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of a Point of Singularity of Curvature to the GaussBonnet Integral is Equal to its Angular Deficit: The Conical Point Case 
The GaussBonnet Theorem in 3D space says that the integral of the Gaussian curvature over a closed smooth surface is equal to 2π times the Euler characteristic of the surface. For a surface S
For example, a sphere of radius R has a constant curvature of 1/R^{2} so the integral of curvature over the surface of the sphere is the constant curvature of 1/R^{2} times the area of the sphere of 4πR^{2} which is equal to 4π. The Euler characteristic of the sphere is 2 so 2πχ(sphere) is 4π. The Euler characteristic of a torus is zero so the integral of the Gaussian curvature over a torus is also zero even though the Gaussian curvature is not identically zero over a torus.
An analog of the above theorem for 2D surfaces embedded in 3D space holds for 1D curves embedded in 2D space. The integral of the curvature around a smooth curve C is equal to 2π; i.e.,
where k is the curvature.
For example, a circle of radius R has a constant curvature of 1/R. The length of the circumference of the circle is 2πR so the integral of curvature around the circle is (1/R)2πR which is 2π.
A curve with angular corners presents a different problem. The curvature at the corner is, in effect, infinite. The total curvature around the curve is the integral over the smooth part plus the sum of the turning angles φ at the corner points. The turning angle at a corner is π less the interior angle at that corner.
This is often called the GaussBonnet formula. Note that the righthand side of the formula fits into the scheme of the GaussBonnet Theorem in that the Euler characteristic of a plane polygon is just 1 because for a polygon there is 1 face and the number of edges and vertices are equal.
For example, a square has zero curvature on the straight parts. The interior angles are all equal to π/2 so are the exterior angles. Thus the sum of the four exterior angles is 2π.
Here conical means cones with circular crosssection perpendicular to their axes. Consider the following figure and the diagram of its crosssection parallel and through its axis.
The radius of the sphere in the figure is unity. The latitude angle Θ is measured from zero at the top of the circle. The relationship between the apex angle β and the angle Θ for the sphere is given by
The area on a sphere included between latitude angle θ=−π/2 and θ=Θ is given by
For a unit sphere the Gaussian curvature is everywhere on it equal to unity. Therefore the integral of Gaussian curvature over the spherical portion of the figure is
On the other hand the Gaussian curvature on the cone part of the figure is zero and hence the integral of Gaussian curvature of the cone part is zero. Let V denote the contribution of the singularity of curvature at the apex of the cone.
The figure satisfies the conditions for the standard GaussBonnet Theorem at all points except the singularity. Its Euler characteristic is 2 and therefore
The circumference of the circle on the cone at a distance r along the surface of the cone is
When the cone out to a distance r is unrolled it fits into a circle of circumference 2πr. Thus the angular deficit φ for the cone is
Thus the contribution of the singularity of curvature at the cone apex is equal to the angular deficit of the cone; i.e.,
The generalization of the GaussBonnet Theorem to surfaces with only conical singularities would be that the total Gaussian curvature (which is equal to the integral of the curvature of the smooth portions of the surface plus the contribution due to the conical points) is equal to 2πχ(S) where χ(S) is the Euler characterisitic of the surface S.
where (φ_{i}: i=1,…, n} is the set of angular deficits of the conical points of the surface.
For more on aspects of the GaussBonnet Theorem.
