Equivalence of $\sigma$-weak topology to another topology

Let $$\mathcal H$$ be a Hilbert space. Define a topology $$\tau_1$$ on $$B(\mathcal H)$$ by the family of seminorms $$x\mapsto |Tr(xa)|,$$ $$a\in L^1(B(\mathcal H)).$$ Here $$B(\mathcal H)$$ denotes the set of all bounded linear maps on $$\mathcal H$$ and $$L^1(B(\mathcal H))$$ denotes the trace class operators. Again define the $$\sigma$$-WOT topology $$\tau_2$$ on $$B(\mathcal H)$$ by pulling back the weak operator topology of $$B(\mathcal H\otimes\ell_2)$$ to $$B(\mathcal H)$$ via the map $$x\mapsto x\otimes 1.$$ How to show that $$\tau_1=\tau_2$$? In many books and lecture notes in von Neumann algebras they have just mentioned that this is true. But I could not find a solid proof.

1 Answer

I think this is just a definition chase, and use of the standard representation of a trace-class operator. Starting with the later, following Chapter II in the first volume of Takesaki, for example, we have that any trace-class operator $$a$$ can be written as $$a(\xi) = \sum_n a_n (\xi|\xi_n) \eta_n \qquad (\xi\in H),$$ where $$(\xi_n),(\eta_n)$$ are orthogonal sequences (perhaps finite) in $$H$$ and $$(a_n)$$ is a sequence of positive reals with $$\sum_n a_n<\infty$$. Further, the trace-class norm of $$a$$ is $$\sum_n a_n$$.

Given such a representative, let $$\alpha_n = a_n^{1/2}\xi_n$$ and $$\beta_n = a_n^{1/2}\eta_n$$ we find that $$a(\xi) = \sum_n (\xi|\alpha_n)\beta_n$$ and $$\sum_n \|\alpha_n\| \|\beta_n\| = \sum_n a_n < \infty$$. Conversely, if $$(\alpha_n), (\beta_n)$$ are any sequences in $$H$$ with $$\sum_n \|\alpha_n\| \|\beta_n\| < \infty$$ we have $$B_0(H)\ni x \mapsto \sum_n (x\alpha_n|\beta_n)$$ defines a bounded linear functional on the compact operators $$B_0(H)$$, and hence there is a trace-class operator $$a:\xi\mapsto \sum_n (\xi|\alpha_n)\beta_n$$ with trace-class norm at most $$\sum_n \|\alpha_n\| \|\beta_n\|$$.

We conclude that the $$\sigma$$-weak topology, $$\tau_1$$, is given by the seminorms $$B(H)\ni x \mapsto \Big|\sum_n (x(\alpha_n)|\beta_n)\Big| \text{ where } \sum_n \|\alpha_n\| \|\beta_n\|<\infty.$$

The WOT is defined by seminorms $$x\mapsto |(x\xi|\eta)|$$. If $$B(H)$$ acts on $$H\otimes\ell_2$$ via $$x\mapsto x\otimes 1$$ then consider the respresentation of vectors in $$H\otimes\ell_2$$ which is $$\xi=\sum_n \xi_n\otimes e_n$$ and $$\eta=\sum_n \eta_n\otimes e_n$$ where $$(e_n)$$ is the standard unit vector basis of $$\ell_2$$ and $$\sum_n \|\xi_n\|^2, \sum_n \|\eta_n\|^2<\infty$$. Hence the induced seminorm on $$B(H)$$ is $$\big| \sum_n (x\xi_n|\eta_n)\big|$$. By Cauchy-Schwarz, $$\sum_n \|\xi_n\| \|\eta_n\| \leq \Big(\sum_n \|\xi_n\|^2, \sum_n \|\eta_n\|^2\Big)^{1/2}<\infty$$. Conversely, if we have $$(\alpha_n),(\beta_n)$$ with $$\sum_n \|\alpha_n\| \|\beta_n\|<\infty$$ then by rescaling, we may suppose that $$\|\alpha_n\|=\|\beta_n\|$$ for each $$n$$; notice that this does not change the seminorm $$x\mapsto \big|\sum_n (x(\alpha_n)|\beta_n)\big|$$. Then $$\sum_n \|\alpha_n\|^2 = \sum_n \|\beta_n\|^2 < \infty$$.

The equivalent between $$\tau_1$$ and $$\tau_2$$ follows.