\chapter[Geometry of transformation groups]{Quaternions, rotations and
\\ the geometry of transformation groups}\index{transformation group}
\label{ch!lie}
\index{Lie group|(}
\index{transformation group|(}
\markboth{\quad\thechapter\ Geometry of transformation
groups\hfill}{}

Chapters~\ref{ch!eucl} to~\ref{ch!proj} discussed transformations such as
Euclidean rotations in the plane that depend continuously on parameters: the
centre of the rotation and its angle. I stressed that composition of
transformations is a natural operation, an idea that led in
Chapter~\ref{ch!gp} to the definition of a geometric transformation group.
Here I focus on groups with a continuous\index{continuous|(} family of
elements, especially some examples arising in geometry where the group of
transformations has an interesting geometry of its own. The discussion is a
first introduction to some of the basic ideas of ``continuous transformation
groups''. The formal definition and a detailed treatment of this type of
``group-manifold'' (or {\em Lie group}) is beyond the scope of this book,
but see \ref{sec!lie} and Segal~\cite{Se}.

As an example, recall that a rotation\index{rotation} of $\E^2$ around a
fixed point $P$ is given by the matrix $\left(\begin{smallmatrix}
\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{smallmatrix}\right)$,
and so depends continuously on the real parameter~$\theta$. This parameter
takes values in a circle. Thus the group of rotations of $\E^2$ around a
fixed point has a geometry of its own, that of the circle, as shown in
Figure~\ref{fig!rot_and_S1}. The relation between rotations in the plane and
the circle can be conveniently expressed in terms of complex numbers, with
the action of rotation by $\theta$ on the column vector
$\left(\begin{smallmatrix}x\\y\end{smallmatrix}\right)$ written as
multiplication of the complex number $x+iy$ by the complex number of
absolute value~1 $\exp(i\theta)$. On the other hand, the set of unit complex
numbers is the circle $S^1$ in the complex plane.

\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}}
\centering \input{lie_rot_S1.pstex_t}
\caption{The geometry of the group of planar rotations}
\label{fig!rot_and_S1}
\end{figure}

A highlight of this chapter is Corollary~\ref{ssec!sph_and_so}, which
applies the homeo\-morphism\index{homeomorphism!criterion} criterion
Theorem~\ref{homeo crit} (one of the main results of Chapter~\ref{top
chap}) to give a description in similar terms of the
topology\index{topology!of $\SO(3)$} of the groups of rotations of $\E^3$
and $\E^4$ around a fixed point. The algebra of complex numbers is
replaced by the algebra of {\em quaternions}
\[
\HH=\bigl\{a+bi+cj+dk\bigr\}\quad \hbox{with $a,b,c,d\in\R$,}
\]
where $i,j,k$ all square to $-1$ and multiply together wisely.
Corollary~\ref{ssec!sph_and_so} describes the topology\index{topology!of
$\SO(3)$} of the group of three- and four-dimensional rotations in terms of
the sphere $S^3$ of unit quaternions.

The group of three dimensional rotations is of basic importance in many
areas of mechanics and physics, describing symmetries of Euclidean space
$\E^3$, a space that old-fashioned empiricists believe we inhabit. The
quantum mechanical treatment of the {\em spin}\index{spin} of the
electron\index{electron} is a pretty illustration of my treatment of the
topology\index{topology!of $\SO(3)$} of the group of three-dimensional
rotations. As most ingredients are at hand already, I cannot resist the
temptation to include a section on this, cribbed more or less directly from
Feynman~\cite{Fn}. The discussion puts together in a very satisfactory way
ideas from algebra (groups, algebra of quaternions), analysis
(topology,\index{topology} compactness),\index{compact} geometry (rotations
of $\E^3$)\index{rotation} and quantum physics (wave function, spin of the
electron).\index{electron} \index{spin}

\section{Topology on groups}\label{sec!top on groups}
\index{topological!group|(}
\markright{\hfill \thesection\ Topology on groups\quad}

A group $G$ is a\index{topological!group} {\em topological group} if it has
a topology\index{topology} defined on it so that multiplication and inverse
are continuous.\index{continuous} In more detail, a topological group is an
object $G$ having two quite different structures: a collection of open
subsets\index{open set} satisfying the axioms for a topology, and a
multiplication map with identity and inverse satisfying the group axioms. I
require the group structure to respect the topological structure in the
sense that
\[
\begin{array}{rcccccccl}
{\rm mult}\colon & G\times G & \to & G &
\quad\hbox{and}\quad & {\rm inv}\colon & G & \to & G \\
& (g,h) & \mapsto & gh & & & g & \mapsto & g\1
\end{array}
\]
are both continuous maps of topological spaces; here $G\times G$ has the
product topology of~\ref{prod top}.\index{product topology}

\paragraph{Example 1} Any finite group $G$ is a topological group under the
discrete topology.\index{discrete topology}

\paragraph{Example 2} The groups $(\R, +)$ and $(\R^*, \times )$ are
topological groups with respect to the usual topology of $\R$. This is just
a fancy way of restating the fact, used all over the place in a first
analysis course, that the four operations addition, subtraction,
multiplication and division are continuous on the reals.

\paragraph{Example 3} A substantial generalisation of the previous example
brings us back to the linear geometries of
Chapters~\ref{ch!eucl}--\ref{ch!proj}. Recall the {\em general linear group}
$\GL(n,\R)$ of $n\times n$ real invertible matrixes. Note that $\GL(n,\R)$
is a subset of the set of real matrixes $M(n\times n,\R)=\R^{n^2}$. This
latter is a metric space, and therefore has a natural metric
topology.\index{metric!topology} Moreover, it is an easy fact that matrix
multi\-plication and inverse are continuous.\index{continuous|)} Hence
$\GL(n,\R)$ is a
topological group. As a consequence, affine transformations $\Aff(n,\R)$
(compare~\ref{sec!afflinA}) also form a topological group.

The group $\R^*$ of constant diagonal matrixes is a subgroup of
$\GL(n+1,\R)$, the {\em centre} of $\GL(n+1,\R)$, that is, the subgroup of
elements commuting with every element $g\in\GL(n+1,\R)$ (see~\ref{PGL}
and~Ex.~\ref{ch!gp}.\ref{ex!cent}). The quotient
\[
\PGL(n+1,\R)=\GL(n+1,\R)/\R^*
\]
is a topological group with the quotient topology.\index{quotient
topology} This is of course the group of projective linear
transformations of $\PP^n$ familiar from~\ref{PGL}, the {\em projective
linear group}.

\paragraph{Example 4} The {\em orthogonal group}
\[
\O(n)=\bigl\{ A\in \GL(n,\R)\bigm| \tr AA=1_n\bigr\},
\]
the group of orthogonal $n\times n$ matrixes,\index{orthogonal!group
$\protect\SO(n)$|(}\index{orthogonal!matrix} is a topological group in
the subspace topology.\index{subspace topology} Hence also $\Eucl(n)$,
the group of Euclidean motions,\index{Euclidean!motion} and the group of
motions of $S^2$ (see~\ref{sec!s2ve2}) are topological groups.

\paragraph{Example 5}\index{hyperbolic!motion}\index{motion}
Hyperbolic motions form a matrix group, the {\em Lorentz group} or group
of Lorentz transformations\index{Lorentz!transformation} (see~\ref{h2d}
for the notation and compare Theorem~\ref{h2d} and \exref{Opq ex})
\[
\O^+(1,2)=\left\{ A\in \GL(3,\R)\left|
\begin{array}{l}
\hbox{$\tr AJA=J$, and $A$ preserves the} \\
\hbox{halves of the cone $q_L(\bv)<0$}
\end{array}
\right\}\right..
\]
This is also a topological group. It and its higher dimensional colleagues
$\O^+(1,n)$ are important in special relativity\index{special!relativity}
and related areas of physics.

\vspace{0.2in}

The topological groups in Examples 2--5 have an interesting ``continuous''
geometry.\index{topological!group|)} Here is a simple example (see
Figure~\ref{fig!rot_and_S1} in the Introduction): recall that $\O(2)$ is the
group of all rotation matrixes\index{rotation!matrix}
$\left(\begin{smallmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta
\end{smallmatrix}\right)$ and reflection matrixes\index{reflection!matrix}
$\left(\begin{smallmatrix}\cos\theta & \sin\theta\\ \sin\theta & -\cos\theta
\end{smallmatrix}\right)$. Thus $\O(2)$ is a union of two connected
components\index{connected!component}, each a copy of the circle $S^1$
parametrised by the angle $\theta$. One aim of this chapter is to generalise
this nice description to some other orthogonal groups.

\section{Dimension counting}\label{sec!parameters}
\markright{\hfill \thesection\ Dimension counting\quad}
\index{dimension!of a Lie group|(}

Here I begin the study of some particular aspects of the geometry of
transformation groups. In this section I want to concentrate on a measure of
their size. Recall that $\O(2)$ can be described geometrically as the union
of two circles. The circle $S^1$ is a one dimensional geometric object in
the sense that its points depend on one real parameter $\theta$; standing at
a point of the circle, there is one direction in which you can move.

Without going into rigorous details, by {\it dimension}\index{dimension} of
a transformation group $G$, denoted $\dim G$, I understand the number of
continuous real parameters needed to characterise an element $g\in G$. The
previous paragraph then shows that $\dim\O(2)=1$. Do not get confused by the
fact that $\O(2)$ has two components; to characterise elements of $\O(2)$, I
need one continuous real parameter (the angle $\theta$) and a discrete
parameter (the choice of one of the components, equivalently the sign of the
determinant, or its value $\pm1$).

