Is the set ${x in mathbb R^n : d(x, M) = c}$ a smooth manifold for a small constant $c$ when $M$ is a smooth manifold embedded in $mathbb R^n$?

Question

I am a beginner of differential geometry. Let $M$ be a smooth manifold embedded in $mathbb R^n$ and consider the
subset
$$
      S = {x in mathbb R^n : d(x, M) = c },
$$
where $d(x, M)$ denotes the distance between a point $x$ and the manifold $M$. I would like to know if $S$ is a smooth (at least $C^1$) manifold for sufficiently small $c > 0$. In one of the simplest cases, that $M$ is a (finite) line segment in $mathbb R^3$, I think $S$ is the union of a cylinder of radius $c$ with its center line $M$ and two half spheres attached to the both ends of the cylinder, and $S$ is smooth.

diracdeltafunk · Answer

Not necessarily; the most important issue is that two points of $M$ which are "far apart inside $M$" might actually be very close together in $mathbb{R}^n$. For example, let $M$ be the image of the smooth embedding $f : mathbb{R} to mathbb{R}^2$ given by $f(t) = (cos(2arctan(t)), sin(2arctan(t)))$; i.e. $M = S^1 setminus {(-1,0)}$. For any $c > 0$, the set ${x in mathbb{R}^2 : d(x,M) = c}$ contains two points which have no open neighborhood homeomorphic to any Euclidean space.
This issue is obliterated if $M$ is compact; then the result does hold (with $S$ being smooth, assuming $M$ is smooth as stated in the question). There is a more general result called the "tubular neighborhood theorem" or the "$varepsilon$-neighborhood theorem" which may be of interest if you care about cases where $M$ is not compact.

Is the set ${x in mathbb R^n : d(x, M) = c}$ a smooth manifold for a small constant $c$ when $M$ is a smooth manifold embedded in $mathbb R^n$?

One Answer

Add your own answers!

Ask a Question