Editing Gauss–Bonnet theorem (section)

== A simple example ==

Suppose {{mvar|M}} is the northern hemisphere cut out from a sphere of radius {{mvar|R}}. Its Euler characteristic is 1. On the left hand side of the theorem, we have <math>K=1/R^2</math> and <math>k_g=0</math>, because the boundary is the equator and the equator is a geodesic of the sphere. Then <math>\int_MK dA=2\pi</math>.

On the other hand, suppose we flatten the hemisphere to make it into a disk. This transformation is a homeomorphism, so the Euler characteristic is still 1. However, on the left hand side of the theorem we now have <math>K=0</math> and <math>k_g=1/R</math>, because a circumference is not a geodesic of the plane. Then <math>\int_{\partial M}k_gds=2\pi</math>.

Finally, take a sphere octant, also homeomorphic to the previous cases. Then <math>\int_MK dA=\frac{1}{R^2}\frac{4\pi R^2}{8}=\frac{\pi}{2}</math>. Now <math>k_g=0</math> almost everywhere along the border, which is a geodesic triangle. But we have three right-angle corners, so <math>\int_{\partial M}k_gds=\frac{3\pi}{2}</math>.