First drafts.

src/op/ii1/fg.tex

@@ -0,0 +1,24 @@
+\section{The Fundamental Group}
+\label{section:vna-fundamental-group}
+
+\begin{definition}[Induced von Neumann Algebra]
+\label{definition:induced-vna-commutant}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $q \in M'$, then $Mq$ is a von Neumann algebra on $qH$, known as the \textbf{induced von Neumann algebra} of $M$ with respect to $q$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:induced-vna-commutant-gymnastics}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $p \in M$, then:
+    \begin{enumerate}
+        \item For any central projection $e \in Z(pMp)$, $e = Z(e)p$. 
+        \item $Z(pMp) = Z(M)p$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.2.1]{AnantharamanPopa}}}. ]
+    (1): Since $(p - e)Me = (p - e)pMpe = 0$, $(p - e)yey^* = 0$ for all $y \in U(M)$, so $(p - e)z(e) = 0$ by \autoref{proposition:central-support-vna}, and $e = z(e)p$. 
+
+    (2): By \autoref{theorem:vn-projection-norm-dense}.
+\end{proof}
+
+
+

src/op/ii1/index.tex

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

src/op/ii1/trace.tex

@@ -0,0 +1,190 @@
+\section{Traces}
+\label{section:tracial-von-neumann-algebra}
+
+\begin{definition}[Trace]
+\label{definition:trace}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\tau: M_+ \to [0, \infty]$, then $\tau$ is a \textbf{trace} on $M$ if:
+    \begin{enumerate}
+        \item For each $x, y \in M$ and $\lambda \ge 0$, $\tau(\lambda x + y) = \lambda \tau(x) + \tau(y)$. 
+        \item For each $x \in M$ and $u \in U(M)$, $\tau(uxu^*) = \tau(x)$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Tracial von Neumann Algebra]
+\label{definition:tracial-von-neumann-algebra}
+    Let $M$ be a von Neumann algebra and $\tau: M \to \complex$ be a faithful tracial state, then $(M, \tau)$ is a \textbf{tracial von Neumann algebra} if $\tau$ is normal. 
+\end{definition}
+
+\begin{definition}[Standard Representation]
+\label{definition:tracial-vna-standard}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then its \hyperref[GNS]{definition:gns-triple} representation on $L^2(M, \tau)$ is the \textbf{standard representation} of $(M, \tau)$.  
+\end{definition}
+
+\begin{proposition}
+\label{proposition:cstar-trace-vna}
+    Let $M$ be a unital $C^*$-algebra, $\tau: M \to \complex$ be a faithful tracial state, and $(L^2(M, \tau), \pi, 1_M)$ be its \hyperref[GNS]{definition:gns-triple} triple, then:
+    \begin{enumerate}
+        \item The uniform structure of pointwise convergence\footnote{Strong operator topology.} on $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ coincides with the uniform structure of $\ol{B_M(0, 1)} \subset L^2(M, \tau)$ induced by the $L^2(M, \tau)$ norm. 
+        \item $\pi(M) \subset B(L^2(M, \tau))$ is a von Neumann algebra if and only if $\ol{B_M(0, 1)}$ is complete in $L^2(M, \tau)$. In which case, $\tau$ is normal. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.6.4]{AnantharamanPopa}}}. ]
+    (1): For any $x, y \in \ol{B_M(0, 1)}$, 
+    \begin{align*}
+    \norm{x - y}_{L^2(M, \tau)}^2 &= \tau((x - y)^*(x - y)) = \dpn{\pi(x - y)^*\pi(x - y)1_M, 1_M}{L^2(M, \tau)} \\
+    &= \dpn{\pi(x - y)1_M, \pi(x - y)1_M}{L^2(M, \tau)} = \norm{\pi(x - y)1_M}_{L^2(M, \tau)}
+    \end{align*}
+
+    so the uniform structure of pointwise convergence is stronger than the uniformity induced by the $L^2(M, \tau)$ norm.
