Let $M$ be an injective von Neumann subalgebra of $B(H)$. For a completely positive map $\phi:B(H)\to B(H)$, let $Mult(\phi)$ denote be the multiplicative domain of $\phi$. For any conditional expectation (CE) $E:B(H)\to M$, the range of $E$ is contained in $Mult(E)$ because $E$ is a bimodule map. Indeed, the image of $E$ is contained in $\cap\{Mult(E): E \text{ is a CE from }B(H)\text{ onto }M\}$.

Question: Is $M=\cap\{Mult(E): E \text{ is a CE from }B(H)\text{ onto }M\}$? I am mainly interested in the answer to this question when $M$ is Type I, or even when $M$ is a masa.

Evidence: not much. If there is a faithful CE $E$ onto $M$ then $M=Mult(E)$; in this case the answer is "yes". So, for example, the claim is true when $M$ is a purely atomic masa. I believe the answer is also "yes" when $M$ is purely atomic (not necessarily abelian). Beyond that I have no idea.


1 Answer 1


This is not true in general. For example, consider an infinite amenable group $\Gamma$ and let $M$ be the group von Neumann algebra $L\Gamma \subset \mathcal B(\ell^2\Gamma)$ endowed with its usual trace $\tau(x) = \langle x \delta_e, \delta_e \rangle$. For $\gamma \in \Gamma$ let $P_\gamma$ denote the rank one projection onto $\mathbb C \delta_\gamma$. Then $\{ P_\gamma \}_{\gamma \in \Gamma}$ forms an orthogonal family of projections and hence for any conditional expectation $E: \mathcal B(\ell^2\Gamma) \to L\Gamma$, and each finite set $F \subset \Gamma$ we have $$ 1 \geq \tau \circ E( \sum_{\gamma \in F} P_\gamma) = \sum_{\gamma \in F} \tau \circ E( \lambda(\gamma) P_e \lambda(\gamma^{-1})) = | F | \tau \circ E(P_e). $$ Since $\tau$ is faithful it then follows that $E(P_e) = 0$ and hence $P_e$ must be in the multiplicative domain of $E$.

A slight generalization of the above argument shows in fact that if $M \subset \mathcal B(\mathcal H)$ is any injective finite von Neumann algebra without minimal projections then the compact operators are always contained in the multiplicative domain of any conditional expectation $E: \mathcal B(\mathcal H) \to M$.

If one considers only type I von Neumann algebras, then the argument above can be adapted to show that $M$ being completely atomic is a necessary condition to have $M = \cap \{ Mult(E) : E {\rm \ is \ a \ CE \ from \ } \mathcal B(\mathcal H) {\rm \ to \ } M \}$. This condition should also be sufficient. When $M$ is completely atomic then there in fact exists a normal conditional expectation and the result shouldn't be too difficult from there.


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