\setcounter{chapter}{0}
\renewcommand{\chaptername}{Appendix}
\renewcommand{\thechapter}{\Alph{chapter}}

\chapter[Metrics]{Metrics}\label{app!metric}
\index{metric|(}\index{metric!space|(}
\markboth{\quad\thechapter\ Metrics\hfill}{}

\begin{defn} A {\em metric} on a set $X$ is a specification of a {\em
distance}\index{distance!function} $d(x,y)$ between any two points $x,y\in
X$, in other words a map $d\colon X\times X\to\R$, required to satisfy the
following axioms for all $x,y,z\in X$:
\begin{enumerate}

\item $d(x,y)\ge0$ and $d(x,y)=0$ if and only if $x=y$;

\item $d(x,y)=d(y,x)$;

\item the triangle inequality:\index{triangle inequality} $d(x,y)\le
d(x,z)+d(z,y)$.
\end{enumerate}
\end{defn}

For example, the real line $\R$ with $d(x,y)=|x-y|$ is a metric
space.\index{metric!space} The epsilon-delta definition of continuity of a
function in a first calculus course uses that $\R$ is a metric space
(compare~\ref{metric motiv}). Theorem~\ref{Rn}, Corollary~\ref{s2trineq} and
Corollary~\ref{h2trig} say that the vector space $\R^n$ and hence Euclidean
space $\E^n$,\index{Euclidean!space $\E^n$} the sphere $S^2$\index{sphere
$S^2$} and the hyperbolic plane\index{hyperbolic!plane $\sH^2$} $\sH^2$ are
all metric spaces with their respective distance
functions.\index{distance!function} The set of complex numbers $\C$ is also
a metric space under the distance function $d(z_1,z_2)=|z_2-z_1|$. Some
frivolous examples show that many distance
functions\index{distance!function} in use in the real world are {\em not}
metrics:

\begin{enumerate}
\item Air fares: let $d(x,y)$ be the price of an airline ticket from $x$ to
$y$; this is usually unsymmetric, and does not satisfy the triangle inequality.
\item The distance you travel by car to go from one point of a town to
another; this is not symmetric, because of one-way traffic systems. However,
it satisfies the triangle inequality, because you take the minimum over
paths\index{minimum over paths}, at least if your taxi driver is honest.
\item For a cyclist, up a hill is of course much further than down.
\end{enumerate}

I use the following simple definition to pass from a metric space to the
slightly more general notion of topological space in Chapter~\ref{top chap}
(see Section~\ref{metric motiv}).

\begin{defn}\label{def!ball} Let $X$ be a metric space, $x\in X$ a point and
$\ep>0$ a real number. The ball in $X$ of radius $\ep$ centred at $x$ is the
subset
\[
B(x,\ep) = \bigl\{ y \in X \bigm| d(x,y) < \ep \bigr\} \subset X.
\]
\end{defn}

For example, if $X=\R$ is the real line, then $B(x,\ep)$ is the open
interval $(x-\ep,x+\ep)$. All the definitions of continuity of $f(x)$ in the
first calculus course can be expressed in terms of these intervals.

\begin{defn} Let $(X,d)$ and $(Y,d_Y)$ be metric spaces. An {\em
isometry}\index{isometry}\index{preserves distances|(} is a bijective
map $f\colon X\to Y$ satisfying the condition
\[
d_Y(f(x), f(y))= d(x,y).
\]
\end{defn}

The meaning of this definition is that the two spaces $(X,d)$ and $(Y,d_Y)$
are ``the same'' as far as their metric properties are concerned. An example
that is used very often is the fact that the complex numbers $\C$ and the
vector space $\R^2$ are isometric under the map $x+iy\mapsto (x,y)$. Note
that seemingly different metric spaces can be isometric under some weird or
ingenious map; see for example Ex.~\thechapter.\ref{ex!polys} and, for a
geometric example, Ex.~\ref{ch!hyp}.\ref{ex!Hidentify}.

A slightly different case of this definition that comes up all the time in
geometry is when $(X,d)=(Y,d')$ and $f$ is a bijection. Then $f$ is viewed
as a selfmap of $X$ ``preserving all the metric geometry''. The motions of
geometries studied throughout this book provide examples.