I proceed to compute the dimension of transformation groups in some
nontrivial cases. The computations will be performed by describing elements
of the groups in a way which makes it possible to count the parameters
involved directly.

\begin{prop} An\index{dimension} element $g\in\Eucl(n)$ depends on
$\displaystyle\binom{n+1}2$ real parameters, so
$\dim\Eucl(n)=\displaystyle\binom{n+1}2$. Further,
\[
\dim\O(n)=\binom n2, \quad
\dim\GL(n,\R)=n^2, \quad
\dim\PGL(n+1,\R)=n(n+2).
\]
\end{prop}

\begin{pf} The language of Euclidean frames from~\ref{Eframe} gives a
way of specifying elements of the Euclidean group. Choose a reference
frame $\{P_0,P_1,\dots,P_n\}$; then by Theorem~\ref{Eframe}, elements of
the Euclidean group $\Eucl(n)$ correspond one-to-one with the set of
Euclidean frames\index{Euclidean!frame}\index{frame!Euclidean}
$\{Q_0,Q_1,\dots,Q_n\}$. Now calculate:
\begin{itemize}

\item $Q_0\in\E^n$ is any point, so depends on $n$ parameters;

\item $Q_1\in\E^n$ is any point with $d(Q_0,Q_1)=1$, that is, it is any
point of the unit sphere $S^{n-1}$\index{sphere $S^n$} with centre $Q_0$,
hence depends on $n-1$ real parameters;

\item writing $\bbe_1=\vec{P_0P_1}$ and
$\bbe_1^\perp=\E^{n-1}\subset\E^n$ for the orthogonal
complement,\index{orthogonal!complement $V^\perp$} $Q_2$ is given by a
point of the unit sphere $S^{n-2}\subset\E^{n-1}$, so depends on $n-2$
real parameters;

\item similarly, $Q_i$ is given by a point of $S^{n-i}$, and hence
depends on $n-i$ real parameters;

\item in particular, $Q_n$ is one of two points, so has no continuous
parameter.
\end{itemize}

Thus a Euclidean frame\index{frame!Euclidean} depends on
\[
\dim\Eucl(n)= n+(n-1)+\cdots+1+0=\binom{n+1}2
\]
parameters.

An element of $\O(n)$ fixes the origin, which I can take to be $P_0=Q_0$
in the above argument. Hence the dimension count is
\[
\dim\O(n) = (n-1)+\cdots+1+0=\binom{n}{2},
\]
agreeing with $\dim\O(2)=1$. Said slightly differently, $\O(n)$ and
$\Eucl(n)$ differ by the translation part (compare Proposition~\ref{sd
prod}), which accounts for $n$ parameters:
\[
\dim\O(n)= \dim\Eucl(n)-n = \binom{n+1}2- n = \binom{n}2.
\]

The dimension of the general linear group can be calculated in exactly
the same way. Elements of $\GL(n,\R)$ correspond to invertible maps of
the vector space $\R^n$. Such a map is determined by the images of the
$n$ usual basis vectors in $\R^n$, parametrised by a total of $n^2$
numbers (the entries of the matrix representing the map). Not all
parametrisations give invertible maps, but most do: I only have to
exclude matrixes with zero determinant. Hence there are $n^2$ real
parameters involved, so
\[
\dim\GL(n,\R)= n^2.
\]
Finally by Theorem~\ref{PGL} there are as many projective transformations
as projective frames of reference. Hence I have to pick $n+2$ general
points in $\PP^n$, leading to
\[
\dim\PGL(n+1,\R)=(n+2)n
\]
parameters. Incidentally, the dimension of the projective group can also
be calculated from its definition $\PGL(n+1,\R)=\GL(n+1,\R)/\R^*$, which
gives
\begin{align*}
\dim\PGL(n+1,\R)&=\dim\GL(n+1,\R)-1 \\
&=(n+1)^2-1=(n+2)n. \QED
\end{align*}
\end{pf}

You can design your own parameter counts for some other groups not mentioned
in the proposition; for example, do and generalise~\exref{soparam}.
\index{dimension!of a Lie group|)}

\section{Compact and noncompact groups} \label{decomp of GL(n)}
\index{compact!Lie group}
\markright{\hfill \thesection\ Compact and noncompact groups\quad}

\begin{prop} The orthogonal group $\O(n)$\index{orthogonal!group
$\protect\SO(n)$} is a compact topological space.
\end{prop}
\begin{pf} This is a simple application of Proposition~\ref{comp}. The
orthogonal group is a matrix group: it is a subspace of the space
$\R^{n^2}$ of real matrixes. Hence it is enough to show that it is
closed and bounded.\index{closed!and bounded} The equation $\tr A A =
1_n$ defines a closed subset of $\R^{n^2}$, so the main issue is
boundedness. However, if $A=(a_{ij})$ is
orthogonal,\index{orthogonal!matrix|(} then its columns form an
orthonormal basis and in particular for every $1\leq k \leq n$,
$\sum_{i=1}^n a_{ik}^2 =1$. Hence
\[
\sum_{i,k=1}^n a_{ki}^2 =n
\]
which just says that every orthogonal matrix $A$ is contained in a
ball\index{ball} of radius $\sqrt{n}$ in $\R^{n^2}$. \QED \end{pf}

A compact\index{compact} space is often much more pleasant to work with
than a noncompact one. However, many transformation
groups\index{transformation group} are visibly noncompact, such as the
additive group $\R$. On the other hand, the topology and geo\-metry of $\R$
are very simple (for example, $\R$ is simply connected,\index{simply
connected} and can be parametrised by a real parameter without overlap).
Most transformation groups are of course more complicated; however, in a
suitable sense they can be {\em topologically decomposed\/} as a compact
group\index{compact!Lie group} times a group homeomorphic to $\R^n$.

\paragraph{Example 1} The simplest example is the multiplicative group
$\R^*$ of nonzero real numbers. There is a
homeomorphism\index{homeomorphism} (in this case, an isomorphism of groups)
\[
\R_+\times \{\pm 1\} \to \R^*;
\]
in plain English, every nonzero number is the product of a positive
number and a sign. The space $\R_+$ is homeomorphic to $\R$; the group
$\{\pm1\}$ is finite so clearly compact.\index{compact!Lie group}

\paragraph{Example 2} Although the next example looks similarly innocent, it
appears in many different guises throughout geometry, Fourier analysis, Lie
groups,\index{Lie group} representation theory, complex analysis and number
theory. Consider the multiplicative group $\C^*$ of nonzero {\em complex}
numbers. This is a topological group; for example, I can view $\C$ as the
plane $\R^2$ and take the subspace topology.\index{subspace topology} The
space $\C^*$ is obviously noncompact. However, there is a
homeomorphism\index{homeomorphism} (even a group iso\-morphism)
\[
\begin{array}{ccc}
S^1\times\R_+ & \to & \C^* \\
(\theta,r)& \mapsto & r\exp(i\theta).
\end{array}
\]
Here $S^1$ is compact\index{compact!Lie group} (and definitely not
homeomorphic to a product of copies of $\R$, which is the essential
content of~\ref{fun thm}, Corollary~1) and $\R_+$ is
homeomorphic\index{homeomorphism} to $\R$.

\paragraph{Example 3} The final example is more substantial, and deals
with the difference between the groups $\GL(n,\R)$ and $\O(n)$. Write
$T_+(n)\subset\GL(n,\R)$ for the set of upper triangular matrixes with
positive diagonal entries:
\begin{align*}
T_+(n)&=\bigl\{M=(m_{ij})\in\GL(n,\R) \bigm|\hbox{$m_{ij}=0$ for all $i>j$, and
$m_{ii}>0$}\bigr\} \\
&= \left\{\left(
\begin{matrix}
+&*&\cdots\\
0&+&*&\cdots\\
0 &\cdots&\ddots&*\\
0&\cdots&0 &+
\end{matrix} \right)\right\}.
\end{align*}
It is easy to see that $T_+(n)\subset\GL(n,\R)$ is a subgroup.

\begin{thm} Every element $A\in\GL(n,\R)$ can be written in a unique
way in the form $A=BC$, where $B\in\O(n)$ is an orthogonal
matrix\index{orthogonal!matrix} and $C\in T_+(n)$ is an upper
triangular matrix with positive diagonal entries. Moreover, $B$ and
$C$ depend continuously on $A$. The map
\[
\GL(n,\R)\to\O(n)\times T_+(n) \quad\hbox{given by}\quad A\mapsto (B,C)
\]
is a homeo\-morphism\index{homeomorphism} (see~\ref{homeo cont}, but not a
group homomorphism!). \end{thm}

