# Introduction to Groups and Geometries

## Section3.2Möbius geometry

Möbius geometry provides a unifying framework for studying planar geometries. In particular, the transformation groups of hyperbolic and elliptic geometries in the sections that follow are subgroups of the group of Möbius transformations.

### Subsection3.2.1Möbius transformations

A Möbius transformation (also called a linear fractional transformation) is a function of the form
$$f(z)= \frac{az+b}{cz+d}\tag{3.2.1}$$
where $$z$$ is a complex variable, $$a,b,c,d$$ are complex constants, and $$ad-bc\neq 0\text{.}$$

#### Checkpoint3.2.1.

1. Show that if $$c=0\text{,}$$ then (3.2.1) defines a one-to-one and onto map $$f\colon \C\to \C$$ by finding an explicit inverse $$g$$ for $$f$$ and showing that $$f\circ g$$ and $$g\circ f$$ are both the identity function on $$\C\text{.}$$
2. Show that if $$c\neq 0\text{,}$$ then (3.2.1) defines a one-to-one and onto map $$f\colon \C\setminus \{-d/c\}\to \C\setminus\{a/c\}$$ by finding an explicit inverse $$g$$ for $$f$$ and showing that $$f\circ g$$ is the identity function on $$\C\setminus\{a/c\}$$ and and $$g\circ f$$ is the identity function on $$\C\setminus \{-d/c\}\text{.}$$
3. For both cases 1 and 2 above, identify precisely where you use the condition $$ad-bc\neq 0\text{.}$$ Show that, in fact, $$f$$ is invertible if and only if $$ad-bc\neq 0\text{.}$$
A Möbius transformation $$f(z)=\frac{az+b}{cz+d}$$ determines a one-to-one and onto map of the extended complex plane $$\extC$$ to itself with the following assignments: if $$c=0\text{,}$$ we define $$f(\infty)=\infty\text{;}$$ if $$c\neq 0\text{,}$$ we define $$f(-d/c)=\infty$$ and $$f(\infty)=a/c\text{.}$$

#### Checkpoint3.2.2.

Show that the composition of two Möbius transformations is a Möbius transformation. Suggestion: First show that the composition has the form $$z\to \frac{rz+s}{tz+u}\text{.}$$ Next, instead of a brute force calculation to check that $$ru-ts\neq 0\text{,}$$ use Checkpoint 3.2.1.

#### Definition3.2.3.

The definitions and Checkpoint exercises above show that the set of all Möbius transformations forms a group under the operation of composition of functions. This group, denoted $$\M$$, is called the Möbius group. Möbius geometry is the pair $$(\extC,\M)\text{.}$$
Note on notational convention: It is customary to use capital letters such as $$S,T,U$$ to denote Möbius transformations. It is also customary to omit the parentheses, and to write $$Tz$$ instead of $$T(z)$$ to denote the value of a Möbius transformation.
There is a natural relationship between Möbius group operations and matrix group operations. The map $${\mathcal T}\colon GL(2,\C)\to \M$$ given by
$$\left[ \begin{array}{cc}a\amp b\\ c\amp d\end{array}\right]\to \left[z\to \frac{az+b}{cz+d}\right]\tag{3.2.2}$$
is a group homomorphism. The kernel of $${\mathcal T}$$ is the group of nonzero scalar matrices.
\begin{equation*} \ker{\mathcal T}=\left\{\left[ \begin{array}{cc}k\amp 0\\ 0\amp k\end{array}\right], k\neq 0\right\}\text{.} \end{equation*}
The quotient group $$GL(2,\C)/\ker{\mathcal T}$$ is called the projective linear group $$PGL(2,\C)\text{.}$$ Thus we have a group isomorphism
\begin{equation*} PGL(2,\C)\approx \M\text{.} \end{equation*}

#### Checkpoint3.2.4.

Show that the map $${\mathcal T}$$ is a group homomorphism. Show that the kernel of $${\mathcal T}$$ is
\begin{equation*} \ker{\mathcal T}=\left\{\left[ \begin{array}{cc}k\amp 0\\ 0\amp k\end{array}\right], k\neq 0\right\}\text{.} \end{equation*}
Homotheties, rotations, translations, and inversions (see Table 3.1.4 in Section 3.1) are special cases of Möbius transformations. These basic transformations can be viewed as building blocks for general Möbius transformations, as follows.

