# Conditional expectation preserving center-valued trace

In the context of finite von Neumann algebras, I see many results involving the choice of a faithful normal tracial state $\tau$. For instance, the standard representation $L^2(R, \tau)$ of a finite von Neumann algebra $R$ is often used to prove results about $R$ which are independent of the choice of the state $\tau$. Whenever I see such a statement it causes me some discomfort which I phrase below.

1. Instead of making a choice of the state $\tau$ which eventually turns out to be immaterial, shouldn't there be a way of working with the unique faithful and normal center-valued trace $Tr$ directly in such arguments? By the universal property of the `trace', any tracial state must factor as a composition of $Tr$ and a tracial state on the center $Z$. If the center is not separable, then there may not be any faithful normal tracial state but the universal kind of proofs would still go through. My feeling is that instead of the GNS construction for $\tau$ which yields the standard representation, one may have to deal with $Z$-module constructions based on the KSGNS construction for $Tr$. Could someone please point me to references discussing this?
2. It is well-known that for a von Neumann subalgebra $S$ of $R$, there is a unique $\tau$-preserving normal conditional expectation from $R$ onto $S$. Can the same be said of $Tr$ ?

Thank you.

• For the first point: Your intuition is correct. You can do a generalized GNS construction and get a Hilbert $Z$-module with left and right representations of $R$. But this does not produce a rich structure in the case of von Neumann algebras. A $C^\ast$- algebraic version of your that construction has been used in: arxiv.org/pdf/1304.3523.pdf – Adrián González-Pérez Apr 26 '18 at 12:45
• Thank you for the reference. In the case of von Neumann algebras, I was just thinking a direct answer to question 2 may be possible using this $Z$-module as a tool (and a generalized Riesz representation theorem?) even though it may not possess a rich structure. – user123735 Apr 26 '18 at 16:08

Let $M$ be a finite factor with tracial state $\tau$. Let $R = M \oplus M$. Then the center-valued trace is given by $Tr(x,y) = (\tau(x),\tau(y))$.
Let $S = \mathbb{C}\cdot1_{M\oplus M}$. Any conditional expectation $\mathbb{E}: R \rightarrow S$ of $R$ onto $S$ must be given by a state $\varphi: M \oplus M \rightarrow \mathbb{C}$.
So $$Tr\circ \mathbb{E} (x,y) = \left(\varphi(x,y), \varphi(x,y)\right).$$ Thus $Tr \circ \mathbb{E} \neq Tr$.
This is ultimately because $R$ and $S$ have different centers. It might make sense to ask your question with the added assumption that $R$ and $S$ have the same center.