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I am trying to study the paper Consistency of a Counterexample to Naimark’s Problem by Charles Akemann and Nik Weaver, and there is a claim in Lemma 1 of the paper that I am stuck at, which is as follows.

Claim: Let $ \mathscr{A} $ be a $ C^{*} $-algebra and $ g: \mathscr{A} \to \mathbb{C} $ a pure state on $ \mathscr{A} $. Let $ q \in \mathscr{A}^{**} $ be the support projection of $ g $. Also, let $ y $ be the central cover of $ q $ in $ \mathscr{A}^{**} $; in other words, $ y $ is the smallest projection $ p \in \mathcal{Z}(\mathscr{A}^{**}) $ such that $ q \leq p $. Then $ y \mathscr{A}^{**} $ is isomorphic to $ B(\mathcal{H}) $ for some Hilbert space $ \mathcal{H} $.

The claim may seem obvious to the many operator theorists more capable than myself, but I am just not getting it. I believe that the GNS construction is needed here (as the claim is about a pure state, which corresponds to an irreducible $ * $-representation of $ \mathscr{A} $ on some Hilbert space $ \mathcal{H} $), but this is the furthest that I have gotten.

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1 Answer 1

up vote 4 down vote accepted

For each state $\phi$ on $A$ let $\pi_\phi: A \to B(H_\phi)$ be the corresponding GNS representation. Let $\pi: A \to B(H)$ be the direct sum of all of these representations. The von Neumann algebra $\pi(A)'' \subseteq B(H)$ is called the enveloping von Neumann algebra of $A$ and it is naturally isomorphic to $A^{**}$. (See C${}^*$-algebras and Their Automorphism Groups by Pedersen.)

For each $\phi$, since $H_\phi \subseteq H$ we have a natural projection from $\pi(A)'' \subseteq B(H)$ onto $\pi_\phi(A)'' \subseteq B(H_\phi)$. The kernel of this projection is a weak* closed ideal of $\pi(A)'' \cong A^{**}$ and hence it has the form $zA^{**}$ for some central projection $z$ in $A^{**}$. Then $y = 1-z$ is the central cover of $\phi$ and $A^{**} = yA^{**} \oplus zA^{**}$, so that $\pi_\phi(A)'' \cong A^{**}/zA^{**} \cong yA^{**}$.

So the general fact is that $\pi_\phi(A)'' \cong yA^{**}$. If $\phi$ is pure then $\pi_\phi$ is irreducible and $\pi_\phi(A)'' = B(H_\phi)$.

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Thank you very much for your answer, Professor Weaver! – user36116 Dec 5 '13 at 6:27
Sure, no problem. – Nik Weaver Dec 5 '13 at 6:43

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