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+\section{Group von Neumann Algebras}
+\label{section:gvna}
+
+\begin{definition}[Group von Neumann Algebra]
+\label{definition:group-von-neumann-algebra}
+    Let $G$ be a locally compact group, 
+    \[
+    \lambda: G \to U(L^2(G; \complex)) \quad x \mapsto L_x
+    \]
+    
+    be the left regular representation of $G$, and
+
+    \[
+    \lambda: L^1(G; \complex) \to B(L^2(G; \complex)) \quad \lambda(f)(g) = (f * g)
+    \]
+
+    be its extension to $L^1(G; \complex)$, then
+    \[
+    L(G) := \lambda(G)'' = \lambda(L^1(G;\complex))''
+    \]
+
+    is the \textbf{group von Neumann algebra} of $G$. Similarly, if $\rho: G \to U(L^2(G; \complex))$ is the right regular representation of $G$, then $R(G) := \rho(G)'' = \rho(L^1(G;\complex))''$.
+\end{definition}
+\begin{proof}
+    Firstly, for each $f \in L^1(G; \complex)$ and $g, h \in L^2(G; \complex)$, by the \hyperref[Fubini-Tonelli Theorem]{theorem:fubini-tonelli}, 
+    \begin{align*}
+        \dpn{\lambda(f)g, h}{L^2(G; \complex)} &= \int_G f(x)\dpn{L_xg, h}{L^2(G; \complex)} dx = \iint_{G \times G}f(x) g(x^{-1}y) \ol{h}(y) dy dx \\
+        &= \iint_{G \times G}f(x) g(x^{-1}y)dx \ol{h}(y) dy  \\
+        &= \int_G (f * g)(y) \ol{h}(y)dy = \dpn{(f * g), h}{L^2(G; \complex)}
+    \end{align*}
+
+    By \autoref{theorem:l1-group-von-neumann}, $\ol{\lambda(G)}^{\text{\small SOT}} = \ol{\lambda(L^1(G;\complex))}^{\text{\small SOT}}$. 
+\end{proof}
+
+\begin{definition}[Left Convolver]
+\label{definition:left-convolver}
+    Let $G$ be a locally compact group and $f: G \to \complex$ be Borel measurable, then $f$ is a \textbf{left convolver} if for each $g \in L^2(G; \complex)$, $\int_G |f(y)||g(y^{-1}x)|dy < \infty$ for almost every $x \in G$ and $f * g \in L^2(G; \complex)$. The mapping $L_f: L^2(G; \complex) \to L^2(G; \complex)$ defined by $g \mapsto f * g$ is the \textbf{left convolution map} by $f$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:left-convolver-cc}
+    Let $G$ be a locally compact group and $f \in C_c(G; \complex)$, $f$ is a lefy convolver. 
+\end{lemma}
+\begin{proof}
+    By \hyperref[Young's Inequality]{theorem:young-fourier}, for each $g \in L^2(G; \complex)$, $\norm{f * g}_{L^2(G; \complex)} \le \norm{f}_{L^1(G; \complex)}\norm{g}_{L^2(G; \complex)}$. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:left-convolver}
+    Let $G$ be a locally compacy group and $f: G \to \complex$ be a left convolver, then:
+    \begin{enumerate}
+        \item $L_f \in B(L^2(G; \complex))$. 
+        \item $f \in L_{\text{loc}}^1(G; \complex)$. 
+        \item If $f * g = 0$ for all $g \in L^2(G; \complex)$, then $f = 0$ locally almost everywhere. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $\seq{g_n} \subset L^2(G; \complex)$, $\seq{f * g_n} \subset L^2(G; \complex)$, and $g, h \in L^2(G; \complex)$ such that $g_n \to g$ and $f * g_n \to h$ in $L^2(G; \complex)$ as $n \to \infty$. After taking a subsequence, assume without loss of generality that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $f * g_n \to h$ almost everywhere.
+        \item $\sum_{n \in \natp}\norm{g_n - g}_{L^2(G; \complex)} < \infty$.
