Predual of von Neumann algebra

Question

I know that each separable finite von Neumann has separable predual. Can some one give me an example of a non-separable finite von Neumann algebra with separable predual?

Answer 1

(Disclaimer: "separable" for the predual means in the norm topology, while for a von Neumann algebra it means in the sot topology)

Any von Neumann algebra with separable predual is separable. If the predual of $M $ is separable, the (countable) direct sum of the (one-dimensional) GNS representations over a dense subset of the unit ball of the normal states gives a faithful normal representation $M\to B (H) $ with $H $ separable. And $B (H) $ is separable (because $K (H) $ is). As the sot is metrizable on bounded sets of $B (H) $, the unit ball of $M $ is separable (in a metric space, subset of separable is separable), and so $M $ is separable.

When $M$ acts on a separable Hilbert space, it has separable predual. This is because it can be shown that normal linear functionals are determined by square-summable sequences.

It is not true that $M$ separable implies separable predual. An example is $M=\ell^\infty(\mathbb R)$. This $M$ cannot act on a separable Hilbert space because it has uncountably many pairwise orthogonal projections. And/or one can easily check that its predual $\ell^1(\mathbb R)$ is not separable (the canonical basis is uncountable and any two elements are at distance $2$). And, being, abelian, $M$ is finite. And $M$ is $\sigma$-weak separable.

In summary, the implications run like this: for $M\subset B(H)$, $$ \text{$H$ separable}\iff \text{ separable predual}\implies \text{$M$ separable.} $$