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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.
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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
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Nov 17, 2016 at 21:36 review First posts
Nov 17, 2016 at 21:43
Nov 17, 2016 at 21:34 history asked Simon Zhu CC BY-SA 3.0