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The Gauss-Bonnet Theorem
and Its Generalization

The Gauss-Bonnet 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

SK(ω)dω = 2πχ(S)

For example, a sphere of radius R has a constant curvature of 1/R2 so the integral of curvature over the surface of the sphere is the constant curvature of 1/R2 times the area of the sphere of 4πR2 which is equal to 4π. The Euler characteristic of the sphere is 2 so the integral of the curvature of 4π is 2π(2). The Euler characteristic of a torus (doughnut) is zero so the integral of the curvature over the torus is also zero.

An analog of the above theorem for surfaces embedded in 3D space holds for 1D curves in 2D space. The integral of the curvature around a smooth curve is equal to 2π; i.e.,

Ck(s)ds = 2π

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.

Ck(s)ds + ΣCφi = 2π

This is often called the Gauss-Bonnet formula. Note that the right-hand side of the formula fits into the scheme of the Gauss-Bonnet 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 of side D has zero curvature on the straight parts. The interior angles are all equal to π/2 so the sum of the four of them is 2π.

There is an analog of the modification of the Theorem for non-smooth curves for non-smooth surfaces but it requires a more complex development.

A Useful Theorem

Many of the surfaces of interest are surfaces of revolution obtained by rotating a curve about an axis. A sphere is generated by rotating a semi-circle about the axis of the circle which connects the end points of the semi-circle. A torus is generated by rotating a full circle about an axis which does not intersect the circle. The theorem below gives a simple formula for the integral of curvature over a surface of revolution. Note that a surface of revolution is not necessarily a closed surface.


Proof: Without significant loss of generality the axis of rotation may be taken to be the x-axis. The profile curve α(t) is then (x(t),y(t)). The surface of revolution is (x(t),y(t)cos(θ),y(t)sin(θ)), where θ is the angle of rotation about the x-axis.

The parameterization of the profile curve can always be chosen such that
(x'(t))2+y'(t))2=1 (the unit speed curve). For a unit speed profile curve the formula for the Gaussian curvature of the surface of revolution reduces to

K(t,θ) = -y"(t)/y(t).

The element of area of the surface of revolution is dt(y(t)dθ). The integral of curvature over the surface of revolution is then

∫∫K(t,θ)y(t)dθdt = ∫∫(-y"(t)/y(t))y(t)dθdt
= -∫∫y"(t)dθdt
= -2π∫y"(t)dt
= -2π[y'(b)-y'(a)]
= 2π[y'(a)-y'(b)].

The angle φ that the tangent to the profile curve makes with the x-axis is given in general by

sin(φ)=y'(t)/[(x'(t))2+y'(t))2]1/2.

Thus for a unit speed profile curve sin(φ)=y'(t). Therefore the integral of the Gaussian curvature over a surface of revolution is given by

2π[sin(φ(a))-sin(φ(b)]


If a surface of revolution is to be smooth and closed then either the generating curve is closed and does not cross the axis of rotation or it crosses it with φ(a) and φ(b) equal to π/2 in magnitude and of opposite signs. In the first case the surface of revolution is topologically equivalent to a torus and in the second case to a sphere. In the case of a closed generating curve φ(a)=φ(b) and thus the total curvature is zero. For a sphere or a smooth sphere-like surface of revolution the total curvature is 2π(+1-(-1))=4π.

The Incorporation of Conical Points
Into the Gauss-Bonnet Theorem

The theorem may be used to determine the effect of non-smooth singularities such as conical points or ridges. Consider the surface of revolution generated by rotatin a curve around the x-axis. The curve starts as a straight line making an angle of β with the x-axis and ends with an arc of a circle intersecting the x-axis at a right angle.

The surface of revolution is depicted below.

The total curvature for this surface is 2π(sin(β)-(-1))=2π(sin(β)+1). If the generating curve started as an arc of a small cirle intersecting the x-axis at a right angle instead of a sharp angle then the total curvature would be 2π(2). Thus the integral of curvature about the conical point is -2π(1-sin(γ)) where γ is the apex angle of the cone.

The generalization of the Gauss-Bonnet Theorem to surfaces with only conical singularities would be that the total 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 where χ is the Euler characterisitic of the surface.

SK(ω)dω + ΣS[-2π[1-sin(γi)]] = 2πχ(S);
or, equivalently
SK(ω)dω = 2πχ(S); + 2πΣS[1-sin(γi)]

where {γi} is the set of apex angles of the conical points of the surface.

The Incorporation of Sharp Ridges
Into the Gauss-Bonnet Theorem

Consider a surface of revolution which is like a lens. The cross section involves two circular arcs which meet at a sharp angle as is the case below.

The surface of revolution is depicted below.

The Gauss-Bonnet Theorem (and the Surface of Revolution Curvature Theorem) says the total curvature for the lens is equal to 4π. The integral of curvature on the smooth surfaces of the lens can be evaluated separately so the contribution of the sharp ridge can be found as the difference. Since the curvature on the smooth surfaces is a constant (1/r2) the integral of curvature can be found by multiplying the curvature times the area. If the arc for a smooth surface subtends an angle of ψ the area is 2πr2[1-cos(ψ/2)] for one side and 4πr2[1-cos(ψ/2)] for the two sides of the lens. When this area is multiplied by the curvature of (1/r2) the result is an integral of curvature of 4π[1-cos(ψ/2)] on the smooth surfaces. Comparing this with the total curvature for the surface of revolution of 4π shows that the contribution of the sharp ridge is 4πcos(ψ/2).

