Injective operator systems

Question

An operator system $\mathcal{S} \subseteq B(\mathcal{H})$ is called injective if given an operator system inclusion $\mathcal{M} \subseteq \mathcal{N} \subseteq B(\mathcal{K})$ each completely positive map $\phi : \mathcal{M} \rightarrow \mathcal{S}$ has a completely positive extension $\phi : \mathcal{N} \rightarrow \mathcal{S}$.

Arveson has shown that $B(\mathcal{H})$ is injective, hence every $C^*-$ algebra is an injective operator system and in particular finite dimension matrices are injective. I'd like to know if there is some method or result to check if a more general operator system is injective.

Answer 1

No, injectivity does not pass to subspaces. This is a tricky subject. A few facts:

• There is no characterization of injective C$^*$-algebras.

• Finite-dimensional C$^*$-algebras are injective.

• Injective C$^*$-algebras are monotone-complete, which implies that no separable infinite-dimensional C$^*$-algebra is injective (properly: an infinite-dimensional injective C$^*$-algebra is non-separable). So all the usual C$^*$-algebras (UHF, Cuntz, reduced C$^*$-algebras of countable groups, $K(H)$, Jiang-Su, etc., etc., etc.) are not injective.

• Injective von Neumann algebras are well-understood. This was Connes monumental Annals paper from 1976. In summary, a von Neumann algebra is injective if and only if it is AFD.

• Via the Choi-Effros product, it can be shown that an injective operator system is completely order isomorphic to a C$^*$-algebra.

• Hamana proved in 1979 that every operator system has an injective envelope, that is a smallest injective operator system that contains the given one.

• In light of all the above, a characterization of injective operator systems is hopeless. As far as I can tell (I haven't thought about this stuff for a while), there is not even any characterization of when a finite-dimensional operator system is injective. The answer is easy for $2$-dimensional operator systems as they are all isomorphic to $\mathbb C^2$ and thus injective. But I don't think there's a way to tell when a 3-dimensional operator system is injective.