\chapter{Projective geometry}\label{ch!proj}
\index{projective!geometry|(}
\markboth{\quad\thechapter\ Projective geometry\hfill}{}
The affine geometry\index{affine!geometry} studied in Chapter~\ref{ch!aff}
provided one possible solution to the problem of inhomogeneous linear
geometry. However, this turns out not to be the only one. This chapter
treats the alternative: it introduces projective space $\PP^n$ as another
equally natural linear geometry. The construction of $\PP^n$ can be
motivated starting from affine geometry in terms of adding ``points at
infinity.''\index{infinity!hyperplane at}.

Projective geometry is simple to study as pure homogeneous linear algebra,
ignoring the motivation; ``linear algebra continued'' or ``more things to do
with matrixes'' would be accurate subtitles for this chapter. In $\PP^n$,
the statement of affine geometry analogous to the dimension of
inter\-section formula of Theorem~\ref{dim inters} holds without the
``in\-homo\-geneous'' conditions of Chapter~\ref{ch!aff}, so that, for
example, two distinct lines $L_1,L_2\subset\PP^2$ meet in a point $P=L_1\cap
L_2$ without exception.

Projective geometry has lots of applications in math and other subjects.
Projective transformations include the perspectivities, or projections from
a fixed viewpoint from one plane to another, that form the foundation of
perspective drawing; the fact that you can readily recognise an object from
any angle, or a photograph taken from any point (and viewed at any angle)
indicates that your brain processes perspectivities automatically and
instantaneously.

\section{Motivation for projective geometry}\label{motiv proj}
\markright{\hfill\thesection\ Motivation for projective
geometry\quad}
\subsection{Inhomogeneous to homogeneous}\label{subsec!inhom} Recall from
Chapter~\ref{ch!aff} that if $E$, $F$ are affine linear
subspaces\index{affine!linear!subspace} of affine space $\aff^n$, then there
is a nice formula~\ref{dim inters} expressing the dimension of their
intersection\index{dimension!of intersection|(} {\em provided} that $E\cap
F\ne\emptyset$. One of the points of projective geometry is to get rid of
this unpleasant condition. The trouble all comes from the in\-homo\-geneity
of the equations: simultaneous inhomogeneous equations include, say, $x_1=0$
and $x_1=1$, where only two equations reduce $\aff^n$ to the empty
set.\index{empty set}

The solution is the following formal trick. Suppose $\sum a_{ij}x_j=b_i$ is
a set of inhomogeneous equations in $n$ unknowns $x_1,\dots,x_n$ defining an
affine linear subspace\index{affine!linear!subspace} $E\subset\aff^n$.
Replace these by homogeneous equations $\sum a_{ij}x_j=b_ix_0$ in $n+1$
unknowns $x_0,x_1,\dots,x_n$. The solutions with $x_0\ne0$ give ratios
$x_1/x_0,\dots,x_n/x_0$ that give a faithful picture of $E\subset\aff^n$.
But there are also the solutions with $x_0=0$, called ``points at
infinity''\index{infinity!point at}. Including these points adds
information to the set of ordinary solutions; namely, information about all
the ways the ratios $x_1:\cdots:x_n$ can behave as the $x_i$ tend to
infinity. A solution $0,\xi_1,\dots,\xi_n$ (with some $\xi_i\ne0$)
corresponds not to a point of $E$, but to an $(n-1)$-dimensional family of
all parallel lines\index{parallel!lines} with slope $\xi_1:\cdots:\xi_n$
satisfying the homogenised equations $\sum a_{ij}\xi_j=0$, that is, parallel
to some line in $E$ (compare Figure~\ref{fig!An_in_Pn}).

The set $E$ together with these extra solutions is a projective linear
subspace\index{projective!linear subspace|(} of projective space; the
intersection\index{dimension!of intersection|)} of projective linear
subspaces is then governed by the formula of~\ref{dim inters} {\em without
exception}. This does not mean that two projective linear subspaces cannot
have empty intersection;\index{empty set} it only means that they have empty
intersection exactly when they have a numerical reason to do so. In modern
language, the quantity $\dim E+\dim F-\dim\Span{E,F}$ on the right-hand side
of formula~\ref{dim inters} is called the {\em expected dimension} of the
intersection of $E$ and $F$; in projective geometry, linear subspaces always
intersect in a subspace whose dimension equals the expected dimension.

\subsection{Perspective}\label{subsec!pers+mot}
\index{perspective}\index{infinity!hyperplane at}
You recognise Figure~\ref{fig!cube} as a plane picture of a cube in $\R^3$.
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=6.8 true cm\epsffile{proj_pers.eps}}
\caption{A cube in perspective}\index{perspective}
\label{fig!cube}
\end{figure}
The way it is drawn, the horizontal parallel edges appear to meet in points
of the plane.

Suppose I fix the origin $O\in\aff^3$ and map points of a plane
$\Pi\subset\aff^3$, to another plane $\Pi'\subset\aff^3$ by taking $P\in\Pi$
into the point of intersection $P'=OP\cap\Pi'$ of the line $OP$ with $\Pi'$.
A map of the kind is called a {\em perspectivity}. It corresponds to putting
your eye at $O$, with $\Pi'$ a glass plate, $\Pi$ behind it with a figure on
it, and drawing faithfully the figure on the glass as you see it (see
Figure~\ref{fig!perspec}).
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=4.7 true cm\epsffile{proj_artist.eps}}
\caption{Perspective drawing}\index{perspective}
\label{fig!perspec}
\end{figure}

I get a map $f\colon\Pi\to\Pi'$ between two planes. It is easy to see that
$f$ maps lines of $\Pi$ to lines of $\Pi'$, and
parallel\index{parallel!lines} or concurrent lines $L,L',L''$, on $\Pi$ to
parallel or concurrent lines $M,M',M''$ on $\Pi'$. Here I am ignoring
practicalities, such as the finite extent of the plane represented by a
physical piece of glass, or the possibility that some of $\Pi$ might poke
out in front of $\Pi'$ rather than behind (see
Ex.~\ref{ch!proj}.\ref{ex!perspec} for details). Strictly speaking, $f$ is
only locally defined, and the conclusions should be qualified by adding
``within the domain of definition''; the activity takes place in the real
world, and set theoretic niceties do not cause us undue discomfort.