\paragraph{Discussion} The space $\O(n)$ is compact\index{compact!Lie group}
by the above Proposition. The space $T_+(n)$ is
homeomorphic\index{homeomorphism} to $\R^N$, where $N={\binom{n+1}{2}}$.
Many geometric questions on $\GL(n,\R)$ reduce to similar questions
on~$\O(n)$; for a simple example, compare Remark~\ref{SOn conn}. Note also
the dimension count:\index{dimension!of a Lie group}
\[
\dim O(n) + \dim T_+(n) = \binom{n}{2} + \binom{n+1}{2} = n^2 = \dim\GL(n,\R).
\]

\begin{pf} I view the $n\times n$ matrix $A$ as a row made up of $n$ column
vectors $\bbf_i$. Thus $\{\bbf_1,\dots,\bbf_n\}$ is a basis of $\R^n$
because $A\in\GL(n,\R)$. If it is an orthonormal basis then there is no
problem: $A\in\O(n)$, and we must take $B=A$ and $C=1$. If $A$ is not
orthogonal to start with, then the Gram--Schmidt process described in the
proof of Theorem~\ref{sec!bases}, (1) produces an orthonormal basis. Set $B$
to be the matrix formed from the new basis vectors as columns, and $C$ to be
the matrix describing the change of basis. Clearly $B\in\O(n)$; I leave you
to check (see~\exref{ex!decomp_gs}) that $C\in T_+(n)$ and that $B,C$ depend
continuously on $A$. Then the map $A\mapsto(B,C)$ is continuous, and its
inverse is matrix multiplication $(B,C)\mapsto BC$. \QED \end{pf}

\section{Components}\label{SOn conn}
\markright{\hfill \thesection\ Components\quad}
\index{connected!component}

Recall from~\ref{subsec!connected} that every topological space can be
decomposed into a number of components, which are themselves connected. I
repeatedly discussed the geometry of $\O(2)$: a union of two circles. A
circle $S^1$ is connected, so $\O(2)$ has two connected components. This is
typical:

\begin{prop} The group $\O(n)$ has two connected
components,\index{connected!component} distinguished by $\det A=\pm1$.
\end{prop}

\begin{rem} One can use Theorem~\ref{decomp of GL(n)} to show that
$\GL(n,\R)$ also has two connected components,\index{connected!component}
that are distinguished by $\det A>0$ and $\det A<0$; see~\exref{GLn ex}. The
group $\O(1,2)$ of all Lorentz matrixes has 4 components, as discussed
in~\exref{Opq ex}. \end{rem}

\begin{pf} An orthogonal matrix\index{orthogonal!matrix|)} has
determinant $\pm 1$. (Compare~\ref{sec!orth}; recall that I called
$A$ direct\index{direct motion} if $\det A=1$ and opposite\index{opposite
motion} if $\det A=-1$.) The function
\[
\det\colon \O(n)\to \{\pm 1\}
\]
is continuous,\index{continuous} so the two possibilities $\det A=\pm 1$
determine two disjoint open\index{open set} and closed\index{closed} sets of
$\O(n)$. It remains to show that each of these sets is path
connected.\index{connected}

Fix a matrix $A\in\O(n)$. By the Normal Form \index{normal form of a
matrix} Theorem~\ref{sec!norm}, $A$ can be written with respect to a
suitable orthonormal basis in the diagonal block form with $2\times2$
diagonal blocks
\[
B_i=\begin{pmatrix}\cos\theta_i&-\sin\theta_i\\
\sin\theta_i&\cos\theta_i \end{pmatrix},
\]
and one optional block $\pm 1$. For $t$ varying from $0$ to $1$, let
$A(t)$ be the matrix with the same block form as $A$, but with blocks
\[
B_i(t)=\begin{pmatrix} \cos t\theta_i&-\sin t\theta_i\\
\sin t\theta_i&\cos t\theta_i \end{pmatrix}.
\]
The rule $t\mapsto A(t)$ gives a\index{continuous} continuous path
$[0,1]\to\O(n)$ joining $A$ either to the identity or to the element
$\diag(1,\dots,1,-1)$. Therefore, the two subsets of $\O(n)$ defined by
$\det A=\pm1$ are both path connected. A path
connected\index{path!connected} space is connected\index{connected} by
Lemma~\ref{subsec!connected}, (2). \QED \end{pf}

The {\em special orthogonal group}\index{special!orthogonal group $\SO(n)$}
is the group\index{orthogonal!group $\protect\SO(n)$}
\[
\SO(n)=\bigl\{ A\in \O(n)\bigm|\det A=1 \bigr\}.
\]
By the Proposition, this is a connected
component\index{connected!component} of $\O(n)$. Since it is the kernel
of a group homomorphism $\det\colon\O(n)\to\{\pm1\}$, it is also a
normal subgroup of index 2 in $\O(n)$.

In the special case $n=3$, the elements of $\SO(3)$ can be described
explicitly. By the Normal Form Theorem~\ref{sec!norm}, any orthogonal
$3\times 3$ matrix\index{orthogonal!matrix} of determinant one has the
form
\[
\begin{pmatrix} \ 1 \\ &\cos\theta &-\sin\theta\\ &\sin\theta&\cos\theta
\end{pmatrix}
\]
in a suitable basis. If $l$ is the line through the origin with
direction vector given by the first basis element, then the
motion\index{motion} of $\E^3$ described by this matrix is
rotation\index{rotation|(} $\Rot(l,\theta)$ around the line $l$. Hence
$\SO(3)$ is the {\em group of rotations of\/ $\E^3$ about axes
passing through $O$}.

\section[The geometry of $\SO(n)$]{Quaternions, rotations and \\ the
geometry of $\SO(n)$}\label{sec!topson}
\index{quaternions|(}
\index{topology!of $\SO(3)$}
\markright{\hfill \thesection\ The geometry of $\SO(n)$\quad}
As I discussed before, for $n=2$ the group $\SO(2)$ is
homeomorphic\index{homeomorphism} to the circle $S^1$. The purpose here is
to find a similar description of the special orthogonal
groups\index{special!orthogonal group $\SO(n)$} $\SO(3)$ and
$\SO(4)$\index{orthogonal!group $\protect\SO(n)$} in terms of the 3-sphere.
I start with a small detour to introduce the quaternions, the main
protagonists in the game. Note that $\SO(n)$ is the group of direct
motions\index{motion} of $\E^n$ with a fixed point, or in other words the
group of rotations of $\E^n$; hence the aim is to find a connection between
quaternions and rotations (for $n=3,4$).

\subsection{Quaternions}\label{quat sec} The {\em algebra of
quaternions} is the real vector space
\[
\HH=\bigl\{a+bi+cj+dk\bigr\}\quad \hbox{with $a,b,c,d\in\R$,}
\]
with the multiplication law
\[
i^2=j^2=k^2=-1,\quad ij=k,jk=i,ki=j,\quad ji=-k, kj=-i,ik=-j.
\]
The cyclic symmetry makes this easy to remember.

Some terminology, similar to the traditional language of complex
numbers: if $q=a+bi+cj+dk$, write $q^*=a-bi-cj-dk$ for the {\em
conjugate} quaternion. We say that $q$ is {\em real} if $b=c=d=0$ and
{\em pure imaginary} if $a=0$.

\begin{prop}
\begin{enumerate} \renewcommand{\labelenumi}{\theenumi} \renewcommand{\theenumi}{(\arabic{enumi})}

\item $\HH$ is an associative noncommutative
$\R$-algebra of dimension $4$ over $\R$.

\item The conjugation $q\mapsto q^*$ is an {\em antiinvolution}, meaning
\[
(pq)^*=q^*p^* \quad\hbox{for all $p,q\in\HH$.}
\]

\item $|q|^2=qq^*=q^*q=a^2+b^2+c^2+d^2$ is a positive definite quadratic
form\index{quadratic form} on $\HH$; therefore for any nonzero
$q\in\HH$, the element
\[
q\1=q^*/|q|^2
\]
is a $2$-sided inverse of $q$. Hence $\HH$ is a {\rm division algebra}
or {\rm skew field}.

\item If $q\in\HH$ and $q\notin\R$, then $q=A+BI$ with $I$ pure
imaginary, $I^2=-1$ and $A,B\in\R$. Hence the subalgebra $\R[q]$ of
$\HH$ generated by $q$ is of the form $\R[q]\iso\C\subset\HH$.

\item If $I$ is pure imaginary with $I^2=-1$, there exists $J,K\in H$
such that $I,J,K$ have the same multiplication table as $i,j,k$, that is
$I^2=J^2=K^2=-1$ and $IJ=K$, etc.
\end{enumerate}
\end{prop}

\begin{pf} (1) Noncommutativity is clear from the multiplication
table:
$ij=k\ne -k = ji$.

Because everything is $\R$-linear, it is enough to check the associative
law $a(bc)=(ab)c$ for the basis elements $a,b,c\in\{1,i,j,k\}$. If any
of $a,b,c$ is 1 then it is OK. By the cyclic symmetry, I can assume that
the first term $a=i$; if only $i$ appears, then I am working in a copy
of $\C$. This leaves only 8 cases to check by brute force:
\[
\begin{array}{rl}
i(ij)=ik=-j=(i^2)j;&
i(ik)=i(-j)=-k=(i^2)k;\\
i(ji)=i(-k)=j=ki=(ij)i;&
i(j^2)=-i=kj=(ij)j;\\
i(jk)=i^2=-1=k^2=(ij)k;&
i(ki)=ij=k=-ji=(ik)i;\\
i(kj)=-i^2=1=-j^2=(ik)j;&
i(k^2)=-i=-jk=(ik)k.
\end{array}
\]
This is of course pure gobbledygook. A much more convincing argument
is to say that $i,j,k$ are maps of something, such that multiplication
coincides with composition of maps, so is associative for a fundamental
reason; see \exref{quat mult ex}.

