+
+    (4): By (2), for each $x, y \in H$, $E_{x, y} = \sum_{i \in I}E_{P_ix, P_iy}$. By (1), $E_{P_ix, P_iy} \ll \mu_i$ for all $i \in I$, so $\mathscr{E} \subset \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. 
+
+    On the other hand, for each $i \in I$, since $\mu_i$ is a Radon measure, $C(\Omega(A); \complex)$ is dense in $L^1(\mu_i; \complex)$ by \autoref{proposition:radon-cc-dense}. As
+    \begin{align*}
+    \mathscr{E} &\supset \bracsn{E_{x, y}|x, y \in \ol{A\xi_i}} \supset \bracsn{fdE_{\xi_i, \xi_i}|f \in C(\Omega(A); \complex)}  \\
+    &= \bracsn{fd\mu_i|f \in C(\Omega(A); \complex)} 
+    \end{align*}
+    
+
+    and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ is closed, $\mathscr{E} \supset \bracsn{f d\mu_i|f \in L^1(\mu_i; \complex)}$. 
+    
+    Finally, given that the above holds for all $i \in I$, $\mathscr{E} = \mathscr{M}([l^1(I); L^1(\mu_i; \complex)])$. By \autoref{theorem:lp-sum-dual} and \autoref{theorem:lp-duality}, $[l^\infty(I); L^\infty(\mu_i; \complex)] = [l^1(I); L^1(\mu_i; \complex)]^*$, so $\mathscr{E}^*$ may be identified with its image under $\mathscr{M}^*$. 
+    
+    (5): Fix $i \in I$, then for any $S, T \in A$ with $S\xi_i = T\xi_i$,
+    \[
+    \Gamma_AS dE_{\xi_i, \xi_i} = E_{S\xi_i, \xi_i} = E_{T\xi_i, \xi_i} = \Gamma_A T dE_{\xi_i, \xi_i}
+    \]
+
+    so $\Gamma_A S = \Gamma_A T$ $\mu_i$-almost everywhere. Thus the mapping
+    \[
+    U_i: \ol{A\xi_i} \to L^2(\mu_i; \complex) \quad T\xi_i \mapsto \Gamma_AT
+    \]
+
+    is well-defined. Moreover, for any $S, T \in A$, 
+    \[
+    \dpn{S\xi_i, T\xi_i}{H} = \int \Gamma_AS \cdot \ol{\Gamma_A T} dE_{\xi_i, \xi_i} = \dpn{\Gamma_A S, \Gamma_A T}{L^2(\mu_i; \complex)}
+    \]
+
+    so $U_i$ extends into an isometry between $\ol{A\xi_i}$ and $L^2(\mu_i; \complex)$. Thus the mapping
+    \[
+    U: H \to [l^2(I); L^2(\mu_i; \complex)] \quad (Ux)_i = U_i(P_ix)
+    \]
+
+    is an isometry between $H$ and $[l^2(I); L^2(\mu_i; \complex)]$ such that $U(Tx) = \Gamma_AT \cdot Ux$ for all $x \in H$ and $T \in A$. 
+    
+    Finally, given that
+    \begin{enumerate}[label=(\roman*)]
+        \item By \hyperref[Goldstine's Theorem]{corollary:weak-dense-unit-ball}, $C(\Omega(A); \complex)$ is weak*-dense in $\mathscr{E}^*$. 
+        \item The weak* topology on $[l^\infty(I); L^\infty(\mu_i; \complex)]$ is equal to the weak operator topology of $[l^\infty(I); L^\infty(\mu_i; \complex)]$ acting on $[l^2(I); L^2(\mu_i; \complex)]$. 
+        \item $A$ is weak-operator dense in $B$. 
+        \item By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, the isomorphism $\phi \mapsto \int \phi dE$ is continuous from the weak* topology on $\mathscr{E}^*$ to the weak operator topology on $B$.
+    \end{enumerate}
+    
+    the mapping $U$ is a unitary equivalence between $\mathscr{E}^*$ acting on $[l^2(I); L^2(\mu_i; \complex)]$ and $B$ acting on $H$. 
+\end{proof}
+
+\begin{corollary}[Representation of Commutative von Neumann Algebras]
+\label{corollary:commutative-von-neumann-linfty}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative von Neumann algebra with $I \in A$, then:
+    \begin{enumerate}
+        \item There exists a LCH space $\Omega$ and a decomposable Radon measure $\mu$ on $\Omega$ such that $A$ is *-isomorphic to $L^\infty(\mu; \complex)$. 
+        \item If $A$ admits a cyclic vector, then $\Omega$ may be taken to be compact. 
+        \item If $H$ is separable, then $\Omega$ may be taken to be compact. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    Let $E$ be the spectral measure on $\Omega(A)$ associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$. By \hyperref[Spectral Theorem I]{theorem:spectral-theorem-vn-1}, $A$ is *-isomorphic to $\mathscr{E}^*$.
+
+    (1): By (3) of \autoref{lemma:spectral-measure-properties} and \autoref{theorem:hilbert-measures-dual}, $A$ is *-isomorphic to $[l^\infty(I); L^\infty(\mu_i; \complex)]$, where $\seqi{\mu} \subset \mathscr{E}$ is a maximal mutually singular family. Let $\Omega = \bigsqcup_{i \in I}\Omega(A)$, then $\Omega$ is a LCH space. For each $i \in I$, let $\Omega_i$ denote the $i$-th copy of $\Omega(A)$, then
+    \[
+    \mu: \cb_\Omega \to [0, \infty] \quad B \mapsto \sum_{i \in I}\mu_i(B \cap \Omega_i)
+    \]
+
+    is the desired decomposable Radon measure. 
+
+    (2): By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton. 
+
+    (3): If $H$ is separable, then so is $\mathscr{E}$. As such, there exists a single positive Radon measure $\mu \in \mathscr{E}$ on $\Omega(A)$ such that $\mathscr{E}$ is absolutely continuous with respect to it. Therefore the index set in (1) can be taken to be a singleton. 
+\end{proof}
+
+
+\begin{remark}
+\label{remark:spectral-theorem-vn-2}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a commutative $C^*$-subalgebra with $I \in A$, $B$ be the von Neumann algebra generated by $A$, $E: \cb_{\Omega(A)} \to B(H)$ be the spectral measure associated with $A$, and $\mathscr{E} \subset M_R(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$.
+
+    By \hyperref[Spectral Theorem II]{theorem:spectral-theorem-vn-2}, there exists a decomposable measure space $\Omega$, corresponding to a number of copies of $\Omega(A)$, such that $\mathscr{E}$ is a quotient of its $L^1$ space, $\mathscr{E}^*$ is a subspace of its $L^\infty$ space, and $H$ is isomorphic to its $L^2$ space. The preceding isomorphisms are all linked by a unitary equivalence between $B$ acting on $H$, and $\mathscr{E}^*$ acting on the $l^2$ direct sum. 
+
+    The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors: 
+    \begin{enumerate}
+        \item The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^1$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see \autoref{theorem:hilbert-measures-dual}.
+        \item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremely large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$. 
+    \end{enumerate}
+
+    More concretely, (1) manifests as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$. 
+
+    By limiting these two sources of complexity, it is possible to remove the need of multiple copies of $\Omega(A)$. In particular, 
+    \begin{enumerate}
+        \item If $B$ admits a cyclic vector, then only one copy of $\Omega(A)$ is required for the construction in the Spectral Theorem \cite[Theorem 23.1]{Zhu}.
+        \item If $H$ is separable, then at most countably many copies of $\Omega(A)$ are required for the construction in the Spectral Theorem. In which case, the measures can be summed such that $B$ is isomorphic to an $L^\infty$ space on $\Omega(A)$ \cite[Page 24]{FollandHarmonic} \cite[Theorem 23.2]{Zhu}. 
+    \end{enumerate}
+\end{remark}
+

