Timeline for Non commutative topological manifolds
Current License: CC BY-SA 3.0
18 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 8, 2014 at 20:26 | comment | added | Yemon Choi | Well, a manifold is a topological space that looks locally like ${\bf R}^k$. So a NC manifold should be something that behaves "locally" (whatever that means) like a NC version of ${\bf R}^k$. Andreas Thom's answer shows that at least in the unital (i.e. compact) case there is nothing gained by trying to glue commutative pieces, so first work out what kinds of NC algebras would be good pieces | |
| Nov 8, 2014 at 20:11 | comment | added | Ali Taghavi | @YemonChoi thanks for this comment. could you more explain and clarify? | |
| Nov 8, 2014 at 20:09 | comment | added | Yemon Choi | Good point. OK, what you really need to do is work out a NC version of affine space - within the class of abelian $C^*$-algebras, what properties single out $C_0({\bf R}^k)$; and then what non-abelian $C^*$-algebras also satisfy those properties. Then you can start to ask about gluing, as in your original question. | |
| Nov 8, 2014 at 20:07 | comment | added | Ali Taghavi | @YemonChoi "Dense subalgebra" you mean smooth function? So what about the algebraic picture of topological manifold without smooth structure? | |
| Nov 8, 2014 at 20:04 | comment | added | Yemon Choi | "In this situation, is not natural to extend to Banach algebra?" Not really, because the class of Banach algebras is much much bigger than the class of C*-algebras and they display much more varied behaviour. Surely you should think about what makes manifolds special kinds of topological spaces, and then try to encode that behaviour into some algebraic data -- but this seems to require some dense *-subalgebra of a $C^*$-algebra, not the use of Banach algebras which might not look remotely like any $C^*$-algebra | |
| Nov 8, 2014 at 19:57 | comment | added | Ali Taghavi | @YemonChoi Any way, if you do not like this terminology, you could read my questions and my comments as follows:"Let $A$ and $B$ satisfies the condition of the definition. Is it true to say $A\otimes B$ satisfies the definition, too? | |
| Nov 8, 2014 at 19:53 | comment | added | Ali Taghavi | @YemonChoi I changed to "banach algebra" because a user edit my question from "Could be defined" to "Is defined". On the other hand we observe that for $C^{*}$ algebra we do not get a non commutative object. In this situation, is not natural to extend to Banach algebra? | |
| Nov 8, 2014 at 19:48 | comment | added | Yemon Choi | By the way, now you have changed this to Banach algebras I think this is not a good way to try to define a NC manifold. Recall that the Gelfand-Naimark correspondence only works for commutative C*-algebras. Also please read about general Banach algebras to see that they behave very very very very very very very very differently from the C*-case and to call a Banach algebra a noncommutative space is IMHO extremely tendentious | |
| Nov 8, 2014 at 19:38 | comment | added | Yemon Choi | I just rolled back an edit which made one unnecessary change and which made one change which alters the meaning of one of the sentences | |
| Nov 8, 2014 at 19:37 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 8, 2014 at 19:37 | history | rollback | Yemon Choi |
Rollback to Revision 2
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| S Nov 8, 2014 at 19:33 | history | suggested | user786 | CC BY-SA 3.0 |
fixed grammer
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| Nov 8, 2014 at 19:31 | review | Suggested edits | |||
| S Nov 8, 2014 at 19:33 | |||||
| Nov 8, 2014 at 19:20 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 8, 2014 at 18:25 | vote | accept | Ali Taghavi | ||
| Nov 8, 2014 at 17:56 | answer | added | Andreas Thom | timeline score: 12 | |
| Nov 8, 2014 at 17:21 | history | asked | Ali Taghavi | CC BY-SA 3.0 |