# If $\omega$ vanishes in a surface, so do $d\omega$

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:

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

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

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 $$d$$,
$$\iota^* (d\omega) = d (\iota^* \omega) = d(0) = 0.$$
Another proof: Recall the well-known formula for exterior derivative $$d\omega(X_1,X_2,X_3,\dots,X_{r+1})=\sum_i(-1)^{i+1} X_i\omega(X_1,\dots,\widehat{X_i},\dots,X_{r+1})+\sum_{i Now if $$\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$$.