# n-1 form inducing normal unit vector field

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 $$M$$?

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 $$x$$.
• Where do you use the fact that $\omega$ is non-vanishing? Is it to say that c is either positive or negative? Moreover, I don't see how you relate the form to $n(x)$, how does $n$ kind of depend on c. And is there a reason why you only added dx1? – AkatsukiMaliki Dec 4 '18 at 8:44
• The fact that $\omega$ does not vanish is used to remark that $c\neq 0$. – Tsemo Aristide Dec 4 '18 at 9:33
• I think c can be zero in this case, take $\omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$. – AkatsukiMaliki Dec 4 '18 at 15:06