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In a comment to this question, John Ma claims that the Gauss-Bonnet theorem can be proven from Stokes's theorem, but does not explain how.

For two dimensions, Stokes's theorem says that for any smooth 2-manifold (i.e. surface) $S$ and one-form $\omega$ defined on $S$,

$$\oint_{\partial S} \omega = \iint_S d\omega.$$

I could vaguely imagine coming up with some kind of one-form $\omega$ that depends on the metric, such that (a) along the boundary curve $\omega$ maps the boundary tangent vector to the geodesic curvature and (b) in the surface interior $\ast d\omega$ equals the Gaussian curvature. (In more concrete vector-field language, this corresponds to a vector field $\vec{\omega}$ defined over the surface such that (a) on the boundary curve $\vec{\omega} \cdot d\vec{l}$ equals the curve's geodesic curvature and (b) in the surface interior $(\vec{\nabla} \times \vec{\omega}) \cdot d\vec{S}$ equals the Gaussian curvature.) This would reproduce part of the Gauss-Bonnet formula, but how could you possibly get out the Euler characteristic term?

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    $\begingroup$ The proof given by Chern in "A simple intrinsic proof of the Gauss-Bonnet formula for closed Riemannian manifolds" is based on Stokes' theorem. $\endgroup$ – JHF Feb 1 at 13:24
  • $\begingroup$ For the 2-dimensional G-B theorem, you can see (most of) the argument on p. 105 of my differential geometry text. The local-to-global argument to get to $\chi(M)$ is the standard triangulation argument. If you recognize $\chi$ as the sum of the indices of a vector field, you can turn the proof into a direct argument, taking out little balls $B$ around each zero and seeing that $\int_{\partial B} \bar\omega_{12}$ gives the index of the vector field. $\endgroup$ – Ted Shifrin Feb 1 at 17:12
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See Differential Forms and Applications (by Manfredo P. Do Carmo ) chapter 6 section 1. The proof of Gauss-Bonnet's Theorem presented by Do Carmo in his text is essentially the same as given by S.S. Chern and use the Stokes' theorem. The original proof of S.S. Chern is found here.

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  • $\begingroup$ Could you gist it for me? I don't have access to the book or the paper. $\endgroup$ – tparker Feb 2 at 2:57
  • $\begingroup$ @tparker I only have access to the Portuguese version of the Do Carmo book I bought recently. $\endgroup$ – MathOverview Feb 2 at 11:42
  • $\begingroup$ For paper see the link. scholar.google.com.br/… $\endgroup$ – MathOverview Feb 2 at 11:45
  • $\begingroup$ I'm afraid this proof is too advanced for me to understand :( $\endgroup$ – tparker Feb 2 at 15:11

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