Skip to content

Reminder to self: (non-)isomorphism of two von Neumann algebras and their preduals

13 May, 2018

So… it’s not clear if regular blogging will ever resume here, but in the meantime here is something to just clear the pipes, as it were. Although the results in this post are not new, occasionally I want to refer to them, and I don’t recall seeing an explicit reference in the literature. The proofs given here are not really proper expositions for those who don’t know Banach space theory; hopefully they will provide sufficiently suggestive outlines whose details can be filled in.

Theorem 1 below is something I noticed in 2016, but whose proof I forgot to write down at the time. Having just spent a half-hour trying to (re)construct a proof, it seems worth quickly writing down an argument here so that I can find it more easily. Theorems 2 and 3 are then natural things to point out, to indicate the context for Theorem 1; in both cases I’ve tried to piece together proofs from various bits of the literature.


Let T denote the space of trace-class operators on a separable infinite-dimensional Hilbert space H. Let V = L1([0,1], T) be the space of Bochner-integrable T-valued functions on [0,1]; alternatively we could define V to be the projective tensor product of T with L1.

Theorem 1. T and V are not isomorphic as Banach spaces.

Theorem 2. The dual spaces T* and V* are not isometrically isomorphic as Banach spaces.

Theorem 3. The dual spaces T* and V* are isomorphic as Banach spaces.

Proof of Theorem 1

It is known that the Banach space T has the Radon-Nikodym Property (RNP). I will not define the RNP here, but all we need to know is that it passes to closed subspaces, and that L1 does not have the RNP. Since V contains a (complemented) closed subspace isomorphic to L1, it follows that V does not have the RNP.  \Box

Question: Is there a simpler proof of Theorem 1? Invoking the RNP feels like overkill.

Proof of Theorem 2

Observe that T*=B(H) and V*= L([0,1], B(H)); we denote this second von Neumann algebra by N for sake of brevity. Suppose that B(H) is isometrically isomorphic (as a Banach space) to N. By a theorem of Kadison

R. V. Kadison, Isometries of operator algebras. Annals of Math. 54 (1951), no. 2, 325&ndas;338

this would imply that there is a Jordan *-isomorphism φ from B(H) onto N. Because B(H) is a factor, Corollary 11 of Kadison’s paper implies that φ must either be a *-isomorphism or a *-anti-isomorphism. But this is impossible since N has non-trivial centre, while B(H) has trivial centre.  \Box

Question: Can we obtain a more direct proof by investigating Kadison’s arguments and specializing them to the case of B(H)?

Proof of Theorem 3

This can be deduced from a more general result of Robertson and Wassermann:

A. G. Robertson, S. Wassermann, Completely bounded isomorphisms of injective operator systems. Bull. London Math. Soc. 21 (1989), 285–290.

However, it seems better to sketch a simpler argument for this particular case, which admittedly uses some of the same ideas, specifically, some form of Pelczynski’s decomposition method.

Observe that T*=B(H) and V*= L([0,1], B(H)). It is easy to construct an isomorphism of Banach spaces between L[0,1] and C ⊕ L[0,1]; a minor variant of this gives an isomorphism of Banach spaces between V* and T*V*.

Similarly, there is an obvious isomorphism of Banach spaces between L[0,1] and L[0,1]⊕L[0,1], and a minor variation of this gives an isomorphism of Banach spaces between V* and V*V*.

The final ingredient in this proof is the observation that T* is isomorphic as a Banach space to FB*. To justify this, note that there is a projection from B(L2 ⊗ H) onto L([0,1],B(H)) = V*, and that the former space is isomoprhic to B(H) since L2⊗ H is isomorphic to H.

Putting things together:

T*FV*F ⊕ (V*V*) ≅ (FV*) ⊕ V*T*V*V*

as required.  \Box

Advertisements
3 Comments leave one →
  1. Matt Daws permalink
    13 May, 2018 9:08 am

    For 2, I think you can argue using Sakai’s theorem. Indeed, let M be a W*-algebra and E a Banach space, and suppose that M is isometrically isomorphic to E*. Then we can transport the C*-algebra structure of M to E* thus turning E* into a W*-algebra. By Sakai’s theorem, E is hence the collection of normal functionals on M.

    So in your case, if T* and V* were isometrically isomorphic, we could identify T with the normal functionals on V*, namely V (or the other way around) thus showing that T and V were (isometrically) isomorphic.

    Notice that this argument is very much an “isometric” statement.

    • 13 May, 2018 4:12 pm

      Thanks Matt – that seems to work.

      Originally I was actually trying to reconstruct something like this argument, but somehow got myself mixed up when trying to use Sakai’s result. (Inevitable consequence of trying to dash off a blog post at 3am.)

Trackbacks

  1. Non-isomorphisms of some commutative Banach algebras | Since it is not ...

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

%d bloggers like this: