Fixed Points and Group Properties of Möbius Transformations

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Fixed Points of a Möbius Transformation

Definition. A point z is called a fixed point of a transformation $T$ if mapping it returns the same point: T(z) = z.

Finding these points is essential because they classify the geometry of the map (e.g., a rotation rotates around fixed points).

We need to solve the equation: $\frac{az + b}{cz + d} = z$. We separate the analysis into two different cases based on whether the denominator is constant (c = 0) or variable ($c \neq 0$).

Case 1. The Linear Case (c = 0). If c = 0, the function is $T(\infin) := \infin$ (infinity map to itself). Check Finite points by solving: $\frac{a}{d}z + \frac{b}{d} = z \leadsto[\text{Subtract z from both sides:}] \frac{a}{d}z - z = -\frac{b}{d} \leadsto[\text{Factor out z:}] (\frac{a}{d}-1)z = \frac{-b}{d}$

Now, we have two sub-cases based on the coefficient $(\frac{a}{d} - 1)$
Sub-case (i). If a = d (Parabolic), then $(\frac{a}{d} - 1)=0$, the equation becomes $0 \cdot z = \frac{-b}{d}$. If $b \ne 0$, this is impossible. Total Fixed Points: Exactly 1 (only $\infty$), e.g., Translation T(z) = z + b where c = 0, a = 1, and d = 1.
Sub-case (ii). If $a \ne d$ (Loxodromic/Elliptic), z’s coefficient is not sero. We can solve for z: $z = \frac{-b/d}{\frac{a}{d} - 1} = \frac{-b/d}{\frac{a-d}{d}} = \frac{-b}{a-d}$. Total Fixed Points: Exactly 2: one at $\infty$, one at $\frac{-b}{a-d}$.

Case 2. The Rational Case ($c \ne 0$). Check infinity $T(\infin):=\frac{a}{c}$. Since $c \neq 0$, $\frac{a}{c}$ is a finite number and $T(\infty) \neq \infty$m, so $\infin$ is not a fixed point.

To check finite points, we need to solve:

$$ \begin{aligned} T(z) &=z \\[2pt] &[\iff] \frac{az +b}{cz +d} = z \\[2pt] &[\text{Multiply by the denominator:}] az + b = z(cz + d) \\[2pt] &[\iff] az + b = cz^2 + dz \\[2pt] &[\iff] cz^2 + (d - a)z - b = 0 \end{aligned} $$

This is a quadratic equation in z. A quadratic $ax^2+bx+c$ with $a\neq 0$ is a degree-2 polynomial. By the Fundamental Theorem of Algebra, any degree-n polynomial over $\mathbb{C}$ has exactly n roots counted with multiplicity. Therefore, a quadratic has exactly 2 roots counting multiplicity, and thus at most 2 distinct solutions in $\mathbb{C}$.

Therefore, if $c \neq 0$, T has either 1 or 2 fixed points. It cannot have 3.

Summary of Fixed Points Counts.

1. Identity Map ($T(z)=z$, a = d = 1, b = c = 0): Fixes every point in the plane ($\infty$ many).
2. Parabolic (c = 0, a = d): Fixes exactly 1 point ($\infty$).
3. Other ($c \neq 0$ or $a \neq d$): Fixes exactly 2 points.

Furthermore, a Möbius transformation that is NOT the identity can have at most 2 fixed points.

The Uniqueness Theorem (“The 3-Point Rule”). If a Möbius transformation $T$ fixes three distinct points, it must be the Identity map ($T(z) = z$).

Proposition: Determination by 3 Points. If two Möbius transformations S and T agree on three distinct points (i.e., they map the same inputs to the same outputs), then they are the exact same function (S = T).

Proof.

Let $z_1, z_2, z_3$ be distinct points in $\mathbb{C} \cup \{ \infin \}$ and suppose: $T(z_1) = w_1, T(z_2) = w_2, T(z_3) = w_2, S(z_1) = w_1, S(z_2) = w_2. S(z_3) = w_3$.

Consider the composite function $M = T^{-1} \circ S$. Since T is Möbius, $T^{-1}$ is also Möbius. Since the composition of Möbius functions is Möbius, M is a Möbius transformation.

Let’s see what $M$ does to our three points: $M(z_1) = T^{-1}(S(z_1)) = T^{-1}(w_1)$. Since $T(z_1) = w_1$, then $T^{-1}(w_1) = z_1$.

Similarly, $M(z_2) = z_2, M(z_3) = z_3$. The transformation $M$ has three fixed points ($z_1, z_2, z_3$). By the “Uniqueness Theorem” (The 3-Point Rules), any Möbius transformation that fixes three distinct points must be the Identity.

$M = Id \leadsto T^{-1} \circ S = Id \leadsto[\text{Apply T to both sides}] S = T.$

The Group Properties of Möbius Transformations

• The composition (the Closure Property) of two Möbius transformations is another Möbius transformation.

Let $f(z) = \frac{a_1 z + b_1}{c_1 z + d_1}$ and $g(z) = \frac{a_2 z + b_2}{c_2 z + d_2}$.

