The Gauss-Bonnet Theorem

The Gauss-Bonnet theorem is one of the most beautiful results in differential geometry, revealing a profound connection between the local geometry (curvature) of a surface and its global topology (Euler characteristic).

Statement

Theorem6.4Gauss-Bonnet theorem

Let $(M, g)$ be a compact oriented Riemannian 2-manifold without boundary. Then

\int_M K \, \mathrm{dvol}_g = 2\pi \chi(M),

where $K$ is the Gaussian curvature and $\chi(M) = 2 - 2g$ is the Euler characteristic ( $g$ = genus).

Theorem6.5Gauss-Bonnet with boundary

If $(M, g)$ is a compact oriented Riemannian 2-manifold with smooth boundary $\partial M$ , then

\int_M K \, \mathrm{dvol}_g + \int_{\partial M} \kappa_g \, ds = 2\pi \chi(M),

where $\kappa_g$ is the geodesic curvature of $\partial M$ and $ds$ is the arc length element.

Proof Sketch

Proof

Step 1: Triangulation. Choose a geodesic triangulation of $M$ (or any smooth triangulation). Let $F, E, V$ denote the numbers of faces, edges, and vertices. By definition $\chi(M) = V - E + F$ .

Step 2: Local Gauss-Bonnet. For each geodesic triangle $T$ with interior angles $\alpha_1, \alpha_2, \alpha_3$ :

\int_T K \, \mathrm{dvol}_g = (\alpha_1 + \alpha_2 + \alpha_3) - \pi.

This is the angle excess formula (for geodesic triangles, the geodesic curvature along edges vanishes, so the boundary term in the local Gauss-Bonnet formula comes only from the exterior angles at vertices).

Step 3: Summing. Sum over all triangles:

\int_M K \, \mathrm{dvol}_g = \sum_{\text{triangles}} (\alpha_1 + \alpha_2 + \alpha_3 - \pi).

The sum of all angles equals $2\pi V$ (at each interior vertex, angles sum to $2\pi$ ). Each triangle contributes $\pi$ to the subtraction, giving $\pi F$ . Each interior edge is shared by two triangles, and each boundary edge by one, giving $3F = 2E$ (for a manifold without boundary). Therefore:

\int_M K \, \mathrm{dvol}_g = 2\pi V - \pi F = 2\pi V - 2\pi E + 2\pi F = 2\pi(V - E + F) = 2\pi \chi(M). \quad \blacksquare

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Applications

ExampleConsequences of Gauss-Bonnet

Sphere: $\chi(S^2) = 2$ , so $\int_{S^2} K = 4\pi$ for any metric. For the round sphere of radius $r$ : $K = 1/r^2$ , area $= 4\pi r^2$ , and indeed $\frac{1}{r^2} \cdot 4\pi r^2 = 4\pi$ .
Torus: $\chi(T^2) = 0$ , so $\int_{T^2} K = 0$ . A torus embedded in $\mathbb{R}^3$ has regions of positive and negative curvature that must cancel.
Genus $g$ surface: $\int K = 2\pi(2 - 2g)$ . No metric on a genus $g \geq 2$ surface can have $K \geq 0$ everywhere.

RemarkTopological invariance

The Gauss-Bonnet theorem shows that $\int_M K \, \mathrm{dvol}_g$ is a topological invariant -- it depends only on the topology of $M$ , not on the choice of metric $g$ . This is remarkable: deforming the metric changes $K$ and $\mathrm{dvol}_g$ locally, but their integral remains constant.