2
$\begingroup$

Let $A$ be a unital C$^*$-algebra such that the linear span of its projections lies norm dense, and such that below each projection there is an atomic projection. Suppose furthermore that $A$ is simple, i.e. that it has no non-trivial central elements.

Does $A$ have to be a von Neumann algebra? Since if this is true $A$ will be a type I von Neumann factor, the question reduces to: is $A$ isomorphic to $B(H)$ for some Hilbert space $H$?

A result by Kaplansky establishes this when $A$ is an AW$^*$-algebra, i.e. when its abelian subalgebras are monotone complete. Because the projections lie norm-dense the abelian subalgebras will be $\sigma$-monotone complete. Is this enough?

$\endgroup$
3
  • 1
    $\begingroup$ Do you mean "simple" (no nontrivial ideals) or "trivial centre"? They are not the same. $\endgroup$ Feb 14, 2018 at 20:08
  • $\begingroup$ I meant trivial centre. I've been trying to find some results about the difference between simple and trivial centre. Could you maybe give me a reference for the difference (and if one implies the other). $\endgroup$
    – John
    Feb 15, 2018 at 17:18
  • $\begingroup$ If $A$ is unital, then simple implies trivial centre. The reverse implication is false, with the easiest example being $B(\ell^2(\mathbb N))$, which has trivial centre and the non-trivial ideal $K(\ell^2(\mathbb N))$. $\endgroup$ Nov 17, 2019 at 0:47

1 Answer 1

2
$\begingroup$

An easy counterexample, if the requirement is to have trivial centre, is to take $H$ infinite-dimensional, and put $A=K(H)+\mathbb C\,I$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .