If the integral of a $k$ form over any $k$ chain is $0$ then the differential form itself is $0$?

Question

Let $\omega$ be a $k$ form on an open set $A\subseteq \mathbb R^n$ such that for any $k$-chain $c$ in $A$ , $\int_c \omega =0$ , then is it true that $\omega =0$ ?

Answer 1

Yes, it's true. First, remember how you would show this when $k=n$. Given the $n$-form $\omega = f(x)dx_1\wedge\dots\wedge dx_n$, if $\omega$ is not everywhere zero, there is a point $a$ with $f(a)\ne 0$, say $f(a)>0$. By continuity, we choose a neighborhood $U$ of $a$ and choose an $n$-chain $\Delta\subset U$. Then $\int_\Delta \omega >0$, contradicting the hypothesis.

The same idea works for general $k$. Write $\omega = \sum\limits_{I \text{ increasing}} f_I dx_I$. Suppose $f_I(a)>0$ for some point $a$, and choose a $k$-chain $\Delta$ contained in $U\cap \Bbb R^I$.