#### Checkpoint3.2.6.

1. Identify the values of the coefficients $$a,b,c,d$$ in a Möbius transformation $$z\to \frac{az+b}{cz+d}$$ that is a homothety, rotation, translation, and inversion, respectively.
2. Write a Möbius transformation that does "clockwise rotation by one-quarter rotation about the point $$2-i$$".

### Subsection3.2.2The Fundamental Theorem of Möbius Geometry

The Fundamental Theorem of Möbius Geometry says that all three-point sets are congruent. It will turn out to be convenient to work with the "standard" three-point set $$\{1,0,\infty\}\text{.}$$ Given any three distinct complex numbers $$z_1,z_2,z_3$$ it is easy to check that the transformation $$T$$ given by
$$Tz=\frac{z-z_2}{z-z_3}\frac{z_1-z_3}{z_1-z_2}\tag{3.2.3}$$
satisfies $$Tz_1=1,Tz_2=0,Tz_3=\infty\text{.}$$

#### Checkpoint3.2.9.

Verify that (3.2.3) satisfies $$Tz_1=1,Tz_2=0,Tz_3=\infty\text{.}$$
It is easy to adapt (3.2.3) to the extended complex plane, where one of the $$z_i$$ may be the point at infinity.

#### Checkpoint3.2.11.

Prove Lemma 3.2.10 by finding explicit formulas for $$T$$ for each of the three cases $$z_1=\infty\text{,}$$ $$z_2=\infty$$ and $$z_3=\infty\text{.}$$
\begin{align*} (z_1=\infty) \amp \hspace{.2in}\amp Tz= \frac{z-z_2}{z-z_3}\\ (z_2=\infty) \amp \amp Tz= \frac{z_1-z_3}{z-z_3}\\ (z_3=\infty) \amp \amp Tz= \frac{z-z_2}{z_1-z_2} \end{align*}
The next Lemma lays lays the groundwork for proving the uniqueness of congruences between three-point sets.

#### Checkpoint3.2.13.

Hint.
Solve $$z=\frac{az+b}{cz+d}\text{.}$$ You will need to consider cases.

### Subsection3.2.3Cross ratio

Given three distinct points $$z_1,z_2,z_3$$ in $$\extC\text{,}$$ we write $$(\cdot,z_1,z_2,z_3)$$ to denote the unique Möbius transformation that satisfies $$z_1\to 1\text{,}$$ $$z_2\to 0\text{,}$$ and $$z_3\to \infty$$ (the existence and uniqueness of this transformation is guaranteed by the Fundamental Theorem of Möbius Geometry). We write $$(z_0,z_1,z_2,z_3)$$ to denote the image of $$z_0$$ under $$(\cdot,z_1,z_2,z_3)\text{.}$$ The expression $$(z_0,z_1,z_2,z_3)$$ is called the cross ratio of the 4-tuple $$z_0,z_1,z_2,z_3\text{.}$$ By (3.2.3), we have the following explicit formula for the cross ratio.
$$(z_0,z_1,z_2,z_3) = \frac{z_0-z_2}{z_0-z_3}\frac{z_1-z_3}{z_1-z_2}\tag{3.2.4}$$
The cross ratio is a useful tool because it is invariant under Möbius transformations. Here is the formal statement with a proof.

#### Proof.

The transformations $$(\cdot,z_1,z_2,z_3)$$ and $$(\cdot,Tz_1,Tz_2,Tz_3)\circ T$$ both send $$z_1\to 1\text{,}$$ $$z_2\to 0\text{,}$$ and $$z_3\to \infty\text{,}$$ so they must be equal, by the Fundamental Theorem. Now apply both transformations to $$z_0\text{.}$$

### Subsection3.2.4Clines (generalized circles)

A Euclidean circle or straight line is called a cline (pronounced "kline") or generalized circle. The propositions and corollaries in this subsection show that the set of all clines is a single congruence class of figures in Möbius geometry.