+    \end{enumerate}
+    
+    In which case, for almost every $x \in G$, 
+    \begin{align*}
+        \limv{N} f * (g_N - g)(x) &= \int_G f(y)(g_N - g)(y^{-1}x)dy \\
+        &\le \int_G|f(y)|\sum_{n \ge N}|(g_n - g)(y^{-1}x)| dy
+    \end{align*}
+
+    where $\int_G|f(y)|\sum_{n \in \natp}|(g_n - g)(y^{-1}x)| dy < \infty$ by (ii). As such, by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $f * g_N \to f * g$ almost everywhere. By (i), $f * g = h$ almost everywhere. Therefore the \hyperref[Closed Graph Theorem]{theorem:closed-graph} implies that $L_f$ is continuous. 
+
+    (2): Let $U \in \cn_G(1_G)$ be relatively compact with $U = U^{-1}$, then $\one_U \in L^2(G; \complex)$, and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for almost every $x \in G$. For each $x_0 \in G$, by \autoref{proposition:haar-measure-charge}, there exists $x \in G$ such that $x_0 \in xU$ and $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$. As such, for any $K \subset G$ compact, there exists $F \subset G$ finite such that $\int_G |f(y)|\one_U(y^{-1}x)dy < \infty$ for each $x \in F$ and $K \subset \bigcup_{x \in F}xU$. In which case, 
+    \[
+    \int_K |f(y)|dy \le \sum_{x \in F}\int_G |f(y)|\one_U(y^{-1}x)dy < \infty
+    \]
+
+    (3): By \autoref{lemma:left-convolver}, $f$ is locally integrable. Let $K \subset G$ be relatively compact. Using \autoref{lemma:lch-compact-neighbour}, assume without loss of generality that $K$ is also open. 
+    
+    Let $g \in C_c(G; \complex)$ with $\supp{g} \subset K^{-1}$, then there exists $V \in \cn_G(1_G)$ such that $\supp{g}^{-1}V \subset K^{-1}$. In which case, for each $x \in V$, 
+    \[
+    (f * g)(x) = \int_G f(y)g(y^{-1}x) dy = \int_K f(y)g(y^{-1}x) dy = (\one_Kf * g)(x)
+    \]
+
+    so $f * g$ is continuous on $V$ by \hyperref[Young's Inequality]{theorem:young-fourier}. In particular, $f * g|_{V} = 0$ almost everywhere implies that $\int_G f(y)g(y^{-1}) dy = 0$. As such, $\int_G fg = 0$ for all $g \in C_c(G; \complex)$ with $\supp{g} \subset K$. By uniqueness from the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $f|_K = 0$ almost everywhere. Therefore $f = 0$ locally almost everywhere. 
+\end{proof}
+
+\begin{definition}[Space of Left Convolvers]
+\label{definition:left-convolver-algebra}
+    Let $G$ be a group, $LC(G) := \bracsn{L_f| G \to \complex\text{ is a left convolver}} \subset B(l^2(G; \complex))$, and $f, g: G \to \complex$ be left convolvers, then:
+    \begin{enumerate}
+        \item $f \in l^2(G; \complex)$.
+        \item $L_f^* = L_{f^*}$, where $f^*(x) = \ol{f(x^{-1})}$ for all $x \in G$.
+        \item $L_{f * g} = L_fL_g$. 
+        \item $LC(G) \subset B(l^2(G; \complex))$ is a von Neumann algebra. 
+    \end{enumerate}
+
+    The set $LC(G)$ is the \textbf{algebra of left convolvers} on $G$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Propositon 1.3.5]{AnantharamanPopa}}}. ]
+    (1): $f * \delta_e = f \in l^2(G; \complex)$. 
