Timeline for Integral homology classes that can be represented by immersed submanifolds but not embedded submanifolds
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2, 2023 at 11:30 | comment | added | mme | In half codimension in a manifold of dimension 4k you can take two embedded representatives, make them transverse, and take connected sums at the intersection points to get an embedded and connected representative. There are no easy tricks for counterexamples! | |
| Aug 2, 2023 at 11:26 | comment | added | RBega2 | The ambient manifold isn't closed, but if $M=T\mathbb{S}^2$, then shouldn't an intersection number argument imply that twice the generator of $H^2(M)$ can't be represented by an embed submanifold. My algebraic/geometric topology is pretty bad, but the internet seems to suggest that the same idea should work for twice the generator of $H^2(\mathbb{CP}^2)$. | |
| Aug 2, 2023 at 10:58 | comment | added | Jason Starr | I missed that the submanifold is allowed to be disconnected! | |
| Aug 2, 2023 at 10:58 | comment | added | mme | @JasonStarr At least one can still represent this by an embedded submanifold, just not a connected one. Following Thom, I would expect OP's examples to first appear in codimension at least 3 in fairly high dimensional manifolds (dim >= 8?) | |
| Aug 2, 2023 at 10:55 | comment | added | Jason Starr | Then you can replace $\mathbf{S}^1$ by $\mathbf{S}^1\times \mathbf{S}^1$ and still take the homology class of twice the homology class of $\mathbf{S}^1\times\{\ast\}$. | |
| Aug 2, 2023 at 10:12 | comment | added | Zhenhua Liu | Thanks for pointing this out. I should have added that the setting is the same as Thom's paper, i.e., codimension always larger than 0. I'll add this to the question. | |
| Aug 2, 2023 at 7:11 | history | answered | Greg Friedman | CC BY-SA 4.0 |