Talk:Von Neumann algebra

message: How 'bout giving some concrete examples?  Michael Hardy 02:07, 31 May 2004 (UTC)
thread_title: Untitled

message: What do you mean bias? The previous version you reverted explicitly said much of the theory is still applicable in general (bicommutant theorem, Kaplansky density theorem etc.) The important fact to note is that hyperfinite von Neumann algebras (essentially classifiiable by COnnes' classification of hyperfinite factors) are all separably realizable.CSTAR 18:30, 3 Jan 2005 (UTC)
thread_title:  Separability bias 

message: I would like to delete the comment that the space of bounded linear transforms on a  B(L^2(X,\mu))  is highly non-commutative since it doesn't seem to mean much except stating that such an algebra is non-commutative. This is not true if we make everything we can trivial.HStel 13:42, 22 April 2006 (UTC)
replies: text: Go ahead, delete.--CSTAR 14:40, 22 April 2006 (UTC)

text: I couldn't find the term "locally compact measure space" anywhere in the article. Are you suggesting we include a definition of it in any case? --CSTAR 19:38, 17 July 2006 (UTC)
thread_title:  High non-commutativity 

message: I'm not sure a footnote is the best way to do it, but in the interest of mathematical correctness, it is important to point  out that L^\infty(X) is not always a von Neumann algebra.  Sure, the counterexamples might be pathological and uninteresting, but that doesn't mean incorrect statements are okay.
For the non-believers:
Let X be an uncountable set and μ a measure on the countable-cocountable algebra -- the counting measure will do fine.  Choose A \subset X uncountable with uncountable complement, and consider a net of operators in L^\infty(X) that converge in the weak or strong topology towards the projection onto A (an element of B(L^2(X)):  but that projection isn't measurable.
(Alternatively, note that the lattice of projections in L^\infty(X) isn't order complete, but the lattice of projections in a von Neumann algebra is).
IIRC, the algebra of locally measurable essentially bounded functions on a locally finite measure space is a von Neumann algebra;  I'm not sure whether there is any measure space not equivalent to a locally finite one, but if there is such a beast, I see no way of constructing a von Neumann algebra out of L^\infty(X) short of considering its closure (but, of course, the resulting von Neumann algebra would be commutative, and thus isomorphic to L^\infty(X) for some nice (i.e. locally finite) measure space).
RandomP 01:11, 13 May 2006 (UTC)
thread_title:  The pathological measure space case 

message: Why is their definition in the von Neumann-article? I'm moving them to the C*-algebras. KennyDC 01:47, 16 September 2006 (UTC)
thread_title:  Type I-C*-algebras 

message: At some point we should say that von Neumann algebras can be characterized abstractly as a C*-algebra  with a predual. We need to decide where to put this in.--CSTAR 17:05, 9 October 2006 (UTC)
replies: text: Oops its already there!--CSTAR 17:12, 9 October 2006 (UTC)
thread_title:  Abstract von Neumann algebras 

message: The section "Modules over a factor" contains this passage:
"[E]very such module H can be given an M-dimension dim_M(H) (not its dimension as a complex vector space) such that modules are isomorphic if and only if they have the same M-dimension. The M-dimension is additive, and a module is isomorphic to a subspace of another module if and only if it has smaller or equal M-dimension."
Shouldn't any initial discussion of a mathematical quantity like "M-dimension" at least state what type of quantity it is?  E.g., to what set does it belong?  I hope someone knowledgeable about the subject can make this a lot clearer.108.245.209.39 (talk) 02:24, 26 September 2018 (UTC)
thread_title:  Unclear M-dimension