%% LyX 1.5.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[oneside,english]{amsart}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{amssymb}

\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
 \theoremstyle{plain}
\newtheorem{thm}{Theorem}[section]
  \theoremstyle{definition}
  \newtheorem{defn}[thm]{Definition}
 \theoremstyle{definition}
 \newtheorem*{defn*}{Definition}
  \theoremstyle{remark}
  \newtheorem*{claim*}{Claim}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\DeclareMathOperator{\Spin}{Spin}

\usepackage{babel}
\makeatother

\begin{document}

\title{The Dirac and Weyl Spinor Representations}


\author{Mike Thvedt}

\maketitle
This paper is concerned with representations of covers of subgroups
of the orthogonal group of relativistic spacetime. Specifically, I
look at the group $\Spin^{+}(3,1)$, the universal cover of the allowed
isometries of relativistic spacetime, and its representations. These
representations have implications in the field of particle physics.
It builds off an earlier paper of mine, \cite{me}, which gave an
elegant connection between paths in $SO(3)$ up to homotopy and the
behavior of the electron under the action of spatial rotation.


\section{Lie Groups and Universal Covering Groups}

Recall that a Lie group is a group and a differentiable manifold such
that the group operations are smooth. Also recall that the a cover
of a topological space is a topological space together with a locally
homeomorphic surjective map called the covering map.

The following information on universal covers and covering groups
was covered in \cite{me}. I go over it again here because this particular
construction of the universal cover is of direct importance.

\begin{defn}
The \emph{universal cover} of a topological space is a connected cover
with the universal property; i.e., the universal cover of $X$ covers
all connected covers of $X$. 
\end{defn}
It is a property of the universal cover that it is simply connected;
furthermore, any simply connected cover is homeomorphic to the universal
cover with homeomorphism the covering map. We require an important
theorem about universal covers: that they can be constructed as a
space of paths.

\begin{thm}
Let $X$ be a connected manifold with base point $x_{0}$. For any
path $\gamma$ starting at $x_{0}$, let $[\gamma]$ be the equivalence
class of paths homotopic to $\gamma$ with endpoints $x_{0}$ and
$\gamma(1)$. Let $U$ be the set of all such equivalence classes.
Then $U$ is the universal cover of $X$ where the covering map $p:U\rightarrow X$
is defined by $[\gamma]\mapsto\gamma(1)$.
\end{thm}
We can construct a base on this topology as follows. For a homotopy
class of paths $[\gamma]\in U$ and an open neighborhood $D\in X$
where $\gamma(1)\in D$, define $[\gamma,D]$ to be the set of all
paths homotopic to any path formed by $\gamma$concatenated to some
path contained entirely in$D$. The sets $[\gamma,D]$ give a base
for $U$. In rougher words, two paths $\alpha$ and $\beta$ are close
to each other if $\alpha$ is {}``almost'' homotopic with fixed
endpoints to $\beta$ and $\alpha(1)$ is close to $\beta(1)$.

To show that $U$ is the universal cover, it needs to be shown that
$U$ is simply connected, and that this implies the universal property.
For the sake of brevity I will not go into this; but the interested
reader might see, for example, \cite{riemann}.

\begin{defn}
The \emph{double cover} of a topological space is a cover such that
the covering map is a $2:1$ map.
\end{defn}
All connected Lie groups have a universal covering group with the
covering map being a map of Lie groups. We can construct this as follows:
For a Lie group $G$ with identity $e$, let the group $C$ be the
universal cover where the fixed point $x_{0}=e$. For $[\gamma],[\delta]\in C$,
define $[\gamma\delta]$ as the equivalence class of $\gamma\delta$,
where$(\gamma\delta)t=\gamma(t)\delta(t)$. Then $C$ is the universal
covering group of $G$. The proof that this construction works and
is a Lie group this rests on the fact that operators in Lie groups
act smoothly on paths in Lie groups.

This construction is the easiest construction that I have found, but
it misses the universal covers for non-connected Lie groups. A more
general construction as quotients of path spaces can be found in \cite{kolk}
pp. 62-68.

It should be pointed out that if a Lie group is not connected then
its universal cover will \emph{not }be unique as a Lie group. A specific
example of this will occur at the end of this paper.


