# Separable C* algebras and type I states

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!

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 $$\pi_\omega(A) \cong \pi_{\psi_1}(A) \oplus \pi_{\psi_2}(A)$$.

• 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.. Aug 28 at 22:12
• Right, exactly. Aug 28 at 22:13