## Section 3.4 Elliptic geometry

Elliptic geometry is the geometry of the sphere (the 2-dimensional surface of a 3-dimensional solid ball), where congruence transformations are the rotations of the sphere about its center.

We will work with three models for elliptic geometry: one based on quaternions, one based on rotations of the sphere, and another that is a subgeometry of Möbius geometry. Using the natural identification \(xi+yj+zk\leftrightarrow (x,y,z)\) of the pure quaternions \(\R^3_\Quat\) with \(\R^3\text{,}\) we will write \(S_\Quat^2\) to denote the set of unit pure quaternions.

\begin{equation*}
S^2_\Quat=\{xi+yj+zk\in \Quat\colon x^2+y^2+z^2=1\}
\end{equation*}

We begin by establishing some basic facts about the relevant transformation groups.

### Subsection 3.4.1 The group of unit quaternions

Recall from Section 1.2 that \(U(\Quat)\) is the set of quaternions of modulus 1. In fact, \(U(\Quat)\) is a group.

####
Checkpoint 3.4.1.

Recall that the map \(M\colon \Quat \to
{\mathcal M}_{\Quat}\) sends \(r=a+bi+cj+dk\) to the matrix

\begin{equation*}
\left[\begin{array}{cc} a+bi \amp c+di\\ -c+di \amp
a-bi\end{array}\right]\text{.}
\end{equation*}

The image of \(U(\Quat)\) under \(M\) is the matrix group \(SU(2)\) , called the

*special unitary*group.
\begin{equation}
SU(2)= \left\{\left[\begin{array}{cc} a \amp b\\ -b^\ast \amp
a^\ast\end{array}\right]\colon a,b\in \C,
|a|^2+|b|^2=1\right\}\tag{3.4.1}
\end{equation}

Restricting the domain of \(M\) to \(U(\Quat)\) and restricting the codomain of \(M\) to \(M(U(\Quat))=SU(2)\) is an isomorphism of groups

\begin{equation*}
U(\Quat)\approx SU(2).
\end{equation*}

####
Checkpoint 3.4.2.

## Hint.

It is not necessary to perform any new computation to show that \(M\) is a homomorphism. Instead, use (1.2.5).

The action of a unit quaternion as a rotation on \(\R^3_\Quat\) (see Proposition 1.2.9) takes the sphere \(S^2_\Quat\) to itself. The action of the group \(U(\Quat)\) on \(S^2_\Quat\) defines a model of elliptic geometry.

#### Definition 3.4.3.

The quaternion model of elliptic geometry is \((S^2_\Quat,U(\Quat))\text{.}\)####
Checkpoint 3.4.4.

\begin{equation*}
r \to \left[ v\to rvr^\ast \right]
\end{equation*}

is a group action.### Subsection 3.4.2 The group of rotations of the 2-sphere

Let \(R_{{v},\theta}\) denote the rotation of \(\R^3\) about the axis given by a unit vector \({v}\) by an angle \(\theta\text{.}\) We use the standard orientation, so that a positive value of \(\theta\) is a counterclockwise rotation of the plane orthogonal to \({v}\text{,}\) as viewed from "above" where \({v}\) points in the "up" direction. See Figure 3.4.5.

*Notation convention:*For readability and convenience, we write \(R_{X,\theta},R_{Y,\theta},R_{Z,\theta}\) to denote rotations by \(\theta\) radians about the standard basis vectors \((1,0,0), (0,1,0), (0,0,1)\text{,}\) respectively.

