Timeline for Bijection of critical points on two manifolds
Current License: CC BY-SA 3.0
11 events
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Jun 14, 2017 at 20:39 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
May 15, 2017 at 19:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Apr 15, 2017 at 18:25 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Mar 16, 2017 at 18:22 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Feb 14, 2017 at 17:14 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Jan 15, 2017 at 15:52 | answer | added | Guangbo Xu | timeline score: 2 | |
Nov 18, 2016 at 4:24 | comment | added | Simon Zhu | Thanks @WillieWong. To make clear, the Hessian of $f$ is computed as a function on $g^{-1}(c)$, thus $g$ can be the coordinates system of the one dimensional manifold of the critical points of $f|_g^{-1}(c)$, thus leads to the bijection. | |
Nov 17, 2016 at 21:59 | comment | added | Willie Wong | Implicit function theorem. Roughly speaking you can choose coordinates such that $g$ is one of the coordinates. Then $\mathrm{d}f|_{g^{-1}(c)}$ is a map $\mathbb{R}^n\to\mathbb{R}^{n-1}$. Its zeroes are locally one dimensional manifolds provided that the second derivatives $D^2 f$ are not degenerate. Note that the Hessian of $f$ in your example, evaluated at $x = y = 0$ is identically zero. | |
Nov 17, 2016 at 21:43 | history | edited | SashaP | CC BY-SA 3.0 |
added 33 characters in body; edited tags; edited title
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Nov 17, 2016 at 21:36 | review | First posts | |||
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Nov 17, 2016 at 21:34 | history | asked | Simon Zhu | CC BY-SA 3.0 |