src/op/vn/topologies.tex

@@ -109,3 +109,14 @@ Now, a few facts about the more familiar operator topologies:
     by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}. Thus $f$ is a uniform limit of strong-operator continuous functions on $B_{B(H)}(0, 1)$, and as such also strong-operator continuous by \autoref{proposition:uniform-limit-continuous}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spatial-isomorphism-sot-continuous}
+    Let $A$ be a $C^*$-algebra, $H_1, H_2$ be a complex Hilbert spaces, $\pi_1: A \to B(H_1)$ and $\pi_2: A \to B(H_2)$ be injective representations of $A$, and $U: H_1 \to H_2$ be an unitary equivalence, then the mapping
+    \[
+    \pi_1(A) \to \pi_2(A) \quad T \mapsto UTU^{-1}
+    \]
+
+    is strong-operator and weak-operator continuous. 
+\end{proposition}
+
+

src/op/vn/type-decomp.tex

@@ -0,0 +1,303 @@
+\section{Type Decomposition}
+\label{section:vna-type-decomposition}
+
+
+\begin{definition}[Finite Projection]
+\label{definition:finite-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{finite} if for any $Q \in \text{Proj}(A)$ with $P \sim Q$ and $Q \le P$, $P = Q$. For any $P \in \text{Proj}(A)$, $P$ is \textbf{infinite} if it is not finite. 
+\end{definition}
+
+\begin{definition}[Abelian Projection]
+\label{definition:abelian-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then $P$ is \textbf{abelian} if $PAP$ is abelian. 
+\end{definition}
+
+\begin{definition}[Minimal Projection]
+\label{definition:minimal-projection}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P \in \text{Proj}(A)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $PAP = \complex P$. 
+        \item There exists no $Q \in \text{Proj}(A)$ with $0 < Q < P$. 
+    \end{enumerate}
+    
+    If the above holds, then $P$ is \textbf{minimal}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $Q \in \text{Proj}(A)$ with $Q \le P$, then $Q = PQP$. As $PAP = \complex P$, either $PQP = 0$ or $PQP = P$. 
+
+    (2) $\Rightarrow$ (1): Given that there exists 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{lemma}
+\label{lemma:projection-types-gymnastics}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, and $P, Q \in \text{Proj}(A)$. 
+    \begin{enumerate}
+        \item If $P$ is minimal, then $P$ is abelian. 
+        \item If $P$ is abelian, then $P$ is finite. 
+        \item If $P$ is finite and $P \sim Q$, then $Q$ is finite. 
+        \item If $P$ is finite and $Q \le P$, then $Q$ is finite. 
+        \item If $P$ is minimal and $P \sim Q$, then $Q$ is minimal. 
+        \item If $P, Q$ are minimal with $P \sim Q$, then for any $U, V \in A$ with $P = U^*U = V^*V$ and $Q = UU^* = VV^*$, there exists $\lambda \in \partial B_{\complex}(0, 1)$ such that $V = \lambda U$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Section 26.1]{Zhu}}}. ]
+    (1): $PAP = \complex P$ is abelian. 
+
+    (2): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, then there exists $V \in A$ such that $R = V^*V$ and $P = VV^*$. Since $V$ has initial space $R(H) \subset P(H)$ and final space $P(H)$, $V = PVP$ and $V^* = PV^*P$. As $PAP$ is abelian, 
+    \[
+    R = V^*V = PV^*PPVP = PVPPV^*P = VV^* = P
+    \]
+
+    (3): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q$. Let $V \in A$ with $Q = V^*V$ and $P = VV^*$, then $V$ is a partial isometry with initial space $Q(H)$ and final space $P(H)$. In which case, $P = VQV^*$, and $VRV^* \le VQV^* = P$. Let $U = (VRV^*)V$, then
+    \begin{align*}
+    U^*U &= (VRV^*V)^*(VRV^*V) = V^*VRV^* \cdot VRV^*V \\
+    &= V^*VRV^*V = QRQ = R
+    \end{align*}
+    
+
+    and as $Q = V^*PV$, 
+    \begin{align*}
+        UU^* &= (VRV^*V)(VRV^*V)^* = VRV^*V \cdot V^*VRV^* \\
+        &= VRV^*PVRV^* = VRQRV^* = VRV^*
+    \end{align*}
+    
+    so $VRV^* \sim R \sim Q \sim P$. Given that $P$ is finite, $VRV^* = P$. Therefore
+    \[
+    R = QRQ = V^*VRV^*V = V^*PV = Q
+    \]
+
+    (4): Let $R \in \text{Proj}(A)$ with $Q \sim R \le Q \le P$, then $P \sim (P - Q) + R \le P$, so $P - Q + R = P$, and $Q = R$. 
+
+    (5): Since $P \sim Q$, there exists $V \in A$ with $P = V^*V$ and $Q = VV^*$. Let $R \in \text{Proj}(A)$ with $0 < R \le Q$, then $0 \le V^*RV \le V^*QV = P$. By minimality of $P$, $V^*RV = P$, so
