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Nov 27, 2023 at 19:25 comment added UtsabrajSarkar @DmitriPavlov Yes, using Double Commutant theorem it is trivial. But in the Kehe Zhu book this theorem came before the statement of the aforementioned theorem. That's why I was looking for a proof without using it.
Nov 27, 2023 at 17:49 comment added Dmitri Pavlov To answer the question as it is stated in the title: a von Neumann algebra coincides with its double commutant, and the double commutant always contains the identity operator.
Nov 27, 2023 at 15:43 history edited UtsabrajSarkar CC BY-SA 4.0
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Nov 27, 2023 at 7:35 review Close votes
Dec 10, 2023 at 3:05
Nov 27, 2023 at 7:33 history edited Martin Sleziak CC BY-SA 4.0
MathJax: \langle, \rangle
Nov 27, 2023 at 7:20 history edited UtsabrajSarkar CC BY-SA 4.0
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Nov 27, 2023 at 7:17 history edited UtsabrajSarkar CC BY-SA 4.0
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Nov 27, 2023 at 7:15 history edited UtsabrajSarkar CC BY-SA 4.0
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S Nov 27, 2023 at 6:59 review First questions
Nov 27, 2023 at 8:16
S Nov 27, 2023 at 6:59 history asked UtsabrajSarkar CC BY-SA 4.0