Gauss-Bonnet Theorem

Total curvature equals 2π times the Euler characteristic.

The idea

Theorem (Gauss-Bonnet theorem).

Let $S$ be a compact orientable surface without boundary, with Gaussian curvature $K$. Then $\iint_{S} K \, dA = 2\pi\,\chi(S),$ where $\chi(S)$ is the Euler characteristic of $S$.

For a surface of genus $g$, the number of holes, the Euler characteristic is $\chi(S) = 2 - 2g$, so $\chi = 2$ for a sphere ($g = 0$) and $\chi = 0$ for a torus ($g = 1$).

The left side is geometric and local: $K$ varies from point to point and responds to every dent in the surface. The right side is topological: a single integer, unchanged by any smooth deformation, that records only the number of holes. The theorem forces the two to agree, so the total curvature is invariant under every deformation that preserves the topology.

Press a dent into a sphere and $K$ changes throughout the dented region, turning negative where the surface becomes saddle-shaped; the integral remains $4\pi$, because the new negative curvature is balanced exactly by additional positive curvature around the rim. Only a change of genus can move the total. And since $K$ is intrinsic — computable from measurements made within the surface — an inhabitant of a closed surface could determine its number of holes without ever leaving it.

Ways to work on it

• Walkthrough. Total curvature of the sphere from both the integral and the Euler characteristic.
• Practice. Total curvature of a closed surface from its genus.
• Hardest. Recover a surface's genus from its measured total curvature.

Not sure where to start? Take the ten-question placement test.