+
+    (2): Let $x, y \in G$, then
+    \begin{align*}
+        \dpn{L_f\delta_x, \delta_y}{l^2(G; \complex)} &= (f * \delta_x)(y) = f(x^{-1}y) \\ 
+    \dpn{\delta_x, L_{f^*}\delta_y}{l^2(G;\complex)} &= \ol{(f^* * \delta_y)(x)} = f^*(y^{-1}x)
+    \end{align*}
+    
+
+    (3): By (1), $f * g \in l^2(G; \complex)$ is defined. For any $h \in l^2(G; \complex)$, by associativity of convolution, 
+    \[
+    L_{f * g}h = (f * g) * h = f * (g * h) = L_fL_gh
+    \]
+
+    (4): Let $T \in \ol{LC(G)}^{\text{\small SOT}}$ and $\angles{f_\alpha}_{\alpha \in A}$ be left convolvers such that $T = \sotlim_{\alpha \in A}L_{f_\alpha}$. Let $f = T\delta_e \in l^2(G; \complex)$, then
+    \[
+    f = T\delta_e = \paren{\sotlim_{\alpha \in A}L_{f_\alpha}}\delta_e = l^2\text{-}\lim_{\alpha \in A}f_\alpha * e
+    \]
+    
+    For any $g \in l^2(G; \complex)$,
+    \[
+    \norm{Tg - f * g}_{l^\infty(G; \complex)} \le \limsup_{\alpha \in A}\norm{g}_{l^2(G; \complex)}\norm{f - f_\alpha}_{l^2(G; \complex)} = 0
+    \]
+
+    by \hyperref[Young's Inequality]{theorem:young-fourier}. Therefore $Tg = f * g$, $T = L_f$, and $LC(G)$ is strong-operator closed. By \hyperref[von Neumann's Bicommutant Theorem]{theorem:bicommutant}, $LC(G)$ is a von Neumann algebra. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:group-algebra-identified}
+    Let $G$ be a group, then $L(G) = LC(G) = R(G)'$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 1.3.6]{AnantharamanPopa}}}. ]
+    ($\subset$): For any $x, y, z \in G$ and $f \in l^2(G; \complex)$, 
+    \[
+    (\lambda(x)\rho(y)f)(z) = (\rho(h)f)(x^{-1}z) = f(x^{-1}zy) = (\lambda(x)f)(zy) = (\rho(y)\lambda(x)f)(z)
+    \]
+
+    so $\lambda(G) \subset R(G)'$. By \autoref{definition:left-convolver-algebra}, $L(G) \subset LC(G) \subset R(G)'$. 
+
+    ($\supset$): Let $T \in R(G)'$ and $f = T\delta_e \in l^2(G; \complex)$, then for each $x \in G$, 
+    \[
+    T\delta_x = T\rho(x^{-1})\delta_e = \rho(x^{-1})T\delta_e = f * \delta_x
+    \]
+
+    As such, $Tg = f * g$ for all $g \in l^2(G; \complex)$, and $T = L_f$. 
+\end{proof}
+
+\begin{definition}
+\label{definition:group-algebra-fourier-series}
+    Let $G$ be a group. For each $x \in G$, let $u_x = \lambda(x)$ and identify it with $\delta_x$, then $\lambda(G) = \bracsn{u_x|x \in G}$ is the \textbf{canonical unitaries} of $L(G)$. 
+
+    For each $f \in l^2(G)$, $f = \sum_{x \in G}f(x)u_x$. In particular, for each $T \in L(G) = LC(G) \subset l^2(G; \complex)$, there exists $\bracsn{T_x}_{x \in G} \subset \complex$ such that $T$ may be written as $T = \sum_{x \in G}T_x u_x$\footnote{This sum does not converge in the strong-operator or the weak-operator topology. }. In which case, the scalars $\bracsn{T_x}_{x \in G}$ are the \textbf{Fourier coefficients} of $T$. 
+\end{definition}
+
+\begin{definition}[ICC]
+\label{definition:icc}
+    Let $G$ be a group, then $G$ has \textbf{infinite conjugacy classes (ICC)} if every non-trivial conjugacy class of $G$ is infinite. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:icc-factor}
+    Let $G$ be a group, then the following are equivalent:
+    \begin{enumerate}
+        \item $L(G)$ is a factor. 
