Timeline for Decomposition of a von Neumann Algebra into Factors
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2017 at 18:09 | comment | added | Sebastien Palcoux | @Fan: if your goal is to define a trace (if it exists) then it is true that the decomposition works (a trace exists iff there is no type ${\rm III}$ component); but note that according to the definition of your von Neumann algebra, it is sometimes possible to define a trace directly. The main reference on von Neumann algebras is Takesaki book (three volumes). I'm not really an expert on von Neumann algebras, so I can't give you a more specific reference. I advice you to complete your post with the explanation of your goal, and then to advertise that to Nik Weaver. | |
| Aug 29, 2017 at 16:37 | comment | added | Fan | @SebastienPalcoux Yes I am aware of that. I have read a bit about the characterization of each type. Given a von Neumann algebra $W$, which is non-trivial. Let's say I want to see whether it is possible to define a trace on $W$, so the idea is to decompose it into factors and define a trace for each factor (if possible) and add them up. How shall I do the first step then? Is there any reference which I can look up to? Thanks for your reading! | |
| Aug 24, 2017 at 6:24 | comment | added | Sebastien Palcoux | @Fan: the word type has a specific meaning in the factor theory: there are type ${\rm I}_n$, ${\rm I}_{\infty}$, ${\rm II}_1$, ${\rm II}_{\infty}$, ${\rm III}_{0}$, ${\rm III}_{\lambda}$, ${\rm III}_{\infty}$. The type does not characterize completely the factor, except in the amenable case (${\rm III}_{0}$ excepted). | |
| Aug 24, 2017 at 1:04 | history | edited | Fan | CC BY-SA 3.0 |
deleted 6 characters in body
|
| Aug 24, 2017 at 1:03 | comment | added | Fan | @NikWeaver Similar to the theorem given in the finite-dimensional case, to identify the type of each factor. | |
| Aug 24, 2017 at 0:49 | comment | added | Fan | @SebastienPalcoux thanks for your comment. I have updated my question. | |
| Aug 24, 2017 at 0:48 | history | edited | Fan | CC BY-SA 3.0 |
added 23 characters in body
|
| Aug 24, 2017 at 0:32 | comment | added | Nik Weaver | I'm puzzled as to what you might mean by "explicitly computing" a factor decomposition. | |
| Aug 23, 2017 at 21:19 | comment | added | Sebastien Palcoux | You should also consider the finite-dimensional case as trivial, otherwise the following gives an answer: en.wikipedia.org/wiki/Finite-dimensional_von_Neumann_algebra | |
| Aug 23, 2017 at 16:37 | history | asked | Fan | CC BY-SA 3.0 |