diff --git a/src/op/vn/commutative.tex b/src/op/vn/commutative.tex
new file mode 100644
index 0000000..40cec64
--- /dev/null
+++ b/src/op/vn/commutative.tex
@@ -0,0 +1,54 @@
+\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}
+    
+\end{corollary}
+
+
diff --git a/src/op/vn/index.tex b/src/op/vn/index.tex
index ae6b71d..d48e537 100644
--- a/src/op/vn/index.tex
+++ b/src/op/vn/index.tex
@@ -4,5 +4,6 @@
 \input{./topologies.tex}
 \input{./cayley.tex}
 \input{./vn.tex}
+\input{./commutative.tex}
 \input{./spec.tex}
 \input{./fc.tex}
\ No newline at end of file
diff --git a/src/op/vn/spec.tex b/src/op/vn/spec.tex
index f4e17aa..becfcfc 100644
--- a/src/op/vn/spec.tex
+++ b/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 $L^\infty(\Omega; \complex)$. 
+        \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 on $\mu \in \mathscr{E}$ 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 $\bracsn{E_{x, y}|x, y \in H}$.