Gauss–Bonnet theorem

Statement

For any compact two-dimensional Riemannian manifold $M$ with piecewise-smooth boundary $\partial M$, Gaussian curvature $K$, geodesic curvature $k_g$ along $\partial M$, exterior turning angles $\theta_i$ at the vertices, and Euler characteristic $\chi(M)$, $\int_M K\,dA + \int_{\partial M} k_g\,ds + \sum_i \theta_i = 2\pi\chi(M)$. In particular, if $M$ is a closed surface without boundary ($\partial M = \varnothing$), then $\int_M K\,dA = 2\pi\chi(M)$.

Why is it true?

On a curved surface $M$, the interior angle sum of a geodesic triangle $\triangle$ differs from $\pi$ by the integral $\int_\triangle K\,dA$ of Gaussian curvature inside it: positive curvature ($K > 0$, as on a sphere) puffs triangles out so their angles sum to more than $\pi$, while negative curvature ($K < 0$) pinches them below $\pi$. When you triangulate an entire closed surface $M$ and add up the angle excesses over all triangles, the angles around each vertex sum to $2\pi$ and the local metric details cancel out completely, leaving $2\pi(V - E + F) = 2\pi\chi(M)$ — so no matter how you dent or stretch the surface, the total curvature $\int_M K\,dA$ is locked to its topological Euler characteristic $\chi(M)$.

Topics that use this theorem

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Step-by-step proofs

No step-by-step proof yet for this theorem.