Formula for action of Möbius transformation on the hyperboloid model

Question

The group of Möbius transformations are isomorphic to the group of orientation-preserving isometries of hyperbolic space. The 3-dimensional hyperboloid model is a model of hyperbolic space. What's the formula for the action of the Möbius transformations on the hyperboloid model (in terms of the four parameters of the Möbius transformations and the four parameters of the hyperboloid model)? I understand that it will be linear in the coordinates of the hyperboloid model, since all of its isometries are.

Answer 1

The space $\mathbb{R}^{2,1}$ is just $\mathbb{R}^3$ with the inner product

$$ \langle v_1,v_2\rangle=x_1x_2+y_1y_2-z_1z_2. $$

(There are different sign conventions.) The hyperboloid is the top half of $\langle v,v\rangle=-1$ with $z\ge1$. The group $SO(2,1)$ acts by isometries on this sheet. (The metric on the sheet basically comes from $\langle\cdot,\cdot\rangle$.)

We can identify the sheet with $2\times2$ real symmetric matrices with determinant $1$:

$$ (x,y,z)\longleftrightarrow \begin{bmatrix} z+x & y \\ y & z-x \end{bmatrix}. $$

$SL_2(\mathbb{R})$ acts on the space $V$ of real symmetric matrices by congruence, i.e. $A\cdot X:=AXA^T$, which is linear and preserves the determinant so is a map $SL_2(\mathbb{R})\to SO(2,1)$. I checked by hand it is

$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \mapsto \begin{bmatrix} \frac{1}{2}(a^2-b^2-c^2+d^2) & ab-cd & \frac{1}{2}(a^2+b^2-c^2-d^2) \\ ac-bd & ad+bc & ac+bd \\ \frac{1}{2}(a^2-b^2+c^2-d^2) & ab+cd & \frac{1}{2}(a^2+b^2+c^2+d^2) \end{bmatrix} \tag{$\ast$} $$

Recall $SL_2(\mathbb{R})$ acts on the upper-half plane by isometries.

Another way to get this homomorphism $(\ast)$, at least up to conjugacy, is to use the fact the Cayley transform $C$ maps the upper-half plane to the unit disk (and conjugating $SL_2(\mathbb{R})$ by it yields $SU(1,1)$), and stereographic projection $S$ to map the disk to the hyperboloid, so from there it's simply a matter of "transport of structure." That is, given $v$ on the hyperboloid, apply $S^{-1}$ to get it into the unit disk $\mathbb{D}$, then apply $C^{-1}$ to get it to the upper half-plane $\mathbb{H}$, then apply whatever Mobius transformation $[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}]$ you want, then go back by applying $C$ and $S$. You should get $(\ast)$, or a formula equivalent to it anyway.