1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Munk Jul 20 '18 at 1:30
  • $\begingroup$ Yes I am interested in vN algebras $\endgroup$ – mathlover Jul 20 '18 at 4:47
  • $\begingroup$ Transported meaning is $\phi=\omega_{\xi}\circ\pi$ $\endgroup$ – mathlover Jul 20 '18 at 5:31
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.