\section{Minkowski Space and the Lorentz Group}

Physical spacetime is (we think) a manifold, but not one with the
usual Riemannian metric. The mathematician Minkowski showed in 1909
that Einstein's theory of special relativity can be elegantly described
if we imagine physical spacetime as what is now called Minkowski space.
To do this we not only allow but require that the distance between
two points can be nonnegative.

\begin{defn}
The metric space \textbf{$\mathbb{R}^{p,q}$} is the vector space
$\mathbb{R}^{p+q}$ with basis $\{e_{1}\ldots e_{p+q}\}$ together
with the metric $\|\vec{x}\|=(x_{1}^{2}+\ldots+x_{p}^{2}-x_{p+1}^{2}-\ldots-x_{p+q}^{2}).$
The metric space $\mathbb{R}^{3,1}$ is called \emph{Minkowski space.}
\end{defn}
This is not a metric space in the usual sense. The condition that
the metric be positive definite has been replaced with the weaker
condition of nondegeneracy. Note that $\mathbb{R}^{p,q}$ and $\mathbb{R}^{q,p}$
are similar as metric spaces (i.e. their metrics differ by a constant),
so they share almost all properties; they can be made the same with
a simple sign flip.%
\footnote{They are essentially the same as metric spaces, but notably, the usual
Clifford algebra is different in these spaces. It turns out that all
I cover in this paper, and tons more, can be constructed with Clifford
algebrae, in which case one must be careful whether one is in $\mathbb{R}^{3,1}$
or $\mathbb{R}^{1,3}$.%
}


\subsection*{Conventions}

For Minkowski space we usually use write the basis $(e_{0},e_{1},e_{2},e_{3})$
such that $\|e_{0}\|=-1$ and $\|e_{1}\|=\|e_{2}\|=\|e_{3}\|=1$.
Vectors in $\mathbb{R}^{3,1}$ are written $(x_{0},x_{1},x_{2},x_{3})$
or $(t,x_{1},x_{2},x_{3})$. The $x$ components with positive index
are called the \emph{spatial components}, and $x_{0}$ or $t$ is
the \emph{timelike component}. Likewise, the metric subspace $\mathbb{R}^{3}$
with the usual Euclidean metric is \emph{space}. The subspace $\mathbb{R}$
with the negative of the Euclidean metric is \emph{time}. Vectors
with a positive norm are called spacelike; vectors with a negative
norm timelike, and vectors with zero norm (excluding $\vec{0}$) are
called lightlike.

Now that we have a new sort of metric, it follows that we will have
new sorts of isometries.

\begin{defn}
The \emph{orthogonal group}\textbf{ $O(p,q)$} is the group of isometries
of $\mathbb{R}^{p,q}$. Elements of $O(p,q)$ are called rotations.
$O(p,q)$ is Lie under the matrix topology. The \emph{special orthogonal
group}\textbf{ $SO(p,q)$} is the subgroup of $O(p,q)$ of rotations
that preserve orientation. The \emph{orthochronous special orthogonal
group}\textbf{ $SO^{+}(p,q)$}is the identity component of $SO(p,q)$as
a Lie group. The group $O(3,1)$ is called the Lorentz group.
\end{defn}
Like $O(4)$, $O(3,1)$ is six-dimensional. It follows from the similarity
of $\mathbb{R}^{p,q}$ and $\mathbb{R}^{q,p}$ that $O(p,q)\cong O(q,p).$

As $\mathbb{R}^{3}$ as a metric space is a subspace of $\mathbb{R}^{3,1}$,
and $SO(3)$ is the identity component of $O(3)$, it follows that
$SO(3)$ will be a subgroup of $SO^{+}(3,1)$. It remains to examine
the rotations that alter the path of time.

