intersection of the commutant of a von Neumann algebra and the ultraproduct of the von Neumann algebra

Question

In the proof of Proposition 4.1, I don't understand why $E_M(vv^*)=\lambda$. Is $E_N$ the expectation from $M^{\omega}$ to $M$? Can we deduce that $M'\cap M^{\omega}=\Bbb C1$?

Answer 1

Yes, $E_M$ is the expectation onto $M$. For any $x\in M$, $$ E_M(vv^*)x=E_M(vv^*x)=E_M(xvv^*)=xE_M(vv^*). $$ Hence $E_M(vv^*)\in M\cap M'=\mathbb C$.

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I cannot make sense of the second equality, as I don't see how $T$ either acts or is an element of $L^2(M)$. It is not used in the rest of the page, though.

For the first equality, since $M$ is dense in $L^2(M)$ we may assume $\xi=z\in M$. Then, thinking of the elements in $M$ as in $M^\omega$, \begin{align} \Big\|\sum_j \theta(x_j)zy_j\Big\|_2^2 &=\tau_\omega\big(\sum_{k,j}y_k^*z^*\theta(x_k)\theta(x_j)zy_j\big)\\[0.3cm] &=\frac1\lambda\,\tau_\omega\big(\sum_{k,j}y_k^*z^*\theta(x_k)\lambda\theta(x_j)zy_j\big)\\[0.3cm] &=\frac1\lambda\,\tau_\omega\big(\sum_{k,j}y_k^*z^*\theta(x_k)E_M(vv^*)\theta(x_j)zy_j\big)\\[0.3cm] &=\frac1\lambda\,\tau_\omega\big(E_M\big(\sum_{k,j}y_k^*z^*\theta(x_k)vv^*\theta(x_j)zy_j\big)\big)\\[0.3cm] &=\frac1\lambda\,\tau_\omega\big(\sum_{k,j}y_k^*z^*\theta(x_k)vv^*\theta(x_j)zy_j\big)\\[0.3cm] &=\frac1\lambda\Big\|\sum_j v^*\theta(x_j)zy_j\Big\|_2^2 \\[0.3cm] &=\frac1\lambda\Big\|\sum_j x_jv^*zy_j\Big\|_2^2 \\[0.3cm] \end{align}