About separability of von Neumann algebras [closed]

Question

Is a von Neumann algebra always separable in the $\sigma$-weak topology? If not, give a counterexample. Under what conditions will it be separable?

Answer 1

You can find, in any decent textbook on von Neumann algebras, a proof that if $A \subseteq B(\mathcal{H})$ is a von Neumann algebra, and the Hilbert space $\mathcal{H}$ is separable, then the unit ball of the von Neumann algebra is separably metrizable in the $\sigma$-weak topology. This implies that $A$ is $\sigma$-weakly separable, because the countable union of separable spaces is separable (easy proof). For a W*-algebra, the property of being isomorphic to a von Neumann algebra on a separable Hilbert space is equivalent to having a norm-separable predual, so I will freely move back and forth between these two notions in the following.

I will give two examples. The first is a W*-algebra that is $\sigma$-weakly separable but does not have norm-separable predual, so this shows that separability in the $\sigma$-weak topology does not imply separability of the Hilbert space a von Neumann algebra is represented on. The second example is a W*-algebra that is not $\sigma$-weakly separable, as requested in the title.

1. Consider $\ell^\infty(\mathbb{R})$ (bounded functions on $\mathbb{R}$ with no requirement of measurability or continuity whatsoever). It has $\ell^1(\mathbb{R})$ as a predual, which is not norm separable because the family of functions $(\chi_{\alpha})_{\alpha \in \mathbb{R}}$ is uncountable and all elements of it are at a distance of 2 from each other.

   The unit ball of $\ell^\infty(\mathbb{R})$ is homeomorphic to $[-1,1]^{\mathbb{R}}$, because the restrictions of $\sigma$-weak topology and the product topology agree on the unit ball. This space is separable by the Hewitt-Marczewski-Pondiczery theorem. (It is also not that hard to find a countable dense subset directly. This can be done with càdlàg functions that jump at a finite set of rational points and are locally constant elsewhere). Therefore $\ell^\infty(\mathbb{R})$ is the countable union of separable subspaces, and therefore separable.

2. This is easy to do, once you realize the following -- every separable Hausdorff space has cardinality $\leq 2^{\mathfrak{c}}$. If $X$ is a separable topological space, with $e : \mathbb{N} \rightarrow X$ bijectively enumerating a countable dense subset of $X$. For each point $x \in X$, let $N_x$ be the neighbourhood filter. Define $M_x$ to be the inverse image filter of $N_x$ with respect to $e$, i.e. $M_x$ is generated by the filter base $\{ e^{-1}(N) \mid N \in N_x \}$ (this is a filter base because $e(\mathbb{N})$ is dense). We have that the direct image filter of $M_x$ with respect to $e$ converges to $x$. As $X$ is Hausdorff, this implies $M_x \neq M_y$ if $x \neq y$, so $x \mapsto M_x$ is an injection from $X$ to the set of filters on $\mathbb{N}$. As these filters are subsets of $\mathcal{P}(\mathbb{N})$, there are at most $2^{2^{\aleph_0}}$ of them, so $|X| \leq 2^{\mathfrak{c}}$.

   All we need to do, therefore, is to find a W*-algebra with cardinality bigger than $2^{\mathfrak{c}}$. For this, either $\ell^\infty(2^{\mathfrak{c}})$ or $B(\ell^2(2^{\mathfrak{c}}))$ will do nicely.