De Rham cohomology and antiderivatives

Question

A couple of recent questions about antiderivatives reminded me of the following, which I can't recall seeing tackled explicitly anywhere to my satisfaction and that I sketched to an ambitious calculus student long ago. I hope it is not completely trivial...

Consider $f: \mathbb{R} \rightarrow \mathbb{R}$ nice. An antiderivative $F$ of $f$ is defined by $f \ dx = dF$. Let $\Omega = [a,b]$. The Stokes theorem is basically that $d$ and $\partial$ are adjoint w/r/t the standard pairing (a/k/a the integral), viz. $\langle dF,\Omega \rangle = \langle F, \partial \Omega \rangle$.

This is obviously just an overly fancy formulation of the fundamental theorem of calculus (à la Spivak, for example). But its generality is a virtue in motivating the following question (which also betrays my long-decayed knowledge of the most trivial bits of algebraic topology): is there a general formulation of the notion of antiderivative that incorporates basic information about de Rham (co)homology? As a followup, is it necessary to bring bundles into the picture?

Answer 1

This probably doesn't answer your question, but maybe it is close.

The (Extended) Poincare lemma says that if U is a contractible smooth $n$-manifold and $\omega$ is a closed differentiable form meaning $d\omega=0$ then there is a form $\phi$ with $d\phi=\omega$. So, $\phi$ is an antiderivative of $\omega$.

More to the point, if $\omega$ is a closed form on the smooth manifold $M$ then $[\omega]\in H^*(M)$, and $\omega$ has an antiderivative if and only if $[\omega]=0$.

More specifically, suppose that $M$ is an oriented smooth $n$-manifold so that integration is defined, and $\omega$ is a smooth $n$-form, then $\omega$ has an antiderivative if and only if $\int_M\omega=0$.

If $M$ is in addition compact, and $\omega$ is a closed $i$-form then $\omega$ has an antiderivative if and only if for all closed $n-i$ forms $\eta$, $\int_M \omega\wedge \eta=0$.

If $\omega$ is not closed it can never have an antiderivative.

Answer 2

A form is exact (by definition) if it has an antiderivative, and a necessary condition for this is that it be closed. de Rham's cohomology, then, measures the failure of the fundamental theorem of calculus, when extended to forms and considered in the large.

Answer 3

If we assume that the higher dimensional generalization of the fundamental theorem of calculus, $$ \int_a^b f'(x)\,dx = f(b) - f(a)$$ is Stokes' theorem, $$ \int_{M} d\omega = \int_{\partial M} \omega, $$ then it appears to me that the higher dimensional generalization of an "antiderivative" is the "anti-exterior-derivative" of an exact $n$-form and the higher dimensional analogue of an "arbitrary constant" is a closed $(n-1)$-form.

So it appears that the $(n-1)$-th deRham cohomology group of $M$ should say something about the space of "arbitrary constants", but I'm not sure what.