Gauss–Bonnet theorem

Remarks

The proof starts from the local case of a small simple region in one coordinate patch. There the formula follows from Green's theorem, applied to the turning of a tangent vector along the boundary. One then takes a triangulation of $R$ fine enough that each triangle lies in one patch, and adds the local formulas. Interior edges are traversed twice in opposite directions, so their geodesic curvature terms cancel. The angle terms then add up to $2\pi(V - E + F) = 2\pi\chi(R)$.

For the round sphere of radius $r$, one has $K = 1/r^2$ and area $4\pi r^2$, so the total curvature is $4\pi = 2\pi \cdot 2$, and indeed $\chi(S^2) = 2$. For any torus in $\R^3$, the total curvature is $0$, since $\chi = 0$. Compactness is needed for the second formula: the plane has $K = 0$ but $\chi = 1$.

The theorem extends to compact two-dimensional Riemannian manifolds, and the Chern–Gauss–Bonnet theorem extends it to all even dimensions.

Going deeper

For a student

The formula is the standard bridge between the local geometry and the topology of surfaces. It shows that a compact surface with $K > 0$ everywhere must have $\chi > 0$, so it is a sphere when orientable. Applied to a geodesic triangle, where $k_g = 0$, it gives $\int_T K \, dA = \alpha + \beta + \gamma - \pi$ for the interior angles $\alpha, \beta, \gamma$. This explains why angle sums exceed $\pi$ on a sphere and fall short of $\pi$ on surfaces of negative curvature. It is also the first case of the theory of characteristic classes, where the Euler class is represented by a curvature form.