Is multiplication $\sigma$-weakly continuous on bounded subsets?

Question

Let $M \subseteq B(H)$ be a von Neumann algebra. Let $S\subseteq M$ be a bounded subset. Is the multiplication map $$S \times S \to M: (x,y) \mapsto xy$$ $\sigma$-weakly (= weak$^*$) continuous?

Attempt: Assume $x_i \to x$ and $y_i \to y$ in the $\sigma$-weak topology where $(x_i)_i$ and $(y_i)_i$ are bounded nets. We must show that $x_iy_i \to xy$ in the $\sigma$-weak topology. Let $z \in L^1(H)$ be a trace class operator. We need to show that $$|Tr(z(xy-x_iy_i)| \to 0$$

I tried $$|Tr(zxy-x_iy_i)| = |Tr(zxy-zxy_i + zxy_i - zx_i y_i)|$$ $$\le |Tr(z(xy-xy_i)| + |Tr(z(xy_i-x_iy_i))|$$

Now I would like to use something like $$|Tr(zx)| \leq |Tr(z)|\|x\|$$ but I'm not sure that holds.

Answer 1

As written the inequality cannot hold (consider the case $x=z^*$, $\operatorname{Tr}(z)=0$). But, with $|z|$, it is standard (it's Hölder for $1,\infty$, if you think about it).

Let $z=v|z|$ be the Polar Decomposition. Using Cauchy-Schwarz, \begin{align} |\operatorname{Tr}(zx)| &=|\operatorname{Tr}(|z|^{1/2}\,(x|z|^{1/2}v))| \leq\operatorname{Tr}(|z|)^{1/2}\,\operatorname{Tr}(v^*|z|^{1/2}x^*x|z|^{1/2}v)^{1/2}\\[0.3cm] &\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(v^*|z|v)^{1/2}\\[0.3cm] &\leq\operatorname{Tr}(|z|)^{1/2}\,\|x\|\,\operatorname{Tr}(|z|^{1/2}v^*v|z|^{1/2})^{1/2}\\[0.3cm] &\leq\operatorname{Tr}(|z|)\,\|x\| \end{align}

Answer 2

Well, I guess Martin has already aptly (as always) answered this question but still let me expand on my previous comment, as suggested by the OP.

The first thing to notice is that the weak operator topology is not something intrinsic of a von Neumann algebra, as it depends on the underlying Hilbert space.

Some may say that von Neumann algebras are inextricably linked to the Hilbert space where they were born, and this is why people don't care at all about representing von Neumann algebras elsewhere, the prevailing attitude being to keep the algebra within its craddle $B(H)$.

Nevertheless, there is a very limited situation in which it might be useful to consider just one more representation, in addition to the defining one.

Given a von Neumann algebra $M\subseteq B(H)$, consider the representation of $M$ on $H\otimes \ell ^2$ given by $$ T\in M\mapsto T\otimes 1\in B(H\otimes \ell ^2). $$ Viewing $H\otimes \ell ^2$ as an infinite direct sum of copies of $H$, one has that $T\otimes 1$ is the (block) diagonal operator, with $T$ repeated down the diagonal infinitely many times.

$M\otimes 1$ is still weakly closed in $B(H\otimes \ell ^2)$, hence a von Neumann algebra, while the passage from $M$ to $M\otimes 1$ preserves all of the intrinsic properties of $M$. However, the weak operator topology may no longer be the same!

The reason is very simple. A vector $\xi $ in $H\otimes \ell ^2$ is actually a sequence $\xi =(\xi _n)_n$ of vectors, with each $\xi _n$ lying in $H$, and such that $\sum_n\|\xi _n\|^2<\infty $.

Observe that the vector state $\varphi _\xi $ defined by a vector $\xi $, as above, may be expressed, for each $T$ in $M$, by $$ \varphi _\xi (T) = \big \langle (T\otimes 1) \xi , \xi \big \rangle = \sum_n\big \langle T \xi _n, \xi _n\big \rangle . $$

Defining $u_n=\xi _n/\|\xi _n\|$, and $\alpha _n = \|\xi _n\|^2$, we then have that $$ \varphi _\xi (T) = \sum_n\alpha _n\big \langle T u_n, u_n\big \rangle = \sum_n\alpha _n\text{tr}(TS_n) = \text{tr}(TS), $$ where $S_n$ is the rank-one, trace-class operator $$ S_n(\xi ) = \big \langle \xi , u_n\big \rangle u_n, $$ and $S$ is the absolutely converging sum of the $\alpha _nS_n$.

Following along these lines one may prove that the vector states on $M\otimes 1$ in fact correspond to the normal states on $M$.

Summarizing, the passage from $M$ to $M\otimes 1$ turns the $\sigma $-weak operator topology into the weak operator topology.