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 $x$.

  • $\begingroup$ 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? $\endgroup$ – AkatsukiMaliki Dec 4 '18 at 8:44
  • $\begingroup$ The fact that $\omega$ does not vanish is used to remark that $c\neq 0$. $\endgroup$ – Tsemo Aristide Dec 4 '18 at 9:33
  • $\begingroup$ But you wrote obly dx1, why not all up to dxn? $\endgroup$ – AkatsukiMaliki Dec 4 '18 at 10:09
  • $\begingroup$ Only* . But i dont see the relation betwenen c and n(x). $\endgroup$ – AkatsukiMaliki Dec 4 '18 at 10:12
  • $\begingroup$ I think c can be zero in this case, take $\omega = dx + dy$ and you could in fact get after doing the wedge : $dxdy + dydx = 0$. $\endgroup$ – AkatsukiMaliki Dec 4 '18 at 15:06

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