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Möbius transformations form a group and identify with the projective linear quotient of GL_2(C)

Statement

Möbius transformations form a group under composition. More precisely, if $A = (\begin{array}{cc} a & b \\ c & d \end{array}) \in G L_{2} (C),$ let $M_{A}$ be the associated fractional linear map. Then $A ⟼ M_{A}$ is a surjective group homomorphism from $G L_{2} (C)$ onto the Möbius group, its kernel is the scalar subgroup $C^{\times} I$ , and therefore $Mob (\hat{C}) ≅ G L_{2} (C) / (C^{\times} I) = P G L_{2} (C) .$

Facts & Assumptions

Given: Invertible complex $2 \times 2$ matrices and their fractional linear maps.

[L1]

A surjective group homomorphism identifies its codomain with the quotient by its kernel (The image of a group homomorphism is a subgroup and its kernel is a normal subgroup, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ ).

Proof

technique · direct

1.1

For $A = (\begin{array}{cc} a & b \\ c & d \end{array})$ and $B = (\begin{array}{cc} α & β \\ γ & δ \end{array})$ , direct algebra gives $M_{A} (M_{B} (z)) = M_{A B} (z)$ wherever both sides are finite, hence on all of $\hat{C}$ . So $A \mapsto M_{A}$ is a homomorphism, and it is surjective by the definition of Möbius transformation.

givenalgebra

1.2

Because $A^{- 1} \in G L_{2} (C)$ , the identity $M_{A} \circ M_{A^{- 1}} = M_{I} = M_{A^{- 1}} \circ M_{A}$ shows that Möbius transformations form a group. The kernel condition $M_{A} (z) = z$ gives $c z^{2} + (d - a) z - b = 0$ for all finite $z$ , so $b = c = 0$ and $a = d \neq 0$ ; hence the kernel is exactly $C^{\times} I$ .

givenalgebra

2.1

Applying [L1] to the surjective homomorphism proved above and the kernel computation of step 1.2 gives $Mob (\hat{C}) ≅ G L_{2} (C) / (C^{\times} I) = P G L_{2} (C)$ .

L1given∎

Depends on

Used by

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 3 §§2.2-3.5 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Ch. 1 §§1.3-1.4 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 8 §§1-2 (standard reference, not scraped)