#### Checkpoint3.2.20.

Explain how Corollary 3.2.19 follows from Proposition 3.2.16.
Hint.
Let $$z_1,z_2,z_3$$ be three points on $$C$$ and let $$w_1=Tz_1,w_2=Tz_2,w_3=Tz_3\text{.}$$ Let $$U=(\cdot,z_1,z_2,z_3)$$ and let $$V=(\cdot,w_1,w_2,w_3)\text{.}$$ Explain why $$T=V^{-1}U$$ and hence that $$T(C)=V^{-1}U(C)=V^{-1}(\extR)$$ must be a cline.

#### Checkpoint3.2.22.

Explain how Corollary 3.2.21 follows from Proposition 3.2.16.
Hint.
Start here: Let $$C,D$$ be two clines. Let $$z_1,z_2,z_3$$ be three points on $$C$$ and let $$w_1,w_2,w_3$$ be three points on $$D\text{.}$$ Let $$U=(\cdot,z_1,z_2,z_3)$$ and let $$V=(\cdot,w_1,w_2,w_3)\text{,}$$ and let $$T=V^{-1}U\text{.}$$ Explain why $$T$$ takes $$C$$ to $$D\text{.}$$

### Subsection3.2.5Symmetry with respect to a cline

Geometrically, the conjugation map $$z\to z^{\ast}$$ in the complex plane is reflection across the real line. This "mirror" symmetry generalizes to symmetry with respect to any cline, as follows. Given a cline $$C$$ that contains $$z_1,z_2,z_3$$ in $$\extC\text{,}$$ let $$T=(\cdot,z_1,z_2,z_3)\text{.}$$ Given any point $$z\text{,}$$ the symmetric point with respect to $$C$$ is
$$z^{\ast C}= (T^{-1}\circ \conj \circ T)(z)\tag{3.2.5}$$
where $$\conj\colon \extC\to \extC$$ is the extension of the conjugation map to the extended complex plane that sends $$\infty\to \infty^\ast=\infty\text{.}$$ The idea is to map $$C$$ to the real line via $$T\text{,}$$ then conjugate, then map the real line back to $$C\text{.}$$ See Figure 3.2.23.

#### Checkpoint3.2.24.

Show that the symmetric point of a symmetric point is the point you started with. That is, show that $$(z^{\ast C})^{\ast C}=z\text{.}$$ This allows us to speak of "a pair of symmetric points" without ambiguity.
Hint.
$$(T^{-1}\circ \conj \circ T)\circ (T^{-1}\circ \conj\circ T) = \Id$$

#### Proof.

Let $$z_1,z_2,z_3$$ be three points on $$C\text{,}$$ so that $$Sz_1,Sz_2,Sz_3$$ are three points on $$S(C)\text{.}$$ Let $$T=(\cdot,z_1,z_2,z_3)$$ and let $$U=(\cdot,Sz_1,Sz_2,Sz_3)\text{.}$$ By invariance of the cross ratio, we have
\begin{equation*} (U\circ S)z=Tz\text{.} \end{equation*}
Thus we have
\begin{align*} (Sz)^{\ast S(C)} \amp = (U^{-1}\circ \conj\circ U)(Sz) \;\; \text{(by definition)}\\ \amp = (S\circ S^{-1}\circ U^{-1}\circ \conj\circ U\circ S)(z)\\ \amp = S (S^{-1}\circ U^{-1}\circ \conj\circ U\circ S)(z)\\ \amp = S(T^{-1}\circ \conj \circ T)(z)\\ \amp = S(z^{\ast C}) \end{align*}
as desired.

