Timeline for Smooth structure on direct product
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 29, 2020 at 23:21 | history | became hot network question | |||
| Sep 29, 2020 at 18:25 | vote | accept | CommunityBot | ||
| Sep 29, 2020 at 18:15 | comment | added | Connor Malin | @MichaelAlbanese You are right about the Pontryagin class. I think this is sufficient to answer my question. Thank you. | |
| Sep 29, 2020 at 18:13 | comment | added | Michael Albanese | @ConnorMalin: Unfortunately I can't address your comments as I lack the background. Hopefully by consulting the argument in Scorpan, you can find the resolution to your question. | |
| Sep 29, 2020 at 18:12 | comment | added | Connor Malin | To be the domain of a normal invariant of the sphere is equivalent to have your Spivak normal bundle be trivial, which is weaker than having the microbundle being stably parallelizable (which for a smoothable manifold means the tangent bundle is trivial as an $\mathbb{R}^n$ bundle), so perhaps I have misunderstood and in fact only the Spivak normal bundle is trivial, not the tangent microbundle. | |
| Sep 29, 2020 at 18:10 | comment | added | Connor Malin | @MichaelAlbanese I definitely could be misunderstanding, but I thought the Milnor manifolds (of which M is the first example) were constructed to show that the surgery obstruction map from the normal invariants of the sphere to $8\mathbb{Z}$ were surjective. Here the surgery obstruction map takes a normal invariant of the sphere to its signature. | |
| Sep 29, 2020 at 18:00 | comment | added | Michael Albanese | @ConnorMalin: I am not that familiar with microbundles, but wouldn't there be a non-zero $p_1$? | |
| Sep 29, 2020 at 17:40 | answer | added | Danny Ruberman | timeline score: 11 | |
| Sep 29, 2020 at 17:35 | comment | added | Connor Malin | @MichaelAlbanese Shouldn't its product with $S^1$ be smoothable? The microbundle of $M$ is stably parallelizable, so the tangent bundle of this product should be trivial, correct? Then smoothing theory shows a parallelizable manifold has at least one smooth structure. | |
| Sep 29, 2020 at 15:53 | comment | added | Michael Albanese | Worth pointing out that $M\times S^k$ is not smoothable for any $k$; see the lemma on page 219 of The Wild World of 4-Manifolds by Scorpan. | |
| Sep 29, 2020 at 15:20 | history | asked | user164740 | CC BY-SA 4.0 |