\section{The Standard Representation} \label{section:standard-representation} \begin{definition}[Standard Representation] \label{definition:tracial-vna-standard} Let $(M, \tau)$ be a tracial von Neumann algebra, then the \hyperref[GNS representation]{definition:gns-triple} \[ (\pi, L^2(M, \tau), \xi_\tau) \] is the \textbf{standard representation} of $(M, \tau)$. For each $x \in M$, identify $\wh x = x\xi_\tau \in L^2(M, \tau)$, so that $M = \wh M \subset L^2(M, \tau)$. \end{definition} \begin{definition}[Canonical Conjugation] \label{definition:tracial-vna-standard-conjugate} Let $(M, \tau)$ be a tracial von Neumann algebra, then the mapping \[ J: L^2(M, \tau) \to L^2(M, \tau) \quad \wh x \mapsto \wh{x^*} \] is a conjugate linear isometry on $L^2(M, \tau)$, known as the \textbf{canonical conjugation}. \end{definition} \begin{definition}[Left/Right Regular Representations] \label{definition:mvn-standard-regular} Let $(M, \tau)$ be a tracial von Neumann algebra and $x \in M$, then \[ L_x: L^2(M, \tau) \to L^2(M, \tau) \quad L_x\wh y = \wh{xy} \] and \[ R_x: L^2(M, \tau) \to L^2(M, \tau) \quad R_x\wh y = \wh{yx} \] are the \textbf{left/right multiplication maps} by $x$, respectively. The mapping \[ \lambda: M \to B(L^2(M, \tau)) \quad x \mapsto L_x \] is the \textbf{left regular representation} of $M$, with $L(M) = \lambda(M)$ being the \textbf{algebra of left multiplications}. Similarly, \[ \rho: M^{op} \to B(L^2(M, \tau)) \quad x \mapsto R_x \] is the \textbf{right regular representation} of $M^{op}$, with $R(M) = \rho(M)$ being the \textbf{algebra of right multiplications}. \end{definition} \begin{proposition} \label{proposition:vna-standard-conjugation-lr} Let $(M, \tau)$ be a tracial von Neumann algebra, then: \begin{enumerate} \item For each $x \in M$, $JL_xJ = R_{x^*}$. \item $JL(M)J = R(M)$. \end{enumerate} \end{proposition} \begin{proof} (1): For each $x, y \in M$, \[ JL_xJ \wh y = JL_x \wh{y^*} = J \wh{xy^*} = \wh{yx^*} = R_{x^*}y \] \end{proof} \begin{definition}[Multiplication by Vector] \label{definition:vna-standard-vector-multiplication} Let $(M, \tau)$ be a tracial von Neumann algebra and $\xi \in L^2(M, \tau)$, then \[ L^0_\xi: \wh M \to L^2(M, \tau) \quad L_\xi^0(\wh y) = R_y(\xi) = \xi y \] and \[ R^0_\xi: \wh M \to L^2(M, \tau) \quad R_\xi^0(\wh y) = L_y(\xi) = y\xi \] are densely defined, closable operators. Let $L_\xi$ and $R_\xi$ be their respective closures, then $L_\xi$ and $R_\xi$ are the \textbf{left/right multiplication operators by $\xi$}. \end{definition} \begin{proof} Let $\seq{x_n} \subset M$ and $\eta \in L^2(M, \tau)$ with $\limv{n}\wh{x_n} = 0$ and $\limv{n}L_\xi^0(\wh{x_n}) = \eta$, then for each $y \in M$, \begin{align*} \dpn{\eta, \wh y}{L^2(M, \tau)} &= \limv{n} \dpn{L_\xi^0(\wh {x_n}), \wh y}{L^2(M, \tau)} = \limv{n} \dpn{R_{x_n}\xi, \wh y}{L^2(M, \tau)} \\ &= \limv{n}\dpn{\xi, R_{x_n}^*\wh y}{L^2(M, \tau)} = \limv{n}\dpn{\xi, R_{x_n^*}\wh y}{L^2(M, \tau)} \\ &= \limv{n}\dpn{\xi, \wh {yx_n^*}}{L^2(M, \tau)} \\ |\dpn{\eta, \wh y}{L^2(M, \tau)}| &\le \limv{n}\norm{\xi}_{L^2(M, \tau)} \normn{\wh {yx_n^*}}_{L^2(M, \tau)} \\ &\le \limv{n}\norm{\xi}_{L^2(M, \tau)} \normn{y}_{M} \normn{\wh {x_n^*}}_{L^2(M, \tau)} = 0 \end{align*} \end{proof} \begin{definition}[Convolvers] \label{definition:vna-standard-convolver} Let $(M, \tau)$ be a tracial von Neumann algebra and $\xi \in L^2(M, \tau)$, then $\xi$ is a \textbf{left convolver} if $L_\xi$ is bounded, and $\xi$ is a \textbf{right convolver} if $R_\xi$ is bounded. The sets of left and right convolvers are denoted $LC(M)$ and $RC(M)$, respectively. \end{definition} \begin{lemma} \label{lemma:vna-convolver-commute} Let $(M, \tau)$ be a tracial von Neumann algebra, then for each $\xi \in LC(M)$ and $\eta \in RC(M)$, $L_\xi R_\eta = R_\eta L_\xi$. \end{lemma} \begin{proof} First suppose that $\xi, \eta \in M$. In which case, for each $x, y \in M$, \[ \dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{\wh{\xi x \eta}, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)} \] Now suppose that $\xi \in M$, then for each $\eta \in RC(M)$, by the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, \begin{align*} |\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{L_\xi}_{B(L^2(M, \tau))} \norm{x}_M \norm{y}_M \cdot \norm{\eta}_{L^2(M, \tau)} \\ |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{x}_M \norm{\xi}_M\norm{y}_M \cdot \norm{\eta}_{L^2(M, \tau)} \end{align*} so by density of $\wh M$ in $L^2(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in M$ and $\eta \in RC(M)$. Finally, suppose that $\xi \in LC(M)$ and $\eta \in RC(M)$, then for each $z \in M$, \begin{align*} \dpn{L_\xi z, \wh y}{L^2(M, \tau)} &= \dpn{\xi z, \wh y}{L^2(M, \tau)} = \dpn{\xi, \wh y z^*}{L^2(M, \tau)} \\ &= \dpn{\xi, \wh y J\wh z}{L^2(M, \tau)} = \dpn{y^*\xi, J\wh z}{L^2(M, \tau)} = \dpn{\wh z, Jy^*\xi}{L^2(M, \tau)} \end{align*} so \begin{align*} \dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} &= \dpn{L_\xi (x \eta), \wh y}{L^2(M, \tau)} = \dpn{x \eta, J y^* \xi}{L^2(M, \tau)} \\ |\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{x \eta}_{L^2(M, \tau)} \norm{y}_M \cdot \norm{\xi}_{L^2(M, \tau)}\\ |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}| &\le \norm{R_\eta}_{B(L^2(M, \tau))} \norm{x}_M \norm{y}_M \cdot \norm{\xi}_{L^2(M, \tau)} \end{align*} By density of $\wh M$ in $L^2(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)} = \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in LC(M)$ and $\eta \in RC(M)$. \end{proof} \begin{theorem} \label{theorem:vna-convolver-identified} Let $(M, \tau)$ be a tracial von Neumann algebra, then: \begin{enumerate} \item $L(M) = LC(M) = R(M)'$. \item $JL(M)J = R(M) = RC(M) = L(M)'$. \item In particular, for each $\xi \in L^2(M, \tau)$, $L_\xi \in B(L^2(M, \tau))$ if and only if $\xi \in \wh M$. \end{enumerate} \end{theorem} \begin{proof} (1): By definition, $L(M) \subset LC(M)$. By \autoref{lemma:vna-convolver-commute}, $LC(M) \subset RC(M)'$, so \[ L(M) \subset LC(M) \subset RC(M)' \subset R(M)' \] and \[ R(M) \subset RC(M) \subset LC(M)' \subset L(M)' \] Using the \hyperref[Bicommutant Theorem]{theorem:bicommutant}, it remains to show that $R(M)' \subset LC(M)$. To this end, let $T \in R(M)'$ and $\xi = T \wh 1$, then for each $x \in M$, \[ T\wh x = TR_x \wh 1 = R_x T \wh 1 = R_x \xi = \xi x = L_\xi^0 \wh x \] so $T|_{\wh M} = L_\xi^0$. By density of $\wh M$ in $L^2(M, \tau)$ and continuity of $T$, the closure of $L_\xi^0$ is $T$, hence $T = L_\xi$. \end{proof}