Definition of normal state on von Neumann algebra I have a question about the notion of normal state on a von Neumann algebra and its relation to a particular representation of the algebra. Let me take from the book by Bratteli and Robinson:
Definition 2.4.20: A state $\omega : \mathfrak{M} \to \mathbb{C}$ on a von Neumann algebra $\mathfrak{M}$ is called normal if $\omega ( \text{l.u.b.}_\alpha ~ A_\alpha ) = \text{l.u.b.}_\alpha ~ \omega (A_\alpha)$, where $\text{l.u.b.}$ is the least upper bound and $\{ A_\alpha \}$ is an increasing net in $\mathfrak{M}_+$ with an upper bound.
Theorem 2.4.21: Let $\omega$ be a state on a von Neumann algebra $\mathfrak{M}$ acting on a Hilbert space $H$. Then $\omega$ is normal if and only if there exists a positive, trace-class operator $\rho$ on $H$ with $\text{Tr} ( \rho) = 1$ such that
$$ \omega (A) = \text{Tr}  ( \rho A ) , \quad \forall A \in \mathfrak{M}  . $$
The definition is clearly independent of any representation of the algebra, while from the theorem it seems that a state might be normal in one representation and non-normal in another. Also, every state is a vector state in its GNS representation, so it's also normal. But then how is it possible that the definition of normal be representation independent?
 A: You make a good observation. But it is mostly irrelevant, and I will try to explain how.
While von Neumann algebras can be characterized abstractly as those C$^*$-algebras that are Banach space duals, this has zero effect in practice. Every single definition and use of a von Neumann algebra you will find there, is of a concrete von Neumann algebra, that is a sot/wot-closed $*$-subalgebra of some $B(H)$ (equivalently, equal to its double commutant). It is possible, though,  to take a von Neumann algebra and represent it non-normally; this happens in several situations:

*

*as you mention, do GNS for a non-normal state. If the state is faithful, you are now representing your algebra in a way that the state became normal, and your algebra is no longer a von Neumann algebra in the new ambient.


*when you embed your von Neumann algebra in its double dual (i.e., in its enveloping von Neumann algebra), it becomes dense in another von Neumann algebra and all states become normal, so again what you get is no longer a von Neumann algebra.


*if your von Neumann algebra is a factor, then every representation is faithful. In particular, if you take an irreducible representation, you will get $\pi(\mathfrak M)\subset B(H_\pi)$ with $\pi(\mathfrak M)''=B(H_\pi)$. If $\mathfrak M$ is of type II or III, you get that $\pi(\mathfrak M)$ is not a von Neumann algebra.
In the end, what matters is there there is usually no reason to take $\mathfrak M$ and putting somewhere else. And when you have to do it, do it with a normal representation so that $\pi(\mathfrak M)$ is still a von Neumann algebra.