+    \[
+    R = QRQ = VV^*RVV^* = VPV^* = Q
+    \]
+
+    (6): Let $R = U^*V$, then since $U$ and $V$ are partial isometries with initial space $P(H)$ and final space $Q(H)$, 
+    \[
+    PRP = U^*U \cdot U^*V \cdot V^*V = U^*QV = U^*V
+    \]
+
+    so $PRP \in PAP = \complex P$. Thus there exists $\lambda \in \complex$ such that $R = \lambda P$. In which case,
+    \[
+    \lambda U = U \cdot \lambda P = UU^*V = QV = V
+    \]
+
+    and
+    \[
+    P = V^*V = \lambda \ol{\lambda} U^*U = |\lambda|^2 P
+    \]
+
+    so $|\lambda| = 1$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:centrally-orthogonal-sum-properties}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a von Neumann algebra, $\seqi{P} \subset \text{Proj}(A)$ be centrally orthogonal, and $P = \sum_{i \in I}P_i$, then
+    \begin{enumerate}
+        \item For each $T \in A$, $PTP = \sum_{i \in I}P_iTP_i$. 
+        \item If $\seqi{P}$ are abelian, then $P$ is also abelian.
+        \item If $\seqi{P}$ are finite, then $P$ is also finite. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 26.2]{Zhu}}}. ]
+    (1): For each $i \in I$, $Z(P_i) \ge P_i$, so $Z(P_i)P_i = P_i$. For any $i, j \in I$ with $i \ne j$, $Z(P_i)$ and $Z(P_j)$ are orthogonal, so $Z(P_i)P_j = Z(P_i)Z(P_j)P_j = 0$. 
+
+    Let $T \in A$, then by \autoref{proposition:central-support-vna}, $Z(P_i)(H) \supset TP_i(H)$ for all $i \in I$. Therefore
+    \begin{align*}
+    PTP &= \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}Z(P_i)P_i \cdot T \cdot Z(P_j)P_j \\
+    &=  \sum_{i, j \in I}Z(P_i)P_i \cdot Z(P_j) \cdot T \cdot Z(P_j)P_j \\
+    &= \sum_{i \in I}Z(P_i)P_i \cdot T \cdot Z(P_i)P_i = \sum_{i \in I}P_i TP_i 
+    \end{align*}
+
+    (2): Let $S, T \in A$, then by (1),
+    \begin{align*}
+        PSP \cdot PTP &= \sum_{i, j \in I}P_iSP_i \cdot P_jTP_j = \sum_{i \in I}P_iSP_i \cdot P_iTP_i \\
+        &= \sum_{i \in I}P_iTP_i \cdot P_iSP_i = PTP \cdot PSP
+    \end{align*}
+
+    (3): Let $R \in \text{Proj}(A)$ with $P \sim R \le P$, and $V \in A$ with $R = V^*V$ and $P = VV^*$, then for each $i \in I$, $Z(P_i)R \sim Z(P_i)P = P_i$, and $Z(P_i)R \le Z(P_i)P = P_i$. As $\seqi{P}$ are finite, $Z(P_i)R = P_i$ for all $i \in I$. Therefore
+    \[
+    R = RP = R\sum_{i \in I}Z(P_i)P_i = \sum_{i \in I}Z(P_i)RP_i = \sum_{i \in I}P_i = P
+    \]
+\end{proof}
+
+\begin{definition}[Type $\vnI$]
+\label{definition:vna-t1}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnI$} if for every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero abelian projection $Q \in \text{Proj}(A)$ with $P \ge Q$. 
+\end{definition}
+% Todo: add a few equivalent characterisations. 
+
+\begin{definition}[Type $\vnII$]
+\label{definition:vna-t2}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnII$} if:
+    \begin{enumerate}
+        \item $A$ has no non-zero abelian projections. 
+        \item For every non-zero central projection $P \in \text{Proj}(Z(A))$, there exists a non-zero finite projection $Q \in \text{Proj}(A)$ with $P \ge Q$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Type $\vnII_1$]
+\label{definition:vna-t21}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_1$} if $I$ is a finite projection. 
+\end{definition}
+
+\begin{definition}[Type $\vnII_\infty$]
+\label{definition:vna-t2inf}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a type $\vnII$ von Neumann algebra, then $A$ is of \textbf{type $\vnII_\infty$} if $A$ has no non-zero finite central projections. 
+\end{definition}
+
+\begin{definition}[Type $\vnIII$]
+\label{definition:vna-t3}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is of \textbf{type $\vnIII$} if $A$ has no non-zero finite projections. 
+\end{definition}
+
+\begin{theorem}[Type Decomposition]
+\label{theorem:vna-type-decomposition}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then there exist unique von Neumann algebras $A_{\vnI}, A_{\vnII_1}, A_{\vnII_{\infty}}, A_{\vnIII} \subset A$\footnote{Not all four types are guaranteed to be present.} of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively, such that
+    \[
+    A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
