$\sigma-$ weakly continuous functionals on Von Neumann Algebra.

Question

Why and in which sense a Von Neumann algebra $\cal{M}$ can be considered as the dual space of $\sigma-$ weakly continuous linear functionals on $\cal{M}$?

Answer 1

This was first understood for $B(H)$. If you consider the compact operators $K (H) $, their dual are the trace-class operators $T(H)$, the via the duality $$\tag1 \hat R(S)=\operatorname{Tr}(SR),\ \ \ \ \ R\in T(H),\ \ S\in K(H). $$ And with the same duality we have $T(H)^*=B(H)$. This is exactly similar to $c_0^*=\ell^1(\mathbb N)$, $\ell^1(\mathbb N)^*=\ell^\infty(\mathbb N)$.

One can characterize the linear functionals in $B(H)$ that come from $T(H)$ as those that are ultraweakly continuous. And then it was found that this approach works in general: given a von Neumann algebra $M$, one can prove that the dual of $M_*$, the space of ultraweakly continuous linear functionals on $M$, is $M$ via the duality $$ \hat x(\phi)=\phi(x),\ \ \ \ x\in M, \ \ \ \phi\in M_*. $$