\chapter{Building Spacetime from Spin (I)}
The Dirac equation
$$
(\dirac\partial+im)\psi =0
$$
describes electrons and, when Dirac discovered it, predicted the
existence of a particle with the same mass and opposite charge, the
positron. The terms of the equation are
\begin{itemize}
\item $\psi$, called a ``spinor field'', is a function
$\psi\colon\R^4\to\C^4$, where $\R^4$ is $(3+1)$-dimensional
Minkowski space and $\C^4$ is the space of ``Dirac spinors'';
\item $m$ is the mass of the spinors described by $\psi$
(electrons/positrons); and
\item $\dirac\partial$ is called the Dirac operator, and has the form
$\dirac\partial=\gamma^\mu\partial_\mu$, where $\gamma^\mu$ are
the Dirac matrices which ensure that $\dirac\partial$ is a
Lorentz-covariant operator.
\end{itemize}
\section{Spinors}
The space of Dirac spinors $\C^4$ carries a representation of the
Lorentz group on $\R^4$, but the relationship between $\R^4$ and
$\C^4$ is more complex that just complexifying spacetime
coordinates. The relationship is most easily explained in terms not of
Dirac spinors but of Weyl spinors in $\C^2$,
$$
\begin{array}{ccc}
\hbox{Dirac spinors}&&\hbox{Weyl spinors}\\
\C^4&\simeq&\C^2\oplus(\C^2)^*
\end{array}
$$
The notation $(\C^2)^*$ denotes the dual of $\C^2$, which is
isomorphic to $\C^2$ as a vector space but carries the dual
representation of the Lorentz group. In a certain sense, $\C^2$ is the
space of left-handed Weyl spinors, and its dual is the space of
right-handed spinors.
To understand Weyl spinors we need to study four symmetry groups that
appear all over the place in physics: $\SO(3)$, $\SU(2)$, $\SO_0(3,1)$
and $\SL(2,\C)$.
\subsection{The rotation group}
The rotation group $\SO(3)$ is the space of linear transformations
$R\colon\R^3\to\R^3$ which leave an inner product and an orientation
on $\R^3$ invariant. In matrix notation, we write the inner product as
$$
g(x,x)=x\cdot x=x^Tx=x_1^2+x_2^2+x_3^2,
$$
where $x=(x_1,x_2,x_3)$ is a point of $\R^3$.
The condition that $R$ preserve the inner product is $R^*R=1$, where
the adjoint of $R$ is defined by $g(x,Ry)=g(R^*x,y)$ for all
$x,y\in\R^3$. In matrix notation,
$$
g(R^*x,y)=g(x,Ry)=x^TRy=x^TR^{TT}y=(R^Tx)^Ty=g(R^Tx,y),
$$
so $R^*=R^T$ and the inner product is preserved if $R^TR=1$. The
orientation is preserved if $\det R=1$. Hence,
$$
\SO(3)=\{R\colon\R^3\to\R^3\mid R^TR=1\quad\mathrm{and}\quad\det
R=1\}.
$$
\subsection{The Lorentz group}
The proper Lorentz group $\SO(3,1)$ is defined similarly, and consists
of all linear transformations $\Lambda\colon\R^4\to\R^4$ preserving
preserving an orientation and the indefinite inner product
$$
g(x,x)=x^T\eta x=x_0^2-x_1^2-x_2^2-x_3^2,
$$
where $\eta=\diag(1,-1,-1,-1)$. The invariance of the inner product is
still equivalent to $\Lambda^*\Lambda=1$ or, in matrix notation,
$\Lambda^T\eta\Lambda=\eta$. Orientation-preserving transformations
satisfy $\det\Lambda=1$, so
$$
\SO(3,1)=\{\Lambda\colon\R^4\to\R^4\mid \Lambda^T\eta\Lambda=\eta\quad\mathrm{and}\quad\det\Lambda=1\}.
$$
We encounter a complication that was absent in the $\SO(3)$ case,
since every element of $\SO(3)$ can be connected to the identity by a
path staying in $\SO(3)$, but not so in $\SO(3,1)$. To see this,
consider the transformation $\Lambda x=-x$ (interpreted as reversal of
time and orientation simultaneosly), which cannot be obtained from the
identity by a continuous transformation.