(2) Again because everything is $\R$-linear, it is enough to check that
$(pq)^*=q^*p^*$ for basis elements $a,b\in\{1,i,j,k\}$. The brute force
method is an easy exercise: $(1i)^*=-i=(i^*)(1^*)$,
$(ij)^*=-k=(-j)(-i)$, etc.; see \exref{quat conj ex}.

(3) On multiplying out the product $(a+bi+cj+dk)(a-bi-cj-dk)$, the terms
$a^2+b^2+c^2+d^2$ appear in the obvious way from the squared terms. The
cross terms all cancel out, either as $(a\times -bi)+(bi\times a)=0$ or
$(bi\times -cj)+(cj\times -bi)=-bc(i\times j+j\times i)=0$.

(4) Note that $q+q^*=2a$ and $qq^*=|q|^2\in\R$, so that $q$ and $q^*$
are the two roots of a quadratic polynomial $x^2-2ax+|q|^2$ with real
coefficients. Also, $q-q^*=2(bi+cj+dk)$ is pure imaginary, and an easy
calculations similar to that in (3) shows that
$(q-q^*)^2=-4(b^2+c^2+d^2)<0$ (because $q\notin\R$), so that this has no
real roots. Thus $q=A+BI$ where $A=a$, $B=\sqrt{(b^2+c^2+d^2)}$ and $I$
is pure imaginary with $I^2=-1$.

(5) is worked out as an exercise in \exref{quat frame ex}.\QED
\end{pf}

\begin{rem} (3) says that the Euclidean
distance\index{Euclidean!distance} on $\R^4=\HH$ is determined by the
algebra structure of $\HH$ together with the antiinvolution $q\mapsto
q^*$. This has various nice corollaries. For example, the direct sum
decomposition
\[
\HH=\{\hbox{real quaternions}\}\oplus
\{\hbox{imaginary quaternions}\}=\R\oplus\R^3
\]
is orthogonal.\index{orthogonal!direct sum} Also, two imaginary vectors $p$,
$q$ anticommute $pq=-pq$ if and only if the corresponding vectors of $\R^3$
are orthogonal.\index{orthogonal!vector} This point is the main reason that
quaternions can be applied to rotations of $\E^3$ and $\E^4$. \end{rem}

\subsection{Quaternions and rotations}\label{quat rot} Set
\[
U=\{\hbox{unit quaternions}\}=\{q\in\HH\mid qq^*=1\}=S^3\subset\R^4
\]
for the unit quaternions. Note that $U$ has two structures: it is a group
under multiplication, and also has its own geometry as the sphere
$S^3$.\index{sphere $S^n$} The two structures are compatible as
in~\ref{sec!top on groups}. The group $U$ generalises the multiplicative
group of complex numbers of modulus $1$, which is the unit circle
$S^1\subset\C$.

For the next theorem, identify $\HH$ and its quadratic
form\index{quadratic form} $|q|$ with $\E^4$ and its Euclidean
distance.\index{Euclidean!distance} The purely imaginary quaternions
form a linear subspace which gets identified with $\E^3$.

\begin{thm} \begin{enumerate}\renewcommand{\labelenumi}{\theenumi} \renewcommand{\theenumi}{(\arabic{enumi})}

\item For any $p\in U$, left multiplication $a_p\colon x\mapsto px$
defines a map $\HH\to\HH$ which is a direct\index{direct motion|(}
motion\index{motion} of
$\HH=\E^4$ fixing the origin; the same holds for right multiplication
$b_q\colon x\mapsto xq^*$.

\item The group homomorphism $\fie\colon U\times U\to\SO(4)$ defined
by
\[
\fie(p,q)=a_p\circ b_q\colon x\mapsto pxq^*
\]
is surjective, and $\fie(p,q)=\id_{\HH}$ if and only if $(p,q)=(1,1)$ or
$(p,q)=(-1,-1)$.

\item For any $q\in U$, the map $r_q\colon x\mapsto qxq^*$ is a
direct motion\index{motion} of $\HH=\E^4$, which is the identity on
real elements of\/ $\HH$ and takes pure imaginary quaternions of\/
$\HH$ to pure imaginary quaternions. Thus it defines a
rotation\index{rotation} of the subspace $\E^3\subset\HH$ of pure
imaginary quaternions.

\item Any $q\in U$ with $q\notin\R$ has a unique expression in the form
$q=\cos\theta+I\sin\theta$, where $I\in U$ is a pure imaginary
quaternion and $\theta\in(0,\pi)$. Then $r_q=\Rot(I,2\theta)$ is the
rotation of $\R^3$ about the directed axis defined by $I$ through the
angle $2\theta$.

\item The group homomorphism $\psi\colon U=S^3\to\SO(3)$ defined by
\[
\psi(q)= r_q
\]
is surjective, and $\psi(q_1)=\psi(q_2)$ if and only if $q_1=\pm q_2$.
\end{enumerate}
\end{thm}

\begin{pf} (1) It is clear that $a_p$ is a motion,\index{motion} since it
fixes $0$ and $|px|^2=|x|^2$. Moreover, it must be a\index{direct motion|)}
direct motion, for example, because $\det(a_q)$ is a continuous map from
the connected\index{connected} set $U=S^3$ to $\pm1$. (Several other proofs
are possible, see \exref{direct ex}.)

I relegate (2) to~\exref{ex!4dim}.

(3) is obvious, since $a\in\R$ commutes with quaternion
multiplication, so $r_q(a)=qaq^*=aqq^*=a$. Also, if $p^*=-p$, then
$r_q(p)=qpq^*$ has $(r_q(p))^*=(qpq^*)^*=qp^*q^*=-qpq^*$, so $qpq^*$
is pure imaginary.

(4) follows from Proposition~\ref{quat sec}, (4): $\R[q]\iso\C$. The
equation $x^2=-1$ has exactly two roots $\pm I$ in $\C$, and choosing
the appropriate sign gives $q=\cos\theta+I\sin\theta$ with
$\theta\in(0,\pi)$. Then $r_q(I)=I$ follows because $\R[q]\iso\C$, so
that $q^*=q\1$ and $qIq\1=I$.

Now let $J,K$ be as in Proposition~\ref{quat sec}, (5). Then
\begin{align*}
qJq^*&=(\cos\theta+I\sin\theta)J(\cos\theta-I\sin\theta)\\
&=(\cos^2\theta-\sin^2\theta)J+(2\sin\theta\cos\theta)K,
\end{align*}
and similarly $qKq^*=-(2\sin\theta\cos\theta)J+
(\cos^2\theta-\sin^2\theta)K$. Thus $r_q$ fixes the directed axis defined by
$I$, and performs a rotation by $2\theta$ in the plane spanned by $J,K$.

Finally (5) follows by (4); every rotation\index{rotation|)}
is hit exactly twice because of the $2\theta$. \QED \end{pf}
\index{quaternions|)}

\subsection{Spheres and special orthogonal groups}
\label{ssec!sph_and_so}
\index{rotation!group}
\index{orthogonal!group $\protect\SO(n)$}
\index{topology!of $S^3$}\index{special!orthogonal group $\SO(n)$}
After all this algebra, come the relations between groups of rotations
and the sphere $S^3$.

\begin{cor} \begin{enumerate}\renewcommand{\labelenumi}{\theenumi} \renewcommand{\theenumi}{(\arabic{enumi})}

\item There is a homeomorphism\index{homeomorphism}
\[
\SO(3)\simeq S^3/\!{\sim},
\]
where $\sim$ is the equivalence relation on $S^3$ that identifies
antipodal points $\bx$ and $-\bx$.

\item There is a homeomorphism\index{homeomorphism}
\[
\SO(4)\simeq(S^3\times S^3)/\!{\approx},
\]
where $\approx$ is the equivalence relation on $S^3\times S^3$ that
identifies $(\bx,\by)$ with $(-\bx,-\by)$.
\end{enumerate}
\end{cor}
\index{orthogonal!group $\protect\SO(n)$|)}

\begin{pf} Both statements are direct corollaries of the previous theorem
together with\index{homeomorphism!criterion} Theorem~\ref{homeo crit} and
the definition of the quotient topology\index{quotient topology} and its
UMP\index{universal mapping property} discussed in \ref{sub and quot}.

In more detail, by Theorem~\ref{quat rot}, (5) there is a continuous
surjective map $\psi\colon S^3\to\SO(3)$, with $\psi(\bx)=\psi(\by)$ if and
only if $\bx=\by$ or $\bx=-\by$. By the Universal Mapping Property~\ref{sub
and quot} of the quotient topology,\index{quotient topology} there is
consequently a continuous map $\psibar\colon (S^3/{\sim}) \to\SO(3)$ that is
clearly a bijection. Now $S^3$ is compact,\index{compact} and therefore so
is $S^3/{\sim}$ by Proposition~\ref{im compact}. Also the subspace
topology\index{subspace topology} of $\SO(3)\subset\R^9=\{\hbox{$3\times3$
matrixes}\}$ is metric\index{metric!topology} and therefore
Hausdorff.\index{Hausdorff} Therefore all the assumptions of
Theorem~\ref{homeo crit} are satisfied, $\psibar$ is a
homeomorphism,\index{homeomorphism} and (1) follows. (2) is proved in
exactly the same way using the map $\fie\colon U\times U\to\SO(4)$ of
Theorem~\ref{quat rot} (2). \QED \end{pf}

\begin{rem} The statements of the corollary generalise for all $n$; namely,
there exists a compact topological group $\Spin(n)$ called the {\em spinor
group}\index{spinor group $\Spin(n)$} with a surjective homomorphism
$\pi\colon\Spin(n)\to\SO(n)$ with
kernel $\Span{\iota}$ of order~2, so that $\pi$ induces an isomorphism of
groups $\Spin(n)/\Span{\iota}\to\SO(n)$ that is also a homeomorphism. The
pleasant thing about low dimensions is the fact that the spinor groups are
spheres or products of spheres: $\Spin(2)\simeq S^1$, $\Spin(3)\simeq S^3$,
$\Spin(4)\simeq S^3\times S^3$. \end{rem}

\section{The group $\SU(2)$}\label{sec!su2}
\markright{\hfill \thesection\ The group $\SU(2)$\quad}

In this brief section, I identify the group $U$ of unit quaternions
of~\ref{sec!topson} as matrix group. This involves more linear algebra over
the complex numbers, a subject that already made a brief but important
appearance in~\ref{sec!norm}.