The map $f\colon P\mapsto P'$ is constructed in linear terms, but is not
actually linear (see \exref{ex!perspec}): choosing coordinates on
$\Pi,\Pi'=\aff^2$, it can be shown that $f$ is {\em fractional linear}, that
is, of the form
\[
f(\bx)=\frac{A\bx+b}{L\bx+c}
\]
where $A$, $b$, $L$ and $c$ are $2\times2$, $2\times1$, $1\times2$ and
$1\times1$ matrixes. Note that these can be assembled into a $3\times 3$
matrix $\left( \begin{smallmatrix} A&b\\L&c \end{smallmatrix}\right)$.

\subsection{Asymptotes} Figure~\ref{fig!conics} depicts the hyperbola
\begin{figure}[t]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}c}
\centering
\leavevmode\hbox{\epsfysize=8 true cm\epsffile{proj_hyp_para.eps}}
\caption{Hyperbola and parabola}
\label{fig!conics}
\end{figure}
$xy=1$ and the parabola $y=x^2$. Viewed from a long way off, the hyperbola
is very close to the line pair $xy=0$. In fact, outside a big circle of
radius $R$, either $|x|>R$ and $|y|<1/R$ or vice versa. One can argue that,
in turn, the parabola is asymptotic to the line $x=0$, in the sense that the
tangent line at the point $(x_0,x_0^2)$ gets steeper and steeper. This
argument is not actually very convincing: when both $x,y\gg0$, all you can
say is $y=x^2\gg x$. Nevertheless, in the theory of conic sections, it is
said, for example, that ``the two branches of the parabola meet at
infinity'',\index{infinity!point at} or that the parabola ``passes through
the point at infinity corresponding to lines parallel to $x=0$.''

The statements on asymptotes are qualitative views of what happens to the
curves when $x$ or $y$ is large (quite vague, even arguable for those in
quotes). But we have not so far said what asymptotic directions or points at
infinity actually are, which is a disadvantage in discussing asymptotes
formally or in calculating with them. Making sense of asymptotes (of
algebraic plane curves), and providing a simple framework for calculating
with them is one thing that projective geometry does very well.

\subsection{Compactification}\label{Pncomp}
\index{compact}
Here I assume that you know some topology; read this section after
Chapter~\ref{top chap} if you prefer.

Affine space $\aff^n$ is not compact; in contrast, projective space
$\PP^n$ is compact, as are its closed subsets, including all
projective algebraic varieties. Compact sets are much more convenient
than noncompact ones in many contexts of geometry, topology, analysis
and algebraic geometry. Given a closed\index{closed} set
$X\subset\aff^n$, you can compactify\index{compactification} it by
extending $\aff^n$ to $\PP^n$; then $X \subset \aff^n
\subset\PP^n$ and the closure $\Xbar\subset\PP^n$ is
compact.\index{compact} The points at infinity of the closure
$\Xbar$ correspond in a very precise sense to the asymptotic lines of
$X$, and are calculated by the same simple trick of adding a
homogenising coordinate $x_0$. For example, the hyperbola $xy=1$ is
compactified to the circle $S^1$ by adding the two points
$(\infty,0)$ and $(0,\infty)$, and the parabola is compactified to
$S^1$ by adding the single point $(0,\infty)$ at which the two
branches are said to meet.

\section{Definition of projective space}\label{Definition of Pn}
\markright{\hfill\thesection\ Definition of projective space\quad}
{\em Provided you forget about the motivation}, the definition is
very simple: introduce the equivalence relation $\sim$ on
$\R^{n+1}\setminus0$ defined by
\[
(x_0,\dots,x_n)\sim(y_0,\dots,y_n)\iff
\renewcommand{\arraystretch}{1.2}
\left\{ \begin{array}{l}
(x_0,\dots,x_n)=\la(y_0,\dots,y_n)\\
\hbox{for some $0\ne\la\in\R$.}
\end{array}\right.
\]
In other words $\bx\sim\by$ if the two vectors $\bx$ and $\by$ are
proportional, or span the same line (1-dimensional vector subspace)
through 0 in $\R^{n+1}$. Then define projective space to be
\[
\PP^n_\R=\PP^n=\bigl(\R^{n+1}\setminus0\bigr){\big/}{\sim}
=\bigl\{\hbox{lines through 0 in $\R^{n+1}$}\bigr\}.
\]
I write $(x_0:\cdots:x_n)$ for the equivalence class of $(x_0,\dots,x_n)$;
this is the usual notion of relative ratios of $n+1$ real numbers.
$x_0,\dots,x_n$ are {\em homogeneous coordinates} on $\PP^n$. For example,
$\PP^1$ is the set of ratios $(x_0:x_1)$. If $x_0\ne0$ you might as well
just consider $x_1/x_0$, but then you are missing one point corresponding to
the ratio $(0:1)$, where $x_1/x_0=\infty$.