+    
+    On the other hand, by the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the uniform structure of pointwise convergence on $\ol{B_M(0, 1)}$ coincides with the uniform structure of pointwise convergence on any dense subset of $L^2(M, \tau)$. In particular, this includes $M \subset L^2(M, \tau)$. For each $x, y \in \ol{B_M(0, 1)}$ and $t \in M$, since $\tau$ is tracial, 
+    \begin{align*}
+        \norm{(x - y)t}_{L^2(M, \tau)}^2 &= \tau(t(x - y)^*(x - y)t) \le \tau(t^*t(x - y)^*(x - y)) \\
+        &\le \norm{t}_M \tau((x - y)^*(x - y)) = \norm{t}_M^2 \norm{x - y}_{L^2(M, \tau)}^2
+    \end{align*}
+
+    so the uniformity induced by the $L^2(M, \tau)$ norm is stronger than the uniform structure of pointwise convergence.
+    
+    (2): Using (1), it is sufficient to show that $\ol{B_M(0, 1)} \subset B(L^2(M, \tau))$ is complete if and only if it is strong-operator closed. By \autoref{proposition:complete-closed}, $\ol{B_M(0, 1)}$ is complete if and only if it is closed in $H^H$ with respect to the product topology. Finally, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that $\ol{B_M(0, 1)}$ is closed in $H^H$ if and only if it is strong-operator closed in $B(L^2(M, \tau))$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-separable-condition}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then the following are equivalent:
+    \begin{enumerate}
+        \item $M$ is separable. 
+        \item $\ol{B_M(0, 1)}$ is strong-operator separable. 
+        \item $\ol{B_M(0, 1)}$ is separable in $L^2(M, \tau)$. 
+        \item $L^2(M, \tau)$ is separable. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $H$ be a separable Hilbert space with $M \subset B(H)$ and $D \subset H$ be countable and dense. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the strong operator topology on $\ol{B_M(0, 1)}$ coincides with the topology of pointwise convergence on $D$. By \autoref{theorem:uniform-metrisable} and \autoref{proposition:separable-product}, $H^D$ is metrisable and separable. Therefore $\ol{B_M(0, 1)} \subset H^D$ is also metrisable and separable.
+
+    (2) $\Rightarrow$ (3): By \autoref{proposition:cstar-trace-vna}. 
+
+    (3) $\Rightarrow$ (4): Since the \hyperref[GNS representation]{definition:gns-triple} is cyclic, $M$ is dense in $L^2(M, \tau)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-von-neumann-factor}
+    Let $(M, \tau)$ be a tracial von Neumann factor, then exactly one of the following holds:
+    \begin{enumerate}
+        \item There exists $n \in \natp$ such that $M \iso M_n(\complex)$. 
+        \item $M$ is of type $\vnII_1$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Suppose that $M$ has a minimal projection. By \autoref{theorem:type1-bh}, there exists a complex Hilbert space $H$ such that $M \iso B(H)$. Let $\seqi{e} \subset H$ be an orthonormal basis, then $e_i \otimes e_i \in \text{Proj}(M)$ is minimal for all $i \in I$, and all such rank one projections are equivalent. As such, for any $j \in I$, 
+    \[
+    \tau(1_M) = \sum_{i \in I}\tau(e_i \otimes e_i) = |I| \cdot \tau(e_j \otimes e_j)
+    \]
+
+    so $I$ must be finite, and $M \iso M_{|I|}(\complex)$. 
+
+    (2): Suppose that $M$ has no minimal projections. By \autoref{lemma:abelian-minimal-factor}, $M$ has no abelian projections. As $\tau$ is a faithful tracial state, $1_M$ is finite. Therefore $M$ is a $\vnII_1$ factor. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:diffuse-factor-subsplit}
+    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exists non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 \le p$.  
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
+    Since $p$ is not minimal, there exists a projection $q \le p$ with $0 \ne q \ne p$. As $M$ is a factor, $Z(q)Z(p - q) = 1_M$. By \autoref{proposition:central-support-mvn}, there exists non-zero projections $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$, $p_1 \le q$, and $p_2 \le p - q$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:diffuse-factor-split}
+    Let $M$ be a diffuse factor, $p \in \text{Proj}(M)$ be a non-zero projection, then there exists $p_1, p_2 \in \text{Proj}(M)$ with $p_1 \sim p_2$ and $p_1 + p_2 = p$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 4.1.1]{AnantharamanPopa}}}. ]
+    Let
+    \[
+    \mathcal{F} = \bracsn{(q, r) \in \text{Proj}(M)^2|q \sim r, q \perp r, q + r \le p}
+    \] 
+
+    and $\mathcal{M} \subset \mathcal{F}$ be a maximal family such that $\bracsn{q, r|(q,r) \in \mathcal{M}}$ are all mutually orthogonal. Let $q = \sum_{(q, r) \in \mathcal{M}}q$ and $r = \sum_{(q, r) \in \mathcal{M}}r$, then by \autoref{lemma:mvn-equivalent-direct-sum}, $q \sim r$, $q \perp r$, and $q + r \le p$. 
