Timeline for Smoothness of the closest point on a submanifold
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 8 at 15:27 | comment | added | Willie Wong | Cannot be Lipschitz. Just take $M = \mathbb{R}^2\setminus \{0\}$ and $S$ the unit circle. Then $\tilde{s}(\epsilon,0) = (1,0)$ and $\tilde{s}(-\epsilon,0) = (-1,0)$. @Learningmath | |
| Mar 7 at 17:16 | comment | added | Learning math | @WillieWong Since the closest function is continuous but not generally differentiable as shown by your example, is it at least uniformly or better, Lipschitz continuous: By the way, I also wonder if the function $\tilde{s}:M\to S$ Lipchitz, i.e. can we say can we say $d_S(\tilde{s}(p), \tilde{s}(p'))\le Kd_M(p,p') \forall p, p'\in M?$? The submanifold $S$ may or may not be compact. | |
| Jun 15, 2016 at 15:02 | comment | added | Willie Wong | @AsafShachar: it is a lot more delicate! Try, for example, to run an abstract argument to prove that your map is differentiable on $M = \mathbb{T}^2$ and $S = \mathbb{T}$. There are some obstructions to running the inverse function theorem argument. One way I see requires identifying the conjugate points, showing it is a manifold, and using it to run some sort of implicit function theorem argument. And we also have to worry now about focal points since the argument breaks. | |
| Jun 13, 2016 at 4:07 | comment | added | Asaf Shachar | Thanks, this is a nice example. I am wondering what is the situation if we restrict $(M,g)$ to be complete. (Of course, this does not imply that every point has a unique minimizig geodesic to $S$). | |
| Jun 12, 2016 at 22:09 | history | answered | Willie Wong | CC BY-SA 3.0 |