In coordinate free language, if $V$ is an $(n+1)$-dimensional vector space
over $\R$, write $\PP(V)$ for the set of lines of $V$ through 0 (that is,
nonzero vectors up to the equivalence $\bv\sim\la \bv$ for $\la\neq 0$). Of
course, $V\iso\R^{n+1}$ (by a choice of basis), so $\PP(V)\iso\PP^n$. A
point $P\in\PP(V)$ is an equivalence class of vectors $\bv\in V$, or a line
$\R \bv$ through 0; several kinds of notation are popularly used to indicate
that $\bv=(x_0,\dots,x_n)$ is a vector in the equivalence class defining
$P$, for example:
\[
P=P_\bv, \quad P=[\bv], \quad \bv=\wave P, \quad
P_\bv=(x_0:\cdots:x_n), \quad
\hbox{etc.}
\]

To return to the motivation, $\PP^n$ contains the subset $(x_0\ne0)$
consisting of ratios that can be written $(1:x_1:\cdots:x_n)$, which is thus
naturally identified with $\aff^n$. The language used for motivating
projective geometry is quite unsuitable for developing the theory
systematically. For example, the terminology of ``points at
infinity''\index{infinity!hyperplane at} is cumbersome and gives a distorted
view of the symmetry of the situation.

The formal language of projective geometry is simply a reinterpretation of
the ideas of linear algebra; the subset with $x_0\ne0$ is not distinguished
in $\PP^n$, and there is no discrimination against points of the complement
(with $x_0=0$).\index{infinity!point at} Working with the definitions of
projective geometry and formal calculations in homogeneous coordinates is in
many ways easier to understand than how it relates to the motivation
discussed in \ref{subsec!inhom}, and I proceed with this, returning to the
motivation in \ref{aff in proj}. So for the time being, I discuss the
geometry of $\PP^n$ in terms of the vector space $\R^{n+1}$, and I advise
you to forget the motivation.

\section{Projective linear subspaces}\label{sec!subsp}
\markright{\hfill\thesection\ Projective linear subspaces\quad}

The only structures enjoyed by $\PP(V)$ are derived from $V$. Thus
all statements or calculations for $\PP(V)$ must reduce to linear
algebra in $V$ and the equivalence relation $\sim$ on
points of $V$.

As a first example, here is the definition of the line $PQ$ through two points
$P=(x_0:\cdots:x_n)$ and $Q=(y_0:\cdots:y_n)$ of $\PP^n$. First lift
to $\R^{n+1}$ by setting $\wave P=(x_0,\dots,x_n)$ and $\wave
Q=(y_0,\dots,y_n)$ (that is, pick values of $x_i$ and $y_i$ in the
given ratio), then set
\[
PQ=\Span{P,Q}=\bigl\{\hbox{ratios $(\la x_0+\mu y_0:\cdots:\la
x_n+\mu y_n)$ for all $(\la,\mu)\ne(0,0)$}\bigr\}.
\]
The point to notice is that $\la P+\mu Q$ is meaningless as a point of
$\PP^n$, because the ratio $(\la x_0+\mu y_0:\cdots:\la x_n+\mu y_n)$
depends on the choice of $\wave P$ and $\wave Q$ within the equivalence
classes of $P$ and $Q$. However, the {\em set of all\/}
$\la\wave P+\mu\wave Q$ is a well defined 2-dimensional vector subspace
of $V=\R^{n+1}$, and ratios in it form the line $PQ$.

Thinking in a purely formal way about vector subspaces of a vector space
$V$ gives the obvious notion of {\em projective linear
subspace}:\index{projective!linear subspace} if $U\subset V$ is a vector
subspace, $\PP(U)$ is the subset $(U\setminus0)/{\sim}\subset\PP(V)$ of
lines through 0 in $U$. In other words, if $U\subset\R^{n+1}$ then $\PP(U)$
is the set of ratios $(x_0:\cdots:x_n)$ with $(x_0,\dots,x_n)\in U$. The
dimension of $\PP(U)$ is defined to be $\dim\PP(U)=\dim U-1$. Thus
$\dim\PP^n=n$.\index{dimension}

A 0-dimensional subspace is a single point; a 1- or 2-dimensional
projective linear subspace is called a {\em line} or {\em plane}; an
$n-1$-dimensional subspace is a {\em hyperplane}\index{hyperplane}.
I sometimes say {\em $k$-plane} to mean $k$-dimensional projective linear
subspace.

Note that the empty set\index{empty set} $\emptyset$ is a projective
linear subspace: the trivial vector subspace $0\subset\R^{n+1}$ has
$\PP(0)=\emptyset\subset\PP^n$. By convention we write
$\dim\emptyset=-1$, to agree with the general definition just given.
As a rule, prudence might suggest that in mathematical arguments, we
avoid attaching excessive weight to mumbo-jumbo concerning the empty
set or the elements thereof, but here the convention
$\dim\emptyset=-1$ has a precise and useful meaning {\em (in the
context of the geometry of linear subspaces only!)}.

\begin{defn} If $\Si\subset\PP(V)$ is a set, write $\wave\Si\subset V$ for
the union of the lines in $\Si$; let $U$ be the vector subspace of $V$ spanned
by $\wave\Si$, and define the {\em span} or {\em linear span} of $\Si$ to be
$\Span\Si=\PP(U)$. This is the smallest projective linear subspace
containing $\Si$.

