Timeline for representatives of the group of homotopy 7-spheres
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2023 at 18:27 | comment | added | The Amplitwist | Reposting the link mentioned in a previous comment so that it appears in the "Linked" questions list: Csar Lozano Huerta's answer to "Your favorite surprising connections in mathematics" | |
| Apr 24, 2023 at 18:24 | comment | added | The Amplitwist |
The link to springerlink.com is broken, but the article can be found at doi:10.1007/BF02412768 (Zbl 0119.18704).
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| May 31, 2012 at 23:47 | vote | accept | Mauricio | ||
| May 16, 2012 at 16:35 | comment | added | John Klein | @Igor: aha, I made an error with respect to what it means to reduce the structure group, namely, if $\Sigma$ denotes and\ exotic $7$-sphere, then a choice of homeomorphism $\Sigma \cong S^7$ together with the Hopf map $S^7 \to S^4$ gives a Top fiber bundle $\Sigma \to S^4$ with fiber $S^3$. The reduction of structure group says that this Top bundle lifts to a smooth one, but the total space of the lift (which is a smooth manifold, possibly $S^7$, if the lift is chosen suitably) is only homeomorphic to $\Sigma$. My mistake was in thinking it might be diffeomorphic. I retract my previous remark. | |
| May 16, 2012 at 16:04 | comment | added | Igor Belegradek | @John: I am not sure how what you say contradicts Eells-Kuiper. Smale conjecture does imply (via Cerf's work) that the inclusion $O(4)\to Homeo(S^3)$ is a weak homotopy equivalence. Thus any topological $S^3$ bundle is topologically isomorphic to a linear $S^3$-bundle. Any homotopy $7$-sphere $\Sigma$ is homeomorphic to $S^7$, so as you say, $\Sigma$ is homeomorphic to the total space of an $S^3$-bundle over $S^4$, e.g. the standard Hopf bundle. This does not imply that $\Sigma$ is diffeomorphic to total space of a smooth $S^3$-bundle over $S^4$. Where is the contradiction? | |
| May 16, 2012 at 14:59 | comment | added | John Klein | @Igor: $\text{Top}(S^3) \simeq S^3 \times PL_3$ (using the Alexander trick). But $PL_3 \simeq O_3$. Also, $\text{Diff}(S^3) \simeq S^3 \times O_3$ by Hatcher. We infer that $\text{Diff}(S^3) \simeq \text{Top}(S^3)$. So, if this argument is correct (?) we see that a fiber bundle with structure group $\text{Top}(S^3)$ admits a reduction of structure group to $\text{Diff}(S^3)$. This would seem to contradict your answer, given that any exotic $7$-sphere fibers topologically over $S^4$ (cf. above). Where's my mistake? | |
| May 16, 2012 at 6:41 | comment | added | Will Sawin | That is what I meant. Sorry for the confusion. | |
| May 16, 2012 at 4:10 | comment | added | Xiaolei Wu | @ Igor, I think Will Sawin means the equation for the Brieskorn spheres. | |
| May 16, 2012 at 3:33 | comment | added | Igor Belegradek | Will Sawin: what formula? | |
| May 16, 2012 at 3:24 | comment | added | Will Sawin | the formula is given in this mathoverflow answer: mathoverflow.net/questions/14574/… | |
| May 16, 2012 at 2:49 | comment | added | John Klein | Yes, it's true all homotopy 7-spheres are Brieskorn spheres. | |
| May 16, 2012 at 1:47 | history | answered | Igor Belegradek | CC BY-SA 3.0 |