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How 'bout giving some concrete examples? Michael Hardy 02:07, 31 May 2004 (UTC)
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)
I would like to delete the comment that the space of bounded linear transforms on a 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)
O boy o boy, I did it! Herman Stel 22:51, 22 April 2006 (UTC)
Someone please include a definition of "locally compact measure space", does it mean a locally compact Hausdorff space with a measure that is a positive linear functional on continuous maps with compact support??? Kuratowski's Ghost 13:39, 16 July 2006 (UTC)
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 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 uncountable with uncountable complement, and consider a net of operators in that converge in the weak or strong topology towards the projection onto A (an element of : but that projection isn't measurable.
(Alternatively, note that the lattice of projections in 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 short of considering its closure (but, of course, the resulting von Neumann algebra would be commutative, and thus isomorphic to for some nice (i.e. locally finite) measure space).
RandomP 01:11, 13 May 2006 (UTC)
The examples section was rather error-ridden: here are the changes, and reasons for them:
Fails in the case of certain pathological measure spaces, see above.
I fail to see why this is discussed before von Neumann group algebras are.
The crossed product article currently mentions only von Neumann algebras, though if memory serves a similar construction for C*-algebras exists. I am unsure whether the two coincide for von Neumann algebras considered as C*-algebras.
What exactly is the point of defining a von Neumann algebra as being "generated" by a certain set of operators? We haven't even explained how this is always possible, nevermind that it's hardly a good way to define the group von Neumann algebra.
Also, what's the point of considering locally compact rather than discrete groups? It makes the example much more complicated. (I believe the von Neumann group algebra of a locally compact group is isomorphic to that of the same group considered as discrete group).
"L2 functions" is also a slightly inaccurate term when locally compact groups are considered: it might not be totally obvious that the Hilbert space is independent of the choice of Haar measure, besides the whole identification-of-equivalent-functions issue.
The tensor product of two von Neumann algebras is most certainly not a von Neumann algebra (except in some very simple cases). It can be completed to be a von Neumann algebra, and the interesting statement is that the result of this completion is independent of certain choices made to define it.
Von Neumann algebras are virtually always non-separable, in the norm topology. This most definitely no longer belongs in the examples section.
What are constructed from what? This isn't even a stub.
If the removed examples can be salvaged, feel free to do so.
RandomP 01:44, 13 May 2006 (UTC)
R.e.b. 21:40, 16 May 2006 (UTC)
Why is their definition in the von Neumann-article? I'm moving them to the C*-algebras. KennyDC 01:47, 16 September 2006 (UTC)
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)
a comment that says, essentially:
was removed with the edit summary claiming it's incorrect. why is it incorrect? Mct mht 05:18, 26 January 2007 (UTC)
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)
![]() | This article is rated B-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
|
How 'bout giving some concrete examples? Michael Hardy 02:07, 31 May 2004 (UTC)
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)
I would like to delete the comment that the space of bounded linear transforms on a 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)
O boy o boy, I did it! Herman Stel 22:51, 22 April 2006 (UTC)
Someone please include a definition of "locally compact measure space", does it mean a locally compact Hausdorff space with a measure that is a positive linear functional on continuous maps with compact support??? Kuratowski's Ghost 13:39, 16 July 2006 (UTC)
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 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 uncountable with uncountable complement, and consider a net of operators in that converge in the weak or strong topology towards the projection onto A (an element of : but that projection isn't measurable.
(Alternatively, note that the lattice of projections in 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 short of considering its closure (but, of course, the resulting von Neumann algebra would be commutative, and thus isomorphic to for some nice (i.e. locally finite) measure space).
RandomP 01:11, 13 May 2006 (UTC)
The examples section was rather error-ridden: here are the changes, and reasons for them:
Fails in the case of certain pathological measure spaces, see above.
I fail to see why this is discussed before von Neumann group algebras are.
The crossed product article currently mentions only von Neumann algebras, though if memory serves a similar construction for C*-algebras exists. I am unsure whether the two coincide for von Neumann algebras considered as C*-algebras.
What exactly is the point of defining a von Neumann algebra as being "generated" by a certain set of operators? We haven't even explained how this is always possible, nevermind that it's hardly a good way to define the group von Neumann algebra.
Also, what's the point of considering locally compact rather than discrete groups? It makes the example much more complicated. (I believe the von Neumann group algebra of a locally compact group is isomorphic to that of the same group considered as discrete group).
"L2 functions" is also a slightly inaccurate term when locally compact groups are considered: it might not be totally obvious that the Hilbert space is independent of the choice of Haar measure, besides the whole identification-of-equivalent-functions issue.
The tensor product of two von Neumann algebras is most certainly not a von Neumann algebra (except in some very simple cases). It can be completed to be a von Neumann algebra, and the interesting statement is that the result of this completion is independent of certain choices made to define it.
Von Neumann algebras are virtually always non-separable, in the norm topology. This most definitely no longer belongs in the examples section.
What are constructed from what? This isn't even a stub.
If the removed examples can be salvaged, feel free to do so.
RandomP 01:44, 13 May 2006 (UTC)
R.e.b. 21:40, 16 May 2006 (UTC)
Why is their definition in the von Neumann-article? I'm moving them to the C*-algebras. KennyDC 01:47, 16 September 2006 (UTC)
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)
a comment that says, essentially:
was removed with the edit summary claiming it's incorrect. why is it incorrect? Mct mht 05:18, 26 January 2007 (UTC)
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)