Let $V$ be a 2-dimensional $\C$-vector space together with a positive
definite Hermitian form\index{Hermitian form}, represented in some basis by
$|z_1|^2+|z_2|^2$, or the matrix
$\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)$
(see~\ref{sec!herm} for more details on Hermitian forms). A complex linear
transformation of $V$ that preserves this form is {\em unitary}: thus a
matrix $A\in\GL(2,\C)$ is unitary\index{unitary!matrix} if it satisfies
$\herm AA=I_n$, where
$\herm A$ is the Hermitian conjugate defined by $(\herm A)_{ij}=\Abar_{ji}$.
The group of all such matrixes is the {\em unitary group} $\U(2)$. I am
interested in its subgroup, the {\em special unitary
group}\index{unitary!group}\index{special!unitary group $\SU(n)$}
\[
\SU(2)= \bigl\{ A \in \U(2) \bigm|\det A=1 \bigr\}.
\]
As matrix groups, both $\U(2)$ and
$\SU(2)$ are topological groups in an obvious way.

\begin{rem} A unitary matrix $A$ has $|\det A|=1$; see
Ex.~\ref{app!q}.\ref{ex!unitarydet}. Thus the set of possible values for the
determinant is the unit circle $S^1$, which is connected.\index{connected}
Thus $\SU(2)$ is a normal subgroup, but not a connected
component\index{connected!component} of $\U(2)$ in the same way as $\SO(2)$
is in $\O(2)$. \end{rem}

I write out explicitly the condition for a matrix $A\in\GL(2,\C)$ to
be special unitary (compare \ref{sec!refl}). If
$A=\left(\begin{smallmatrix} a&b\\c&d\end{smallmatrix}\right)$, the
equations are
\begin{equation}
\begin{array}{rcl}
a\abar+c\cbar&=&1,\\ a\bbar+c\dbar&=&0,\\ b\bbar+d\dbar&=&1
\end{array}
\quad\hbox{and}\quad \det A=ad-bc=1.
\label{eq!aabar}
\end{equation}
One solves these equations more-or-less as in \ref{sec!refl} to get
$d=\abar$ and $c=-\bbar$, where $a\abar+b\bbar=1$; see
\exref{ex!su2}. Thus
\[
\SU(2)= \left.\left\{ \begin{pmatrix} a & b \\ -\bbar & \abar
\end{pmatrix} \right| a,b\in\C, \ |a|^2 +|b|^2 = 1 \right\}.
\]
This description has an important corollary.
\begin{cor} The map $\left(\begin{smallmatrix} a&b \\ -\bbar&\abar
\end{smallmatrix}\right)\mapsto a+bj$ defines an isomorphism from
$\SU(2)$ to the group $U$ of unit quaternions of~\ref{quat rot}.
\end{cor}
\begin{pf} Write $a=a_1+a_2i$ and $b=b_1+b_2i$. Then $a+bj=
a_1+a_2i+b_1j+b_2k$ using quaternion multiplication. The condition $|a|^2
+|b|^2 = 1$ becomes $|a_1|^2 +|a_2|^2 +|b_1|^2 +|b_2|^2 = 1$ hence $a+bj$
has quaternion norm~1. The map $\SU(2)\to U$ is clearly a bijection. It
remains to check that the map respects multiplication, so that it becomes a
group isomorphism; this a special case of~\exref{sumult}. \QED \end{pf}

Theorem~\ref{quat rot}(5) on the description of $\SO(3)$ can thus be
reformulated as saying that there exists a two-to-one surjective group
homomorphism $\SU(2)\to\SO(3)$ (compare also~\exref{soparam}). The two
groups are now matrix groups (over different fields), but the existence of
the two-to-one map is by no means obvious from the matrix description: the
most convincing way of going from complexes to reals is via quaternions.

\section{The electron spin in quantum mechanics}\label{sec!spin}
\markright{\hfill \thesection\ The electron spin in quantum mechanics\quad}
\index{electron|(} \index{spin}

This section relates the geometry of $\SO(3)$ to a fundamental attribute of
elementary particles: their spin. All the mathematics needed is at hand
already; however, there is no space in the present book to introduce all the
necessary background from quantum mechanics. For more information and
insight, see Feynman's classic~\cite{Fn}, Chapters~1--3.

\subsection{The story of the electron spin}\label{stern}

The story begins in 1925. Two Dutch doctoral students George Uhlenbeck and
Samuel Goudsmit, halfway through their PhD program, noted that the electron
inside the atom appeared to have, besides the three known ``quantum
numbers'' associated with the position of the electron, its angular
momentum\index{angular!momentum} around the nucleus and its magnetic field,
an extra degree of freedom. They postulated the existence of an extra
``quantum number'', which they called the {\em electron spin}. This new
quantum number seemed to behave in many ways like angular momentum, so they
gave the interpretation that it corresponds to some kind of intrinsic
rotational motion.\index{motion} However, the quantum number appeared to
have just {\em two} possible values \plusspin\ and \minusspin, and the
rotation\index{rotation} seemed not to have a definite axis; strange facts
for a ``spinning'' particle. Their advisor Paul Ehrenfest is said to have
commented: ``You are both young enough to be able to afford a stupidity!''\
(he realised soon afterwards though that his students had in fact made an
important discovery).

Unknown to Uhlenbeck and Goudsmit, the experimental verification of their
discovery had been around for three years in the form of the Stern--Gerlach
experiment. In 1922 the German scientists Otto Stern and Walther Gerlach
built the device illustrated schematically in Figure~\ref{sec!spin}a. The
source emits a beam of silver atoms. The beam is directed between the poles
of a magnet, which produces a magnetic field
orthogonal\index{orthogonal!magnetic field} to the direction of the path. As
the atoms are electrically neutral, they are not expected to experience
force; they should thus pass through the device without any change in their
direction. However, a screen on the other side of the device reveals that
the atoms are in fact deflected by the magnetic field, and moreover that
they follow one of two possible paths.

\begin{figure}[hbt] \begin{center}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\leavevmode\hbox{\epsfysize=3in\epsffile{sterna.eps}}
%\input{modsterna.pstex_t}
\caption{The Stern--Gerlach experiment}
\end{center} \end{figure}

The experiment can only be understood in terms of the notion of
spin.\index{spin} A silver atom has an electron on an outer shell, whose
intrinsic spin interacts with the magnetic field. Atoms whose outer electron
is in the \plusspin\ spin state follow a different path from those in the
\minusspin\ spin state.

The mid 1920s was of course the time when quantum mechanics was invented.
Soon after Uhlenbeck and Goudsmit's proposal, Pauli and Dirac incorporated
electron spin into the quantum mechanical theory of the electron, also known
as the Schr\"odinger equation. Since this is not a course about the
electron, I do not need to worry unduly with the details.

\subsection{Measuring spin: the Stern--Gerlach device}

In the following, I assume a modified form of the Stern--Gerlach (SG)
device, illustrated in Figure~\ref{sec!spin}b. This is only a thought
experiment\footnote{The experiment cannot be carried out as described here:
the electron's wave function is too fuzzy because of quantum mechanical
effects, and the separation into two rays is not apparent. The point about
the silver atom featuring in the original Stern--Gerlach experiment is that
it is electrically neutral, but has a relatively free electron on an outer
shell; its motion between magnets is thus governed by the spin of the outer
electron. In the text I stick to the thought experiment involving free
electrons.}, explained in detail in \cite{Fn}, pp.~5--1 and~5--2. An
electron beam arrives from the left, and separates inside the device $S$
into two beams according to its spin under the action of the left hand
``magnet''. A combination of other ``magnets'' forces the electrons back
into their horizontal path; the outcoming beam still consists of a mixture
of electrons in the two spin states.

\begin{figure}[hbt] \begin{center}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\leavevmode\hbox{\epsfysize=2in\epsffile{sternb.eps}}
\caption{The modified Stern--Gerlach device}
\end{center} \end{figure}

Assume now that I block the path of one of the beams inside the device, as
in the case of device $S$ of Figure~\ref{sec!spin}c. Then the electrons
leaving the device $S$ are all in a definite spin state \plusspin. In this
sense, I have now ``measured'' the spin of this beam of electrons: I know
precisely what state they are in. (Unfortunately, I have lost about half
my electrons along the way, but that seems to be unavoidable in this kind
of game. Compare with a large accountancy firm hired to count your money.)
In particular, if I attach another SG device $S'$ {\em in the same
position} after the first as in Figure~\ref{sec!spin}c, then I know the
path of all the electrons inside the device; blocking the other
path then makes no difference.

\begin{figure}[hbt] \begin{center}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering \input{sternc.pstex_t}
\caption{Two identical SG devices}
\end{center} \end{figure}

However, let us now put another SG device $T$ {\em in a different spatial
position} in the path of my uniform spin electron ray; see
Figure~\ref{sec!spin}d. The ray now {\em separates again}; the electrons
choose two different paths in a specific ratio (which can be measured again
by blocking one or other of the paths) depending on the position of
\begin{figure}[hbt] \begin{center}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}d}
\centering \input{sternd.pstex_t}
\caption{Two different SG devices}
\end{center} \end{figure}
the new SG device. Hence knowing that the electron is in spin state
\plusspin\ {\em in one direction} does not mean that it is in spin state
\plusspin\ in {\em all directions}. It registers as spin \plusspin\ or
\minusspin\ in some different direction following, it seems, a fixed dress
code.

