Timeline for Is there a good definition of (topological) K-Theory over arbitrary spaces?
Current License: CC BY-SA 3.0
15 events
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| Apr 5 at 7:16 | comment | added | Mark Grant | @AlexBogatskiy Paracompact and infinite dimensional might cause problems. If you try and work out the SW classes of a stable inverse to the canonical line bundle on $\mathbb{RP}^\infty$, you'll see what I mean. | |
| Apr 4 at 16:37 | comment | added | Alex Bogatskiy | @MarkGrant according to Bredon (Theorem II.14.2 in his book), stable inverses exist over all smooth manifolds. So isn't paracompactness really what's needed? | |
| Sep 11, 2011 at 22:43 | answer | added | Theo Johnson-Freyd | timeline score: 6 | |
| Aug 30, 2011 at 15:58 | answer | added | Alain Valette | timeline score: 5 | |
| Aug 30, 2011 at 15:21 | comment | added | Georges Elencwajg | @pudin: thanks for your editing. The reason abbreviations like "plz" are not welcome here is that they look childish, buddy-buddy, unprofessional and obviously serve no useful purpose.You are requesting help from (among others) some of the best mathematicians on earth, at least four Fields medalists, and from mature people many of whom are middle-aged or more. Since you have never met (most of) them, a modicum of reserve and civility is expected on this site. This little point settled, I'm sure you meant well and we are very happy to welcome you in our community. | |
| Aug 30, 2011 at 14:36 | history | edited | old account | CC BY-SA 3.0 |
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| Aug 30, 2011 at 13:51 | comment | added | old account | "try to prove"... | |
| Aug 30, 2011 at 13:48 | comment | added | old account | ad 1: in my definition of KO(X) there's no need for the existence of inverses they are just added formally. your comment enters only when you try to proof that every class in the \emph{reduced} K-group may be represented by an actual bundle, right? Then first of all restricting to bundles that have a stable inverse is the same as restricting to bundles that have a finite numerable atlas which is the same as restricting to bundles that pull back from a finite grassmannian, correct? and for a noncompact space this seems to me far smaller than $[-,BGL]$. | |
| Aug 30, 2011 at 12:39 | comment | added | Donu Arapura | We like our vwls. | |
| Aug 30, 2011 at 11:52 | comment | added | old account | edited but may I ask why? | |
| Aug 30, 2011 at 11:48 | history | edited | old account | CC BY-SA 3.0 |
added 6 characters in body
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| Aug 29, 2011 at 19:13 | comment | added | Georges Elencwajg | Please, don't use the spellings "plz", "Thx". | |
| Aug 29, 2011 at 15:12 | answer | added | Neil Strickland | timeline score: 9 | |
| Aug 29, 2011 at 14:59 | comment | added | Mark Grant | As I understand, the compact Hausdorff condition is needed to ensure stable inverses, ie for every bundle $\zeta$ there is a bundle $\xi$ such that $\zeta\oplus\xi$ is trivial. What if you only consider iso classes of stably invertible bundles in the definition of $V(X)$? Do you still end up with $[X,Z\times BGL]$? | |
| Aug 29, 2011 at 14:49 | history | asked | old account | CC BY-SA 3.0 |