##### Lemma7.2.1

Assume \(k \gt 0\text{.}\) A lune in \((\mathbb{P}^2_k,{\cal S}_k)\) with interior angle \(\alpha\) has area \(2\alpha/k\text{.}\)

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 \(k \gt 0\text{,}\) we may construct a sphere with constant curvature \(k\text{.}\) According to Example 7.1.5, the sphere centered at the origin with radius \(1/\sqrt{k}\) 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 \(k > 0\text{.}\)

Consider the sphere \(\mathbb{S}^2_k\) centered at the origin of \(\mathbb{R}^3\) with radius \(1/\sqrt{k}\text{.}\) Define stereographic projection \(\phi_k: \mathbb{S}^2_k \to \mathbb{C}^+\) - just as we did in Section 3.3 - to obtain the formula \begin{equation*} \phi_k(a,b,c) = \begin{cases}\frac{a}{1-c\sqrt{k}}+\frac{b}{1-c\sqrt{k}}i \amp \text{ if \(c \neq \frac{1}{\sqrt{k}}\); } \\ \infty \amp \text{ if \(c=\frac{1}{\sqrt{k}}\). } \end{cases} \end{equation*}

Diametrically opposed points on \(\mathbb{S}^2_k\) get mapped via \(\phi_k\) to points \(z\) and \(z_a\) that satisfy the equation \(z_a =\frac{-1}{k\overline{z}}\text{,}\) by analogy with Lemma 6.1.2. We call two such points in \(\mathbb{C}^+\) antipodal with respect to \(\mathbb{S}^2_k\text{,}\) or just antipodal points if the value of \(k\) is understood.

Our model for elliptic geometry with curvature \(k\) has space \(\mathbb{P}^2_k\) equal to the closed disk in \(\mathbb{C}\) of radius \(1/\sqrt{k}\text{,}\) 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 \({\cal S}_k\text{,}\) consists of those Möbius transformations that preserve antipodal points with respect to \(\mathbb{S}^2_k\text{.}\) That is, \(T \in {\cal S}_k\) if and only if the following holds: \begin{equation*} \text{if}~ z_a = -\frac{1}{k\overline{z}}~~ \text{then}~~ T(z_a) = -\frac{1}{k\overline{T(z)}}. \end{equation*}

The geometry \((\mathbb{P}^2_k,{\cal S}_k)\) with \(k \gt 0\) is called *elliptic geometry with curvature \(k\)*. Note that \((\mathbb{P}^2_1,{\cal S}_1)\) is precisely the geometry we studied in Chapter 6.

The transformations of \(\mathbb{C}^+\) in the group \({\cal S}_k\) correspond precisely with rotations of the sphere \(\mathbb{S}^2_k\text{.}\) One can show that transformations in \({\cal S}_k\) have the form \begin{equation*} T(z) = e^{i\theta}\frac{z-z_0}{1+k\overline{z_0}z}. \end{equation*}

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 \(z_a = -\frac{1}{k\overline{z}}\text{.}\) These lines correspond precisely to great circles on the sphere \(\mathbb{S}^2_k\text{.}\)

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 \(\vec{r}\) in \(\mathbb{P}^2_k\) is \begin{equation*} {\cal L}(\vec{r}) = \int_a^b \frac{2|\vec{r}\hskip.04in^\prime(t)|}{1 + k|\vec{r}(t)|^2}~dt. \end{equation*}

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 \((\mathbb{P}^2_k,{\cal S}_k)\) turns out to be \(\pi/(2\sqrt{k})\text{.}\)

The area of a region \(R\) given in polar form is computed by the formula \begin{equation*} A(R) = \iint_R \frac{4r}{(1+kr^2)^2}dr d\theta. \end{equation*}

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 \((\mathbb{P}^2_k,{\cal S}_k)\text{.}\)

Assume \(k \gt 0\text{.}\) A lune in \((\mathbb{P}^2_k,{\cal S}_k)\) with interior angle \(\alpha\) has area \(2\alpha/k\text{.}\)

We remark that the lune with angle \(\pi\) actually covers the entire disk of radius \(1/\sqrt{k}\text{.}\) Thus, the area of the entire space \(\mathbb{P}^2_k\) is \(2\pi/k\text{,}\) which matches half the surface area of a sphere of radius \(1/\sqrt{k}\text{.}\) We often call \(s =
1/\sqrt{k}\) 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 (as a function of \(r\)): \begin{equation*} \int \frac{4r}{(1+kr^2)^2}dr = \frac{-2}{k(1+kr^2)} + C. \end{equation*} This fact may speed up future integral computations.

