# Stokes' theorem proof without FTC

To present time I have not found a proof of Stokes' Theorem for manifolds that does not involve the Fundamental Theorem of Calculus in some way. Is it possible to prove the general theorem without employing its famous special case? Or does it necessarily have to rely on the FTC?

• Stoke's Theorem reduces to the fundamental theorem when you do it on the line. It's basically the biggest generalization (at least that I know of) of the fundamental theorem of calculus. I can't imagine a proof that doesn't somehow use this fact, unless maybe you can imitate the proof of the fundamental theorem in coordinates and stitch together. – Alfred Yerger Dec 16 '17 at 1:54
• I cannot prove this, but I firmly believe that you have to prove it by invoking the FTC and Fubini's Theorem. Or else you rig things by defining $d$ abstractly so that Stokes's Theorem holds but you have no idea how to define it in coordinates. – Ted Shifrin Dec 16 '17 at 2:39
• @AlfredYerger it can be (slightly) generalized to covariant exterior derivatives. – NAC Dec 16 '17 at 2:39
• @TedShifrin Are you referring to the coordinate-free definition such that, for $\omega \in \Omega^p(M)$, we set $$d\omega(X_0,\dots,X_p) = \sum_{i=0}^p (-1)^i X_i \omega(X_0,\dots, \hat X_i, \dots, X_p) + \sum_{i < j} (-1)^{i+j} \omega([X_i,X_j], X_0, \dots, \hat X_i, \dots, \hat X_j, \dots, X_p)$$ or do you mean defining $d$ through its properties and then showing that it satisfies the above formula? At any rate, couldn't the usual coordinate representation of $d\omega$ be derived from such an abstract definition? – giobrach Dec 16 '17 at 10:15
• No, I was referring to the definition given, for example, by Hubbard and Hubbard in their text. At least in the editions I am acquainted with, they defined the exterior derivative by what is needed to make an infinitesimal Stokes's Theorem. – Ted Shifrin Dec 16 '17 at 18:01