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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:

$$\sum_{I}a_I(u)du_I(v_1,\dots,v_r)=0\implies \sum_{I}da_I(u)\wedge du_I(v_1,\dots,v_r).$$

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  • $\begingroup$ More generally, if $\omega$ is constant on a surface, then $d\omega$ vanishes along that surface $\endgroup$
    – becko
    Jul 29 '20 at 8:23
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    $\begingroup$ @becko What does constant on a surface means? $\endgroup$ Jul 29 '20 at 8:24
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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 $d$,

$$ \iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$

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Another proof: Recall the well-known formula for exterior derivative $$ d\omega(X_1,X_2,X_3,\dots,X_{r+1})=\sum_i(-1)^{i+1} X_i\omega(X_1,\dots,\widehat{X_i},\dots,X_{r+1})+\sum_{i<j}(-1)^{i+j}\omega([X_i,X_j],X_1,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_{r+1}). $$ Now if $\omega$ vanishes on $M$, then for all $X_1,\dots,X_{r+1}\in\mathcal{X}(M)$, we have $[X_i,X_j]\in\mathcal{X}(M)$ and hence $\omega([X_i,X_j],X_1,\dots,\widehat{X_i},\dots,\widehat{X_j},\dots,X_{r+1})=0$ by supposition, $X_i\omega(X_1,\dots,\widehat{X_i},\dots,X_{r+1})$ is differentiating constant $0$ so is $0$, and so $d\omega$ vanishes on $M$.

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