In elliptic geometry with curvature \(k\text{,}\) the area of a triangle with angles \(\alpha, \beta,\) and \(\gamma\) is \begin{equation*} A = \frac{1}{k}(\alpha+\beta+\gamma -\pi). \end{equation*}

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 \((\mathbb{P}^2_k, {\cal S}_k)\) where \(k = 1/6375^2~ \text{km}^{-2}\text{.}\) The area of a triangle on the Earth's surface having angles \(\alpha, \beta,\) and \(\gamma\) is \begin{equation*} A = \frac{1}{k}(\alpha + \beta + \gamma - \pi). \end{equation*} 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.

Prove that for \(k > 0\text{,}\) any transformation in \({\cal S}_k\) has the form \begin{equation*} T(z) = e^{i\theta}\frac{z-z_0}{1+k\overline{z_0}z}, \end{equation*} where \(\theta\) is any real number and \(z_0\) is a point in \(\mathbb{P}^2_k\text{.}\) Hint: Follow the derivation of the transformations in \({\cal S}\) found in Chapter 6.

Verify the formula for the stereographic projection map \(\phi_k\text{.}\)

Assume \(k \gt 0\) and let \(s = 1/\sqrt{k}\text{.}\) Derive the following measurement formulas in \((\mathbb{P}^2_k,{\cal S}_k)\text{.}\)

a. The length of a line segment from \(0\) to \(x\text{,}\) where \(0 \lt x \leq s\) is \begin{equation*} d_{k}(0,x) = 2s\arctan(x/s). \end{equation*}

b. The circumference of the circle centered at the origin with elliptic radius \(r \lt \pi/(2\sqrt{k})\) is \(C=2\pi s\sin(r/s).\)

c. The area of the circle centered at the origin with elliptic radius \(r \lt \pi/(2\sqrt{k})\) is \(\displaystyle A = 4\pi s^2 \sin^2\bigg(\frac{r}{2s}\bigg).\)

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 \(k \to 0^+\text{.}\)

a. Show that the elliptic distance \(d_{k}(0,x)\) from \(0\) to \(x\text{,}\) where \(0 \lt x \leq s,\) approaches \(2x\) as \(k \to 0^+\) (twice the usual notion of Euclidean distance).

b. Show that the elliptic circumference of a circle with elliptic radius \(r\) approaches \(2\pi r\) as \(k \to 0^+\text{.}\)

c. Show that the elliptic area of this circle approaches \(\pi r^2\) as \(k \to 0^+\text{.}\)

*Triangle trigonometry in \((\mathbb{P}^2_k,{\cal S}_k)\text{.}\)*

Suppose we have a triangle in \((\mathbb{P}^2_k,{\cal S}_k)\) with side lengths \(a,b,c\) and angles \(\alpha, \beta, \gamma\) as pictured in Figure 6.3.14.

a. Prove the elliptic law of cosines in \((\mathbb{P}^2_k,{\cal S}_k)\text{:}\) \begin{equation*} \cos(\sqrt{k}c)=\cos(\sqrt{k}a)\cos(\sqrt{k}b)+\sin(\sqrt{k}a)\sin(\sqrt{k}b)\cos(\gamma). \end{equation*}

b. Prove the elliptic law of sines in \((\mathbb{P}^2_k,{\cal S}_k)\text{:}\) \begin{equation*} \frac{\sin(\sqrt{k}a)}{\sin(\alpha)}=\frac{\sin(\sqrt{k}b)}{\sin(\beta)}=\frac{\sin(\sqrt{k}c)}{\sin(\gamma)}. \end{equation*}