# The double dual of the unitization of a $C^*$-algebra

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?

## 1 Answer

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.

• 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. – JustDroppedIn Aug 25 '20 at 16:31
• Hint: a weak*-closed ideal would itself be a von Neumann algebra and hence must have a unit. – Nik Weaver Aug 25 '20 at 21:17