 # Introduction to Groups and Geometries

### 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\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\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)\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)\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} \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)\tag{3.6.6}\\ \cos(2\arctan u) \amp = \frac{1-u^2}{1+u^2}\tag{3.6.7} \end{align}
2. The Hyperbolic Pythagorean Theorem. Show that
\begin{equation} \cosh(c)=\cosh(a)\cosh(b)\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)\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.