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

Question

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.

Answer 1

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.