Timeline for Handle decompositions subordinate to an open cover
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 24 at 19:33 | answer | added | Danny Ruberman | timeline score: 4 | |
| Oct 8 at 11:48 | comment | added | Bruno Martelli | To be honest, also interpreting a smooth handle decomposition as a partition of $M$ is something that should be carefully explained... I prefer to see a handle decomposition as an operation. So even the question itself is not completely well-posed IMHO (and this makes it even more interesting). | |
| Oct 8 at 8:18 | answer | added | Ryan Budney | timeline score: 2 | |
| Oct 7 at 23:23 | comment | added | Ryan Budney | I think so, but I don't recall exactly where to find this. I believe there's a fairly cheap and easy argument. I'll think about this on the way home from the office... You can get a piecewise smooth "Morse" function that does everything you want considering the distance to the barycentres of the top-dimensional simplices. The issue is how do you smooth that to a proper Morse function. | |
| Oct 7 at 22:37 | comment | added | Stefan Friedl | the phrase "thicken the simplices into a handle decomposition" makes me a little nervous. As I wrote above: "In pictures this looks reasonable, but I am not sure about the technicalities". Is there any type of reference for such a statement? I don't know how to make this idea work. | |
| Oct 7 at 22:34 | history | edited | Stefan Friedl |
edited tags
|
|
| Oct 7 at 6:30 | comment | added | Ryan Budney | Yes, such things exist. The simplest approach is to take a smoothly-compatible triangulation to your manifold, and subdivide it until it is subordinate to your open cover. Then you thicken the simplices into a handle decomposition. I suppose this is your Morse theory question, but skipping the Morse theory. I would imagine these kinds of arguments go back to Morse. | |
| Oct 7 at 6:13 | history | asked | Stefan Friedl | CC BY-SA 4.0 |