I am trying to understand the Von Neumann decomposition, according to which every Von Neumann Algebra can be uniquely decomposed as integral (or direct sum) of factors. More specifically, I am trying to connect (if possible) the Von Neumann decomposition to the GNS theorem. Can someone please help me answer the following questions?
Given a normal state, is it true that its GNS representation is always a factor? The answer would be different if the dimension of the algebra is either finite or infinite?
Is it possible that the algebra obtained by applying GNS theorem on a irreducible state is not a factor?
From the quantum mechanics point of view, why is the Von Neumann decomposition important?
Hopefully someone more comfortable with quantum mechanics can answer your question 3. But I can help with 1 and 2.
Being a factor is a property of a von Neumann algebra, not a representation. So when you ask whether a GNS representation $\pi_\phi: M \rightarrow B(H_\phi)$ associated to a state $\phi$ is a factor, I assume you’re asking whether the von Neumann algebra generated by the image, $\pi_\phi(M)’’$, is a factor. For normal states, you can drop the double commutant, since the image will already be a von Neumann algebra.
The answer to question 1 is no. It’s helpful to consider the opposite end of the spectrum from factors. A factor is an algebra with the smallest possible center ($\mathbb{C}1$). An abelian algebra is an algebra with the largest possible center (the entire algebra). In general, an abelian von Neumann algebra (on a separable Hilbert space) is isomorphic to $L^\infty(\mu)$ for some probability measure $\mu$, which can be taken to be a measure on $\mathbb{R}$. If you take your algebra to be $L^\infty(\mu)$ and take your normal state to be integration against the probability measure $\mu$, then the GNS representation ends up being equivalent to the usual action of $L^\infty(\mu)$ on $L^2(\mu)$ by multiplication. This type of example shows that the image of the GNS representation doesn’t have to be a factor. It can be $L^\infty(\mu)$, which is abelian. If you take $\mu$ to be an average of a few point masses, then $L^\infty(\mu)$ is finite dimensional. If you take $\mu$ to be Lebesgue measure on some interval, then $L^\infty(\mu)$ is infinite dimensional. Therefore, whether you’re in finite or infinite dimensions, the image of the GNS representation isn’t guaranteed to be a factor by default.
The answer to question 2 is no, for goofy reasons. You can prove that a representation $\pi: M \rightarrow B(H)$ is irreducible if and only if $\pi(M)’ = \mathbb{C}1$. So if $\phi$ is a pure state (meaning its GNS representation is irreducible), then $\pi_\phi(M)’ = \mathbb{C}1$, which implies that the von Neumann algebra generated by the image, $\pi_\phi(M)’’$, is all of $B(H_\phi)$, which is a factor. This happens for every irreducible representation - not just the ones coming from the GNS construction. Their images will always be (or, more accurately, generate) type I factors.
I’ll end by saying: I suspect it would be better for you to try to connect the direct integral decomposition to the spectral theorem instead of trying to involve the GNS construction. The direct integral decomposition (resp. spectral theorem) uses central (resp. spectral) projections to chop up the action of your algebra (resp. operator) into much simpler pieces. You don’t really need the GNS construction to explore those connections. See, for example, the direct integral section of the spectral theorem wikipedia page for more.