We will write \(\Rot(S^2)\) to denote the set of all rotations.

\begin{equation*}
\Rot(S^2)=\{R_{{v},\theta}\colon
{v}\in\R^3,|{v}|=1,\theta\in\R\}
\end{equation*}

To see why the set \(\Rot(S^2)\) is a group

under the operation of composition, consider the map \(U(\Quat)\to \Rot(S^2)\) given by \(r\to R_r\) established by Proposition 1.2.9. The fact that \(R_r\circ R_s = R_{rs}\) (see Exercise 1.2.6.2) implies that the composition of two rotations is a rotation. The remaining group properties are straightforward. Once we have proved that \(\Rot(S^2)\) is a group, the same equation \(R_r\circ R_s = R_{rs}\) shows that the map \(r\to R_r\) is a homomorphism of groups \(U(\Quat)\to \Rot(S^2)\text{.}\) The kernel of this homomorphism is \(\{\pm 1\}\text{.}\) This establishes an isomorphism

^{ 1 }

For the purpose of a self-contained exposition based on elementary geometry, using only complex and quaternion algebra, we do not utilize the fact that \(\Rot(S^2)\) is the group \(SO(3)\) of orthogonal transformations of \(\R^3\) with determinant 1.

\begin{equation*}
U(\Quat)/\{\pm 1\} \approx \Rot(S^2).
\end{equation*}

####
Checkpoint 3.4.6.

## Hint.

Use Proposition 1.2.9.

#### Definition 3.4.7.

The spherical model of elliptic geometry is \((S^2,\Rot(S^2))\text{.}\)We conclude with a useful fact about constructing arbitrary rotations by composing rotations from a specific set elementary types, namely, rotations about the \(z\)-axis by arbitrary angles, and rotations about the \(x\)-axis by \(\pi/2\) radians. We start with a Lemma that shows how to do this for \(y\)-axis rotations.

#### Lemma 3.4.8. Rotations about the \(y\)-axis.

An arbitrary rotation about the \(y\)-axis is a composition of a rotations about the \(x\)-axis by \(\pi/2\) radians with a rotation about the \(z\)-axis. Specifically, we have the following.
\begin{equation}
R_{Y,\theta}= R_{X,\pi/2}^{-1}\circ R_{Z,\theta}\circ R_{X,\pi/2}\tag{3.4.2}
\end{equation}

#### Proof.

#### Proposition 3.4.9. Generators for \(\Rot(S^2)\).

The set
\begin{equation*}
\{R_{Z,\theta}\colon \theta\in
\R\}\cup \{R_{X,\pi/2}\}
\end{equation*}

is a generating set for \(\Rot(S^2)\text{.}\) This means that any rotation may be written as a composition of rotations about the \(z\)-axis and rotations about the \(x\)-axis by \(\pi/2\) radians.#### Proof.

\begin{equation*}
R,R_{Z,\theta_1},R_{Y,\theta_2},R_{Z,\theta_3}
\end{equation*}

takes the north pole \(N\) through the sequence
\begin{equation*}
N\to R(N)\to N' \to N''=N\to N\text{.}
\end{equation*}

Meanwhile, \(G\) traces the path
\begin{equation*}
G\to R(G)\to
G'\to G''\to G
\end{equation*}

while the great circle \(C\) is transformed in the sequence
\begin{equation*}
C\to R(C) \to C' \to C''\to C.
\end{equation*}

This leads to the decomposition \(R=R_{Z,-\theta_1}\circ R_{Y,-\theta_2}\circ R_{Z,-\theta_3}\text{.}\) ### Subsection 3.4.3 The elliptic subgroup of the Möbius group

Let \(\S\) denote the group of transformations of \(\extC\) that are conjugate to rotations of \(S^2\) via the stereographic projection.

\begin{equation*}
\S=\{s\circ R\circ s^{-1} \;\;\colon \;\;R\in
\Rot(S^2)\}
\end{equation*}

The group \(\S\) is called the elliptic group. It is easy to check that the map \(\Rot(S^2)\to
\S\) given by \(R\to s\circ R\circ s^{-1}\) is an isomorphism of groups, so we have

\begin{equation*}
\S\approx \Rot(S^2).
\end{equation*}

####
Checkpoint 3.4.11.

Exercises in Section 1.3 show that \(s\circ
R_{Z,\theta}\circ s^{-1}\) the Möbius transformation \(T_{Z,\theta}\) given by \(z\to e^{i\theta}z\) (see Exercise 1.3.3.4) and that \(s\circ
R_{X,\pi/2}\circ s^{-1}\) is the Möbius transformation \(T_{X,\pi/2}\) given by \(z\to \frac{z+i}{iz+1}\) (see Exercise 1.3.3.6). The fact (Proposition 3.4.9) that rotations of the form \(R_{Z,\theta},R_{X,\pi/2}\) generate \(\Rot(S^2)\) implies that the Möbius transformations \(T_{Z,\theta}\) and \(T_{X,\pi/2}\) generate \(\S\text{.}\) Therefore \(\S\) is in fact a subgroup of the Möbius group.

