Suppose that $\Gamma$ is a discrete group. For an extremal character $\tau$ on $\Gamma$, we denote by $M$ the corresponding GNS vNa (which is a tracial factor). This is a consequence of a well-known theorem of Thomma which says the following:
Theorem (Thomma) Let $\Gamma$ be a discrete group. There is a one to one correspondence between:
- Equivalence classes of unitary representations $\pi : \Gamma \to \mathcal{U}(M)$, where $M$ is a finite von Neumann algebra with a given normal faithful trace $\tau$ , and such that $\pi(\Gamma)'' = M$, and
- Characters $\phi : \Gamma \to \mathbb{C}$, which is given by $\phi(g) = \tau(\pi(g))$.
Moreover, $M$ is a factor if and only if $\phi$ is an extreme point in the space of characters.
I have a question about the standard form of a von Neumann algebra $M$ (obtained from an extremal character via the GNS construction). Using the Theorem of Thomma we know that $(M,\tau)$ is a tracial von Neumann algebra, $\tau$ is a faithful normal trace on $M$. Now, if we do the GNS construction of $M$ again with respect to $\tau$ (via the inner product $\langle x,y\rangle=\tau(y^*x))$, we obtain a representation $\pi_{\tau}$, a separable Hilbert space $L^2(M)$ and a cyclic vector $\xi_{\tau}$.
Define $J: \pi_{\tau}(M)(\xi_{\tau})\to L^2(M)$ by $J(\pi_{\tau}(x)\xi_{\tau})=\pi_{\tau}(x^*)\xi_{\tau}$. Observe that for all $x,y \in M$, we have \begin{align*}\langle J(\pi_{\tau}(x)\xi_{\tau}),J(\pi_{\tau}(y)\xi_{\tau})\rangle&=\langle\pi_{\tau}(x^*)\xi_{\tau}, \pi_{\tau}(y^*)\xi_{\tau}\rangle\\&=\tau(yx^*)\\&=\tau(x^*y)\\&=\langle (\pi_{\tau}(y)\xi_{\tau}),(\pi_{\tau}(x)\xi_{\tau})\rangle\end{align*} Hence, $J$ is a conjugate linear unitary with $J^2=1$.
Question: If $JMJ=M'$, then every $T\in B(L^2(M))$ which commutes with $J\pi(g)J$ must be in M?
My thinking was that if $J\pi(g)JT=TJ\pi(g)J$, then $JxJT=TJxJ$ for all $x\in\text{Span}\left\{\pi(g):g\in \Gamma\right\}$ and hence, $T\in (JMJ)'=(M)''=M$.
Is this correct?
Thanks for your help!!