+        \item $G$ is ICC. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 1.3.9]{AnantharamanPopa}}}. ]
+    (2) $\Rightarrow$ (1): Let $T \in Z(L(G))$, then for each $x \in G$, 
+    \[
+    T\delta_e = \lambda(x)T\lambda(x^{-1}) = \lambda(x)T\rho(x) = \lambda(x)\rho(x)(T\delta_e)
+    \]
+
+    so $T\delta_e$ is invariant under conjugation, and hence constant on conjugacy classes of $G$. 
+    
+    Since $G$ is ICC and $T\delta_e \in l^2(G; \complex)$, there exists $\lambda \in \complex$ with $T\delta_e = \lambda \delta_e$. As such, $T = \lambda I$, and $Z(L(G)) = \complex I$. 
+
+    $\neg$ (2) $\Rightarrow$ $\neg$ (1): Let $C \subset G$ be a non-trivial finite conjugacy class, then for each $x \in G$, $\delta_x * \one_C = \one_{xC} = \one_{Cx} = \one_C * \delta_x$, so $L_{\one_C}$ is a non-trivial element of $Z(L(G))$. 
+\end{proof}
+
+\begin{definition}[Wreath Product]
+\label{definition:wreath-product}
+    Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, $H^{(I)} = \bigoplus_{i \in I}H$, and $\Gamma$ act on $H^{(I)}$ by left translation ($(\gamma \xi)_i = \xi_{\gamma^{-1}i}$), then the \textbf{generalised wreath product} $H \wr_I \Gamma$ is the semidirect product $H^{(I)} \rtimes \Gamma$. The \textbf{wreath product} $H \wr \Gamma$ is the special case where $I = \Gamma$, and $\Gamma$ acts by left translation. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:wreath-product-icc}
+    Let $H$ be a non-trivial group, $\Gamma$ be a group acting on a set $I$, and $\Gamma_f \subset \Gamma$ be the subgroup of elements whose conjugacy class is finite. If
+    \begin{enumerate}[label=(\alph*)]
+        \item The orbits of $\Gamma \curvearrowleft I$ are finite.
+        \item The restricted action $\Gamma_f \curvearrowright I$ is faithful. 
+    \end{enumerate}
+
+    then the wreath product $H \wr_I \Gamma$ is ICC. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition 1.3.10]{AnantharamanPopa}}}. ]
+    For each $g \in H \wr_I \Gamma$ and subgroup $K \subset H \wr_I \Gamma$, let $g^K = \bracsn{kgk^{-1}|k \in K}$. 
+
+    Let $g = (\xi, \gamma) \in H \wr_I \Gamma$. First suppose that $\xi \ne e_{H^{(I)}}$, then the support of $\xi$ is non-empty. By (a), the support of $\xi$ has infinite orbit under $\Gamma$, so $g^G \supset g^\Gamma$ is infinite. 
+
+    Now suppose that $\xi = e_{H^{(I)}}$ and $\gamma \ne e_{\Gamma}$. If $\gamma \in \Gamma \setminus \Gamma_f$, then the conjugacy class of $\gamma$ is infinite, and the conjugacy class of $g$ is also infinite. Now suppose that $\gamma \in \Gamma_f$. By (b), there exists $i \in I$ with $\gamma i_0 \ne i_0$. As $H$ is non-trivial, there exists $\xi_0 \in H^{(I)} \setminus \bracsn{e_{H^{(I)}}}$ with support $\bracsn{i_0}$. In which case, if $g_0 = (\xi_0, e_{\Gamma})$, then
+    \[
+    g_0^{-1}gg_0 = (\xi_0^{-1}\gamma\xi_0, \gamma)
+    \]
+
+    and hence $g$ has infinite conjugacy class because $\xi_0^{-1}\gamma\xi_0 \ne e_{H^{(I)}}$. Therefore $g^G$ is also infinite. 
+\end{proof}