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.



Clarification on predual on existence of separating vector
So I *think* you are asking:



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> 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 $M\_\*$?
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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 $M$.



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


