relation of supp of form $\omega$ and $d\omega$ Given $n$ dimension smooth manifold,and smooth $k$ (where $k\le n-1$) form $\omega$.
Assume we know $\text{supp}\ \omega \subset U$ where $U$ is open subset of $M$
Can we say anything about support for $\omega$ and $d\omega$,for example the proporsitions below is true or false:


*

*$\text{supp} (d\omega) = \text{supp}(\omega)$

*$\text{supp}(d\omega) \subset \text{supp}(\omega)$

(I try to show for example when $\omega$ is 0-form,and $\text{supp}(d\omega)\subset \text{supp}\omega\ $ i.e. denote $Z(\omega) = \{p:\omega_p \ne 0\}$ it's sufficient to show $Z(d\omega) \subset Z(\omega)$ but we can't say if $\omega_p =0$ then $(d\omega)_p = 0$ ?since $(d\omega)_p(X_p)$ is determined by the neighborhood value of $\omega$ around $p$ not only a single point?
 A: Here is an argument similar to the one mentioned by Kajelad in the comments above.
We show that the exterior derivative $d:\Omega^k(M)\to \Omega^{k+1}(M)$ decreases support in that $\operatorname{supp}(d \omega)\subseteq \operatorname{supp}(\omega)$.

Proof: Suppose $p\not\in \operatorname{supp}(\omega)$. Then there exists an open set $U\ni p$ such that $\omega|_U\equiv0$. So, for any $q\in U$, $(d\omega)_q=0$ and therefore $p\not\in \operatorname{supp}(d\omega)$. I.e. $\operatorname{supp}(\omega)^c\subseteq \operatorname{supp}(d\omega)^c$ and hence $\operatorname{supp}(d\omega)\subseteq \operatorname{supp}(\omega)$.

In the above argument, we used that if $\omega|_U\equiv 0$, then viewing it in local coordinates as
$$\sum_{i_1<\cdots<i_k}f_{i_1,\ldots, i_k}dx^{i_1}\wedge\cdots dx^{i_k}$$
we must have $f_{i_1,\ldots, i_k}(q)=0$ for all $q\in U$. Hence, because $f_{i_1,\ldots, i_k}$ are constant in $U$ their partial derivatives vanish. I.e. $(d\omega)|_U\equiv 0$. By the way, this also allows you to construct a counterexample where the containment is strict. For instance, take $f\in \Omega^0(\Bbb{R})$ which is constantly equal to $1$. $df\equiv 0$.
You can also modify this to get an example for compactly supported forms, but I'll leave that to you.