+
+    If $q + r \ne p$, then by \autoref{lemma:diffuse-factor-subsplit} applied to $p - (q + r)$, there exists non-zero projections $(q', r') \in \text{Proj}(M)$ with $q' \sim r$, $q' \perp r'$, and $q + r \le  p - (q + r)$, which contradicts the maximality of $\mathcal{M}$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:diffuse-factor-dyadic}
+    Let $M$ be a diffuse factor, then for each $n \in \natp$, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. If $\tau: M \to \complex$ is a tracial state, then $\tau(p_j) = 2^{-n}$ for all $1 \le j \le 2^n$.
+\end{corollary}
+\begin{proof}
+    By \autoref{lemma:diffuse-factor-split}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-simple}
+    Let $M$ be a factor, then $M$ admits at most one tracial state. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.3]{AnantharamanPopa}}}. ]
+    Using \autoref{proposition:tracial-von-neumann-factor}, assume without loss of generality that $M$ is diffuse.
+    
+    Let $p \in \text{Proj}(M)$ with $0 \le p < 1_M$. Let $n \in \natp$, then by \autoref{corollary:diffuse-factor-dyadic}, there exist mutually orthogonal equivalent projections $\bracsn{p_j}_1^{2^n} \subset \text{Proj}(M)$ such that $\sum_{j = 1}^{2^n}p_j = 1$. By the \hyperref[corollary:vna-factor-comparability]{theorem:vna-comparability}, there exists a unique $0 \le k \le 2^n - 1$ such that
+    \[
+    \sum_{j = 1}^k p_j \preceq p \prec \sum_{j = 1}^{k+1}p_j
+    \]
+
+    As such, for any tracial state $\tau: M \to \complex$, 
+    \[
+    \frac{k}{2^n} \le \tau(p) \le \frac{k+1}{2^n}
+    \]
+
+    so $\tau(p)$ is independent of the choice of $\tau$. By \autoref{theorem:vn-projection-norm-dense}, $\tau$ is unique.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-unique-trace}
+    Let $(M, \tau)$ be a tracial von Neumann algebra, then $M$ is a factor if and only if $\tau$ is the unique normal faithful tracial state on $M$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.4]{AnantharamanPopa}}}. ]
+    If $M$ is a factor, then $\tau$ is unique by \autoref{proposition:tracial-factor-simple}. Suppose that $M$ is not a factor, then $M$ admits a non-trivial central projection $z \in Z(M)$. In which case, for any $\alpha \in (0, 1)$, 
+    \[
+    \tilde \tau(x) = \frac{\alpha}{\tau(z)} \tau(xz) + \frac{1 - \alpha}{1 - \tau(z)}\tau(x(1 - z))
+    \]
+
+    is also a normal faithful tracial state. In particular, 
+    \[
+    \tilde \tau(z) = \frac{\alpha}{\tau(z)} \tau(z) + \frac{1 - \alpha}{1 - \tau(z)}\tau(z(1 - z)) = \alpha
+    \]
+
+    so if $\alpha \ne \tau(z)$, then $\tilde \tau \ne \tau$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:tracial-factor-calkin}
+    Let $(M, \tau)$ be a tracial factor, then $M$ admits no non-trivial two-sided ideal. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.5]{AnantharamanPopa}}}. ]
+    Let $I \subset M$ be a non-zero two-sided ideal. By \autoref{proposition:vna-two-sided-ideal}, $I$ admits a non-zero projection $e \in I$. Let $\mathcal{P} \subset \text{Proj}(P)$ be a maximal family of mutually orthogonal projections equivalent to $e$. Since $\tau$ is faithful, $\tau(p) > 0$ for all $p \in \mathcal{P}$, so $\mathcal{P}$ is finite. 