\begin{defn}
A \emph{Lorentz boost }(or simply \emph{boost}) is a rotation $X\in SO^{+}(3,1)$
that fixes a plane in the spacelike subspace $\mathbb{R}^{3}$. 
\end{defn}
For a vector $\vec{x}=(t,x_{1},x_{2},x_{3})$, we can write a Lorentz
boost $L_{3}$ in the plane $(t,x_{3})$ through the hyperbolic angle
$\varphi$ as follows:\[
L_{3}=\left(\begin{array}{cccc}
\cosh\varphi & 0 & 0 & \sinh\varphi\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
\sinh\varphi & 0 & 0 & \cosh\varphi\end{array}\right)\]


\begin{defn}
The orthochronous Lorentz group $SO^{+}(3,1)$ is generated by the
spacelike rotations in $SO(3)$ and the Lorentz boosts. We have seen
that no rotation, be it boost, spatial or both, is orientation reversing.
Furthermore, it is plain from the description of Lorentz boosts that
no boost can reverse the direction of time in a timelike vector. For
a vector with timelike component $t$ and spacelike components $\vec{x},$
written $(t,\vec{x})$, call $T$ the rotation with $t\mapsto-t$
and fixing $\vec{x}$, and $S$ the spacelike rotation in $O(3)$
with $\vec{x}\mapsto-\vec{x}$. The rotation $ST$ is orientation
preserving, so adding $ST$ to $SO^{+}(3,1)$ gives $SO(3,1).$ Adding
$S$ or $T$ to $SO(3,1)$ gives us $O(3,1)$. It turns out these
are the only rotations we can add to $SO^{+}(3,1)$. Then $O(3,1)$
has four connected components, and $O(3,1)/SO^{+}(3,1)\cong V$, the
Klein four-group.
\end{defn}

\section{Relativity}

It remains to be shown that $\mathbb{R}^{3,1}$ is spacetime in special
relativity. Einstein postulated that the laws of physics are invariant
under all allowed transformations of space. Importantly, Einstein
required that physical laws be the same for both stationary and moving
observers. In other words, all motion is to be relative, but the speed
of light absolute. How is this accomplished?

Einstein proposed we consider two coordinate systems to be equivalent,
and that observers%
\footnote{An observer is anything that {}``observes'' the laws of the universe.
Just about everything is an observer, including you and me.%
} see themselves as stationary in their own coordinate systems. Suppose
these two coordinate systems have the same origin, and that the velocity
difference between them is $\vec{v}$. He showed that one coordinate
system can be transformed to the other by a Lorentz boost in the $\vec{v}$
direction with angle $\varphi$ where $\|\vec{v}\|/c=\tanh\varphi$.
Or you can think about it this way: giving yourself a velocity $\vec{v}$
(with respect to your previous coordinate frame) is the same as giving
yourself a Lorentz boost of the same velocity, and then remaining
stationary. (Of course, both of these require the same energy and
force.)

$c\approx299,700,000$ meters per second, but we can set $c=1$ if
we make our units the right size. Now introduce a metric such that
the distance covered by any ray of light will be $0$. This is precisely
the metric of Minkowski space. It follows that any isometry of Minkowski
space will preserve the speed of light, which is what we wanted. (Note
that the magnitude of a lightlike vector will remain $0$ under any
Lorentz transformation in the Minkowski metric, but will generally
not retain its magnitude in the Euclidean metric.) We have the poetic
result that adding velocity to a coordinate system is the same as
a rotation in time.

Recall from high school physics or your favorite science fiction television
show that we can get closer and closer to the speed of light but not
equal or surpass it by normal means of propulsion.%
\footnote{Unless you live in the Star Wars universe.%
}%
\footnote{Recently there have been reports in mainstream media of the speed
of light {}``being broken''. This is merely a consequence of mainstream
media being bad at science reporting. The speed of light has never
been observed to have been broken.%
} This implies that a sequence of Lorentz boosts, formed by applying
the same boost over and over, will take an observer arbitrarily close
to, but never directly at, the speed of light. So we have the following:

\begin{thm}
$SO^{+}(3,1)$ is not compact.
\end{thm}
As we shall soon see, the Lorentz group is not enough to describe
all the ways to rotate a particle. We also need to study the covers
of $SO^{+}(3,1)$.

\begin{defn}
\textbf{$\Spin(n)$} is the double covering group of $SO(n)$. \textbf{$\Spin(p,q)$}
is a double covering group of $SO(p,q)$. A group $\Spin(p,q)$ might
not be unique.
\end{defn}
$\Spin(n)$ is a double covering group of $SO(n).$ For $n>2$, $\Spin(n)$
is a universal covering group, and is simple and compact. In the theory
of so-called semisimple Lie groups, these and the other simple, compact,
simply connected Lie groups are the building blocks of compact Lie
groups; analogous to how simple finite groups are the building blocks
of finite groups.