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

\item\label{isom ex} Let $X$ be a metric space and $t\colon X\to X$
a map that preserves distances\index{preserves
distances|)}\index{distance!function} $d(t(x),t(y))=d(x,y)$. Prove that
$t$ is injective. Give an example in which $t$ is not bijective; in
other words, $X$ can be isometric to a strict subset of itself, just
as in set theory, an infinite set can be in bijection with a strict
subset. \begin{hint} Think of ``Hilbert's hotel''.\end{hint}

\item\label{ex!setdist}
Let $S=[1,\dots,n]$ be a set containing $n$ elements, and $X$ the set of
all subsets of $S$. For $x,y\in X$, write $d(x,y)$ for the size of the
symmetric difference of $x$ and $y$ (the number of elements of $S$
contained in one of $x,y$ but not the other). Show that $d$ is a metric
on the set $X$. What happens to the construction if $S$ is infinite?
What happens if $S$ is infinite but I insist that $X$ consists only of
the finite subsets of $S$?

\item\label{ex!polys}
Let $P$ be the set of polynomials in one variable with coefficients
in $\Z/2$; remember, this means that we work over the field $\{0,1\}$
with two elements where the addition law includes $1+1=0$. If $f$ and
$g$ are two polynomials, let $d(f,g)$ be the number of nonzero terms
in the difference $f-g$. Show that $d$ is a metric on $P$. Show also
that $P$ with this metric is isometric to some metric space appearing
in the previous exercise.

\item Prove that a metric space with exactly 3 points is isometric to a
subset of $\E^2$.

\item Let $X=\{A,B,C,D\}$ with $d(A,D)=2$, but all the other distances
equal to~1. Check that $d$ is a metric. Prove that the metric space
$X$ is not isometric to any subset of $\E^n$ for any $n$. Can you
realise $X$ as a subset of a sphere $S^2$\index{sphere $S^2$} of
appropriate radius, with the spherical ``great circle''
metric?\index{spherical!geometry}\index{curvature} \begin{hint} I am
sure you know the riddle: An explorer start out from base camp, walks
10 miles due South, meets a bear, runs 10 miles due West, then 10
miles due North and finds himself back at base
camp.\index{spherical!triangle} {\em What colour was the
bear?} Compare Figure~\ref{fig!bear}.\end{hint}

\setcounter{figure}{\value{section}}
\begin{figure}[b]
\centering
\leavevmode\hbox{\epsfysize=3.5 true cm\epsffile{polar-bear.ps}}
\caption{The bear}
\label{fig!bear}
\end{figure}

\end{enumerate}
\index{metric|)}\index{metric!space|)}

\chapter{Linear algebra}\label{app!q}
\markboth{\quad\thechapter\ Linear algebra\hfill}{}

The distance\index{distance!function} function in $\R^n$ is given by the
norm $|\bx|^2=\sum x_i^2$, which comes from the standard inner
product\index{bilinear form} $\bx\cdot\by=\sum x_iy_i$. The ideas here are
familiar from Pythagoras' theorem and the equations of conics in plane
geometry, and from the vector manipulations in $\R^3$ used in applied math
courses. A quadratic form\index{quadratic form} in variables $x_1,\dots,x_n$
is simply a {\em homogeneous} quadratic function in the obvious sense. For
clarity I recall the formal definitions and results from linear algebra.

\section{Bilinear form and quadratic form}
\markright{\hfill \thesection\ Bilinear form and quadratic form\quad}
\label{bilin}
\index{bilinear form|(}

\begin{defn}
Let $V$ be a finite dimensional vector space over $\R$. A {\em symmetric
bilinear form} $\fie$ on $V$ is a map $\fie\colon V\times V\to\R$ such that
\begin{enumerate}
\renewcommand{\labelenumi}{(\roman{enumi})}

\item $\fie$ is linear in each of the two arguments, that is
\[
\fie(\la \bu +\mu \bv,\bw)=\la\fie(\bu,\bw)+\mu\fie(\bv,\bw)
\]
for all $\bu,\bv,\bw\in V$, $\la,\mu\in\R$, and similarly for the
second argument.