+    \]
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 26.3]{Zhu}}}. ]
+    ($\vnI$): By Zorn's lemma, there exists a maximal family $\seqi{P} \subset \text{Proj}(A)$ of centrally orthogonal abelian projections. Let $P = \sum_{i \in I}P_i$, then $P$ is abelian by (2) of \autoref{lemma:centrally-orthogonal-sum-properties}. 
+
+    Let $P_{\vnI} = Z(P)$, then $A_{\vnI} := P_{\vnI}AP_{\vnI}$ is a von Neumann algebra with identity $P_{\vnI}$. Let $R \in \text{Proj}(Z(A_{\vnI})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnI}$, $RP \le R$ is a non-zero abelian projection. Therefore $A_{\vnI}$ is of type $\vnI$. 
+
+    ($\vnII$): Assume without loss of generality that $I \in A$. Since $\seqi{P}$ is maximal and $(I - P_{\vnI}) \in Z(A)$, $(I - P_{\vnI})A(I - P_{\vnI})$ has no non-zero abelian projections. 
+
+    By Zorn's lemma, there exists a maximal family $\seqj{Q} \subset \text{Proj}((I - P_{\vnI})A(I - P_{\vnI}))$ of centrally orthogonal finite projections. Let $Q = \sum_{j \in J}Q_j$, then $Q$ is finite by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}. 
+
+    Let $P_{\vnII} = Z(Q)$ and $A_{\vnII} = P_{\vnII}AP_{\vnII}$, then $A_{\vnII}$ is a von Neumann algebra with identity $P_{\vnII}$. Let $R \in \text{Proj}(Z(A_{\vnII})) \setminus \bracs{0}$, then since $0 < R \le P_{\vnII}$, $RQ \le R$ is a non-zero finite projection by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $A_{\vnII}$ is of type $\vnII$. 
+
+    ($\vnIII$): Let $P_{\vnIII} = I - P_{\vnI} - P_{\vnII}$ and $A_{\vnIII} = P_{\vnIII}AP_{\vnIII}$. Since $P_{\vnIII} \in Z(A)$ and $\seqi{P}$, $\seqj{Q}$ are maximal, $A_{\vnIII}$ has no non-zero finite projections. Therefore $A_{\vnIII}$ is of type $\vnIII$, and $A = A_{\vnI} \oplus A_{\vnII} \oplus A_{\vnIII}$. 
+
+    ($\vnII_1$): By Zorn's lemma, there exists a maximal family $\bracsn{R_k}_{k \in K} \subset \text{Proj}(A_{\vnII})$ of orthogonal central finite projections. Let $P_{\vnII_1} = \sum_{k \in K}R_k$, then $P_{\vnII_1}$ is a central finite projection by (3) of \autoref{lemma:centrally-orthogonal-sum-properties}. Hence $A_{\vnII_1} = P_{\vnII_1}AP_{\vnII_1}$ is of type $\vnII_1$. 
+
+    ($\vnII_\infty$): Let $P_{\vnII_\infty} = P_{\vnII} - P_{\vnII_1}$ and $A_{\vnII_\infty} = P_{\vnII_\infty}AP_{\vnII_\infty}$, then $P_{\vnII_\infty}$ is a central projection. By maximality of $\bracsn{R_k}_{k \in K}$, $A_{\vnII_\infty}$ admits no non-zero finite central projections. Therefore $A_{\vnII_\infty}$ is of type $\vnII_\infty$, $A_{\vnII} = A_{\vnII_1} \oplus A_{\vnII_\infty}$, and
+    \[
+    A = A_{\vnI} \oplus A_{\vnII_1} \oplus A_{\vnII_{\infty}} \oplus A_{\vnIII}
+    \]
+    
+    (Uniqueness): Let $A = A_{\vnI}' \oplus A_{\vnII_1}' \oplus A_{\vnII_{\infty}}' \oplus A_{\vnIII}'$ be a decomposition of $A$ into von Neumann algebras of type $\vnI$, $\vnII_1$, $\vnII_\infty$, and $\vnIII$, respectively. 
+
+    Let $P_{\vnI}'$, $P_{\vnII_1}'$, $P_{\vnII_\infty}'$, and $P_{\vnIII}'$ be the identity elements of $A_{\vnI}'$, $A_{\vnII_1}'$, $A_{\vnII_{\infty}}'$, and $A_{\vnIII}'$, respectively, then
+    \[
+    I = P_{\vnI}' \oplus P_{\vnII_1}' \oplus P_{\vnII_\infty}' \oplus P_{\vnIII}'
+    \]
+
+    is an orthogonal direct sum, and
+    \begin{enumerate}
+        \item[($\vnI$)] Let $P_1 = P'_{\vnI}(I - P_{\vnI})$, then by construction of $P_{\vnI}$, there exists no non-zero abelian projection $R \in \text{Proj}(A)$ with $R \le P_1$. As both $P_{\vnI}'$ and $(I - P_{\vnI})$ are central, $P_1 \in A_{\vnI}'$, so $P_1 = 0$ because $A_{\vnI}'$ is of type $\vnI$. Thus $P_{\vnI}' \le P_{\vnI}$. By symmetry, $P_{\vnI} = P_{\vnI}'$ and $A_{\vnI} = A_{\vnI}'$.
+        \item[($\vnII$, $\vnIII$)] Let $P'_{\vnII} = P_{\vnII_1}' \oplus P_{\vnII_\infty}'$, $A_{\vnII}' = A_{\vnII_1}' \oplus A_{\vnII_\infty}'$, and $P_2 = P'_{\vnII}(I - P_{\vnI} - P_{\vnII})$. By construction of $P_{\vnII}$, there exists no non-zero finite projection $R \in \text{Proj}(A)$ with $R \le P_2$. Since $P_2 \in A_{\vnII}'$ and $A_{\vnII}'$ is of type $\vnII$, $P_2 = 0$ and $P_{\vnII}' \le P_{\vnII}$. By symmetry, $P_{\vnII} = P_{\vnII}'$. Thus $P_{\vnIII} = P_{\vnIII}'$, $A_{\vnII} = A_{\vnII}'$, and $A_{\vnIII} = A_{\vnIII}'$.
