TITLE: Arveson's extension for normal completely positive maps
QUESTION [7 upvotes]: My question deals with a version of Arveson's extension theorem (for the standard version, see, e.g., Paulsen's book Completely Bounded Maps and Operator Algebras). Let $\mathcal A$ be a von Neumann algebra, $\mathcal R\subset\mathcal A$ an operator system, and $\mathcal H$ a Hilbert space. If $\Phi_0:\mathcal R\to\mathcal L(\mathcal H)$ is a linear normal completely positive unital map, does there exist a linear normal completely positive unital extension $\Phi:\mathcal A\to\mathcal L(\mathcal H)$ for $\Phi_0$ (complete positivity for a linear map on an operator system defined as in Paulsen's book)?
I am aware that if one gives certain restrictions for the operator system, the extension can be carried out. E.g., when $\mathcal A$ is a type-I factor and $\mathcal R$ is contained in the ultraweak closure of the set of its compact operators, the Arveson-like extension result for normal maps holds. Does anyone know whether the normal extension exists in general or whether there are some weaker versions of Arveson's theorem for normal maps? I am mostly interested in the case where $\mathcal A$ is a type-I factor. Thank you in advance.

REPLY [8 votes]: The answer is NO and here's a counterexample.
Let $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the $\ell_\infty$-direct sum of the full matrix algebras ${\mathbb M}_n$, and let $X\subset\prod_{n\in{\mathbb N}}{\mathbb M}_n$ be the weak$^*$-closed subspace defined by $$X=\{ (x_n)_{n=1}^\infty : \Theta_{m,n}(x_m)=x_n \mbox{ for all }m>n\},$$ where $\Theta_{m,n}\colon{\mathbb M}_m\to{\mathbb M}_n$ is the compression to the upper-left corner. Then, $\Phi\colon X\to{\mathbb B}(\ell_2)$, $\Phi((x_n)_{n=1}^\infty) = \lim_n x_n$ is a normal unital completely positive map (in fact, an weak$^*$-homeomorphic complete order isomorphism). The map $\Phi$ does not extends to a normal ucp map on $\prod_{n\in{\mathbb N}}{\mathbb M}_n$. (In case one wants a type I factor, use the embedding $\prod_{n\in{\mathbb N}}{\mathbb M}_n\subset{\mathbb B}(\bigoplus\ell_2^n)$.)
Indeed, suppose $\Psi\colon\prod_{n\in{\mathbb N}}{\mathbb M}_n\to{\mathbb B}(\ell_2)$ is any normal ucp map which maps the closed unit ball $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ onto the closed unit ball $({\mathbb B}(\ell_2))_1$. Then, $x \in (\prod_{n\in{\mathbb N}}{\mathbb M}_n)_1$ such that $\Psi(x)$ is unitary is in the multiplicative domain $\mathrm{mult}(\Psi)$ of $\Psi$. Since ${\mathbb B}(\ell_2)$ is spanned by unitary elements, $\Psi$ restricted to the $\mathrm{C}^*$-subalgebra $\mathrm{mult}(\Psi)$ is a surjective $*$-homomorphism. If $\Psi$ were normal, $\mathrm{mult}(\Psi)$ would become a von Neumann subalgebra of $\prod_{n\in{\mathbb N}}{\mathbb M}_n$ and $\Psi$ would become a normal surjective $*$-homomorphism from $\mathrm{mult}(\Psi)$ onto ${\mathbb B}(\ell_2)$, which is absurd. (Although it's overkill, there's also a result of Sukochev and Haagerup--Rosenthal--Sukochev saying that the predual ${\mathbb B}(\ell_2)_*$ does not Banach embed to the predual $(\prod_{n\in{\mathbb N}}{\mathbb M}_n)_*$.)