\item $\fie(\bu,\bv)=\fie(\bv,\bu)$ for all $\bu,\bv\in V$.
\end{enumerate}

A {\em quadratic form}\index{quadratic form} $q$ on $V$ is a map
$q\colon V\to\R$ such that
\[
q(\la \bu+\mu \bv)= \la^2 q(\bu) + 2\la\mu\fie(\bu,\bv) + \mu^2 q(\bv)
\]
for all $\bu,\bv\in V$, $\la,\mu\in\R$, where $\fie(\bu,\bv)$ is a
symmetric bilinear form.
\end{defn}

\begin{prop}
A quadratic form is determined by a symmetric bilinear form and vice-versa by
the rules
\[
q(x)=\fie(\bx,\bx) \quad\hbox{and}\quad
\fie(\bx,\by)=\half\bigl(q(\bx+\by)-q(\bx)-q(\by)\bigr).
\]
\end{prop}

Choosing a basis $\bbe_1,\dots,\bbe_n$ of $V$, a quadratic form $q$ or its
associated symmetric bilinear form $\fie$ are given by
\[
q(\bx)=\sum_{i,j} a_{ij}x_ix_j = \tr\bx K \bx, \quad
\fie(\bx,\by)=\sum_{i,j} a_{ij}x_iy_j = \tr\bx K \by.
\]
Here $\bx=\tr(x_1,\dots,x_n)=\sum x_i\bbe_i$, $\by=\tr(y_1,\dots,y_n)=\sum
y_i\bbe_i$ and $K=(k_{ij})$ is a symmetric matrix whose entries are given by
$k_{ij}=\fie(\bbe_i, \bbe_j)$.

\section{Euclid and Lorentz} \label{sec!euclandlor}
\markright{\hfill \thesection\ Euclid and Lorentz\quad}
\index{Lorentz!form $q_L$}
There are two special bilinear forms that are useful in geometry. To see the
first, let $V=\R^n$ be the vector space with the standard basis
\[
\bbe_1=\tr(1,0,\dots,0),\dots,\bbe_n=\tr(0,\dots,0,1).
\]
The {\em Euclidean inner product}\index{Euclidean!inner product} corresponds
to the matrix
\[
I=\diag(1,1,\dots, 1).
\]
It is the familiar
\[
\fie_E(\bx,\by)=\bx\cdot\by=\tr\bx I \by = \sum_i x_iy_i,
\]
with corresponding quadratic form
\[
q_E(\bx)=|\bx|^2=\sum_ix_i^2.
\]
As you know, an {\em orthonormal basis} of $\R^n$ is a set of $n$
vectors $\bbf_1,\dots,\bbf_n\in\R^n$ such that
\[
\bbf_i\cdot \bbf_j=\de_{ij}=
\begin{cases}
0&\hbox{for $i\ne j$,}\\
1&\hbox{for $i=j$.}
\end{cases}
\]
The model for this definition is the usual basis
$\bbe_i=(0,\dots,1,0,\dots)$ of $\R^n$ (with 1 in the $i$th place). The
inner product\index{Euclidean!inner product} $\fie_E$ expressed in terms of
an orthonormal basis $\bbf_1,\dots, \bbf_n$ of $V$ still has matrix $I$.

For the indefinite case, it is convenient to change notation slightly, so
let $V=\R^{n+1}$ be the vector space with the standard basis
$\bbe_0,\dots,\bbe_n$. The {\em Lorentz dot product}\index{Lorentz!dot
product $\cdot_L$} is the symmetric bilinear form given by the matrix
\[
J=\diag(-1,1,\dots,1).
\]
If $\bx=(t,x_1,\dots, x_n)$ and $\by=(s,y_1,\dots, y_n)$ then
\[
\fie_L(\bx,\by)=(t,x_1,\dots,x_n)\cdot_L(s,y_1,\dots,y_n)=-ts+\sum
x_iy_i.
\]
The {\em Lorentz norm}\index{Lorentz!norm} is the associated quadratic
form $q_L\colon V\to\R$, defined by
\[
q_L(t,x_1,\dots,x_n)=-t^2+\sum x_i^2.
\]
A {\em Lorentz basis}\index{Lorentz!basis} $\bbf_0,\bbf_1,\dots,\bbf_n$
is a basis of $V$ as a vector space, with respect to which $q_L$ has the
standard diagonal matrix~$J$; that is,
\[
q_L(\bbf_0)=-1, \quad q_L(\bbf_i)=1 \quad \hbox{for $i\ge1$}
\quad\hbox{and}\quad \bbf_i\cdot_L \bbf_j=0 \quad\hbox{for
$i\ne j$.}
\]

