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Fixed Points of a Möbius Transformation

Definition. A point $z$ is called a fixed point of a transformation $T$ if the transformation maps the point exactly to itself: $T(z) = z$.

Geometric Significance: Finding these points is essential because they classify the fundamental geometry of the map. For instance, a pure rotation rotates fixed its center, and a pure dilation scales outward from a fixed point. Fixed points act as the “anchors” of the transformation.

To find the fixed points of $T(z) = \frac{az + b}{cz + d}$, we must 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).

Because $c = 0$, we must have $d \neq 0$ (otherwise $ad - bc = 0$). The transformation simplifies to $T(z) = \frac{a}{d}z + \frac{b}{d}$. By our definition of the extended complex plane, linear transformations always fix infinity: $T(\infty) = \infty$.

To find finite fixed points, we solve:

$$ \begin{aligned} \frac{a}{d}z + \frac{b}{d} &= z \leadsto[\text{Subtract z from both sides:}] \\[4pt] \frac{a}{d}z - z &= -\frac{b}{d} \leadsto[\text{Factor out z:}] \\[4pt] \left(\frac{a}{d} - 1\right)z &= -\frac{b}{d} \end{aligned} $$

We have two sub-cases based on the coefficient $\left(\frac{a}{d} - 1\right)$:

Sub-case (i) $a = d$ (Parabolic Translation): The coefficient becomes $0$, yielding the equation $0 \cdot z = -\frac{b}{d}$.
If b = 0, the transformation becomes T(z) = z, the trivial map. It fixes every point in the extrended complex plane (infinitely many fixed points).
If $b \neq 0$, this equation has no finite solutions. Total Fixed Points: Exactly 1 (only $\infty$). (Example: $T(z) = z + b$ simply translates the plane; only infinity stays put).
Sub-case (ii) $a \neq d$ (Loxodromic/Elliptic): The coefficient is non-zero, allowing us to divide: $$ z = \frac{-b/d}{\frac{a-d}{d}} = \frac{-b}{a-d} $$ Total Fixed Points: Exactly 2 (one at $\infty$, one at $\frac{-b}{a-d}$).

Case 2. The Rational Case ($c \ne 0$).

First, we check infinity. By definition, $T(\infty) = \frac{a}{c}$. Since $c \neq 0$, $\frac{a}{c}$ is a finite number, meaning $T(\infty) \neq \infty$. Therefore, $\infty$ is not a fixed point.

To check finite points, we solve:

$$ \begin{aligned} \frac{az + b}{cz + d} &= z \\[4pt] az + b &= z(cz + d) \\[4pt] az + b &= cz^2 + dz \\[4pt] cz^2 + (d - a)z - b &= 0 \end{aligned} $$

Because $c \neq 0$, this is a true quadratic equation in $z$. By the Fundamental Theorem of Algebra, any degree-2 polynomial over $\mathbb{C}$ has exactly 2 roots (counting multiplicity). Therefore, this equation yields either 1 distinct solution (a double root) or 2 distinct solutions. It can never have 3.

Summary of Fixed Point Counts

Identity Map ($T(z) = z$, $a=d=1, b=c=0$): Fixes every point in the plane (infinitely many).
Parabolic: Fixes exactly 1 point (either $\infty$ or a finite double root).
Loxodromic/Elliptic: Fixes exactly 2 distinct points.

This leads to a highly restrictive and powerful geometric rule: Any Möbius transformation that is NOT the Identity map 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 $M = T^{-1} \circ S$. Because the inverse and composition of Möbius transformations are Möbius, $M$ is a valid 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.

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

The Group Properties of Möbius Transformations

Möbius transformations are not just isolated functions; they form a perfect algebraic structure known as a Group. Furthermore, this group perfectly mirrors the algebra of $2 \times 2$ matrices and their geometry is encoded by their action on the Riemann sphere.

Closure under composition. The composition 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. Because matrix multiplication is strictly associative ($(AB)C = A(BC)$), the composition of Möbius transformations is associative: $(f \circ g) \circ h = f \circ (g \circ h)$.

This also matches matrix multiplication, which is associative.

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

The Isomorphism to Matrix Groups

Because of these properties, there is a perfect structural mapping (a homomorphism) between invertible $2 \times 2$ complex matrices —the group $\text{GL}_2(\mathbb{C})$— and Möbius transformations. In other words, 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 $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

The Projective Linear Groups: $\text{PGL}_2(\mathbb{C})$ and $\text{PSL}_2(\mathbb{C})$

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: $GL_2(\mathbb{C}) \to \mathbb{M}(\mathbb{C}_{\infin})$ whose kernel is the set of non-zero scalar matrices. Consequently, $\mathbb{M}(\hat{\mathbb{C}})$ is isomorphic to the quotient group $\text{PGL}_2(\mathbb{C})$. 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 $\mathrm{PGL}_2(\mathbb{C})$.

Normalization and $\text{PSL}_2(\mathbb{C})$

Instead of using any invertible matrix, we can restrict our attention strictly to matrices with a determinant of $1$. This group is called the Special Linear Group, $\text{SL}_2(\mathbb{C})$.

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

Linear Algebra Fact: For an $n \times n$ matrix, $\det(cM) = c^n \det(M)$. Therefore, for our $2 \times 2$ matrix, $\det(M') = \left(\frac{1}{\sqrt{\Delta}}\right)^2 \det(M) = \frac{1}{\Delta} \cdot \Delta = 1$.

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 shrinks to exactly two matrices: $\{ I, -I \}$.

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 $\mathrm{PSL}_2(\mathbb{C})$. It demonstrates that every geometric Möbius mapping has exactly two determinant-$1$ matrix representations, which differ only by a minus sign.