+
+    Let $q = 1_M - \sum_{p \in \mathcal{P}}p$, then by the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $q \preceq e$ or $e \prec q$. If $e \prec q$, then there exists a subprojection $p \le q$ with $e \sim p$, which contradicts the maximality of $\mathcal{P}$. Thus $q \preceq e$. 
+
+    Let $\mathcal{Q} = \mathcal{P} \sqcup \bracsn{q}$, then $p \preceq e$ for all $p \in \mathcal{Q}$ and $1_M = \sum_{p \in \mathcal{Q}}p$. For each $p \in \mathcal{Q}$, let $v_p \in M$ be a partial isometry such that $v_pv_p^* = p$ and $v_p^*v_p \le e$. As such, $p = v_pev_p^* \in I$ for all $p \in \mathcal{Q}$. Therefore $1_M = \sum_{p \in \mathcal{Q}}p \in I$, and $I = M$.  
+\end{proof}
+
+\begin{proposition}
+\label{proposition:ii1-trace-projection-equivalence-classes}
+    Let $M$ be a $\vnII_1$ factor with tracial normal state $\tau: M \to \complex$, then the mapping $p \mapsto \tau(p)$ induces a bijection from the equivalence classes of $\text{Proj}(M)$ to $[0, 1]$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 4.1.6]{AnantharamanPopa}}}. ]
+    (Injective): Let $p, q \in \text{Proj}(M)$ with $\tau(p) = \tau(q)$. By the \hyperref[comparability theorem]{corollary:vna-factor-comparability}, either $p \prec q$, $p \sim q$, or $q \prec p$. Since $\tau$ is faithful, $p \prec q$ implies that $\tau(p) < \tau(q)$, and $q \prec p$ implies that $\tau(q) < \tau(p)$. Therefore the only possiblity is $p \sim q$. 
+
+    (Surjective): By \autoref{lemma:diffuse-factor-split}, there exists a family $\bracsn{p_r|r \in \mathbb D \cap [0, 1]} \subset \text{Proj}(M)$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item For each dyadic rational number $r \in \mathbb D \cap [0, 1]$, $\tau(p_r) = r$. 
+        \item For each $r, s \in \mathbb D \cap [0, 1]$ with $r \le s$, $p_r \le p_s$. 
+    \end{enumerate}
+
+    For each $t \in [0, 1]$, let $p_t = \sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}$, then since $\tau$ is normal, 
+    \begin{align*}
+    \tau(p_t) &= \tau(\sup\bracsn{p_r|r \in \mathbb D \cap [0, 1], r \le t}) \\
+    &= \sup\bracsn{\tau(p_r)|r \in \mathbb D \cap [0, 1], r \le t} = t
+    \end{align*}
+    
+    
+\end{proof}
+

src/op/vn/weak.tex

@@ -0,0 +1,210 @@
+\section{Normal Maps}
+\label{section:ultraweak}
+
+\begin{lemma}[Jordan Decomposition]
+\label{lemma:weak-operator-jordan}
+    Let $H$ be a complex Hilbert space, $M \subset H$ be a von Nuemann algebra, and $\omega: M \to \complex$ be a weak-operator continuous linear functional, then there exists positive weak-operator continuous functionals $\bracsn{\omega_k}_1^4$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$. 
+\end{lemma}
+\begin{proof}
+    Assume without loss of generality that there exists $\xi, \eta \in H$ such that $\dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H}$ for all $x \in M$. In which case, by \hyperref[polarisation]{proposition:polarisation-complex}, 
+    \[
+    \dpn{x, \omega}{M} = \dpn{x\xi, \eta}{H} = \frac{1}{4}\sum_{k = 0}^3 i^k \dpn{x(\xi + i^k\eta), \xi + i^k\eta}{H}
+    \]
+    
+    for all $x \in M$. 
