# Projections from Tubular Neighborhoods

1. Aug 24, 2012

### Goklayeh

Hello! Could anybody give me some hint with the following problem? Consider a smooth, compact embedded submanifold $M = M^m\subset \mathbb{R}^n$, and consider a tubular neighborhood $U = E(V)\supset M$, where $E: (x, v) \in NM \mapsto x + v \in M$ is a diffeomorphism from a open subset of the normal bundle $NM$ of the form $V = \left\{(x, v) \in NM \: : \: \left| v \right| < \delta\right\}$. We know that $r:=\pi \circ E^{-1}$ is a smooth retraction, where $\pi:(x, v) \in NM \mapsto x \in M$ is the projection. How can I prove that if $y \in U$, where $U$ is a sufficiently small tubular neighborhood, then $r(y)$ realizes the minimum of the distance from the points of $M$? I just proved, following Lee's hints, that if $y \in \mathbb{R}^n$ has a closest point $x \in M$, then $y - x \in N_x M$, but I can't realize how to use this information!

2. Aug 24, 2012

### Goklayeh

Maybe, I've solved: let $y = x + v$, for some $(x, v) \in V$, and consider any other point $p \in M$, together with a curve $\gamma: (-\epsilon, \epsilon) \to M$ such that $\gamma(0) = x, \dot{\gamma}(0) = v$ joining $x, p$. Without loss of generality, $y = 0$. If $f(t):=\frac{1}{2}\left|\gamma(t)\right|^2$, then
$$\dot{f}(0) = \left<\gamma(0), \dot{\gamma}(0)\right> = \left<x, v\right> = 0$$
since $y - r(y) = x + v - x = v \in N_x M$. Since $f$ is convex, $0$ is a minimum, hence the thesis for the arbitrariness of the curve $\gamma$. Am I wrong?