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Let $(M,g)$ be a Riemannian manifold and let $N$ be a subset of $M$.

On one hand, it is well known that if $N$ is an embedded submanifold of $M$, then it admits a tubular neighborhood, and, consequently, a closest-point projection from such a neighborhood onto $N$ (for the case $M=\mathbb{R}^n$, see https://mathoverflow.net/a/283477). On the other hand, the existence of a tubular neighborhood is not a necessary condition for the existence of a closest-point projection, as Willie Wong pointed out at https://mathoverflow.net/a/242030.

In this context, my question is whether are known conditions on $N$ which are weaker than supposing that it is an embedded submanifold of $M$ that still assure that $N$ admits a neighborhood over which we can define its closest-point projection over $N$.

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  • $\begingroup$ $N$ being a closed convex set in $\mathbb R^2$ is sufficient, so being an embedded submanifold is certainly very far from necessary. $\endgroup$ Commented Oct 9, 2023 at 12:19
  • $\begingroup$ @MikhailKatz If I'm not mistaken, closed convex subsets of $\mathbb{R}^n$ are such that we can define the closest-point projection from all $\mathbb{R}^n$ onto those sets (see [math.stackexchange.com/questions/1065880/…, for instance). I was looking for the existence of the closest-point projection only on a neighborhood of $N$. Furthermore, I believe the analog to a "closed convex subset of $\mathbb{R}^n$" in our context would be "closed strongly geodesically convex", which is different from being an embedded submanifold... $\endgroup$
    – gpr1
    Commented Oct 9, 2023 at 12:24
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    $\begingroup$ Take the closed region bounded by a Jordan curve $J$ oriented counterclockwise. If the geodesic curvature of $J$ is uniformly bounded below, then there is a neighborhood where the nearest-point projection is defined. $\endgroup$ Commented Oct 9, 2023 at 12:28
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    $\begingroup$ They are called sets with positive reach utgjiu.ro/math/sma/v03/p07.pdf $\endgroup$ Commented Oct 9, 2023 at 12:39

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