7
$\begingroup$

I am studying the proof that if $A$ is a $C^*$-algebra such that $A^{**}$ is a semidiscrete vN algebra, then $A$ has the completely positive approximation property (CPAP). I was able to handle the unital case, but I am stuck in the non-unital setting: The authors (N. Brown and N. Ozawa) suggest that one should prove that if $A^{**}$ is semidiscrete then so is $(\tilde{A})^{**}$ and then conclude by proving that if $\tilde{A}$ has the CPAP then so does $A$.

My problem is this: I can't prove that the double dual of the unitization will be semidiscrete. I cannot understand the double dual of the unitization in relevance to the double dual of $A$ at all. The authors state that $(\tilde{A})^{**}\cong A^{**}\oplus\mathbb{C}$ and mention that it is furthermore true that if $B$ is any $C^*$-algebra with a (closed, two-sided) ideal $J$, then $B^{**}\cong J^{**}\oplus(B/J)^{**}$. First of all, does $\cong$ mean as vector spaces or as $C^*$-algebras? How can one prove this isomorphism? Extra bonus question: If all double duals involved are endowed with their ultraweak topologies, is $\cong$ a homeomorphism?

$\endgroup$
9
$\begingroup$

Believe it or not, these are $*$-isomorphisms as C${}^*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{**}$ is a weak* closed two-sided ideal of $B^{**}$, and every weak*-closed two-sided ideal of a von Neumann algebra is a direct summand. I suppose these are good exercises. The supremum in $B^{**}$ of an approximate unit for $J$ will be a central projection $p$ such that $pB^{**} = J^{**}$.

Bonus, any $*$-isomorphism of von Neumann algebras is automatically a weak* homeomorphism. That is because it preserves order and hence must be normal.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your answer. Much appreciated, I will try the exercises you are pointing out, even though they seem a little difficult to me. $\endgroup$ – JustDroppedIn Aug 25 '20 at 16:31
  • 1
    $\begingroup$ Hint: a weak*-closed ideal would itself be a von Neumann algebra and hence must have a unit. $\endgroup$ – Nik Weaver Aug 25 '20 at 21:17

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.