If $P_0,\dots,P_s$ are $(s+1)$ points then $\dim\Span{P_0,\dots,P_s}\le
s$; equality holds if and only if the vectors $\wave P_0,\dots,\wave
P_s\in\R^{n+1}$ are linearly independent. In this case,
$P_0,\dots,P_s$ are said to be {\em linearly independent}\/ in $\PP^n$.
\end{defn}

\section{Dimension of intersection}\label{proj dim inters}
\index{dimension!of intersection|(}
\markright{\hfill\thesection\ Dimension of intersection\quad}
\begin{thm} Let $E,F\subset\PP^n$ be projective linear subspaces.
Then
\begin{equation}
\dim E\cap F= \dim E+\dim F-\dim\Span{E,F};
\label{dim eq1}
\end{equation}
here the convention $\dim\emptyset=-1$ is in use. \end{thm}

\begin{pf} Write $\wave E,\wave F\subset\R^{n+1}$ for the vector
subspaces overlying $E$ and $F$. Then $E\cap F=\PP(\wave E\cap\wave
F)$ and $\Span{E,F}=\PP(\wave E+\wave F)$. By the linear algebra
formula \ref{dim inters}, (\ref{eq!dim.int}), we have
\begin{equation}
\dim(\wave E\cap\wave F)
=\dim\wave E+\dim\wave F-\dim(\wave E+\wave F),
\label{dim eq2}
\end{equation}
and since $\dim\PP(U)=\dim U-1$ for every vector subspace
$U\subset\R^{n+1}$, (\ref{dim eq1}) follows by subtracting 1 from
both terms on the left and right hand sides of (\ref{dim eq2}). \QED
\end{pf}
\index{projective!linear subspace|)}
\index{dimension!of intersection|)}

\section[$\protect\PGL(n+1)$ and projective frames of reference]{Projective
linear transformations and\\ projective frames of reference}
\label{PGL}
\markright{\hfill\thesection\ $\protect\PGL(n+1)$ and projective
frames of reference\quad}

A nonsingular linear map $\R^{n+1}\to\R^{n+1}$ represented by an
invertible matrix $A$ acts in an obvious way on the set of
lines of $\R^{n+1}$ through 0: namely, it takes the line $\R \bv$ to
$\R(A\bv)$ for every $0\ne \bv\in\R^{n+1}$. A map
$T\colon\PP^n\to\PP^n$ is a {\em projective transformation}
(also called {\em projectivity} or {\em projective linear map}) if it
arises in this way from a linear map. In other words, if we write
$P_\bv\in\PP^n$ for the point represented by $\bv\in\R^{n+1}$, then $T$ is
a projective transformation if there is an invertible matrix $A$ such that
\[
T(P_\bv)=P_{A\bv}\qquad\hbox{for all $\bv\in\R^{n+1}$.}
\]
Here $A\bv$ is the product of $A$ and $\bv$, viewed as a column vector. The set
of all projective transformations is written $\PGL(n+1)$.

Because $\bv$ and $\la \bv$ represent the same point of $\PP^n$, a
scalar matrix $\la\cdot\id=\diag(\la,\dots,\la)$ with $\la\ne 0$ acts
as the identity. Moreover, if $A$ is an invertible matrix
and $\la\in\R$ and $\la\ne0$, then $A$ and the product $\la A$ have
exactly the same effect on every point of $\PP^n$. Thus the set of
projective transformations is
\[
\PGL(n+1)=\bigl\{\hbox{invertible $(n+1)\times (n+1)$ matrixes}\bigr\}/\R^*
\]
where $\R^*=\bigl\{\la\cdot\id\mid0\ne\la\in\R\bigr\}$.

The following definition, which may seem unexpected at first, is quite
characteristic of projective geometry.

\begin{defn} A {\em projective frame of
reference}\index{frame!projective} (or {\em simplex of reference}) of
$\PP^n$ is a set $\{P_0,\dots,P_{n+1}\}$ of $n+2$ points such that
any $n+1$ are linearly independent, that is, span $\PP^n$.

This means
\begin{enumerate}

\item there exists a basis $\bbe_0,\dots,\bbe_n$ of $\R^{n+1}$ such
that $P_i=P_{\bbe_i}$ for $i=0,\dots,n$;

\item the final point $P_{n+1}$ is $P_{\bbe_{n+1}}$, where
\[
\bbe_{n+1}=\sum_{i=0}^n\la_i\bbe_i,\qquad\hbox{with $\la_i\ne0$ for
every $i$.}
\]
\end{enumerate}
Indeed, the first $n+1$ points $P_0,\dots,P_n$ are linearly
independent, and the final point $P_{n+1}$ is not contained in any of
the $n+1$ hyperplanes\index{hyperplane} $\{x_i=0\}$. The {\em standard
frame of reference}\index{frame!projective} is
\begin{equation}
\renewcommand{\arraystretch}{1.4}
\begin{matrix}
P_i=(0:\cdots:1:\cdots:0)\quad\hbox{(with 1 in the $i$th place)}\\
\hbox{and }P_{n+1}=(1:1:\cdots:1).
\end{matrix}
\end{equation}
That is, $\bbe_i$ for $i=0,\dots,n$ is the standard basis of
$\R^{n+1}$ and $\bbe_{n+1}=\sum_{i=0}^n \bbe_i$. The final point
$P_{n+1}=(1:\cdots:1)$ is there to ``calibrate'' the coordinate
system.
\end{defn}

\begin{thm} Let $\{P_0,\dots,P_{n+1}\}$ be the standard frame of
reference. Then there is a one-to-one correspondence between
projective transformations and frames of reference, defined by $T\mapsto
T(P_0),\dots,T(P_{n+1})$. \end{thm}

\begin{pf} Write $\bbe_0,\dots,\bbe_n$ for the standard basis of
$\R^{n+1}$, and set $\bbe_{n+1}=\sum_{i=0}^n \bbe_i$. Now
let $\{Q_0,\dots,Q_{n+1}\}$ be a different frame of reference, and choose
representatives $\bbf_0,\dots,\bbf_n$, $\bbf_{n+1}\in\R^{n+1}$ of the
points $Q_0,\dots,Q_{n+1}$.

