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I have a question about commutative direct summands of $C$*-algebras.

Let $A$ be a $C$*-algebra (with unit) and suppose that its bidual $A^{**}$ has a commutative direct summand, that is, $A^{**}=B\oplus C$ such that $B$ is non-zero and commutative. Does this force $A$ to have a non-zero commutative direct summand as well?

Thank you very much in advance for your feedback!

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Let $A$ be the continuous functions $f$ from $[0,2]$ into $M_2(\mathbb{C})$ such that $f(t)$ is diagonal for $0 \leq t \leq 1$. Then $A$ has no commutative direct summand, but the atomic part of its bidual should be equal to the bounded functions $f$ from $[0,2]$ into $M_2(\mathbb{C})$ such that $f(t)$ is diagonal for $0 \leq t \leq 1$, which contains plenty of commutative direct summands.

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    $\begingroup$ It is not clear to me that your description of the bidual is accurate since $C[0,1]^{**}$ is much larger than $\ell_\infty[0,1]$. Also, what is the gain of going from $C[0,1]$ to matrices? $\endgroup$
    – Tomasz Kania
    Commented Oct 24, 2018 at 17:28
  • $\begingroup$ The bidual of $C[0,1]$ is indeed much bigger, but its atomic part is $\ell_\infty[0,1]$. And the atomic part is a direct summand, so a direct summand of the atomic part is a direct summand of the bidual. I have to go to matrices in order to make sure that $A$ has no commutative direct summand; the problem with $C[0,1]$ is that it is commutative, so it has a commutative direct summand. $\endgroup$
    – Marten Wortel
    Commented Oct 25, 2018 at 6:14
  • $\begingroup$ If you only require a complemented subalgebra then the diagonal matrices are such a summand too. $\endgroup$
    – Tomasz Kania
    Commented Oct 25, 2018 at 11:25
  • $\begingroup$ I assumed the original question referred to algebra direct summands. $\endgroup$
    – Marten Wortel
    Commented Oct 26, 2018 at 6:33
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No. Take the algebra of continuous functions on some connected space so that it does not have non-trivial direct summands. On the other hand, in the second dual you will find minimal projections coming from point evaluations that give rise to non-trivial, one-dimensional summands.

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  • $\begingroup$ Thanks for your feedback! Do you mean if the second dual has central atoms? How would you prove that? $\endgroup$
    – Mark Roelands
    Commented Oct 23, 2018 at 8:21
  • $\begingroup$ @MarkRoelands, please see Section 3 here google.pl/url?sa=t&source=web&rct=j&url=https://… $\endgroup$
    – Tomasz Kania
    Commented Oct 23, 2018 at 8:44
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    $\begingroup$ Thank you! I edited the question to only allowing a non-zero direct summand. Would that also be true in general? $\endgroup$
    – Mark Roelands
    Commented Oct 23, 2018 at 13:22
  • $\begingroup$ @MarkRoelands, yes. Even more is true, please see Lemma 1 here: msp.org/pjm/1974/50-1/pjm-v50-n1-p11-s.pdf $\endgroup$
    – Tomasz Kania
    Commented Oct 23, 2018 at 17:48

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