+        \item[($\vnII_1$, $\vnII_\infty$)] Let $Q_2 = P'_{\vnII_1}(P_{\vnII} - P_{\vnII_1})$, then there exists no non-zero finite central projection $R \in \text{Proj}(A)$ with $R \le Q_2$. However, since $A_{\vnII_1}'$ is of type $\vnII_1$, $P'_{\vnII_1}$ is itself a finite projection, and every subprojection of $P'_{\vnII_1}$ is finite by (4) of \autoref{lemma:projection-types-gymnastics}. Thus $Q_2 = 0$ and $P_{\vnII_1}' \le P_{\vnII_1}$. By symmetry, $P_{\vnII_1}' = P_{\vnII_1}$. Therefore $P_{\vnII_\infty}' = P_{\vnII_\infty}$, $A_{\vnII_1}' = A_{\vnII_1}$, and $A_{\vnII_\infty}' = A_{\vnII_\infty}$.
+    \end{enumerate}
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:abelian-minimal-factor}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, and $P \in \text{Proj}(A)$ be abelian, then $P$ is minimal. 
+\end{lemma}
+\begin{proof}
+    Let $Q \in \text{Proj}(A)$ with $0 \le Q \le P$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $Q \preceq P - Q$ or $P - Q \preceq Q$. Assume without loss of generality that $Q \preceq P - Q$. 
+    
+    Let $V \in A$ such that $Q = V^*V$ and $VV^* \le P - Q$, then $V$ is a partial isometry with initial and final spaces contained in $P(H)$. Since $P$ is abelian, $V \in PAP$, and $Q = V^*V = VV^* \le P - Q$. Therefore $Q = 0$, and $P$ is minimal. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:type1-bh-matrix}
+    Let $H$ be a complex Hilbert space, $A \subset B(H)$ be a factor, $T \in A$, $\seqi{P} \subset \text{Proj}(A)$ be non-zero minimal projections such that $I = \sum_{i \in I}P_i$, and $\bracsn{V_{i, j}}_{i, j \in I} \subset A$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$ for all $i, j \in I$, then there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i, j}V_{i, j}$. 
+\end{lemma}
+\begin{proof}
+    Let $i, j \in I$, then since $P_i$ and $P_j$ are minimal, 
+    \begin{align*}
+        (P_iTP_j)^*(P_iTP_j) &= P_jT^*P_iTP_j \in \complex P_j \\
+        (P_iTP_j)(P_iTP_j)^* &= P_iTP_jT^*P_i \in \complex P_i
+    \end{align*}
+    
+
+    If $P_iTP_j \ne 0$, then $P_jT^*P_iTP_j$ and $P_iTP_jT^*P_i$ are positive, and there exists $\lambda > 0$ such that $\lambda P_iTP_j$ is a partial isometry with initial space $P_j(H)$ and final space $P_i(H)$. By \autoref{lemma:projection-types-gymnastics}, there exists $\mu_{i, j} \in \complex$ such that $P_iTP_j = \mu_{i, j} V_{j, i}$. Therefore
+    \[
+    T = \sum_{i, j \in I}P_iTP_j = \sum_{i, j \in I}\mu_{i, j}V_{j, i}
+    \]
+\end{proof}
+
+
+\begin{theorem}[Classification of Type $\vnI$ Factors]
+\label{theorem:type1-bh}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a factor, then the following are equivalent:
+    \begin{enumerate}
+        \item $A$ is of type $\vnI$. 
+        \item There exists a minimal projection $P \in \text{Proj}(A)$. 
+        \item For every non-zero $P \in \text{Proj}(A)$, there exists a non-zero minimal projection $Q \in \text{Proj}(A)$ such that $P \ge Q$. 
+        \item There exists a complex Hilbert space $K$ and a *-isomorphism $\pi: A \to B(K)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite{TownesType1}}}. ]
+    (1) $\Rightarrow$ (2): Since $A$ is of type $\vnI$, $A$ admits a non-zero abelian projection $Q \in \text{Proj}(A)$. By \autoref{lemma:abelian-minimal-factor}, $Q$ is minimal. 
+
+    (2) $\Rightarrow$ (3): Let $P \in \text{Proj}(A)$ and $Q \in \text{Proj}(A)$ be a minimal projection. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P \preceq Q$ or $Q \preceq P$. 
+
+    If $P \preceq Q$, then there exists $R \in \text{Proj}(A)$ with $P \sim R \le Q$. In which case, $R$ is minimal, and $P$ is also minimal by (5) of \autoref{lemma:projection-types-gymnastics}. 
+
+    If $Q \preceq P$, then there exists $R \in \text{Proj}(A)$ with $Q \sim R \le P$. By (5) of \autoref{lemma:projection-types-gymnastics}, $R$ is minimal with $R \le P$. 
+
+    (3) $\Rightarrow$ (4): By Zorn's lemma, there exists a maximal orthogonal family $\seqi{P} \subset \text{Proj}(A)$ of minimal projections. Since $\seqi{P}$ is maximal and every non-zero projection admits a non-zero minimal subprojection, $I = \sum_{i \in I}P_i$, and $H = \bigoplus_{i \in I}P_iH$.
+