The reason spin is important to us in this paper is because of the
connection between particle physics and spin. Isometries of spacetime
do not actually describe all the possible ways to rotate a particle.
In particular, a rotation of $2\pi$ is not homotopic to a rotation
of $0$, and somehow particles know this. An electron rotated $2\pi$
in $\mathbb{R}^{3}$ will end up with the opposite phase\cite{griffiths}.
Particles with this property are called \emph{spinning}, and a spinning
particle said to have spin $n/2$ will be invariant under a $4\pi/n$
rotation.


\section{Representations and Spinors}

\begin{defn}
A \emph{representation} of a group $G$ is a vector space m together
with a homomorphism $\rho:G\rightarrow Aut(V)$, the automorphism
group of $V$. When $\rho$ is obvious (and often even when it is
not) it is simply said that $V$ is a representation of $G$, or $V$
represents $G$. Confusingly, it is also common to refer to the map
$\rho$ itself as a representation. Also, the image $\rho(G)$ is
sometimes called a representation.
\end{defn}
Recall that a group action of $G$ on a set $X$ can be defined as
a map $\rho:G\rightarrow S_{X}$, the symmetric group of $X$, and
that $S_{X}$ is precisely the group of automorphisms of $X$ in the
category of sets. Then a representation is simply a group action in
the category of vector spaces. So for a representation $V$ of $G$
we can speak of $G$ acting on $V$ or vectors in $V$, and $gV$
can be shorthand for $\rho(g)(V)$.

\begin{defn}
A representation $V$ of $G$ is \emph{reducible} if there exists
some subspace $W\subset V$ such that $gW\subset W$ for $\forall g\in G$.
An \emph{irreducible representation}, or \emph{irrep}, is one that
is not reducible.
\end{defn}
Some representations are more or less the same:

\begin{defn}
Two representations $\rho_{1}:G\rightarrow V$ and $\rho_{2}:G\rightarrow W$
(over the same field) are said to be \emph{equivalent} if there exists
an isomorphism $\alpha:V\rightarrow W$ such that $G$ acts the same
on both; i.e., $\alpha\circ\rho_{1}(g)\circ\alpha^{-1}=\rho_{2}(g)$.
\end{defn}
It turns out that any finite-dimensional representation of a compact
semisimple Lie group {[}we need to define semisimple here] can be
decomposed into the direct sum of irreps \cite{hall}. Unfortunately
for us, we are interested in representations of covers of $SO^{+}(3,1)$,
which is not compact. Nevertheless, there exist clever tricks for
finding the finite-dimensional irreps that we need. The representations
we need have a name:

\begin{defn*}
A representation of a spin group is called a \emph{spinor representation}.
Elements of this representation are called \emph{spinors}\textbf{\emph{.}}
\end{defn*}
In \cite{me} I touched upon spinor representations of $\Spin(3)$.
There is an isomorphism $\Spin(3)\cong SU(2)$, and $SU(2)$ acting
on $\mathbb{C}^{2}$ is of course a representation of $\Spin(3)$.
This representation consists of two-dimensional column spinors with
complex entries. These spinors are used to describe the states of
electrons in three dimensions. To an electron, we assign an \emph{angular
momentum vector} that describes the spin and orientation of a particle
in space. In particular, I showed the following:

\begin{thm}
Action on spinors of electrons in 3-space by rotation is described
completely by $\Spin(3)$, the group of rotation paths up to fixed-endpoint
homotopy in $SO(3)$. Furthermore, each rotation path in $\Spin(3)$
has precisely the same effect on a particle's angular momentum vector
as it does on the angular momentum vector of an ordinary spinning
object in classical physics.
\end{thm}
This theorem follows from the two-valued action of $SO(3)$ on spinors
by factoring through to $\Spin(3)$, and that a rotation of $2\pi$
is not the same as the identity.

According to \cite{lonesuto}, the group $\Spin(3,1)$ acts on particles
in Minkowski space in the expected way, which is all we need to infer
that this theorem extends to particles in spacetime.


\section{The Weyl spinor representations of $\Spin^{+}(3,1)$}

First we need a mathematically convenient description of $\Spin^{+}(3,1)$,
the universal cover of $SO^{+}(3,1)$.

\begin{claim*}
$\Spin^{+}(3,1)\cong SL(2,\mathbb{C})$.
\end{claim*}
\begin{proof}
Proof by exhibition.

