# Clarification on predual on existence of separating vector

We know predual of a von Neumann algebra $M$ as a Banach space is independent of Hilbert space where the $M$ is represented. Now the question is if we represent $M$ in $B(\mathcal{H})$, where $M$ has separating vector, then is the predual is same as weak operator dual, finally boiled down the question is weak*-topology and weak operator topology on $M$ are same?? If they are same why we need to study complicated elements of predual as weak operator dual elements are nice to see.

So I think you are asking:

Let $(M,H)$ be a von Neumann algebra with a separating vector $\xi\in H$. Is the weak operator topology on $M$ the same as the weak$^*$-topology coming from $M_*$?

The answer is of course "yes". As you allude to, if $(M,H)$ has a separating vector, then every $\omega\in M_*^+$ is of the form $\omega = \omega_\eta$ where $\omega_\eta(x) = (x\eta|\eta)$ for $x\in M$. (This is a result from the comparison theory of projections; for example Corollary 1.12 in Chapter V of Takesaki).

From the polar decomposition of normal functionals, if follows that each $\omega\in M_*$ has the form $\omega = \omega_{\alpha,\beta}$ for some $\alpha,\beta\in H$, and from this it follows that the WOT and the weak$^*$-topology agree on $M$.

Given this, why do we bother with the weak$^*$-topology?

Well, because not every von Neumann algebra has a separating vector! If $(M,H)$ is with $H$ separable, then consider $M$ acting on $H\otimes\ell^2$ as $M\otimes 1$. Let $(\xi_n)$ be a dense sequence in $H$, and set $$\xi = \sum_n 2^{-n/2} \|\xi_n\|^{-1} \xi_n \otimes e_n$$ where $(e_n)$ is an orthonormal basis for $\ell^2$. Then $\xi$ is a unit vector, and if $x\in M$ with $(x\otimes 1)\xi=0$ then $x\xi_n=0$ for all $n$ and so $x=0$. Thus, as the cost of replacing $H$ with $H\otimes\ell^2$ we have found a separating vector. (Of course, getting from weak$^*$ to WOT using this construction can be done in a more direct way).

Notice this same argument would work for any $(M,H)$ where $H$ contains a countable "separating set" for $M$. This is equivalent to $M$ being $\sigma$-finite, which is equivalent to there being $\omega \in M_*^+$ which is faithful (so the GNS construction provides a separating vector).

Moral of the story: arguments involving separating vectors don't really extend beyond the $\sigma$-finite case, and may need you to adjust the Hilbert space you act upon.