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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.

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  • $\begingroup$ $\lambda(g)\delta_h = \delta_{gh}$, I guess. $\endgroup$
    – user42761
    Apr 15, 2017 at 15:06
  • $\begingroup$ You are right, I have fixed it! Somehow I mixed it with $(\lambda(g)\delta_h)(k)=\delta_h(g^{-1}k)$, i.e. the discrete fourier transform definition... $\endgroup$ Apr 15, 2017 at 15:12

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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.

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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 $P=\overline{\{a\Omega:\ a\geq0\}}$.

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  • $\begingroup$ Thanks for your answer, I didn't know this definition yet, only the one I have introduced above (but I just realized that mine was (probably only?) for finite factors ...) Is it obvious to call it "the" standard form? As I have introduced it above, it was clear to me, since the trace is unique and GNS representation is up to unitary equivalence. In the case of group algebras, do we actually need the tracial property? Since we already know that $\delta_e$ is cyclic and separating? $\endgroup$ Apr 15, 2017 at 15:39

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