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Nov 17, 2020 at 10:03 history edited dohmatob CC BY-SA 4.0
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Nov 17, 2020 at 8:42 comment added dohmatob On second thought, if $M$ is closed and sufficiently smooth, then it can be embedded into some euclidean space of dimension $m=O(n^3)$. Then one can lift $F$ from $\mathbb R^n$ to $\mathbb R^m$ in an obvious way. Then the tentative definition in my comment (based on inverse retractions) should reduce to the attempt in my question above, modulo an $\epsilon$ which we later limit to $0^+$. What do you think ?
Nov 17, 2020 at 8:41 comment added user35593 Yes its always possible but it might be difficult to find a concrete embedding for a specific manifold. Also it might be favorable to have a method that is independent of the embedding. Therefore the exponential map.
Nov 17, 2020 at 8:36 comment added dohmatob BTW, I don't understand you remark about "isometrically embedded". Do you mean embedded in euclidean space of any dimension (this is always possible if $M$ is closed, thanks to the Nash embedding theorem) or embedded in $R^n$ (i.e a constraint on the embedding dimension) ?
Nov 17, 2020 at 8:22 comment added dohmatob Thanks for the suggestion. Definnition 2.1 is referring to differentiability. I guess you where referring to something like $$ \partial f(w) := \left\{w^\star \in T_wM \mid \liminf_{x,y \to w,\; x \ne y}\frac{F(x) - F(y) -\langle w^\star, \exp^{-1}_w(y)-\exp^{-1}_w(x)\rangle}{\|\exp_w^{-1}(y) - \exp_w^{-1}(x)\|} \ge 0\right\} $$
Nov 17, 2020 at 8:08 comment added user35593 That works if the manifold is isometrically embedded in Euclidean space. If not you might use Definition 2.1. of researchgate.net/publication/…
Nov 17, 2020 at 7:07 history edited dohmatob CC BY-SA 4.0
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Nov 17, 2020 at 2:21 history edited LSpice CC BY-SA 4.0
Redundant accent: \'é -> é, and minor proofreading
Nov 17, 2020 at 2:11 history asked dohmatob CC BY-SA 4.0