11
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ May be worth to add link to theorem or say what is it about .... then you will get my + 1 :) $\endgroup$ Jan 6, 2013 at 16:33
  • $\begingroup$ @a.chernov done! $\endgroup$
    – Koushik
    Jan 6, 2013 at 16:56
  • $\begingroup$ 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? $\endgroup$
    – Yemon Choi
    Jan 6, 2013 at 18:27
  • $\begingroup$ @YemonChoi that there is a fair number elements in the topology for a particular purpose? $\endgroup$
    – user123124
    May 10, 2017 at 11:38

2 Answers 2

13
$\begingroup$

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 ${\cal A}$.

$\endgroup$
2
  • 1
    $\begingroup$ This is a nice description, but I don't really see how it's an explanation... $\endgroup$
    – Yemon Choi
    Jan 6, 2013 at 18:28
  • 3
    $\begingroup$ How about "an explanation of the intuitive meaning"? $\endgroup$
    – Nik Weaver
    Jan 6, 2013 at 20:20
-2
$\begingroup$

Weak closure is more universal than norm closure. Commutative C* algebra is an algebra of continuous functions on spectrum, which could be any compact space. Von Neumann algebras deal with measure, and measure space is universal - the interval [0,1] with Lebesgue measure (provided measure of a point equals to 0). Check the V.A.Rokhlin theorem: https://en.wikipedia.org/wiki/Standard_probability_space

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.