# Minimal projections and pure states in scattered $C^*$ algebras

The question I have came up when reading a paper of Ghasemi and Koszmider https://arxiv.org/abs/1611.00221 on scattered $C^*$ algebras and the non-commutative Cantor-Bendixson derivatives. Throughout the paper, the parallels between scattered $C^*$ algebras and $C(K)$ spaces, $K$ a scattered, compact, Hausdorff topological space, are discussed.

Of particular importance are minimal projections in scattered $C^*$ algebras. In this context, a projection $p$ in the $C^*$ algebra is minimal if $pAp=\mathbb{C}p$. In the $C(K)$ case, the minimal projections are indicators of singletons $1_{\{\varpi\}}$, where $\varpi\in K$ is isolated.

If $p$ is a minimal projection in $A$, then for any $a\in A$, $pap=\lambda_p(a) p$ for some $\lambda_p(a)\in \mathbb{C}$. Then $a\mapsto \lambda_p(a)$ defines a pure state. In the $C(K)$ case, if $p=1_{\{\varpi\}}$ is the singleton indicator at the isolated point $\varpi$, $\lambda_p(f)=f(\varpi)$. So $\lambda_p=\delta_{\varpi}$, the Dirac evaluation functional.

As mentioned above, the map $\Phi$ which takes $p\mapsto \lambda_p$ takes the minimal projection $p$ to the pure state $\lambda_p$. What properties does $\Phi$ have? Is it injective? What is the range (that is, which pure states arise as $\lambda_p$ for some $p$)?

• It's certainly injective --- if $\lambda_p = \lambda_q$ then $\lambda_p(q) = 1$ since $\lambda_q(q) = 1$. But that means $pqp = p$, and similarly we get $qpq = q$. This can only happen if $p = q$. I don't think there's any meaningful answer to "what is the range". – Nik Weaver May 5 '17 at 18:15
• Could you tell me why you think there's no meaningful answer to that question? – user496750 May 5 '17 at 18:26
• I just don't think there's anything more informative one can say than "the pure states of the form $\lambda_p$". – Nik Weaver May 6 '17 at 4:07