### ExercisesExercises

###### 1.Euclidean subgroup of the Möbius group.

Let $\E$ denote the subgroup of the Möbius group $\M$ generated by rotations and translations, that is, transformations of the type $z\to e^{it}z$ for $t\in \R$ and $z\to z+b$ for $b\in C\text{.}$ The geometry $(\C,\E)$ is sometimes called "Euclidean geometry". Is $(\C,\E)$ equivalent to the Euclidean geometry defined in Subsection 3.1.1? Why or why not?

###### 2.Elliptic geometry and spherical geometry.

Is elliptic geometry $(\extC,\S)$ equivalent to spherical geometry defined in Subsection 3.1.1? Why or why not?

###### 3.Parallel displacements in hyperbolic geometry.

Let $T$ be an element of the hyperbolic group $\H\text{,}$ with a single fixed point $p$ and normal form

\begin{equation*} \frac{1}{Tz-p} = \frac{1}{1-p}+\beta. \end{equation*}

Show that $p\beta$ must be a pure imaginary number, that is, there must be a real number $k$ such that $p\beta=ki\text{.}$

###### 4.
Prove that all elements of the elliptic group $\S$ are elliptic in the normal form sense, i.e., we have $|\alpha|=1$ in the normal form expression
\begin{equation*} \frac{Tz-p}{Tz-q}= \alpha \frac{z-p}{z-q}. \end{equation*}
Suggestion: First find the fixed points $p,q\text{,}$ then put $z=\infty$ in the normal form equation and solve for $\alpha\text{.}$
###### 5.Alternative derivation of the formula for elliptic group elements.

To obtain an explicit formula for elements of the elliptic group, we begin with a necessary condition. Let $R=s^{-1}\circ T\circ s$ be the rotation of $S^2$ that lifts $T$ via stereographic projection. If $P,Q$ are a pair of endpoints of a diameter of $S^2\text{,}$ then $R(P),R(Q)$ must also be a pair of endpoints of a diameter. Exercise 1.3.3.7 establishes the condition that two complex numbers $p,q$ are stereographic projections of endpoints of a diameter if and only if $pq^\ast = -1\text{.}$ Thus we have the following necessary condition for $T\text{.}$

\begin{equation} pq^\ast = -1 \;\;\text{ implies }\;\; Tp(Tq)^\ast = -1\label{diamendpreservecond}\tag{3.6.1} \end{equation}

Now suppose that $Tz=\frac{az+b}{cz+d}$ with $ad-bc=1\text{.}$ Solving the equation $Tp = \frac{-1}{(T(\frac{-1}{p^\ast}))^\ast}$ leads to $c=-b^\ast$ and $d=a^\ast\text{.}$ Thus we conclude that $T$ has the following form.

\begin{equation} Tz=\frac{az+b}{-b^\ast z+a^\ast},\;\;|a|^2+|b|^2=1\label{elliptictranssu2form}\tag{3.6.2} \end{equation}

Carry out the computation to derive (3.6.2). Explain why there is no loss of generality by assuming $ad-bc=1\text{.}$

###### 6.Identifications of $U(\Quat)$ and $\S$ with $\Rot(S^2)$.

The discussion of elliptic geometry (Section 3.4) establishes two ways to construct rotations from matrices. The purpose of this exercise is to reconcile these identifications. Given $a,b\in \C$ with $|a|^2+|b|^2=1\text{,}$ let us define the following objects, all parameterized by $a,b\text{.}$

\begin{align*} M_{a,b} \amp =\left[\begin{array}{cc}{a}\amp {b}\\{-b^\ast}\amp {a^\ast}\end{array}\right]\\ r_{a,b}\amp = \re(a)+\im(a)i+\re(b)j+\im(b)k \\ R_{r_{a,b}}\amp = \left[u\to r_{a,b}ur_{a,b}^\ast\right]\;\; \text{for }u\in S^2_\Quat\\ T_{a,b} \amp = \left[z\to \frac{az+b}{-b^\ast z+a^\ast}\right] \;\;\text{for }z\in \extC\\ R_{T_{a,b}} \amp = s^{-1}\circ T_{a,b}\circ s \end{align*}

The above objects are organized along two sequences of mappings. The rotation $R_{r_{a,b}}$ is at the end of the "quaternion path"

\begin{align} SU(2)\amp \to U(\Quat) \to \Rot(S^2_\Quat)\label{quatpath}\tag{3.6.3}\\ M_{a,b} \amp \to r_{a,b} \to R_{r_{a,b}}\notag \end{align}

and the rotation $R_{T_{a,b}}$ is at the end of the "Möbius path"

\begin{align} SU(2)\amp \to \S \to \Rot(S^2)\label{mobiuspath}\tag{3.6.4}\\ M_{a,b} \amp \to T_{a,b} \to R_{T_{a,b}}.\notag \end{align}

This problem is about comparing the rotations $R_{r_{a,b}}$ and $R_{T_{a,b}}$ (see Table 3.6.1) and reconciling the difference. The angles of rotation are the same, but the axes are different, but only by a reordering of coordinates and a minus sign.