$(f \circ g)(z) = \frac{a_1 \left( \frac{a_2 z + b_2}{c_2 z + d_2} \right) + b_1}{c_1 \left( \frac{a_2 z + b_2}{c_2 z + d_2} \right) + d_1}$.

To clear the fractions, multiply the numerator and the denominator by $(c_2 z + d_2)$:

$(f \circ g)(z) = \frac{a_1(a_2 z + b_2) + b_1(c_2 z + d_2)}{c_1(a_2 z + b_2) + d_1(c_2 z + d_2)}$

Next, expand and group the terms by z: Numerator: $(a_1 a_2 z + a_1 b_2) + (b_1 c_2 z + b_1 d_2) = (a_1 a_2 + b_1 c_2)z + (a_1 b_2 + b_1 d_2)$, and Denominator: $(c_1 a_2 z + c_1 b_2) + (d_1 c_2 z + d_1 d_2) = (c_1 a_2 + d_1 c_2)z + (c_1 b_2 + d_1 d_2)$

So the new coefficients $A, B, C, D$ are: $(f \circ g)(z) = \frac{(a_1 a_2 + b_1 c_2)z + (a_1 b_2 + b_1 d_2)}{(c_1 a_2 + d_1 c_2)z + (c_1 b_2 + d_1 d_2)}$

The matrix multiplication of the coefficients perfectly match the coefficients we derived algebraically. Since the determinant of a product is the product of determinants ($\det(AB) = \det(A)\det(B)$), and the original determinants were non-zero, the new determinant is also non-zero. $\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 + b_1 c_2 & a_1 b_2 + b_1 d_2 \\ c_1 a_2 + d_1 c_2 & c_1 b_2 + d_1 d_2 \end{pmatrix}$. Therefore, the composition is indeed a Möbius transformation.

• Inverse (The Inverse Property).

The inverse of $f(z) = \frac{az+b}{cz+d}$ is found by solving for z: $w = \frac{az+b}{cz+d} \Rightarrow z = \frac{dw - b}{-cw + a}$ and the new coefficients are $A = d, B = -b, C = -c, D = a$

Thus: $f^{-1}(w) = \frac{dw - b}{-cw + a}$ which is again of the Möbius form. We need to verify the Determinant Condition to ensure it is a valid Möbius transformation. $\det(f^{-1}) = AD -BC = (d)(a) - (-b)(-c) = da - bc \ne 0$ (f is a valid Möbius transformation).

This corresponds to the inverse matrix formula: $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. In Möbius transformations, the scalar factor $\frac{1}{ad-bc}$ cancels out because it appears in both numerator and denominator, so we can ignore it.

Associativity. Since matrix multiplication is associative ($(AB)C = A(BC)$), and function composition is associative, Möbius transformations are associative. $(f \circ g) \circ h = f \circ (g \circ h)$

Identity Element, $I(z) = \frac{1z + 0}{0z + 1} = z$. Matrix: $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ (The Identity Matrix). Determinant: $1(1) - 0(0) = 1 \neq 0$

There is a mapping between 2x2 matrices and these functions. Every Möbius transformation corresponds to a matrix in $GL(2,\mathbb{C}): M=\left( \begin{matrix}a&b\\ c&d\end{matrix}\right), ad-bc\neq 0$ where composition corresponds to matrix multiplication and inverse corresponds to matrix inverse.

Möbius Property	Matrix Property
T(z)	Matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$
Composition $T_1 \circ T_2$	Multiplication $M_1 \times M_2$
Inverse $T^{-1}$	Inverse $M^{-1}$
Identity $I(z)=z$	Identity $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

Since $GL(2,\mathbb{C})$ is closed under multiplication and inversion, Möbius transformations inherit this closure. The set of Möbius transformations forms a group under composition, often denoted: $\mathrm{PGL}_2(\mathbb{C})=GL(2,\mathbb{C})/\{ \lambda I:\lambda \in \mathbb{C^{\mathnormal{\times }}}\}$. This captures the fact that scalar multiples of the matrix represent the same transformation.

Theorem. Let $\mathbb{M}(\mathbb{C}_{\infin})$ denote the set of Möbius transformations $\mathbb{M}(\mathbb{C}_{\infin}) = \{f: \mathbb{C}_{\infin} \to \mathbb{C}_{\infin} | f(z) = \frac{az + b}{cz + d}, ad - bc \ne 0 \}$. The set $\mathbb{M}(\mathbb{C}_{\infin})$ is a group under composition of functions and there is a surjective group homomorphism Υ: $GL_2(\mathbb{C}) \to \mathbb{M}(\mathbb{C}_{\infin})$ with kernel the diagonal matrices. The group $GL_2(\mathbb{C})/kI$ is denoted as $\mathrm{PGL}_2(\mathbb{C})$.

Proof.

We defined a map $\Upsilon$ that translates matrices into functions: $GL_2(\mathbb{C}) \to \mathbb{M}(\mathbb{C}_{\infin})$ given by $(\begin{smallmatrix}a & b\\\ c & d\end{smallmatrix}) \to \frac{az+b}{cz+d}$.

To prove this is a group homomorphism, we must show it preserves the group operation (multiplication corresponds to composition).

