Timeline for Topological structure of SO(n) as a product
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2015 at 6:03 | comment | added | user51223 | @NajibIdrissi We known there is a fibration such as $SO(n-1)\to SO(n)\to S^{n-1}$. If you assume that the decomposition holds, then your input data are: the existence of the fibration and the decomposition of its total as the product of fibre and base, by which you to find a section easily. This situation is slightly different from what you have asked and the upcoming answers afterwards. | |
| Oct 6, 2015 at 13:17 | comment | added | Achim Krause | Nope, not true in general: Take a countable union of copies of $S^1$ and define a map $\cup S^1\rightarrow S^1$ by the double cover $S^1\rightarrow S^1$ on each summand. The fiber is countably many points, the total space is a product of countably many points with $S^1$, but there's no section. | |
| Oct 6, 2015 at 12:46 | comment | added | Najib Idrissi | This is probably a naive question, but is it always true that if $C \cong A \times B$, then any fibration $A \to C \to B$, but where the maps involved are not necessarily the canonical ones, always admits a section? | |
| Oct 6, 2015 at 12:28 | history | answered | user51223 | CC BY-SA 3.0 |