### Subsection3.2.6Normal forms

We conclude this section on Möbius geometry with a discussion of the normal form of a Möbius transformation. We begin with a Lemma.
Now suppose that a Möbius transformation $$T$$ has two fixed points, $$p$$ and $$q\text{.}$$ Let $$S$$ be given by $$Sz = \frac{z-p}{z-q}\text{.}$$ Let $$w=Sz$$ and let $$U=S\circ T\circ S^{-1}$$ be the transformation of the $$w$$-plane that is conjugate to $$T$$ via $$S$$ (see Exercise Group 1.3.4.3–6). It is easy to check that $$U$$ has exactly two fixed points $$0$$ and $$\infty\text{.}$$ Applying the previous Lemma, we have $$Uw=\alpha w$$ for some nonzero $$\alpha\in \C\text{.}$$ Applying both sides of $$S\circ T=U\circ S$$ to $$z\text{,}$$ we have the following normal form for $$T\text{.}$$
$$\frac{Tz-p}{Tz-q}= \alpha \frac{z-p}{z-q}\tag{3.2.6}$$
The transformation $$T$$ is called elliptic, hyperbolic, or loxodromic if $$U$$ is a rotation ($$|\alpha|=1$$), a homothety ($$\alpha \gt 0$$), or neither, respectively.
Finally, suppose that a Möbius transformation $$T$$ has exactly one fixed point at $$p\text{.}$$ Let $$S$$ be given by $$Sz = \frac{1}{z-p}\text{.}$$ Again, let $$w=Sz$$ and let $$U=S\circ T\circ S^{-1}\text{.}$$ This time, $$U$$ has exactly one fixed point at $$\infty\text{.}$$ Applying the Lemma, we have $$Uw=w+\beta$$ for some nonzero $$\beta\in \C\text{.}$$ Applying both sides of $$S\circ T=U\circ S$$ to $$z\text{,}$$ we have the following normal form for $$T\text{.}$$
$$\frac{1}{Tz-p}= \frac{1}{z-p}+\beta\tag{3.2.7}$$
A Möbius transformation of this type is called parabolic. Here is a summary of the classification terminology associated with normal forms.

### Subsection3.2.7Steiner circles

The discussion of normal forms show that any non-identity Möbius transformation is conjugate to one of two basic forms, $$w\to \alpha w$$ or $$w\to w+\beta\text{.}$$ The natural coordinate system for depicting the action of $$w\to \alpha w$$ is standard polar coordinates. See Figure 3.2.29. A homothety is a flow along radial lines and a rotation is a flow around polar circles. The natural "degenerate" coordinate system for depicting a translation $$w\to w+\beta$$ is a family of lines parallel to the line that contains the origin and $$\beta\text{.}$$ A translation by $$\beta$$ is a flow along these parallel lines. See Figure 3.2.30.
Pulling the polar and degenerate coordinate grids back to the $$z$$-plane by $$S^{-1}$$ leads to coordinate grids called Steiner circles.
1
The convention for which Steiner circles are considered "first" or "second" kinds is not universal. Here we follow the convention used by Ahlfors [1] and Henle [4].
In the case where $$T$$ has two fixed points $$p,q\text{,}$$ the conjugating map $$Sz=\frac{z-p}{z-q}$$ takes $$p\to 0\text{,}$$ $$q\to \infty\text{.}$$ Therefore $$S^{-1}$$ maps $$0\to p$$ and $$\infty \to q\text{.}$$ The transformation $$S^{-1}$$ maps radial lines in the $$w$$-plane to clines in the $$z$$-plane that contain $$p$$ and $$q$$ called Steiner circles of the first kind and $$S^{-1}$$ maps polar circles in the $$w$$-plane to clines in the $$z$$-plane called Steiner circles of the second kind or circles of Apollonius. See Figure 3.2.29.
In the case where $$T$$ has one fixed point $$p\text{,}$$ the conjugating map $$Sz=\frac{1}{z-p}$$ sends $$p\to \infty\text{,}$$ so $$S^{-1}$$ maps $$\infty \to p\text{,}$$ and $$S^{-1}$$ maps lines in the $$w$$-plane that are parallel to the line through $$0$$ and $$\beta$$ to clines in the $$z$$-plane that contain $$p\text{.}$$ Every cline in this family is tangent to every other cline in this family at exactly the one point $$p\text{.}$$ Clines in this family are called degenerate Steiner circles. See Figure 3.2.30. Table 3.2.31 summarizes the graphical depiction of Möbius transformations.