Let $A$ be the set of 2x2 Hermitian matrices. $A$ is a metric space
over $\mathbb{R}$ (but not $\mathbb{C})$ with norm the determinant.
Define an isomorphism $\sigma:\mathbb{R}^{3,1}\cong A$ as follows:\[
\left(\begin{array}{c}
t\\
x_{1}\\
x_{2}\\
x_{3}\end{array}\right)\mapsto\left[\begin{array}{cc}
t-x_{3} & x_{2}-ix_{1}\\
x_{2}+ix_{1} & t+x_{3}\end{array}\right]\]


A simple calculation shows that, for $\vec{v}\in\mathbb{R}^{3,1}$,
$\det(\sigma(\vec{v}))=\|\vec{v}\|$, so $\sigma$ is an isometry
of metric spaces. Isometric metric spaces are indistinguishable as
metric spaces.

Let $SL(2,\mathbb{C})$ act on $A$ not in the usual way but in the
following way: \[
a\mapsto sas^{*}\]


where $s\in SL(2,\mathbb{C})$. Call such a transformation $g_{s}$.
$(sas^{*})^{*}=sa^{*}s^{*}=sas^{*}$, so this action preserves hermitianivity.
It is clear that $g_{st}=g_{s}g_{t}$, and that the action of all
these $g$'s is smooth. So $SL(2,\mathbb{C})$ acts smoothly on $A$,
and on $\mathbb{R}^{3,1}$ through the isometry $\sigma$. This gives
a map of Lie groups $\rho:SL(2,\mathbb{C})\rightarrow O(3,1)$.

Let $k\in ker(\rho)$. Then $a=kak^{*}$ for all $a\in A$. Then $I=kk^{*}$
implying $k^{*}=k^{-1}$, so $a=kak^{-1}$. This means that $k$ is
in the center of $SL(2,\mathbb{C})$, so $k=\pm I$. So $\rho$ is
a $2:1$ map.

Now $SL(2,\mathbb{C})$ is simply connected. Then the image of $\rho$
is also connected, and can only be the identity component of $O(3,1)$.
Hence $SL(2,\mathbb{C})\cong Spin^{+}(3,1)$ , the universal (and
double) cover of $SO^{+}(3,1)$.
\end{proof}
Then $SL(2,\mathbb{C})$ acting on $\mathbb{C}^{2}$ is naturally
a representation of $\Spin^{+}(3,1)$. We also have the conjugate
representation $\bar{\rho}:\Spin^{+}(3,1)\rightarrow SL(2,\mathbb{C})$
acting on $\mathbb{C}^{2}$ defined by $\bar{\rho}(s)=\overline{\rho(s)}$.
Let $\bar{\rho}(\Spin^{+}(3,1))$ be called $\overline{SL(2,\mathbb{C})}$.

The representations of $SU(2)$ and $SL(2,\mathbb{C})$ are closely
related. In fact, it turns out there is a complete classification
of the finite-dimensional representations of the former that leads
to a convenient classification on the latter, but that is beyond the
scope of this paper. This is covered in many texts on the representation
theory of Lie groups; for example, \cite{semisimple}. The text \cite{groups reps physics}
provides a physics-based approach.

The representation $\mathbb{C}^{2}$ acted upon by its special linear
group is irreducible since $SL(2,\mathbb{C})$ clearly fixes no subspace
of $\mathbb{C}^{2}$. Its conjugate representation (sometimes called
the dotted or right-handed representation) is also irreducible. These
are called the \emph{Weyl spinor representations}, and they describe
the states of neutrinos \cite{groups reps physics}.


