\chapter{Building Spacetime from Spin(II)}
We have seen that $\SL(2,\C)$ acts on
\begin{itemize}
\item $\C^2$, the space of ``Weyl spinors'' via
$$
g\colon\psi\mapsto g\psi,
\quad
\psi\in\C^2;
$$
and
\item $H=\{2\times 2\quad\hbox{hermitian matrices}\}$ via
$$
g\colon X\mapsto gXg^\dagger,
\quad
X\in H.
$$
\end{itemize}
The space $H$ is another way to look at Minkowski spacetime because
the action of $\SL(2,\C)$ on $H$ preserves $\det X$ and, if we
consider the basis
$$
\sigma_0=\pmatrix{1&0\cr
0&1},
\quad
\sigma_1=\pmatrix{0&1\cr
1&0},
\quad
\sigma_2=\pmatrix{0&-i\cr
i&0},
\quad
\sigma_3=\pmatrix{1&0\cr
0&-1}
$$
so that all $X\in H$ are of the form
$X=x_0\sigma_0+x_1\sigma_1+x_2\sigma_2+x_3\sigma_3$, then
$$
\det X=x_0^2-x_1^2-x_2^2-x_3^2
$$
is the metric on Minkowski space. In this way we get a $2\colon 1$ and
onto homomorphism $p\colon\SL(2,\C)\to\SO_0(3,1)$.
Also, $H_0=\{X\in H\mid\tr X=0\}$ is isomorphic to $3$-dimensional
Euclidean space and $\SU(2)\subseteq\SL(2,\C)$ is the subgroup mapping
$H_0$ to itself.
We studied the topology of $\SO(3)$ and saw that one good way to
visualize this group is as a sphere of radius $\pi$ with opposite
points of the boundary identified.
$$
\xy
(0,0)*\xycircle(15,15){};
(0,0)*\dir{*}="0";
(-9,-12)**\dir{-} ?(1)*\dir{>};
"0";(9,12)**\dir{-} ?(1)*\dir{>};
(-15,-12)*\hbox{$\SO(3)$};
\endxy
$$
In this way, we see that $\SO(3)\simeq\R P^3$, the three-dimensional
real projective space. We have the following $2\colon 1$ maps
$$
\xymatrix{\SU(2)\ar[d]^{2\colon 1}&S^3\ar[d]^{2\colon 1}\\
\SO(3)\ar[r]^{\sim}&\R P^3}
$$
This suggest that we should try to visualize $\SU(2)$ as a
$3$-dimensional sphere, and one way to do this is to realize that
$\SU(2)$ is the space of unit quaternions!
\begin{exercise}
Check that
$$
\SU(2)=\{x\in\SL(2,\C)\mid x^\dagger x=1, \det x=1\}
$$
is equal to the space of unit quaternions
$$
\{x_0\sigma_0+i(x_1\sigma_1+x_2\sigma_2+x_3\sigma_3)\mid x_0^2+x_1^2+x_2^2+x_3^2=1\}.
$$
\end{exercise}
A way to visualize the $2\colon 1$ map from $\SU(2)$ to $\SO(3)$ is to
visualize $S^3$ as $\R^3$ with a single ``point at infinity'', and
consider the orbits of
$$
x\mapsto -\pi^2{x\over|x|^2}.
$$
It is clear that each point of the interior of the sphere of radius
$\pi$ maps to one and only one exterior point (the origin mapping to
the ``point at infinity'') and that opposite points of the surface map
to ech other. This is the ``antipodal map'' on $S^3$ and identifying
pairs of antipodal points leads to a the representation of $\R P^3$ as
a ball in $\R^3$ with antipodal boundary points identified.
\section{Understanding spinors}
So, $\SL(2,\C)$ acts on $\C^2$ (spinors) and on $H$ (Minkowski
spacetime).
\subsection{States of Weyl spinors}
In quantum mechanics the space of states of a system is the space of
unit vectors in a complex Hilbert space, ``modulo phase'' (i.e., up to
multiplication by a complex scalar of norm $1$). In our case, if we
give Weyl spinors $\psi={\psi_1\choose\psi_2}$ the Hilbert space norm
$\|\psi\|^2=|\psi_1|^2+|\psi_2|^2$, unit spinors form a
$3$-dimensional sphere $S^3$. Hence,
$$
\{\hbox{states of a Weyl spinor}\}\simeq S^3/\U(1).
$$
\subsection{The complex projective line}
Each $1$-dimensional subspace of $\C^2$ can be associated uniquely to
a unit vector modulo phase. In other words, each state of a Weyl
spinor uniquely determines the $1$-dimensional subspace of all spinors
proportional to it. Weyl spinors can then be interpreted as
homogeneous coordinates in the projective space $\C P^1$. We say that
two nonzero Weyl spinors ${\psi_1\choose\psi_2}$ and
${\phi_1\choose\phi_2}$ represent the same state if, and only if,
$\psi_1\phi_2=\psi_2\phi_1$. This is an equivalence relation.