\section{Complements and bases} \label{sec!bases}
\markright{\hfill \thesection\ Complements and bases\quad}

Let $(V,\fie)$ be a vector space with bilinear form.\index{bilinear form|)}

\begin{defn} For a vector subspace $W\subset V$, define the {\em
complement}\index{orthogonal!complement $V^\perp$}
of $W$ with respect to $\fie$ to be
\[
W^\perp=\bigl\{\bx\in V\bigm| \fie(\bx, \bw)=0 \quad
\hbox{for all $\bw\in W$}\bigr\}.
\]
\end{defn}

In general, complements need not have any particularly nice properties;
notice for example that the zero inner product (with matrix $K=0$) gives
$W^\perp=V$ for all subspaces $W$. However, for ``nice'' inner products the
situation is completely different. I write this section explicitly with the
minimal generality needed for the geometric applications; all this can be
souped up to obtain the general Gram--Schmidt process, Sylvester's law of
inertia, etc.

\begin{thm} Let $\fie$ be the Euclidean inner
product\index{Euclidean!inner product} on $V=\R^n$. Let $W$ be a subspace
of $\R^n$. Then
\begin{enumerate}

\item $W$ has an orthonormal basis $\bbf_1,\dots, \bbf_k$.

\item Any vector $\bv\in\R^n$ has a unique expression $\bv=\bw+\bu$ with
$\bw\in W$ and $\bu\in W^\perp$; in other words, $\R^n$ is the direct sum
$W\oplus W^\perp$.
\end{enumerate}
\end{thm}
\begin{pf} Suppose that $W$ is not the zero vector space, take a nonzero
$\bv_1\in V$ and let $\bbf_1=\bv_1/|\bv_1|$ be a vector with unit length
in the direction of $\bv_1$. If $\bbf_1$ spans $W$ then I am home. If
not, take $\bv_2$ outside the span of ${\bbf_1}$ and let $\bbf_2$ be a
unit vector in the direction of $\bv_2-(\bv_2\cdot\bbf_1)\bbf_2$. Then,
as you can check, the cunning choice of the direction of $\bbf_2$
ensures that it is orthogonal\index{orthogonal!vector} to $\bbf_1$, and
it lies in $W$. Now continue this way by induction. Either the
constructed $\bbf_1, \dots, \bbf_k$ generate $W$, or you can find
$\bv_{k+1}\in W$ outside their span, and then a unit vector in the
direction of $\bv_{k+1} - \sum (\bv_{k+1}\cdot \bbf_i)\bbf_i$ can be
added to the collection.

For the second statement, find an orthonormal basis $\bbf_1,\dots,
\bbf_k$ of $W$, and extend it using the same method to an orthonormal
basis $\bbf_1,\dots,\bbf_n$ of $\R^n$. Then every vector $v\in \R^n$
has a unique expression
\[
\bv=\sum_{i=1}^n \la_i\bbf_i
\]
and then
\[
\bw=\sum_{i=1}^k \la_i\bbf_i, \ \ \bu=\sum_{i=k+1}^n \la_i\bbf_i
\]
is the only possible choice. \QED \end{pf}

The procedure of the proof is algorithmic, so lends itself easily to
calculations; to make sure that you understand it, do
Ex.~\thechapter.\ref{ex!calcobasis}.

\begin{thm} Let $V=\R^{n+1}$ with the Lorentz dot product\index{Lorentz!form
$q_L$}\index{Lorentz!dot product $\cdot_L$} and form.
\begin{enumerate}
\addtocounter{enumi}{2}

\item Let $\bv\in\R^{n+1}$ be any vector with $q_L(\bv)<0$. Then
$q_L(\bw,\bw)>0$ for $\bw$ a nonzero vector in the Lorentz
complement\index{Lorentz!complement} $\bv^\perp$.

