Gauss-Bonnet theorem

If a flat surface disc is replaced by a smooth disc agreeing with the original surface outside the same boundary, their boundary geodesic curvatures agree. Gauss-Bonnet theorem therefore forces the replacement’s total Gaussian curvature to be zero. If its curvature is everywhere nonnegative, continuity forces it to vanish everywhere. Thus a positive-curvature bulge cannot be glued this way without some negative curvature or a loss of smoothness.

A geodesic triangle on the unit sphere with interior angles $\alpha ,\beta ,\gamma$ has areaFor a sphere of radius $R$, multiply by $R^2$. Triangulation gives a convex spherical $n$-gon area ${\sum }_i{\alpha }_i-(n-2)\pi$.

In the curvature-minus-one Poincare half-plane model, a geodesic polygon with $n$ sides and interior angles ${\alpha }_1,\dots ,{\alpha }_n$ has areaThis is the polygonal Gauss-Bonnet theorem.

A simple closed curve of length $L$ on the unit sphere enclosing the smaller area $A$ satisfiesEquality holds for a circle. A variational proof uses Gauss--Bonnet to turn area maximization at fixed length into geodesic-curvature minimization, proves that an extremizer has constant geodesic curvature and is planar, and evaluates its spherical cap.

For a standard ring torus, Gaussian curvature is positive outside, negative inside, and zero along the top and bottom circles.

A ring torus with major radius $R$ and minor radius $r<R$ has parametrizationand Gaussian curvature $K(\theta )=\cos \theta /[r(R+r\cos \theta )]$.

The total Gaussian curvature of a closed surface is two pi times its Euler characteristic.

The topology and compactness of a one-point closure do not determine the total curvature when the added point need not be a regular point of the surface. Rotate a smooth simple arc in the half-plane about its boundary axis. Let its arc-length parameter be $s\in [0,L)$, let its radius be $r(s)>0$ for $0<s<L$, and arrangeas $s\uparrow L$, where $0\le a<1$. The initial end closes smoothly, while the terminal point is omitted. The resulting surface is diffeomorphic to a sphere minus one point and has compact one-point closure. Since $K=-r^{\prime \prime }/r$ and $dA=rdsd\theta$,which need not equal $4\pi$.

Two disjoint simple closed curves on a sphere bound an annulus $A$. If both are geodesics, its boundary geodesic-curvature terms vanish, whereas $\chi (A)=0$. Gauss-Bonnet would givewhich is impossible when $K>0$ everywhere.