Nuclear C*-algebra

Remarks

There is always a smallest C*-norm $\norm{\cdot}_{\min}$ and a largest $\norm{\cdot}_{\max}$, so $A$ is nuclear when $\norm{\cdot}_{\min} = \norm{\cdot}_{\max}$ on $A \odot B$ for all $B$. By theorems of Choi–Effros and Kirchberg, $A$ is nuclear if and only if it has the completely positive approximation property. This means that there are nets of completely positive contractions $\varphi_\lambda \colon A \to M_{n_\lambda}(\C)$ and $\psi_\lambda \colon M_{n_\lambda}(\C) \to A$ with $\psi_\lambda(\varphi_\lambda(a)) \to a$ in norm for every $a \in A$. A C*-algebra is nuclear if and only if it is an amenable Banach algebra.

Example: $M_n(\C)$ is nuclear. Take $\varphi_\lambda = \psi_\lambda$ equal to the identity map of $M_n(\C)$, which is completely positive and contractive. Then $\psi_\lambda(\varphi_\lambda(a)) = a$, so the completely positive approximation property holds. Commutative C*-algebras and AF algebras are also nuclear.

Non-example: let $\mathbb{F}_2$ be the free group on two generators. For a discrete group $G$, the reduced group C*-algebra $C^*_r(G)$ is nuclear if and only if $G$ is amenable; since $\mathbb{F}_2$ is not amenable, $C^*_r(\mathbb{F}_2)$ is not nuclear. The algebra $B(H)$ of all bounded operators on an infinite-dimensional Hilbert space is not nuclear either.