For a discrete group $\Gamma$ we let $\lambda: \Gamma \rightarrow B(l^{2}(\Gamma))$ denote the left regular representation and $\rho$ denote the right regular representation. The reduced C*-algebra of $\Gamma$, denoted $C^{*}_{\lambda}(\Gamma)$,is the completion of $\mathbb{C}[\Gamma]$ with respect to the norm $$||x||_{r}=||\lambda(x)||_{B(l^{2}(\Gamma))}$$ Similarily, we can define $C^{*}_{\rho}(\Gamma)$.
My question is: do we have that $C_{\lambda}^{*}(\Gamma)'=C^{*}_{\rho}(\Gamma)$? Here, $C_{\lambda}^{*}(\Gamma)'$ denote the commutant of $C_{\lambda}^{*}(\Gamma).$
The double commutant of $C^*_\lambda(\Gamma)$ is usually called the group von Neumann algebra of $\Gamma$, and denoted by $L(\Gamma)$. Similarly, the double commutant of $C^*_\rho(\Gamma)$ is denoted (I don't think this notation is as standard as the previous one) by $R(\Gamma)$.
It is well-known that $L(\Gamma)'=R(\Gamma)$, and it immediately follows that $R(\Gamma)'=L(\Gamma)$. As the triple commutant agrees with the first one, you have $$ C^*_\rho(\Gamma)'=C^*_\lambda(\Gamma)'',\ \ C^*_\lambda(\Gamma)'=C^*_\rho(\Gamma)'' $$