Stokes' Theorem asserts that for a compactly-supported differential form $\omega$ of degree $n-1$ on a smooth oriented $n$-dimensional manifold $M$ we have the marvellous equation $$\int_M d\omega = \int_{\partial M} \omega.$$ Doesn't that look like a naturality condition in the sense of category theory? Somehow, integration is natural with respect to boundaries (or vice versa?). Can we make this precise?

What I have tried so far: If $\Omega_0^k(M)$ denotes the vector space of compactly-supported differential forms of degree $k$ on $M$, and $d : \partial M \hookrightarrow M$ denotes the inclusion of the boundary, Stokes' Theorem says that the diagram $$ \require{AMScd} \begin{CD} \Omega_0^{n-1}(M) @>{d}>> \Omega_0^n(M) \\ @Vd^*VV @VV{\int_{M}}V \\\ \Omega_0^{n-1}(\partial M) @>{\int_{\partial M}}>> \mathbb{R} \end{CD} $$

commutes. Is that correct? (I'm not sure about the $d^*$). This looks more like dinaturality, but I am not sure how to make a precise connection. Perhaps the cobordism category will be useful?

Any other categorical interpretation of Stokes' Theorem would also be appreciated. Notice that such interpretations are by no means useless, a priori, and could perhaps even lead to more conceptual proofs. See for instance

  • Roeder, David. "Category theory applied to Pontryagin duality." Pacific Journal of Mathematics 52.2 (1974): 519-527.

  • Hartig, Donald G. "The Riesz representation theorem revisited." American Mathematical Monthly (1983): 277-280.

Ali Caglayan
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Martin Brandenburg
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    Very nice question! – magma Apr 10 '15 at 01:28
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    [Here's](http://math.stackexchange.com/a/145201/120540) a fun blast from MSE past; I just happened upon it today, and couldn't help but be reminded of your question. – pjs36 Apr 11 '15 at 01:40
  • Can someone give a reference to a text where the pullback of differential forms with respect to smooth maps between manifolds **with boundary** is discussed? I couldn't find such a text. Notice that I need this to talk about the pullback $d^*$. And I am not still not 100% sure if integrating a form on $M$ over $\partial M$ means that we pull it back to $\partial M$ and then integrate over $\partial M$. It is very reasonable, though. – Martin Brandenburg Apr 11 '15 at 10:52
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    The pullback is defined pointwise, and the tangent space at a boundary point is defined in an essentially identical manner, so the standard definition of pullback will work in this generality as well. Your interpretation of $\int_{\partial M}$ is correct. –  Apr 11 '15 at 16:12
  • @MikeMiller: Thank you. Meanwhile, I have found that John Lee's book on Smooth Manifolds discusses all the details. – Martin Brandenburg Apr 11 '15 at 20:26

3 Answers3


Here's an attempt to make the "dinaturality" observation (more) precise. I will be leaving out many details that I haven't worked out, so I may go wrong somewhere; I hope the general outline makes sense though.

  • Let $\mathbb N$ be the poset of natural numbers in the usual ordering (or $\mathbb N$ could be the universal chain complex, i.e. category with the same objects given by natural numbers, $\mathrm{Hom}(n,m) = \mathbb{R}$ if $m \in \{n,n+1\}$, $0$ else, and all composites with nonidentity maps equal to zero. Then all the functors here are $\mathbb R$-linear).
  • Let $\mathcal V$ be the category of topological vector spaces or some suitable similar category.
  • Fix a manifold $X$ (or some other sort of smooth space).

Then we have functors

  • $C: \mathbb N^\mathrm{op} \to \mathcal V$ where $C_n$ is the vector space freely generated by smooth maps $Y \to X$ where $Y$ is a compact, $n$-dimensional, oriented manifold with boundary, and the induced map $\partial: C_{n+1} \to C_n$ is the boundary map.
  • $\Omega: \mathbb N \to \mathcal V$ is the de Rham complex; $\Omega_n = \Omega_n(X)$ is the space of $n$-forms on $X$ and the induced map $\mathrm d: \Omega_n \to \Omega_{n+1}$ is the exterior derivative.

Assuming that $\mathcal V$ has a suitable tensor product defined, we obtain a functor

  • $C \otimes \Omega: \mathbb N ^\mathrm{op} \times \mathbb N \to \mathcal V$.

while there is also the constant functor

  • $\mathbb R: \mathbb N ^\mathrm{op} \times \mathbb N \to \mathcal V$

Then Stokes' theorem says that we have an extranatural transformation

  • $\int : C \otimes \Omega \to \mathbb R$ which, given a map $Y \to X$ and a form $\omega$ on $X$, pulls the form back to $Y$ and integrates it (returning 0 if it's the wrong dimension).

Interestingly, this means that integration should descend to a map out of the coend $\int : \int^{n \in \mathbb N} C_n \otimes \Omega_n \to \mathbb R$ (that first integral means integration of differential forms while the second means a coend). I'm not sure what the value of this coend is or how much it depends on the details I've left ambiguous. I suppose it probably has something to do with the de Rham cohomology of $X$?

