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I have looked at several elementary proofs (i.e., using only basic calculus, and not using differential form or manifold) of Stokes' theorem in books and Wikipedia, and all seem to use the fact that the mixed second partials of the parametrization of the surface are equal ($\psi_{uv}=\psi_{vu}$). But the statement of Stokes' theorem assumes that the surface is smooth, which only requires $C^1$ parametrization (and nonvanishing Jacobian). Is it possible that in Stokes' theorem, we need that the surface is $C^2$, not $C^1$ ?

ashpool
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  • I believe that $C^2$ is in fact required, even with differential forms and pullbacks. Where are you seeing a careful treatment that says $C^1$? – Ted Shifrin Dec 05 '23 at 01:45
  • @TedShifrin Wikipedia (https://en.wikipedia.org/wiki/Stokes%27_theorem#Proof) and Petrovic's Advanced Calculus, p. 450. – ashpool Dec 05 '23 at 02:00
  • I don’t see any specification in Wiki of what “smooth” means. They do specify that the vector field is $C^1$, which is correct. I have no access to Petrovic and know nothing about it. – Ted Shifrin Dec 05 '23 at 02:13
  • I think I saw the $C^1$-surface proof in Kaplan's Advanced Calculus, too, although the book is in library and has no online access. It had the same error of using the mixed derivative theorem. – ashpool Dec 05 '23 at 04:42
  • Peter Lax's Multivariable Calculus with Application also has the $C^1$-surface proof using mixed derivative test, on p.356. – ashpool Dec 05 '23 at 04:45
  • Even Stewart and Thomas calculus states Stokes' theorem with $C^1$ surface. – ashpool Dec 05 '23 at 04:49
  • It is far from elementary, but you can deduce the $C^1$ case by uniformly approximating by $C^2$ functions. Indeed, the theorem holds more generally if one ventures into the realm of currents. See DeRham’s book and books on geometric measure theory. Calculus books are out of their depth here. – Ted Shifrin Dec 05 '23 at 07:21
  • @TedShifrin I don't know what step of the proof would require the surface to be $C^2$? If we have a $C^1$ surface $M\subset \mathbb{R}^{n+m}$ which is the graph of a $C^1$ map $\phi:\mathbb{H}^n=\mathbb{R}^{n-1}\times[0,+\infty)\to\mathbb{R}^m$, then if $\omega$ is a $C^1$ $n$-form, then $\int_M\omega$ is given by $\int_{\mathbb{H}^n} \phi^\omega$, and $\phi^\omega$ is $C^1$ (by definition) and the fundamental theorem of calculus does the rest? – Martin Frenzel Aug 08 '24 at 09:33
  • @MartinFrenzel No, $\phi^*\omega$ will not be $C^1$ by definition. It will be only $C^0$, unfortunately. – Ted Shifrin Aug 09 '24 at 16:58
  • @TedShifrin Yes, this is true. I think you can fix this by a decomposition $\omega=\sum_i\omega_i$ and a chart $\phi_i$ for each $\omega_i$, such that $\phi_i^*\omega_i$ is $C^1$. Essentially if $M\subset\mathbb R^{n+m}$ is the graph of $f:\mathbb H^n\to\mathbb R^m$, and if $$\omega=a\cdot dx_1\wedge\cdots\wedge dx_n\in\Omega(\mathbb R^{n+m}),$$ then the pullback to $\mathbb H^n$ is $$a\circ (Id,f)\cdot dx_1\wedge\cdots\wedge dx_n,$$ which is $C^1$. I've written this up here. I hope it's right! – Martin Frenzel Aug 10 '24 at 13:01
  • @MartinFrenzel But what about $\omega = a\cdot dx_{n+1}$? (Add more to the wedge product as you wish.) – Ted Shifrin Aug 12 '24 at 12:10
  • @TedShifrin Yes this is addressed in the answer. When your form is $a\cdot dx_{i_1}\wedge\cdots\wedge dx_{i_n}$ then you write $M$ as the graph of a function from the $x_{i_1},\cdots ,x_{i_n}$ hyperplane and perform the analysis there. You can assume that this is always possible by rotating $M$ and making it smaller such that the projection tho all these hyperplanes is a diffeomorphism. Since LHS and RHS in Stokes' theorem are additive you can prove it for each of these components of $\omega$ separately. – Martin Frenzel Aug 13 '24 at 12:24

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