+    For each $i, j \in I$, by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $P_i \preceq P_j$ or $P_j \preceq P_i$. In both cases, since both projections are minimal, $P_i \sim P_j$. Thus there exists a partial isometry $V_{i, j} \in A$ with initial space $P_i(H)$ and final space $P_j(H)$ such that $P_i = V_{i, j}^*V_{i, j}$ and $P_j = V_{i, j}V_{i, j}^*$. By fixing a particular family\footnote{The partial isometries need not be unique. }, assume without loss of generality that $V_{i, j}^* = V_{j, i}$ for all $i, j \in I$. 
+
+    Fix $i_0 \in I$, let $H_0 = P_{i_0}H$, and define
+    \[
+    U: H \to l^2(I; H_0) \quad (Ux)_i = V_{i, i_0}P_ix
+    \]
+
+    then since $H = \bigoplus_{i \in I}P_iH$ and each $V_{i, i_0}$ is a partial isometry, $U$ is an isometry with inverse
+    \[
+    U^{-1}: l^2(I; H_0) \to H \quad U^{-1}x = \sum_{i \in I}V_{i_0, i}x_i
+    \]
+
+    For each $i \in I$, denote $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then for every $x \in l^2(I; H_0)$, 
+    \[
+    UP_iU^{-1}x = U P_i\sum_{j \in I} V_{i_0, j}x_j = e_{i}V_{i, i_0}P_iV_{i_0, i}x_i = e_iP_{i_0}x_i = e_ix_i
+    \]
+    
+    
+    so $UP_iU^{-1}$ is the projection onto the $i$-th component of $l^2(I; H_0)$. 
+    
+    Let $T \in A$, then by \autoref{lemma:type1-bh-matrix}, there exists $\bracsn{\mu_{i, j}}_{i, j \in I} \subset \complex$ such that $T = \sum_{i, j \in I}\mu_{i,j}V_{i, j}$. In which case, for any $x \in l^2(I; \complex)$ and $v \in H_0$, 
+    \[
+        UTU^{-1}(xv) = UT \sum_{i \in I}V_{i_0, i}x_i v = U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v
+    \]
+
+    For each $i, j \in I$, there exists $\lambda_{i, j} \in \partial B_\complex(0, 1)$ such that $V_{i, j}V_{i_0, i} = \lambda_{i, j}V_{i_0, j}$ by (6) of \autoref{lemma:projection-types-gymnastics}. For every $i \in I$, let $e_i = \one_{\bracs{i}} \in l^2(I; \complex)$, then
+    \begin{align*}
+        UTU^{-1}(xv) &=U\sum_{i, j, k \in I}\mu_{i, j}x_k \cdot V_{i, j}V_{i_0, k} \cdot v =  U\sum_{i, j \in I}\mu_{i, j}x_i \cdot V_{i, j}V_{i_0, i} \cdot v \\
+        &= U\sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot V_{i_0, j} \cdot v = \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot V_{j, i_0}P_jV_{i_0, j} \cdot v \\
+        &= \sum_{i, j \in I}\lambda_{i, j}\mu_{i, j}x_i \cdot e_j \cdot v \in l^2(I; \complex v)
+    \end{align*}
+
+    Moreover, if $v \ne 0$, then $UTU^{-1}(xv) = 0$ for all $x \in l^2(I; \complex)$ implies that $\mu_{i, j} =0 $ for all $i, j \in I$, and $T = 0$. Thus for any $v \in H_0 \setminus \bracs{0}$, the mapping
+    \[
+    \pi: A \to B(l^2(I; \complex v)) \quad T \mapsto UTU^{-1}|_{l^2(I; \complex v)}
+    \]
+
+    is an injective $*$-homomorphism. 
+
+    Finally, since $\ol{B_A(0, 1)}$ is weak-operator compact and $U$ is an isometry, $\pi(\ol{B_A(0, 1)})$ is also weak-operator compact, and $\pi(A) \subset B(l^2(I; \complex v))$ is a von Neumann algebra. Let $i, j \in I$, then for each $x \in l^2(I; \complex)$, 
+    \begin{align*}
+    \pi(V_{i, j})(xv) &= x_i \cdot e_j \cdot V_{j, i_0}V_{i, j}V_{i_0, i} \cdot v = \lambda_{i, j}x_i \cdot e_j \cdot V_{j, i_0}V_{i_0, j} \cdot v \\
+    &= \lambda_{i, j}x_i \cdot e_j \cdot v
+    \end{align*}
+    
+
+    so $\pi(V_{i, j}) = \lambda_{i, j} e_jv \otimes e_iv$. As $\bracsn{e_iv}_{i \in I}$ is an orthonormal basis for $l^2(I; \complex v)$, $\pi(A) = B(l^2(I; \complex v))$. 
+
+    (4) $\Rightarrow$ (1): Identify $A = \pi(A) = B(K)$, and let $T \in Z(A)$. For each $v \in K$ with $\norm{v}_K = 1$, $T(v \otimes v) = (v \otimes v)T$, so $Tv = T(v \otimes v)v = (v \otimes v)Tv$, and there exists $\lambda \in \complex$ such that $Tv = \lambda v$. 
+    
+    For any $w \in K$ linearly independent from $v$, there exists $\mu \in \complex$ with $Tw = \mu w$, and $\rho \in \complex$ with $T(v + w) = \rho(v + w)$. In which case, $\rho v + \rho w = \lambda v + \mu w$, so $\lambda = \rho = \mu$, and $T = \lambda I$. Therefore $A$ is a factor. 
+
+    For any $v \in K$ with $\norm{v}_K = 1$, $v \otimes v$ is a minimal, and hence abelian projection by (1) of \autoref{lemma:projection-types-gymnastics}. As $v \otimes v \le I$, $A = B(K)$ is of type $\vnI$. 
+\end{proof}
+
+