We don't want our symmetry groups to have disconnected components, so
we define $\SO_0(3,1)$ to be the connected component of the identity
in $\SO(3,1)$. We then have the inclusion
$$
\xymatrix{
\SO(3)\ar@{^(->}[r] &\SO_0(3,1)\\
R\ar@{|->}[r]&\pmatrix{1&0\cr
0&R\cr}\\}
$$
\subsection{The unitary group}
There is an amazing physical fact about electrons, and that is that
they can distinguish a rotation by $360^\circ$ from no rotation at
all, but a rotation by $720^\circ$ is indistinguishable from no
rotation.
\paragraph{The coffee-cup trick} Yadda, yadda, yadda.
\paragraph{Topology of $\SO(3)$} An explanation of the coffee-cup trick
or of the behaviour of electrons under $360^\circ$ rotations is that
there are certain special curves in $\SO(3)$ that join the identity to
itself but cannot be contracted to a point.
We can find a convenient representation of the rotation group $\SO(3)$
in terms of axes and angles. It is well-known that every rotation
fixes an axis in space. Every rotation can be described by a direction
in space and an angle in $[0,\pi]$ of counterclockwise rotation about
the given direction. All rotations of angle $0$ are equivalent and
represent the identity and, more importantly, rotations of $\pi$ about
opposite directions represent the same rotation.
Therefore, the rotation group can be represented by a solid
$3$-dimensional sphere of radius $\pi$, with opposite points of the
boundary identified. The following curve represents all the rotations
of angle $0$ to $2\pi$ about a single axis.
$$
\xy
(0,0)*\xycircle(15,15){};
(0,0)*\dir{*}="0";
(-9,-12)**\dir{-} ?(.5)*\dir{<};
"0";(9,12)**\dir{-} ?(.5)*\dir{>};
(-15,-12)*\hbox{$\SO(3)$};
\endxy
$$
Because of the identification of the opposite points of the boundary,
it is impossible to ``detach'' the curve from the boundary by
deformations, and therefore this curve is noncontractible. On the
other hand, the sequence of rotations of angles from $0$ to $4\pi$ is
contractible, as shown by the following diagram.
$$
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
(-6,-8)**\dir{=} ?(.5)*\dir3{<};
"0";(6,8)**\dir{=} ?(.5)*\dir3{>};
\endxy
\quad
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
(-6,-8)**\dir{-} ?(.5)*\dir{<};
"0";(6,8)**\dir{-} ?(.5)*\dir{>};
"0";(8,6)**\dir{-} ?(.5)*\dir{>};
"0";(-8,-6)**\dir{-} ?(.5)*\dir{<};
\endxy
\quad
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
(-6,-8)**\dir{-} ?(.5)*\dir{<};
"0";(6,8)**\dir{-} ?(.5)*\dir{>};
"0";(10,0)**\dir{-} ?(.5)*\dir{>};
"0";(-10,0)**\dir{-} ?(.5)*\dir{<};
\endxy
\quad
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
(-6,-8)**\dir{-} ?(.5)*\dir{<};
"0";(6,8)**\dir{-} ?(.5)*\dir{>};
"0";(0,10)**\dir{-} ?(.5)*\dir{<};
"0";(0,-10)**\dir{-} ?(.5)*\dir{>};
\endxy
\quad
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
(-6,-8)**\dir{-} ?(.5)*\dir{<};
"0";(0,-10)**\dir{-} ?(.5)*\dir{>};
(0,10);(6,8)**\crv{(0,0)} ?(.3)*\dir{>};
\endxy
\quad
\xy
(0,0)*\xycircle(10,10){};
(0,0)*\dir{*}="0";
"0";**\crv{(0,-10)&(-6,-8)} ?(.2)*\dir{>};
\endxy
$$
\paragraph{The Pauli spin matrices} When Pauli was trying to reconcile
Quantum mechanics with the geometry of $3$-dimensional space, he found
the spin matrices associated to $\SU(2)$, the double cover of
$\SO(3)$. Similarly, Dirac was led to the group $\SL(2,\C)$, the
double cover of $\SO_0(3,1)$, when discovering the Dirac equation. The
relationship between the four groups is summarised in the following
diagram:
$$
\xymatrix{
&\mathrm{space}&\mathrm{spacetime}\\
\mathrm{quantum}&\SU(2)\ar@{^(->}[r]\ar[d]^{2:1}&\SL(2,\C)\ar[d]^{2:1}\\
\mathrm{classical}&\SO(3)\ar@{^(->}[r]&\SO_0(3,1)\\}
$$
Now, $\SU(2)$ is the group of complex linear transformations
$U\colon\C^2\to\C^2$ with unit determinant and preserving the complex
inner product
$$
\langle z,z\rangle=z^\dagger\cdot z=\bar z^Tz=\bar z_1z_1+\bar z_2z_2+\bar z_3z_3,
$$
where $\bar z$ denotes the complex conjugate, and $z^\dagger$ the
conjugate transpose, of $z$. That $U$ preserves the inner product is
equivalent to $U^*U=1$ or, in matrix notation, $U^\dagger U=\bar
U^TU=1$. In other words,
$$
\SU(2)=\{U\colon\C^2\to\C^2\mid U^\dagger U=1\quad\hbox{and}\quad\det
U=1\}.
