﻿ The Partition of a Closed Simple Manifold Satisfying the Conditions of the Gauss-Bonnet Theorem into Three or More Parts
San José State University

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The Partition of a Closed Simple
Manifold Satisfying the Conditions
of the Gauss-Bonnet Theorem into
Three or More Parts

The Gauss-Bonnet Theorem says that for a manifold topologically equivalent to a sphere the surface integral of the Gaussian curvature is equal to 4π. In symbols

∫∫MKdA = 4π

Since the properties being considered are topological it is reasonable to evaluate them for a simple, tractible case. Consider a unit sphere divided by latitude circles at Θ1 and Θ2 with Θ12.

The more general version of the Gauss-Bonnet Theorem says that for a manifold M with a boundary ∂M the surface integral of the Gaussian curvature plus the line integral of the geodesic curvature for the boundary equals 2π times the Euler characteristic of the manifold. In symbols

∫∫MKdA + ∫∂Mkgds = 2πχ(M)

One portion of the partitioned sphere is that part between −π/2 and Θ1. Another is between Θ1 and Θ2. The third one is between Θ2 and π/2.

Since the Gaussian curvature of a unit sphere is everywhere unity the integral of the Gaussian curvature of a portion of a unit sphere is equal to its area. For any two latitude circles the area between them is 2π times the difference in the sines of the latitude angles. This means that for the three portions of the sphere the areas are:

2π(1+sin(Θ1)) 2π(sin(Θ2)−sin(Θ1)) 2π(1−sin(Θ2))

It has been determined that the line integral of the geodesic curvature of a latitudinal circle is −2πsin(Θ). Note that the term is positive if Θ is negative. Thus for the first portion

2π(1+sin(Θ1)) −2πsin(Θ1) = 2π

The complement of the portion of the sphere below Θ is equivalent to the portion below (π−Θ) and thus the value of the line integral of geodesic curvature for it is 2πsin(Θ). More simply the line integral of geodesic curvature for one side of a latitude circle is the opposite in sign for its value on the other side of the latitude circle. This is as it should be because when two parts of the same manifold are brought together the line integrals for the geodesic curvature of the two parts should cancel out. Thus for the middle portion

2π(sin(Θ2)−sin(Θ1)) −2πsin(Θ2) + 2πsin(Θ1which reduces to zero

For the third portion

2π(1−sin(Θ2)) + 2πsin(Θ2) = 2π

The first and third portions are topologically equivalent to a disk and therefore their Euler characteristics are both 1. The middle portion is topologically equivalently is a torus and therefore its Euler characteristic is zero.