 \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!aff} to~\ref{ch!proj} discussed transformations such as
Euclidean rotations in the plane that depend continuously on parameters:
the center 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 family of elements,
especially some examples arising in geometry that have an interesting
geometry of their 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 on
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} of the 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).

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

A group $G$ is a {\em topological group} if it has a
topology\index{topology} defined on it so that multiplication and
inverses are 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
 \[
 \begin{array}{rcccccccl}
 {\rm mult}\colon & G\times G & \to & G &
 \quad\hbox{and}\quad & {\rm inv}\colon & G & \to & G \\
 & (g,h) & \mapsto & g\circ h & & & g & \mapsto & g\1
 \end{array}
 \]
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!aff}--\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. 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})
 \[
 \O^+(1,2)=\left\{ A\in \GL(3,\R)\left|
 \begin{array}{l}
 \hbox{$\tr AJA=J$, and $A$ preserves} \\
 \hbox{the cone $q_L(\bv)<0$}
 \end{array}
 \right\}\right..
 \]
 This is also a topological group. It and its higher dimensional
colleagues $\O^+(1,n)$ are important in special relativity 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, 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\ Counting parameters\quad}

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 shows then 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 value of the determinant).

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}.

 \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 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.

 \paragraph{Example 2} Although the next example looks similarly
innocent, it appears in many different guises throughout geometry,
Fourier analysis, Lie group theory,\index{Lie group} 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
isomor\-phism)
 \[
 \begin{array}{ccc}
 S^1\times\R_+ & \to & \C^* \\
 (\theta,r)& \mapsto & r\exp(i\theta).
 \end{array}
 \]
 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} (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:
\[\dim O(n) + \dim T_+(n) = \binom{n}{2} + \binom{n+1}{2} = n^2 = \dim\GL(n,\R).
\]

 \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 it to 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}

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}. Components of the group
$\O(1,2)$ of all Lorentz matrixes are 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 if $\det A=1$ and opposite if $\det A=-1$.) The function
 \[
 \det\colon \O(n)\to \{\pm 1\}
 \]
is 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 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} 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
$\SO(3)$ 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
$(bi\times -cj)+(cj\times -bi)=0$.

 (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 why 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 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 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 $S^3$}
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 $\psibar\colon
S^3/{\sim}\ \to\SO(3)$ which is clearly bijective. 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 property} 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 group $\Spin(n)$ called the {\em spinor group}
together with a fixed-point free automorphism (self-map)
$\iota$ of order 2 such that
$\SO(n)\simeq \Spin(n)/\langle \iota\rangle$. 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 will identify the group $U$ of unit quaternions
from~\ref{sec!topson} as a
group of matrixes. 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 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}
 \[
 \SU(2)= \bigl\{ A \in \U(2) \bigm|\det A=1 \bigr\}.
 \]
 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 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}

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. 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}
 \markright{\hfill \thesection\ Measuring spin\quad}

In the following, I assume a modified form of the Stern--Gerlach (SG)
device, illustrated on 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 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 $S$ on 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 $S^\prime$
{\em in the same
position} after the first as on 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 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.

 \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}

 \subsection{The spin operator}
 \markright{\hfill \thesection\ The spin operator\quad}

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 $S^\prime$
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
$S^\prime$ is determined.

 \subsection{Rotate the device}
 \markright{\hfill \thesection\ Rotate the device\quad}

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.

\subsection{The solution} \label{sec! spin and rot}
 \markright{\hfill \thesection\ The solution\quad}

I am going to be dishonest at this point by pulling a rabbit out of a
hat. It is possible to derive the expressions
for $\psi_T^\pm$ in terms of $\psi_S^\pm$ from first principles; a
beautiful and self-contained account of this is given by Feynman~\cite{Fn},
pp.~6--1 to 6--14. The answer is that 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 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 on 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 sign
change; 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 sign choice
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, after a turn of $4\pi$ I return to the identity matrix in
$\SU(2)$!

This thought experiment with paths reflects the topological fact that
the fundamental group\index{fundamental!group} of $\SO(3)$ has order
two; this is best left to more advanced topology texts. It is also
responsible for the somewhat enigmatic statement, often read in
physics texts, that ``the wave function of the electron is not
invariant under rotation by $2\pi$, rather it is multiplied 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.

 \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 $\SL(n)$ (the group of invertible $n\times n$ matrixes
with determinant $1$), the 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). See \ref{SOn conn}
for $\O(n)$ and \exref{Opq ex} for the group $\O^+(1,2)$.

 \item[Maximal compact subgroup]\index{maximal compact subroup} 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~\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}.
\cite{FH} is recommended for further study.

 \item[Symmetry groups in physics] Lie groups\index{Lie group}
commonly appear as symmetry groups of interesting physical systems. The
mathematics of the group and the physics of the system are often related
in a beautiful and nontrivial way. 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~\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 $\protect\Aff(n)$} \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}

 \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 $\det
A=\pm 1$. \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 $BO_{48}$ 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}.

\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

