Intuitive meaning of Double Commutant Theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:06:33Z http://mathoverflow.net/feeds/question/118208 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118208/intuitive-meaning-of-double-commutant-theorem Intuitive meaning of Double Commutant Theorem Koushik 2013-01-06T16:02:01Z 2013-01-06T18:05:44Z <p>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. <a href="http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem</a></p> http://mathoverflow.net/questions/118208/intuitive-meaning-of-double-commutant-theorem/118209#118209 Answer by Nik Weaver for Intuitive meaning of Double Commutant Theorem Nik Weaver 2013-01-06T16:26:54Z 2013-01-06T18:05:44Z <p>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 ${\cal A}$.</p>