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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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    $\begingroup$ 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
    Commented Jan 9, 2019 at 14:08
  • $\begingroup$ @WillieWong what is the counter example? $\endgroup$
    – Amir Sagiv
    Commented Jan 9, 2019 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
    Commented Jan 9, 2019 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
    Commented Jan 9, 2019 at 20:09
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    $\begingroup$ @DeaneYang Thanks. Is there a reference to it? $\endgroup$
    – Amir Sagiv
    Commented Jan 10, 2019 at 7:48

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