(Also called fractional linear or bilinear transformations. Also Mobius transformations)
They are maps defined on a open subset of $\mathbb{C}$ of the form
$$ f(z)=\frac{a z+b}{c z+d} $$where a, b, c, d are any complex numbers satisfying ad − bc ≠ 0. If ad = bc, the rational function defined above is a constant since
$$ f(z)=\frac{a z+b}{c z+d}=\frac{a(c z+d)}{c(c z+d)}-\frac{a d-b c}{c(c z+d)}=\frac{a}{c} $$and is thus not considered a Möbius transformation.
These maps can be extended to the Riemann sphere $\bar{\mathbb{C}}$. If $c\neq 0$
$$ f\left(\frac{-d}{c}\right)=\infty \text { and } f(\infty)=\frac{a}{c} $$and if $c=0$
$$ f(\infty)=\infty. $$Thus a Möbius transformation is always a bijective holomorphic function from the Riemann sphere to the Riemann sphere, and they constitute a group. As stated here, they corresponds exactly to the conformal maps on the Riemann sphere.
The group of Moebius transformations $\mathcal{M}$ is, in fact, the same as the projective linear group $\mathbb{P}GL(2,\mathbb{C})=GL(2,\mathbb{C})/\{\lambda I\}$ acting over $\mathbb{C}\mathbb{P}^1$. This identification comes this way: the Moebius transformation
$$ z\mapsto\frac{a z+b}{c z+d} $$with $ad-bc\neq 0$ corresponds to
$$ \left[\left(\begin{array}{ll} a & b \\ c & d \end{array}\right)\right] \in GL(2,\mathbb{C})/\{\lambda I\} $$with nonzero determinant.
In other words, the complex vector space $\mathbb{C}^2$ gives rise to the complex projective line (aka the Riemann sphere). The linear transformations of $\mathbb{C}^2$ give rise to Moebius transformations.
But several linear transformations give rise to the same Moebius transformation, so we take equivalence classes. Even if we restrict to the singular matrices with unit determinant, we have a 2-1 relation. That is, we have a 2-1 surjection from $SL(2,\mathbb{C})$ to the group of Moebius, $\mathcal{M}$. That is,
$$ \mathcal{M}\equiv GL(2,\mathbb{C})/\{\lambda I\} \equiv SL(2,\mathbb{C})/\{\pm I\} $$so we can express the Moebius transformation (abusing of notation) by
$$ \pm \left(\begin{array}{ll} a & b \\ c & d \end{array}\right) $$with $ad-bc=1$.
The subgroup of $\mathcal{M}$ consisting of Moebius transformations with real coefficients, i.e. the special linear real group
$$SL(2,\mathbb{R})=\{A\in M_2(\mathbb{R}): det(A)=1\}$$
acts on the upper half complex plane (Poincare half plane) $\mathbb{H}=\{z=x+yi: y>0\}$ in the following way:
$$\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \cdot z=\frac{a z+b}{c z+d}.$$But very surprisingly: the isometries of the hyperbolic 3-space (see hyperbolic geometry) $\mathbb{H}^3$ is the full group of Moebius transformations!! ([Needham 2021] page 81)
And moreover. According to [Needham 2021] page 72: all three geometries of constant curvature (not only hyperbolic) have symmetry groups that are subgroups of the group of Moebius transformations of the Riemann sphere. I haven't "seen" yet, but it is clear from what it is said here about all the three geometries living inside hyperbolic 3-space.
[Needham 2021] page 77 :
Every Möbius transformation of C yields a unique Lorentz transformation of spacetime. Conversely, it can be shown (Penrose and Rindler, 1984, Ch. 1) that to every Lorentz transformation there corresponds a unique (up to sign) Moebius transformation.
Maybe this has to do with the discussion at Minkowski space#Relation of Minkowski plane and hyperbolic plane.
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Author of the notes: Antonio J. Pan-Collantes
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