A smooth (n-1)-surface $S \subset \mathbb {R}^n$ is orientable $ \implies S $ has a continuous field of normal vectors to it. I would like to be able to show that: if a (n-1)-surface $S \subset \mathbb {R}^n$ of class $ C ^ 1 $ is orientable $ \implies S $ has a continuous field of normal vectors to it.
Below I show where I have arrived and where I stop.
$ S $ is a surface of class $ C ^ 1 $, so for each of its points $ x \in S $ there exists the tangent space to $S$ at that point, $ TS_x $. A basis for each of these spaces can be found by taking the columns of the Jacobian (evaluated in $ x $) of any chart that contains $ x $ in its domain of action. Given the basis, therefore, we can find the two normals to $ S $ in $ x $, that is the only two vectors with unit norm that are orthogonal to all vectors of the first basis found. Between these two normals, one can be chosen using the following criterion: the basis of $ \mathbb {R} ^ n $ formed by the (n-1) vectors of the basis of $ TS_x $ previously found plus the normal, must be oriented as the canonical basis of $ \mathbb {R} ^ n $. At this point it remains only to show that the normal thus selected, varies continuously on $S$. I have to show that:
$ \lim_ {S \ni x \to x_0 \in S} \mathbf {n} (x) = \mathbf {n} (x_0) $
Thanks to the fact that we are interested in the limit value, we can forget that $ S $ is described by several charts and consider any chart $ \phi \in \mathcal {A} $, fixed, which contains $ x_0 $ in its domain of action.
I feel that somehow I have to tie in the fact that $ J _ {\phi} (x) $ and $ J _ {\phi} (x_0) $ are $ \delta $-close (in any matrix norm) in order to conclude that the normal $ \mathbf {n} (x) $ and $ \mathbf {n} (x_0) $ are $\epsilon $-close, but I can't formalize this argument.
Can anyone help me, please?
EDIT: Proof attempt
$S$ is a surface of class $C^1$, therefore for each of its points $x\in S$ there exists the space tangent to it in that point, $TS_x$, of dimension $n-1$. Since the orthogonal complement of $TS_x$ in $\mathbb{R}^n$ has dimension 1, we can find the two normals to $TS_x$, ie the only two vectors with unit norm that are orthogonal to all vectors of a basis of $TS_x$. Between these two normals one can choose one using the following criterion: the basis of $\mathbb{R}^ n$ formed by the (n-1) vectors of the basis of $TS_x$ previously found and by the normal, must be oriented as the canonical basis of $\mathbb{R}^ n$. Such normal can be found practically as:
$$\mathbf{n}(x)=\frac{\det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\ 
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\ 
\mathbf{e}_n & | & | & |
\end{pmatrix}}{\left | \det\begin{pmatrix}
\mathbf{e}_1 & |  & | & | \\ 
... & \mathbf{v}_1 & ... & \mathbf{v}_{n-1}  \\ 
\mathbf{e}_n & | & | & |
\end{pmatrix} \right |_{\mathbb{R}^n}}$$
where $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$ are the columns of the Jacobian in $x$ of any chart that contains $x$ in its domain of action (they are a basis of $TS_x$).
We must therefore show that $\mathbf{n}(x)$ thus defined is such that:

*

*$\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ is oriented like $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$;

*The above formula for $\mathbf{n}(x)$ always gives the same vector if you use another map of the orienting atlas of $S$ (covering the point $x$), i.e. a different set $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)\}$;

*$\mathbf{n}(x)$ is continuous $\forall x\in S$.