Since $\bbe_0,\dots,\bbe_n$ and $\bbf_0,\dots,\bbf_n$ are two bases of
$\R^{n+1}$, the usual result of linear algebra is that there is a
uniquely determined linear map $A\colon \R^{n+1}\to\R^{n+1}$ such that
$A\bbe_i=\bbf_i$ for $i=0,\dots,n$. If $\bbf_0,\dots,\bbf_n$ are
column vectors, $A$ is the matrix with the given columns $\bbf_i$.
However, that is not what is given, and not what is required! If you
understand that, you have understood the proof.

Indeed, the $\bbf_i$ are determined only up to scalar multiples.
Start again: for any nonzero multiples $\la_i\bbf_i$ of $\bbf_i$ (for
$i=0,\dots,n$), there is a uniquely determined linear map
$A\colon\R^{n+1}\to\R^{n+1}$ such that $A\bbe_i=\la_i\bbf_i$ for
$i=0,\dots,n$, given by the matrix $A$ with columns $\la_i\bbf_i$.
Using the assumption that $\bbf_0,\dots,\bbf_n$ is a
basis,\index{frame!projective} I choose the $\la_i$ such that
$\bbf_{n+1}=\sum_{i=0}^n\la_i\bbf_i$. Then, because $Q_0,\dots,Q_{n+1}$
is a frame of reference, $\la_i\ne0$ for $i=0,\dots,n$, and
$A\bbe_{n+1}=\bbf_{n+1}$ by choice of $A$. Since
$A\colon\R^{n+1}\to\R^{n+1}$ is a linear map with
$\bbe_i\mapsto\la_i\bbf_i$ and $\bbe_{n+1}\mapsto \bbf_{n+1}$, it
defines a projective linear map $T\colon \PP^n\to\PP^n$ taking
$P_i\mapsto Q_i$ for $i=0,\dots,n+1$.

For the uniqueness, let us look back through the construction: first,
the condition $T(P_i)=Q_i$ for $i=0,\dots,n$ determines the columns
of $A$ up to multiplying each column by a scalar $\la_i$; so far, any
$\la_i$ will do (possibly different choices for different columns).
Next, the condition $T(P_{n+1})=Q_{n+1}$ fixes the $\la_i$ up to a
common scalar factor: because we must send $\bbe_{n+1}=\sum \bbe_i$
into a multiple of $\bbf_{n+1}=\sum_{i=0}^n\la_i\bbf_i$, we have to
choose these values of $\la_i$. The only remaining choice in $A$
would be to multiply the whole thing through by a scalar. Thus $T$
is uniquely determined. \QED \end{pf}

\section[Projective linear maps of $\protect\PP^1$ and the
cross-ratio]{Projective linear maps of $\protect\PP^1$ and the cross-ratio}
\label{pers cross-ratio}
\index{cross-ratio|(}
\markright{\hfill\thesection\ $\protect\PP^1$ and the cross-ratio\quad}

\begin{cor} There exists a unique projective linear transformation of\/
$\PP^1$ taking any $3$ distinct points $P,Q,R\in\PP^1$ to any other $3$.
\end{cor}

\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=4.2 true cm\epsffile{proj_phimap.eps}}
\caption{The 3-transitive action of $\protect\PGL(2)$ on $\protect\PP^1$}
\label{3-transitive}
\end{figure}

Since any 3 distinct points go into any other 3 points, I can say that
projective linear transformations act {\em $3$-transitively} on $\PP^1$.
This means that there can be no nontrivial function $d(P,Q)$ of 2 points or
$\si(P,Q,R)$ of 3 points that is invariant under these transformations.

However, there is a function of 4 distinct points invariant under projective
linear transformations, namely their {\em cross-ratio} $\{P,Q;R,S\}$. To
define it, note that any choices of representatives
$\bp,\bq\in\R^2\setminus0$ of $P,Q$ form a basis. Choosing this basis gives
\begin{equation}
P=(1:0),\quad Q=(0:1),\quad R=(1:\la)\quad\hbox{and}\quad S=(1:\mu).
\label{rs eq}
\end{equation}
for some $\la,\mu$. Set $\{P,Q;R,S\}=\la/\mu$.
\begin{figure}
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=9.4 true cm\epsffile{proj_cross_ratio.eps}}
\caption{The cross-ratio $\{P,Q;R,S\}$}
\label{fig!x-ratio}
\end{figure}

Changing the representative $\bq\mapsto\mu \bq$ sets $\mu=1$ so that
$S=(1:1)$. Thus the definition amounts to taking $P,Q,S$ as the frame of
reference of $\PP^1$,\index{frame!projective} and then defining
$\{P,Q;R,S\}=\la$, where $R=(1:\la)$. Since by Theorem~\ref{PGL}, the
projective transformation taking $P,Q,S$ to $(1:0),(0:1),(1:1)$ is unique,
$\{P,Q;R,S\}$ is well defined, and invariant under transformations in
$\PGL(2)$.

