2 closed differential $k$-forms are in the same (de Rham) cohomology class iff their difference is an exact differential $k$-form.

Let $f:A\subseteq\mathbb{R}\longrightarrow\mathbb{R}$ be integrable.

The set of all antiderivatives of $f$ is precisely the set of all functions $F:A\subseteq\mathbb{R}\longrightarrow\mathbb{R}$ whose derivative is $f$. Now, if $F$ and $G$ are 2 antiderivatives of $f$, their difference $F-G$ is a constant function.

I suspect these constant functions are exact differential forms (since one can get a constant function by differentiating a polynomial of degree $1$). This would mean that $F$ and $G$ differ by an exact form, and so they're in the same cohomology class, but only if they're closed forms, which they can't be since their derivative is $f$ which is in general not zero.

Still, the idea that $F$ and $G$ differ by a (presumably) exact differential form, and the involvement of the differential operator, gives the whole thing a cohomological flavor.

  1. Are the antiderivatives of $f$ in the same cohomology class?
  2. Do they form the set of all representatives of that class?
  3. Is it even profitable to use the language of cohomology classes to think of antiderivatives? (For example, in multivariable calculus, cohomological ideas are helping me understand the $\text{grad}$, $\text{curl}$ and $\text{div}$ operators a lot better, and put disparate integral/differential theorems (Green, Gauss, Stokes, Poincaré, etc.) into a coherent "big picture". After cohomology, one would even say calculus is beautiful.)

(Also, please correct any technical mistakes/imprecisions.)

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    $\begingroup$ No, no, no! If $F$ and $G$ are antiderivatives of $f$, their difference $F-G$ is a locally constant function, not necessarily a constant function! These locally constant functions are precisely the closed $0$-forms. $\endgroup$ Commented Aug 4, 2017 at 15:36
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    $\begingroup$ The sheaf of antiderivatives of $f$ is a torsor over the constant sheaf $\mathbb{R}$, and the set of equivalence classes of such torsors is isomorphic to the sheaf cohomology $H^1(A, \mathbb{R})$. I think maybe a comparison theorem would make this in turn isomorphic to $H^1_{DR}(A)$. $\endgroup$ Commented Aug 4, 2017 at 18:17
  • $\begingroup$ Thanks! Super helpful. In particular: 1) the locally constant functions are precisely the closed $0$-forms (I suppose this works for scalar fields in $\mathbb{R}^n$, too?), and 2) the antiderivatives of $f$ form a sheaf which is also an $\mathbb{R}$-torsor $\endgroup$ Commented Aug 4, 2017 at 20:43
  • $\begingroup$ @DanielSchepler What do you mean by "the set of equivalence classes of such torsors"? Do you mean we take the sheaf of antiderivatives of every (integrable) function $f$ (each of which is a torsor), and then define an equivalence relation on this set of sheaves? And then the set of equivalence classes forms the 1st sheaf cohomology group (with coefficients in $\mathbb{R}$? $\endgroup$ Commented Aug 4, 2017 at 20:50
  • $\begingroup$ I guess I should have said isomorphism classes of torsors. (So for example, any $\mathscr{G}$-torsor which has a global section is isomorphic to the trivial torsor $\mathscr{G}$.) $\endgroup$ Commented Aug 4, 2017 at 21:03

1 Answer 1


There are a lot of incorrect ideas here. First, antiderivatives of $f$ aren't even in a cohomology class at all. A cohomology class consists of a coset of the set of closed forms by the vector subspace of exact forms. In particular, for a form (in this case, a $0$-form) to be in a cohomology class, it must be a closed form! In order for a function $F$ to be a closed $0$-form, its derivative must be $0$. So the antiderivatives $F$ and $G$ themselves are not elements of any cohomology class (unless $f=0$), though the differences $F-G$ are.

As for whether the differences $F-G$ are all elements of the same cohomology class, the answer is no. Two closed $0$-forms are in the same cohomology class iff they differ by an exact $0$-form. By definition, an exact $0$-form is the differential of a $(-1)$-form. But there is no such thing as a $(-1)$-form (or rather, we formally say the vector space of $(-1)$-forms is the trivial vector space), so the only exact $0$-form is $0$. So two closed $0$-forms are in the same cohomology class only if they are equal. This renders your question (2) rather trivial, since any degree $0$ cohomology class has only one element.

Also, it is not correct that $F-G$ is constant: it is only locally constant. If $A$ is disconnected, $F-G$ can take different values on different connected components.

Here, then, is the correct story. The set of closed $0$-forms consists of the set of all smooth functions whose derivative vanishes. These are exactly the locally constant functions. Since the only exact $0$-form is $0$, the closed $0$-forms are the same thing as the $0$th de Rham cohomology.

On the other hand, given any smooth function $f$ which has an antiderivative $F$, you can find another antiderivative $G$ of $f$ by adding any locally constant function to $F$, and all other antiderivatives of $f$ can be found in this way. That is, the ambiguity in defining "the" antiderivative of $f$ is determined by the set of all locally constant functions, aka the $0$th de Rham cohomology.

As for your third question, is this formulation in terms of de Rham cohomology enlightening? Maybe, maybe not. As I see it, its value is that it gives directions to generalize the phenomenon of non-uniqueness of antiderivatives. For instance, if instead of an open subset $A\subseteq\mathbb{R}$ you had an arbitrary manifold, the correct generalization would be to talk about antiderivatives of $1$-forms on your manifold, not antiderivatives of functions (the differential of a function on a manifold is a $1$-form, not another function).

And of course, you can also consider de Rham cohomology in higher dimensions as another generalization. Note though that there is an important difference in higher dimensions: exact forms no longer have to be trivial, so closed forms (which, as the forms with differential $0$, literally capture the non-uniqueness of "antiderivatives") are no longer the same as cohomology classes. Instead, cohomology classes capture the non-uniqueness of antiderivatives modulo the "trivial" reasons that antiderivatives should not be unique (those coming from exact forms). Another way to think of it is that they capture the part of the non-uniqueness of antiderivatives that comes from the global topology of your manifold, rather than just from the formal local properties of your manifold. In the case of dimension $0$, the global topology in question is how many connected components your manifold has, which determines how many locally constant functions there are.

  • $\begingroup$ By the $0$-th de Rham cohomology do you mean the $0$-th de Rham cohomology group, ie. $H^0_{dR}(A, \mathbb{R})$, ie. $\displaystyle {\text{Ker }d_0 \over \text{Img }d_{-1}}$, ie. closed $0$-forms (locally constant functions) modulo exact $0$-forms (the zero function)? This would mean that the $0$-th de Rham cohomology group is precisely the set of all closed $0$-forms (ie. $\text{Ker }d_0$) for the sole reason that the image of $d_{-1}$ is trivial. Further, this should be the only de Rham cohomology group with this property. $\endgroup$ Commented Aug 4, 2017 at 21:32
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    $\begingroup$ Yes, that's correct. $\endgroup$ Commented Aug 4, 2017 at 21:33
  • $\begingroup$ Thank you. Also, your answer was at the precise level of abstraction that I was hoping for. It's helped me out a ton! $\endgroup$ Commented Aug 5, 2017 at 0:24

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