It seens like an easy question, but I am newbie in differential forms.
Let $\omega$ a $C^\infty$ $r$-form in an open set $U\subset\Bbb{R}^n.$ If $\omega$ vanishes over the tangent vectors of a surface contained in $U$, so do $d\omega$.
First things first, I want to write down everything.
Let $U\subset\Bbb{R}^n$ open, and $\omega:U\rightarrow\mathcal{A}_r(\Bbb{R}^n)$ a $C^\infty$ differential form. So, I can write
$$\omega(x)(v_1,\dots,v_r)=\sum_{I}a_I(x)dx_I(v_1,\dots,v_r).$$
Here, the sum extends over all ascendent $r$-uples $I=\{i_1,\dots,i_r\}$ of integers less than $n$, $a_I:U\rightarrow\Bbb{R}$ is a $C^\infty$ function, and $dx_I=dx_{i_1}\wedge\dots\wedge dx_{i_r}$.
Let $M\subset U$ a surface, $\dim M=m$, and let $\varphi:V_0\subset\Bbb{R}^m\rightarrow V\subset M$ a parametrization in $M$, with $\varphi(u)=x\in V.$
I'm trying to right $\omega$ in $M$, but I'm confused, because the functions $a_I$ take vectors in $\Bbb{R}^n$, not in $\Bbb{R}^m$, and I want to write in the parametrization $\varphi.$
Ok, besides this problem, basically I need to prove something like:
$$\sum_{I}a_I(u)du_I(v_1,\dots,v_r)=0\implies \sum_{I}da_I(u)\wedge du_I(v_1,\dots,v_r).$$
Best Answer
Let $S$ be that surface and $\iota : S\to U$ be the inclusion. The fact that $\omega$ vanishes over the tangent vectors of $S$ is the same as saying that $\iota^*\omega = 0$.
With that in mind, the proof is trivial: Since pullback $\iota^*$ commutes with exterior differentiation $d$,
$$ \iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$