Goal
----


I'm interested in describing Möbius transformations in the plane, and I'd like to define them in terms of three points and their images.


What I have tried
-----------------


I know that a projective transformation in $\mathbb{CP}^1$ can be seen as a Möbius transformation, and in that setup I'm well used to finding the corresponding transformation matrix, generalizing [this approach](https://math.stackexchange.com/a/339033/35416).


But I also know that Möbius geometry can be described as an interpretation of $\mathbb{P}(\mathbb R^{3,1})$, as I'll describe in more detail below. My problem is, I don't know how to express a Möbius transformation defined by three points in that world. I know that apart from the points themselves, the circumcircle of the defining points has to map to the circumcircle of their images. But that's still only four constraints, while I believe I need five (since this is a three-dimensional projective space, similar to $\mathbb{RP}^3$).


The last constraint is probably that the light cone must be preserved, but that appears to be a non-linear condition, so I don't know how to incorporate that. Some knowledge about Lorentz transformations might help here, but I don't know much about those. In particular, I don't see how to characterize Lorentz transformations using *linear* conditions.


I also considered doing the $\mathbb{CP}^1$ approach to find a transformation (described as a matrix in $\mathbb C^{2\times 2}$), but trying to translate that into a $\mathbb R^{4\times 4}$ matrix for Möbius geometry I failed again: an intermediate representation had quadratic terms I couldn't get rid of.


Question
--------


How can I find a matrix in Möbius geometry, modeled as $\mathbb{P}(\mathbb R^{3,1})$, if my input is three points in the plane and their images?


Möbius geometry
---------------


Here is my understanding of Möbius geometry in the plane. My primary source for this is currently section 3.2 of [this paper](http://arxiv.org/abs/math/0608291v2) (although I did change a sign in my vector below since theirs seemed to be incorrect). It suggests [Introduction to Möbius Differential Geometry](http://www.dmg.tuwien.ac.at/hertrich-jeromin/imdg/imdg.pdf) by Hertrich-Jeromin as well as [Vorlesungen über Differentialgeometrie III: Differentialgeometrie der Kreise und Kugeln](http://dx.doi.org/10.1007/978-3-642-50823-3). The former I find rather hard to read in a cookbook style, while the latter is not available from our library just now.


I imagine $\mathbb{P}(\mathbb R^{3,1})$ similar to $\mathbb{RP}^3$, i.e. projective three-space, but with the following bilinear form:


$$ \langle a,b\rangle = a\_1b\_1 + a\_2b\_2 + a\_3b\_3 - a\_4b\_4 $$


A circle with center $(x,y)$ and radius $r$ in this world is a vector


$$ \begin{pmatrix}2x \\ 2y \\ x^2+y^2-r^2-1 \\ x^2+y^2-r^2+1 \end{pmatrix} $$


or a multiple thereof. Two circles $a$, $b$ intersect orthogonally if and only if $\langle a,b\rangle = 0$ since


$$\langle a,b\rangle = 2\left(r\_a^2 + r\_b^2 - (x\_a-x\_b)^2 - (y\_a-y\_b)^2\right)$$


A circle of radius zero is a point. This kind of circle is the only one which is orthogonal to itself. Hence the light cone $\left\{p\in\mathbb{P}(\mathbb R^{3,1})\;\middle|\;\langle p,p\rangle = 0\right\}$ characterizes the set of all points. Since a Möbius transformation maps points to points, it has to preserve that light cone.



Möbius transformation by 3 points in the Minkowski model
In fact, you are looking for an imbedding
$\text{PGL}\_2{\mathbb C}=\text{PSL}\_2{\mathbb C}\hookrightarrow\text{PO}({\mathbb R};3,1)=\text{Isom}\,{\mathbb H}\_{\mathbb R}^3$
that provides the subgroup of all orientation-preserving isometries of
${\mathbb H}\_{\mathbb R}^3$.


Consider the $4$-dimensional ${\mathbb R}$-linear space $H$ formed all hermitian $2\times 2$-matrices. It is easy to verify that $q(h):=\det h\in{\mathbb R}$ defined for all $h\in H$ is a quadratic form of signature $(3,1)$.


The group $\text{SL}\_2{\mathbb C}$ acts on $H$ by the rule $g\cdot h:=ghg^\*$ (on the right-hand side, we use the matrix multiplication and the hermitian transpose). Obviously, this action preserves the quadratic form $q$ and, therefore, provides the desired imbedding.


The imbedding $g\mapsto i(g)$ is "quadratic" : the coefficients of $i(g)$ are quadratic homogeneous expressions in coefficients of $g$ (they also involve the complex conjugation). This "quadratic" feature is inescapable because, in a sense, an imbedding in question is unique. So, if you need, you can extract from it all particular formulae you may want to.


