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Let $A$ be a unital $C^{\ast}$-algebra and let $P(A)$ be a space of pure states on $A$ (a state $\omega$ is called pure it is an extreme point in space of states).

a) Is the space $P(A)$ compact with respect to the weak-$\ast$ topology? In the commutative case it appers that $P(A)$ is the space of nonzero linear characters and this space is compact, is this also true in the noncommutative case?

b) In concrete situation, $A=B(H)$, the space of bounded linear operators on the Hilbert space is it true that $P(A)$ is extremely disconnected? I would be grateful if anyone could help.

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  • $\begingroup$ In case b) the pure states are simply the orthogonal projections of rank 1, or ? And this space is connected. $\endgroup$ – jjcale Jul 14 '12 at 16:13
  • $\begingroup$ No, there are oher pure states besides these. For example, the pullbacks of the pure states of the Calkin algebra $B(H)/K(H)$, where $K(H)$ is the set of compact operators. $\endgroup$ – Yulia Kuznetsova Jul 15 '12 at 20:58
  • $\begingroup$ Maybe another formulation of this comment: as vector states of an irreducible representations induce pure states and are unitarily equivalent, the previous counterexample can be stated as the existence of a disjoint irreducible representation in that particular case. $\endgroup$ – Noix07 Feb 18 '16 at 10:39
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a) The set of pure sates may not be *-closed in the dual space of $A$, see for example a paper of J.Glimm: http://www.ams.org/journals/tran/1960-095-02/S0002-9947-1960-0112057-5/ So of course it may be not compact.

Maybe it is reasonable to consider the spectrum of $A$ insead, i.e. the space $\hat A$ of all classes of irreducible representations of $A$. Look at Dixmier "C*-algebras and their representations", chapter 3. In particular, 3.1.8 tells you that if $A$ is unital then $\hat A$ and the space of primitive ideals are both compact but maybe not Hausdorff.

b) For $A=B(H)$, he set $P(A)$ is much bigger than simply the set of states of tpe $f_x(a)=(ax,x)$ for $x\in H$, $a\in A$. But the set of these $f_x$ is arc connected, because the unit sphere in $H$ is arc connected. So $P(A)$ cannot be totally disconnected.

[Hope I didn't write anything silly at this late hour]

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    $\begingroup$ To complement Julia's answer, maybe it's worth to add that what is usually called the pure state space of A, that is the weak* closure of the space of pure states $P(A)$, is of course compact, since it is closed in the (compact) state space. This is an elementary comment, but maybe there can be some confusion at times between the pure state space and the space of pure states. $\endgroup$ – Fabien Besnard Apr 16 '13 at 12:25

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