#### Definition 3.4.12.

The Möbius subgeometry model of elliptic geometry is \((\extC,\S)\text{.}\)We can say more about the specific form of elements in \(\S\) in terms of the group homomorphism \({\mathcal
T}\colon GL(2,\C)\to \M\) that sends the matrix \(\left[\begin{array}{cc} a\amp b\\ c\amp
d\end{array}\right]\) to the Möbius transformation \(z\to \frac{az+b}{cz+d}\) (see (3.2.2)). Observe that the transformations

\begin{align}
T_{Z,\theta} \amp= {\mathcal T}\left(\left[\begin{array}{cc}
e^{i\theta/2} \amp 0\\ 0 \amp
e^{-i\theta/2}\end{array}\right]\right)\tag{3.4.3}\\
T_{X,\pi/2} \amp= {\mathcal T}\left(\left[\begin{array}{cc}
\frac{1}{\sqrt{2}}\amp \frac{i}{\sqrt{2}}\\
\frac{i}{\sqrt{2}}\amp \frac{1}{\sqrt{2}}\end{array}\right]\right)\tag{3.4.4}
\end{align}

are images of elements of the group \(SU(2)\) (see (3.4.1)). Because \(T_{Z,\theta},T_{X,\pi/2}\) generate \(\S\text{,}\) it follows that every element of \(\S\) is the image under \({\mathcal T}\) of an element of \(SU(2)\text{.}\)

####
Checkpoint 3.4.13.

Thus we have proved the following explicit formula for elements of \(\S\text{.}\)

#### Proposition 3.4.14. Formula for transformations in the elliptic group.

A map \(T\) is an element of \(\S\) if and only if \(T\) may be written in the form \(Tz=\frac{az+b}{-b^\ast z+a^\ast}\) for some \(a,b\in \C\) with \(|a|^2+|b|^2=1\text{.}\)

Setting \(e^{i\theta}=\frac{a}{a^\ast}\) and \(z_0=\frac{-b}{a}\text{,}\) we have the following alternative form for \(T\in \S\text{.}\)

\begin{equation}
Tz=e^{i\theta}\frac{z-z_0}{z_0^\ast z+1}\tag{3.4.5}
\end{equation}

### Subsection 3.4.4 Circles in \(S^2\) and clines in \(\extC\)

A

*circle*in \(S^2\) is a circle in a plane intersecting \(S^2\text{.}\) A great circle is the intersection of \(S^2\) with a plane through the origin. In elliptic geometry, a great circle is called an elliptic straight line because the path of shortest length connecting two given points in \(S^2\) is an arc of a great circle. Circles in \(S^2\) that are not great circles are called elliptic cycles. Elliptic straight lines and elliptic cycles in the Möbius subgeometry model \((\extC,\S)\) are stereographic projections of elliptic straight lines and elliptic cycles in the spherical model. It turns out that elliptic straight lines and elliptic cycles in \(\extC\) are in fact clines. Here is the statement and proof.#### Proposition 3.4.15. Stereographic projection takes circles to clines.

Let \(C\) be a circle that is the intersection of \(S^2\) with a plane in \(\R^3\text{.}\) If \(C\) contains the north pole \((0,0,1)\) of \(S^2\text{,}\) then \(s(C\setminus\{(0,0,1)\})\) is a Euclidean straight line in \(\C\text{.}\) Otherwise, \(s(C)\) is a circle in \(\C\text{.}\)#### Proof.

