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?

  • $\begingroup$ I have to admit I don't understand the question. I mean, I understand all the words, and I know that the definition of antiderivative is very much a local definition, whereas the existence of such depends on global considerations. Is your goal to somehow undermine the local nature of antiderivativeness? But it is inherently local, and I don't get why you think it could be made otherwise. I suppose it might just be my lack of imagination that is speaking. $\endgroup$ – Harald Hanche-Olsen Mar 30 '10 at 15:37
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    $\begingroup$ I think it would be helpful to note that antiderivatives are not unique (and you shouldn't use the word "the" above). This nonuniqueness arises from the existence of nonzero de Rham coboundaries. $\endgroup$ – S. Carnahan Mar 30 '10 at 15:40
  • $\begingroup$ @Scott: Well said; done. $\endgroup$ – Steve Huntsman Mar 30 '10 at 15:41
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    $\begingroup$ Oops, replace coboundary with cocycle. It is the kernel of d that creates the ambiguity. $\endgroup$ – S. Carnahan Mar 30 '10 at 15:42
  • $\begingroup$ This is more interesting in the case of complex variables. There are formulations of Cauchy's theorem incorporating homology and cohomology. $\endgroup$ – Regenbogen Mar 30 '10 at 15:46

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.

  • $\begingroup$ Can you re-specialize your answer to $U=M=\mathbb{R}$ and explain what you get? If have a suspicion that you are making some unwarranted assumptions (more smoothness than strictly necessary) that would be easier to see if you used simpler language. $\endgroup$ – Jacques Carette Mar 30 '10 at 20:47
  • $\begingroup$ You only need C^1. $\endgroup$ – Charlie Frohman Mar 30 '10 at 23:07

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.

  • $\begingroup$ If I read this correctly you are saying that an antiderivative is a form plus a cohomology group. $\endgroup$ – Steve Huntsman Mar 30 '10 at 15:53
  • $\begingroup$ Not really. I am only stating the only sensible "general formulation of the notion of antiderivative that incorporates basic information about de Rham (co)homology" I can think of. As Scott observed, antiderivatives are determined---when they exist---up to cocycle. This is not the same as being determined up to a cohomoloy class (consider, for example, the situation in $\mathbb R^n$ where there are very few cohomology classes...), and I do not know what "a form plus a cohomology group" means. $\endgroup$ – Mariano Suárez-Álvarez Mar 30 '10 at 15:58
  • $\begingroup$ I should have said "plus the kernel of $d$". $\endgroup$ – Steve Huntsman Mar 30 '10 at 16:15
  • $\begingroup$ ...recall, e.g., $\int F' dx = F + C$, where really $C$ means an arbitrary constant and could be rewritten as $\mathbb{R}$. $\endgroup$ – Steve Huntsman Mar 30 '10 at 16:18

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.

  • $\begingroup$ The consensus seems to be to just add (a generic element of) $\ker d$. This obviously gives something sensible, but it's unsatisfying to me. Seems like I was looking for something where there's nothing. $\endgroup$ – Steve Huntsman Mar 30 '10 at 23:52

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