+\end{proof}
+
+
+\begin{lemma}[Radon-Nikodym]
+\label{lemma:vna-radon-nikodym}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, $\omega: M \to \complex$ be a positive linear functional, and $\xi \in H$ such that $\omega(x) \le \dpn{x\xi, \xi}{H}$ for all $x \in M_+$, then there exists $x' \in M_+'$ such that $\omega(x) = \dpn{x(x'\xi), x'\xi}{H}$ for all $x \in M$. 
+\end{lemma}
+\begin{proof}
+    By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for each $x, y \in M$, 
+    \[
+    |\omega(y^*x)|^2 \le \omega(x^*x)\omega(y^*y) \le \norm{x\xi}_H^2 \norm{y\xi}_H^2
+    \]
+
+    so the mapping $(x, y) \mapsto \omega(y^*x)$ is a bounded sesquilinear form on $M\xi$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $z \in B(\ol{M\xi})$ such that $\omega(y^*x) = \dpn{zx\xi, y\xi}{H}$ for all $x, y \in M$. In particular, $0 \le \omega(x^*x) = \dpn{zx\xi, x\xi}{H}$ for all $x \in M$, so $z \ge 0$. Moreover, for any $x, y, t \in M$, 
+    \[
+        \dpn{ztx\xi, zy\xi}{H} = \omega(y^*tx) = \omega((t^*y)^*x) = \dpn{zx\xi, t^*y\xi}{H} = \dpn{tzx\xi, y\xi}{H}
+    \]
+
+    so $t \in M'_+$. Finally, let $p: H \to \ol{M\xi}$ be the orthogonal projection onto $\ol{M\xi}$, and $x' = (zp)^{1/2} \in M'$, then for each $x \in M$, 
+    \[
+        \omega(x) = \dpn{zx\xi, \xi}{H} = \dpn{zpx\xi, \xi}{H} = \dpn{x(zp)^{1/2}\xi, (zp)^{1/2}\xi}{H}
+    \]
+    
+\end{proof}
+
+\begin{lemma}
+\label{lemma:completely-additive-projection-choose}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\varphi: M \to \complex$ be a positive linear functional. If for any mutually orthogonal projections $\mathcal{P} \subset M$, $\varphi\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in P}\varphi(p)$, then there exists a non-zero projection $p \in M$ and $\xi \in pH$ such that $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$. 
+\end{lemma}
+\begin{proof}[Proof, {{\cite[Lemma 2.5.6.]{AnantharamanPopa}}}. ]
+    Let $\eta \in H$ with $\varphi(1_M) < \dpn{\eta, \eta}{H}$. Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections with $\varphi(p) \ge \dpn{p\eta, \eta}{H}$ for all $p \in \mathcal{P}$. By assumption, 
+    \begin{align*}
+    \varphi\paren{\sum_{p \in \mathcal{P}}p} &\ge \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H \\
+        \varphi\paren{1_M - \sum_{p \in \mathcal{P}}p}&= \varphi(1_M) - \varphi\paren{\sum_{p \in \mathcal{P}}p} \\
+        &< \dpn{\eta, \eta}{H} - \angles{\sum_{p \in \mathcal{P}}p\eta, \eta}_H  \ge 0
+    \end{align*}
+
+    Let $p := 1_M - \sum_{q \in \mathcal{P}}q > 0$, then by maximality of $\mathcal{P}$, $\varphi(q) < \dpn{q\eta, \eta}{H}$ for all subprojections $q \le p$. 
+    
+    Let $x \in (pMp)_+$ with $0 \le x \le 1$. By \autoref{theorem:vn-projection-norm-dense}, there exists projections $\seq{q_n} \subset pMp$ such that $x = \sum_{n = 1}^\infty \frac{q_n}{2^n}$. In which case\footnote{Here, only continuity of $\varphi$ is used.}, 
+    \[
+    \dpn{x\eta, \eta}{H} = \sum_{n = 1}^\infty 2^{-n}\dpn{q_n\eta, \eta}{H} \ge \sum_{n = 1}^\infty \frac{\varphi(q^n)}{2^n} = \varphi(x)
+    \]
+
+    Therefore $\dpn{x\eta, \eta}{H} \ge \varphi(x)$ for all $x \in (pMp)_+$. By the \hyperref[Radon-Nikodym theorem for positive functionals]{lemma:vna-radon-nikodym}, there exists $\xi \in pH$ with $\varphi(x) = \dpn{x\xi, \xi}{H}$ for all $x \in pMp$. 
