A "parallel manifold" is always orientable

Question

I want to solve the following problem from Spivak's Calculus on Manifolds:

Let $M$ be an $(n-1)$ dimensional manifold in $\mathbb{R}^n$. Let $M(\varepsilon)$ be the set of end points of normal vectors (in both directions) of length $\varepsilon$ and suppose $\varepsilon$ is small enough so that $M(\varepsilon)$ is also an $(n-1)$ dimensional manifold. Show that $M(\varepsilon)$ is orientable (even if $M$ is not).

I'm aware that the question was already asked here, but I'm not completely satisfied with the answer, that basically defines a normal field on $M(\varepsilon)$ by aiming back for $M$. As I've said in the comment, what if a point $q \in M(\varepsilon)$ is the end point of several normal vectors in $M$? (if this can always be avoided by taking $\varepsilon$ sufficiently small, I would like a proof of that). Also, why is the resulting normal field really is continuous?

Alternate solutions are also welcome. Thank you!

Answer 1

Here's how Guillemin & Pollack explain it: Define the normal bundle $$N(M) =\{(x,v) \in M \times \mathbb{R}^n \mid v \perp T_x M \}.$$ Its straightforward to see that $N(M)$ is a manifold. Now define $h: N(M) \to \mathbb{R}^n$ by $h(x,v)=x+v$. This map is regular at $M_0 =\{(x,0) \in N(M)\}$ and restricts to a diffeomorphism $h|_{M_0}: M_0 \to M$. Then a generalization of the Inverse Function Theorem implies that $h$ must be a diffeomorphism on a neighborhood $U\subset N(M)$ of $M_0$. For convenience, let $V=h(U)$, the corresponding neighborhood of $M$ in $\mathbb{R}^n$.

Next, define $\sigma: N(M) \to M$ by $(x,v)\mapsto x$ and $\pi: V \to M$ by $\pi=\sigma \circ h^{-1}$. Then $\pi$ is our desired projection that takes every point in a neighborhood of $M$ and sends it to a point $y\in M$ such that $y-x$ is normal to $M$. If $M$ is compact, there is a constant $\epsilon>0$ such that $$N_\epsilon(M) = \{ (x,v) \in N(M) \mid \|v\|=\epsilon\}$$ is contained in $U$, hence $M_\epsilon=h(N_\epsilon (M))$ is contained in $V$. (Just let $\epsilon$ be the minimal value of the distance function from $M_0$ to $N(M) \setminus U$.) If $M$ is not compact, we can cover it with open sets $U_\alpha$ whose closures $\overline U_\alpha$ are compact, do the above calculation to find a constant $\epsilon_\alpha$ on each $\overline U_\alpha$, then use a partition of unity to get a function $\epsilon: M \to (0,\infty)$ such that $M_\epsilon$ lies inside $V$, as desired. The restriction of $\pi$ to $M_\epsilon$ does exactly what we want: Define a normal vector field at $y\in M_\epsilon$ by $y-\pi(y)$. This is well-defined, smooth, and nonvanishing, so the hypersurface $M_\epsilon$ is orientable.

Answer 2

The trouble is that Spivak's statement of the problem is imprecise. If we take his statement literally, then it's not true. For example, we could start with an embedded Möbius band $B\subseteq\mathbb R^3$, and let $M$ be the set of points at a small distance $\epsilon$ from $B$. Then $M$ is an embedded submanifold diffeomorphic to a cylinder. In this case, $M(\epsilon)$ is also a $2$-dimensional manifold, so you'd have to say that "$\epsilon$ is small enough." But one connected component of $M(\epsilon)$ is $B$, which is not orientable.

What Spivak evidently had in mind was to choose $\epsilon$ small enough that the map $h\colon N(M)\to\mathbb R^3$ defined in @squirrel's answer is a diffeomorphism on an open subset containing $\{(x,v)\in N(M): \|v\|\le \epsilon\}$. In that case, @squirrel's argument shows that $M(\epsilon)$ is orientable.