Timeline for Is this invariant of an exotic 4-sphere trivial?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2020 at 13:45 | comment | added | skupers | Replace "vector bundle" by "topological microbundle" in your construction. You get an $n$-dimensional oriented topological microbundle on $S^{n-1} \times S^1$, and for $n=4$ this is depends on an element of $\pi_3(STop(4))$ (giving (1), only depending $M$ as a topological manifold).This is the same microbundle as obtained from your vector bundle on $S^{n-1} \times S^1$ by only remembering it is a microbundle (so (2) holds). | |
| Jan 28, 2020 at 12:36 | comment | added | Alex Gavrilov | @skupers. Can you elaborate on the conclusion? We need (1) An analogous invariant in $\pi_3(STOP(4))$ (even though it would be trivial in the end) and (2) a commutative diagram. I am not sure about either. | |
| Jan 26, 2020 at 18:14 | comment | added | skupers | Since the map $Top(4)/O(4) \to Top/O$ is 5-connected by Freedman-Quinn, the map $\pi_3(SO(4)) \to \pi_3(STOP(4))$ is injective. So this invariant only depends on $M$ as a topological manifold. | |
| Jan 26, 2020 at 14:54 | comment | added | Lev Soukhanov | Can not we obtain non-trivial invariant from a standard $S^4$ by choosing some elaborate homeomorphism: $S^4 \rightarrow S^4$? | |
| Jan 26, 2020 at 11:20 | history | asked | Alex Gavrilov | CC BY-SA 4.0 |