Timeline for Relative commutants of abelian von Neumann algebras
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2019 at 17:49 | vote | accept | Jesse Peterson | ||
| Feb 4, 2019 at 13:06 | answer | added | Jiang | timeline score: 4 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Aug 9, 2013 at 18:52 | comment | added | Jesse Peterson | @Jiang: Yes, that's correct. This example is far from typical though. | |
| Aug 9, 2013 at 14:18 | comment | added | Jiang | @jesse, following the last comment, let $N$ be the Von Neumann algebra generated by the Bernoulli action $G$ on $\{0,1\}^G$, let $Y$ denotes all the functions that vanish on the identity element of $G$, then denote $A$ to be $L^{\infty}(Y)$, is $A'\cap N=L^{\infty}(\{0,1\}^G)$? | |
| Jan 17, 2012 at 2:34 | comment | added | Jesse Peterson | That's a good point. I am unaware of any sort of general existence result for such abelian subalgebras though. Perhaps it is possible to construct such an algebra by hand, but it seems to be a non-trivial result. | |
| Jan 16, 2012 at 15:47 | comment | added | Jiang | One comment, if we could choose one abelian von Neumann subalgebra $A$ such that there exists one unitary $u$ in the normalizer of $A'\cap N$ but not in the normalizer of $A$, then set $B=uAu^*$, this would give us one counterexample. | |
| Jan 9, 2012 at 22:23 | comment | added | Jesse Peterson | Yes, it is important that $A$ and $B$ contain the same unit as $N$ (I think in most books this is part of the definition of a von Neumann subalgebra). Otherwise this property is not even satisfied for $\mathcal B(\mathcal H)$, since if we consider $A \subset \mathcal B(\mathcal H)$ any self-adjoint subalgebra, then we have $A' = (A'')'$ and $A''$ always contains the identity operator. | |
| Jan 9, 2012 at 20:49 | comment | added | Martin Argerami | Sorry, I should read with more care! And your proof is better than mine, as my argument used non-unital subalgebras (which you are probably not considering). | |
| Jan 9, 2012 at 12:01 | comment | added | Jesse Peterson | That's correct. I mentioned it above. | |
| Jan 9, 2012 at 10:06 | comment | added | Martin Argerami | As a naive comment, the condition implies that $N$ is a factor. | |
| Jan 6, 2012 at 22:13 | history | asked | Jesse Peterson | CC BY-SA 3.0 |