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+\chapter{Von Neumann Algebras}
+\label{chap:von-neumann-algebras}
+
+\input{./topologies.tex}
diff --git a/src/op/vn/topologies.tex b/src/op/vn/topologies.tex
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+\section{Topologies on $B(H)$}
+\label{section:topologies-on-bh}
+
+Let $H$ be a complex Hilbert space. Thanks to its self-duality, there is a natural dual pairing
+\[
+B(H) \times (H \otimes H) \quad \dpn{T, \phi \otimes x}{B(H)} =  \dpn{Tx, \phi}{H}
+\]
+
+Depending on the topology placed on $H \otimes H$, and the corresponding completion, a handful of different topologies arise on $B(H)$. In fact, the above duality produces a predual for $B(H)$, being the trace class operators: 
+
+\begin{definition}[Ultraweak Topology]
+\label{definition:bh-ultraweak-topology}
+    Let $H$ be a complex Hilbert space, then the dual of $H \wh \otimes_\pi H$ is $B(H)$, and the $\sigma(B(H), H \wh \otimes_\pi H)$-topology is the \textbf{ultraweak}/\textbf{$\sigma$-weak} topology on $B(H)$. 
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:projective-tensor-product-dual}. 
+\end{proof}
+
+Seeing that $B(H)$ is a dual Banach space, the following fact is immediate:
+
+\begin{proposition}
+\label{proposition:bh-ultraweak-bounded}
+    Let $H$ be a complex Hilbert space, then every bounded subset of $B(H)$ is relatively compact with respect to the ultraweak topology. 
+\end{proposition}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. 
+\end{proof}
+
+Now, a few facts about the more familiar operator topologies:  
+
+\begin{proposition}
+\label{proposition:bh-operator-topologies-facts}
+    Let $H$ be a complex Hilbert space, then:
+    \begin{enumerate}
+        \item The dual of $B(H)$ with respect to its strong and weak operator topologies is $H \otimes H$. 
+        \item Every bounded subset of $B(H)$ is relatively compact in the weak operator topology. 
+        \item The composition map $(S, T) \mapsto ST$ is separately continuous in the strong and weak operator topologies. 
+        \item For any bounded subset $B \subset B(H)$, the composition map $(S, T) \mapsto ST$ restricted to $B \times B(H)$ is continuous in the strong operator topology. 
+        \item The adjoint map $T \mapsto T^*$ is continuous in the weak operator topology and the ultraweak topology. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): By \autoref{proposition:bh-ultraweak-bounded}. 
+
+    (4): Let $\angles{S_\alpha}_{\alpha \in A} \subset B$, $\angles{T_\alpha}_{\alpha \in A} \subset H$, and $S, T \in H$ such that $S_\alpha \to S$ and $T_\alpha \to T$ in the strong operator topology. Since $\{S_\alpha| \alpha \in A\}$ is equicontinuous, for any $x \in H$,
+    \[
+    \lim_{\alpha \in A} S_\alpha T_\alpha x = \lim_{\alpha \in A}S_\alpha Tx = \lim_{\alpha \in A}STx
+    \]
+\end{proof}
+
+
+