src/op/vn/vn.tex

@@ -6,7 +6,7 @@
     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 family of projections $\mathcal{P} \subset A_{sa}$, $\sup(\mathcal{P}) \in A$ is the projection onto $\ol{\bigcup_{P \in \mathcal{P}}P(H)}$. 
+        \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}
 
     and
@@ -46,7 +46,7 @@
 
     for all $S \in \cf$ with $S \ge T$. As such a $T$ exists for all $\eps > 0$, $R = \sotlim_{T \in \cf}T$. 
 
-    (2): Assume without loss of generality that $\mathcal{P}$ is directed. By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
+    (2): By (1), $\sup(\mathcal{P})$ exists in $A$. Since the set of projections in $B(H)$ is strong-operator closed, $\sup(\mathcal{P}) = \sotlim_{P \in \mathcal{P}}P$ is also a projection. For each $x \in \bigcup_{P \in \mathcal{P}}P(H)$, there exists $P \in \mathcal{P}$ with $Px = x$. In which case, 
     \[
     \dpn{\sup(\mathcal{P})x, x}{H} \ge \dpn{Px, x}{H} = \dpn{x, x}{H} = \norm{x}_H^2
     \]
@@ -89,7 +89,7 @@
 
     By (3) applied to $TT^*/\norm{TT^*}_{B(H)}$, the orthogonal projection onto $\ol{T(H)}$ is in $A$. 
 
-    (5): Let $\mathcal{P}$ be the set of all projections in $A$, then $\mathcal{P} \subset A_{sa}$ is bounded and directed. By (2), $P = \sup_{Q \in \mathcal{P}}Q \in A$, which is the maximum projection in $A$. 
+    (5): Let $\mathcal{P}$ be the set of all projections in $A$, and $\cf \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$. For each $F \in \cf$, let $P_F$ be the projection onto $\braks{\sum_{P \in F}P}(H)$, then $P_F \ge P$ for all $P \in F$ and $P_F \in A$ by (4). Since $\bracsn{P_F}_{F \in \cf}$ is a bounded and directed family of projections, $\sup_{F \in \cf}P_F \in A$ by (1). As $\sup_{F \in \cf}P_F \in \mathcal{P}$, it is the maximum projection in $A$. 
 