\subsection{The Riemann sphere}
The complex projective line $\C P^1$ is well known to be isomorphic to
$\C$ with a point at infinity, an extension of the complex numbers
known as the Riemann sphere.
To see this, consider a $1$-dimensional complex subspace of $\C^2$
$$
\psi=\bigl\{\alpha{\textstyle{\psi_1\choose\phi_2}}\colon\alpha\in\C\bigr\}.
$$
If $\psi\neq 0$, we can write this as
$$
\psi=\bigl\{\alpha{\textstyle{\psi_1/\psi_2\choose 1}}\colon\alpha\in\C\bigr\},
$$
and we can associate the complex number $\psi_1/\psi_2$ to
$\psi$. Otherwise,
$$
\psi=\bigl\{\alpha{\textstyle{1\choose 0}}\colon\alpha\in\C\bigr\},
$$
corresponding to $\psi_1/\psi_2=\infty$.
A plane with a single point at infinity is topologically equivalent to
a sphere, as can be borne out by the following construction (the
stereographic projection)
$$
\xy
(0,0)*\xycircle(15,15){};
(-12,9)*{};(-30,9)**\dir{-};
(-50,-20)**\dir{-};
(30,-20)**\dir{-};
(50,9)**\dir{-};
(12,9)**\dir{-};
(0,11)*\dir{*}="0" +(0,2)*{\infty};
"0";(25,-15)**\dir{-} ?(1)*\dir{*} +(2,0)*{z} ?(.4)*\dir{*} +(-4,-2)*{f(z)};
\endxy
$$
\subsubsection{The Hopf fibration}
We have seen that the states of a Weyl spinor can be interpreted as
points on a $2$-dimensional sphere, so we have $S^3/\U(1)\simeq
S^2$. This is called the Hopf fibration
$$
\xymatrix{S^1\ar[d]\ar@{^(->}[r]&S^3\ar[d]\\
\{*\}\ar@{^(->}[r]&S^2\\}
$$
which shows that it is possible to fill the $3$-dimensional sphere
$S^3$ with an $S^2$ worth of circles.
\begin{exercise}
Draw the Hopf fibration in $\R^3\cup\{\infty\}$.
\end{exercise}
\subsection{Spin directions}
Intuitively, the Riemann sphere is just the set of all directions in
which a unit vector can point.
There are ``angular momentum operators'' acting on $\C^2$, qhich take
the form
$$
J_i={1\over 2}\sigma_i
\quad(i=1,2,3)
$$
The factor of $1\over 2$ is there so the commutation relations come
out to be
$$
[J_1,J_2]=J_3
\quad\hbox{(and cyclic permutations)}
$$
Given a unit spinor $\psi$, we say that its expected angular momentum
is the vector with components $\bracket\psi{J_i\psi}$. This is in no
way special about spinors, it is just the way quantum mechanics
works. Note that $\bracket\psi{J_i\psi}$ does not change if we
multiply $\psi$ by a phase, which explain why states are considered
modulo phase.
\begin{exercise}
Show that the length of the expected angular momentum vector is
$\sqrt{{1\over 2}({1\over 2}+1)}$.
\end{exercise}
In this way we can identify unit spinors modulo phase with directions
in space.
\subsection{$1$-dimensional projectors in $\C^2$}
Each point on the complex projective line represents a $1$-dimensional
subspace of $\C^2$. We can associate to each $1$-dimensional subspace
a projector, that is, a linear map $\rho\colon\C^2\to\C^2$ such that
$\rho^2=\rho$ and $\rho=\rho^\dagger$. The fact that $\rho$ projects
onto a $1$-dimensional subspace is equivalent to the condition
$\tr\rho=1$.
\begin{exercise}
Show that $\rho\colon\C^2\to\C^2$ is a hermitian $1$-dimensional
projector if, and only if, $\rho=\ket\psi\bra\psi$ for some state
$\psi$.
\end{exercise}
\subsection{The celestial sphere}
The equation $\rho^2=\rho$ and $\rho\neq 0,1$ implies that
$\det\rho=0$, so each $1$-dimensional projector is a hermitian matrix
with unit trace and vanishing determinant. In other words, it is an
element of $H$ (Minkowski space) such that
$x_0^2-x_1^2-x_2^2-x_3^2=0$, or a point of the light cone. The sphere
of spinor states is a set of null vectors. The light cone is the union
of the light rays through the origin, and we will see that in a
precise sense the $S^2$ of spinor states is the sphere of all
directions of light rays through the origin, also know as ``the sky''
or the ``celestial sphere''.