+\end{proof}
+
+
+
+\begin{definition}[Normal Linear Functional]
+\label{definition:vna-normal-functional}
+    Let $H$ be a complex Hilbert space, $M \subset B(H)$ be a von Neumann algebra, and $\omega: M \to \complex$ be a positive linear functional, then the following are equivalent:
+    \begin{enumerate}
+        \item For any bounded, upward-directed family $A \subset M$, $\omega(\sup(A)) = \sup(\omega(A))$.
+        \item For any mutually orthogonal projections $\mathcal{P} \subset M$, $\omega\paren{\sum_{p \in \mathcal{P}}p} = \sum_{p \in P}\omega(p)$. 
+        \item There exists $\seq{\xi_n} \in l^2(\natp; H)$ such that $\omega(x) = \sum_{n \in \natp}\dpn{x\xi_n, \xi_n}{H}$ for all $x \in M$. 
+        \item $\omega$ is the norm limit of a sequence of weak-operator continuous positive linear functionals. 
+        \item $\omega$ is weak-operator continuous on bounded subsets of $M$.
+        \item $\omega$ is strong-operator continuous on bounded subsets of $M$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 2.5.5]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (3): Let $\mathcal{P} \subset M$ be a maximal family of mutually orthogonal projections such that for each $p \in \mathcal{P}$, there exists $\xi_p \in pH$ such that $\omega(x) = \dpn{x\xi_p, \xi_p}{H}$ for all $x \in pMp$. Let $q = \sum_{p \in \mathcal{P}}p$, then by maximality of $\mathcal{P}$ and \autoref{lemma:completely-additive-projection-choose} applied to $(1 - q)M(1 - q)$, $\sum_{p \in \mathcal{P}}p = 1_M$. 
+
+    For each $x \in M$ and $p \in \mathcal{P}$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, 
+    \[
+    |\omega(xp)|^2 \le \omega(1) \cdot \omega(px^*xp) = \omega(1) \cdot \norm{x\xi_p}_H^2
+    \]
+
+    so the mapping $y \xi_p \mapsto \omega(yp)$ extends into a bounded linear functional on $\ol{M\xi_p}$. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-hilbert}, there exists $\eta_p \in \ol{M\xi_p}$ such that $\omega(xp) = \dpn{x\xi_p, \eta_p}{H}$ for all $x \in M$. 
+
+    Let $\mathcal{F} \subset 2^{\mathcal{P}}$ be the collection of all finite subsets of $\mathcal{P}$, directed by inclusion. For each $F \in \mathcal{F}$, let $q_F = \sum_{p \in F}p$ and $\omega_F \in M^*$ be defined by $\omega_F(x) = \omega(q_Fxq_F)$, then for each $x \in M$, 
+    \[
+    \omega_F(x) = \sum_{p \in F}\omega(q_Fxp) = \sum_{p \in F}\dpn{q_Fx\xi_p, \eta_p}{H}
+    \]
+
+    so $\omega_F$ is weak-operator continuous. Finally, for any $x \in M$, 
+    \begin{align*}
+        |\omega(x) - \omega_F(x)| &= |\omega(x) - \omega(xq_F)| + |\omega(xq_F) - \omega(q_Fxq_F)| \\
+        &= |\omega(x(1_M - q_F))| + |\omega((1 - q_F)xq_F)| \\
+        &\le 2\norm{x}_M\omega(1)^{1/2} \cdot \omega(1 - q_F)^{1/2}
+    \end{align*}
+    
+    By (2), $\lim_{F \in \mathcal{F}}\omega(1 - q_F)^{1/2} = 0$. Therefore $\omega = u\text{-}\lim_{F \in \mathcal{F}}\omega_F$. 
+\end{proof}
+
+\begin{lemma}
+\label{lemma:normal-functional-bound}
+    Let $M$ be a von Neumann algebra, $\varphi, \omega \in M^*$ be positive linear functionals. If $\omega$ is normal and $\varphi \le \omega$, then $\varphi$ is also normal. 
+\end{lemma}
+\begin{proof}
+    By (2) of \autoref{definition:vna-normal-functional}. 
