n-1 form inducing normal unit vector field

Question

Suppose we have a $n-1$ dimensional manifold $M \subset \mathbb{R}^n$ and a non-vanishing $n-1$ form $\omega$ on $M$. How would this imply the existence of a normal unit vector field on $M$?

Answer 1

Let $x\in M$, there exists a neighborhood $U$ of $x$ in $\mathbb{R}^n$, a submersion $f:U\rightarrow \mathbb{R}$ such that $M\cap U=f^{-1}(0)$, $T_xM=\{u\in T_x\mathbb{R}^n:df_x(u)=0\}$. Write $df_x=(\partial f_1(x),...,\partial f_n(x))$. You can identify $(\partial f_1(x),...,\partial f_n(x))$ with a vector $u_x$ of $T_x\mathbb{R}^n$ such that $Vect(u_x)$ is a supplementary space to $T_xM$. Let $\Omega$ be the canonical volume form $dx_1\wedge...\wedge dx_n$, $(\partial f_1(x)dx_1+...+\partial f_n(x)dx_n)\wedge \omega =c\Omega$, if $c>0$, define $n(x)={1\over{\|u_x\|}}u_x$ if $c<0$, define $n(x)=-{1\over{\|u_x\|}}u_x$. Remark that $n(x)$ is well defined in a neighborhood of $x$ since it does not depend of the choice of $f$. In fact $u_x$ is a unit vector orthogonal to $T_xM$ relatively to the usual scalar product and it continuously depends of $x$.