Moebius transformations
(Also called fractional linear or bilinear transformations. Also Mobius transformations)
They are maps defined on a open subset of
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
and is thus not considered a Möbius transformation.
These maps can be extended to the Riemann sphere
and if
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.
Projective complex line approach
The group of Moebius transformations
with
with nonzero determinant.
In other words, the complex vector space
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
so we can express the Moebius transformation (abusing of notation) by
with
Relation to hyperbolic geometry
The subgroup of
acts on the upper half complex plane (Poincare half plane)
But very surprisingly: the isometries of the hyperbolic 3-space (see hyperbolic geometry)
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.
Relation to Special Relativity
[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.