The circumference of the sharp ridge is 2πrsin(ψ/2) so the curvature per unit length of the ridge is:

4πcos(ψ/2)/[2πrsin(psi;/2)] = (2/r)cot(ψ/2)
or, equivalently,
2cos(ψ/2)/r1
 

where r1 is the radius of curvature of the line of the ridge.

The diagram below shows that the angle ψ subtended by the arc is the same as the angle of the ridge.

The above equation which make the contribution of the ridge line to the curvature integral inversely proportional to the radius of curvature of the ridge line suggests that a straight ridge line, one for which the radius of curvature is infinite, would make no contribution to the curvature integral. This in fact can be established by separate arguments. Consider the surface of revolution created previously to examine the contribution of a conical point. Imagine this surface split in half and the two parts separated and then connected by a line, two planes and part of a cylinder. This extended surface would still have a total curvature of 4π. The planes have zero curvature so the contribution of the straight ridge line must be exactly counterbalanced by the contribution of the segment of the cylinder. But the Gaussian curvature of a cylinder is zero at all points so the contribution of the segment of the cylinder is zero and hence also that of the straight ridge line. An additional argument for the contribution of straight ridge lines to the curvature integral of a surface being zero comes from a consideration of polyhedra. If a polyhedron is scaled up the lengths of the ridge lines are also scaled up but the curvature integral remains constant at 4π. Thus the contribution of the ridge lines to the curvature integral is zero. Finally the Gaussian curvature for a straight ridge line is zero for the same reason that the Gaussian curvature of a cylinder; i.e., the Gaussian curvature is the product of the curvatures in the two principal directions for the surface and for the cylinder and the straight ridge the curvature is zero in one principal direction.

Since the faces of the polyhedra are flat and have zero Gaussian curvature the entire curvature integral is due to the contribution of the vertices.

For a regular polyhedron having V vertices the curvature integral of each vertex is then 4π/V. It would be nice to have this contribution to the curvature integral expressed as a function of the angular deficit of the vertex; i.e., the difference between 2π and the sum of the angles at the vertex. For example, the angles at the corner of a cube are three angles of π/2 radians each. Thus the angular deficit for a vertex of a cube is (2π-3π/2)=π/2. The contribution of each of the eight vertices of a cube to the curvature integral is 4π/8=π/2.

For a tetrahedron at each vertex there are three angles of π/3 so the angular deficit is π. The contribution of each vertex of the tetrahedron is 4π/4=π. A regular rhomboid has six vertices in which four equilateral triangles come together. Thus the angular deficit is 2π-4π/3=2π/3. The contribution of each vertex of the six vertices is 4π/6=2π/3. So again the contribution to the curvature integral of a vertex is equal to its angular deficit.

The survey of this proposition for regular polyhedra continues. First note the interior angle of a regular polygon of n sides is (π-2π/n). Thus for a pentagon the interior angles are equal to 3π/5. The dodecahedron has three pentagons coming together at each vertex, which thus has an angular deficit of π/5. The number of vertices of a dodecahedron is 20 so the contribution of each to the curvature integral is 4π/20=π/5.

The iscosahedron has five triangles coming together at each of its twelve vertices. Thus the angular deficit at each vertex is (2π-5π/3)=π/3. Each vertex contributes 4π/12=π/3 to the curvature integral. The proposition thus holds for all the regular polyhedra.

The contribution of a conical point of 2π(1-sin(γi) can now be interpreted as an angular deficit. If a cone with apex angle of γ is cut along a generating line and unrolled the angle covered by the unrolled cone is 2πsin(γ) so the angular deficit is 2π-2πsin(γ) or 2π(1-sin(γ)). Therefore the conjecture to be proven in general is:

The proof of this conjecture would mean that the generalization of the Gauss-Bonnet Theorem is that the integral of the Gaussian curvature over the smooth portions of a closed surface plus the sum of the angular deficits of the singular points is equal to 2π times the Euler characteristic of the surface.

For the Gauss-Bonnet formula for geodesic curvature in the 2D plane the contribution at a corner point is expressed in terms of the turning angle φ. The turning angle is just the angular deficit at such a corner point π-η, where η is the interior angle at the corner.

The Inclusion of Indented Points and
Sharp Creases in the Analysis

It was found above that the integral of curvature about the conical point is -2π(1-sin(γ)) where γ is the apex angle of the cone. If γ is greater than π/2 the surface has not a protusive point but an indented feature. The effect of such indented points on the curvature integral of a surface is thus already included within the previous analysis for points.

Likewise a sharp crease, such would be part of the surface of revolution of for a profile curve shown below, is included in the analysis for the effect of a sharp ridge with the angle subtended by the arc being greater than π/2. If the crease has a straight edge it contribution to the curvature integral of a surface is zero, as is the case of a ridge with a straight edge.

(To be continued.)

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