# Can the Gauss-Bonnet theorem be proven from Stokes's theorem?

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?

• The proof given by Chern in "A simple intrinsic proof of the Gauss-Bonnet formula for closed Riemannian manifolds" is based on Stokes' theorem.
– JHF
Feb 1 '19 at 13:24
• 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. Feb 1 '19 at 17:12