3
$\begingroup$

Let $A$ be a C*-algebra. According to operator algebraists, it is well known that $A$ embeds into the atomic part of its double dual in the following sense: if $z$ is the central projection in $A^{**}$ onto its atomic part, then the composition of the embedding $A \hookrightarrow A^{**}$ with multiplication by $z$ yields an embedding $A \hookrightarrow zA^{**}$.

I see how one can prove this, but I need this result in my paper which is not aimed at operator algebraists, so I would like to give a reference. However, I cannot find this in the literature. I looked at Takesaki, Blackadar, Dixmier, Sakai, Kadison-Ringrose, and the closest I got was Definition III.6.35 in Takesaki where he defines the universal atomic representation, but I don't see how the above result immediately follows from this definition or its subsequent results.

Does anyone have a reference for this?

$\endgroup$

1 Answer 1

3
$\begingroup$

If you are only interested in seeing the statement written down in the literature, then see the first paragraph of the paper A Gelfand representation theory for C$^*$-algebras by C. Akemann. This in turn refers to Dixmier's text Les C$^*$-algebres et leurs représentations, page 39, but when I look at the results on that page in the text (about existence of representations of C*-algebras), it would not be apparent how to derive the claim unless one is already familiar with the theory of C$^*$-algebras.

I also just stumbled (via Akemann's earlier paper The general Stone-Weierstrass problem) upon the following:

On the Borel structure of C-algebras* by E.B. Davies

Theorem 3.2 of the paper is just a bit more general than what you are asking for, showing that a certain envelope of each C$^*$-algebra embeds in the reduced atomic representation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.