Show that algebraic direct sum is $\sigma$-weakly dense.

Question

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!

Answer 1

Lemma: Let $(X_i)_{i\in I}$ be a collection of Banach spaces indexed by a set. Then there exists an isometric isomorphism $$\bigoplus_{i\in I}^{\ell^\infty}(X_i)^*\cong\bigg(\bigoplus_{i\in I}^{\ell^1}X_i\bigg)^* $$

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.

Answer 2

I am posting one more answer, relevant to the double commutant theorem. I review the essentials first and then try to apply it, reaching a conclusion that does not have to do with DCT. You can read the last three paragraphs independently of the rest of the answer.

Theorem: (Von Neumann's DCT) Let $A\subset\mathcal{B(H)}$ be a $*$-algebra that contains the identity operator. Then $\overline{A}^{SOT}=A''$

Note that convex subsets of $\mathcal{B(H)}$ have the same strong and weak closure, so the DCT also says that $\overline{A}^{WOT}=A''$.

Also, if $A$ is a $*$-subalgebra, then the ultraweak closure of $A$ is equal to the strong closure, which is equal to the weak closure.

This follows from Kaplansky's density theorem: since the ultraweak topology is stronger than the weak topology, it is immediate that $\overline{A}^{w^*}\subset\overline{A}^{WOT}$. On the other hand, let $x\in \overline{A}^{WOT}$. By Kaplansky's density theorem we can find a norm-bounded net $(a_i)$ in $A$ so that $a_i\to x$ in the weak topology. But the weak topology coincides with the ultraweak topology on norm-bounded subsets, so $a_i\to x$ ultraweakly, thus $x\in\overline{A}^{w^*}$.

Note that we can also improve the DCT a little bit and require only that the identity operator belongs to the strong closure of $A$, because then $$\overline{A}^{SOT}=\overline{A+\mathbb{C}1_\mathcal{H}}^{SOT}=(A+\mathbb{C}1_{\mathcal{H}})''=A''.$$

Corollary: If $A\subset\mathcal{B(H)}$ is a $*$-subalgebra and $1_\mathcal{H}$ belongs to the strong/ weak/ ultraweak closure of $A$, then $\overline{A}^{SOT}=\overline{A}^{WOT}=\overline{A}^{w^*}=A''$.

Now we adopt the notation that OP uses in the post. We can take a concrete representation of $M$, so we can find a Hilbert space $\mathcal{K}$ so that $M\subset\mathcal{B(K)}$ and $M=\overline{M}^{SOT}$ and $1_\mathcal{K}\in M$. Note that $1_\mathcal{K}$ will be the same as the unit of $M$, which is the element $(1_{\mathcal{H}_i})_{i\in I}$.

The thing now is this: Obviously $N$ is a $*$-subalgebra of $\mathcal{B(K)}$; if we can show that $1_\mathcal{K}$ belongs to the strong/weak/ultraweak closure of $N$, then we can conclude that $\overline{N}^{w^*}=N''$ and then all we must do is verify that $N''=M$, which I guess is going to be easy.

BUT! Why bother with all this? Observe that $N$ is an ideal in $M$. The ultraweak closure of $N$ is thus an ultraweakly closed ideal in $M$. Therefore, if we manage to show that $1_M$ belongs to the ultraweak closure of $N$, then we immediately have that $\overline{N}^{w^*}=M$.

Conclusion: The essence of what we are trying to prove lies in showing that the unit of $M$ can be approximated ultraweakly by elements of $N$. This is essential in every approach (DCT, the one in the other answer and the one with the ideal observation). I think the technology required to show this is what I present in my other answer. Sure, one can take a concrete representation of $M$ and try to approximate the unit of $M$ through $N$ in the weak topology which might be simpler, but I doubt that it is going to be essentially simpler.

PS: obviously, the net that is going to approximate $(1_{\mathcal{H}_i})$ (the unit of $M$) in any case is the net $(x_\lambda)\subset N$ where $\Lambda$ is the set of finite subsets of $I$ and $x_\lambda$ is the sequence having $0$ everywhere except the slots $i\in\lambda$ where it has $1_{\mathcal{H}_i}$