Consider the abstract von Neumann algebra
$$M:= \ell^\infty-\bigoplus_{i \in I} B(H_i)$$
which consists of elements $(x_i)_i$ with $\sup_i \|x_i\| < \infty$ and $x_i \in B(H_i)$.
Let $N$ be the algebraic direct sum $\bigoplus_i B(H_i)$, thus it consists of elements $(x_i)_i$ such that only finitely many $x_i$ are non-zero. I want to show that $N$ is $\sigma$-weakly dense in $M$.
I guess my main problem is that I don't understand the $\sigma$-weak topology on $M$. By a result of Sakai, it is the unique topology on $M$ coming from a weak$^*$-topology when we realise $M$ as the dual of some other Banach space.
Maybe, we can do the following
$$M \cong \ell^\infty-\bigoplus_{i \in I} T(H_i)^* \cong \left(\ell^1-\bigoplus_{i \in I} T(H_i)\right)^*$$
But this does not seem very practical! Any insight in the matter will be appreciated!
Best Answer
I leave the proof to you, but we can discuss the details if there is any problem.
By Sakai's theorem the predual is unique and thus the $\sigma$-weak topology is canonical, in the sense that it is an intrinsic object that is independent of the concrete representation of the von Neumann algebra. More specifically, if $M_i=\mathcal{B(H}_i)$, then $$\bigoplus_{i\in I}^{\ell^\infty}M_i=\bigoplus_{i\in I}^{\ell^\infty}(M_{i*})^*\cong(\bigoplus_{i\in I}^{\ell^1}M_{i*})^*$$ where $A_*$ denotes the predual of the von Neumann algebra $A$. In other words, the predual of $M$ is the same as the $\ell^1$ direct sum of the preduals of the $M_i$, as you correctly guessed. You believe that this is not practical and you are wrong about that!
Fix $x=(x_i)_{i\in I}\in M$. We wish to find a net of elements such that only finitely many coordinates are non-zero that converges ultraweakly to our point $x$. Let $\Lambda$ be the set of finite subsets of $I$ and we direct $\Lambda$ by the usual inclusion, so $\Lambda$ becomes a directed set. We denote the elements of $\Lambda$ by $\lambda$, so $\lambda$ is simply a finite subset of $I$. For $\lambda\in\Lambda$ we define $x_\lambda=(x_\lambda^{(i)})_{i\in I}$ where $x_\lambda^{(i)}=0$ if $i\in I\setminus\lambda$ and $x_\lambda^{(i)}=x_i$ if $i\in\lambda$. So the net $(x_\lambda)_{\lambda\in\Lambda}$ lies in the subspace $N$ you have specified. We will show that $x_\lambda\to x$ ultraweakly.
We fix $\varepsilon>0$ and a functional $\Phi$ in the predual of $M$ and we want to find $\lambda_0\in\Lambda$ so that for any $\lambda\geq\lambda_0$ (here $\geq$ refers to the direction we gave to $\Lambda$ of course) we have that $|\Phi(x)-\Phi(x_\lambda)|<\varepsilon$.
Edit: Wait a minute; how am I fixing a functional in the predual, how does that make any sense? Well, recall that the predual of a von Neumann algebra $A$ is identified with the set of normal linear functionals on $A$ and therefore a net $(a_\mu)\subset A$ converges ultraweakly to $a\in A$ if and only if $\phi(a_\mu)\to\phi(a)$ for all normal linear functionals $\phi$. Now that we have cleared this out, let's go on with our proof:
By the identification of the preduals that we explained in the first paragraph, $\Phi$ is identified with a sequence $(\phi_i)_{i\in I}$ where $\phi_i\in M_{i*}$ such that $\sum_{i\in I}\|\phi_i\|<\infty$. Since the generalized series converges, we can find a finite subset $F\subset I$ so that $\sum_{i\in I\setminus F}\|\varphi_i\|<\frac{\varepsilon}{\|x\|}$. Set $\lambda_0=F$ and suppose that $\lambda\geq\lambda_0$, i.e. $\lambda\supset\lambda_0$. Then
$$|\Phi(x)-\Phi(x_\lambda)|=|\sum_{i\in I}\phi_i(x_i)-\sum_{i\in I}\phi_i(x_\lambda^{(i)})|=|\sum_{i\in I\setminus\lambda}\phi_i(x_i)|\leq\sum_{i\in I\setminus\lambda}|\phi_i(x_i)|\leq\sum_{i\in I\setminus F}|\phi_i(x_i)|\leq$$ $$\leq\|x\|\cdot\sum_{i\in I\setminus F}\|\phi_i\|<\varepsilon $$
and we are done.