\subsection{The spin operator}

As both experiment and speculation confirm, the electron spin takes two
possible values $+1$ and $-1$, where I ignore unnecessary constants. In the
framework of quantum mechanics, such a two-state system is modelled on a
2-dimensional complex vector space $V$ with a definite Hermitian
form\index{Hermitian form} on it, which I denote by bracket $(\ ,\,)$. Every
electron in this simple model is described by its {\em wave function}
$\psi\in V$, which we normalise to unit length $(\psi,\psi)=1$.

An SG device $S$ in a fixed spatial position corresponds to a linear
operator $\opo_S\colon V\to V$. The possible spin states with respect to
this spatial direction correspond to the different {\em eigenvalues} of this
map. In the present case, the eigenvalues must therefore be $\pm1$. There
are corresponding normalised eigenvectors $\psi_S^+$ and $\psi_S^-$:
\[
\opo_S(\psi_S^+)=\psi_S^+, \quad \opo_S(\psi_S^-)=-\psi_S^-.
\]
Quantum mechanics postulates that the operator $\opo_S$ is {\em Hermitian}
(\exref{herm_ex}). It follows that the eigenvectors are orthogonal
$(\psi_S^+, \psi_S^-) = 0$. Thus $\{\psi_S^+, \psi_S^-\}$ is a Hermitian
basis in the 2-dimensional vector space~$V$.

The electron with wave function $\psi_S^+$ is in the \plusspin\ spin state
and that wave function $\psi_S^-$ in the \minusspin\ spin state. These
electrons are in {\em eigenstates} of the spin operator $\opo_S$. An
arbitrary electron has a wave function $\psi\in V$ which is a linear
combination of the basis vectors:
\[
\psi=\al\psi_S^+ + \be\psi_S^-.
\]
Such a state is referred to as a {\em mixed state}.

An electron in a mixed state $\psi=\al\psi_S^+ + \be\psi_S^-$ arriving at
our SG device $S$ passes along the \plusspin\ or \minusspin\ path in the
device with probability $|\al|^2$ or $|\be|^2$ respectively. These numbers
are called {\em probability amplitudes}. Because both basis vectors
$\psi_S^\pm$ and the vector $\psi$ are normalised to unit length, $|\al|^2 +
|\be|^2 =1$; thus these probabilities add to one.

Once we block the \minusspin\ path, the outcoming electrons are all in the
\plusspin\ eigenstate: their wave function is the eigenvector $\psi_S^+\in
V$. This explains their behaviour in a next SG device $S\prime$ in the same
spatial position as $S$, pictured in Figure~\ref{sec!spin}c. The operator
corresponding to the device $S'$ is $\opo_{S'}=\opo_S$, and the electrons
are all in the \plusspin\ eigenstate of this operator. So they choose the
two paths with probability $|\al|^2=1$, respectively $|\be|^2=0$; in other
words, their path through $S'$ is determined.

\subsection{Rotate the device}

To perform our next thought experiment, imagine a beam of electrons leaving
a device in one of the definite eigenstates, and arriving at another device
in a different spatial position as in Figure~\ref{sec!spin}d. The new SG
device $T$ corresponds to an operator $\opo_T$ and hence to a new Hermitian
basis $\{\psi_T^+, \psi_T^-\}$ of $V$ consisting of eigenvectors of
$\opo_T$.

I wish to study an electron ray in one of the spin eigenstates
$\psi_S^\pm$, when it passes through $T$.
The experiment says that electrons will follow one of two possible paths in
$T$, and I want the probability of its taking one or other of the paths.
According to the rule spelled out in the last section, I should write the
vector $\psi_S^+$ (and also $\psi_S^-$) in terms of the new basis
$\{\psi_T^+,\psi_T^-\}$ to find the probability amplitudes. This is
simply a change of basis, given by a $2\times 2$ matrix $A_{S\to T}$,
an element of $\GL(2,\C)$ (in fact $\U(2)$ as both bases are
Hermitian). The task is to find $A_{S\to T}$ from $S$, $T$.

To proceed, I need to make precise the geometry of an SG device in 3-space.
Note that an SG device in physical space $\E^3$ determines {\em two
distinguished orthogonal directed lines};\index{orthogonal!line} namely,
there is the distinguished direction of the electron beam, and the
distinguished direction of the magnetic field
orthogonal\index{orthogonal!magnetic field} to it; see
Figure~\ref{sec!spin}a. I can think of these directed lines as two
coordinate axes in a coordinate system, and there is a unique way of adding
a third directed coordinate axis orthogonal to the first two to make a
right-handed coordinate system in 3-space. The new system $T$ determines in
the same way a new right-handed coordinate system in $\E^3$. The
transformation which gets me from $S$ to $T$ is a direct
motion\index{motion} of $\E^3$, and thus a rotation\index{rotation} $g\in
\SO(3)$. (Note that only directions matter in this discussion; the origin of
the coordinate system is not important, and I ignore
translations.)\index{translation}

According to the earlier discussion, I need a recipe associating an element of
$\GL(2,\C)$ with a transformation $S\to T$, presumably in a continuous manner.
In other words, I needed a map
\[
A\colon \SO(3)\to \GL(2,\C).
\]
It can also be argued from basic principles of quantum mechanics that the
map $A$ should respect composition; after all, $S\to T$ followed by $T\to R$
should be the same as $S\to R$. Hence the map $A$ should be a group
homomorphism. This however presents a puzzle: there is no obvious way to map
$\SO(3)$ to the group of linear maps on a 2-dimensional $\C$-vector space
(apart from the map which takes every rotation to the identity matrix, which
would contradict the experimentally observed fact that spin does depend on
direction). In fact there is absolutely no such map at all. \clearpage

\subsection{The solution} \label{sec! spin and rot}

Although the expressions for $\psi_T^\pm$ in terms of $\psi_S^\pm$ and the
rotation taking $S$ to $T$ can be derived from first principles, I cannot
improve on Feynman's beautiful and self-contained account (in pp.~6--1 to
6--14 of~\cite{Fn}), and I just state the result: namely, although there is
no map $A\colon\SO(3)\to \GL(2,\C)$, there is an obvious map
\[
\wA\colon\SU(2)\to\GL(2,\C)
\]
from the group $\SU(2)$ to $\GL(2,\C)$; a $2\times 2$ unitary
matrix\index{unitary!matrix} is
certainly invertible, so the inclusion map will do. On the other hand,
$\SU(2)$ is not too different from $\SO(3)$; by
Corollary~\ref{ssec!sph_and_so}, they are related by a two-to-one map. Thus
$\wA$ can be thought of as a two-valued function on $\SO(3)$.

Up to a knowledge of the explicit form of the map $\SU(2)\to \SO(3)$
that can easily be derived from the expressions in~\ref{quat rot},
this answers the original question of how to compute the ratio of electrons
following the two paths of Figure~\ref{sec!spin}d: $S\to T$ is given by an
element of $\SO(3)$, and there are {\em two} possible changes of basis
\begin{eqnarray*}
\psi_T^+ &=& \al_+\psi_S^+ + \be_+\psi_S^-\\
\psi_T^- &=& \al_-\psi_S^+ + \be_-\psi_S^-
\end{eqnarray*}
for matrixes $\begin{pmatrix} \al_+&\al_-\\\be_+&\be_-\end{pmatrix}\in
\SU(2)$ which differ from each other only in a change of sign; the
eigenvectors are in any case determined only up to sign, and the physical
meaning is only carried by the amplitudes $|\al_\pm|$ and $|\be_\pm|$ which
are independent of the choice of signs made.

One way to think of the process is to start with an SG device $S$ and then
starting to turn it around a fixed axis. This determines a path in the group
$\SO(3)$ starting from the identity. Starting from the identity matrix in
$\SU(2)$, I can follow this path in $\SU(2)$, and see what happens to the
transformation matrix. It turns out that after a full turn by $2\pi$ of my
device, that is, after a loop\index{loop} in $\SO(3)$ returning to the
identity, my path\index{path} in $\SU(2)$ takes me to the {\em negative} of
the identity matrix. Following the loop\index{loop} in $\SO(3)$ once again,
I can continue my path\index{path} in $\SU(2)$, and lo and behold! a turn of
$4\pi$ returns me to the identity matrix in $\SU(2)$.

