Hopf map $S^3 \to S^2$ stereographic projection I'm reading up on this topic and I've seen lots of literature suggest that the complex number representation of the stereographic projection of $S^2$ can be expressed as 
$\frac{\zeta^1 + i\zeta^2}{1-\zeta^3} = \frac{x^1 + ix^2}{x^3 +ix^4}$
where
$\zeta^1 = 2(x^1x^3 + x^2x^4)\\
\zeta^2 = 2(x^2x^3 - x^1x^4)\\
\zeta^3 = (x^1)^2 + (x^2)^2 - (x^3)^2 - (x^4)^2$
and
$S^3: (x^1)^2 + (x^2)^2 + (x^3)^2 + (x^4)^2 = 1\\
S^2: (\zeta^1)^2 + (\zeta^2)^2 + (\zeta^3)^2 = 1.$ 
I understand how to get to the LHS, but the RHS doesn't seem clear to me. Any help would be much appreciated.
 A: Write points of $S^3$ as things in $\Bbb C^2$, i.e., pairs $(z, w)$ where $|z|^2 + |w|^2 = 1$. Then the Hopf-map (to $\Bbb C \cup \{ \infty \}$) is defined by
$$
(z, w) \mapsto z/w
$$
Now the codomain (the extended complex plane) isn't exactly $S^2$, so you need to stereographically project a point $p + q i$ up to the unit 2-sphere. There are multiple stereographic projections from the extended complex plane to the 2-sphere, so things are a little ambiguous. But the first step isn't so bad: the point $(z, w) \in S^3$, where $z = a + bi, w = c + di$, gets sent to 
\begin{align}
u + iv 
&= \frac{z}{w} \\
&= \frac{a + bi}{c + di} \\
&= \frac{a + bi}{c + di}\frac{c - di}{c - di} \\
&= \frac{(a + bi)(c-di)}{c^2 + d^2}\\
&= \frac{(ac + bd) + (bc -ad)i}{c^2 + d^2}\\
&= \frac{(ac + bd)}{c^2 + d^2} + \frac{(bc -ad)i)}{c^2 + d^2}
\end{align}
so that $u = \frac{(ac + bd)}{c^2 + d^2}$ and $v = \frac{(bc -ad)}{c^2 + d^2}$. (You can see that the numerators of these things look a lot like $\zeta^1$ and $\zeta^2$!)
The point $u + iv$ is sent, by stereographic projection from the south pole (say) to the point on the line from 
$$
(0,0,-1) \to (u,v,0)
$$
whose squared coordinates sum to $1$, i.e., the point 
$$(p,q,r) = (1-t)(0,0,-1) + t(u, v, 0).$$
Here I'm parameterizing the line through those two points as 
$$
L(t) = (1-t) (0, 0, -1) + t (u, v, 0) = (1-t)A + t B
$$
which is an expression linear in $t$, so if we look at
$$
H = \{ L(t) \mid t \in \Bbb R \}
$$
we get a straight line in $3$-space. When $t = 0$, we're at $A$ (i.e., the south pole); when $t = 1$, we get $B$, i.e., the point $(u,v, 0)$, so $H$ really is the line we want. 
For other values of $t$, we get points on the segment between $A$ and $B$ (if $0 < t < 1$) or outside of the segment (for other $t$ values). 
When we ask "For what value of $t$ is $L(t)$ on the unit sphere?", we get a quadratic in $t$ (as you'll see below); one solution to this quadratic is $t = 0$, corresponding to the point $(0,0,-1)$, the south pole. The nonzero root of this quadratic will be the one that interests us in general. When $u$ and $v$ are both $\infty$, the quadratic degenerates into a linear expression with its only root at $t = 0$, i.e., the "point at infinity" maps to the south pole of $S^2$. 
with the property that $p^2 + q^2 + r^2 = 1$, i.e., 
\begin{align}
1 &= t^2 u^2 + t^2 v^2 + (1-t)^2\\
1 &= t^2 u^2 + t^2 v^2 + 1-2t + t^2\\
0 &= t^2 u^2 + t^2 v^2 + -2t + t^2\\
0 &= t^2 (u^2 + v^2 + 1) + -2t\\
2t &= t^2 (u^2 + v^2 + 1) \\
2 &= t (u^2 + v^2 + 1) \\
t &= \frac{u^2 + v^2 + 1}{2}
\end{align}
From this, you can compute $(p,q,r)$, and I'll bet that they look a whole lot like $\zeta^1, \zeta^2, \zeta^3$.
