Gauss-Bonnet theorem

Radially project a convex polyhedron from an interior point onto the unit sphere and triangulate each face without adding vertices. There are $T=2E-2F$ spherical triangles. The Gauss-Bonnet theorem gives each area as angle sum minus $\pi$. Summing angles around the projected vertices gives $2\pi V$, while the total area is $4\pi$. Thus $4\pi =2\pi V-\pi T$, proving the formula. Convexity guarantees that the projected triangles cover the sphere without overlaps of interiors.

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 Poincaré 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.

On the unit sphere, write $x=A(V)$ and $s=4\pi$. The expression equals $|xA(D\setminus V)-(s-x)A(D\cap V)|$, at most $x(s-x)$. A constant-version spherical isoperimetric inequality therefore gives the displayed bound, uniformly in nonempty open $D$. A smooth positive area density $\theta$ gives a second metric; integrating this bound over $D_t=\{\theta >t\}$ by the layer cake representation bounds the discrepancy between the two normalized area fractions by $\kappa (\max \theta -\min \theta )L^2/(s\tilde{A}(S))$.

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 $K(\theta )=\cos \theta /[r(R+r\cos \theta )]$.

A solid torus of major radius $A$ and minor radius $R<A$, centred on the horizontal plane, has volume $2{\pi }^{2}A{R}^2$. In circular torus coordinates, $z=r\cos \alpha$. Integrating $z^2$ with the Jacobian determinant $r(A+r\sin \alpha )$ gives ${\int }_V{z}^{2}dV=({\pi }^2/2)A{R}^4$; the part odd in $\sin \alpha$ vanishes. Thus the mean squared height is $R^2/4$. The identity is useful for moments of inertia and flux calculations using the divergence theorem.

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.