Notice: This material will be included in a forthcoming (summer 2000) book with the tentative title Experiencing Geometry in Two- and Three-Dimensional Spaces. This new book will be an expanded and updated version of Experiencing Geometry on Plane and Sphere. This material is in draft form and may not be duplicated or quoted without the author's written permission, except for purposes of review or trying out the material with students. As always comments are welcome and will affect the final draft. Send comments to dwh2@cornell.edu.
Chapter 7
In this chapter we will find a formula for the area of triangles on spheres and hyperbolic planes. We will then investigate the connections between area and parallel transport, a notion of local parallelism that is definable on all surfaces. We will also introduce the notion of holonomy which has many applications in modern differential geometry and engineering.
Problem 7.1. The Area of a Triangle on a Sphere
A biangle or lune is any one of the four connected regions determined by two (not coinciding) great circles. The two angles of the lune are congruent.
a. The two sides of each interior angle bi of a triangle D on a sphere determines two congruent lunes with lune angle bi. Show how the three pairs of lunes determined by the three interior angles covers the sphere with some overlap. (What is the overlap?)
b. Each exterior angle ai of the triangle D determines a single lune that is exterior to the triangle. What is the union of the three such lunes determined by the three exterior angles of D?
c. Find a formula for the area of a lune with lune angle q in terms of q and A, the (surface) area of the sphere.
d. Find a formula for the area of triangle D on a sphere.
e. Consider a very tiny triangle on a sphere with very large radius. How does your formula in Part d relate to what you know about the area and angles of a triangle on the plane?
Suggestions
This is one of the problems that you almost certainly must do on an actual sphere. There are simply too many things to see, and the drawings we make on paper distort lines and angles too much. The best way to start is to make a small triangle on a sphere, and extend the sides of the triangle to complete great circles. Then look at what you've got. You will find an identical triangle on the other side of the sphere, and you can see several lunes that extend out from the triangles. Use Parts a-c to do Part d. After that, it is simply a matter of adding everything up properly. The key to this problem is to put everything in terms of areas that you know. We will see later (Problem 16.2) that the area of the whole sphere is A = 4pr2, or you may find a derivation of this formula in a multivariable calculus text, or you can just leave your answer in terms of A.
Problem 7.2. Area of Hyperbolic Triangles
We can try to mimic the derivation of the area of spherical triangles, but of course there are no lunes and the area of the hyperbolic plane is evidently infinite. Nevertheless, if we focus on the exterior angles of a hyperbolic triangle and look at the regions formed, we obtain in Figure 7.1 a picture of the situation in the annular hyperbolic plane.
Figure 7.1. Triangle with an ideal triangle & three 2/3-ideal triangles.
In Figure 7.1, a triangle is drawn with its interior angles, bi, and exterior angles, ai. The three extra lines are geodesics which are asymptotic at both ends to an extended side of the triangle. We call the region enclosed by these three extra geodesics an ideal triangle. In the annular hyperbolic plane these are not actually triangles since their vertices are at infinity. In Figure 7.1 we see that the ideal triangle is divided into the original triangle and three "triangles" which have two of their vertices at infinity. We call a "triangle" with two vertices at infinity a 2/3-ideal triangle. You can use this decomposition to determine the area of a hyperbolic triangle in much the same way you determined the area of a spherical triangle. So, first we must determine the areas of ideal and 2/3-ideal triangles.
It is impossible to picture the whole of an ideal triangle in an annular hyperbolic plane, but it is easy to picture ideal triangles in the upper half plane model. In the upper half plane model an ideal triangle is a triangle with all three vertices either on the x-axis or at infinity. See Figure 7.2.
Figure 7.2. Ideal triangles in the upper half plane model.
At first glance it appears that there must be many different ideal triangles; however:
a. Prove that all ideal triangles on the same hyperbolic plane are congruent.
[Hint: Review your work on Problem 4.2. Perform an inversion that takes on the vertices to infinity and the two sides from that vertex to vertical lines. Then apply a similarity to the upper half plane taking the standard ideal triangle with vertices (-1,0), (0,1), and ¥.]
b. Show that the area of an ideal triangle is pr2. (Remember this r is the radius of the annuli.)
[Hint: Since the distortion dist(z)(a,b) is r/b, the desired area is
.]
We now picture in Figure 7.3 2/3-ideal triangles in the upper half plane model.
Figure 7.3. 2/3-ideal triangles in the upper half plane model.
c. Prove that all 2/3-ideal triangles with angle q are congruent and have area (p-q)r2.
[Hint: Show, using Problem 4.2 that all 2/3-ideal triangles with angle q are congruent to the standard one at the right of Figure 7.3 and show that the area is the double integral:
.]
d. Find a formula for the area of a hyperbolic triangle.
[Hint: Look at Figures 7.1 and 7.4. Then use Parts c and d.]
Figure 7.4. Finding the area of a hyperbolic triangle.
Introducing Parallel Transport and Holonomy
Imagine that you are walking along a straight line or geodesic carrying a stick that makes a fixed angle with the line you are walking on. If you walk along the line maintaining the direction of the stick relative to the line constant, then you are performing a parallel transport of that "direction" along the path.
Figure 7.5. Parallel transport.
To express the parallel transport idea, it is common terminology to say that:
Parallel transport has become an important notion in differential geometry, physics, and mechanics. One important aspect of differential geometry is the study of properties of spaces (surfaces) from an intrinsic point of view. As we have seen, in general, it is not possible to have a global notion of direction from which we are able to determine when a direction (or vector) at one point is the same as a direction (or vector) at another point. However, we can say that they have the same direction with respect to a geodesic g if they are parallel transports of each other along g. Parallel transport can be extended to arbitrary curves as we shall discuss at the end of this chapter. With this notion it is possible to talk about how a particular vector quantity changes intrinsically along a curve (covariant differentiation). In general, covariant differentiation is useful in the areas of physics and mechanics. In physics, the notion of parallel transport is central to some of the theories that have been put forward as possible candidates for a "Unified Field Theory," a hoped for but as yet unrealized theory that would unify all known physical laws about forces of nature.
Now let us explore what happens when we parallel transport a line segment around a triangle. For example, consider on a sphere an isosceles triangle with base on the equator and opposite vertex on the North Pole (see Figure 7.6). Note that the base angles are right angles. Now start at the North Pole with a vector (a directed geodesic segment) and parallel transport it along one of the sides of the triangle until it reaches the base. Then parallel transport it along the base to the third side. Then parallel transport back to the North Pole along the third side. Notice that the vector now points in a different direction then it did originally. Now you can follow a similar story for the right hyperbolic triangle represented in Figure 7.7 and see that here also there is a difference between the starting vector and the ending parallel transported vector. This difference is called the holonomy of the triangle. Note that the difference angle is counterclockwise on the sphere and clockwise in the hyperbolic plane.
Figure 7.6. The holonomy of a double-right triangle on a sphere.
Figure 7.7. Holonomy of a hyperbolic triangle.
This works for any small triangle (i.e., a triangle that is contained in an open hemisphere) on a sphere and for all triangles in a hyperbolic plane. We can define the holonomy of a (small, if on a sphere) triangle, H(D), as follows:
If you parallel transport a vector (a directed geodesic segment) counterclockwise around the three sides of a small triangle, then the holonomy of the triangle is the smallest angle measured counterclockwise from the original position of the vector to its final position.
For the spherical triangle in Figure 7.6 we see that the holonomy is positive and equal to the upper angle of the triangle. For the hyperbolic triangle in Figure 7.7 we see that the holonomy is negative (clockwise).
Holonomy can also be defined for large triangles on a sphere but it is more complicated because of the confusion as to what angle to measure. For example, what should be the holonomy when you parallel transport around the equator — 0 radians or 2p radians?
Problem 7.3. The Holonomy of a Small Triangle
Find a formula that expresses the holonomy of a small triangle on a sphere, and a formula that expresses the holonomy of any triangle on a hyperbolic plane.
Suggestions
What happens to the holonomy when you change the angle at the North Pole of the triangle in Figure 7.6? What happens if you parallel transport around the triangle a vector pointing in a different direction? Parallel transport vectors around different triangles on your model of a sphere. Try it on triangles that are very nearly the whole hemisphere and try it on very small triangles. What do you notice? Try this also on your models of the hyperbolic plane, again for different size triangles.
A good way to approach the formula for general triangles is to start with any geodesic segment at one of the angles of the triangle, and follow it as it is parallel transported around the triangle. Keep track of the relationships between the angles this segment makes with the sides and the exterior angles. See Figure 7.8 which is drawn for spherical triangles; the reader should be able to draw an analogous picture for a general hyperbolic triangle.
Figure 7.8. Holonomy of a general triangle.
Pause, explore, and write out your ideas for this problem before reading further.
The Gauss-Bonnet Formula for Triangles
In working on Problem 7.3 you should find (among other things) that:
The holonomy of a (small, if on a sphere) triangle is equal to 2p minus the sum of the exterior angles or equal to the sum of the interior angles minus p.
Let b1, b2, b3 be the interior angles of the triangle and a1, a2, a3 the exterior angles. Then algebraically the statement above can be written as:
H(D) = 2p - (a1 + a2 + a3) = (b1 + b2 + b3) - p.
The quantity [ å bi - p ] = [ 2p - S ai ] is also called the excess of D, and when the excess is negative, the positive quantity [ p - å bi ] = [ S ai - 2p ] is called the defect of D.
If you haven't already seen it, note now the close connection between the holonomy, the excess, and the area of a triangle. Note that the holonomy is positive for triangles on a sphere and negative for triangles in a hyperbolic plane. One consequence of this formula is that the holonomy does not depend on either the vertex or the vector we start with. This is to be expected since parallel transport does not change the relative angles of any figure.
Following Problems 7.1, 7.2, and 7.3 we can write the result for triangles on a sphere with radius r in this form:
sphere: H(D) = 2p - (a1 + a2 + a3) = A(D) 4p/A = A(D) r-2.
For a hyperbolic plane made with annuli with radius r we get:
hyperbolic: H(D) = 2p - (a1 + a2 + a3) = -A(D) r-2.
The quantity r-2 is traditionally called the Gaussian curvature or just plain curvature of the sphere and - r-2 is called the (Gaussian) curvature of the hyperbolic plane. If K denote the (Gaussian) curvature as just defined, then the formula
2p - (a1 + a2 + a3) = A(D) K
is called the Gauss-Bonnet Formula (for triangles).
Can you see how this result gives the bug an intrinsic way of determining the quantity K and thus also determining the extrinsic radius r?
The Gauss-Bonnet Formula not only holds for triangles in an open hemisphere or in a hyperbolic plane but can also be extended to any simple (that is, non-intersecting) polygon (that is, a closed curve made up of a finite number of geodesic segments) contained in an open hemisphere or in a hyperbolic plane.
Problem 7.4. Gauss-Bonnet Formula for Polygons
Definition. The holonomy of a simple polygon, H(G), in an open hemisphere or in a hyperbolic plane is defined as follows:
If you parallel transport a vector (a directed geodesic segment) counterclockwise around the sides of the simple polygon, then the holonomy of the polygon is the smallest angle measured counterclockwise from the original position of the vector and its final position.
If you walk around a polygon with the interior of the polygon on the left, the exterior angle at a vertex is the change in the direction at that vertex. This change is positive if you turn counterclockwise and negative if you turn clockwise. (See Figure 7.9.)
Figure 7.9. Exterior angles.
We will first look at convex polygons because this is the only case we will need later and it is easier to understand. A region is called convex if every pair of points in the region can be joined by a geodesic segment lying wholly in the region.
a. Show that if G is a convex polygon in an open hemisphere or in a hyperbolic plane, then
H(G) = 2p - S ai = S bi - (n - 1)p = Area(G) K,
where Sai is the sum of the exterior angles, S bi is the sum of the interior angles, and K is the Gaussian curvature.
[Hint: Divide the convex polygon into triangles as in Figure 7.10. Now apply 7.3 to triangle and carefully add up the results. You can check directly that H(G) = 2p - S ai.].
Figure 7.10. Dividing a convex polygon into triangles.
*b. Prove that every simple polygon on the plane or on a hemisphere or on a hyperbolic plane can be dissected into triangles without adding extra vertices.
Suggestions
Look at this on the plane, hemispheres, and hyperbolic planes. The difficulty in this problem is coming up with a method that works for all polygons, including very general or complex ones, like the polygon in Figure 7.11.
Figure 7.11. General polygon.
You may be tempted to try to connect nearby vertices to create triangles, but how do we know that this is always possible? How do you know that in any polygon there is even one pair of vertices that can be joined in the interior? The polygon may be so complex that parts of it get in the way of what you're trying to connect. So you might start by giving a convincing argument that there is at least one pair of vertices that can be joined by a segment in the interior of the polygon. In order to see that there is something to prove here, Figure 7.12 shows an example of a polyhedron in 3-space which has no pair of vertices that can be joined in the interior. The polyhedron consists of eight triangular faces and six vertices. Each vertex is joined by an edge to four of the other vertices and the straight line segment joining it to the fifth vertex lies in the exterior of the polygon. Therefore it is impossible to dissect this polyhedron into tetrahedra without adding extra vertices. This example and some history of the problem are discussed in [Tx: Eves, page 211] and [Z: Ho].
Note that there is at least one convex vertex (a vertex with interior angle less than p) on every polygon (in fact it is not too hard to see that there must be at least three such vertices). To see this, pick any geodesic in the exterior of the polygon and parallel transport it towards the polygon until it first touches the polygon. It is easy to see that the line must now be intersecting the polygon at a convex vertex.
Figure 7.12. A polyhedron with vertices not joinable in the interior.
*c. Show that if G is a simple polygon in an open hemisphere or in a hyperbolic plane, then
H(G) = 2p - S ai = S bi - (n - 1)p = Area(G) K,
where Sai is the sum of the exterior angles, S bi is the sum of the interior angles, and K is the Gaussian curvature.
[Hint: Start by applying Part b. Then proceed as in Part a, but for this part you may find it easier to show that the holonomy of the polygon is the sum of the holonomies of the triangles by removing one triangle at a time. Again, you can check directly that H(G) = 2p - S ai.].
Gauss-Bonnet Formula for Polygons on Surfaces
The above discussion of holonomy is in the context of an open hemisphere and a hyperbolic plane, but the results have a much more general applicability and constitute a major aspect of differential geometry. In particular, we can extend this result even further to general surfaces, even those of non-constant curvature. In fact, Gauss defined the (Gaussian) curvature K(p) at a point p on any surface to be,
K(p) = limD®p H(D) / A(D),
where the limit is taken over a sequence of small (geodesic) triangles that converge to p. The reader can check that the Gaussian curvature of a sphere (with radius r) is 1/r2 and that the Gaussian curvature of a hyperbolic plane (with radius r, the radius of the annular strips) is -1/r2. This definition leads us to another formula, namely,
The Gauss-Bonnet Formula for Polygons on Surfaces
On any smooth surface (2-manifold), if G is a (geodesic) polygon that can be continuously deformed to a point in its interior, then
H(G) = 2p - S ai = òòI(G) K dA,
where the integral is the (surface) integral over I(G), the interior of the polygon.
The proof of this formula involves dividing the interior of G into many triangles, each so small that the curvature K is essentially constant over its interior, and then applying the Gauss-Bonnet Formula for spheres and hyperbolic planes to each of the triangles.
In addition, all of the versions of the Gauss-Bonnet Formula given thus far can be extended to arbitrary, simple, piece-wise smooth, closed curves. If g is such a curve, then we can define the holonomy H(g) = lim H(gi), where the limit is over a sequence (which converges point-wise to g) of geodesic polygons {gi} whose vertices lie on g. Using this definition, the Gauss-Bonnet formula can be extended even further:
The Gauss-Bonnet Formula for Curves Which Bound a Contractible Region
H(g) = A(g) K (on a sphere or hyperbolic plane)
or,
H(g) = òòI(g)K dA (on general surfaces),
where I(g) is the interior of the region bounded by g.