1

Let $H$ be an infinite dimensional (possibly separable) Hilbert space. What can we say about $B(H)^{**}$?

Clearly, it is very large (Dual and bidual spaces of B(H) in norm topology , Double dual of the space of bounded operators on Hilbert space).

To be more specific: is $B(H)^{**}$ a factor? Is $B(H)^{**}$ type I?

I'd also be interested in some references discussing biduals of von Neumann algebras.

Mogget
  • 845
  • 4
  • 10

1 Answers1

3

Leaving aside the finite-dimensional case, an enveloping von Neumann algebra can never be a factor, since it contains every representation as a quotient; so it has lots and lots of central projections.

And it cannot possibly be type I, because every normal representation of a type I von Neumann algebra is type I and the enveloping von Neumann algebra of $B(H)$ has at least a type III representation. This because the enveloping von Neumann algebra has a normal representation onto the double commutant of every representation; so one can go $B(H)\xrightarrow{\ \ \ } B(H)/K(H)$, and it is known that the Calkin algebra has a type III factor representation.

As a general comment, precisely because universal objects are universal, there are usually insanely big, and they cannot be concretely characterized other than by its universal properties.

Key results to see the arguments above are Lemma III.2.2 and Theorem III.2.4 in Takesaki volume I. The existence of the type III representation was shown by Sakai in 1967.

Martin Argerami
  • 217,281