The Predual of a von Neumann algebra

Question

Let $M$ $\subseteq$ $B(H)$ be a von Neumann algebra. I am wondering how does $M_*$ sit inside $B(H)_*$ upto isometry. Note - $M_*$ denotes the predual of $M$. Thanks for any help.

Answer 1

I don't think it does. The natural map of which the inclusion $\iota : M \to B(H)$ is the dual would be a projection of $B(H)_*$ onto $M_*$. Namely, each element of $B(H)_*$, considered as an ultraweakly continuous linear functional on $B(H)$, maps to its restriction to $M$.

Answer 2

Robert is right.

To provide a counter-example, consider any embedding $L_\infty[0,1]\subset B(L_2[0,1])$. Certainly, $L_1[0,1]$ is the (unique) predual of $L_\infty[0,1]$. Now, the predual of $B(L_2[0,1])$ is a (separable) dual of the algebra of compact operators, hence it has the Radon–Nikodym property (and this property passes to closed subspaces). On the other hand it is well-known that $L_1[0,1]$ lacks the Radon–Nikodym property. This can be easily generalised to the following setting.

Suppose that $\mathcal{M}\subset B(H)$ is an infinite-dimensional, finite von Neumann algebra. Then $\mathcal{M}_*$ does not embed into $B(H)_*$ as a Banach space.

See Chapter VII of

J. Diestel and J. J. Uhl Jr., Vector Measures, volume 15 of Math. Surveys. AMS, Providence, RI, (1977).

for an excellent treatment of the Radon–Nikodym property.

Answer 3

Assume the inclusion $\iota:M_*\to B(H)_*$ is an isometry. Taking dual, $\pi=\iota^*$ forms a w*-w* continuous *-homomorphism from $B(H)$ onto $M$. It implies that, $M$ is isomorphic to $B(H)/K$ where $K$ is a two-sided ideal in $B(H)$.

For example if $H$ is separable then $M$ should be (isomorphic to) the Calkin algebra. Firstly, the Calkin algebra is not von Neumann algebra. Secondly, if the assertion is true then there is just one (non-trivial) von Neumann algebra in $B(H)$ which is clearly impossible.