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.
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 $T_{r_0} \Bbb S^3 \leftrightarrow \Bbb R^3$.)