     Let $T \in A$, then by (4), $P$ is greater than the projection onto $\ol{T(H)}$, so $PT = T$. On the other hand, since $PT^* = T^*$, $TP = T$ as well. Therefore $P$ is the multiplicative identity in $A$. 
 \end{proof}
@@ -181,6 +181,12 @@
     Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a $C^*$-subalgebra, then $A$ is a \textbf{von Neumann algebra acting on $H$} if $A$ is closed in the strong operator topology. 
 \end{definition}
 
+\begin{definition}[Factor]
+\label{definition:vna-factor}
+    Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a von Neumann algebra, then $A$ is a \textbf{factor} if $Z(A) = \complex I$. 
+\end{definition}
+
+
 \begin{theorem}[Kaplansky Density Theorem]
 \label{theorem:kaplansky-density}
     Let $H$ be a Hilbert space, $A \subset B(H)$ be a $C^*$-subalgebra, and $B$ be the strong-operator closure of $A$, then:
@@ -225,7 +231,7 @@
     \]
 
     then $T' \in \ol{B_{M_2(B)_{sa}}(0, 1)}$. By (1), there exists a net $\angles{(R_\gamma, S_\gamma, T_\gamma)}_{\gamma \in C} \subset A^3$ such that:
-    \begin{enumerate}
+    \begin{enumerate}[label=(\roman*)]
         \item For each $\gamma \in C$, 
         \[
         \norm{\begin{bmatrix} R_\gamma & T_\gamma \\ T^*_\gamma & S_\gamma \end{bmatrix}}_{B(H^2)} \le 1

src/topology/main/baire.tex

@@ -16,7 +16,7 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: Let $\seq{A_n} \subset 2^X$ be closed with empty interior, then $\seq{A_n}$ are nowhere dense. Hence $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
 
-    $(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_n = U_n^c$, then $A_n$ is closed. For any $\emptyset U \subset A_n$ open, $U \cap U_n \ne \emptyset$ by density of $U_n$, so $A_n$ has empty interior.
+    $(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_n = U_n^c$, then $A_n$ is closed. For any $\emptyset \ne U \subset A_n$ open, $U \cap U_n \ne \emptyset$ by density of $U_n$, so $A_n$ has empty interior.
 
     Suppose that $\bigcap_{n \in \natp}U_n$ is not dense, then there exists $\emptyset \ne V \subset X$ open such that $\bigcup_{n \in \natp}A_n \supset V$, which contradicts the fact that $\bigcup_{n \in \natp}A_n$ has non-empty interior.
 

src/topology/main/cube.tex

@@ -69,7 +69,7 @@
     \mathcal{S} = \bracsn{(E, F) \in \mathcal{B}^2 | \ol{E} \subset F}
     \]
 
-    By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f|_E = 1$ and $f|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
+    By \hyperref[Urysohn's Lemma]{lemma:urysohn}, for each $(E, F) \in \mathcal{S}$, there exists $f_{EF} \in C(X; [0, 1])$ such that $f_{EF}|_E = 1$ and $f_{EF}|_{F^c} = 0$. For any $x \in X$ and $U \in \cn^o_X(x)$, there exists $E, F \in \mathcal{B}$ such that $x \in E \subset \ol{E} \subset F \subset U$. Thus $f_{EF}(x) = 1$ and $f_{EF}|_{U^c} = 0$. Therefore
     \[
     \cf = \bracsn{f_{EF}|(E, F) \in \mathcal{S}} \subset C(X; [0, 1])
     \]

src/topology/notation.tex

@@ -26,6 +26,14 @@
     $f \prec U$ & $f \in C_c(X; [0,1])$ with $\mathrm{supp}(f) \subset U$. & \autoref{definition:compactly-supported-01} \\
     $C_0(X; E)$ & Continuous functions vanishing at infinity. & \autoref{definition:vanish-at-infinity} \\
     $BC(X; E)$ & Bounded continuous functions $X \to E$. & \autoref{definition:bounded-continuous-function-space} \\
+    % ---- $L^p$ Spaces ----
+    $B^\infty(X; E)$ & Bounded $E$-valued strongly measurable functions on $X$. & \autoref{definition:bounded-borel-function} \\
+    $B^\infty(X)$ & Bounded $\complex$-valued Borel measurable functions on $X$. & \autoref{definition:bounded-borel-function} \\
+    $\mathcal{L}^p(X; E)$, $\mathcal{L}^p(\mu; E)$, $\mathcal{L}^p(X, \cm, \mu; E)$ & $E$-valued $p$-integrable functions on $X$. & \autoref{definition:lp-unequivalence} \\
+    $\norm{f}_{L^p(X; E)}$ & $L^p$ norm of $f$: $\braks{\int \norm{f}_E^p d\mu}^{1/p}$. & \autoref{definition:lp-unequivalence} \\
+    $\mathcal{L}^\infty(X; E)$, $\mathcal{L}^\infty(\mu; E)$, $\mathcal{L}^\infty(X, \cm, \mu; E)$ & $E$-valued essentially bounded functions on $X$. & \autoref{definition:esssup} \\
+    $\norm{f}_{\mathcal{L}^\infty(X; E)}$ & Essential supremum of $f$. & \autoref{definition:esssup} \\
+    $L^p(X, \cm, \mu; E)$ & $E$-valued $L^p$ space on $(X,\cm,\mu)$; quotient of $\mathcal{L}^p$ by a.e.-equality. & \autoref{definition:lp} \\
 
     % DST 
     $\mathscr{N}$ & The Baire space. & \autoref{definition:the-baire-space} \\