Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

By definition, a von Neumann algebra is a C*‑algebra A that admits a predual, i.e., a Banach space Z such that Z* is isomorphic to the underlying Banach space of A. (We require that isomorphisms in the category of Banach spaces preserve the norm.)

Moreover, a morphism of von Neumann algebras is a morphism f: A→B of the underlying C*‑algebras that admits a predual, i.e., a morphism of Banach spaces g: Y→Z such that g*: Z*→Y* is isomorphic to f in the category of morphisms of Banach spaces.

A theorem by Sakai states that all preduals of A induce the same weak topology on A (the ultraweak topology). In particular, every predual is canonically isomorphic to the dual of A in the ultraweak topology and the predual is unique up to unique isomorphism.

The same is true for morphisms: The unique predual of a morphism f: A→B of von Neumann algebras is its dual f*: B*→A* in the ultraweak topology. A morphism of the underlying C*‑algebras has a predual if and only if it is continuous in the ultraweak topology.

Thus the predual can be seen as a functor L1 that sends a von Neumann algebra to its dual in the ultraweak topology and likewise for morphisms. The functor L1 is faithful, which boils down to the fact that an element of a von Neumann algebra that vanishes on every element of the predual must be zero.

Whenever we have a faithful functor F: C→D it is natural to try to factor it as a composition of two functors G: C→E and H: E→D, where the objects of E are the objects of D equipped with some additional structures, the morphisms of E are the morphisms of D that preserve these structures, H is the functor that forgets these structures, and G is an equivalence of categories. In our case we are looking for additional structures on the predual such that morphisms of preduals that preserve these structures are precisely morphisms that come from morphisms of von Neumann algebras. Since the functor G is an equivalence of categories, this amounts to an alternative definition of von Neumann algebras in terms of preduals.

Two such structures on the predual are easy to identify. The first one is given by dualizing the unit map 1: k→A. Here k is the field of scalars over which all von Neumann algebras and Banach spaces are defined. The dual map is the famous Haagerup trace tr: L1(A)→k.

The second structure is given by dualizing the conjugate-linear involution *: A→A. The dual is the modular conjugation *: L1(A)→L1(A), which is important in Tomita-Takesaki theory. Finally, the trace commutes with the conjugation.

All morphisms of preduals that come from morphisms of von Neumann algebras preserve these two structures. Dualizing all morphisms of preduals that preserve these two structures gives us ultraweakly continuous maps of the original von Neumann algebras that preserve the unit and the involution.

However, there is no guarantee that these maps preserve the multiplication, at least I am not aware of any proof or refutation of this statement. We can naïvely try to dualize the multiplication map A⊗A→A and get a map of the form L1(A)→L1(A)⊗L1(A). However, it is unclear what kinds of tensor products we should use (see a question on this matter) and whether dualizing A⊗A actually gives us L1(A)⊗L1(A).

Question 1: Is there an example of a map of preduals such that its dual map preserves the unit and the involution but does not preserve the multiplication? If yes, what kind of additional structure maps between preduals should preserve to ensure that they actually come from morphisms of von Neumann algebras? In particular, can we dualize the product on a von Neumann algebra using some kind of tensor product?

Question 2: What Banach spaces (possibly equipped with a trace, an involution, and perhaps some other structures) arise as preduals of von Neumann algebras?

share|improve this question
    
There's a generic answer to the abstract question about faithful functors: an object of E is an object x of D equipped with an object y of C and an isomorphism to x from F(y), or (if you prefer) equipped with an equivalence class of such structures. One then looks for more satisfying answers. –  Toby Bartels Aug 12 '12 at 15:02
    
I would have answered the original question thus: a von Neumann algebra (or better, since the arrows go the other way, a noncommutative localisable measurable space) is a Banach space equipped with a C*-algebra structure on its dual. This seems intrinsic enough; it includes, for example, a trilinear operation (the dual's multiplication) that assigns a scalar to a vector and two continuous linear functionals. (It gets complicated, but so is the extended Haagerup tensor product.) –  Toby Bartels Aug 12 '12 at 15:46
1  
All that said, Matthew Daws's answer is clearly better. That's because the best question that you asked is "In particular, can we dualize the product on a von Neumann algebra using some kind of tensor product?". And that's what he answered. –  Toby Bartels Aug 12 '12 at 15:55

3 Answers 3

up vote 13 down vote accepted

In particular, can we dualize the product on a von Neumann algebra using some kind of tensor product?

The answer to this is explored in a number of papers. AFAIK, it was first considered by Quigg in "Approximately periodic functionals on C*-algebras and von Neumann algebras.", http://www.ams.org/mathscinet-getitem?mr=806641 Here he considered the Banach space projective tensor product, but this only works for subhomogeneous algebras.

The general case was answered by Effros and Ruan in "Operator space tensor products and Hopf convolution algebras", http://www.ams.org/mathscinet-getitem?mr=2015023 If $M$ is a von Neumann algebra with predual $M_*$ then there is a "tensor product", called the Extended Haagerup Tensor Product $M_* \otimes_{eh} M_*$, and a (complete) contraction $\delta:M_* \rightarrow M_* \otimes_{eh} M_*$ which is the predual (in some slightly technical sense) of the multiplication map $M\otimes M\rightarrow M$. (My scare quotes are because the algebraic tensor product $M_*\otimes M_*$ is not (norm) dense in $M_* \otimes_{eh} M_*$).

share|improve this answer

Let $ M $ and $ N $ be von Neumann algebras.

  1. It is known that the preduals $ M_{*} $ and $ N_{*} $ are isometric as Banach spaces if and only if $ M $ and $ N $ are Jordan $ * $-isomorphic. Please see David Sherman’s paper Noncommutative $ L^{p} $ structure encodes exactly Jordan structure.

  2. If $ M $ is infinite-dimensional, hyperfinite and semifinite, then its predual $ M_{*} $ is isomorphic to one of a list of thirteen Banach spaces. Please see the Haagerup-Rosenthal-Sukochev paper Banach Embedding Properties of Non-Commutative $ L^{p} $-Spaces for this result and its complements.

share|improve this answer

Why can't one consider $\ell_\infty$ and $M_2(\ell_\infty)$? They are isometrically isomorphic as Banach spaces, the first one is commutative, while the latter one is not. It seems that the theorem quoted by BigBill is not necessarily true, as these two guys are not Jordon isomorphic.

share|improve this answer
    
Why do you say they are isometrically isomorphic? –  Yemon Choi Jul 20 '12 at 23:39
    
Well, $\ell_\infty$ is isometric to its square, therefore to any its power; here: $\ell_\infty^{\oplus 4}$. –  RadekM Jul 21 '12 at 8:25
    
But the C-star norm on $M_2\otimes \ell^\infty$ is not that of $\ell^\infty^{\oplus 4}$, just as the C-star norm on $M_2$ is not that of $\ell_\infty^4$. –  Yemon Choi Jul 21 '12 at 8:40
    
What is the norm on the predual of $M_2\otimes\ell_\infty$? –  RadekM Jul 21 '12 at 8:44
1  
(Look, if you think there is an isometry, write it down explicitly) –  Yemon Choi Jul 21 '12 at 23:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.