# Folium in GNS construction and von Neumann algebras

The GNS construction allows one to represent a $C^*$-algebra as the algebra of bounded operators on a Hilbert space when a state is fixed, this state being represented as a vector on the Hilbert space. Another state will be represented as a vector on a different Hilbert space. I am trying to find back the usual quantum picture: states are vectors or more generally positive trace class operators on a fixed Hilbert space.

Given a state $\omega$ on the $C^*$-algebra $\mathfrak{A}$, associated to a representation $(\pi_{\omega},\mathcal{H})$, one can define $\omega$-normal states: $\mu$ is an $\omega$-normal state iff there exists a positive trace class operator $\rho$ on $\mathcal{H}$ verifying $\mu(A)=\operatorname{tr}(\rho\pi_{\omega}(A))$ for all $A \in \mathfrak{A}$.

The folium of $\omega$ is the set of $\omega$-normal states, which corresponds basically to states which "live" on the same Hilbert space as $\omega$.

$(\ast)$: One can also show that the folium of a faithful state is *-weak-topology dense among the set of states.

That's the furthest I could get in understanding all this.

What I want is one folium for all states, so I probably need to put some conditions on my states ensemble.

One brute force way to do this is to ask all states to be positive trace class operators so there is a chance they are $\omega$-normal states for some specific $\omega$. von Neumann algebras are duals of trace class operators ensembles, so I get the feeling that at some point one has to take the weak closure of the $C^*$-algebra if we want to achieve this "one folium" goal, but then I can't find any clear explanation of how we do this, and if the positivity condition can't restrict us to something more specific than a von Neumann algebra.

Also, are there any weaker conditions that do the job so we don't need von Neumann algebras? For example using proposition $(\ast)$?

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For example, if your C*-algebra $A$ contains an element with spectrum equal to $[0,1]$, you'll never find one folium for all states. This is essentially asking for a single measure $\mu$ on $[0,1]$ so that all regular Borel measures are absolutely continuous with respect to $\mu$ (such a measure, of course, doesn't exist).
Yemon, yes. But I'm sure if $A$ has something with spectrum $[0,1]$ it must also have something hermitian with spectrum $[0,1]$ –  Caleb Eckhardt Aug 22 '13 at 1:51