Homogeneous (projective) coordinates and spinors When a complex number is considered as the stereographic projection from a sphere to the Argand plane, and then is represented by two “homogeneous coordinates” (in order to allow for a “point at infinity” corresponding to the point of projection) such that ζ=ξ/η, what are the rules for translating expressions involving the original complex number to expressions in the complex pair? I need to understand, in particular, why the transformation
ζ’ = (αζ + β)/(γζ+ δ)
is equivalent to
ξ’ = αξ + βη,   η’ = γξ +δη
and why
[using * for complex conjugation]
x = (ζ + ζ*)/(ζζ* + 1)
turns into
x = (ξη* + ηξ*)/(ξξ* +  ηη*)
It would of course help to see an example of a (ξ,η) corresponding to a null (t,x,y,z), so that I could work out the products for myself, and then see how
ξξ*=t+z  (or 1+z in the t=1 hyperplane)
ηη*=t-z  (or 1-z  in the t=1 hyperplane)
but (ξξ* +  ηη*), the trace of the matrix representing the 4-vector, doesn’t seem to equal 2 (or 2t)
rather it equals (ζζ* + 1), which reduces to (1+z)/(1-z)
((This question is an expansion and restatement of a question I asked before. It relates to Spinors and Spacetime by Penrose & Rindler.))
 A: To appreciate the discussion a bit more I would recommend comparing their discussion with the discussion (and main reference) here.
You can check that re-interpreting
$$\zeta' = \frac{a \zeta + b}{c \zeta + d}$$
as a matrix equation
$$\begin{bmatrix} \zeta' \\ 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \zeta \\ 1 \end{bmatrix} = \begin{bmatrix} a \zeta + b \\ c \zeta + d \end{bmatrix}$$
applied to a composition of two such transformations
$$\begin{bmatrix} \zeta'' \\ 1 \end{bmatrix} = \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix} \begin{bmatrix} \zeta' \\ 1 \end{bmatrix} =  \begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix}  \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \zeta \\ 1 \end{bmatrix} $$
will agree with the matrix interpretation of the direct composition of two such transformations:
$$\zeta'' = \frac{a' \zeta' + b'}{c' \zeta' + d'} = \frac{a' \frac{a \zeta + b}{c \zeta + d} + b'}{c' \frac{a \zeta + b}{c \zeta + d} + d'} = ...$$
To be clear you just need to multiply out the two matrices then apply that to the $[\zeta,1]^T$ vector, and then compare the result to this last equation where you're going to just multiply top and bottom by $c \zeta + d$.
The map
$$\zeta' = \frac{a \zeta + b}{c \zeta + d} \to \begin{bmatrix} \zeta' \\ 1 \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \zeta \\ 1 \end{bmatrix} = \begin{bmatrix} a \zeta + b \\ c \zeta + d \end{bmatrix}$$
has the issue that it acts as
$$\zeta' = \frac{(\lambda a) \zeta + (\lambda b)}{(\lambda c) \zeta + (\lambda d)} = \frac{a \zeta + b}{c \zeta + d} \to \begin{bmatrix} \lambda  a & \lambda  b \\ \lambda  c & \lambda  d \end{bmatrix} \begin{bmatrix} \zeta \\ 1 \end{bmatrix} $$
but we can use this redundancy to always choose $\lambda$ so that the determinant is one, $ad - bc = 1$. This still leaves a $\lambda = \pm 1$ reduncancy so the map after this restriction is still one-to-two.
Setting $\zeta = \frac{\xi}{\eta}$ in $\zeta' = \frac{a \zeta + b}{c \zeta + d}$ we have
$$\frac{\xi'}{\eta'} = \frac{a \frac{\xi}{\eta} + b}{c \frac{\xi}{\eta} + d} = \frac{a \xi + b \eta}{c \xi + d \eta}$$
so again we can re-interpret it as a matrix equation
$$\begin{bmatrix} \xi' \\ \eta' \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} = \begin{bmatrix} a \xi+ b \eta \\ c \xi + d \eta \end{bmatrix}$$
and check the composition rule still holds from both perspectives, in other words we have
$$\xi' = a \xi + b \eta , \\ \eta' = c \xi + d \eta,$$
from the previous result.
Since
$$x = \frac{\zeta + \zeta^*}{1 + \zeta \zeta^*}$$
we have on using $\zeta = \frac{\xi}{\eta}$ that
$$x = \frac{\zeta + \zeta^*}{1 + \zeta \zeta^*} = \frac{\frac{\xi}{\eta}+ \frac{\xi^*}{\eta^*}}{1 + \frac{\xi}{\eta} \frac{\xi^*}{\eta^*}} = \frac{\eta^* \xi+ \eta \xi^*}{\eta \eta^* + \xi \xi^*} $$
If we now interpret $x$ as
$$x = \frac{X}{T}$$
we can set
$$x = \frac{X}{T}  = \frac{\eta^* \xi+ \eta \xi^*}{\eta \eta^* + \xi \xi^*} $$
and so set
$$X = \frac{1}{\sqrt{2}}(\eta^* \xi+ \eta \xi^*) , \\ T = \frac{1}{\sqrt{2}} (\eta \eta^* + \xi \xi^*),$$
where the $\frac{1}{\sqrt{2}}$ are just a convention (obviously they cancel out from the numerator and denominator).
To try and find $\xi$ and $\eta$ in terms of $(T,X,Y,Z)$ we can take $\begin{bmatrix} \xi \\ \eta \end{bmatrix}$ and form
$$\begin{bmatrix} \xi \\ \eta \end{bmatrix} \begin{bmatrix} \xi^* & \eta^* \end{bmatrix} = \begin{bmatrix} \xi \xi^* & \xi \eta^* \\ \eta \xi^* & \eta \eta^* \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} T + Z & ... \\ ... & T - Z \end{bmatrix} $$
we can see the trace is
$$\frac{1}{\sqrt{2}}[(T + Z) + (T - Z)] = \sqrt{2} T = (\xi \xi^* + \eta \eta^* )$$
and you can find the $...$'s.