### Subsection 3.4.5 Angles and orientation on \(S^2\)

The standard orientation for angles on \(S^2\) (see Subsection 3.4.2) is also called the

*outward-pointing normal*orientation. The standard orientation measures angles from the viewpoint of an observer standing on the*outside*of the sphere. The*inward-pointing normal*orientation is the reverse orientation that measures angles from the viewpoint of an observer walking on the inside of the sphere. See Figure 3.4.16.#### Corollary 3.4.17. Stereographic projection is conformal.

Stereographic projection preserves oriented angles with respect to the inward-pointing normal orientation.

#### Proof.

### Subsection 3.4.6 Elliptic length and area

The distance between points \(P,Q\) on \(S^2\) is the length of the arc of a great circle that connects them. Because the sphere has radius 1, the arc length is the same as the radian measure of the angle \(\angle
POQ\text{,}\) where \(O\) is the origin. From vector calculus, we have the following dot product formula.

\begin{equation*}
(\overrightarrow{OP})\cdot (\overrightarrow{OQ})=|\overrightarrow{OP}||\overrightarrow{OQ}|\cos(\angle
POQ)
\end{equation*}

Solving for \(\cos(\angle POQ)\text{,}\) we obtain the formula for the distance \(d_{S^2}(P,Q)\) between points \(P,Q\) in \(S^2\text{.}\)

\begin{equation}
d_{S^2}(P,Q)=\cos^{-1}\left((\overrightarrow{OP})\cdot (\overrightarrow{OQ})\right)\tag{3.4.6}
\end{equation}

To "transfer" the metric (3.4.6) to \(\extC\) by stereographic projection means that we define the elliptic metric \(d_{\extC}\) on \(\extC\) by the following.

\begin{equation}
d_{\extC}(p,q):=d_{S^2}(s^{-1}(p),s^{-1}(q))\tag{3.4.7}
\end{equation}

#### Proposition 3.4.19.

The elliptic metric (3.4.7) is invariant under the action of the elliptic group. That is, we have

\begin{equation}
d_{\extC}(p,q) = d_{\extC}(Tp,Tq)\tag{3.4.8}
\end{equation}

for all \(p,q\in \extC\) and for all \(T\in\S\text{.}\)

In order to obtain a formula for computing \(d_{\extC}(p,q)\text{,}\) we follow the same procedure for hyperbolic distance. First, we find the distance \(d_{\extC}(0,u)\text{,}\) where \(0\leq u\leq
1\text{.}\) Let \(S=(0,0,-1)=s^{-1}(0)\) and let \(U=s^{-1}(u)\text{.}\) Let \(0=(0,0,0)\text{,}\) let \(N=(0,0,1)\text{,}\) let \(\alpha =
\angle SNU\) and let \(\theta = \angle SOU\) (see Figure 3.4.20). It is a simple exercise to show that \(\alpha = \theta/2\text{,}\) so that we have

\begin{equation}
d_{\extC}(0,u)=d_{S^2}(S,U)=\theta = 2\alpha=2\arctan u. \tag{3.4.9}
\end{equation}

####
Checkpoint 3.4.21.

For the general case, let \(z_1,z_2\in \extC\text{.}\) Because and let \(Tz=e^{it}\frac{z-z_1}{z_1^\ast z+1}\) be the transformation in \(\S\) (using the form (3.4.5)) that sends \(z_1\to 0\) and \(z_2\to u\geq 0\text{.}\) Applying (3.4.9), we have the elliptic distance formula in \(\extC\text{.}\)

\begin{equation}
d_{\extC}(z_1,z_2)=2\arctan
\left|\frac{z_2-z_1}{z_1^\ast z_2 + 1}\right|\tag{3.4.10}
\end{equation}

Now let \(\gamma\) be a parametric curve \(z(t)=x(t)+iy(t)\) in \(\extC\text{.}\) Using the same argument as in the paragraph preceding the hyperbolic length integral formula (3.3.6), using the first order Taylor approximation \(\arctan u \approx u\) and making the appropriate changes, we arrive at the elliptic length integral formula.

\begin{equation}
\text{Length}(\gamma)=2\int_a^b \frac{|z'(t)|}{1+|z(t)|^2}\;dt\tag{3.4.11}
\end{equation}

####
Checkpoint 3.4.22.

####
Checkpoint 3.4.23.

Find the length of the elliptic cycle parameterized by \(z(t) = \alpha e^{it}\) for \(0\leq t\leq 2\pi\text{,}\) where \(0\lt \alpha\leq 1\) .

