Projection from $SO^+(1,4)$ into $\mathbb{S}^3$ This question is related with the projective light cone model of the conformal geometry of $\mathbb{S}^3$.
Let me denote the Minkowski space $\mathbb{R}_1^5$ as $\mathbb{R}^5$ equipped with a Lorentzian metric
$$<x,y>=-x_0y_0+x_1y_1+x_2y_2+x_3y_3+x_4y_4=x^tI_{1,4}y,$$
where $I_{1,4}=\text{diag}(-1,1,1,1,1)$.
Let $\mathcal{C}_+^4$ be the forward light cone of $\mathbb{R}_1^5$, i.e. for any $x\in\mathcal{C}_+^4$, we have that $x_0>0$. One can see that the projective light cone
$$\mathcal{Q}^3=\left\{\left[x\right]\in\mathbb{R}P^4|x\in\mathcal{C}_+^4\right\},$$
with the induced conformal metric, is conformally equivalent to $\mathbb{S}^3$, and the conformal group of $\mathcal{Q}^3$ is exactly the orthogonal group $O(1,4)/\{\pm 1\}$ of $\mathbb{R}_1^5$, acting on $\mathcal{Q}^3$ by $T(\left[x\right])=\left[Tx\right]$, for $T\in O(1,4)$. 
Denote by $SO^+(1,4)$ the connected component of $O(1,4)$ containing $I$, that is for any $T\in SO^+(1,4)$ we have that $\det T=1$ and $T$ preserves the signature of the first coordinate of any $x\in\mathbb{R}_1^5$ (i.e., preserves the time direction).

Does anyone know an explicit expression for the projection that arises after this construction, say $\pi:SO^+(1,4)\to\mathbb{S}^3$? Or any idea on how to derive it?
Or, in case not, any reference to consult for this topic?
I am also interested in the projection $\pi^\prime:SO^+(1,4)\to\mathbb{R}^3$, which I can construct if I know the previous one or the other way around, so if this one is known by anyone this is also really helpful for me.
 A: First, observe that each null line in $\mathcal Q$ intersects the reference plane $\Pi := \{x_0 = 1\}$ in precisely one point, so that $\mathcal Q \cap \Pi \cong \Bbb S^3$: The points $p = (1, r)$ in this intersection are precisely those that satisfy $0 = \langle (1, r) , (1, r) \rangle = -1 + || r ||^2 ,$ that is, $$||r||^2 = 1$$ (here, $||\cdot||$ is the usual Euclidean norm on $\Bbb R^4 \cong \Pi$).
Now, pick a reference line $\ell_0 \in \mathcal Q$, which we can write as $[ p_0 ]$ for some unique $p_0 = (1, r_0) \in \mathcal Q \cap \Pi$. Then, $A \in \textrm{SO}^+(1, 4)$ maps $\ell_0$ to
$$A \cdot \ell_0 = [ A \cdot p_0] ,$$
and scaling $A \cdot p_0$ by the reciprocal of its first entry gives us a point in $\mathcal Q \cap \Pi$.
A little more explicitly, with respect to the basis used to define the bilinear form, given an element $$A = \pmatrix{\lambda & y^{\top} \\ x & B} \in \textrm{SO}(1, 4),$$ we have
$$A \cdot p_0 = \pmatrix{\lambda & y^{\top} \\ x & B} \pmatrix{1\\r_0} = \pmatrix{\lambda + y^{\top} r_0 \\ x + B r_0} . $$ Then, the intersection of the line $[A \cdot p]$ with $\Pi$ is some $(1, r')$, and solving for $r'$ gives the explicit map $\textrm{SO}(1, 4) \to \Bbb S^3$:
$$A = \pmatrix{\lambda & y^{\top} \\ x & B} \mapsto (\lambda + y^{\top} r_0)^{-1} (x + B r_0) = \frac{x + B r_0}{||x + B r_0||}.$$
We can make this more explicit but less symmetric by picking a convenient explicit point $r_0$. (We can also use the fact that $A$ preserves the bilinear form to write $y$ in terms of $A$, $x$, $\lambda$.)
I do not know which map is meant by $\pi' : \textrm{SO}^+(1, 4) \to \Bbb R^3$. (That said, the differential of the map $\pi$ is a map $\mathfrak{so}(1, 4) \to T_{r_0} \Bbb S^3$, and if we take $r_0$ to be a canonical basis vector, we get a more-or-less canonical identification $T_{r_0} \Bbb S^3 \leftrightarrow \Bbb R^3$.)
