DrDu said:
rubi, I think at least in ordinary QM vector representations of mixed states are reducible, while those of pure states are irreducible. Is this true for all C* algebras?
This is a little bit subtle. What is true is the following: The vector representation ##(\mathcal H_\omega,\pi_\omega,\Omega_\omega)## given by the GNS construction of a state ##\omega## is irreducible if and only if ##\omega## is pure. However, that doesn't mean that all vector representations are irreducible. For example, we can take the irreducible GNS representation ##(\mathcal H_\omega,\pi_\omega,\Omega_\omega)## of a pure state ##\omega## and specify the following data: ##\mathcal H=\mathcal H_\omega\oplus\mathcal H_\omega##, ##\pi=\pi_\omega\oplus\pi_\omega## and ##\Omega=\frac{1}{\sqrt{2}}\Omega_\omega\oplus\Omega_\omega##. Define a state ##\xi(A)=\left<\Omega,\pi(A),\Omega\right>_{\mathcal H}##. We then find $$\xi(A) = \frac{1}{2}\left<\Omega_\omega\oplus\Omega_\omega,\pi_\omega(A)\Omega_\omega\oplus\pi_\omega(A)\Omega_\omega\right>_{\mathcal H}=\frac{1}{2}\left<\Omega_\omega,\pi_\omega(A)\Omega_\omega\right>_{\mathcal H_\omega}+\frac{1}{2}\left<\Omega_\omega,\pi_\omega(A)\Omega_\omega\right>_{\mathcal H_\omega}=\frac{1}{2}\omega(A)+\frac{1}{2}\omega(A)=\omega(A) \text{.}$$ Thus ##\xi=\omega##. However, ##(\mathcal H,\pi)## is clearly reducible, because ##\mathcal H\oplus 0## is an invariant subspace. So we have found a reducible vector representation of a pure state ##\omega##. This is of course possible, because only the GNS representation of ##\omega## needs to be irreducible. Our representation ##(\mathcal H,\pi,\Omega)## is not the GNS representation of ##\omega##. So even in QM, vector representations needn't be irreducible.
So let's look at type ##III_1## factors ##\mathfrak A## now. Can there be irreducible vector representations? I think what is going on is the following: At first, the fact that every state on ##\mathfrak A## must be mixed seems to invalidate this. However, what it really means is the following: Let's take the GNS representation ##(\mathcal H_\omega,\pi_\omega,\Omega_\omega)## induced by a state ##\omega## on ##\mathfrak A##. Clearly, it must be reducible due to the theorem I stated in the beginning. However, that means that there is an invariant subspace of ##\mathcal H_0\subseteq \mathcal H_\omega## and I can take it to be minimal, i.e. ##(\mathcal H_0,\pi_\omega\rvert_{\mathcal H_0})## is irreducible and I can take any vector ##\Psi\in\mathcal H_0##. Then I can define the state ##\xi(A)=\left<\Psi,\pi_\omega(A)\Psi\right>_{\mathcal H_0}## and it clearly defines a vector state in an irreducible representation on ##\mathfrak A##. How can this be? The answer is that the data ##(\mathcal H_0, \pi_\omega\rvert_{\mathcal H_0},\Psi)## does not arise as the GNS data of some state on ##\mathfrak A##. If I apply the GNS construction to the state ##\xi##, I will end up with a reducible representation, because Yngvason tells us that ##\xi## must be a mixed state and our theorem tells us that mixed states produce reducible GNS representations.
(However, I still find it fishy. The states should form a convex set and as such, it should have extremal points, which should correspond to pure states. I need to think about this more.)