Global Gauss–Bonnet by triangulation

Try

A round sphere of any radius has total Gaussian curvature 4π: its area scales by R² while K scales by 1/R². A torus has total curvature zero even though an ordinary embedded torus has positive and negative curvature regions.

These are topological constraints on the integral, not pointwise descriptions. Global Gauss–Bonnet states that a compact oriented surface without boundary satisfies ∫K dA=2πχ, where χ is its Euler characteristic.

Predict

Predict how local boundary terms can disappear when a surface is divided into triangles. Each interior edge belongs to two pieces with opposite induced orientations, so their geodesic-curvature integrals cancel.

Predict whether the corner angles cancel in the same way. They must be counted carefully: angles around a smooth interior vertex total 2π, while each triangular exterior turn is π minus its interior angle.

Experiment

Check standard closed-surface totals.

Surface type	Euler characteristic	Total Gaussian curvature
Sphere	2	4π
Torus	0	0
Oriented genus-two surface	−2	−4π

The genus formula χ=2−2g is a topological input. For a particular triangulation, compute χ=V−E+F directly. Different triangulations give the same invariant, but that invariance is not established by counting one mesh.

Understand

Take a finite smooth triangulation of a closed oriented surface, fine enough that each face lies in a chart where local Gauss–Bonnet applies. Existence of such triangulations is a stated topology/manifold theorem. The edges need not be geodesic; their curvature integrals cancel pairwise because the induced directions are opposite.

Summing the local right sides gives 2πF. Across all triangular corners, exterior turns sum to 3πF minus the sum of interior angles. The latter is 2πV because the triangles surround each vertex once. Thus ∫M KdA+3πF−2πV=2πF, or ∫M KdA=2πV−πF. Every edge belongs to two triangular faces, so 3F=2E. Substitution gives 2π(V−E+F)=2πχ, proving the closed formula.

For a compact surface with piecewise smooth boundary, the same bookkeeping cancels only interior edge terms. The surviving boundary integrals and original corner turns produce ∫M KdA+∫∂M k_g ds+Σ exterior turns=2πχ(M). The boundary orientation places the surface to the left relative to its chosen normal.

Apply

A torus cannot carry a metric with K strictly positive everywhere. A continuous positive function on a compact positive-area surface has positive integral, contradicting χ=0. This does not force K to vanish pointwise; sign-changing curvature can integrate to zero.

Repair “Zero total curvature means the surface is locally flat.” An embedded torus is the counterexample. The theorem constrains one global integral, not every local value.

For optional depth, orientation is convenient for the boundary proof, while an unoriented closed-surface version can be obtained through the orientation double cover. That extension requires a covering-space argument not developed here. The present proof supplies the complete cancellation mechanism once triangulation and the local theorem are available.

Continue

Use total curvature as a consistency check on calculations and numerical meshes, while keeping discretization error separate from the theorem. Continue to global surfaces and index for further consequences and a careful accounting of what local geometry does not determine globally.