Timeline for Smoothness of the closest point on a submanifold
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2016 at 13:36 | comment | added | Willie Wong | The final statement in the previous comment can be found in most textbooks on differential geometry of curves and surfaces, and boils down to some multivariable calculus. | |
| Jun 13, 2016 at 13:34 | comment | added | Willie Wong | A high powered way to do this is to use what Deane wrote + my comment thereof. Roughly speaking the condition $d_S(p) = 1/\kappa$ is implied by "being a focal point". An alternative way to see what was written here is to note that the normal exponential map on $S$ is a diffeomorphism to its image when restricted to $U$ in the normal bundle of $S$ where for $s\in S$ we have the section $U_s \subset B(s; 1 / \kappa)$ where $\kappa$ is the largest principal curvature at $s$. | |
| Jun 13, 2016 at 2:31 | comment | added | Asaf Shachar | OK, is it easy to see then why in all the other points $\tilde s$ is indeed differentiable? | |
| Jun 12, 2016 at 22:58 | comment | added | Willie Wong | @AsafShachar: I think it is meant that $\tilde{s}$ is $C^1$ except possibly at points $p$ where ... There are obviously points that satisfy the condition $d_s(p) = 1 / \kappa(\tilde{s}(p))$ where $\tilde{s}$ is still $C^1$; for example, points on the convex side of the surface $S$. | |
| Jun 12, 2016 at 20:20 | comment | added | Asaf Shachar | Thanks. Can you please elaborate on why $\tilde s$ is not differentiable at the points you mentioned? | |
| Jun 12, 2016 at 19:04 | history | edited | Mohammad Safdari | CC BY-SA 3.0 |
added 131 characters in body
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| Jun 12, 2016 at 18:11 | history | answered | Mohammad Safdari | CC BY-SA 3.0 |