Timeline for Morse theory Vs degree theory
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2014 at 10:51 | comment | added | Liviu Nicolaescu | Let us continue this discussion in chat. | |
| Sep 12, 2014 at 20:56 | comment | added | Liviu Nicolaescu | To see that there are only two critical points you don't need Morse theory. It is a very, very simple exercise involving Lagrange multipliers. In any case a point $p$ is critical point for the altitude if and only if the tangent plane at that point horizontal. There are only two such points. | |
| Sep 12, 2014 at 15:44 | comment | added | Vrouvrou | ok, and how we use morse inequality to say that there is only two critical point | |
| Sep 12, 2014 at 15:17 | comment | added | Liviu Nicolaescu | There is no third critical point on $S^2$. | |
| Sep 12, 2014 at 14:45 | comment | added | Vrouvrou | and how to fined the thired critical point ? | |
| Sep 12, 2014 at 14:28 | comment | added | Liviu Nicolaescu | Let $S^2$ be the unit sphere in $\mathbb{R}^3$, $x^2+y^2+z^2=1$. The function $f: S^2\to\mathbb{R}$ that associates to a point $p\in S^2$ its altitude $z=z(p)$ has exactly two critical points, a maximum (the North Pole $(0,0,1)$ and a minimum (the South Pole $(0,0,-1)$. This can be seen by observing that these are the only two points on the sphere where the tangent plane is horizontal, i.e., parallel with the $xy$-plane. | |
| Sep 12, 2014 at 13:59 | comment | added | Vrouvrou | @JeanVanSchaftingen please can you tell me how you defin $f:S^2\rightarrow \mathbb{R}$ and why it has two critical point ? thank you | |
| Sep 12, 2014 at 13:20 | comment | added | Jean Van Schaftingen | @LiviuNicolaescu I am assuming in my example that I have already two minimum points, and I am searching for additional critical points. | |
| Sep 12, 2014 at 13:01 | comment | added | Liviu Nicolaescu | @ Jean I think that in your example you need to replace $S^2$ with a surface $\Sigma$ of positive genus so that $H_1(\Sigma)\neq 0$. On $S^2$ there exist Morse functions with exactly two critical points, a minimum and a maximum and no saddle points. (Take for example the height function.) | |
| Sep 12, 2014 at 12:30 | comment | added | Jean Van Schaftingen | With degree theory, you cannot prove the existence of any additional solution, whereas with Morse theory you prove the existence of two additional solutions. | |
| Sep 12, 2014 at 10:55 | comment | added | Vrouvrou | What about the number of solutions please ? "better estimate of the number of solutions " | |
| Sep 12, 2014 at 10:37 | history | answered | Jean Van Schaftingen | CC BY-SA 3.0 |