\item Let $\bbf_0\in \R^{n+1}$ be a vector with $q_L(\bbf_0)=-1$. Then
$\bbf_0$ is part of a Lorentz basis\index{Lorentz!basis} $\bbf_0,\dots,
\bbf_n$ of $\R^{n+1}$.
\end{enumerate}
\end{thm}

\begin{pf} For (1), suppose that $\bv=(t,x_1,\dots,x_n)$ and
$\bw=(s,y_1,\dots,y_n)$ satisfy $q_L(\bv)<0$ and $\bv\cdot_L\bw=0$, that is
\begin{equation}
-t^2+\sum_{i=1}^n x_i^2<0
\label{eq!h1}
\end{equation}
and
\begin{equation}
-st+\sum_{i=1}^n x_iy_i=0
\label{eq!h2}
\end{equation}
Then (\ref{eq!h1}) and (\ref{eq!h2}) give that
\begin{eqnarray*}
\Bigl(-s^2+\sum_{i=1}^n y_i^2\Bigr)t^2
&=&-s^2t^2+t^2\Bigl(\sum_{i=1}^n y_i^2\Bigr)\\
&>&-\Bigl(\sum_{i=1}^n x_iy_i\Bigr)^2
+\Bigl(\sum_{i=1}^n x_i^2\Bigr)\Bigl(\sum_{i=1}^n y_i^2\Bigr),
\end{eqnarray*}
provided that the $y_i$ are not all~0. But we know that the last line is
$\ge0$ (in fact it is equal to $\sum(x_iy_j-x_jy_i)^2$, compare \ref{Rn}),
so
\[
-s^2+\sum_{i=1}^n y_i^2>0
\]
which is the statement.

For (2), pick $\bv_1\in\R^{n+1}$ linearly independent of $\bbf_0$ and
set
\[
\bw_1=\bv_1+(\bbf_0\cdot_L\bv_1)\bbf_0.
\]
Then $\bw_1$ is a nonzero element of $\bbf_0^\perp$, so by (1) it has
positive Lorentz norm.\index{Lorentz!norm} Hence I can set
$\bbf_1=\bv_1/\sqrt{q_L(\bv_1)}$. Then by construction $\bbf_0, \bbf_1$
are part of a Lorentz basis.\index{Lorentz!basis} Now continue with the
inductive method used in the proof of the previous theorem. \QED
\end{pf}

\section{Symmetries}\label{sec!fiesymm}
\markright{\hfill \thesection\ Symmetries\quad}
Return to the case of a general symmetric bilinear form $\fie$ on the vector
space~$V$, and its associated quadratic form $q$.

\begin{prop} Let $\al\colon V\to V$ be a linear map. Then equivalent
conditions:
\begin{enumerate}

\item $\al$ preserves $q$, that is, $q(\al(\bx))=q(\bx)$ for all
$\bx\in V$.

\item $\al$ preserves $\fie$, that is, $\fie(\al(\bx),\al(\by))=\fie(\bx,\by)$ for all $\bx,\by\in V$.
\end{enumerate}
\end{prop}
\begin{pf} The equivalence simply follows from the fact that $q$ is determined
by $\fie$ and conversely, $\fie$ is determined by $q$ from
Proposition~\ref{bilin}.
\QED \end{pf}

Now identify $V$ with $\R^n$ using the standard basis $\bbe_1,\dots,\bbe_n$.
Let $K=\{\fie(e_i,e_j)\}$ be the matrix of $\fie$.

\begin{prop} $\!\!\!${\rm (continued)} \
Let $A$ be the $n\times n$ matrix representing $\al$ in the given basis.
Then the previous two conditions are also equivalent to
\begin{enumerate}
\addtocounter{enumi}{2}

\item $A$ satisfies the matrix equality $\tr A K A = K$.
\end{enumerate}
\end{prop}

\begin{pf} Recall $\fie(\bx, \by)=\tr \bx A \by$. Hence
\[
\fie(\al(\bx),\al(\by))=\fie(\bx,\by) \iff \tr (A\bx) K (A\by) = \tr\bx \tr A K A \by = \tr\bx K \by
\]
and the latter holds for all $\bx$ and $\by$ if and only if $\tr A K A=K$.
\QED \end{pf}

A useful observation is the following:
\begin{lem} If $\det K\ne 0$ (we say that the form $\fie$ is
{\em nondegenerate}) then the equivalent conditions above imply
$\det A=\pm 1$.
\end{lem}
\begin{pf} From (3) and properties of the determinant it follows that
\[
(\det A)^2\det K = \det K.
\]
If $\det K\ne 0$ then I can divide by it.
\QED \end{pf}