This thought experiment with paths reflects the topological fact that the
fundamental group\index{fundamental!group} of $\SO(3)$ is $\Z/2$, and its
universal cover is the map $S^3\to\SO(3)$ of \ref{sec!topson}--\ref{sec!su2}
(see a first course in topology for the language). It is also responsible
for the mysterious statement turning up frequently in physics texts, that
``rotation by $2\pi$ does not leave the wave function of the electron
invariant, but multiplies it by $(-1)$''. As I am told, this can be directly
demonstrated by experiment.

As a final comment, note that in this chapter I dealt with spin for a
``spin~$\frac{1}{2}$'' particle such as the electron, whose spin can take
two values \plusspin\ or \minusspin. There are also ``spin~$1$'' particles
such as the heavy particles $Z, W^\pm$ which are responsible for nuclear
forces. Their spin can take the values \plusspin, $0$ or \minusspin. Much of
the discussion of this chapter applies to such three-state systems;
compare~\cite{Fn}, Chapter~5. Their spin can be measured by a three-way SG
device. The vector space $W$ representing spin states is now 3-dimensional
over $\C$, and the transformation $S\to T$ between SG devices corresponds to
a map $B\colon \SO(3)\to\GL(3,\C)$. In this case, there is no great mystery:
this map is, up to conjugation, the obvious inclusion map, where I think of
a $3\times 3$ real orthogonal matrix\index{orthogonal!matrix} as a $3\times
3$ complex invertible matrix (the ``vector representation''). For this
reason spin~1 particles are often called ``vector
particles''.\index{electron|)}

\section{Preview of Lie groups}\label{sec!lie}
\index{Lie group}
\markright{\hfill \thesection\ Preview of Lie groups\quad}

The topological groups\index{topological!group} $\GL(n,\R)$ and $\O(n)$ are
examples of {\em Lie groups},\index{Lie group} groups whose elements depend
on a finite number of continuous parameters. Examples of Lie groups include
the Euclidean group $\Eucl(n)$,\index{Euclidean!group $\protect\Eucl(n)$}
the Lorentz group\index{Lorentz!group} $\O^+(1,2)$, the special linear group
$\SL(n)$\index{special!linear group $\SL(n)$} (the group of invertible
$n\times n$ matrixes with determinant $1$), the\index{spinor group
$\Spin(n)$} spinor groups $\Spin(n)$,
and groups defined using the complex numbers such as the group $\GL(n,\C)$
of invertible matrixes over~$\C$. Here is a list of features of general Lie
group theory that made an appearance in this chapter:
\begin{description}\index{dimension}
\item[Dimension] The geometry of the group around any point can be described
by $d$ parameters, where the number $d$ is independent of the point chosen,
and is called the {\em dimension} of the group. Examples from
Proposition~\ref{sec!parameters}:
\[
\dim\O(n)=\binom n 2\quad\hbox{and}\quad\dim\Eucl(n)=\binom{n+1}2.
\]

\item[Components] A Lie group\index{Lie group} $G$ has a number of connected
components\index{connected!component} (finite or infinite), all of them
geometrically the same (homeomorphic). The component containing the identity
is a normal subgroup, and the other components are its cosets. See \ref{SOn
conn} for $\O(n)$ and \exref{Opq ex} for the group $\O^+(1,2)$.

\item[Maximal compact subgroup]\index{maximal compact subgroup} A connected
Lie group\index{Lie group} $G$ is homeomorphic\index{homeomorphism} to a
product $H\times\R^N$ of a compact Lie group\index{compact!Lie group} $H$
and a space $\R^N$ in which all loops are contractible (compare~\ref{wind}).
The examples of~\ref{decomp of GL(n)} are typical: {\em compactness} is
achieved by imposing a {\em positive definite} orthogonal or Hermitian
form\index{Hermitian form}.

\item[The universal cover] A connected\index{connected!Lie group} Lie
group\index{Lie group} $G$ has a cover $\wG\to G$ by a simply
connected\index{simply connected} Lie group $\wG$ (possibly $G$ itself). The
typical examples are the exponential map $\C\to\C^*$ and the two-to-one
spinor covers $S^3\to\SO(3)$ and $S^3\times S^3\to\SO(4)$ discussed
in~\ref{ssec!sph_and_so}.

\item[Complexification and real forms] The group $\GL(n,\C)$ is the {\em
complexification} of the group $\GL(n,\R)$: the latter is a matrix group,
and I can simply take complex instead of real entries. Conversely, we say
that $\GL(n,\R)$ is a {\em real form} of $\GL(n,\C)$. Along the same lines,
the group $\O(n,\C)$ of $n\times n$ complex matrixes, which leave the
standard quadratic~(!) form $\sum_i x_i^2$ invariant, is a complexification
of the group $\O(n)$. However, $\O(n)$ is not the only real form: over the
complex numbers, there is no difference between the forms $\sum_i x_i^2$ and
$-x_1^2+\sum_{i>1}x_i^2$. Thus the Lorentz group $\O(1,n-1)$ is also a real
form of~$\O(n,\C)$.

\item[Linear representations] Just as finite groups, Lie groups\index{Lie
group} are often studied via their linear (matrix) representations. In plain
language, we associate to every group element $g\in G$ an $n\times n$
(complex) matrix $A_g$ so that $A_hA_g=A_{hg}$. In fancier language, this is
nothing but a group homomorphism $G\to \GL(n,\C)$; one familiar example is
the map $\tilde A\colon \SU(2)\to \GL(2,\C)$ from~\ref{sec! spin and rot}. I
recommend Fulton and Harris \cite{FH} for further study.

\item[Symmetry groups in physics]\index{physics} Lie groups\index{Lie group}
commonly appear as symmetry\index{symmetry} groups of interesting physical
systems. The mathematics of the group and the physics of the system are
often related in beautiful and nontrivial ways. The interaction occurs on
two levels: ``classical'' (meaning Newtonian mechanics and Maxwell
electromagnetic theory) and ``modern'' (meaning relativity theory or quantum
mechanics, possibly both). The story of the electron in~\ref{sec! spin and
rot} is the starting point of the ``quantum'' level of this interaction; for
more discussion, turn to~\ref{sec!geom&phys} and Sternberg~\cite{St}.
\end{description}

\section*{Exercises to Chapter~\thechapter}
\markright{\hfill Exercises to Chapter~\thechapter\quad}
\addcontentsline{toc}{section}{Exercises to Chapter~\thechapter}
\begin{enumerate}
\renewcommand{\labelenumi}{\thechapter.\theenumi.}

\item \index{affine!group $\Aff(n)$} \index{Euclidean!group
$\protect\Eucl(n)$} How much bigger is the affine group $\Aff(n)$ than
the Euclidean group $\Eucl(n)$? \begin{hint} compare $\GL(n)$ and
$\O(n)$ in~\ref{decomp of GL(n)}.\end{hint}

\item \label{ex!count_ecul_params}
\begin{enumerate}
\index{dimension!of a Lie group}

\item Show that rotations, translations,\index{translation} reflections and
glides of $\E^2$ (Theorem~\ref{sec!motE2}) depend respectively on 3, 2, 2 and 3
parameters.

\item Count parameters for each of the types of motion\index{motion} of
Theorem~\ref{sec!motE3}. (Answers: (1) Translation: 3; (2) Rotation: 5;
(3) Twist: 6; (4) Reflection: 3; (5) Glide: 5; (6) Rotary reflection: 6.
For example, a rotation\index{rotation} is specified by a line of
3-space, which depends on 4 parameters, plus an angle.) \end{enumerate}

\item \label{soparam}
Count the number of {\em real} parameters for the groups $\SO(3)$ and
$\SU(2)$; verify that they depend on the same number of parameters,
as you would expect from the two-to-one cover discussed in~\ref{sec!su2}.
\begin{hint} use Proposition~\ref{sec!parameters}, respectively the
results of~\ref{sec!su2}.\end{hint}

\item\label{GLn ex} Determine the connected
components\index{connected!component} of $\GL(n,\R)$ using
Theorem~\ref{decomp of GL(n)} and Proposition~\ref{SOn conn}.

\item\label{Opq ex}\index{Lorentz!group}\index{Lorentz!matrix}
Let
\[
\O(1,2)=\bigl\{ A\in \GL(3,\R)\bigm|\tr AJA=J\bigr\}
\]
be the group of all Lorentz matrixes, which contains the Lorentz group
$\O^+(1,2)$ introduced in~\ref{sec!top on groups}, Example~5. Show that this
group has {\em four} connected components,\index{connected!component}
distinguished by the fact whether a matrix preserves the cone $q_L(\bv)<0$
or maps it to $q_L(\bv)>0$ (that is, whether it is in $\O^+(1,2)$), and
$\det A=\pm1$. \begin{hint} imitate the proof of Theorem~\ref{SOn conn},
using the Lorentz normal form statement of
Ex.~\ref{app!q}.\ref{ex!lor_normalform}. Distinguish carefully between four
types of possible diagonal matrixes arising as end products.\end{hint}

\item\label{ex!decomp_gs} Let $A\in\GL(n,\R)$ be a matrix with columns
$\bbf_i$. Following the proof of Theorem~\ref{sec!bases}, (1) carefully,
show that it is possible to construct an orthonormal basis $\{\bbe_i\}$ of
$\R^n$, so that in each step
\[
\bbe_i= c_{i1}\bbf_1+\cdots+c_{ii}\bbf_i
\]
with $c_{ii}>0$. Let $C=(c_{ij})$ and $B$ the matrix with columns $\bbe_i$;
check that $A=BC$ and that $B\in\O(n)$, $C\in T_+(n)$ (compare~\ref{decomp
of GL(n)}). Check also that the entries of $B$ and $C$ depend continuously
on those of $A$.

