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Is there a way to equip every C*-algebra A with a functorial topology such that the canonical map A→A** is an isomorphism if and only if A is a von Neumann algebra? Here A** denotes the dual of A* in some other functorial topology.

Of course if we restrict our attention to von Neumann algebras from the beginning, then the σ-weak topology on A and the norm topology on A* answer the question.

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Just to be precise: by "functorial topology" do you mean something like: a functor from the category of $C^*$-algebras and *-homomomorphisms, to the category of (locally convex) topological vector spaces and continuous linear maps? –  Yemon Choi Feb 5 '10 at 7:14
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(I'll add my obligatory grumble about NCG, but each to their own;) ) –  Yemon Choi Feb 5 '10 at 7:15
    
Yes, this is one possible definition. But I will not object if somebody comes up with a different way to define a functorial topology. –  Dmitri Pavlov Feb 5 '10 at 7:17
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3 Answers 3

This isn't quite an answer, but it might lead to one. Takesaki, III, Thm 3.16 shows that a C*-algebra A is a W*-algebra (i.e. non-spacial version of a von Neumann algebra) if and only if A is monotone closed and admits sufficiently many normal positive linear functionals.

Monotone closed == bounded increasing net of self-adjoints has a supremum.

Normal == Positive functional which respects the supremum. I guess we can then define normal to mean: positive and negative parts etc. are normal.

So on the class of Monotone Closed C*-algebras, we could let A* be the space of normal functionals, and give A* to norm topology. Then I think A=A** only when A is a von Neumann algebra (or otherwise A won't even inject into A**).

What I don't see is how to extend this to all C*-algebras.

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Trying to use normal functionals was my first guess, but as you say it's not clear how to get the notion to make sense for all $C^*$-algebras. Perhaps there is a decent notion of "order completion"? –  Yemon Choi Feb 5 '10 at 18:13
    
Well, my guess is that one can denote by A* the linear span of positive normal functionals on A (here a positive normal functional is a positive functional that preserves all supremums that exist) and by A** the dual of A* as a Banach space. Then the canonical map A→A** is an isomorphism if and only if A is a von Neumann algebra (Sakai's theorem). However, it is unclear to me whether A* is the dual of A in some topology for all C*-algebras A (and not just von Neumann algebras). –  Dmitri Pavlov Feb 6 '10 at 2:59
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Here's a topology that will work for separable von Neumann algebras:

Fix a separable Hilbert space H. The topology on A is obtained by pulling back the σ-weak topology on B(H) along all possible C*-algebra homomorphisms A→ B(H).

Similarly, the topology on A* is obtained by pushing forward the norm topology on
l1B(H) = (B(H)),σ-weak)* along the maps l1B(H) → A* induced by the above homomorphisms A→ B(H).


If you want this to work for all C*-algebras, then the answer to your question is probably ''no''. Indeed, a non-normal representation of a von Neumann algerba A→ B(H) is a morphism of C*-algebras, and in particular should be continuous for your topology (that's how I interpret your requirement "functorial topology"). This looks impossible to me.

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Consider the monomorphism A→A** of Banach spaces. Here A** denotes the second dual of A as a Banach space. The Banach space A** is a von Neumann algebra with the predual being A*. See Section 1.17 in Sakai's C*‑algebras and W*‑algebras.

We have a commutative square of Banach spaces consisting of morphisms A→A**→B** and A→B→B**. Thus we can pull back the ultraweak topology on A** to A and obtain a functorial topology on C*‑algebras due to the commutativity of the square above.

Henceforth denote by A* the dual of A in the new topology and by A** the dual of A* in the norm topology

If the canonical morphism A→A** is an isomorphism, then A has a predual, therefore it is a von Neumann algebra.

Unfortunately, if A is a von Neumann algebra, then the functorial topology does not coincide with the ultraweak topology and the canonical morphism A→A** is not an isomorphism.

We can fix this problem by composing the monomorphism A→A** with the multiplication by a certain central projection. However, the definition of this central projection relies on the fact that A is a von Neumann algebra and I don't see any way to extend it to arbitrary C*‑algebras.

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I'm being slow here, but when you say "pull back the ultraweak topology on $A^{**}$ to $A$" do you mean something different from "restriction to the embedded copy of $A$"? –  Yemon Choi Dec 24 '10 at 23:11
    
@Yemon: I mean that we pull back the ultrweak topology on A** along the morphism A→A**. In other words, we restrict the ultraweak topology on A** to the embedded copy of A. –  Dmitri Pavlov Dec 24 '10 at 23:31
    
But unless I have misunderstood terminology: isn't the ultraweak topology on $A^{**}$ the $\sigma(A^{**},A^*)$-topology, so the restriction of this topology to (the emebdded copy of) $A$ would be the $\sigma(A,A^*)$-topology, a.k.a. the weak topology on $A$? –  Yemon Choi Dec 25 '10 at 0:15
    
@Yemon: Yes, the ultraweak topology on A** is the σ(A**,A*)-topology. I now see a potential problem with this approach: If the embedding A→A** is not ultraweakly continuous, then we cannot say that the ultraweak topology pulls back to itself. –  Dmitri Pavlov Dec 25 '10 at 2:03
    
@Dmitri Pavlov. would like to suggest the following. Consider all finite or infinite partitions of unity on $ A $. Now take the closed subspace of $A'$ consisting of those functionals which behave well over all of these partitions. I claim (tentatively) that the dual of the latter does what you want. If the family of partitions is finite then the condition is vacuous and you simply get the double dual. If $A$ is already a von Neumann algebra, you get the normal functonals, i.e., the predual. If A is equal to the dual, then it is, of course, a von Neumann algebra. –  jbc Sep 30 '12 at 10:22
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