Möbius transformation by 3 points in the Minkowski model

Question

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.

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 (although I did change a sign in my vector below since theirs seemed to be incorrect). It suggests Introduction to Möbius Differential Geometry by Hertrich-Jeromin as well as Vorlesungen über Differentialgeometrie III: Differentialgeometrie der Kreise und Kugeln. 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.

Answer 1

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.

Answer 2

It is fun to see in an invariant way where the embedding $\text{PGL}_2{\mathbb C}=\text{PSL}_2{\mathbb C}\hookrightarrow\text{PO}({\mathbb R};3,1)$ which Sasha speaks of `comes from'. Take $V$ to be a two-dimensional complex vector space so that ${\mathbb C} {\mathbb P}^1$ is the quotient of $V \setminus 0$ by the group ${\mathbb C} \setminus 0$ of nonzero complex scalars. Now split the group up as $ {\mathbb R}^+ \times S^1 = $ (scalings)x (rotations) and form the same quotient but in two steps: first by rotation, then by scaling.

Divide $V$ by the rotation group $S^1$ using invariant theory as follows.

A) LEMMA: The vector space $Q$ of real quadratic $S^1$-invariant functions on $V$ forms a 4-dimensional real vector space. Any $S^1$-invariant polynomial on $V$ is a function of these quadratic invariants. There is a single quadratic relation $R$ amongst the invariants.

STANDARD REALIZATION OF LEMMA. Choose linear coordinates $z_1, z_2$ for $V$. A basis for $Q$ is then $|z_1|^2, |z_2|^2, Re(z_1 \bar z_1), Im(z_1 \bar z_2)$. The relation $R$ comes from $|z_1|^2|z_2|^2 = |z_1 \bar z_2|^2$. EXERCISE: as a real quadratic form on $Q$ the relation $R$ has signature $3,1$.

B) Let $Q^*$ be the real dual vector space to $Q$. Let $\delta V \to Q^*$ be evaluation map $\delta(v) (P) = P(v)$, for $P \in Q$. It follows from the lemma that this map defines an isomorphism $V/S^1$ onto the image of $\delta$. The image of $\delta$ is the ``positive half'' of the light cone in $Q^*$
relative to the dual quadratic form to the quadratic form $R$ on $Q$. (The origin is sent to the cone point.) In terms of the standard realization we can write
$\delta(v) = v v^*$ where $v= (z_1,z_2)$. We are then identifying $Q^*$ with the space of Hermitian matrices on $V$. The components of the rank 1 Hermitian matrix $v v^*$ are the basis of invariants. The map $\delta$ is a $Gl(V)=\text{GL}_2{\mathbb C}$ equivariant map (and $S^1$-invariant) map from $V$ to $Q^*$.

Next, divide by scalings $t \in {\mathbb R}^+$. Use $\delta(tv) = t^2 \delta (v)$. Thus dividing by scalings over on the image of $\delta$ side corresponds to dividing out the positive light cone by positive scalings. We get as a result a $\text{GL}_2{\mathbb C} : \text{O}({\mathbb R};3,1)$-equivariant embedding of ${\mathbb C} {\mathbb P}^1$ into real projective 3-space whose image is the projectivized light cone: a real projective conic.

Where are the circles of Mobius in this real conic? Three distinct points (or rays) on the light cone determine a unique 3-plane in $Q^*$. The intersection of this 3-plane with the light cone, viewed real-projectively is the corresponding circle connecting the three points.