+\end{proof}
+
+\begin{definition}[Normal Positive Map]
+\label{definition:normal-linear-map}
+    Let $M, N$ be von Neumann algebras and $\phi: M \to N$ be a positive linear map, then the following are equivalent:
+    \begin{enumerate}
+        \item For any bounded, upward-directed family $A \subset M$, $\phi(\sup(A)) = \sup(\phi(A))$. 
+        \item For any normal, positive linear functional $\omega \in N^*$, $\omega \circ \phi$ is a normal, positive linear functional.
+        \item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the weak operator topologies. 
+    \end{enumerate}
+
+    If $\phi$ is in addition a *-homomorphism, then the above are also equivalent to
+    \begin{enumerate}[start=3]
+        \item $\phi$ restricted to bounded subsets of $M$ is continuous with respect to the strong operator topologies. 
+    \end{enumerate}
+    
+    If the above holds, then $\phi$ is \textbf{normal}.
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 2.5.8]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (3): Let $\omega \in M^*$ be a weak-operator continuous functional. By \autoref{lemma:weak-operator-jordan}, there exists positive weak-operator continuous functionals $\bracsn{\omega_k}_1^4$ such that $\omega = \sum_{k = 0}^3 i^k \omega_k$. By assumption, $\omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. In which case, $\omega \circ \phi = \sum_{k = 0}^3 i^k \omega_k \circ \phi$ is weak-operator continuous on bounded subsets of $M$. Therefore $\phi$ is weak-operator continuous on bounded subset of $M$.
+
+    (3) $\Rightarrow$ (1), (4) $\Rightarrow$ (1): By (1) of \autoref{theorem:existence-of-projections-vna}.
+
+    (3) $\Rightarrow$ (4)\footnote{Central here is the observation that $x_\alpha \to 0$ in the weak operator topology i.f.f. $x_\alpha^*x \to 0$ in the strong operator topology. }: Suppose that $M$ acts on the Hilbert space $H$, and $N$ acts on the Hilbert space $K$. Let $\angles{x_\alpha}_{\alpha \in A} \subset M$ be a bounded net converging to $x \in M$, then for each $\xi \in H$, 
+    \[
+    \lim_{\alpha \in A}\dpn{(x - x_\alpha)^*(x - x_\alpha)\xi, \xi}{H} = \lim_{\alpha \in A}\normn{(x - x_\alpha)\xi}_H^2 = 0
+    \]
+
+    By \hyperref[polarisation]{proposition:polarisation-complex}, $\wotlim_{\alpha \in A}(x - x_\alpha)^*(x - x_\alpha) = 0$. As $\angles{x_\alpha}_{\alpha \in A}$ is bounded, 
+    \[
+    \wotlim_{\alpha \in A}\phi((x - x_\alpha)^*(x - x_\alpha)) = 0
+    \]
+
+    as well. In particular, for any $\xi \in K$, 
+    \[
+    \lim_{\alpha \in A}\norm{\phi(x - x_\alpha)}_K^2 = \lim_{\alpha \in A}\dpn{\phi(x - x_\alpha)^*\phi(x - x_\alpha)\xi, \xi}{K} = 0
+    \]
+
+    Therefore $\phi$ restricted to bounded subsets of $M$ is strong-operator continuous.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:vna-isomorphism-normal}
+    Let $M, N$ be von Neumann algebras, then every isomorphism $\phi: M \to N$ is normal. In particular, the weak and strong operator topologies on $M$, restricted to bounded subsets, is independent of its representation. 
+\end{proposition}
+\begin{proof}
+    As $\phi$ is an isomorphism, it is an order isomorphism. By (1) of \autoref{definition:normal-linear-map}, $\phi$ is normal. By (3) and (4) of \autoref{definition:normal-linear-map}, its restriction to bounded subsets of $M$ is both weak-operator and strong-operator continuous. 