Using the same argument as in the paragraph preceding the hyperbolic area integral formula (3.3.10), using the elliptic length differential \(ds=\frac{2|z'(t)|\;dt}{1-|z(t)|^2}\) in place of the hyperbolic length differential and making the appropriate changes, we obtain the elliptic area integral formula.

\begin{equation}
\text{Area}(R)=\iint_R
dA = \iint_R \frac{4r\;dr\;d\theta}{(1+r^2)^2}.\tag{3.4.12}
\end{equation}

####
Checkpoint 3.4.24.

Find the area of the elliptic disk \(\{|z|\leq
\alpha\}\text{,}\) for \(0\lt \alpha\lt 1\text{.}\)

### Exercises 3.4.7 Exercises

#### 1.

Show that the three models of elliptic geometry are equivalent.#### 2.

Prove Proposition 3.4.19.

#### Area of an elliptic triangle..

The following sequence of exercises establishes the area formula for elliptic triangles.

##### 3. Area of an elliptic 2-gon.

An elliptic 2-gon is a figure with two vertices connected by two elliptic line segments. In \(\extC\text{,}\) any 2-gon is congruent to a set of the form \(R_{\alpha} :=\{z\in \C\colon
0\leq \arg z \leq \alpha\}\cup \{\infty\}\) for some \(\alpha\) in the range \(0\leq \alpha\lt
2\pi\text{.}\) See Figure 3.4.25. Use an area integral in \(\extC\) to show that the area of \(R_{\alpha}\) is \(2\alpha\text{.}\) Verify that this is the right answer using a picture of \(S^2\text{.}\)

##### 4. Area of an elliptic triangle.

Let \(\triangle ABC\) be an elliptic triangle. Let \({\mathcal C}_{AB},{\mathcal
C}_{AC},{\mathcal C}_{BC}\) denote the great circles that extend the sides \(AB,AC,BC\) of the triangle. See Figure 3.4.26.

- Explain why \(A,A'\) are endpoints of the same diameter, that is, endpoints of a diameter of \(S^2\) or their stereographic projections in \(\extC\text{.}\)
- Explain why \(\triangle A'B'C'\) has the same area as \(\triangle ABC\text{,}\) even though the two triangles are not necessarily congruent! (Note that the interior of \(\triangle A'B'C'\) is the
*exterior*of the three great circles, that is, on the side that contains the point \(\infty\text{.}\)) Hint: What does part (a) of this problem imply about the relationship between points \(X\) and \(X'\) on \(S^2\) for \(X=A,B,C\text{?}\) - Let \(R\) denote the interior of the region shown in the figure on the right in Figure 3.4.26. Explain why the area of \(R\) is\begin{equation*} 2\angle A + 2\angle B + 2\angle C - 2\text{ Area}(\triangle ABC). \end{equation*}Suggestion: Decompose \(R\) using overlapping 2-gons.
- Let \(R'\) denote the exterior of the region \(R\text{,}\) that is, \(R'=\extC\setminus R\text{.}\) Explain why the area of \(R'\) is\begin{equation*} 2\angle A + 2\angle B + 2\angle C - 2\text{ Area}(\triangle A'B'C'). \end{equation*}Suggestion: Decompose \(R'\) into overlapping 2-gons.
- Explain why the area of elliptic triangle \(\triangle ABC\) is\begin{equation} \text{Area}(\triangle ABC)= (\angle A +\angle B + \angle C)-\pi.\tag{3.4.13} \end{equation}