The Homogeneous Space $SO^+(1,3)/ \text{Sim}(2)$ I'm trying to understand the homogeneous space $SO^+(1,3)/\text{Sim}(2)$.
In wikipedia it is said that "$SO^+(1,3)/\text{Sim} (2)$ is the Kleinian geometry that represents conformal geometry on the sphere $S^2$", but I couldn't find much information that explain this statement.
I tried to construct the homogeneous space as the space of left cosets of the corresponding subgroup of $SL(2,\mathbb{C})$ (the universal covering group of $SO^+(1,3)$).
Identifying a point $x = (x_0, x_1, x_2, x_3)$ in $\mathbb{R}^{1,3}$ with the hermitian $2\times 2$ matrix
$$ x \leftrightarrow X,\quad X = \begin{pmatrix} x_0 + x_3 & x_1 - i x_2 \\ x_1 + ix_2 & x_0 - x_3 \end{pmatrix}, \quad \mathrm{det} X = x_0^2 - x_1^2 - x_2^2 - x_3^2$$
a matrix $A \in SL(2,\mathbb{C})$transforms $X$ via
$$X' = AXA^\dagger.$$
The subgroup $H$ stabilizing a null line ($\mathrm{det}X = 0$), e.g. $\left\{ \begin{pmatrix} \lambda & 0 \\ 0 & 0 \end{pmatrix}, \lambda \in \mathbb{R} \right\} $, are matrizes of the form
$$A_{H} = \begin{pmatrix} a & b \\ 0 & \frac{1}{a} \end{pmatrix}; a,b \in \mathbb{C}, a \neq 0$$
since
$$\begin{pmatrix} a & b \\ 0 & \frac{1}{a} \end{pmatrix} \begin{pmatrix} \lambda & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a^* & 0 \\ b^* & \frac{1}{a^*} \end{pmatrix} = \begin{pmatrix} \lambda |a|^2 & 0 \\ 0 & 0 \end{pmatrix}$$
The subgroup $H$ is isomorphic to $Sim(2)$.
For a fixed element of $SL(2,\mathbb{C})$ the left coset is
$$\begin{pmatrix} e & f \\ g & h \end{pmatrix} H = \left\{ \begin{pmatrix} e & f \\ g & h \end{pmatrix} \begin{pmatrix} a & b \\ 0 & \frac{1}{a} \end{pmatrix}; a,b \in \mathbb{C}, a \neq 0 \right\} \\
\qquad \qquad \quad = \left\{ \begin{pmatrix} ea & eb + \frac{f}{a} \\ ga & gb + \frac{h}{a} \end{pmatrix}; a,b \in \mathbb{C}, a \neq 0 \right\}.$$
For me, this looks like the whole $SL(2,\mathbb{C})$, but it should result in a partition. Maybe this is not the way to construct this coset space. On dimensional grounds, since $H$ is $4$ dimensional, the homogeneous space is 2D.
 A: Here's a more direct (and probably more common) way of approaching this construction. Pick an concrete bilinear form of Lorentzian signature $(1, 3)$ and realize $SO^+(1, 3)$ and $Sim(2) < SO(1, 3)^+$ (or rather, as we'll see, their Lie algebras) explicitly. It's convenient to take the bilinear form given in coordinates $(x^0, x^1, x^2, x^\infty)$ to have the block-antidiagonal form
$$\Sigma := \pmatrix{\cdot&\cdot&1\\\cdot&I_2&\cdot\\1&\cdot&\cdot} ,$$ in which case $$SO(1, 3) = \{A \in GL(4, \Bbb R) : A^T \Sigma A = \Sigma\} ,$$ and $SO^+(1, 3)$ is the subgroup that preserves space and time orientations.
It's convenient here to pass to the point of view of Lie algebras; since $SO^+(1, 3)$ is just a connected component of $SO(1, 3)$, their Lie algebras are identical. Computing explicitly, we find that in the given coordinates
\begin{multline}\mathfrak{so}(1, 3)
= \{A \in \mathfrak{gl}(4) : A^\top \Sigma + \Sigma A = 0 \} \\
= \left\{\pmatrix{b&Y&\cdot\\X&B&Y^\top\\\cdot&X^\top&-b} : B \in \mathfrak{so}(2); X \in \Bbb R^2; Y \in (\Bbb R^2)^*; b \in \Bbb R \right\} .\end{multline}
Here, $$\mathfrak{so}(2) = \left\{\pmatrix{\cdot&-\beta\\\beta&\cdot} : \beta \in \Bbb R\right\} .$$
Since the first column vector $\partial_0$ is isotropic (an advantage of our block-antidiagonal choice of $\Sigma$), we may as well nominate the isotropic line $$\ell := \Bbb R \cdot \partial_0 = \pmatrix{\ast\\0\\\vdots\\0}$$ and compute its stabilizer $\mathfrak{p}$.
Computing gives, $$A \cdot \partial_0 = \pmatrix{b&Y&\cdot\\X&B&Y^\top\\\cdot&X^\top&-b} \pmatrix{\ast\\0\\\vdots\\0} = \pmatrix{\ast\\X\\\cdot} ,$$
so $A$ fixes $\ell$ iff $X = 0$, that is, we can realize $\mathfrak{p}$ as the block-upper-triangular Lie subalgebra
$$
\pmatrix{b&Y&\cdot\\\cdot&B&Y^\top\\\cdot&\cdot&-b}
$$
Notice that the block diagonal elements (those with $Y = 0$) comprise a subalgebra isomorphic to $\mathfrak{so}(2) \oplus \Bbb R \cong \mathfrak{cso}(2)$, the algebra of conformal (angle-preserving) infinitesimal linear transformations of the plane. This subalgebra acts on the complementary subspace (in fact, subalgebra) $\mathfrak{g_+} := \{B = 0; b = 0\} \cong \Bbb R^2$ via $(B, b) \cdot Y^\top = (B + bI) Y^\top$, but this is exactly the action of a conformal (infinitesimal) linear transformation $(b, B) \in \Bbb R \oplus \mathfrak{so}(2)$ on an (infinitesimal) translation $Y^\top \in \Bbb R^2$ in the plane. In summary, $\mathfrak{p} = \Bbb R^2 \rtimes \mathfrak{so}(2) = \mathfrak{sim}(2)$.
Incidentally, there's nothing particularly special about $Sim(2)$ here---the same construction works for $Sim(n)$, $n \geq 2$.
