Timeline for Is there always a universal bundle over a classifying space?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2016 at 13:18 | vote | accept | Jan Steinebrunner | ||
| Nov 13, 2016 at 13:17 | vote | accept | Jan Steinebrunner | ||
| Nov 13, 2016 at 13:18 | |||||
| Nov 6, 2016 at 21:22 | vote | accept | Jan Steinebrunner | ||
| Nov 13, 2016 at 13:17 | |||||
| Nov 5, 2016 at 20:50 | answer | added | Peter May | timeline score: 5 | |
| Nov 5, 2016 at 19:27 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 6, 2016 at 19:24 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 6, 2016 at 19:11 | answer | added | Jan Steinebrunner | timeline score: 2 | |
| Sep 6, 2016 at 18:46 | comment | added | Denis Nardin | If you have found a compelling answer to your question, by all means post it. | |
| Sep 6, 2016 at 18:37 | comment | added | Jan Steinebrunner | As it turns out the two universal properties are equivalent for the most types of bundles! This is Assertation 8.3 of Kirby and Siebenmann's "Foundational Essay IV". I think this is pretty surprising/amazing, should I give this as an answer to the question? And should I edit the question such that it fits this answer better? | |
| Aug 27, 2016 at 12:13 | comment | added | Jan Steinebrunner | Later on a PL structure on a TOP bundle $\xi$ is defined as a TOP morphism $\phi:\xi\to\gamma_{PL}$, this is somehow problematic, it there are multiple morphisms even to $\gamma_{TOP} $ . I will have a close look at the consequences later. Thank you a lot ! | |
| Aug 27, 2016 at 12:06 | comment | added | Denis Nardin | Either way I feel confident enough to state that 2.2-2.4 in that paper are wrong, but I venture a guess that if you use the correct definitions the rest paper won't be really affected. | |
| Aug 27, 2016 at 11:58 | comment | added | Jan Steinebrunner | Yes, but I think there is an additional problem in 2.4: a morphism to the universal bundle can only be thought of as a trivialisation, if its base map is constant and the fibre it's mapping to has a preferred trivialisation | |
| Aug 27, 2016 at 11:51 | comment | added | Denis Nardin | Right, that sounds stronger. In general you'll have to twist the inclusion $\xi_A\to \xi$ by an automorphism of $\xi_A$. See for example remark 2.4: he seems to say that all trivializations of a trivial bundle are homotopic, but that's simply false: the set of trivialization up to homotopy is isomorphic to the set $[X,Top_N]$ | |
| Aug 27, 2016 at 11:47 | comment | added | Jan Steinebrunner | I think Proposition 2.3 states that the morphism is unique up to homotopy. | |
| Aug 27, 2016 at 11:43 | comment | added | Jan Steinebrunner | Sorry, I unintentionally posted it before I was finished. What he requires then has to be a much stronger condition. I am not even sure, whether it is still true like this | |
| Aug 27, 2016 at 11:37 | comment | added | Jan Steinebrunner | @DenisNardin Ok, then this is the reason for my confusion. I took the definition from arxiv.org/abs/math/0105047 Where Rudyak defines a universal fibre bundle like this on page 22. Theorem-Definition 2.2 | |
| Aug 27, 2016 at 11:27 | comment | added | Denis Nardin | I think you are misunderstandiing the definition of universal bundle. Usually only what you call the "base morphism" is required to be unique: as you correctly note the universal vector bundle of rank $n$ does have automorphisms. Basically the definition of universal bundle is rigged so to correspnd to a natural isomorphism $F(X)\cong [X,B]$. | |
| Aug 27, 2016 at 11:03 | history | asked | Jan Steinebrunner | CC BY-SA 3.0 |