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I was wondering if there is a characterisation for $C^*$-algebras (unital) for which the bidual does not have any central atoms. It is not sufficient for example to demand that the $C^*$-algebra does not have a non-zero commutative algebra direct summand:

Consider the $C^*$-algebra $$ A:=\bigl\{f\in C([0,2];\mathrm{M}_2(\mathbb{C}))\mid \ f \mbox{ is diagonal on } [0,1]\bigr\}. $$ Then $A$ does not have a non-zero commutative algebra direct summand, but the atomic part of the bidual should be $$ A^{**}_\mathrm{atomic}=\bigl\{f\in\ell^\infty([0,2];\mathrm{M}_2(\mathbb{C}))\mid f \mbox{ is diagonal on } [0,1]\bigr\} $$ which has plenty of central atoms. See this question.

It is true that the bidual of a $C^*$-algebra $A$ has no central atoms if and only if the state space $S(A)$ of $A$ has no singleton split faces. This is Corollary 5.33 and Corollary 5.34 in Geometry of State Spaces of Operator Algebras by Alfsen and Shultz. This characterises these algebras in terms of the state space, but knowing a condition on $A$ would be very interesting.

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    $\begingroup$ Why the tag jordan-algebra (a Jordan-algebra is a [non-necessarily-associative] commutative algebra satisfying the axiom $(xy)(xx)=x(y(xx))$)? $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 8:44
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    $\begingroup$ Because the self-adjoint part of a $C$*-algebra is a Jordan algebra (in fact it is a JB-algebra) with the product $x\circ y:=\frac{1}{2}(xy+yx)$. $\endgroup$
    – Mark Roelands
    Commented Oct 27, 2018 at 9:00
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    $\begingroup$ OK thanks, but how is this connected to the question? $\endgroup$
    – YCor
    Commented Oct 27, 2018 at 9:37
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    $\begingroup$ You could ask the same question for JB-algebras: Is there a characterisation for those JB-algebras that have no central projections in the bidual? $\endgroup$
    – Mark Roelands
    Commented Oct 27, 2018 at 9:43
  • $\begingroup$ I'm not sure what kind of condition you are looking for. But if you just translate '$A^{* *}$ has a projection that is both central and minimal' into a statement about $A$ you obtain '$A$ has a one-dimensional quotient.' $\endgroup$
    – Caleb Eckhardt
    Commented Oct 29, 2018 at 16:19

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