8
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.