\item Write the following matrixes in the form $BC$ of Theorem~\ref{decomp
of GL(n)} with $B\in\O(n)$ and $C\in T_+(n)$:
\[
\left(
\begin{matrix}
1&1+\sqrt3\\
\sqrt3&-1+\sqrt3
\end{matrix}
\right),
\quad
\left(
\begin{matrix}
1&3\\
1&4
\end{matrix}
\right),
\quad
\left(
\begin{matrix}
1&0&3\\
2&-1&4\\
2&1&2
\end{matrix}
\right).
\]

\item[] Exercises on quaternions

\item \label{quat mult ex} Show that 4 complex matrixes
\[
1= \begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix},\quad
I= \begin{pmatrix} i&0\\ 0&-i\\ \end{pmatrix},\quad
J= \begin{pmatrix} 0&1\\ -1&0\\ \end{pmatrix},\quad
K= \begin{pmatrix} 0&i\\ i&0\\ \end{pmatrix}
\]
multiply together by the same rules as the 4 basic quaternions
$1,i,j,k$. Since matrix multiplication is associative, use this to
give a better proof of Proposition~\ref{quat sec}, (1).

\item \label{quat conj ex} Complete the proof by brute force of
$(pq)^*=q^*p^*$ for quaternion conjugation (Proposition~\ref{quat sec},
(2)). Give a better proof along the lines of the previous exercise.

\item Study the group $G_8=\{\pm1,\pm i,\pm j,\pm k\}$ of unit
quaternions. Write out the group multiplication table, and find a
convincing reason (or failing that, any reason) why $G_8$ is not isomorphic to
the dihedral group $D_8$ appearing in Ex.~\ref{ch!gp}.\ref{ex!dihed}.

\item If $p=ai+bj+ck$ and $q=di+ej+fk$ are two pure imaginary
quaternions, calculate $pq+qp$ directly using the definition of
quaternion multiplication.

\item \label{quat frame ex} Prove that a pure imaginary quaternion
$p$ satisfies $p^2=-|p|^2$. Also if $p,q$ are pure imaginary then
$pq+qp=0$ if and only if they are orthogonal\index{orthogonal!vector}
with respect to the quadratic form $a^2+b^2+c^2+d^2$. \begin{hint}
Orthogonal w.r.t.\ a quadratic form $Q$ is expressed in terms of the
associated bilinear form\index{bilinear form}
$\fie(p,q)=Q(p+q)-Q(p)-Q(q)$; apply this with
$Q(q)=qq^*=-q^2$.\end{hint}

Deduce that 3 vectors $I,J,K\in\HH$ have the same multiplication
table as the quaternion basis $i,j,k$ if and only if they are an
oriented orthonormal frame of $\R^3$. Prove Proposition~\ref{quat
sec}, (5).

\item Show how to express $\C$ in terms of $2\times2$ matrixes over
$\R$ of the form $\left( \begin{smallmatrix} a&b\\-b&a
\end{smallmatrix}\right)$.

\item\label{sumult} Show that the algebra of $2\times2$ matrixes
over $\C$ of the form $\left(\begin{smallmatrix} a&b\\-\bbar&\abar
\end{smallmatrix}\right)$ is an algebra isomorphic to the quaternions
$\HH$. \begin{hint} consider the basis given in \exref{quat mult ex}
and compare also~\ref{sec!su2}.\end{hint}

\item\label{direct ex} Consider left multiplication by
$M=\left(\begin{smallmatrix} a+ib&c+id\\-c+id&a-ib \end{smallmatrix}\right)$
acting on $\C^2$. Write out the action of $M$ on $\C^2=\R^4$ in terms
of the $\R$-basis $(1,0)$, $(i,0)$, $(0,1)$, $(0,i)$ of $\C^2$.
% Answer
% \[
% \begin{pmatrix} a&b&-c&d\\-b&a&-d&-c\\c&d&a&-b\\-d&c&b&a \end{pmatrix}.
% \]
Prove that the determinant of the map on $\R^4$ is
$(a^2+b^2+c^2+d^2)^2$. Use this to give another proof that $a_q$ is
direct in Theorem~\ref{quat rot}, (1).

\item Prove that $2\times2$ matrixes over $\R$ of the form $\left(
\begin{smallmatrix} a&b\\b&a \end{smallmatrix}\right)$ form an algebra $B$,
and study its properties. Why is it not very interesting? \begin{hint} show
that $B$ is closed under addition and multiplication of matrixes. Find a
basis over $\R$, and write out the multiplication table.\end{hint}

\item By analogy with the previous question, investigate the algebra of
$2\times2$ matrixes over $\C$ of the form $\left( \begin{smallmatrix}
a&b\\-b&a \end{smallmatrix}\right)$.

\item Use the argument of Theorem~\ref{quat rot} to find a unit quaternion
$q$ so that the rotation $r_q\colon x\mapsto qxq^*$ is $(x,y,z)\mapsto
(y,-x,z)$.

\item Find a unit quaternion $q$ so that the rotation $r_q\colon x\mapsto
qxq^*$ is $x\mapsto y\mapsto z\mapsto x$. \begin{hint} the effort-intensive
method is to use brute force. The thinking person's method is to represent
$x\mapsto y\mapsto z$ as a rotation through angle $\theta$ about directed
axis $L$, then use Theorem~\ref{quat rot}.\end{hint}

\item\label{ex!su2} By analogy with \ref{sec!refl}, solve the relations
(\ref{eq!aabar}) of \ref{sec!su2} to get $d=\abar$, $c=-\bbar$. \begin{hint}
For example, do $\hbox{second line}\times d-\hbox{third line}\times c$, then
substitute $ad-bc=1$ on the right-hand side.\end{hint}

\item (Harder) Using the results of the two preceding exercises, show how to
find a subgroup $\mathrm{BO}_{48}$ of the unit quaternions which has a
surjective two-to-one map to the group of rotations of the cube in $\SO(3)$.

\item\label{ex!4dim} (Harder) Complete the proof of Theorem~\ref{quat rot}
(2). \begin{enumerate}

\item Prove that $\fie(p,q)=\id_{\HH}$ if and only if $(p,q)=(1,1)$ or
$(p,q)=(-1,-1)$. \begin{hint} $p1q^*=1$ if and only if $p=q$, and $pip^*=i$
if and only if $pi=ip$ if and only if $p=a+bi$, etc.\end{hint} Deduce that
$\fie$ induces an injective map $(S^3\times S^3)/{\pm1}\to\SO(4)$.

\item Prove that $\fie$ is surjective. \begin{hint} find a suitable
$\fie(p,q)$ to send 1 to a given unit vector $r\in\HH$. Now compose with
$r^*$ to assume that $1\mapsto 1$, and apply Theorem~\ref{quat rot}
(4).\end{hint}
\end{enumerate}

\item \label{ex!octonions}
(Harder) Consider the algebra $\mathbb O$ of $2\times2$
matrixes over the quaternions $\HH$ of the form $\left(
\begin{smallmatrix} a&b\\-b^*&a^* \end{smallmatrix}\right)$ where
$a^*$ is the quaternion conjugate of $a$ as in \ref{quat sec}.
\begin{enumerate}
\item Show that $\mathbb O$ is an 8-dimensional division algebra (algebra
with two-sided multiplicative inverses for nonzero elements) over $\R$.
Find an explicit basis for $\mathbb O$ and write out some of the multiplication
table.
\item Show that multiplication in $\mathbb O$ is not associative,
but it satisfies the identity
\[
x(xy)=(xx)y\mbox{ for } x,y\in\mathbb O.
\]
\item Contemplate on the possibility of doing projective geometry over the
division algebra $\mathbb O$ (compare the end of~\ref{Axiomatic projective
geometry}). \end{enumerate} $\mathbb O$ is the algebra of {\em Cayley
numbers} or {\em octonions}. For much more on this, see Conway and Smith
\cite{ConSmi}.

\begin{hint} You get a division algebra by introducing an octonion
conjugate $\widehat a$ such that $a\widehat a=|a|^2$ is positive definite,
as in~\ref{quat sec}. It is easy to find examples of nonassociative
octonion multiplication; to prove the weaker identity, one possibility is
to use your basis for $\mathbb O$ over $\R$ in a brute-force
proof similar to that of Proposition~\ref{quat sec} (1) given in the text.
To do projective geometry, you have to start by thinking about the
relation $\bx\sim\la\bx$ used to define projective space. Do not be surprised
if you run into difficulty.
\end{hint}

\item[] Hermitian matrixes

\item\label{herm_ex} An $n\times n$ complex matrix $A$ is called
{\em Hermitian}, if ${\herm A}=A$. (See~\ref{sec!su2} for the
Hermitian conjugate $\herm A$.) Show that
\begin{enumerate}
\item every eigenvalue of a Hermitian matrix is real;
\item eigenvectors for different eigenvalues are orthogonal with
respect to the Hermitian form\index{Hermitian form} on $\C^n$
(compare Step 3 in the proof of Theorem~\ref{ssec!normf}!).
\end{enumerate}
\end{enumerate}

\index{Lie group|)}
\index{transformation group|)}

\endinput

