Prove there exists a outward unit normal field on the boundary this manifold

Question

Let $M$ be a compact subset of $\mathbb{R}^3 $ with the standard orientation $\mu =[e_1,e_2,e_3] $ and let $S = \partial{M}$ is its smooth boundary with the induced orientation from $M$. Prove there exists a smooth normal unit field $\textbf{n}(x) = (n_1(x),n_2(x),n_3(x))$ for $x \in S$ such that if {$v_1,v_2$} is a basis for $T_xS$ with [$v_1,v_2$] = $\partial{\mu}$ then [$v_1,v_2,\textbf{n}(x)$] = $\mu$. I know that given $v_1$ and $v_2$, I can construct a unit vector $v_3$ orthogonal to $v_1$ and $v_2$ using the cross product. I think this would imply that $n_1 = v_1$, $n_2 = v_2$ and $n_3 = v_3$ but how does this imply exactly that [$v_1,v_2,\textbf{n}(x)$] = [$v_1,v_2,v_3$] = $\mu$?

Answer 1

By your description, I assume your $M$ is a smooth orientable $3$-manifold. Then I think you can drop the compactness hypothesis. Indeed, since your $M$ is a manifold with boundary $\partial M=S$, then $S$ is a $2$-manifold without boundary in $\mathbb{R}^3$, or a usual surface. As you pointed out, it is orientable as it can be put on the boundary orientation inherited from $M$. But you don't have to specify orientation, just now translate your orientability to charts. One well-known definition of orientation on a manifold is that you have an oriented atlas, i.e a compatible atlas $\{(\phi,U)\}$ such that whenever $U\cap V\neq\emptyset$ for a pair of carts $(\phi,U)$, $(\psi,V)$ in the atlas, $\det(\psi\circ \phi^{-1})|_{\phi(U\cap V)}>0$. Use this definition for your surface $S$. You can use parametrizations (inverses of charts) in this case, as they may be more useful as maps from open subsets of $\mathbb{R}^2$ into $S\cap \mathbb{R}^3$.
Whenever $(x,A)$ is parametrization of $U=x(A)\subset S$ and $p\in U$, define $\eta(p)=\frac{x_u\times x_v}{|x_u\times x_v|}(x(p))$, where $u,v$ are the coordinates in $A\subset \mathbb{R}^2$. This is well defined at $p$ because since $x$ is a parametrization, its differential is one-one and hence $x_u\times x_v\neq 0$. Clearly it is unit and normal to $T_pS=span\{x_u, x_v\}$. It's also smooth on $U$ by definition. So, you are only left with proving that it doesn't depend on the parametrization used around $p$.
For this, observe that if $(y,B)$ is any other parametrizations of $S$ with $p\in x(A)\cap y(B)$ and $\eta^{'}(p)=\frac{y_u\times y_v}{|y_u\times y_v|}(y(p))$, since $\eta(p), \eta^{'}(p)$ are both orthogonal to $T_pS$ and unit, it can only happen $\eta(p)=\pm \eta^{'}(p)$. The plus sign is going to happen precisely because $(y_u\times y_v)(y(p))=x_u\times x_v (x(p))\cdot \det(y^{-1}\circ x)|_{x^{-1}(p)}$, and this $\det$ is positive by the choice of your parametrizations, or charts.
$\textbf{Added}$: this last step you can prove it by observing that the $dx_{x^{-1}(p)}, dy_{y^{-1}(p)}: \mathbb{R}^2\rightarrow T_pS\subset \mathbb{R}^3$ are isomorphisms, the cross product in $\mathbb{R}^3$ is bilinear and antisymmetric, and that the only (up to scalars) bilinear and antisymmetric map in $\mathbb{R}^2$ is $\det$. It takes some details, but that's an idea to prove it. Another is just brute force calculation.