+\end{proof} 
+
+\begin{proposition}
+\label{proposition:vna-normal-bound}
+    Let $M, N$ be von Neumann algebras, $\omega \in M^*$ and $\psi \in N^*$ be normal positive linear functionals with $\psi$ being faithful, and $\phi: M \to N$ be a positive linear map with $\psi \circ \phi \le \omega$, then $\phi$ is normal. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 2.5.11]{AnantharamanPopa}}}. ]
+    Let $\varphi = \psi \circ \phi \le \omega$, then $\varphi$ is normal by \autoref{lemma:normal-functional-bound}. For any bounded, upward-directed family $A \subset M$, 
+    \[
+         \psi(\sup(\phi(A))) =\sup(\psi(\phi(A))) = \sup(\varphi(A)) = \varphi(\sup(A)) = \psi(\phi(\sup(A)))
+    \]
+
+    As $\psi$ is faithful, $\sup(\phi(A)) = \phi(\sup(A))$. Therefore $\phi$ is normal by (1) of \autoref{definition:normal-linear-map}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:normal-image-vna}
+    Let $M$ be a von Neumann algebra, $H$ be a complex Hilbert space, and $\pi: M \to B(H)$ be a normal unital *-homomorphism, then $\pi(M)$ is a von Neumann algebra. 
+\end{proposition}
+\begin{proof}
+    Let $N = \pi(M)''$ be the von Neumann algebra generated by $\pi(M)$. By the \hyperref[Kaplansky Density Theorem]{theorem:kaplansky-density}, $\ol{B_N(0, 1)}^{B(H)} = \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}}$. By \autoref{proposition:compact-extensions} and \autoref{proposition:bh-ultraweak-bounded}, $\ol{B_M(0, 1)}^M$ is weak-operator compact, so $\pi(\ol{B_M(0, 1)}^M)$ is weak-operator compact and contains $\pi(B_M(0, 1))$. Thus
+    \[
+    \ol{B_{\pi(M)}(0, 1)}^{B(H)} \supset \pi(\ol{B_M(0, 1)}^M) \supset \ol{B_{\pi(M)}(0, 1)}^{\text{\small WOT}} \supset \ol{B_N(0, 1)}^{B(H)}
+    \]
+
+    and $\ol{B_{\pi(M)}(0, 1)}^{B(H)} = \ol{B_{N}(0, 1)}^{B(H)}$, so $\pi(M) = \ol{\pi(M)}^{B(H)} = \pi(N)$ by \autoref{theorem:cstar-quotient}.
+\end{proof}
+
+
+
+
+\begin{lemma}
+\label{lemma:tracial-von-neumann-gns}
+    Let $M$ be a von Neumann algebra, $\varphi \in S(M)$ be a state, and $(H, \pi, \xi)$ be the \hyperref[GNS triple]{definition:gns-triple} associated with $\varphi$, then $\pi$ is normal if and only if $\varphi$ is normal. In which case, $\pi(M) \subset B(H)$ is a von Neumann algebra. 
+    \begin{enumerate}
+        \item $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)} \to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous. 
+        \item $\pi(M) \subset B(H)$ is a von Neumann algebra. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Suppose that $\varphi$ is normal. Since $\pi(M)\xi$ is dense in $H$, the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} implies that the weak operator topology on $\ol{B_{\pi(M)}(0, 1)}$ coincides with the topology of pointwise convergence on $\bracsn{\eta \otimes \theta|\eta, \theta \in \pi(M)\xi}$. For each $x, y, z \in M$, 
+    \[
+    \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H} = \dpn{\pi(z^*)\pi(x)\pi(y)\xi, \xi}{H} = \varphi(z^*xy)
+    \]
+
+    By \hyperref[separate continuity of composition]{proposition:bh-operator-topologies-facts} in the weak operator topology, the functional $x \mapsto \dpn{\pi(x)\pi(y)\xi, \pi(z)\xi}{H}$ is weak-operator continuous. As such, $\pi|_{\ol{B_M(0, 1)}}: \ol{B_M(0, 1)} \to \ol{B_{\pi(M)}(0, 1)}$ is weak-operator continuous. 
+    
+    
+    (2): By \autoref{proposition:normal-image-vna}.
+\end{proof}
+
+\begin{definition}[Separable]
+\label{definition:separable-vna}
+    Let $M$ be a von Neumann algebra, then $M$ is \textbf{separable} if there exists a separable Hilbert space $H$ and a faithful normal representation $\pi: M \to B(H)$. 
+\end{definition}
+
+