Timeline for Is there an irreducible, noncompact commuting, nonnormal operator, with spectrum strictly continuous?
Current License: CC BY-SA 4.0
15 events
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| Mar 15, 2023 at 17:00 | history | edited | Glorfindel | CC BY-SA 4.0 |
broken link fixed, cf. https://meta.mathoverflow.net/q/5301/70594
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jul 30, 2013 at 11:20 | history | edited | Sebastien Palcoux |
I replace the tag "reference-request" by "c-star-algebras".
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| Jul 24, 2013 at 11:08 | vote | accept | Sebastien Palcoux | ||
| Jul 23, 2013 at 21:40 | comment | added | Sebastien Palcoux | It's ok @BillJohnson, thank you for the exhortation. | |
| Jul 23, 2013 at 19:12 | comment | added | Bill Johnson | Definitely a question for MO. Don't cross post unless you say so and give the URL for the other post. | |
| Jul 23, 2013 at 17:29 | answer | added | Ollie | timeline score: 6 | |
| Jul 21, 2013 at 14:07 | comment | added | Sebastien Palcoux | Hi Julien, I did not know, thank you for this information. I hesitated between the two. I have posted it on MO first, and then on MSE. Where do you think is best for it ? | |
| Jul 21, 2013 at 13:43 | comment | added | Julien | Cross-posted on MSE. Hi Sébastien, you are supposed to let people know when you ask the same question on both sites. | |
| Jul 20, 2013 at 7:54 | comment | added | Sebastien Palcoux | Could the "downvoter" and "closer" indicate what's wrong ? Thank you. | |
| Jul 20, 2013 at 0:20 | review | Close votes | |||
| Jul 20, 2013 at 9:50 | |||||
| Jul 19, 2013 at 20:12 | comment | added | Sebastien Palcoux | @YemonChoi : Thank you for this comment. What's not clear with irreducibility ? If the group $\Gamma$ is ICC (infinite conjugacy classes), it generates a $II_{1}$ factor, and there are plenty of projections in its commutant, so there is no irreducibility for a single element. Next, for any discrete group $\Gamma$, if we take an irreducible representation $H$, the von Neumann algebra it generates is the whole $B(H)$. | |
| Jul 19, 2013 at 19:39 | history | edited | Sebastien Palcoux | CC BY-SA 3.0 |
I add precision about irreducibility and also the tag von neumann algebra.
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| Jul 19, 2013 at 18:56 | comment | added | Yemon Choi | One more possible example (though not an answer to your question): I think that if you take any non-zero element of the von Neumann algebra of a discrete group $\Gamma$, thought of as an operator on $\ell^2(\Gamma)$, then I think you get noncompact commuting and strictly continuous spectrum, and clearly you can arrange for non-normal. But irreducibility is not clear to me right now | |
| Jul 19, 2013 at 16:11 | history | asked | Sebastien Palcoux | CC BY-SA 3.0 |