\begin{rem} To see the point of cross-ratio, it is useful to compare the
invariant quantities in $\aff^1$ and in $\PP^1$. In $\aff^1$, to be able to
measure, you need to fix the points 0 and 1, then any other point $P$ is
fixed by $\la=(x-0)/(1-0)$. In $\PP^1$ you need also to fix the point at
infinity.\index{infinity!point at} \end{rem}

\begin{prop} Consider four distinct lines of\/ $\R^2$ through $O=(0,0)$ that
are the equivalence classes of\/ $P,Q,R,\linebreak[2] S$, and let $L$ be any
line of\/ $\R^2$ not through the origin intersecting these four lines in
$\bp,\bq,\br,\bs$ respectively (see Figure~\ref{fig!x-ratio}). Then
\begin{equation}
\{P,Q;R,S\}=\frac{\bp-\br}{\bp-\bs}\cdot\frac{\bq-\bs}{\bq-\br}.
\label{cross-sect eq}
\end{equation}
Here the quotients on the right-hand side are ratios of vectors along $L$. You
could equally take them as ratios of $x$-coordinates or $y$-coordinates of the
points; or equally, the ratio of (signed) lengths
$\frac{\pm |\bp-\br|}{\pm |\bp-\bs|}\cdot
\frac{\pm |\bq-\bs|}{\pm |\bq-\br|}$.
\end{prop}

\begin{pf} As in the definition of $\{P,Q;R,S\}$, choose $\bp$ and $\bq$ as
the standard basis of $\R^2$. Then $L$ is given by $x+y=1$. If $\la,\mu$ are as in
(\ref{rs eq}) then $\br\in\R^2$ is in the equivalence class of $(1:\la)$ and
is on $L$, so that necessarily
\begin{equation}
\br=\frac{(1,\la)}{1+\la};\quad\hbox{similarly}\quad
\bs=\frac{(1,\mu)}{1+\mu}.
\end{equation}
The remaining calculation is very easy:
\begin{equation}\left.
\renewcommand{\arraystretch}{1.4}
\begin{matrix}
\bp-\br=\frac{\la}{1+\la}(1,-1), & \bq-\br=\frac{-1}{1+\la}(1,-1)\\
\bp-\bs=\frac{\mu}{1+\mu}(1,-1), & \bq-\bs=\frac{-1}{1+\mu}(1,-1)
\end{matrix}
\right\}\implies
\frac{\bp-\br}{\bp-\bs}\cdot\frac{\bq-\bs}{\bq-\br}=\frac{\la}{\mu}.
\end{equation}
This proves the proposition. \QED \end{pf}

\section[Perspectivities]{Perspectivities}
\label{sec!pers}
\markright{\hfill\thesection\ Perspectivities\quad}

Let $\Pi,\Pi'$ be hyperplanes\index{hyperplane} in $\PP^n$ and let $O$ be a
point outside $\Pi$ and $\Pi'$. The {\em perspectivity} $f\colon\Pi\to\Pi'$
from $O$ is obtained by mapping $P\in \Pi$ to the point of intersection
$f(P)$ of the projective line $OP$ with $ \Pi'$. Note that since $O$ is not
on $\Pi'$, the line $OP$ cannot be contained in $\Pi'$, and hence the
intersection of $OP$ with $\Pi'$ is a single point by the dimension of
intersection \index{dimension!of intersection} formula Theorem~\ref{proj dim
inters}. The case $n=3$, perspectivity between two planes in 3-space, was
described in \ref{subsec!pers+mot} and illustrated on
Figure~\ref{fig!perspec}. As opposed to the example in \ref{subsec!pers+mot}
(compare~Ex.~\ref{ch!proj}.\ref{ex!perspec}), the map $f$ is everywhere
defined, since new points have been added to affine space to form projective
space; this will be discussed further below.

It is easy to write a perspectivity in terms of coordinates. Choose
coordinates $(x_0:x_1:\cdots:x_n)$ so that $ \Pi=\{x_0=0\}$,
$\Pi'=\{x_1=0\}$ and $O=(1,1,0,\dots,0)$. Then for a point
$P=(0:x_1:\cdots:x_n)$ of $\Pi$, the line $OP$ is the set of points
$\{(\la:\la+\mu x_1:\mu x_2:\cdots:\mu x_n)\}\subset\PP^n$ (compare the
first paragraph of~\ref{sec!subsp}). The intersection point with $\Pi'$ is
then at $(\la:\mu)=(-x_1:1)$, so
\[
f\colon (0:x_1:\cdots:x_n) \mapsto (-x_1:0:x_2:\cdots:x_n).
\]
In particular, you can view the perspectivity $f$ as a projective
transformation from $\Pi=\PP^{n-1}$ with coordinates $(x_1:\cdots:x_n)$ to
$\Pi'=\PP^{n-1}$ with coordinates $(x_0:x_2:\cdots:x_n)$
given by the matrix $A=\diag(-1,1,\dots,1)$.

\begin{prop} The cross-ratio of four points on a line is invariant
under perspectivities; namely, if $L$ is a line in $\Pi$ and $P, Q, R, S\in L$
are four points on the line, then\index{perspective}
\[
\{P,Q;R,S\}= \{f(P),f(Q);f(R),f(S)\}.
\]
\end{prop}
\begin{pf} First of all, the right hand side of this expression is defined,
since the image of $L$ is a line in $\Pi'$; this follows from the fact that
$f$ is a projective transformation, but as an exercise you can check that it
also follows from the definition of $f$ and the dimension of intersection
formula. Then $f\colon L \to f(L)$ is a
projective transformation between
lines; the cross-ratio is preserved under projective transformations
of $\PP^1$, so it is preserved under perspectivities also.
Note that the equality of cross-ratios also follows from Figure~\ref{fig!x-ratio}
and the discussion of Proposition~\ref{pers cross-ratio},
once you restrict the discussion to the plane
$\PP^2\subset\PP^n$ spanned by $O$ and $L$, and interpret $O$ in
Figure~\ref{fig!x-ratio} as a point of this $\PP^2$ rather than
the affine origin $(0,0)\in\R^2$.
\QED\end{pf}
\index{cross-ratio|)}