The exercises outlined below verify the values for $v,\theta$ in Table 3.6.1.

1. Use Proposition 1.2.9 to justify the values for $v$ and $\theta$ for $R_{r_{a,b}}$ in Table 3.6.1.
2. Solve $T_{a,b}z=z$ to show that one of the fixed points of $T_{a,b}$ is
\begin{equation*} p=-ib\left( \frac{\sqrt{1-(\re(a))^2} + \im(a)}{|b|^2}\right). \end{equation*}
3. Show that $s\left(\frac{(\im(b),-\re(b),\im(a))}{\sqrt{1-(\re(a))^2}}\right)=p\text{.}$
4. Show that
\begin{equation} T_{a,b} = s\circ h\circ R_{r_{a,b}}\circ h \circ s^{-1} \label{trhcommute}\tag{3.6.5} \end{equation}
where $h\colon \R^3\to \R^3$ is given by $(x,y,z)\to (z,-y,z)\text{.}$ Here's one way to do this: evaluate both sides of (3.6.5) on the three points $p,0,\infty\text{.}$ Explain why this is sufficient! Use quaternion multiplication to evaluate $R_{r_{a,b}}\text{.}$ For example, $R_{r_{a,b}}(1,0,0)=r_{a,b}ir_{a,b}^\ast$ under the natural identification $\R^3\leftrightarrow \R^3_\Quat\text{.}$
5. Here is one way to reconcile the quaternion path (3.6.3) with the Möbius path (3.6.4). Let $H=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1\amp 1 \\ 1\amp -1\end{array}\right]$ (the matrix $H$ is sometimes called the Hadamard matrix) and let $C_{iH}$ denote the map $M\to (iH)M(iH)^{-1}\text{.}$ Show that the diagram in Figure 3.6.2 commutes. Hint: Notice that $iH\in SU(2)$ and that $Q(iH)=\frac{1}{\sqrt{2}}(i+k)\text{,}$ and that $R_{Q(iH)}=h\text{.}$ Figure 3.6.2. The map $C_{iH}$ is given by $C_{iH}(T)=(iH)T(iH)^{-1}\text{.}$ The column of maps on the left is the "Möbius path", and the column of maps on the right is the "quaternion path". Figure 3.6.3. Right triangle $\triangle ABC$
###### 7.Pythagorean Theorems.

Let $\triangle ABC$ be a right triangle with right angle $\angle C$ with side lengths $a=d(B,C)\text{,}$ $b=d(A,C)\text{,}$ and $c=d(A,B)$ so that the length of the hypotenuse is $c\text{.}$ See Figure 3.6.3.

1. Prove the following identities.
\begin{align} \cosh \left(\ln\left(\frac{1+u}{1-u}\right)\right) \amp = \frac{1+u^2}{1-u^2}\;\;(0\lt u\lt 1)\label{coshhyperbolicdistidentity}\tag{3.6.6}\\ \cos(2\arctan u) \amp = \frac{1-u^2}{1+u^2}\label{cosellipticistidentity}\tag{3.6.7} \end{align}
2. The Hyperbolic Pythagorean Theorem. Show that
\begin{equation} \cosh(c)=\cosh(a)\cosh(b)\label{hyperbolicpythagthm}\tag{3.6.8} \end{equation}
if $T$ is a hyperbolic triangle.
3. The Elliptic Pythagorean Theorem. Show that
\begin{equation} \cos(c)=\cos(a)\cos(b)\label{ellipticpythagthm}\tag{3.6.9} \end{equation}
if $T$ is an elliptic triangle.

Suggestion: Use a transformation to place $C$ at $0$ in $\D$ or $\extC\text{,}$ with $A$ is real and $B$ pure imaginary. Use the formula $d(p,q) = \ln((1+u)/(1-u))$ with $u=\left|\frac{q-p}{1-p^\ast q}\right|$ for hyperbolic distance. Use the formula $d(p,q) = 2\arctan(u)\text{,}$ with $u=\left|\frac{q-p}{1+p^\ast q}\right|$ for elliptic distance. The identities from part (a) will be useful.