\chapter{Building Spacetime from Spin(III)}
The following spaces are the same:
\begin{itemize}
\item states of a Weyl spinor;
\item the complex projective line;
\item the Riemann sphere;
\item the angular momentum vectors of a spin-$1/2$ particle;
\item the hermitian $1$-dimensional projectors in $\C^2$;
\item the celestial sphere.
\end{itemize}
Let us see in detail how the last two are isomorphic. Hermitian
$1$-dimensional projectors are linear maps $\rho\colon\C^2\to\C^2$
such that
$$
\rho=\rho^\dagger,
\qquad
\tr\rho=1,
\qquad\hbox{and}\quad
\rho^2=\rho.
$$
The equation $\tr\rho=1$ rules out the solutions $\rho=0,1$ of the
equation $\rho^2=\rho$. Therefore, the rank of $\rho$ is $1$ and
$\det\rho=0$. This means that hermitian projectors correspond to
elements of $H$ of the form
$X=x^0\sigma_0+x^1\sigma_1+x^2\sigma^2+x_3\sigma^3$, and since $1=\tr
X=2x^0$ we have $x^0={1\over 2}$. This is a circle that intersects
each light ray through the origin precisely once, so we have a unique
light ray associated to each projector.
Conversely, we can show that every $\rho=\rho^\dagger$ such that
$\tr\rho=1$ and $\det\rho=0$ is a $1$-dimensional projector. Since
$\det\rho=0$ and $\tr\rho=1$, the rank of $\rho$ is $1$ and we can
find vectors $\phi,\psi$ such that $\rho=\ket\phi\bra\psi$ and
$\bracket\psi\phi=\tr\rho=1$. But then
$\rho^2=\bracket\psi\phi\rho=\rho$ automatically, and
$\rho=\rho^\dagger$ implies $\phi=e^{i\theta}\psi$.
Now, since the space of all light rays through the origin can be
described in terms of $H$ and $\det$, the group $\SL(2,\C)$ acts on
it.
\begin{exercise}
Show that if we represent points in the celestial sphere by the
corresponding points on the Riemann sphere, then the action of
$\SL(2,\C)$ is
$$
\eta\mapsto{a\eta+b\over c\eta+d}
\qquad\hbox{with}\quad
\pmatrix{a&b\cr
c&d}\in\SL(2,\C).
$$
\end{exercise}
In complex analysis one learns that these are all the conformal
transformations of the Riemann sphere. So, if one accelerates to high
speed, the constellations will appear distorted by an angle-preserving
transformation.
Also, if we replace $\C$ by another normed division algebra, we get
similar facts:
$$
\begin{array}{rclc}
&\hbox to 0pt{\hss Groups\hss}&&\hbox{Weyl spinors}\\
\SL(2,\R)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(2,1)&\R^2\\
\SL(2,\C)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(3,1)&\C^2\\
\SL(2,\Qua)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(5,1)&\Qua^2\\
\SL(2,\Oct)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(9,1)&\Oct^2\\
\end{array}
$$
[Do this with a $\backslash\texttt{tabular}$ environment and the text
``$\C$ Weyl spinors are a representation of $\SL(2,\C)$, which is a
double cover of $SO_0(3,1)$''.]
There is a nice way to visualize the correspondence between the
directions of space and the Riemann sphere. Represent each spinor
${\psi_1\choose\psi_2}$ by the complex number
$\psi=\psi_2/\psi_1$. Then, the projection operator corresponding to
$\psi\in\C$ is $\rho_\psi={1\over
1+|\psi|^2}\left({1\atop\psi}{\psi^*\atop|\psi|^2}\right)$. This is
obviously hermitian, and it is easy to see that this has unit trace
and vanishing determinant. Also, in the limit $\psi\to\infty$ we get
$\rho_\infty=\left({0\atop 0}{0\atop 1}\right)$. Now, observe that
$$
\rho_\psi={1\over 2}\sigma_0+{\real\psi\over
1+|\psi|^2}\sigma_1+{\imag\psi\over
1+|\psi|^2}\sigma_2+{1-|\psi|^2\over 2(
1+|\psi|^2)}\sigma_3,
\qquad\hbox{and}\quad
\rho_\infty={1\over 2}(\sigma_0-\sigma_3),
$$
so we get the correspondence
$$
\begin{array}{cc}
\ket\uparrow & 0\\
\ket\downarrow & \infty\\
\ket\rightarrow & 1\\
\ket\leftarrow & -1\\
\ket\otimes & i\\
\ket\odot & -i\\
\end{array}
$$
[Draw a picture of the sphere.]
Similarly, if we did this for $\Qua$ we would get $5$ orthogonal axes
for the $\Qua\cup\{\infty\}$ $4$-dimensional sphere, $9$ orthogonal
axes for the $\Oct\cup\{\infty\}$ $8$-dimensional sphere and $2$
orthogonal axes for the $\R\cup\{\infty\}$ $1$-dimensional sphere.