Let $\mathcal{A}$ be a von Neumann algebra. A GNS-construction of state $\omega:\mathcal{A}\rightarrow \mathbb{C}$ is the representation $\pi_\omega$ on the Hilbert space $\mathcal{H}^{} _ {\omega}$(which is a completion of the quotient space $\mathcal{A}/\mathfrak{J}$, where $\mathfrak{J}$ is a Gelfand ideal whose elements satisfying $\omega(X^*X)=0$), defined as $\pi _ \omega (A) [B] = [AB] $ for the equivalence class $[A]$. A state $\omega$ is primary if $\pi _ \omega$ is primary, i.e. $\pi _ \omega(\mathcal{A})''$ is a factor(or has trivial center).
A state $\omega_1$ is in the folium of representation $\pi _ \omega$ (or $\pi _ \omega$-normal) when there is a positive linear form $\rho$ on $\mathcal{H} _ \omega$ such that $\omega_1=\DeclareMathOperator{\Tr}{Tr}\Tr(\rho \pi_\omega (A))$.
In Local Quantum Physics by R. Haag, Theorem 2.2.17 in Chapter III says the following:
(...) (iii) Every state in the folium of a primary representation is primary.
If we consider the folium of a representation is a set of "physical" states(which is given when we fix a single "physical" state), and primary representation as a "representation containing only one irreducible representation type", then the Theorem above makes sense: If a representation basically contains only one irreducible representation, then every "physical" states related to it also need to have the same irreducible representations only, hence it must be primary.
- Is my understanding here correct? Can I understand the folium of representation and primary representation as above?
- If so, How can I show this statement rigorously? To show this, I need to take a $\pi_\omega-$normal state $\omega_1$ then show $\pi_{\omega_1}$ is primary. But it is hard to find the relation between the representations $\pi_\omega$ and $\pi_{\omega_1}$ for me.
The following is a rigorous proof of the statement:
Let $\omega$ be primary and $\omega_1$ be given by $\omega_1(A) = \textrm{Tr}(\rho\pi_\omega(A))$ where $\rho$ is a positive trace-class operator on $H_\omega$ of trace 1. Let $\mathbb{B}_2(H_\omega)$ be the space of Hilbert-Schmidt operators on $H_\omega$. There is a natural left action of $\mathbb{B}(H_\omega)$ on this Hilbert space. We observe that this left action induces a representation of $\mathcal{A}$ on $\mathbb{B}_2(H_\omega)$, which we shall denote by $\pi_2$. We note that $\pi_2(\mathcal{A})'' \simeq \pi_\omega(\mathcal{A})''$ since $\pi_2$ is the restriction of the normal left action of $\mathbb{B}(H_\omega)$ on $\mathbb{B}_2(H_\omega)$ and therefore extends to a normal representation of $\pi_\omega(\mathcal{A})''$. In particular, $\pi_2(\mathcal{A})''$ is a factor.
Now, consider the map $i: \mathcal{A} \rightarrow \mathbb{B}_2(H_\omega)$ given by $i(A) = \pi_\omega(A)\rho^{1/2}$. We observe that,
$$\omega_1(A^*A) = \textrm{Tr}(\rho\pi_\omega(A)^*\pi_\omega(A)) = \textrm{Tr}((\pi_\omega(A)\rho^{1/2})^*(\pi_\omega(A)\rho^{1/2})) = ||\pi_\omega(A)\rho^{1/2}||_{HS}^2$$
Hence, $i$ induces an isometric embedding $H_{\omega_1} \hookrightarrow \mathbb{B}_2(H_\omega)$. We observe that this embedding is covariant under the action of $\mathcal{A}$, so we may consider $H_{\omega_1}$, together with the action of $\mathcal{A}$ on it, as a subspace of $\mathbb{B}_2(H_\omega)$. This subspace is invariant under $\mathcal{A}$, so there exists some projection $p \in \pi_2(\mathcal{A})'$ s.t. $H_{\omega_1} = p\mathbb{B}_2(H_\omega)$. This implies that $\pi_{\omega_1}(\mathcal{A})'' = \pi_2(\mathcal{A})''p$. As $\pi_2(\mathcal{A})''$ is a factor, so is $\pi_2(\mathcal{A})''p$, and thus $\pi_{\omega_1}(\mathcal{A})''$ is a factor.
The following are some clarifications on why $\pi_2(\mathcal{A})'' \simeq \pi_\omega(\mathcal{A})''$:
We observe that $\mathbb{B}(H_\omega)$ acts on the Hilbert space $\mathbb{B}_2(H_\omega)$ by $S \cdot T = ST$ where $S \in \mathbb{B}(H_\omega)$ and $T \in \mathbb{B}_2(H_\omega)$. One easily check that this is a normal faithful representation so it induces a normal embedding $\pi: \mathbb{B}(H_\omega) \hookrightarrow \mathbb{B}(\mathbb{B}_2(H_\omega))$. (In fact, $\mathbb{B}_2(H_\omega) \simeq H_\omega \otimes \overline{H_\omega}$ and $\pi$ is simply given by $T \mapsto T \otimes 1$.) $\pi_2$ is then simply $\pi \circ \pi_\omega$ and $\pi_2(\mathcal{A}) = \pi(\pi_\omega(\mathcal{A}))$. Since $\pi_\omega(\mathcal{A})$ is weakly dense in $\pi_\omega(\mathcal{A})''$ and $\pi$ is normal, we have $\pi(\pi_\omega(\mathcal{A})'') \subseteq \pi_2(\mathcal{A})''$. But since $\pi_\omega(\mathcal{A})''$ is a von Neumann algebra and $\pi$ is normal, the range must be a von Neumann algebra. The range clearly contains $\pi_2(\mathcal{A})$, so the range must contain the vNa generated by $\pi_2(\mathcal{A})$, i.e., $\pi_2(\mathcal{A})''$. Hence, $\pi(\pi_\omega(\mathcal{A})'') = \pi_2(\mathcal{A})''$. Since $\pi$ is injective, this means it restricts to an isomorphism between $\pi_\omega(\mathcal{A})''$ and $\pi_2(\mathcal{A})''$.