Timeline for What is $\hat{A}=\{[\pi]:\pi$ is a irreducible representation of $A$} ( $A$ is a $C^*$-algebra)?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 30, 2015 at 18:57 | comment | added | Rasmus | That is true. $ $ | |
| Sep 30, 2015 at 17:55 | answer | added | Sabrina Gemsa | timeline score: 3 | |
| Sep 28, 2015 at 19:01 | comment | added | Sabrina Gemsa | yes you are right, but my argumentation with the gelfand space (=character space) was wrong | |
| Sep 28, 2015 at 18:34 | comment | added | Rasmus | ... which is basically $(0,1]$, isn't it? | |
| Sep 28, 2015 at 18:18 | review | Close votes | |||
| Sep 30, 2015 at 5:31 | |||||
| Sep 28, 2015 at 18:18 | comment | added | Sabrina Gemsa | @RasmusBentmann I did a mistake I think, $I$ is not commutative and $\widehat{C_0(X,M_2)}$ must be $X\times \hat{M_2}=(0,1]\times \{[id]\}$. | |
| Sep 28, 2015 at 18:09 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
deleted 21 characters in body
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| Sep 28, 2015 at 18:03 | comment | added | Johannes Hahn | The coproducts are wrong here. These spaces do not decompose into smaller pieces usually. In commutative case it is particularly easy to see, since $X \cong \widehat{C(X)}$ and $X$ can be connected. In particular $\widehat{C([0,1])}$ is not $\{0\} \sqcup (0,1]$. What you meant is a decomposition of the underlying sets of those spaces, not of the spaces themselves. | |
| Sep 28, 2015 at 11:17 | comment | added | Sabrina Gemsa | ok, I will do it:) (in a few hours) | |
| Sep 28, 2015 at 9:35 | comment | added | Rasmus | You're welcome! If everthing's clear now, I encourage you to write a detailed answer! :) | |
| Sep 28, 2015 at 9:32 | comment | added | Sabrina Gemsa | ok, thanks, I understand! You helped me a lot! (if you want you can write an answer. ) | |
| Sep 28, 2015 at 9:29 | comment | added | Rasmus | $\mathbb C^2=C_0(\{a,b\},\mathbb M_1)$. | |
| Sep 28, 2015 at 9:26 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
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| Sep 28, 2015 at 9:14 | comment | added | Sabrina Gemsa | yes, thank you, I remember! therefore $\hat{I}$ must be homeomorphic to $(0,1]$. I know that $\widehat{C_0(X,M_2)}$ equals the Gelfand space of $C_0(X,M_2)$ and the Gelfand space of $C_0(X,M_2)$ is homeomorphic to $X$, the homeomorphism is the point-evaluation-map. But I still don't know what $\widehat{A/I}$ is. It must be a discrete set with 2 elements, but I don't see why... | |
| Sep 28, 2015 at 7:42 | comment | added | Rasmus | Indeed, $\widehat{C_0(X,\mathbb M_n)}$ is homeomorphic to $X$. Do you see a natural map? (This gives you both $\widehat{A/I}$ and $\hat{I}$. It is not hard to describe the topology on $\hat A=\hat I\sqcup\widehat{A/I}$.) | |
| Sep 27, 2015 at 22:10 | history | edited | Sabrina Gemsa | CC BY-SA 3.0 |
added 81 characters in body; edited title
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| Sep 27, 2015 at 21:56 | history | asked | Sabrina Gemsa | CC BY-SA 3.0 |