: for $s \in G$, $\lambda_s$ and $\rho_s$ are unitaries on $L^2(G)$ given by $\lambda_s \xi(t) = \xi(s^{-1}t)$ and $\rho_s \xi(t) = \sqrt{\Delta(s)} \xi(ts)$. The (left) von Neumann algebra of $G$ is defined by ${\mathcal{L}}G = \lambda(G)''$. Its commutant is $\rho(G)''$, the right von Neumann algebra of $G$. If $f \in L^1(G)$, the formulas $\langle \lambda(f)\xi,\eta\rangle = \int f(s) \langle \lambda_s \xi,\eta\rangle ds$ and $\langle \rho(f)\xi,\eta\rangle = \int f(s) \langle \rho_s \xi,\eta\rangle ds$ define operators $\lambda(f) \in {\mathcal{L}}G$ and $\rho(f) \in \rho(G)''$ that are integral operators on $L^2(G)$ with kernel $$\label{eq=kernel_lambdaf} (\lambda(f))_{s,t} = \Delta(t^{-1}) f(st^{-1})\ ,\ (\rho(f))_{s,t} = \sqrt{\Delta(s^{-1}t)} f(s^{-1}t).$$ As a consequence, $\lambda(f)^\ast= \lambda(f^\ast)$ and $\rho(f)^\ast=\rho(f^\ast)$ where $f^\ast \in L^1(G)$ is given by $$\label{eq=def_of_xistar}f^\ast(s) = \overline{f(s^{-1})} \Delta(s^{-1}).$$

When $G$ is discrete, ${\mathcal{L}}G$ is finite and carries a natural trace. More generally when $G$ is unimodular ${\mathcal{L}}G$ carries a natural semifinite trace. In general, although ${\mathcal{L}}G$ might be semifinite, it is not equipped with a natural trace, but rather with its natural weight, given by $\varphi (x^*x) = \|f\|_{L^2G}^2$ if there is $f \in
L^2G$ such that $x \xi = f \ast \xi$ for all $\xi \in L^2G$ and $\varphi(x^*x)=\infty$ otherwise. (it is an easy exercice to check that $\varphi$ 