3
$\begingroup$

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!

$\endgroup$

2 Answers 2

1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$ 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. $\endgroup$ Oct 25, 2018 at 6:14
  • $\begingroup$ If you only require a complemented subalgebra then the diagonal matrices are such a summand too. $\endgroup$ Oct 25, 2018 at 11:25
  • $\begingroup$ I assumed the original question referred to algebra direct summands. $\endgroup$ Oct 26, 2018 at 6:33
4
$\begingroup$

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.

$\endgroup$
4

Your Answer

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

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