$$
To understand how $\SU(2)$ can represent rotations, we have to use the
fact that $\R^3$ with its usual inner product is isomorphic to
$2\times 2$ complex, hermitian, traceless matrices:
$$
\xymatrix{
(x_1,x_2,x_3)\ar[r]\ar[d]^\cdot&\pmatrix{x_3&x_1-ix_2\cr
x_1+ix_2&-x_3\cr}\ar[ld]^{-\det}\\
x_1^2+x_2^2+x_3^2}
$$
We denote
$$
H_0=\Bigl\{\textstyle{\left(x_3\atop x_1+ix_2\right.\left.x_1-ix_2\atop
x_3\right)}\colon (x_1,x_2,x_3)\in\R^3\Bigr\}
$$
($H$ is for hermitian and $0$ for traceless). We now need an action of
$\SU(2)$ on $H_0$ preserving the determinant and the trace of elements
of $H_0$. Such an action is given by
$$
g\colon X\mapsto gXg^{-1}
$$
where $X=\left(x_3\atop x_1+ix_2\right.\left.x_1-ix_2\atop x_3\right)$
is the element of $H_0$ associated to $x=(x_1,x_2,x_3)$. To show that
this is the appropriate action of $\SU(2)$ on $H_0$, we calculate the
adjoint, trace and determinant of $gXg^{-1}$:
\begin{itemize}
\item $\det(gXg^{-1})=\det g\det X\det g^{-1}=\det X$;
\item $\tr(gXg^{-1})=\tr(Xg^{-1}g)=\tr X$; and
\item
$(gXg^{-1})^\dagger=(gXg^\dagger)^\dagger=g^{\dagger\dagger}X^\dagger
g^\dagger=gXg^\dagger$, where we have used the fact that $X$ is
Hermitian ($X^\dagger=X$) and $g$ is unitary $(g^\dagger=g^{-1})$.
\end{itemize}
There is just one detail left, and that is that there are two elements
of $\SU(2)$ corresponding to each rotation.
\begin{exercise}
Check that $gXg^\dagger=X$ has exactly two solutions in $\SU(2)$,
namely $g=\pm1$, so that the map $\SU(2)\to\SO(3)$ is $2\colon 1$.
\end{exercise}
\subsection{The special linear group}
We can illustrate the $2\colon 1$ map $\SL(2,\C)\to\SO_0(3,1)$ by very
similar arguments to the above. In short, $\R^4$ with the Minkowski
metric is isomorphic to the space of Hermitian $2\times 2$
matrix. Indeed,
$$
\det\pmatrix{x_0+x_3&x_1-ix_2\cr
x_1+ix_2&x_0-x_3\cr}=x_0^2-x_1^2-x_2^2-x_3^2.
$$
Now, we can define an action of $SL(2,\C)$ on the space $H$ of
Hermitian matrices by $X\mapsto gXg^\dagger$. The proof that this maps
$H$ to itself and preserves $\det X$ is the same as for $SU(2)$ acting
on $H_0$. It is because there is no need to preserve the trace of $X$
that the condition $g^\dagger=g^{-1}$ can be dropped.
\paragraph{Other dimensions}
This ``trick'' using complex matrices only works in $4$-dimensional
spacetime. It is a coincidence that Pauli and Dirac were inclined to
use complex numbers because they wanted to interpret the components of
spinors as complex probability amplitudes, and were thus led to these
formulations in terms of small matrices. Similar constructions exist
for the other real division algebras (the real numbers $\R$,
quaternions $\Qua$ and octonions $\Oct$), according to the following
table
$$
\begin{array}{rclc}
&\hbox to 0pt{\hss Groups\hss}&&\hbox{Weyl spinors}\\
\SL(2,\R)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(2,1)&\R^2\\
\SL(2,\C)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(3,1)&\C^2\\
\SL(2,\Qua)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(5,1)&\Qua^2\\
\SL(2,\Oct)&\stackrel{2\colon 1}{\longrightarrow}&\SO_0(9,1)&\Oct^2\\
\end{array}
$$