What can we say about differential forms that have the same differential?

Question

If two smooth functions $\mathbb{R} \to \mathbb{R}$ have the same derivative, then they differ by a constant, being primitives of the same function. How does this statement generalize to manifolds and smooth differential forms? If I know that $d \omega = d \eta$, can I say anything about the relationship between $\omega $ and $\eta$? I know we cannot expect such a strong and general statement as in the one-dimensional real case, but feel free to take all the necessary restrictions in order to get a "nice" statement. I am particularly interested in $1-$forms. I am sure this will somehow be related to Stokes' and/or de Rham's theorems, but I did not manage to state anything clearly so far.

Answer 1

Notice that if $d\omega=d\eta$ for two differential $p$-forms $\omega$ and $\eta$, then $d(\omega-\eta)=0$ implies that $\omega-\eta$ is in the kernel of the linear map $d:\Omega^p\to\Omega^{p+1}$. In particular, this means that $\omega-\eta$ is a closed differential $p$-form.

To summarize what this means: $d\omega=d\eta$ if and only if there is a differential form $\alpha$ (of the same degree as $\omega$ and $\eta$) with $d\alpha=0$ and $\omega=\eta+\alpha$.

In particular, for $0$-forms (i.e. smooth functions), we get the usual result that $df=dg$ if $f$ and $g$ differ by a constant. For $1$-forms, such an $\alpha$ has the property that $d\alpha(X,Y)=X\alpha(Y)-Y\alpha(X)-\alpha([X,Y])=0$ for every vector field $X$ and $Y$.