### Exercises3.2.8Exercises

#### 1.Decomposition of Möbius transformations into four basic types.

1. Explain why a transformation of the form $$z\to az\text{,}$$ with $$a$$ any nonzero complex constant, is a composition of a homothety and a rotation.
2. Explicitly identify each homothety, rotation, translation, and inversion in (3.2.8) to (3.2.11) in the derivation below for the case $$c\neq 0\text{.}$$
\begin{align} z \amp \to cz+d\tag{3.2.8}\\ \amp\to \frac{1}{cz+d}\tag{3.2.9}\\ \amp\to \frac{bc-ad}{cz+d} + a\tag{3.2.10}\\ \amp\to \frac{1}{c}\left(\frac{bc-ad}{cz+d} + a\right)\tag{3.2.11}\\ \amp = \frac{az+b}{cz+d}\tag{3.2.12} \end{align}
3. Write your own decomposition for the case $$c=0\text{.}$$

#### 2.

Prove the Fundamental Theorem of Möbius Geometry (Proposition 3.2.14).
Hint.
Start by using Lemma 3.2.10 to get a Möbius transformation $$U$$ that maps $$z_1,z_2,z_3$$ to $$1,0,\infty\text{,}$$ and there is a Möbius transformation $$V$$ that maps $$w_1,w_2,w_3$$ to $$1,0,\infty\text{.}$$

#### 3.

Find Möbius transformations that make the following assignments.
1. $$\displaystyle 1\to a, 0\to b, \infty\to c$$
2. $$\displaystyle a\to d, b\to e, c\to f$$

#### 4.

Prove Proposition 3.2.16. Suggestion: Let $$Tz=\frac{az+b}{cz+d}\text{,}$$ then manipulate $$Tz=(Tz)^\ast$$ to an equation with $$|z|^2,z,z^\ast$$ terms and coefficients involving $$a,b,c,d$$ and their conjugates. For the case when the coefficient of $$|z|^2$$ is not zero, use "complex completing the square" (see (1.1.3)) to derive the equation of a circle. Peek at the first part hint below if you need to, and use it to work partially forwards from $$Tz=(Tz)^\ast\text{,}$$ and partially backwards from the equation in the hint. For the case when the coefficient of $$|z|^2$$ is zero, derive the equation for a line. For the "furthermore" statement, you will need to show that any three points in the plane determine a unique circle or straight line. This is a Euclidean statement, and it is straightforward to prove this using Euclidean methods (peek at the second part of the hint below if you need to).
Hint.
Equation for the circle:
\begin{equation*} \left| z-\left(\frac{a^\ast d-bc^\ast}{ac^\ast -a^\ast c}\right)\right|^2 = \left|\frac{ad-bc}{ac^\ast -a^\ast c}\right|^2 \end{equation*}
Why do three noncollinear points $$P,Q,R$$ determine a unique circle? The center $$C$$ of the circle must be the intersection of the perpendicular bisectors of segments $$PQ\text{,}$$ $$QR\text{.}$$ The radius must be the distance from $$C$$ to any one of $$P,Q,R\text{.}$$

