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The reach of a set $X\subseteq \mathbb{R}^d$ is the supremum of all $r \geq 0$ such that for all $y\in X^c$ with $dist(y,X)<r$ there is a unique $x\in X$ with $dist(y,x)= dist(y,X)$.

My question: If $X$ is an $m$-dimensional $C^{\infty}$ Riemannian manifold, is there an intrinsic definition of the reach, i.e., one that does not involve an ambient space?

I know that the reach is bounded from below by the inverse of the curvature. Furthermore, by Whitney embedding theorem, $X$ is embedded in $\mathbb{R}^{2m}$, so there is always an ambient space to be found. However, it is not clear whether all embeddings will have the same reach.

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  • $\begingroup$ @BenMcKay you were right my definition was faulty. I've edited so it is in accord with Wikipedia. $\endgroup$ – Amir Sagiv Jan 9 at 13:21
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    $\begingroup$ Regarding the final question: for any $r > 0$ I can give you an isometric embedding of $\mathbb{R}^k$ to $\mathbb{R}^{k+d}$, $k,d$ arbitrary, so that the reach of $X = \mathbb{R}^k$ is precisely $r$. So no. $\endgroup$ – Willie Wong Jan 9 at 14:08
  • $\begingroup$ @WillieWong what is the counter example? $\endgroup$ – Amir Sagiv Jan 9 at 14:19
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    $\begingroup$ The simplest higher dimensional counterexample is that of an open subset of $\mathbb{R}^n$ with the flat metric. In general, if the codimension of the embedding is sufficiently large, the embedding becomes flexible, and it is possible, if laborious, to show that, for any Riemannian manifold, $r$ can be made arbitrarily small. $\endgroup$ – Deane Yang Jan 9 at 15:58
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    $\begingroup$ @DeaneYang: Take $M$ and embed it into $\mathbb{R}^k$ by Whitney. Fix a point $p$ on $M$ and a tangent vector $v\in T_pM$. "Fold" $\mathbb{R}^k$ in $\mathbb{R}^{k+1}$ so that $p$ lies "on the fold" and the fold is "perpendicular to $v$". So you just need to add 1 more dimension to the ambient space to do the construction. $\endgroup$ – Willie Wong Jan 9 at 20:09

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