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Let $M$ be a compact submanifold of dimension $n$ in $\mathbb{R}^{n+1}$ show that exists a normal unitary smooth vector field in $M$

First i think to these is equivalent to show that $M$ is orientable, then how $\dim(M)=n$ and $M$ is compact then i try to work with the Jordan - Brouwer separation theorem because these gives me a open set $A$ such that $\overline{A}$ is a compact submanifold and such that $\partial (\overline{A})=M$ but i not sure to these is correct.

Any hint or help i will be very grateful

Nick_W
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    The existence of a unit normal smooth vector field is indeed equivalent to orientability. So, the word "orientable" should appear in the hypothesis. So, assuming that you have orientability, you could use the Collar Neighborhood Theorem which gives an embedding of the normal bundle of $M$ into the ambient space. – Laz Dec 09 '22 at 03:03
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    Conversely, if you have a unit normal vector field, then the tangent bundle to $\mathbb{R}^{n+1}$ restricted to $M$, which is trivial, is the Whitney sum of $TM$ and $T^{\perp} M$, it's normal bundle. The lat one is trivial by assumption. Now use the fact that, in a Whitney sum, orientability of two terms implies the orientability of the third one. Actually, you can use the last argument on both directions, but the Collar Neighborhood Theorem is something useful to know. – Laz Dec 09 '22 at 03:08
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    @Laz thanks these aclare me the equivalence about the orientability and the existence of normal unitary vector field – Nick_W Dec 09 '22 at 03:13
  • I try to use these equivalence in my problem trying to show that $M$ is orientable – Nick_W Dec 09 '22 at 03:14
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    To avoid any circular logic (some proof of the NCT might first go through the fact that the normal bundle to $M$ is trivial), I would use the other argument. Both conditions guarantee that $T\mathbb{R}^{n+1}|_M$ is trivial. – Laz Dec 09 '22 at 03:16
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    @Laz thanks! I have a doubt because the hint of the book is to use the Jordan Brouwer theorem and these confused me – Nick_W Dec 09 '22 at 03:21
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    @Laz Actually, orientabiliy needs not be an assumption. The fact that $M$ is compact, without boundary and with codimension 1 ensures the orientability: this is a consequence of Jordan-Brouwer theorem – Didier Dec 11 '22 at 12:37
  • @JL_Conte You need to assume $M$ being closed (compact without boundary), OR orientable for this result to be true. The second assumption isn't really interesting (it is really easy to show that orientable and unit normal are equivalent) so I guess you should have the first in your assignment / lecture notes – Didier Dec 11 '22 at 12:39
  • @Didier, I could argue in a dual way about both answers posted. Although the OP mentioned J-B Separation Theorem as a possible argument, nowhere in the hypothesis it says that $M$ is closed, so there is no reason to assume it. Orientability iff unit normal vector field is a perfectly reasonable exercise, whether you consider it easy or not. More context is needed about why the OP mentioned J-B. Plus, your extra hypothesis is way stronger than mine, which is an equivalent one. I don't see why is more interesting to assume stronger hypothesis. – Laz Dec 11 '22 at 17:59
  • @Laz A closed codimension 1 submanifold of an euclidean space is orientable and this is a consequence of Jordan-Brouwer's Theorem. This with the fact that OP refers to this theorem is the reason why I think OP forgot to mention this. I should have chosen other words for "not interesting", sorry if you felt my comment was rude. I did not meant to – Didier Dec 11 '22 at 18:25
  • The OP's mention of J-B is not enough to believe that there was a missing hypothesis in the question. Again, more context in needed. My point is that you both assumed an extra hypothesis, that although reasonable, is not mentioned is the question per se. That's not different from what I did. Don't worry, I try to evaluate the veracity of statements, no feelings involved. – Laz Dec 11 '22 at 18:31
  • @Laz Let's just wait for OP's reply that will clarify a bit. – Didier Dec 11 '22 at 18:34

2 Answers2

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It seems like people suggested in the comments that orientability should be a hypothesis. I don’t think the problem assumes orientability. In fact, it seems like the purpose is to prove that a codimension $1$ submanifold of $\mathbb{R}^{n+1}$ must be orientable.

Assume for a contradiction $M$ is not orientable. Then there exists a loop $\gamma$ on $M$ such that it is not possible to choose consistently oriented bases on all coordinate patches simultaneously that intersect $\gamma$. By picking an arbitrary orientation on some chart, moving along the loop by marking the tip of a unit vector at each point given by this orientation and connecting the two ends of this path once we loop around $\gamma$ once, we can create a loop $\eta$ intersecting $M$ at a single point $p$. Equivalently, this is a section of the normal bundle that vanishes at a single point.

Since $\mathbb{R}^{n+1}$ is contractible, the loop $\eta$ must be the boundary of a disc $D$ in $\mathbb{R}^{n+1}$. We can assume that $\partial D = \eta$ and $M$ are transverse. We can also assume that $D$ and $M$ are transverse after a perturbation that fixes $\partial D$. Hence the intersection $K = D\cap M$ is a compact $1$-dimensional manifold, possibly with boundary. Indeed, $\partial K = \partial(D\cap M) = \partial D \cap M$, thus its boundary is non-empty. But then $K$ must be a union of paths. In particular, its boundary contains at least $2$ points but $\partial D \cap M = \eta \cap M = \{p\}$. This is a contradiction.

In the above argument, I proved orientability because it seems like you observed that the original statement is equivalent to orientability. If it is not clear why this is the case and you can't figure out why, feel free leave a comment.

CeyhunElmacioglu
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    I agree with you: orientability need not be assumed. But $M$ should be assumed closed (without boundary) – Didier Dec 11 '22 at 12:41
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    @Didier Yes, you are right. I implicitly used $M$ is closed and I believe it is essential. – CeyhunElmacioglu Dec 11 '22 at 12:50
  • "In fact, it seems like the purpose is to prove that a codimension 1 submanifold of $\mathbb{R}^{n+1}$ must be orientable." Aside from the fact that thus written this is false, where did you get it from? J-B Separation Theorem is mentioned by the OP as a possible argument, that's it. – Laz Dec 11 '22 at 18:15
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Here is a sketch of another proof, using Jordan-Brouwer's Theorem, which relies on the fact that $M$ is closed (that you did not explicitly mentioned in your question, but this is needed to apply this Theorem). The other is already very good.

Let $A$ be the bounded component of $\Bbb R^{n+1}\setminus M$. Consider the signed distance $r\colon \Bbb R^{n+1}\to \Bbb R$ defined as $$ r(x) = \begin{cases} -d(x,M) & \text{if } x \in A,\\ d(x,M) & \text{if } x \notin A. \end{cases} $$ Then:

  1. Show that $r$ is smooth in a neighbourhood of $M$,
  2. Show that $\nabla r |_{M}$ is an outward unit normal vector field on $M$.
Didier
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