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

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

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

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up vote 9 down vote accepted

There are probably several answers to this. Here's my take.

Two things make the classification of von Neumann algebras interesting and useful, in my view:

1) After you define the types, the abundance of projections allows you to show that any von Neumann algebra is a direct sum of subalgebras of some of the types.

2) There are many cases where the type information on its own tells you a big deal about the algebra: I'm thinking of results like:

  • Type I factors can be completely classified;
  • Type II$_1$ factors always carry a faithful normal tracial state;
  • Type II$_\infty$ factors are always a tensor product of a II$1$ and a I$\infty$;
  • Type III factors are a crossed product of a II$_\infty$.
  • AFD factors can be completely characterized for all types.

For C $\!\!^*$-algebras, one can try to play the same game (for example, "simple" could play the role of "factor", Type I C $\!\!^*$-algebras, purely infinite versus finite, AFD, etc.), but one is immediately hampered by the (eventual) lack of projections, that forbids to always have a C $\!\!^*$-algebra as a direct sum of simpler ones.

As a final word, "classification" is also used as in Elliott's Classification Program. In this setting, it is not clear at all that von Neumann algebras are on better footing that C$^*$-algebras. Of course type I von Neumann algebras can be completely classified, and rather easily; but, for example, a complete classification of all II$_1$ factors is considered completely hopeless by all experts.

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Thank you,I learn a lot.But why there is "good" projection in the W*algebra? – Strongart Feb 23 '12 at 11:00
The fact that a W* algebra is equal to its double commutant guarantees that it has lots of projections: concretely, any W* is the norm closure of the span of its projections. Moreover, intersections and unions of projections stay within the algebra, and this allows to construct projections in the algebra via maximality arguments: that's why one can decompose in types via central projections. – Martin Argerami Feb 23 '12 at 14:18
I see,thank you again. – Strongart Feb 24 '12 at 11:14

This is somewhere between an answer and a comment. Just think about the dichotomy between the two in the abelian case. In this case classifying von Neumann algebras is fairly simple. However, classifying all $C^*$-algebras essentially encompasses the entire field of topology.

From this point of view $C^*$-classification is hopeless.

Also a final note, which isn't an aspect that make proof easier but might be viewed as an extension of the above observation to the noncommutative case. In some sense there just are fewer von Neumann algebras so they are easier to classify. What I mean by this is that given a single von Neumann algebra there might be many weakly dense $C^*$-subalgebras that are not isomorphic.

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I'm not sure I agree with this. You can do a broad classification of C$^*$-algebras in terms of type I, purely infinite, etc., and this is more or less similar to the von Neumann classification (although less fancy and probably less useful). But with von Neumann algebras you get the same kind of problems: in the same sense that the classification of C$^*$-algebras is hopeless, so is the classification of von Neumann algebras: no one has any expectation of classifying all II$_1$-factors, say. – Martin Argerami Jul 22 '15 at 22:14

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