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An Extension of the Gauss-Bonnet Theorem to Include
Geometric Objects with Sharp Edges and Conical Vertices

The Gauss-Bonnet Theorem in 3D space says that the integral of the Gaussian curvature over a smooth surface S with boundary ∂S plus the integral of the geodesic curvature around that boundary is equal to 2π times the Euler characteristic of the surface. For a surface S

SK(p)dA + ∫∂Skgds = 2πχ(S)

The Gauss-Bonnet Theorem is usually stated for a manifold which means there are no points where the Gaussian curvature is not defined. It can be extended to include not only surfaces but also sharp edges and vertices. The statement of the extension is as follows.

An Extended Version of the Gauss-Bonnet Theorem

Let X be a geometric object without a boundary, consisting of a set S={si; i=1, …, n} of surfaces where Gaussian curvature is everywhere defined and a set of edges E={ek, k=1, …, m} where two surfaces come together at a sharp angle and a set V={vp; p=1,…, q} of vertices of a conical nature. The sets E and V may be empty. Let χ(X) be the Euler characteristic of this object. Then

ΣSsiKdAi + Σk[∫kgidsk + ∫kgjdsk] + Σpαvp = 2πχ(X)

where [∫kgidsk + ∫kgjdsk] represents the geodesic curvature of edge k evaluated in the two surfaces i and j whose meeting forms edge k. K represents the Gaussian curvature at a point of a surface and αvp represents the angular deficit at vertex vp.

A conical vertex is one that is generated by a straight line sweeping around a cross-section curve such as a circle or polygon. Pyramidal points are of a conical nature. An edge may project away from the object or project inward as for a crease.

Special cases of this theorem are the polyhedra where Gaussian curvature on the flat faces is zero, geodesic curvature on the straight edges is zero so only the angular deficits of the vertices are involved. And, of course, the conditions for the standard Gauss-Bonnet Theorem satisfy the extended theorem. The theorem has been shown to hold true for the special cases of the objects below.

  

The case of an object with a boundary is easily included. A boundary curve is simply a one-sided edge and the integral of geodesic curvature needs only to be evaluated for the surface in which the boundary lies.

A Further Extended Version of the Gauss-Bonnet Theorem

Let X be a geometric object with a boundary ∂X. with X consisting of a set S={si; i=1, …, n} of surfaces where Gaussian curvature is everywhere defined and a set of edges E={ek, k=1, …, m} where two surfaces come together at a sharp angle and a set V={vp; p=1,…, q} of vertices of a conical nature. The sets E and V may be empty. Let χ(X) be the Euler characteristic of this object. Then

ΣSsiKdAi + Σk[∫kgidsk + ∫kgjdsk] + Σpαvp + ∫∂Xkgds = 2πχ(X)

What is needed is a term to represent the LHS of the above equation in the same way that net turning represents the analogous term for 1D curves in 2D space. Such a term would have to incorporate the 2D, 1D and 0D aspects of the curvature of a figure. The term net envelopment captures some aspects of curvature and the angular deficit of the vertices. The Gaussian curvature is the rate at which the 2D surface envelopes 3D space. The geodesic curvature is the rate at which a 1D curve envelopes 2D space. And finally the angular deficit of a conical point is a rate at which 3D space is enveloped by the 2D surface of the conical point.

Proof:

(Under construction.)

Lemmas concerning pyramids

Lemma 1: Let P be an n-cornered (and n-sided) polygon which is the base of a pyramid. Then the sum of the angular deficits of the vertices of this pyramid is equal to 4π.

Proof: Since the pyramid is a polyhedron this is just a special case of Descartes' Closure Theorem and hence needs no proof. But the proof given here is for the purpose of extending the result to (half) cones with cross sections which are not necessarily polygons.

Let the angles in the triangles impinging upon the apex of the pyramid be labled {ai, bi, ci: i= 1, …, n} with ai being the angle at the apex of the pyramid. Let the interior angles of the polygonal base of the pyramid be labled {di: i=1, … n}. The exterior angles of the polygonal base are given by ei=π−di.

The angular deficit A at the apex of the pyramid is given by

A = 2π − Σai

At i-th corner of the base the angular deficit Bi is given by

Bi = 2π − (ci-1+ bi+di) for i=2, … n
and B1 = 2π − (cn+ b1+d1)

Summing the angular deficits of the corners of the base gives

ΣBi = n(2π) − (Σbi + Σci + Σdi)

This sum combined with the angular deficit A at the apex gives

A + ΣBi = (n+1)(2π) − (Σai + Σbi + Σci + Σdi)

Since (ai+bi+ci) is equal to π for all i

Σ(ai+bi+ci) = nπ
and hence
A + ΣBi = (n+1)(2π) − (nπ + Σdi)
A + ΣBi = (n + 2)π − Σdi

Since ei=π−di

Σei = nπ − Σdi

But Σei is equal to 2π. Therefore

Σdi = (n−2)π

This means that

A + ΣBi = (n + 2)π − (n−2)π = 4π

The term 4π is equal to 2π times the Euler characteristic of the pyramid of 2.

Pyramids with a More General Base

Let the base of a pyramid be a figure whose boundary C is a curve with n corners but the edges between corners are not necessarily straight lines. Then its envelopment is equal to 4π.

The Gaussian curvature on a cone is zero everywhere zero everywhere except at the apex and there it is not defined. Let a sequence of points on the path be selected, say {x(sj, y(sj: j=1, 2, …, m} with the corner points of C being included in the sequence. That sequence generates a polyhedral pyramid and, by Lemma 1, the net sum of the angular deficits is 4π. The deviations of the polygonal base of that pyramid from the curve C are bounded. A more refined division of the path will further reduce the deviation from C. As the maximum size of the intervals on the path goes to zero the polygonal approximation goes to the curve C. The net sum of the angular deficits for all of the approximating pyramids is 4π; therefore its limit is 4π. The angular deficit at the cone apex is always the same so is the same in the limit.

Pyramids without a Base

For further work the figure that must be analyzed is an open-ended general cone. This figure is topologically equivalent to a disk or solid polygon. Their Euler characteristic is 1 since there is one face and the number of vertices and edges are equal. For the extended version of the Gauss-Bonnet Theorem to hold the integral of the Gaussian curvature plus the integral of the geodesic curvature over the open edge must be equal to 2π.

First consider an open pyramid C in which the open edge is an n-cornered polygon P. Let the angles of the triangular sides be labled {ai, bi, ci: i= 1, …, n}. The angular deficit of apex is given by

A = 2π − Σai

The geodesic curvature of a curve in a plane is the turning of the normal to the curve. For a polygon the normal turns an amount at a corner equal to the exterior angle at that corner.

When the open-ended pyramid is slit along a generator line it can be unfolded into a plane. Any generator may be used but it is more convenient to use the generator that goes through the apex point and the midpoint of the first side of the polygonal base of the pyramid. When the pyramidal cone is unfolded and flattened the internal angles of the bent line are given by

ci-1 + bi
and therefore
the exterior angles are
ei = π − (ci-1 + bi)

The total turning along the bent line is therefore

Σei = nπ − Σ(ci + bi)

This combined with the angular deficit at the apex is

A + B = 2π − Σai + nπ − (Σci+Σbi)
which reduces to
A + B = (n+2)π − Σ(ai + bi + ci)

But

ai + bi + ci = π for all i

Therefore the expression for A+B reduces to

A + B = (n+2)π − nπ = 2π

The Euler characteristic of an open-ended pyramidal cone is 1 so 2πχ(C) is equal to 2π.

The Constancy of the Envelopment of
a Cross-Section Curve of a Cone

Consider a truncated cone C.

This object satisfies the Gauss-Bonnet Theorem and its Euler characteristic is zero. Let B1 and B2 be the two boundary curves of the truncated cone. Then

CKdA + EB1 + EB2 = 0
where
EBi = ∫Bikgds + ΣBiαj

Because of their opposite orientation

EB1 = − EB2

If EB2 is held fixed and the truncation level changed the same value of EB2 is found. Therefore the same value of EB is found for any cross section curve of the cone.

The Extension of the Gauss-Bonnet
Theorem to Geometric Objects which
Have a Finite Number of Conical Points

Let G be a geometric figure which satisfies the conditions of the Gauss-Bonnet Theorem except for n conical points. Let M be the figure which results from clipping away all of the conical points. M then has an additional boundary consisting of n holes. Let H be the sum of the net turning angles of the holes. Since M satisfies the Gauss-Bonnet Theorem

MKdA + H = 2πχ(M)

Let αi be the angular deficit of the i-th point. For the excised conical point surfaces with boundaries identical to the holes of M except with the opposite orientation

Σαi − H = 2πn

The recombination of the conical points with the M then gives

MKdA + H + Σαi − H= 2πχ(M) + 2πn
which reduces to
MKdA + Σαi = 2π(χ(M) + n)

The Gaussian curvature on the surface of a cone is zero. Therefore

MKdA = ∫GKdA

The relationship between the Euler characteristics of M and G is

χ(M) = χ(G) − n
or, equivalently
χ(G) = χ(M) + n

Therefore

GKdA + Σαi = 2πχ(G)

where χ(G) is equal to χ(M) plus n, the number of conical points of G.

As an example, consider the G as the following object

For this object M is a truncated sphere in which the surfaces of the cone and the sphere are tangent at the circle of intersection. The integral of the Gaussian curvature over the surface of the truncated sphere is 2π(1 + sin(γ)) where γ is half the apex angle of the cone. The angular deficit of the cone is 2π(1 − sin(γ)). Thus

GKdA + Σαi = 2π(1 + sin(γ)) + 2π(1 − sin(γ)) = 4π

The Euler characteristic of the truncated sphere is 1 because it is topologically equivalent to a disk. There is only one conical point so the Euler characteristic of G is 1+1=2 and hence 2πχ(G) is equal to 4π. Of course, the object itself is topologically equivalent to a sphere and hence has Euler characterisitic of 2. However, before hand it was not certain that the singularity of the point would not affect the Euler characteristic of the object.

As another illustration consider any polyhedron. The Gaussian curvature on the faces of the polyhedron is zero. The Euler characteristic of any polyhedron without holes is 2. Therefore

Σαi should be equal to 2π(2).
and it is

This is known as Descartes' Closure Rule. This illustrates that the effect of the straight edges of a polyhedron is zero. The issue of sharp edges must be dealt with next.

(To be continued.)

For more on aspects of the Gauss-Bonnet Theorem.


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