Let $f(z) = \frac{a_1 z + b_1}{c_1 z + d_1}$ and $g(z) = \frac{a_2 z + b_2}{c_2 z + d_2}$ be Möbius transformations. Matrix side multiplication: $\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \cdot \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 + b_1 c_2 & a_1 b_2 + b_1 d_2 \\ c_1 a_2 + d_1 c_2 & c_1 b_2 + d_1 d_2 \end{pmatrix}$. As proven in the previous section, the composition $(f \circ g)(z)$ yields a fraction with exactly those same coefficients: $(f \circ g)(z) = \frac{(a_1 a_2 + b_1 c_2)z + (a_1 b_2 + b_1 d_2)}{(c_1 a_2 + d_1 c_2)z + (c_1 b_2 + d_1 d_2)}$

$\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} = \begin{pmatrix} a_1 a_2 + b_1 c_2 & a_1 b_2 + b_1 d_2 \\ c_1 a_2 + d_1 c_2 & c_1 b_2 + d_1 d_2 \end{pmatrix} \to \frac{(a_1 a_2 + b_1 c_2)z + (a_1 b_2 + b_1 d_2)}{(c_1 a_2 + d_1 c_2)z + (c_1 b_2 + d_1 d_2)} = (f \circ g)(z)$. Since the coefficients match perfectly, it means that under the map Υ: $GL_2(\mathbb{C}) \to \mathbb{M}(\mathbb{C}_{\infin})$ matrix multiplication corresponds to composition of functions, which means that it is a homomorphism, $\Upsilon(AB) = \Upsilon(A) \circ \Upsilon(B)$. The map preserves the structure.

The map is surjective (onto). For any Möbius transformation $f(z) = \frac{az+b}{cz+d}$ with $ad-bc \neq 0$, the matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ exists and is in $GL_2(\mathbb{C})$ (because its determinant is non-zero). Therefore, every possible Möbius function comes from at least one matrix or is hit by the matrix with the same entries.

The Kernel (Ker $\Upsilon$). The kernel is the set of all matrices that map to the Identity Function $f(z) = z$ which is the identity in the group $\mathbb{M}(\mathbb{C}_{\infin})$

We solve: $\frac{az + b}{cz + d} = z \leadsto az + b = cz^2 + dz \leadsto cz^2 + (d-a)z - b = 0$

For this to be true for all z (the identity function must work everywhere), every coefficient of the polynomial must be zero: c = 0, b = 0, and d -a = 0, so d = a, hence the matrix must look like: $\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix} = a \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = aI$.

Since the matrix must be invertible ($a \neq 0$), the kernel is the set of all non-zero scalar multiples of the identity matrix, $\text{Ker}(\Upsilon) = \{ kI \mid k \in \mathbb{C}^* \}$.

The First Isomorphism Theorem states that for any group homomorphism $\phi :G\rightarrow G'$, the quotient group $G/\ker (\phi )$ is isomorphic to the image of $\phi$. In mathematical notation: $G/\ker (\phi )\; \cong \; \mathrm{Im}(\phi )$.

By The First Isompormism Theorem, Υ induces a group isomorphism: $GL_2(\mathbb{C}) / \{kI\} \cong \mathbb{M}(\mathbb{C}_{\infty})$. This quotient group is called the Projective General Linear Group and is denoted as $\mathrm{PGL}_2(\mathbb{C})$.

Theorem. Let $\mathrm{SL}_2(\mathbb{C})$ denotes the group of complex matrices with determinant one. There is a surjective group homomorphism Υ: $SL_2(\mathbb{C}) \to \mathbb{M}(\mathbb{C}_{\infin})$ with kernel $\pm I$, so one gets an isomorphism $SL_2(\mathbb{C})/I \to \mathbb{C}_{\infin}$. This just says that, we can represent any Möbius transformation using a determinant-1 matrix. In other words, instead of the condition ad − bc = 0, we could just as well have used ad − bc = 1 to simplify our algebra.

Proof

If you have a matrix M with determinant $\Delta \neq 0$, you can normalize it by dividing the matrix by $\sqrt{\Delta}, M' = \frac{1}{\sqrt{\Delta}} M$.

The new determinant is: $\det(M') = \left(\frac{1}{\sqrt{\Delta}}\right)^2 \det(M) = \frac{1}{\Delta} \cdot \Delta = 1$

In general, $\det (cM)=c^n\det (M)$ for an $n\times n$ matrix.

Since M and M’ differ only by a scalar constant, they produce the same Möbius transformation. This means the map $\Upsilon$ is still surjective even if we restrict the inputs to $SL_2(\mathbb{C})$.

In $SL_2(\mathbb{C})$, the kernel matrices kI must satisfy the determinant condition: $\det(kI) = 1 \leadsto k^2 = 1 \implies k = 1 \text{ or } k = -1$. So the kernel contains only two elements: $\{ I, -I \}$.

Conclusion: $L_2(\mathbb{C}) / \{\pm I\} \cong \mathbb{M}(\mathbb{C}_{\infty})$. This quotient is called the Projective Special Linear Group and is denoted as $\mathrm{PSL}_2(\mathbb{C})$.