Definition 2.1.2 If $A$ is a C*-algebra and $N$ is a von Neumann algebra, a map $\theta:A \rightarrow N$ is called weakly nuclear if there exist c.c.p. maps $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ and $\psi:M_{k(n)}(\mathbb{C}) \rightarrow N$ such that $\psi_{n} \circ \phi_{n} \rightarrow \theta$ in the point-ultraweak topology.

Definition 2.3.3 A con Neumann algebra $M$ is called semidiscrete if the identity map $id_{M}:M \rightarrow M$ is weakly nuclear.

If $A$ and $B$ are both semidiscrete, then $A\oplus B$ is also semidiscrete?


When I prove this question above, I met with a problem. That is,

If $T_{i}$ and $S_{i}$ are point-weak convergent to $T$ and $S$ respectively, then how to verify $T_{1}\oplus T_{2}$ is point-weak convergent to $T\oplus S$?


Yes. You can factorise the identity map on $A\oplus B$ in the point-ultraweak topology taking direct sums of the respective approximations for $A$ and $B$. Then you pretend that $M_{k_A(n)}$ and $M_{k_B(n)}$ live in some bigger full matrix algebra and you are done. In other words, you can replace the matrix algebras $M_{k(n)}$ in the definition of semidiscreteness by arbitrary finite-dimensional C*-algebras.

| cite | improve this answer | |
  • $\begingroup$ Yeah, I did like this and I met with a problem in the proof. Please see the addition of question above (I have edited my question). $\endgroup$ – Yan kai Apr 2 '14 at 1:42

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.