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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!

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  • $\begingroup$ It would have been a good idea to ask why one can choose $\epsilon$ small so that what worries does not occur instead of reasking the same question! (In order to elucidate what my comment on my answer on the old question means) $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 18:31
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    $\begingroup$ There are non-compact submanifolds for which $M(\epsilon)$ is not a manifold for any $\epsilon$, for example. Consider any curve in the plane which contains both the $x$ axis and the part of the graph of $\exp(-x)$ in the first quadrant. $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 18:34
  • $\begingroup$ @MarianoSuárez-Alvarez you're right, sorry. I just thought that I shouldn't bother you on someone else's question. $\endgroup$ – user1337 Jun 13 '15 at 15:00
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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.

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  • $\begingroup$ I'm not sure if I get this right, but why does $M(\epsilon)$ contain a Möbius band component? Doesn't it look something like this? $\endgroup$ – user1337 Jun 13 '15 at 15:03
  • $\begingroup$ It's easier to see using this picture, taken from the same Wikipedia animation as the one you linked to. The surface in the picture is $M$; the set $M(\epsilon)$ consists of two connected components, one that you get to by moving a distance $\epsilon$ from the red side, and another that you get to by moving a distance $\epsilon$ from the blue side. If you choose $\epsilon$ just right, the red-side component will be a Möbius band, each point of which is exactly $\epsilon$ away from two different points on the red side. $\endgroup$ – Jack Lee Jun 14 '15 at 19:04
  • $\begingroup$ @user1337: Just to add to Jack Lee's response, let me point out that if $M$ is the set of points a distance $\epsilon$ from $B$, then $B$ is contained in the set of points a distance $\epsilon$ from $M$. That's why $M(\epsilon)$ contains $B$, a nonorientable surface. $\endgroup$ – Kyle Jun 16 '15 at 16:29
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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.

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  • $\begingroup$ If $M$ is not compact the subset $M(\epsilon)$ need not be a manifold. $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 18:35
  • $\begingroup$ @MarianoSuárez-Alvarez: What's your concern? Self-intersection? $\endgroup$ – Kyle Jun 11 '15 at 18:41
  • $\begingroup$ $\epsilon$ is a number. The concern of the OP is self-intersection, in fact. $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 18:47
  • $\begingroup$ @MarianoSuárez-Alvarez: As pointed out in your observation and also in Jack Lee's observation about $M(\epsilon)$ being nonorientable in some cases, we have to clarify Spivak's question. Since Spivak is looking for a positive result, it seems reasonable to me to allow $\epsilon$ to vary, since that includes the special case where $\epsilon$ is constant --- always an option when $M$ is compact --- as well as opening up the possibility of $M$ being noncompact. $\endgroup$ – Kyle Jun 11 '15 at 18:58
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    $\begingroup$ Surely, the positive result is the one wanted. I am just emphasizing that what the OP wants to know how to do —how do we choose $\epsilon$ small enough?— is simply not doable in some cases. To get around this, you have changed his objective, and that is the sensible thing to do in that circumstance $\endgroup$ – Mariano Suárez-Álvarez Jun 11 '15 at 19:38

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