Möbius Transformations

Algebraic Structures

Möbius Transformations

The set of Möbius transformations is given by where for any , where is the general linear group over . We will often write , since the other two cases are compatible with the limits of the fraction.

Lemma

Any Möbius transformation is a composite of translations, dilations, and inversions.

Proof Case , we can write Otherwise, , which is a dilation followed by a translation.

Proposition

Any Möbius transformation is biholomorphic.

Proof Case . Then , which is linear, and the inverse is also linear, so it suffices to show that is holomorphic on . The only thing that needs to be checked is -differentiability at . Since , we have to consider at . That is , since (as is invertible), this is -differentiable at by the quotient rule. Case . Notice that when , , which is holomorphic. By the above lemma, is a composition of holomorphic functions, hence holomorphic.

Proposition

forms a group under function composite.

Proposition

Let . Then . Indeed forms a surjective group homomorphism.

Proof Suppose and . Then $$(\lambda_{X_{1}}\circ\lambda_{X_{2}})(z)=\lambda_{X_{1}}(a_{2}z+b_{2})=\frac{a_{1}(a_{2}z+b_{2})+b_{1}}{c_{1}(a_{2}z+b_{2})+d_{1}}=\frac{(a_{1}a_{2}+b_{1})z+b_{1}d_{2}}{(c_{1}a_{2}+d_{1})z+c_{1}d_{2}}=\lambda_{X_{1}X_{2}}(z)$$$\square$

Proposition

Two Möbius transformations are equal if and only if their matrices are scalar multiples of each other:

Proof This is clear from the expression of .

Corollary The kernel of is the set of scalar matrices in :

Corollary The Möbius group is isomorphic to , the projective general linear group over :

Lemma

Let , . Then has exactly one or two fixed points in .

Proof Suppose . First consider the case when . Then , which has one fixed point at . For , is a fixed point if and only if . If , this has a unique solution, giving a total of two fixed points in ; If , we cannot have (otherwise is the identity), so there is exactly one fixed point . Now case . Then , so we only need to consider fixed points in . Moreover, cannot be a fixed point because . Thus is a fixed point if and only if which has one or two solutions in .

Proposition

Let be distinct points, and let be distinct points. Then there exists a unique Möbius transformation such that for .

Proof Suppose both satisfy the conditions. Then fixes more than two points, and hence is the identity by the above lemma. Thus . Indeed, it is enough to show existence by the case when , , and , since once we have maps , such that then the composite will be a Möbius transformation mapping . So without loss of generality, we assume , , and . Case , then does the job. The other cases are just imposing some of to be . For example, if , , then does the job.

Comment

Now we can roughly guess a Möbius transformation mapping the unit disk to the upper half-plane by considering the Möbius transformation mapping to : Actually we are guessing it maps the circle to the real line, and the disk to the upper half plane. And we need to introduce the concept of generalized circles to make this precise.

Generalized Circles

Line

A line in is the set of solutions to where and .

Circle

A circle in is the set of solutions to where and with .

Generalized Circle

A generalized circle in is either a circle in or a line in plus the point at infinity. In fact, on the Riemann sphere, these all correspond to circles.

Theorem

A Möbius transformation maps generalized circles to generalized circles.

Proof By the above lemma, any Möbius transformation is a composite of ones of the form and . Affine maps take circles to circles, lines to lines and infinity to infinity, so it is enough to show that maps generalized circles to generalized circles. Case is a line and , then we can write for some and , . Then which is a circle as and so . Case is a line and , then for some , thus which is a line. Case is a circle and , then for some and , . Then which is a circle. Case is a circle and , then for some , thus which is a line.

Warning

If and are circles, may not map the center of to the center of .

Proposition

For all sets of three distinct points , there is a unique generalized circle containing them.

Proof Existence: By the proposition, there is a Möbius transformation mapping , , to , , . Then is a generalized circle containing , , . Uniqueness: Suppose and are two generalized circles containing , , . There is a Möbius transformation such that , , . Then and are generalized circles containing , , . Since the real axis is the unique line containing and , we have , and hence .

Corollary

If for a generalized circle, and , then is the generalized circle containing , , .

Proof Direct consequence of the above proposition.

Generalised Disk

A generalized disk is a connected component of , where is a generalized circle.

Each generalized circle bounds two generalized disks. If a Möbius transformation maps to , it must map to , and since it is a continuous bijection, it maps connected components to connected components. That is, Möbius transformations map generalized disks to generalized disks. So the Möbius transformation maps to or . We can decide which one by testing at a single point. We have , so is impossible, and we are left to conclude .