\section[Orthogonal and Lorentz matrixes]{Orthogonal and Lorentz matrixes}\label{sec!o&lm}
\index{orthogonal!matrix}\index{Lorentz!orthogonal!matrix}
\markright{\hfill \thesection\ Orthogonal and Lorentz matrixes\quad}

Consider $\R^n$ with the Euclidean inner product,\index{Euclidean!inner
product} and let $\bbe_1,\dots,\bbe_n$ with $\bbe_i=(0,\dots,1,0,\dots)$
be the usual basis. If $\bbf_1,\dots,\bbf_n\in\R^n$ are any $n$ vectors,
there is a unique linear map $\al\colon\R^n\to\R^n$ such that
$\al(\bbe_i)=\bbf_i$ for $i=1,\dots,n$. Namely write $\bbf_j$ as the
column vector $\bbf_j=(a_{ij})$; then $\al$ is given by the matrix
$A=(a_{ij})$ with columns the vectors $\bbf_j$. Now, by
Proposition~\ref{sec!fiesymm} and by direct inspection, the following
conditions are equivalent:
\begin{enumerate}

\item $\bbf_1,\dots,\bbf_n$ is an orthonormal basis;

\item the columns of $A$ form an orthonormal basis;

\item $\tr AA=I$;

\item $\al$ preserves the Euclidean inner product.\index{Euclidean!inner product}
\end{enumerate}
We say that $\al$ is an {\em orthogonal
transformation}\index{orthogonal!transformation} and $A$ an
{\em orthogonal matrix}\index{orthogonal!matrix} if these conditions hold. We get the following result.

\begin{prop} $\al\mapsto(\al(\bbe_1),\dots,\al(\bbe_n))$ establishes a
one-to-one correspondence
\[
\left\{
\begin{matrix}
\hbox{orthogonal transformations}\\
\hbox{$\al$ of $\R^n$}
\end{matrix}
\right\}
\bij
\left\{
\begin{matrix}
\hbox{orthonormal bases}\\
\hbox{$\bbf_1,\dots,\bbf_n\in\R^n$}
\end{matrix}
\right\}.
\]
\end{prop}

\vspace{0.2in}

If $(V,\fie)$ is Lorentz,\index{Lorentz!space} a matrix $A$ satisfying
the condition $\tr AJA=J$ of Proposition~\ref{sec!fiesymm} (3) is
called {\em a Lorentz matrix}.\index{Lorentz!matrix} I leave you to
formulate the analogous correspondence between Lorentz
bases\index{Lorentz!basis} and Lorentz matrixes.

\section{Hermitian forms and unitary matrixes}\label{sec!herm}
\markright{\hfill \thesection\ Hermitian forms and unitary
matrixes\quad}\index{unitary!matrix|(}\index{Hermitian form|(}

This section discusses a slight variant of the above material, for
vector spaces over the field $\C$ of\index{complex number} complex
numbers. Let $V$ be a finite dimensional vector space over $\C$. A {\em
Hermitian form} $\fie\colon V\times V \to \C$ is a map satisfying the
conditions
\[
\fie(\la \bu +\mu \bv,\bw)=\labar\fie(\bu,\bw)+\mubar\fie(\bv,\bw)
\]
and
\[
\fie(\bu, \la \bv +\mu \bw)=\la\fie(\bu,\bv)+\mu\fie(\bu,\bw),
\]
where $\la,\mu\in\C$; note the appearance of the complex conjugate in
the first row. The corresponding {\em Hermitian norm} $q$ on $V$ is
\[
q(\bv) = \fie(\bv, \bv).
\]
The relation between $\fie$ and $q$ is slightly more complicated than
in the real case; I leave you to check the rather daunting looking
identity
\[
\fie(\bu, \bv) = \frac{1}{4}\Big(q(\bu+\bv)- q(\bu-\bv)+i
q(\bu+i\bv)-i q(\bu-i\bv)\Big).
\]
The terms in the identity are not so important; what is important is
the fact that $q$ gives back $\fie$.

Since I am only interested in a special case, I choose a basis
$\{\bbe_1,\dots,\bbe_n\}$ of $V$ straight away and assume that
\[
\fie(\la_1\bbe_1+\cdots+\la_n\bbe_n,\mu_1\bbe_1+\cdots+\mu_n\bbe_n)
=\labar_1\mu_1+\cdots+\labar_n
\mu_n.
\]
Such a form is called a {\em definite} Hermitian form. Under $\fie$,
$\bbe_1,\dots,\bbe_n$ form a Hermitian or orthonormal basis:
$\fie(\bbe_i,\bbe_j)=\de_{ij}$.

