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Is there any intuitive explanation of the Double Commutant Theorem for Von Neumann Algebras? By intuitive I mean in terms of Quantum Mechanics. For example, duality of states and observables in the case of the Gelfand-Naimark Theorem. http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem

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May be worth to add link to theorem or say what is it about .... then you will get my + 1 :) –  Alexander Chervov Jan 6 '13 at 16:33
@a.chernov done! –  Koushik Jan 6 '13 at 16:56
Thank you! +1:) –  Alexander Chervov Jan 6 '13 at 17:40
It seems that what you want (as supplied by Nik Weaver) is an intuitive description rather than an intuitive explanation. I zmean, what is an intuiive explanation of something being WOT-closed? –  Yemon Choi Jan 6 '13 at 18:27

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up vote 9 down vote accepted

Okay, here's an explanation in terms of quantum mechanics. Let ${\cal A}$ be a family of observables, modeled as self-adjoint operators on some Hilbert space, and let ${\cal U}$ be the group of all unitary transformations that leave every observable in ${\cal A}$ invariant. You can consider ${\cal U}$ to be a kind of symmetry group. Mathematically it is the set of unitaries in the first commutant ${\cal A}'$ of ${\cal A}$, and the set of all observables left invariant by ${\cal U}$ is the double commutant of ${\cal A}$. So the double commutant theorem says that the set of all observables left invariant by every transformation that leaves every observable in ${\cal A}$ invariant, is the self-adjoint part of the von Neumann algebra generated by ${\cal A}$.

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This is a nice description, but I don't really see how it's an explanation... –  Yemon Choi Jan 6 '13 at 18:28
How about "an explanation of the intuitive meaning"? –  Nik Weaver Jan 6 '13 at 20:20

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