Using Stokes' theorem to define the exterior derivative operator

In the excellent paper "Differential forms and integration" by Terence Tao, the author has mentioned that "... one can view Stokes' theorem as a definition of the derivative operation $\omega\rightarrow d\omega$, thus differentiation is the adjoint of the boundary operation".

My question is how exactly (rigorously) can one define the exterior derivative via Stokes' theorem? Where can I read about that in detail?

Solution 1:

This approach is championed in Arnold's book on classical mechanics. The exterior derivative of a $k$-form $\omega$ is supposed to be a $(k+1)$-form, defined as follows: given tangent vectors $v_1,\dots,v_{k+1}$ at some point $x$ of the manifold, pick a coordinate chart, and let $\Lambda_\varepsilon$ be a parallelogram spanned by $\varepsilon v_1,\dots,\varepsilon v_{k+1}$ in that chart. Then, define $$ d\omega(v_1,\dots,v_{k+1}):=\lim_{t\to0}\varepsilon^{-k-1}\int_{\partial \Lambda_\varepsilon}\omega. $$ The area of each face of $\partial \Lambda_\varepsilon$ is of order $\varepsilon^k$. If we expand the coefficients of $\omega$ in a Taylor series in our coordinate chart, then the only contribution to the limit will come from the linear terms: the constant terms give zero contribution to the integral because of cancellation between the opposite faces, and terms of order $O(\varepsilon^2)$ will give contribution of order $\varepsilon^{k+2}$. And the contribution from the linear terms is straighforward to compute, recovering the usual algebraic definition of the exterior derivative.

Now, assume that we replace the parallelograms $\Lambda_\varepsilon$ by their images $\tilde{\Lambda}_\varepsilon$ under a diffeomorphism whose differential at $x$ is the identity map. The constant terms still gives zero contribution, since a constant form is exact (this is a crucial point; if we tried to differentiate, e. g., vector fields or metric tensors on a smooth manifold without additional structure, we would fail here.) The contribution of linear terms will be the same for $\Lambda_\varepsilon$ and $\tilde{\Lambda}_\varepsilon$, up to a negligible correction. This shows, in effect, that the whole construction was independent of the coordinate chart.

Solution 2:

On page 136 of Models for Smooth Infinitesimal Analysis, Ieke Moerdijk & Gonzalo E. Reyes define the exterior derivative this way in the context of synthetic differential geometry (SDG), which is a truly beautiful and intuitive theory that makes all those old infinitesimal arguments rigorous. As they discuss in the text, the Dubuc topos provides a model for SDG which comes with a canonical fully faithful embedding of the classical category of smooth manifolds, and this gives us the classical exterior derivative.

However, if you prefer to work strictly in the classical setting, then you can adapt the SDG argument using limits (as V. I. Arnold does), or use the proof of de Rham's theorem. In the latter case, integration of a form over smooth singular chains gives you a smooth singular cochain, and you just need to check that the usual coboundary for singular cochains takes differential forms to forms. I think Warner presents Stokes' theorem for singular chains in his text.