\section[Affine space as a subset of projective space]{Affine space $\aff^n$ as
a subset of \\ projective space $\PP^n$}
\label{aff in proj}
\markright{\hfill\thesection\ Affine space in projective
space\quad}

A hyperplane\index{hyperplane} $H\subset\PP^n$ corresponds to an
$n$-dimensional subspace $W\subset\R^{n+1}$, the kernel of a linear form
$\al\colon\R^{n+1}\to\R$. Then $\PP^n\setminus H$ can be naturally
identified with $\aff^n$, and $H=\PP^{n-1}$ with sets of parallel
lines in $\aff^n$. The point is very simple: given $\ell$, I can
choose coordinates in $\R^{n+1}$ so that
\[
\al(x_0,\ldots, x_n)=x_0
\]
is the first coordinate. Then
\begin{align*}
\PP^n\setminus H
&\hbox{}=\bigl\{\hbox{ratios }(x_0:\cdots x_1:\cdots:x_n)\bigm| x_0 \ne0 \bigr\} \\
&\hbox{}=\Bigl\{\hbox{$n$-tuples }
\Bigl(\frac{x_1}{x_0},\dots,\frac{x_n}{x_0}\Bigr)\Bigr\}=\aff^n.
\end{align*}

In Figure~\ref{fig!An_in_Pn}, $P$ is a point with $x_0\ne0$, so its
equivalence class contains a unique point in the affine
hyperplane\index{hyperplane} $\aff^n$ defined by $(x_0=1)$. A point $Q$ with
$x_0=0$ does not correspond to any actual point of $\aff^n$; instead, it
corresponds to all the lines of $\aff^n$ parallel\index{parallel!lines} to
$\bv=\wave Q$.

Note that this discussion reverses the process of ``going from inhomogeneous
to homogeneous'' sketched in~\ref{subsec!inhom}; the points of the
hyperplane $H\subset\PP^n$ are at infinity\index{infinity!hyperplane at}
when viewed from the affine space $\aff^n$ defined by $(x_0=1)$. However,
splitting points into ``finite'' and ``infinite'' is not intrinsic to
projective space, but depends on the choice of $H$ (or the linear form
$\al$).

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}
\centering \input{proj_an_pn.pstex_t}
\caption{The inclusion $\protect\aff^n\subset\protect\PP^n$}
\label{fig!An_in_Pn}
\index{affine!space $\aff^n$!in projective space}
\end{figure}

\section{Desargues' theorem} \label{Desargues} \index{Desargues' theorem|(}
\markright{\hfill\thesection\ Desargues' theorem\quad}
\begin{thm}[Desargues' theorem] Let $\triangle PQR$ and $\triangle P'Q'R'$
be two triangles in\/ $\PP^n$ with $n\ge2$. Suppose that $\triangle
PQR$ and $\triangle P'Q'R'$ are in perspective\index{perspective}
from some point $O\in\PP^n$ (that is, $OPP'$, $OQQ'$ and $ORR'$ are lines).
Then the corresponding sides of $\triangle PQR$ and $\triangle
P'Q'R'$ meet in $3$ collinear points. In other words,
\begin{equation}
\left.
\begin{array}{l}
\hbox{$QR$ and $Q'R'$ meet in $A$,}\\
\hbox{$PR$ and $P'R'$ meet in $B$,}\\
\hbox{$PQ$ and $P'Q'$ meet in $C$}
\end{array}
\right\}
\quad
\hbox{and $A$, $B$, $C$ are collinear}
\label{eq!Des}
\end{equation}
(see Figure~\ref{fig!Des}). The converse also holds: condition
$(\ref{eq!Des})$ implies that $\triangle PQR$ and $\triangle P'Q'R'$ are
in perspective\index{perspective} from some point $O$.
\end{thm}

\begin{pf}
If the two triangles are in perspective\index{perspective} from $O$, the linear
subspaces $\Span{O,P,Q,P',Q'}$ and $\Span{O,P,R,P',R'}$ are planes
that have at least the line $\Span{O,P,P'}$ in common. Hence
\[
\dim\Span{O,P,Q,R,P',Q',R'}=\hbox{$2$ or $3$}
\]
by Theorem~\ref{proj dim inters}. Also, the construction of
$A$, $B$, $C$ in (\ref{eq!Des}) makes sense: the two lines $PQ$ and
$P'Q'$ are coplanar (contained in the plane $\Span{O,P,Q,P',Q'}$), so
meet in a unique point $C$, and similarly for the other pairs of sides.

\begin{figure}[t]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering
\leavevmode\hbox{\epsfysize=9.2 true cm\epsffile{proj_desargues3.eps}}
\caption{The Desargues configuration in $\protect\PP^2$ or $\protect\PP^3$}
\label{fig!Des}
\end{figure}

\paragraph{Step 1} Suppose first that $P,Q,R$ and $P',Q',R'$ span a
3-dimensional space $\PP^3$ so are not in any $\PP^2$, and that they are in
perspective\index{perspective} from $O$.
Set $L=\Span{P,Q,R}\cap\Span{P',Q',R'}$. This is the intersection of
two distinct planes in $\PP^3$, and is therefore a line by
Theorem~\ref{proj dim inters}.\index{dimension!of intersection} But
by construction, $A\in L$ since $A=QR\cap Q'R'$. The same applies to
$B$ and $C$, so that also $B,C\in L$ and the 3 points are collinear.