As for the first point, it follows from:
$$\det\begin{pmatrix}
|  & | & | & |\\ 
 \mathbf{v}_1 & ... & \mathbf{v}_{n-1} &\mathbf{n}  \\ 
 | & | & |&|
\end{pmatrix} = C_1^2+...+C_n^2>0$$
where the determinant has been developed along the last column (the normal) and the various $C_i$ are the cofactors of the Laplace development (which are also the components of $\mathbf{n}$, as the normal was first defined).
Since the possible normals for each $TS_x$ are only two, even changing the map used to define $\mathbf{n} (x)$, by point 1 one would always obtain the one capable of orienting $\{\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x),\mathbf{n}(x)\}$ as $\{\mathbf{e}_1 (x),…, \mathbf{e}_{n} (x)\}$, so point 2 is proved.
Finally, the continuity $\forall x \in S$ can be inferred by taking any card covering x (thanks to point 2 and thanks to the fact that continuity is a local property) and noting that the single components of the various $\mathbf{v}_1 (x),…, \mathbf{v}_{n-1} (x)$  are continuous, being the chart of class $C^1$, and that $\det$ and $|\cdot |_(\mathbb{R}^n)$ are also continuous functions.
However, it seems that this 'proof' still does not go well, because if I am not mistaken it can still be applied to each step to the mobius strip. Still, what @neal told me to do I think I managed to prove it in point 2 above.
 A: Your method of selecting a normal vector field is correct. Continuity will follow from continuity of the determinant and, as you intuit, can be handled in each individual coordinate chart, as it is a local argument.
What's missing is actually the crux of the argument, which is that the normal field thus selected is consistent between coordinate charts. You need to start with the fact that $S$ has an orienting atlas, and then you can use its properties to show the normal field is invariant under coordinate change.
A: Here I will use the definition (or characterization) that a manifold $M$ is orientable if (and only if) the line bunde $\Lambda^{\dim M} TM$ is trivial. It is a well-known fact and is not hard to show it is equivalent to having an orientation preserving atlas. I find it easier to use because there is no calculation, we just use the vocabulary of vector bundles.
All the following vector bundles will be $C^0$ because $S$ is $C^1$.
Let $i : S \to \mathbb{R}^n$ be the inclusion map. Let $i^*(T\mathbb{R}^n)$ be the pullback bundle of the tangent bundle of $\mathbb{R}^n$ on $S$ by $i$. Using the euclidean structure of $\mathbb{R}^n$, one can decompose $i^*\left(T\mathbb{R}^n\right)=TS \oplus N_p$ where $TS$ is the tangent bundle of $S$ and $NS$ is the normal bundle of $S$ in $\mathbb{R}^n$:
\begin{align}
NS = \left\{ (p,v) ~|~ v \in {T_pS}^{\perp}\right\}
\end{align}
The normal bundle of a hypersurface is a line bundle. We want to show that there exists a section of $NS$ that always have norm $1$. Notice that $i^*(T\mathbb{R}^n)$ is a trivial rank $n$ vector bundle, and $S$ is orientable, thus $\Lambda^n\left(i^*T\mathbb{R}^n\right)$ and $\Lambda^{n-1}\left(TS\right)$ are trivial line bundles. But it is easy to show, using the decomposition, that
\begin{align}
\Lambda^n\left(i^*T\mathbb{R}^n \right) = \Lambda^{n-1}\left(TS\right) \otimes \Lambda^1\left(NS\right)
\end{align}
and using the triviality of the two first line bundles show that $\Lambda^1\left(NS\right)$ is also a trivial line bundle over $S$: indeed, as the LHS is trivial, there exist a nowhere vanishing section $s\in \Lambda^n\left(i^*T\mathbb{R}^n\right)$. By the above decomposition, there exist $a_i,b_i$ sections of $\Lambda^{n-1}(TS)$ and $\Lambda^1(NS)$ respectively and $f_i$ $C^1$ functions with
\begin{align}
s = \sum_{i=1}^k f_i\cdot( a_i\otimes b_i)
\end{align}
As $\Lambda^{n-1}(TS)$ is trivial, there exist a nowhere vanishing section $a_0$ and smooth functions $g_i$ such that $a_i =g_ia_0$, and then
\begin{align}
s = \sum_{i=1}^k f_i\left(g_ia_0\right)\otimes b_i = a_0\otimes \left(\sum_{i=1}^kf_ig_ib_i \right)
\end{align}
The fact that $s$ and $a_0$ are nowhere zero shows that $b_0 = \sum_{i=1}^k (f_ig_i)b_i$ is a nowhere zero section of $\Lambda^1(NS)$. Consequently, $\Lambda^1(NS)$ is trivial as a line bundle.
But $\Lambda^1\left(NS\right)=NS^* \simeq NS$ (recall that $NS$ is of rank $1$), so $NS$ is a trivial line bundle over $S$: there exists a non-vanishing $C^1$ section $X \in \Gamma(NS)$.
To conclude, notice that $\nu = \frac{X}{\|X\|}$ is a $C^0$ normal unitary vector field over $S$.
Edit: another proof Let $S$ be an orientable $C^1$ hypersurface of $\mathbb{R}^n$, that is endowed with an orientable atlas $(U_{\alpha},\phi_{\alpha})$ - that is with $\det(\mathrm{d}(\phi_{\alpha}\circ{(\phi_{\beta}}^{-1}))>0$ if $U_{\alpha}\cap U_{\beta}\neq 0$. Let $\phi_{\alpha}: U_{\alpha} \to \phi_{\alpha}(U_{\alpha})\subset\mathbb{R}^{n-1}$ be such a chart. Let $V_1,\ldots,V_{n-1}$ be the pullback vector fields of the canonical basis of $\mathbb{R}^{n-1}$, that is
\begin{align}
V_i(p) = \mathrm{d}{\phi_{\alpha}}(p)^{-1} e_i,~ p \in U_{\alpha}
\end{align}
Then $(V_1(p),\ldots,V_{n-1}(p))$ is a basis of $T_pS$. By the Gram-Schmidtt process, one can construct on $U_{\alpha}$ an orthonormal frame $(E_1,\ldots,E_{n-1})$ of $TU_{\alpha}$, that is at each $p\in S$, $(E_1(p),\ldots,E_{n-1}(p))$ is an orthonormal basis of $T_pS$.
Notice that since the charts are $C^1$, the orthonormal frame above is $C^0$ ($\mathrm{d}\phi$ is $C^0$ and the Gram-Schmidtt algorithm is smooth).
Now, let $(E_1,\ldots,E_{n-1})$ be an orthonormal frame on $U_{\alpha}$. Fix $p\in U_{\alpha}$ and define $N_p$ to be the unique solution in $\mathbb{R}^n$ of the system:
\begin{align}
\forall i\in \{1,\ldots,n-1\},~ \langle E_i(p),N_p\rangle& = 0 & \det\left(E_1(p),\ldots,E_{n-1}(p),N_p\right) = 1
\end{align}
Then $N_p$ is the unit normal vector we are looking for. To conslude, we have to show two things:

