GNS construction and separability.

Question

Start with a $C^*$-algebra $\mathcal{A}$ and take an state (positive normalized linear functional). Then build its GNS representation and get the GNS Hilbert space $\mathcal{H}_{GNS}$. The questions are:

Q1) If $\mathcal{A}$ is separable (in the $C^*$ norm topology), then $\mathcal{H}_{GNS}$ is separable?

Q2) If we start with other Hilbert space $\mathcal{H}$, and consider the algebra of all bounded operators $\mathcal{A} = \mathcal{B}(\mathcal{H})$ and a (normalized) vector state $\Psi \in \mathcal{H}$. It's know that $\mathcal{B}(\mathcal{H})$ is not separable. But, is $\mathcal{H}_{GNS}$ separable?

I will find very helpful any bibliography (theorems, examples, counterexamples) about separability of the GNS representation? For example, a challenger question:

Q3) If the original $C^*$-algebra was indeed a $W^*$-algebra (or a von Neumann algebra), is there any relation between the type of algebra (I, II, III, etc.) and/or the fact the state is/isn't normal, with the separability of $\mathcal{H}_{GNS}$?

Thanks,

D

Answer 1

1. Yes, the Hilbert space is separable. Suppose that $\{b_n\}$ is a dense sequence in $\mathcal A$. Note that $\mathcal H$ is constructed as the closure of (a quotient of ) $\mathcal A$. Given $h\in \mathcal H$, by construction there exists a sequence $\{a_n\}\subset \mathcal A$ with $\hat a_n\to h$. Now let $b_{k_n}\in \mathcal A$ with $\|a_n-b_{k_n}\|<1/2^n$. Then $$\tag1 \|h-\hat b_{k_n}\|\leq\|h-\hat a_n\|+\|\hat a_n-\hat b_{k_n}\|, $$ and $$\tag2\|\hat a_n-\hat b_{k_n}\|^2=f((a_n-b_{k_n})^*(a_n-b_{k_n}))\leq\|a_n-b_{k_n}\|^2 $$ Then $(1)$ and $(2)$ together imply $$ \limsup\|h-\hat b_{n_k}\|\leq\limsup 1/2^n=0. $$

2. If you take a vector state $\psi$, you have $\psi=\langle \cdot \,e_1,e_1\rangle$ for some $e_1\in\mathcal H$ with $\|e_1\|=1$. Complete to an orthonormal basis $\{e_n\}$. You have $$\psi(A)=\text{Tr}\,(E_{11}A)=A_{11}.$$ When you do GNS, you need to quotient by the subspace $$ \mathcal H_0=\{A\in \mathcal{B(H)}:\ \psi(A^*A)=0\}. $$ The condition $\psi(A^*A)=0$ is the same as $AE_{11}=E_{11}A=0$. So $B-A\in \mathcal H_0$ if and only if $E_{11}B=E_{11}A$, $BE_{11}=AE_{11}$. That is, $\mathcal{H_{\psi}=B(H)/H_0}$ is spanned by the classes of $E_{11}, E_{12}, E_{13},\ldots$, and $E_{21}, E_{31}, E_{41},\ldots$. Being spanned by a countable set, $\mathcal H_\psi$ is separable.

3. When $\psi$ is any normal state, if $M$ is separable then so is $\mathcal H_\psi$. Suppose that $X=\{b_n\}\subset M$ is dense and countable. Using Kaplansky, we may tweak $X$ so that any $a\in M$ is a limit of a bounded net in $X$ (basically, write $X=\bigcup_nX_n$, where $X_n$ is the set obtained by applying the Kaplansky function to the elements of $X$ so that they approach those elements of $M$ with norm less than $n$). Now, given $a\in M$, there exists a net $\{b_j\}$ in $X$ with $b_j\to a$ in the $\sigma$-weak topology. Now \begin{align} \|b_j-a\|^2&=\psi((b_j-a)^*(b_j-a))=\psi(b_j^*b_j)+\psi(a^*a)-2\operatorname{Re}a^*b_j \to 0, \end{align} where the limit works because $\{b_j\}$ is bounded. So $\hat X$ is dense in $\mathcal H_\psi$.

I don't think type has anything to do with it.