\section{The Dirac spinor representation}

For electrons the picture is more complicated. We require that the
spinors of electrons be representations of the entire spin group,
$\Spin(3,1)$ \cite{groups reps physics}. We can achieve this with
\emph{Dirac spinors.}

To do this, we simply take the direct sum of the two Weyl representations.
Let $\chi_{1},\chi_{2}$ be two Weyl spinors, and $\left(\begin{array}{c}
\chi_{1}\\
\chi_{2}\end{array}\right)$ a Dirac spinor. An element $s\in\Spin^{+}(3,1)$ acts on Dirac spinors
as the matrix $\left(\begin{array}{cc}
s & 0\\
0 & \bar{s}\end{array}\right)$ . To get a representation of the entire spin group we add the matrix
$j$:

\[
j=\left(\begin{array}{cccc}
-i & 0 & 0 & 0\\
0 & -i & 0 & 0\\
0 & 0 & i & 0\\
0 & 0 & 0 & i\end{array}\right)\]
This matrix represents the total reflection sending a vector $\vec{u}$
to $-\vec{u}$. $j$ commutes with every $\left(\begin{array}{cc}
s & 0\\
0 & \bar{s}\end{array}\right)$; also, $j^{2}=-1$ which represents a $2\pi$ rotation. So $j$ has
all the necessary properties we need for it to represent total reflection.

It would also be possible to pick a properly commuting $j$ such that
$j^{2}=1$, representing the identity rotation. Such a representation
would also give a double covering of $SO(3,1)$. It follows that there
are two choices of the nonorthochronous group $\Spin(3,1)$.

In fact, we can do even better. We can add the matrices

\[
s=\left(\begin{array}{cccc}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\end{array}\right)\]


and

\[
t=\left(\begin{array}{cccc}
0 & 0 & i & 0\\
0 & 0 & 0 & i\\
-i & 0 & 0 & 0\\
0 & i & 0 & 0\end{array}\right)\]


in order to get a double covering representation of $O(3,1)$ \cite{shapiro}.
I won't prove it here. This is called a pin representation%
\footnote{This joke, due to Serre (see \cite{abs}) is known by many but fully
appreciated by few; it is said that 'pin' is a vulgar expression in
Serre's native language.%
}. This is only one of several ways to have a pin representation consisting
of Dirac spinors. There are in fact two nonisomorphic Pin groups,
due to the nonconnectedness of $O(3,1)$; ones in which two reflections
are $2\pi$ are called $Pin_{-}$, and the others $Pin_{+}$.

\begin{thebibliography}{10}
\bibitem{humphreys}Humphreys, J.E. \emph{Introduction to Lie Algebras
and Representation Theory.} Springer-Verlag, 166pp.

\bibitem{semisimple}Knapp, Anthony W. \emph{Representation Theory
of Semisimple Groups.} Princeton University Press, 773pp.

\bibitem{onischick}Onishchik, A.L, ed. \emph{Lie Groups and Lie Algebras..}
Vol. 1. Springer-Verlag, 235pp.

\bibitem{symmetries}Fuchs, Juergen and Schweigert, Christoph. \emph{Symmetries,
Lie Algebras and Representations.} Cambridge University Press, 438pp.

\bibitem{hall}Hall, Brian C. \emph{Lie Groups, Lie Algebras, and
Representations.} Springer-Verlag, 351pp.

\bibitem{groups reps physics}Jones, H. F. \emph{Groups, Representations
and Physics.} Institue of Physics Publishing, 326pp.

\bibitem{lonesuto}Lonesuto, Pertti. Clifford Algebras and Spinors.
Cambridge University Press, 306 pp.

\bibitem{riemann}Beardon, A.F. A Primer on Riemann Surfaces. Cambridge
University Press, 188 pp.

\bibitem{griffiths}Griffiths, David J. Introduction to Quantum Mechanics.
2nd ed. Pearson Prentice Hall, 468 pp.

\bibitem{me}Thvedt, Michael. \emph{Rotations, Rotation Paths, and
Quantum Spin. {[}Available online at} http://www.math.uchicago.edu/\textasciitilde{}may/VIGRE/VIGREREU2006.html].

\bibitem{shapiro}Gel'Fand, Minlos and Shapiro. \emph{Representations
of the roation and Lorentz groups and their applications. Pergamon
Press, 366pp.}

\bibitem{steinberg}Steinberg, S. \emph{Group Theory and Physics.}
Cambridge University Press, 429pp.

\bibitem{abs}M. F. Atiyah, R. Bott, A. A. Shapiro, Clifford modules,
Topology 3 (1964), 3-38.

\bibitem{kolk}Duistermaat, J. J., and Kolk, J. A. C. Lie Groups.
Springer-Verlag, 344 pp.
\end{thebibliography}

\end{document}
