construct a faithful normal conditional expectation on the ultraproduct of a von Neumann algebra

Question

The Lemma is from the paper “structure of bicentralizer algebras and inclusions of type III”. When reading the proof of the second part of the Lemma, I met with some problems.

1. The conditional expectation can be constucted as following； $$\forall x \in M^\omega, \quad E(x)=\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})$$.

Is it natrual to think of this construction? It is easy to check that $$\forall y \in \theta(F)'\cap M^\omega, E(y)=y \quad and \quad E(axb)=aE(x)b \quad \forall a,b\in \theta(F)'\cap M^\omega, \forall x\in M^\omega.$$ How to check that $E$ is positive?

2. According to the above construction, we have

$(\varphi \circ E_n)^\omega((x_n)^\omega)=\lim_{n\to \omega}\varphi \circ E_n(x_n)=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$

$\varphi^\omega \circ E((x_n)^\omega)=\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))$.

How to verify that $\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$.

Answer 1

The proof seems to be using that $\theta$ is unital, and it mixes $n$ as the size of $F$ and as the index of the sequences.

With $\theta$ being unital, the matrix units $\{e_{kj}\}$ of $F$ are taken to matrix units $\{\theta(e_{kj})\}$ of $M^\omega$ with $\sum_{k=1}^n\theta(e_{kk})=1$. Then \begin{align} E(x) &=\psi(1)1E(x)=\sum_i\psi(e_{ii})\sum_k\theta(e_{kk})E(x)=\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki}e_{jk})E(x)\\[0.3cm] &=\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})E(x)\theta(e_{jk})\\[0.3cm] &=E\Big(\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\Big). \end{align} For $e_{st}\in F$, we have \begin{align} \theta(e_{st})\sum_{i,k,j}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk}) &=\sum_{i,k,j}\psi(e_{ij})\theta(e_{st}e_{ki})x\theta(e_{jk})\\[0.3cm] &=\sum_{i,j}\psi(e_{ij})\theta(e_{si})x\theta(e_{jt})\\[0.3cm] &=\sum_{i,j,k}\psi(e_{ij})\theta(e_{si})x\theta(e_{jk}e_{st})\\[0.3cm] &=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\,\theta(e_{st}).\\[0.3cm] \end{align} Hence $\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\in\theta(F)'\cap M^\omega$ and thus $$ E(x)=E\Big(\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk})\Big)=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x\theta(e_{jk}). $$ Not only is $E$ positive but it is completely positive, as every conditional expectation is. But if you want to check the positivity directly here, since $\psi$ is a state on the matrices it is of the form $$ \psi(z)=\operatorname{Tr}(Hz)=\sum_{s=1}^nw_s^*zw_s $$ \begin{align} \sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x^*x\theta(e_{jk}) &=\sum_{i,j,k}\psi(e_{ij})\theta(e_{ki})x^*x\theta(e_{jk}) \\[0.3cm] &=\sum_{i,j,k,s} \theta(e_{ki})x^*w_s^*e_{ij}w_sx\theta(e_{jk})\\[0.3cm] &=\sum_{i,j,k,s} \theta(e_{ki})x^*w_s^*e_{i1}e_{1j}w_sx\theta(e_{jk})\\ &=\sum_{k,s} \Big(\sum_i\theta(e_{ki})x^*w_s^*e_{1i}^*\Big)\Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)\\[0.3cm] &=\sum_{k,s} \Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)^*\Big(\sum_je_{1j}w_sx\theta(e_{jk})\Big)\geq0. \end{align}

How to verify that $$\varphi^\omega(\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})(x_n)^\omega\theta_n(e_{jk})))=\lim_{n\to \omega}\varphi (\sum_{1\leq i,j,k \leq n} \psi(e_{ij})\theta_n(e_{ki})x_n\theta_n(e_{jk}))$$

The definition of $\varphi^\omega$ is that $$ \varphi^\omega((x_n)^\omega)=\lim_{n\to\omega}\varphi(x_n). $$