0

Many operator algebra books discuss the classifiation of W*algebras (von Neumann algebras), but not the C*algebras. Why?

I think a direct reason is that we havre the projection comparison theorem in a W*algebra, so we can compare projections in the factors.

But I want know some basic reason, going back to the original definition, from which part, the W*algebra is more rich than the C*algebra, so it can be classified.

flag
2 
 What do you mean by classification? The type decomposition? I find this question rather vague and ill-defined. – Yemon Choi Feb 23 2012 at 11:11
For instance, K-theory distinguishes the reduced group C*-algebras of free groups of different rank, but the group von Neumann algebras of such groups are notoriously not distinguished by any invariants... – Yemon Choi Feb 23 2012 at 11:13
1 
 pm, classification is not so much in progress as an industry, going back arguably to Elliott's result for the AF algebras. I fear that this kind of discussion is not what the OP was after, but then I can't work out what precisely the OP actually is after... – Yemon Choi Feb 23 2012 at 11:33
1 
 By the way, I thought that in a semisimple category you want decomposition of modules as direct sums. Is there a rigorous version that allows for direct integrals? – Yemon Choi Feb 23 2012 at 11:35
2 
 You might think of the analogue in the commutative world : classification of (standard, say) measured spaces is easier than that of (locally compact, say) topological spaces. I admit this is somewhat primitive, though. – BS Feb 23 2012 at 12:41
show 4 more comments

closed as not a real question by Yemon Choi, Matthew Daws, Marc Palm, Dmitri Pavlov, S. Carnahan♦ Feb 28 2012 at 0:31

Browse other questions tagged or ask your own question.