Standard form of group von Neumann algebra For a group $G$ we may define its left regular representation on $\ell^2(G)$ via $\lambda(g)\delta_h=\delta(gh)$ where $(\delta_g)_{g\in G}$ is the canonical basis. Then $M(G)=\lambda(G)^{''}$ is the generated group von Neumann algebra. By $\tau(x)=\langle x\delta_e, \delta_e\rangle$ it is given a faithful, normal state on $M(G)$. W.r.t. $\tau$ we may build the GNS construction of $M(G)$, this is called the standard form. In lecture notes by Vaughan Jones from 2009 he states that the group von Neumann algebra is already in standard form. This is not clear to me. I see that both the construction of the group von Neumann algebra as well as the GNS construction use left regular representations, but the GNS Hilbert space is then the completion of $M(G)$ w.r.t. tracial norm $x\mapsto \|x\delta_e\|$. Is this already complete? Do we then identify $x$ with $x\delta_e$ (which is possible since $\delta_e$ is separating)? It would be nice if someone could clarify what is meant with this assertion.
 A: I guess one way to see it is by showing directly that $\mathbb C G$ is dense not only in $\ell^2(G)$ but also in $L^2(M(G),\tau)$ and those two norms, $\|\cdot\|_2$ and $\|\cdot\|_{\tau}$, do coincide on $\mathbb C G$.
Another approach I would suggest is through the notion of von Neumann dimensions. Since $\delta_e\in \ell^2(G)$ is a cyclic and separating vector for $M(G)$, the von Neumann dimension $dim_{M(G)} \ell^2(G)=1$; on the other hand, $L^2(M(G),\tau)$ is in standard form and thus $dim_{M(G)} L^2(M(G),\tau)=1$. As $dim_{M(G)} L^2(M(G),\tau)=dim_{M(G)} \ell^2(G)$, they are unitarily equivalent $M$-modules and hence may be viewed as one space. For detailed discussions on von Neumann dimensions, see Vaughan Jones' planar algebras lecture notes.
A: The standard form of a von Neumann algebra $M$ is a representation $M\subset B(H)$ that has a conjugate linear isometric involution $J$ and and a self-dual cone $P\subset H$ with


*

*$JMJ=M'$

*$J\xi=\xi$ for all $\xi\in P$

*$aJaJ\subset P$ for all $a\in M$

*$JcJ=c^*$ for all $c\in\mathcal Z(M)$
It is the "correct" representation to do Tomita-Takesaki theory. A sufficient condition to have such a representation is that $M$ has a cyclic and separating vector $\Omega$; in the case of the left regular representation of $G$, the additional property of having the trace allows one to define $J$ by 
$$
Ja\Omega=a^*\Omega,
$$
and $P=\overline{\{a\Omega:\ a\geq0\}}$.
