One half of De Rham's theorem says that if $\omega$ is a closed $p$-form on a smooth manifold that integrates to zero against all $p$-cycles, then $\omega$ must be exact.
My question is slightly different: whether for a fixed $\omega$ the integrates-to-zero-against-all-cycles condition, in itself, implies that $\omega$ is closed. Call the integrates-to-zero hypothesis the "vanishing condition."
Certainly this statement is true for $1$-forms on an open subset $U$ of $\mathbb{R}^n$. In that case, the vanishing of $\omega$ against all $1$-cycles implies that $\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$ when $\gamma_1$ and $\gamma_2$ are $1$-chains with the same endpoints; therefore we can define an explicit primitive of $\omega$ on each connected component of $U$ by $f(x)=\int_y^x\omega$ for some fixed point $y$. Thus for this special case the vanishing condition implies not only that $\omega$ is closed but also that $\omega$ is exact.
To what extent does the conclusion that $\omega$ is closed generalize to $p>1$ and/or beyond open subsets of Euclidean space? I know I'm not thinking very clearly about this. My guess is that the result won't generalize; if it did, we wouldn't need to assume the closure of $\omega$ in the hypothesis of the half of de Rham's theorem I stated above. But I can't think of a counterexample.
Yes. Pick a point $p$ and a coordinate chart around it; we may as well work in $\Bbb R^n$ around zero. Fix a $(p+1)$-plane $H$ and let $S(\varepsilon)$ be the $\varepsilon$-sphere in it, and $D(\varepsilon)$ the $\varepsilon$-ball. Stokes' theorem says that $\int_{S(\varepsilon)} \omega = \int_{D(\varepsilon)} d\omega$. If $d\omega$ evaluates nontrivially on some $(p+1)$-plane, choose that to be $H$. Then the latter integral must be nonzero for sufficiently small $\varepsilon$ - it's the integral of a nowhere zero function - and thus the first integral is nonzero. So if $\omega$ evaluates to zero against every $p$-cycle it is closed.