# GNS representation doubt

Let $M$ be a von Neumann algebra inside $B(\mathcal{H})$, if we take $\xi$ $\in \mathcal{H}$, consider vector state $\omega_{\xi}$,

$(i)$ Does there exist GNS Hilbert space coming from $M$ such that whose transported state will be $\omega_{\xi}$?
$(ii)$ Furthermore are GNS Hilbert spaces are always separable?
$(iii)$ what is the importance of doing GNS construction just to only get cyclic vector, can anybody help me out connecting more ideas with GNS construction?

• For (ii), von Neumann algebra almost never represent onto separable Hilbert spaces (finite-dimensionality is necessary I reckon). (i) I am unsure of what the questions is, what is the transported state you mentioned? (iii) Are you only interested in von Neumann algebras or C*-algebras in general? – Munk Jul 20 '18 at 1:30
• Yes I am interested in vN algebras – mathlover Jul 20 '18 at 4:47
• Transported meaning is $\phi=\omega_{\xi}\circ\pi$ – mathlover Jul 20 '18 at 5:31

If $M$ is sot/wot separable (i.e., most known and used von Neumann algebras out there) and $\phi$ is normal, then the Hilbert space $H_\phi$ is separable. To see this, let $\{x_n\}\subset M$ be a dense sequence; then for any $x\in M$ there exists a bounded subnet $\{x_{n_j}\}$ ($\|x_{n_j}\|\leq c$) with $\psi(x_{n_j})\to \psi(x)$ for all normal states $\psi$ (the "bounded" is obtained via Kaplansky on the balls of radius $n$). Fix $\varepsilon>0$ and $\eta\in H_\phi$. Then there exists $x\in M$ with $\|\eta-\hat x\|<\varepsilon$. So \begin{align} \limsup_j \|\hat x_{n_j}-\eta\|_\phi &\leq\limsup_j\|\hat x_{n_j}-\hat x\|_\phi+\|\hat x-\eta\|_\phi\\ \ \\ &=\limsup_j\phi((x_{n_j}-x)^*(x_{n_j}-x))+\varepsilon\\ \ \\ &\leq (c+\|x\|)\limsup_j\phi(x_{n_j}-x)+\varepsilon\\ \ \\ &=\varepsilon. \end{align} As $\varepsilon$ was arbitrary, $\hat x_{n_j}\to\eta$, and $H_\phi$ is separable.

The original importance of the GNS construction is that it allows you to start with an abstract C$^*$-algebra, and construct a representation on a Hilbert space. By adding the GNS representations over all (classes of) states, one gets a faithful representation of your abstract C$^*$-algebra as bounded operators on a Hilbert space. And the construction is explicit; in particular, having a cyclic vector allows one to define operators explicitly; a nice example of this is the proof that a non-degenerate representation from an ideal extends to the whole algebra (see Lemma I.9.14 in Davidson's C$^*$-Algebra By Example for details).

As for the "transport" question, I'm not sure what you are after. If you start with any state and do GNS, you get a vector state. If you start with a vector state and do GNS with it, nothing too remarkable happens.

• More specifically can we say any two cyclic representations of $C^*$ algebra is unitary equivalent as a consequence of GNS @Martin? – mathlover Jul 21 '18 at 9:24
• No, not at all. That would make all GNS representations equivalent. – Martin Argerami Jul 21 '18 at 12:30