# A question on nuclearity

Definition 2.1.1. If $A$, $B$ are C*-algebra, a map $\theta: A\rightarrow B$ is called nuclear if there exist contractive completely positive maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi_{n}: M_{k(n)}(\mathbb{C})\rightarrow B$ such that $\psi_{n}\circ\phi_{n}\rightarrow\theta$ in the point-norm topology for all $a\in A$.

Definition 2.1.2. If $A$ is a C*-algebra and $N$ is a von Neumann algebra, a map $\theta: A\rightarrow N$ is call weakly nuclear if there exist contractive completely positive maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi_{n}: M_{k(n)}(\mathbb{C})\rightarrow B$ such that $\psi_{n}\circ\phi_{n}\rightarrow\theta$ in the point-ultraweak topology: $$\eta(\psi_{n}\circ\phi_{n}(a))\rightarrow\eta(\theta(a)),$$ for all $a\in A$ and all normal functionals $\eta\in N_{*}$.

My question is : If $\phi: A\rightarrow B\subset B^{**}$ is weakly nuclear, can we conclude that $\phi$ is nuclear?

Since the definition of weakly nuclear requires the codomain to be a von Neumann algebra, I will assume $\phi:A\to B^{**}$.
Let $A=B(H)$, $B=K(H)$. Then $B^{**}=B(H)$. Then you can take $\phi$ to be the identity map, which is not nuclear (because $B(H)$ is not nuclear) but is weakly nuclear (by 2.1.4 in Brown-Ozawa).
• $B(H)$ lacks nuclearity for a very strong reason - it lacks the Banach-space approximation property (this is due to Szankowski). Actually, the only nuclear von Neumann algebras are the finite sums of matrix algebras over abelian von Neumann algebras. – Tomasz Kania Jul 13 '14 at 17:13