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The Partition of Manifolds Satisfying the Gauss-Bonnet Theorem |
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Here is a statement of the Gauss-Bonnet Theorem.
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 k_{g} 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.
There is the matter of the sign of the integral of geodesic curvature that needs to be considered and this is best elucidated with the relatively easy topic of spheres.
The Gaussian curvature of a sphere of radius R is 1/R² 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 k_{g} on this circle in the unit sphere is given by
The integral of geodesic curvature around the circle of intersection is then
Geodesic curvature is exasperatingly complex concept. It has do not with just a curve, but a curve and the surface it is contained in. But it also depends upon whether the surface it is contained in is opening up like a flower or a bowl or closing in like spherical vase. In a sense the geodesic curvature represents the deviation of a curve from a geodesic of the surface it is in. 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 Gaussian curvature for a sphere of radius R is a constant 1/R². The Gaussian curvature for a unit sphere is then everywhere equal to 1. The integral of Gaussian curvature is over any spherical region is just the area of that region.
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
Hence
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 characteristic of the manifold.
Note however if a unit sphere is to be divided into two parts it is a matter of choosing a value for the cutoff parameter a. The smaller part corresponds to a negative value of a and the larger part to the corresponding positive value for a. When the two parts of the sphere are brought together the integrals of the geodesic curvature on their boundary are opposite in sign and cancel out.
There is a crying need for a term to denote the sum of the two integrals for a manifold. The term envelopment can be used on the basis that curvature is the rate at which the figure is enveloping the next higher dimensional space. That is to say, Gaussian curvature is the rate at which the 2D surface is enveloping 3D space and the geodesic curvature is the rate at which the 1D curve is enveloping 2D space.
A closed surface which has an Euler characteristic of 2 and hence has a total envelopment of 4π.The partition of the surface results in two parts each of which has an envelopment of 2π. This equal division of the envelopment occurs no matter how unequal the division of the area.
For more on aspects of the Gauss-Bonnet Theorem.
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