How to see this is a normal vector field?

Question

Suppose $M\subseteq\mathbb{R}^{n+1}$ is an $n$-dimensional submanifold, and let $\omega$ be an $n$-form on $M$. Let $\{e_1,...,e_n\}$ be the standard basis of $\mathbb{R}^{n+1}$ and $p$ be the orthogonal projection from the tangent space of $\mathbb{R}^{n+1}$ to the tangent space of $M$ at each point. How does one see that $$e_1\omega(p(e_2),...,p(e_{n+1}))+e_2\omega(p(e_3),...,p(e_{n+1}),p(e_1))+...+e_{n+1}\omega(p(e_1),...,p(e_n))$$ is a normal vector field on $M$?

Answer 1

The assertion is trivial if $\omega = 0$. So we assume $\omega\neq 0$. By normalizing, we assume that $$\tag{1} \omega (v_1, \cdots,v_n) = 1$$ for some orthonormal bassis $\{v_1, \cdots v_n\}$ of the tangent plane. Let $v_{n+1}$ be the unit normal vector. Then

$$\tag{2} e_j = \sum_{\alpha=1}^{n+1} P_{j\alpha} v_{\alpha}.$$

Note that we have

$$ p(e_j) = \sum_{\alpha=1}^n P_{j\alpha} v_{\alpha}. $$

This implies $$\begin{split} \omega (p(e_{i_1}), \cdots, p(e_{i_n})) &= \sum_{\alpha_1, \cdots, \alpha_n=1}^nP_{i_1\alpha_1} \cdots P_{i_n\alpha_n}\ \omega(v_{\alpha_1}, \cdots, v_{\alpha_n})\\ &= \sum_{\alpha \in S_n}\text{sgn}(\alpha) P_{i_1\alpha_1} \cdots P_{i_n\alpha_n}, \end{split}$$

where $\text{sgn}(\alpha)$ is the sign of the permutation $(1, \cdots, n)\to (\alpha_1, \cdots, \alpha_n)$. In particular,

$$\hat \omega_i : =\omega (p(e_{i+1}), \cdots p(e_{n+1}), p(e_1), \cdots p(e_{i-1})) = (-1)^{n+1} (-1)^i \det M_i,$$

where $M_i$ is the minor of $P$ with respect to $(i, n+1)$. So

$$\begin{split} \omega &:= e_1 \hat\omega_1 + \cdots e_{n+1} \hat\omega_{n+1} \\ &= \sum_{i=1}^{n+1} e_i \hat \omega_i \\ &= \sum_{i=1} ^{n+1} \sum_{\alpha=1}^{n+1} P_{i\alpha} v_\alpha \hat\omega_i \\ &=(-1)^{n+1} \sum_{\alpha=1}^{n+1} \left(\sum_{i=1}^{n+1}(-1)^i P_{i\alpha} \det M_i\right) v_\alpha \\ &= \pm(-1)^{n+1} v_{n+1}. \end{split}$$

The last equality holds since for $\alpha \neq n+1$, we have

$$\tag{3} \sum_{i=1}^{n+1}(-1)^i P_{i\alpha} \det M_i = \det (A_i) = 0,$$

where $A_i$ is the matrix by replacing the $(n+1)$-th column of $P$ by the $\alpha$-th column (thus it has two identical columns). While when $\alpha =n+1$, the term on the left hand side of $(3)$ equals $\det P$, which is $\pm 1$ since $P$ is orthogonal.

Thus $\omega $ is up to a sign the unit normal vector $v_{n+1}$.