Timeline for Is the Milnor construction contractible
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2019 at 16:49 | answer | added | Qayum Khan | timeline score: 5 | |
| Dec 4, 2013 at 15:33 | comment | added | Oldřich Spáčil | @O.Straser By Proposition 14.4.6 in T. tom Dieck: Algebraic Topology the space $E_{G}$ is indeed contractible and the assumption on $G$ is only that it is a topological group (as far as I can understand). The argument is pretty much the same as the one mentioned in the above comment by few_reps. | |
| Dec 4, 2013 at 9:49 | comment | added | few_reps | I can't see why Andrew Stacey's argument (see mathoverflow.net/questions/198/… ) would not work ... if $G$ is a subspace of some vector space $V$ over $\mathbf R$ you can see the Milnor construction as the convex hull of the individual copies of $G$ in the product $V\times V\times ...$ and apply the shift on that ... or do I miss something ? | |
| Dec 4, 2013 at 9:44 | comment | added | Oliver Straser | Do you know any example? | |
| Dec 4, 2013 at 9:26 | comment | added | johndoe | If I'm not mistaken, you only need $E_G$ to be weakly contractible in order to view it as a universal $G$-bundle, $G$ arbitrary. I guess $E_G$ might indeed be non-contractible for wild $G$. | |
| Dec 4, 2013 at 8:58 | history | edited | Oliver Straser | CC BY-SA 3.0 |
deleted 6 characters in body
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| Dec 4, 2013 at 8:41 | comment | added | johndoe | perhaps do you mean contractible? | |
| Dec 4, 2013 at 8:29 | history | asked | Oliver Straser | CC BY-SA 3.0 |