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

@@ -283,3 +283,13 @@
  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/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/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/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/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/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/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} \\
@@ -16,11 +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:spectr}
+    $E_{x, y}$ & $E_{x, y}(B) = \dpn{E(B)x, y}{H}$. & \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/index.tex

@@ -4,5 +4,8 @@
 \input{./topologies.tex}
 \input{./cayley.tex}
 \input{./vn.tex}
+\input{./commutative.tex}
 \input{./spec.tex}
-\input{./fc.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

@@ -199,9 +199,9 @@
 
 \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 $I \in A$, $B \subset B(H)$ 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:
+    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 $\mu_i$ 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 $i \in I$, 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
@@ -288,6 +288,31 @@
     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 $\bracsn{E_{x, y}|x, y \in H}$.

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: