7.2: Elliptic Geometry with Curvature k > 0
( \newcommand{\kernel}{\mathrm{null}\,}\)
One may model elliptic geometry on spheres of varying radii, and a change in radius will cause a change in the curvature of the space as well as a change in the relationship between the area of a triangle and its angle sum.
For any real number k>0, we may construct a sphere with constant curvature k. According to Example 7.1.1, the sphere centered at the origin with radius 1√k works. Geometry on this sphere can be modeled down in the extended plane via stereographic projection. This geometry will be called elliptic geometry with curvature k>0.
Consider the sphere S2k centered at the origin of R3 with radius 1√k. Define stereographic projection ϕk:S2k→C+, just as we did in Section 3.3, to obtain the formula
ϕk(a,b,c)={a1−c√k+b1−c√ki if c≠1√k; ∞ if c=1√k .
Diametrically opposed points on S2k get mapped via ϕk to points z and za that satisfy the equation za=−1k¯z, by analogy with Lemma 6.1.1. We call two such points in C+ antipodal with respect to S2k, or just antipodal points if the value of k is understood.
Our model for elliptic geometry with curvature k has space P2k equal to the closed disk in C of radius 1√k, with antipodal points of the boundary identified. This space is simply a scaled version of the projective plane from Chapter 6.
The group of transformations, denoted Sk, consists of those Möbius transformations that preserve antipodal points with respect to S2k. That is, T∈Sk if and only if the following holds:
if za=−1k¯z then T(za)=−1k¯T(z).
The geometry (P2k,Sk) with k>0 is called elliptic geometry with curvature k. Note that (P21,S1) is precisely the geometry we studied in Chapter 6.
The transformations of C+ in the group Sk correspond precisely with rotations of the sphere S2k. One can show that transformations in Sk have the form
T(z)=eiθz−z01+k¯z0z.
We define lines in elliptic geometry with curvature k to be clines with the property that if they go through z then they go through za=−1k¯z. These lines correspond precisely to great circles on the sphere S2k.
The arc-length and area formulas also slip gently over from Chapter 6 to this more general setting.
The arc-length of a smooth curve r in P2k is
L(r)=∫ba2|r′(t)|1+k|r(t)|2 dt.
As before, arc-length is an invariant, and the shortest path between two points is along the elliptic line through them. In the exercises we derive a formula for the distance between points in this geometry. The greatest possible distance between two points in (P2k,Sk) turns out to be π(2√k).
The area of a region R given in polar form is computed by the formula
A(R)=∬R4r(1+kr2)2drdθ.
To compute the area of a triangle, proceed as in Chapter 6. First, tackle the area of a lune, a 2-gon whose sides are elliptic lines in (P2k,Sk).
Assume k>0. A lune in (P2k,Sk) with interior angle α has area 2αk.
- Proof
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Without loss of generality, we may consider the vertex of our lune to be the origin. As before, elliptic lines through the origin must also pass through ∞, so our two lines forming the lune are Euclidean lines. After a convenient rotation, we may further assume one of these lines is the real axis, so that the lune resembles the one in Figure 6.3.4. To compute the area of the lune, compute the integral
A=2∫α0∫1/√k04r(1+kr2)2drdθ.
Letting u=1+kr2 so that du=2krdr, the bounds of integration change from [0,1√k] to [1,2]. Then,
A=2∫α02k∫21duu2dθ=4k∫α012dθ=2αk.
Thus, the angle of a lune with interior angle α is 2αk.
We remark that the lune with angle π actually covers the entire disk of radius 1√k. Thus, the area of the entire space P2k is 2πk, which matches half the surface area of a sphere of radius 1√k. We often call s=1√k the radius of curvature for the geometry; it is the radius of the disk on which we model the geometry.
Also, the integral computation in the proof of Lemma 7.2.1 reveals the following useful antiderivative:
∫4r(1+kr2)2dr=−2k(1+kr2)+C.
This fact may speed up future integral computations.
In elliptic geometry with curvature k, the area of a triangle with angles α,β, and γ is
A=1k(α+β+γ−π).
- Proof
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As in the case k=1, the area of any triangle may be determined from the area of three lunes and the total area of P2k, as depicted in Example 6.3.1.
Example 7.2.3: Triangles on the Earth.
The surface of the Earth is approximately spherical with radius about 6375 km. Therefore, the geometry on the surface of the Earth can be reasonably modeled by (P2k,Sk) where k=163752 km−2. The area of a triangle on the Earth's surface having angles α,β, and γ is
A=1k(α+β+γ−π).Can you find the area of the triangle formed by Paris, New York, and Rio? Use a globe, a protractor, and some string. The string follows a geodesic between two points when it is pulled taut.
Exercises
Prove that for k>0, any transformation in Sk has the form
T(z)=eiθz−z01+k¯z0z,
where θ is any real number and z0 is a point in P2k.
- Hint
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Follow the derivation of the transformations in S found in Chapter 6.
Verify the formula for the stereographic projection map ϕk.
Assume k>0 and let s=1√k. Derive the following measurement formulas in (P2k,Sk).
- The length of a line segment from 0 to x, where 0<x≤s is dk(0,x)=2sarctan(xs).
- The circumference of the circle centered at the origin with elliptic radius r<π(2√k) is C=2πssin(rs).
- The area of the circle centered at the origin with elliptic radius r<π(2√k) is A=4πs2sin2(r2s).
In this exercise we investigate the idea that the elliptic formulas in Exercise 7.2.3 for distance, circumference, and area approach Euclidean formulas when k→0+.
- Show that the elliptic distance dk(0,x) from 0 to x, where 0<x≤s, approaches 2x as k→0+ (twice the usual notion of Euclidean distance).
- Show that the elliptic circumference of a circle with elliptic radius r approaches 2πr as k→0+.
- Show that the elliptic area of this circle approaches πr2 as k→0+.
Triangle trigonometry in (P2k,Sk).
Suppose we have a triangle in (P2k,Sk) with side lengths a,b,c and angles α,β,γ as pictured in Figure 6.3.5.
- Prove the elliptic law of cosines in (P2k,Sk): cos(√kc)=cos(√ka)cos(√kb)+sin(√ka)sin(√kb)cos(γ).
- Prove the elliptic law of sines in (P2k,Sk): sin(√ka)sin(α)=sin(√kb)sin(β)=sin(√kc)sin(γ).