## Section7.2Elliptic Geometry with Curvature $k \gt 0$

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 $\phi_k: \mathbb{S}^2_k \to \mathbb{C}^+\text{,}$ 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 $\boldsymbol{r}$ in $\mathbb{P}^2_k$ is

\begin{equation*} {\cal L}(\boldsymbol{r}) = \int_a^b \frac{2|\boldsymbol{r}^\prime(t)|}{1 + k|\boldsymbol{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 $(\mathbb{P}^2_k,{\cal S}_k)\text{.}$

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 $\infty\text{,}$ 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 6.3.10. To compute the area of the lune, compute the integral

\begin{equation*} A = 2 \int_0^\alpha \int_0^{1/\sqrt{k}}\frac{4r}{(1+kr^2)^2}drd\theta. \end{equation*}

Letting $u = 1 + kr^2$ so that $du = 2krdr\text{,}$ the bounds of integration change from $[0,1/\sqrt{k}]$ to $[1,2]\text{.}$ Then,

\begin{equation*} A = 2 \int^\alpha_0 \frac{2}{k}\int_1^2 \frac{du}{u^2}d\theta = \frac{4}{k}\int_0^\alpha \frac{1}{2} d\theta =\frac{2\alpha}{k}. \end{equation*}

Thus, the angle of a lune with interior angle $\alpha$ is $2\alpha/k\text{.}$

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:

\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.

As in the case $k = 1\text{,}$ the area of any triangle may be determined from the area of three lunes and the total area of $\mathbb{P}^2_k\text{,}$ as depicted in Example 6.3.11.

###### Example7.2.3Triangles 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 $(\mathbb{P}^2_k, {\cal S}_k)$ 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.

### SubsectionExercises

###### 1

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.

###### 2

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

###### 3

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 $\displaystyle A = 4\pi s^2 \sin^2\bigg(\frac{r}{2s}\bigg).$

###### 4

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 $k \to 0^+\text{.}$

###### 5

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 6.3.13.

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*}