*

*the definition of $N_p$ does not depend on the chart $U_{\alpha}$, that is, if $p\in U_{\alpha}\cap U_{\beta}$, then the above construction does not depend on the orthonormal frame defined by $\phi_{\alpha}$ or $\phi_{\beta}$

*$p \mapsto N_p$ is continuous

First, to show that $N_p$ is well defined, notice that if $V=(V_1,,\ldots,V_{n-1})$ and $W=(W_1,\ldots,W_{n-1})$ are the frames defined by $\phi_{\alpha}$ and $\phi_{\beta}$ respectively, then they define the same orientation on $U_{\alpha}\cap U_{\beta}$ because $\det_VW= \det\left(\mathrm{d}(\phi_{\alpha}\circ{\phi_{\beta}}^{-1})\right)>0$ (maybe it is the inverse of its determinant but the sign is the same, I did not make the full calculation). Thus, if $E$ and $F$ are the frame given by the ram-Schmidtt process from $V$ and $W$, they still define the same orientation, and thus,the definition of $N$ does not depend on $U_{\alpha}$ and $U_{\beta}$ of the chart $\phi_{\alpha}$ or $\phi_{\beta}$, but only on the orientation they define.
Secondly, to show that $N$ is continuous, it suffices to show that it is locally continuous, i.e that it is continuous on all charts $U_{\alpha}$ (recall that continuity in $\mathbb{R}^n$ is a local property). On $U_{\alpha}$, let $E_1,\ldots,E_{n-1}$ be the local orthonormal frame defined above. Let $f_{\alpha} : U_{\alpha}\times\mathbb{R}^n \to \mathbb{R}^n$ defined by
\begin{align}
f(p,X) = \left(\langle X,E_1(p)\rangle,\ldots,\langle X,E_{n-1}(p)\rangle, \det(E_1(p),\ldots,E_{n-1}(p),X)-1 \right)
\end{align}
One can show (easy but annoying calculations) that $f$ satisfies the hypothesis of the implicit function theorem around $f^{-1}(0)$ ($f$ is smooth in the second variable). Therefore, there exists a continuous function $p \in U_{\alpha} \mapsto X_p\in \mathbb{R}^n$ such that
\begin{align}
f(p,Y) = 0 \iff Y = X(p)
\end{align}
(maybe one has to restrict to a subset of $U_{\alpha}$ that is still open, this does not affect continuity as it is a local notion)
Consequently, $N = X$ and $N$ is continuous.