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  • This looks quite good. I will have to check some things. For now, I have the following questions: 1) Why do you want to put a topology on the vector spaces? 2) The definition of $C_n$ misses the word "$n$-dimensional", right? 3) $\partial : C_{n+1} \to C_n$ is defined by mapping $Y \to X$ to $\partial Y \to Y \to X$? 4) There are some set-theoretic difficulties for the definition of the basis of $C_n$, I think. How to solve this properly? – Martin Brandenburg Apr 10 '15 at 22:53
  • @Martin (1) Aren't infinite-dimensional discrete vector spaces "in poor taste":-)? Actually, though, going discrete would simplify things -- in particular the issue of figuring out what tensor product we want. (2) Good catch, I'll fix that. (3) Exactly, and the orientation is the induced one from $Y$. There are definitely some puzzling things about this formulation -- foremost is why we should need to do everything relative to a space $X$ when the statement of Stokes' theorem doesn't need it; I think there are still more insights to be had. – tcamps Apr 10 '15 at 23:01
  • Basically I agree, but on the other hand, fixing some base $X$ may pave the way for some kind of "relative Stokes' Theorem", which is perhaps even more interesting than the "global Stokes' Theorem". This is just speculation, based on the development of Grothendieck-Riemann-Roch. – Martin Brandenburg Apr 10 '15 at 23:06
  • 5) The linear variant for $\mathcal{N}$ looks quite natural, in particular when defining $\Omega$. But for $C$ it doesn't work yet, because $(Y \to X)$ is mapped by $\partial\partial$ to $(\emptyset \to X)$. So we should better identify $\emptyset \to X$ with the zero in the vector space. Perhaps further identifications should be made, also possibly for solving problem 4). – Martin Brandenburg Apr 10 '15 at 23:14
  • Tim, this is wonderful; I'd be happy to include your answer in my "coend" note! – fosco Dec 21 '16 at 11:55
  • Please do! I'll be interested to see it -- I think there are a number of ways one could nail down some of the details here. – tcamps Dec 21 '16 at 15:34
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    I'm confused. What if $X$ is closed and $Y$ is, well, something with a compatible dimension? Then you always get $\int_X d \omega = 0$, but it may not always be that $\int_Y f^\ast \omega = 0$ where $f:Y \to X$ (or $\int_{\partial Y} f^\ast \omega = 0$, I don't really understand how you define $\partial$). – Piotr Mar 02 '17 at 18:09

There is some discussion here:


and a reformulation. I must confess I didn't spend a lot of time on it, as I don't know what are $(\infty, 1)$-categories, etc.

To me, the greatest thing about Stokes' theorem is that it paves the way for de Rham's theorem. Indeed we can't even state the latter without the former. The de Rham theorem is very classical and important in geometry.

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    Wow, the section on "Abstract formulation in cohesive homotopy-type theory" is really ... I don't understand any word :-). But the structure indicates that this is not really about a categorical formulation (their version of Stokes' theorem is still some equation of integrals), but rather a generalization where the manifold is replaced by some more abstract space (objects of a cohesive $\infty$-topos?). So this is probably not what I'm looking for. – Martin Brandenburg Apr 09 '15 at 21:56
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    Martin writes "...I don't understand any word". Indeed : whenever I finish browsing nlab I have the feeling that I know less than when I started! I find it reassuring that even someone as categorically minded as Martin isn't enthusiastic about the linked post . – Georges Elencwajg Apr 10 '15 at 22:29
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    @GeorgesElencwajg: Perhaps it's like reading EGA. You have to get used to it. Years ago, only some pages at the nlab were useful for me. Today, almost every page is very useful for me, and often it is a pleasure to read. But I have no previous knowledge about cohesive topoi, higher differential geometry, physics and related stuff, which explains why I don't understand "any" word in the mentioned section. – Martin Brandenburg Apr 10 '15 at 22:59

In definition of dinatural transformation you have linked, there is the following:

Let $F, G: C^{op} \times C \to D$ be functors. A dinatural transformation from F to G (...)

Honestly I can't find such vast functors. But as for standard naturality first we need two parallel functors $(F,G:C\rightarrow D)$. In our case there is one firmly settled contravariant functor $\Omega: Smooth\rightarrow DGA,$ where $DGA$ is category of Differential graded algebras, but problem lies in pointing second one. We may try to construct second to be the composion $\Omega\circ\partial,$ where $\partial:Smooth\rightarrow Smooth$ takes $M$ to $\partial M.$ But is it make sense? Not at all! Cause we don't know how to contruct $\partial(f): \partial M\rightarrow \partial N$ from $f:M\rightarrow N.$

I see differently the diagram you have plotted. If we have two pairings on sets $<>_1:A_1\times B_1\rightarrow C, <>_2:A_2\times B_2\rightarrow C$ and three functions $F:A\rightarrow A_1,G:A\rightarrow A_2,$ and $H:B_1\rightarrow B_2$ such that for each $a\in A, b\in B_1$ $$<F(a),b>_1=<G(a),H(b)>_2$$ then for fixed $b\in B_1$ we can plot this as \begin{array}{cc}\phantom{\dfrac{a}{b}}A & \xrightarrow{\Large F} & A_1\phantom{\dfrac{a}{b}}\\ G \downarrow ~~~~~ && ~~~~~\downarrow <*,b>_1 \\ A_2 & \xrightarrow{ \Large<*,H(b)>_2 }& C\end{array}

Functoriality (not mentioning of naturality) does not seems to work here, but the fact is that in the case of smooth manifolds $H$ induces $G,$ which is just pullback of inclusion.

Fallen Apart
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