Timeline for Topological cobordisms between smooth manifolds
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2020 at 17:54 | comment | added | Connor Malin | Yes sorry Gomez-Lopez is the other author | |
| Aug 15, 2020 at 15:39 | comment | added | Oscar Randal-Williams | 1) You neglected one of the authors of that paper. 2) Cobordism categories seem to have their uses, but not I think for this question. 3) Though smoothing theory indeed says that framed topological manifolds can be smoothed, which answers the framed variant of the question. | |
| Aug 15, 2020 at 14:28 | comment | added | Connor Malin | You might be able to extract it from Kupers’ paper on the topological cobordism category: arxiv.org/pdf/1810.05277.pdf . In it he mentions that his smoothing theory arguments actually imply an equivalence of the nerves of the cobordism categories in the stable framed case. | |
| Dec 13, 2019 at 7:57 | answer | added | Dennis Sullivan | timeline score: 14 | |
| Jun 26, 2018 at 19:53 | comment | added | Arun Debray | @Victor: on a closed smooth manifold, the Wu formula describes how to recover the Stiefel-Whitney classes from Poincaré duality and the Steenrod module structure of mod 2 cohomology. The Steenrod module structure and Poincaré duality are present on closed topological manifolds, so one can use them in the same way to define Stiefel-Whitney classes. Then Stiefel-Whitney numbers can be obtained by evaluating on the fundamental class as usual. | |
| Jan 4, 2018 at 1:58 | comment | added | Victor | I can see that the Pontryagin classes are defined for non-smooth manifolds (more generally for any oriented topological $R^n$-bundle) because the map $BSO\to BSTOP$ is known to be a rational homotopy equivalence. But why Stiefel-Whitney numbers (or classes?) can be defined for non-smooth manifolds? | |
| Mar 3, 2017 at 20:47 | comment | added | Tom Goodwillie | What about the unoriented case? Is there a direct proof that if a smooth manifold bounds a topological manifold (or even a Poincare complex) then it bounds a smooth manifold? | |
| Jun 23, 2015 at 14:35 | comment | added | Oscar Randal-Williams | No: it is not the quality of the map that affects the outcome of the Pontrjagin--Thom construction, but the quality of the bundle. | |
| Jun 12, 2015 at 12:05 | comment | added | user51223 | Is this analogous, or equivalent through Thom-Pontrjagin construction, to asking whether or not within the homotopy class of a continuous map between two smooth manifolds, there is a choice of a smooth map, which vaguely I would imagine would follow from Weierstrass theorem?! | |
| Jun 5, 2015 at 8:08 | history | asked | Oscar Randal-Williams | CC BY-SA 3.0 |