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Let $A$ be a separable $C^*$-algebra and let $\omega$ be a state on $A$. Then there is an "orthogonal" probability measure $\mu$ on the pure state space $P(A)$ of $A$ such that $\omega(x) = \int_{P(A)} \psi(x) \, d\mu(\psi)$ [Takesaki 1, IV.6.28]).

If I understand correctly the orthogonality of $\mu$ means that the GNS representation of $\omega$ is (unitarily equivalent to) a direct integral: $$ (H_\omega,\pi_\omega) = \int_{P(A)}^\oplus (H_\psi,\pi_\psi) \, d\mu(\psi) $$ and $\Omega_\omega = \int_{P(A)}^\oplus \Omega_\psi\,d\mu(\psi)$ (see for example [Takesaki 1, IV.8.31].

But doesn't this imply that the von Neumann algebra $\pi_\omega(A)''$ takes the form $$\pi_\omega(A)'' = \int_{P(A)}^\oplus\pi_\psi(A)'' \,d\mu(\psi) = \int_{P(A)}^\oplus B(H_\psi) \,d\mu(\psi)$$ (because $\psi$ is pure one has $\pi_\psi(A)''=B(H_\psi)$). This would imply that $\pi_\omega(A)''$ is a type I von Neumann algebra since it can be written as a direct integral of type I factors, right? The argument must be wrong since not every state on a separable $C^*$-algebra is a type I state (see the answer to this question Factor states on C*-algebras).

Any help would be much appreciated!

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1 Answer 1

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I don't think $\pi_\omega(A)''$ has that form. For example, take $A = M_2$ and let $\omega$ be the normalized trace. Then $\omega = \frac{1}{2}(\psi_1 + \psi_2)$ where $\psi_i(x) = \langle xe_i, e_i\rangle$, for $x \in M_2$ and $\{e_1,e_2\}$ the standard basis of $\mathbb{C}^2$. That is, $\omega$ is the integral $\int \psi\, d\mu(\psi)$ where $\mu = \frac{1}{2}(\delta_{\psi_1} + \delta_{\psi_2})$. Then $\pi_\omega(M_2)'' \cong \pi_{\psi_1}(M_2)'' \cong \pi_{\psi_2}(M_2)'' \cong M_2$, so $\pi_\omega(M_2)'' \not\cong \pi_{\psi_1}(M_2)'' \oplus \pi_{\psi_2}(M_2)''$.

We do have an isomorphism between $H_\omega$ and $H_{\psi_1} \oplus H_{\psi_2}$ which takes $\pi_\omega(x)$ to $\pi_{\psi_1}(x) \oplus \pi_{\psi_2}(x)$, but of course that doesn't make $\pi_\omega(A) \cong \pi_{\psi_1}(A) \oplus \pi_{\psi_2}(A)$.

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    $\begingroup$ Thanks a ton! Now that I see it, it was a rather stupid mistake. Of course we only get the diagonal action in the direct sum.. $\endgroup$
    – Lauritz
    Aug 28 at 22:12
  • $\begingroup$ Right, exactly. $\endgroup$
    – Nik Weaver
    Aug 28 at 22:13

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