Von Neumann algebra decomposition as integral of factors and mixed state decomposition as sum of irreducible states

Question

My question is the following. It is known that any Von Neumann algebra can be uniquely decomposed as integral over algebra factors. It is also know that any mixed state can be uniquely expressed as sum of irreducible states. There is any connection between these two decompositions? Maybe by using the GNS theorem?

Answer 1

I think I found the solution to my own question, but please let me know if you think this answer is not good or not complete. I think the answer can be found in Theorem 4.4.7 of Bratteli - Robinson Vol.I, page 447. The theorem says the following: Let $U$ be a separable C-* algebra, $\pi$ be a nondegenerate representation of the algebra on a separable Hilbert space $H$, and $B$ be an abelian Von Neumann algebra contained in the commutant of $U$. It follows that there is a measure space $Z$, a positive bounded measure $\mu$ on $Z$, a measurable family of Hilbert spaces $H (z)$ with $z \in Z$, a measurable family of representations $\pi (z)$ of the algebra $U$ on the Hilbert spaces $H (z)$ and a unitary operator $U$ such that $U B U^{*}$ is the set of diagonalizable operators on $\int_{Z}^{\oplus} \mu (z) H (z)$, and such that for every element $A \in U$, $ U \pi (A) U^{*} = \int_{Z}^{\oplus} \mu (z) \pi (z) (A)$. Moreover, if $B$ is equal to the center of the algebra, then the decomposition of the theorem is the decomposition over factors, and if $B$ is the maximal abelian in $\pi (U)^{'}$ then the decomposition is over pure states.