Timeline for The difference between $l^1(G)$ and the reduced group $C^*$ algebra $C_r^*(G)$
Current License: CC BY-SA 2.5
19 events
| when toggle format | what | by | license | comment | |
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| May 5, 2011 at 6:16 | vote | accept | yeshengkui | ||
| May 1, 2011 at 17:12 | answer | added | Alain Valette | timeline score: 6 | |
| Aug 8, 2010 at 9:22 | answer | added | Andreas Thom | timeline score: 4 | |
| May 31, 2010 at 19:28 | history | edited | Dmitri Pavlov |
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| Mar 29, 2010 at 3:07 | vote | accept | yeshengkui | ||
| May 5, 2011 at 6:16 | |||||
| Mar 29, 2010 at 1:53 | comment | added | yeshengkui | To Ulrich: It's a good idea to enlarge $C$ to be an ideal. Thanks for your comments. But why $A/I$ is not vanishing in general? | |
| Mar 28, 2010 at 19:00 | comment | added | Jonas Meyer | I agree with Yemon. To elaborate, every concrete C*-algebra has a weak closure that is a von Neumann algebra, but that does not make all questions about concrete C*-algebras also about von Neumann algebras. I agree that NG is related to CG and other completions of CG, but NG doesn't appear to be part of your question. I really do not want to make a big deal out of this, but you deserve to know where I was coming from when I retagged. I'll intervene no further in the tags of this question. | |
| Mar 28, 2010 at 18:14 | comment | added | Yemon Choi | I remain to be convinced that there is a connection to von neumann algebras, unless the answer uses machinery from that topic, or if the question is motivated by an application to group von Neumann algebras. But I prefer a minimalist approach to tags; other people's preferences may vary. | |
| Mar 28, 2010 at 14:27 | comment | added | Ulrich Pennig | If $I$ denotes the commutator ideal, then you have a map $A/C \to A/I$ of Banach spaces (since $C \subset I$). No $g \in G$ belongs to the class of the zero element in $A/I$ since this is an algebra and $g$ is invertible there. Therefore $g$ is not represented by the zero class in $A/C$ either. | |
| Mar 28, 2010 at 13:25 | comment | added | yeshengkui | Ulrich, you are right. | |
| Mar 28, 2010 at 13:17 | comment | added | Ulrich Pennig | But the set of all commutators is not closed with respect to products, is it? So it is itself not an algebra. | |
| Mar 28, 2010 at 12:46 | comment | added | yeshengkui | OK. Actually, there is some relation with the group von Neumann algebra. All the $l^1(G)$ and $C_r^*(G)$ are subalgbra of the group von Neumann algebra $NG$, which is the weak operator norm completion of $CG$. Furthermore, $T(NG)$ is just its center $Z(NG).$ | |
| Mar 28, 2010 at 7:04 | history | edited | Jonas Meyer |
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| Mar 28, 2010 at 6:06 | comment | added | Jonas Meyer | Funny, but I honestly did hesitate on that one, because I know that some have a different view of what NCG should mean. But because I was already removing the apparently irrelevant von Neumann algebras tag, it seemed best. @yeshengkui: I hope that you don't mind the tag edit. If there are connections to von Neumann algebras or NCG that you have in mind for these specific questions, please let us know. | |
| Mar 28, 2010 at 5:44 | comment | added | Yemon Choi | But Jonas, what will people do now you have taken a c-star algebra question/topic and taken the NCG tag off? how will we all cope? :) | |
| Mar 28, 2010 at 5:18 | history | edited | Jonas Meyer |
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| Mar 28, 2010 at 5:13 | answer | added | Yemon Choi | timeline score: 8 | |
| Mar 28, 2010 at 5:06 | history | edited | Yemon Choi | CC BY-SA 2.5 |
Fixed LaTeX
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| Mar 28, 2010 at 3:29 | history | asked | yeshengkui | CC BY-SA 2.5 |