Relationship between $\operatorname{supp} w$ and $\operatorname{supp} dw$

Question

I would like to show that $\operatorname{supp} dw\subset \operatorname{supp}w$, where differential $m$-form in a $\Sigma$ surface of dimension $m+1$. If

$$w(u)=\Sigma_i (-1)^ia_i(u) du_1\wedge ...\wedge \widehat{du_i}\wedge ...\wedge du_{m+1}$$

and

$$dw(u)=\Sigma_i \frac{\partial a_i}{\partial u_i}(u) du_1\wedge ...\wedge du_{m+1},$$

I need to show that $\frac{\partial a_i}{\partial u_i}(u)=0$, since that $u\in \operatorname{supp}a_i$, all right? I tried to do by definition

$$\frac{\partial a_i}{\partial u_i}(u)=\lim_{t\to 0}\frac{a_i(u+tu_i)-a_i(u)}{t}=\lim_{t\to 0}\frac{a_i(u+tu_i)}{t}$$

but I stopped here.

Can someone help-me?

Answer 1

This is actually easier. Let $C = \text{supp}(\omega)$ and $U = \Sigma \setminus C$. Then $\omega |_U = 0$ and so $d(\omega |_U) = 0$. But as $d\omega)|_U = d(\omega|_U)$, we have $d\omega = 0$ on $U$ and so $\text{supp}\omega \subset C$.