Timeline for Connectivity of the group of invertible elements of $C(S^{2})\otimes A$
Current License: CC BY-SA 3.0
12 events
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| Mar 30, 2016 at 17:18 | comment | added | hänsel | @Ali: If A is stabilized by the compact Operators, i.e. $A=(B \otimes K)^+$, then the answer is true if and only if $K_1(C(S^2) \otimes A)=0$. Bott periodicity Needs $C(R^2)$ instead of $C(S^2)$. | |
| S Nov 15, 2015 at 13:05 | history | bounty ended | CommunityBot | ||
| S Nov 15, 2015 at 13:05 | history | notice removed | CommunityBot | ||
| Nov 9, 2015 at 15:17 | comment | added | Ali Taghavi | @user75274 Is it obvious that the connected component of GL(A) has laways trivial second homotopy? | |
| Nov 8, 2015 at 19:05 | comment | added | user75274 | Perhaps I am missing something, but isn't it the case that for a Banach algebra $B$, $GL(B)$ is connected iff it is path-connected? I think it's then clear that $GL(C(S^2)]\otimes A)) = C(S^2, GL(A))$ is path-connected iff $GL(A)$ is? | |
| S Nov 7, 2015 at 11:39 | history | bounty started | Ali Taghavi | ||
| S Nov 7, 2015 at 11:39 | history | notice added | Ali Taghavi | Draw attention | |
| Nov 2, 2015 at 16:26 | comment | added | Ali Taghavi | @PaulSiegel I think that the following two, are not equivalent: 1.unitary group of A is connected. 2. $K_{1}(A)$ is trivial, right? | |
| Nov 2, 2015 at 16:15 | comment | added | Paul Siegel | You can generally replace "invertible" with "unitary" up to homotopy when working with C* algebras, so by Bott periodicity I think you just want to look at $K_1(A)$. | |
| Nov 2, 2015 at 15:04 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 2, 2015 at 6:52 | history | edited | Ali Taghavi |
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| Nov 2, 2015 at 6:05 | history | asked | Ali Taghavi | CC BY-SA 3.0 |