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Illustration of the Gauss-Bonnet |
Theorem Using Truncated Spheres
Let M be a compact two-dimensional Riemannian manifold with boundary ∂M. Let K(p) be the Gaussian curvature of M at point p, and let kg be the geodesic curvature of ∂M. Then
where dA is the element of area of the surface, ds is the line element along the boundary of M, and χ(M) is the Eulers-Poincaré characteristic of M.
The matter of the direction of traversal of the ∂M needs to be specified. That direction of traversal is counterclockwise when M is on the left.
The Gaussian curvature of a sphere of radius R is 1/R at each point of its surface. Therefore a sphere of unit radius has Gaussian curvature equal to unity at all points. Consider the unit radius sphere given by
The intersection is the circle given by
The geodesic curvature kg on this circle in the unit sphere is given by
The integral of geodesic curvature around the circle of intersection is then
In a sense the geodesic curvature represents the deviation of a curve from a geodesic of the surface. A geodesic on a sphere is a great circle, a circle whose center of curvature is the same as the center of the sphere. If a=0 then the circle of intersection is an equatorial circle, a geodesic, and, according to the above formula, the integral of the geodesic curvature is equal to zero.
The integral of Gaussian curvature on the section of the sphere between z=−1 and a is, for the unit sphere, just equal to the area of the sphere between −1 and a. This is given by the integration of the latitutde angle θ from −π/2 to its value for z=a. That value is such that sin(Θ)=a/1 or Θ=sin−1(a).
The infinitesimal of area is the circumference of the circle at θ times dθ.
The radius r of that circle is just cos(θ) so
Since sin( Θ)=a, the above formula reduces to
The integral of the geodesic curvature around the bounding circle was found to be
Therefore the contribution of integral of geodesic curvature is positive when a is negative and the truncated sphere is less than half of a sphere. It is negative when a is positive and the truncated sphere is more half of a sphere. It is zero when a is equal to zero which is the case of a hemisphere.
Thus the terms of the Gauss-Bonnet Theorem are
A truncated sphere is topologically equivalent to a disk and its Euler-Poincaré characteristic is 1.
Therefore the integral of the Gaussian curvature over the manifold plus the integral of the geodesic curvature over the boundary of the manifold does reduce to 2π times the Euler-Poincaré characteristic of the manifold.
For more on aspects of the Gauss-Bonnet Theorem.
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