#### 5.Symmetry with respect to a cline.

1. Prove Proposition 3.2.25 by completing the details in the following outline. Let $$C=T^{-1}(\extR)$$ be a cline that contains points $$z_1,z_2,z_3\text{,}$$ where $$T=(\cdot,z_1,z_2,z_3)\text{.}$$ First, consider the case when $$C$$ is a circle, say, with equation $$|z-a|=r\text{,}$$ where $$a,r$$ are the center and radius, respectively, of $$C$$ (see (1.1.1)).
1. Square both sides of $$|z-a|=r$$ and solve for $$z^\ast$$ to get
\begin{equation*} z^\ast = \frac{r^2}{z-a}+a^\ast\text{.} \end{equation*}
2. We have
\begin{align*} (\conj\circ T)(z) \amp = (z^\ast,z_1^\ast,z_2^\ast,z_3^\ast)\\ \amp = \left(z^\ast, \frac{r^2}{z_1-a}+a^\ast,\frac{r^2}{z_2-a}+a^\ast,\frac{r^2}{z_3-a}+a^\ast\right)\\ \amp \stackrel{(1)}{=} \left(\frac{r^2}{z^\ast - a^\ast}+a,z_1,z_2,z_3\right)\\ \amp = T\left(\frac{r^2}{(z-a)^\ast}+a\right). \end{align*}
Explain how invariance of the cross ratio is used to justify the equality (1) in this derivation.
3. Explain how we can conclude that $$z^{\ast C}= \frac{r^2}{(z-a)^\ast}+a\text{,}$$ and that indeed this does not depend on the choice of $$z_1,z_2,z_3\text{.}$$
4. Now suppose that $$C$$ is a line, say, with equation $$\im(az+b)=0$$ (see (1.1.2)). This is the same as $$az+b=(az+b)^\ast\text{.}$$ Solve to get
\begin{equation*} z^\ast = \frac{az+b-b^\ast}{a^\ast}. \end{equation*}
5. We have
\begin{align*} (\conj\circ T)(z) \amp = (z^\ast,z_1^\ast,z_2^\ast,z_3^\ast)\\ \amp = \left(z^\ast,\frac{az_1+b-b^\ast}{a^\ast},\frac{az_2+b-b^\ast}{a^\ast}, \frac{az_3+b-b^\ast}{a^\ast}\right)\\ \amp \stackrel{(1)}{=} \left(\frac{z^\ast a^\ast + b^\ast - b}{a},z_1,z_2,z_3\right)\\ \amp = T\left(\frac{z^\ast a^\ast + b^\ast - b}{a}\right). \end{align*}
Explain how invariance of the cross ratio is used to justify the equality (1) in this derivation.
6. Explain how we can conclude that $$z^{\ast C}=\frac{z^\ast a^\ast + b^\ast - b}{a}\text{,}$$ and that indeed this does not depend on the choice of $$z_1,z_2,z_3\text{.}$$
2. Show that $$z,z'$$ are symmetric with respect to a circle $$C$$ with center $$a$$ and radius $$r$$ if and only if $$|z-a||z'-a|=r^2$$ and $$z,z'$$ lie on the same ray emanating from $$a\text{.}$$ See Figure 3.2.32.
Hint.
The equation for $$z^{\ast C}$$ in part iii gives $$(z'-a)(z-a)^\ast = r^2\text{.}$$ For the first statement, take the norm of both sides. For the second statement, take the argument of both sides.
3. Show that $$z,z'$$ are symmetric with respect to a line $$C$$ if and only if $$z,z'$$ are reflections of one another across $$C\text{.}$$ See Figure 3.2.32.

#### Normal forms and Steiner circles.

##### 7.
Find the normal form and sketch a graph using Steiner circles for the following transformations.
1. $$\displaystyle z\to \frac{1}{z}$$
2. $$\displaystyle z\to \frac{3z-1}{z+1}$$
##### 8.
Let $$p$$ be the single fixed point of a Möbius transformation that is conjugate to $$w\to w+\beta$$ via $$Sz=\frac{1}{z-p}\text{.}$$ Show that the single line in the degenerate Steiner clines through $$p$$ is parallel to the direction given by $$\beta^\ast\text{.}$$
Hint.
Show that $$S^{-1}w=\frac{pw+1}{w}\text{,}$$ so $$S^{-1}$$ takes $$0,\beta,\infty$$ to $$\infty,\frac{p\beta+1}{\beta},p\text{.}$$ Thus the single degenerate Steiner straight line through $$p$$ is in the direction given by $$\frac{p\beta+1}{\beta}-p =\frac{1}{\beta}\propto \beta^\ast\text{.}$$
##### 9.
Show that a (generalized) circle of Apollonius (a Steiner circle of the second kind) is characterized as the set of points of the form
\begin{equation*} C=\left\{P\in \C\colon \frac{d(P,A)}{d(P,B)}=k\right\} \end{equation*}
for some $$A,B\in \C$$ and some real constant $$k\gt 0\text{.}$$