The following is completely analogous to Proposition~\ref{sec!fiesymm}:

\begin{prop} Let $\al\colon V\to V$ be a linear map represented by the
$n\times n$ matrix $A$ in the given basis. Then the following are
equivalent:
\begin{enumerate}

\item $\al$ preserves the norm $q$;

\item $\al$ preserves the Hermitian form $\fie$;\index{Hermitian form|)}

\item $A$ satisfies $\herm A A = I_n$, where $\herm A$ is the {\em
Hermitian conjugate} defined by $\herm A = \tr \Abar$; that is,
$(\herm A)_{ij}=\Abar_{ji}$.
\end{enumerate}
\end{prop}

The transformation $\al$ or the matrix $A$ representing it is {\em
unitary} if it satisfies these conditions; the set of $n\times n$
unitary matrixes is denoted $\U(n)$. Unitary transformations
(possibly on infinite dimensional spaces) have many pleasant
properties which makes them ubiquitous in mathematics. They are also
the basic building blocks of quantum mechanics and hence presumably
nature; in this book I discuss one tiny example of this
in~\ref{sec!spin}.

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

\item\label{ex!calcobasis}
Let $\bbf_1=(2/3,1/3,2/3)$ and $\bbf_2=(1/3,2/3,-2/3)\in\R^3$; find all
vectors $\bbf_3\in\R^3$ for which $\bbf_1,\bbf_2,\bbf_3$ is an orthonormal
basis.

\item By writing down explicitly the conditions for a $2\times 2$
matrix to be Lorentz,\index{Lorentz!matrix} show that any such matrix
has the form
\[
\renewcommand{\arraystretch}{1.2}
\begin{pmatrix} \cosh s & \sinh s\\ \sinh s & \cosh s \end{pmatrix}
\quad\hbox{or}\quad
\begin{pmatrix} \cosh s & -\sinh s\\ \sinh s & -\cosh s \end{pmatrix}.
\]

\item \label{ex!lor_normalform} This exercise is a generalisation of
the previous one; it shows that any Lorentz
matrix\index{Lorentz!matrix} can be put in a simple normal
form\index{normal form of a matrix} in a suitable Lorentz
basis;\index{Lorentz!basis} the Euclidean case is included in the
main text in~\ref{sec!norm}. Let $\al\colon \R^{n+1}\to
\R^{n+1}$ be a linear map given by a Lorentz matrix
$A$.\index{Lorentz!matrix} Prove that there exists a Lorentz
basis\index{Lorentz!basis} of $\R^{n+1}$ in which the matrix of $\al$
is
\[
B=
\left( \begin{smallmatrix}
\pm 1 \\[3pt]
&I_{k^+}\\[4pt]
&&-I_{k^-}\\[4pt]
&&& B_1\\
&&&&\ddots\\
&&&&& B_l
\end{smallmatrix}\right)
\quad\hbox{or}\quad
B=
\left( \begin{smallmatrix}
B_0\\[3pt]
&I_{k^+}\\[4pt]
&&-I_{k^-}\\[4pt]
&&& B_1\\
&&&&\ddots\\
&&&&& B_{l-1}
\end{smallmatrix}\right)
\]
where $B_0=\pm\left(\begin{smallmatrix}\cosh\theta_0&\sinh\theta_0\\
\sinh\theta_0&\cosh\theta_0 \end{smallmatrix}\right)$, $B_i=
\left(\begin{smallmatrix}\cos\theta_i&-\sin\theta_i\\
\sin\theta_i&\cos\theta_i \end{smallmatrix}\right)$ for $i>0$, and
$I_{k^{\pm}}$ are identity matrixes. \begin{hint} argue as in the
Euclidean case in \ref{ssec!normf}; the only extra complication is
that you have to take into account the sign of the Lorentz
form\index{Lorentz!form $q_L$} on the eigenvectors. The statement
follows by sorting out the cases that can arise.\end{hint}

\item \label{ex!unitarydet} Prove that a unitary matrix has
determinant $\det A\in \C$ of absolute value~1.\index{unitary!matrix|)}
\end{enumerate}
