Is Stokes' Theorem natural in the sense of category theory?

Question

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.

Answer 1

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 $X$?

Answer 2

There is some discussion here:

http://ncatlab.org/nlab/show/Stokes+theorem

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.

Answer 3

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.