Timeline for Normal bundle information in the surgery structure set
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 10, 2020 at 20:39 | comment | added | Igor Belegradek | Ranicki's book is nice but I would start studying surgery with the original papers by Milnor-Kervaire, Browder, and Novikov, who take the time to explain things. In particular, Browder's textbook on simply-connected surgery is a good start. | |
| Sep 10, 2020 at 19:50 | comment | added | Connor Malin | @IgorBelegradek Thank you, I was not sure if I could upgrade the equivalence of spherical fibrations to one of vector spaces. | |
| Sep 10, 2020 at 19:43 | comment | added | Igor Belegradek | I do not quite follow your terminology but the claim "homotopy equivalence can always be covered by fiberwise isomorphism of [stable] normal bundle" in incorrect. Do a web search "non-tangential homotopy equivalences". What is true is that homotopy equivalence (of closed manifolds) pulls back the stable spherical fibrations. | |
| Sep 10, 2020 at 19:41 | comment | added | archipelago | I suggest to look for examples of homotopy equivalent closed smooth manifolds with different first Pontryagin class by considering linear $S^3$ bundles over $S^4$. | |
| Sep 10, 2020 at 18:41 | comment | added | Connor Malin | @archipelago Thanks this answers my questions. So I’m not sure about something I said which is that all tangent/normal bundles are stably equivalent no matter which smooth structure we pick, at least if we are compact. I mostly believed this because I thought it was necessary for the maps to make sense in the exact sequence, but now you’ve explained it it needn’t be true. Do you know a counter example? | |
| Sep 10, 2020 at 18:30 | comment | added | archipelago | You pull back the tangent bundle of $N$ along a homotopy inverse. All this can be reformulated in terms of normal bundles, but I prefer the tangential perspective. | |
| Sep 10, 2020 at 18:27 | comment | added | Connor Malin | @archipelago Ah, so then is the map just if we have a homotopy equivalence we can push the normal bundle of the domain manifold forward and have a canonical normal map? | |
| Sep 10, 2020 at 18:14 | comment | added | archipelago | Your definition of $N(M)$ is off: a class in $N(X)$ for a Poincaré complex $X$ is represented by a degree 1 map $N\rightarrow X$ from a manifold covered by a bundle map $TN\oplus \varepsilon^k\rightarrow \xi$ for some $k\ge0$ and \emph{some} bundle $\xi$ over $X$. If $X$ happens to be a manifold itself, $\xi$ need not agree with $TX$. | |
| Sep 10, 2020 at 16:01 | history | asked | Connor Malin | CC BY-SA 4.0 |