# How rough can differential form, manifold and chain be for Stokes theorem to hold?

When I first learned Stokes theorem, everything is assumed to be smooth to prevent any strange things happen. But to apply to more cases, I may need to use a version of Stokes theorem that holds for rougher forms, chains and manifolds. For instance, when I learned Cauchy integral theorem, the paths and the analytic functions are only assumed to be $C^1$.

Does Stokes theorem hold for merely $C^1$ forms, chains and manifolds? I think so because we only exterior differentiate once in the equation of Stokes theorem, and the pullback by the chain (parametrisation of surfaces) only uses the first derivative, while the continuity of first derivative is added to ensure the exterior derivative ($d\omega$) and the pullback form ($(\partial c)^*\omega$) is integrable.

Edit: I see that Stokes theorem holds for manifolds with corners. I suppose these are equivalent to piecewise smooth surfaces, lines, etc. However, it is not all of what I am asking about. I am asking whether Stokes theorem holds when the differential form, manifold and chain are simply assumed to be $C^1$.

• In page 137 of Spivak's Calculus on Manifolds (essentially the last page), he briefly comments that we actually can state Stokes' theorem for the so called manifolds-with-corners, modeled in open sets of "quadrants" (e.g., a $k$-manifold-with-corners is locally a piece of $\Bbb R^k$ bounded by pieces of $(k-1)$-planes). I have never seen the details written anywhere, though. – Ivo Terek Nov 16 '17 at 3:15

In Lang's Real Analysis you'll find Stokes's Theorem stated for a $C^2$ manifold and $C^1$ form $\omega$. He goes on to discuss what to do with singularities.

Although Lang asserts existence of a $C^2$ partition of unity using the $C^2$ hypothesis, a $C^1$ partition of unity suffices. However, as @MatheinBoulomenos pointed out to me in chat, you need a $C^2$ chart in order to be sure that the pullback of the $C^1$ form (bumped off by the $C^1$ partition of unity) is still a $C^1$ form. (Lang is very sloppy about paying attention to details in this regard.) Then it's the usual computation.

• But do we really need the form to be $C^1$ when we integrate it? What I have in mind is that when we integrate a real-valued function over a compact rectangle, only the continuity of the function suffices for integrability. So I think maybe we can weaken the condition of the manifold to be $C^1$ so that we obtain that the pullback form is merely continuous, while Stokes formula still holds. – edm Nov 17 '17 at 12:20
• Don't forget that you need to differentiate the form before you integrate over the rectangle! – Ted Shifrin Nov 17 '17 at 15:04
• To be more precise about what I am saying, in the formula $\int_Md\omega=\int_{\partial M}\omega$, I would hope that both $M$ and $\omega$ only need to be $C^1$, so that both $d\omega$ and $\varphi^*\omega$ will be continuous forms, where $\varphi$ is a $C^1$ chart. And of course I would hope that we can integrate any continuous forms. – edm Nov 17 '17 at 15:14
• Nevermind, I see what happened. By charting $M$ with a function $\varphi$, we may have to exhaust the differentiability of $d\omega$. – edm Nov 17 '17 at 16:03

Stokes' theorem holds for manifolds with corners. The following is Theorem $16.25$ on page $419$ in John M. Lee's book $\textit{Introduction to Smooth Manifolds}$.

$\textbf{Theorem}$ (Theorem $16.25$, [Lee]). Let $M$ be an oriented smooth $n$-manifold with corners, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$\int_M d\omega = \int_{\partial M}\omega.$$

The author proves the above theorem in the book. Furthermore, Lee also discusses and proves Stokes' theorem for surface integrals (see Theorem $16.34$ on page $427$).

$\textbf{Theorem}$ (Theorem $16.34$, [Lee]). Let $M$ be an oriented Riemannian $3$-manifold with or without boundary, and let $S$ be a compact oriented $2$-dimensional smooth submanifold with boundary in $M$. For any smooth vector field $X$ on $M$, $$\int_S \langle \text{curl } X,N \rangle_g \:dA = \int_{\partial S}\langle X,T\rangle_g \: ds,$$ where $N$ is the smooth unit normal vector field along $S$ that determines its orientation, $ds$ is the Riemannian volume form for $\partial S$ (with respect to the metric and orientation induced from $S$), and $T$ is the unique positively oriented unit tangent vector field on $\partial S$.