\paragraph{Step 2} We reduce to the first case. Thus suppose that
$P,Q,R$ and $P',Q',R'$ are in the plane $\Pi=\Span{OPQRP'Q'R'}$. Let
$M\in\PP^3\setminus\Pi$ be any point, and lift $R,R'$ off the plane:
pick $S,S'$ as in Figure~\ref{des lift} in
perspective\index{perspective} from $O$ such that $S$ and $R$ are in
perspective\index{perspective} from $M$, and $S'$ and $R'$ are
likewise in perspective\index{perspective} from $M$. Then $\triangle
PQS$ and $\triangle P'Q'S'$ are as in Step~1. So the 3 points
\[
QS\cap Q'S'=\wave A,\quad PS\cap P'S'=\wave B\quad \hbox{and}\quad PQ\cap P'Q'=C
\]
are collinear in $\PP^3$, so lie on a line $\wave L\subset\PP^3$. But
it is easy to see from the construction that $\wave A$, $\wave B$ lie
above $A, B$ in perspective\index{perspective} from $M$, so $A$, $B$,
$C$ are collinear.

For proofs of the converse see \exref{ex!Des1}--\ref{ex!Des2} and
\ref{dual}. \QED \end{pf}

\begin{figure}[t]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering\input{proj_desarg_lift.pstex_t}
% \leavevmode\hbox{\epsfysize=5 true cm\epsffile{proj_desargues4.eps}}
\caption{Lifting the Desargues configuration to $\PP^3$}
\label{des lift}
\end{figure}

It is interesting to note exactly what is used in the proof of
Desargues' theorem just given. It is pure {\em incidence
geometry}\index{incidence of lines} in $\PP^n$ with $n\ge3$, in the
sense that it uses nothing beyond particular cases of formula
(\ref{dim eq1}) of Theorem~\ref{proj dim inters}: two distinct points
of $\PP^n$ span a line, two concurrent lines span a plane, two
distinct lines in a plane intersect in a point, two distinct planes
of $\PP^3$ intersect in a line, etc. The final part of the proof,
Step~2 assumes also that there exists a point not in the plane $\Pi$
(that is, that we are in $\PP^n$ with $n\ge3$), and that the two
lines $MR$ and $MR'$ each have at least one point in addition to
$M,R$ and $M,R'$.\index{Desargues' theorem|)}

\section{Pappus' theorem}\label{Pappus}\index{Pappus' theorem|(}
\markright{\hfill\thesection\ Pappus' theorem\quad}
\begin{thm}[Pappus' theorem] Let $L$, $L'\subset\PP^2$ be two lines
and
\[
P,Q,R\subset L\qquad\hbox{and}\qquad P',Q',R'\subset L'
\]
two triples of distinct points on $L$ and $L'$ (not equal to $L\cap L'$). Then
the $3$ points
\[
QR'\cap Q'R=A,\quad PR'\cap P'R=B\quad\hbox{and}\quad PQ'\cap P'Q=C
\]
are collinear (see Figure~\ref{pap fig}). Notice that the figure is a
configuration of $9$ lines and $9$ points with $3$ lines through each point and
$3$ points on each line. \end{thm}

\setcounter{figure}{\value{section}}
\addtocounter{figure}{-1}
\begin{figure}[t]
\centering \input{proj_pappus.pstex_t}
\caption{The Pappus configuration}
\label{pap fig}
\end{figure}

\begin{pf} This can also be proved via a lifting to $\PP^3$, but this
requires a bit more information about $\PP^3$ (specifically, quadric
surfaces in $\PP^3$ and properties of lines on them). I sketch the easy
proof in coordinates.

By Theorem~\ref{PGL}, I can choose homogeneous coordinates $(x:y:z)$
such that
\[
P=(1:0:0),\quad Q=(0:1:0),\quad P'=(0:0:1)\quad \hbox{and }Q'=(1:1:1).
\]
Then
\[
L=PQ:\{z=0\}, \ P'Q:\{x=0\}, \ L'=P'Q':\{x=y\} \ \hbox{and }PQ':\{y=z\}.
\]
Therefore
\[
C=P'Q\cap PQ'=(0:1:1).
\]
Now let $R=(1:\be:0)$ and $R'=(1:1:\ga)$. Then easy calculations give
\[
PR':\{z=\ga y\} \qquad
P'R:\{y=\be x\}
\]
so that $B=(1:\be:(\be\ga))$ and
\[
QR':\{z=\ga x\} \qquad
Q'R:\{y-z=\be(x-z)\}
\]
so that $A=(1:(\be+\ga-\be\ga):\ga)$. Finally, $A$, $B$, $C$ are all on the line
\[
\{y-z=\be(1-\ga)x\}. \QED
\]
\end{pf}\index{Pappus' theorem|)}

\section{Principle of duality}
\label{dual}
\index{duality|(}
\markright{\hfill\thesection\ Principle of duality\quad}
Projective duality is based on the idea that the space $(\R^{n+1})^*$ of
linear forms $\al\colon\R^{n+1}\to\R$ is also isomorphic to $\R^{n+1}$.
Namely, if $\bbe_0,\dots,\bbe_{n+1}$ is a basis of $\R^{n+1}$ then the dual
basis is given by the linear form
\[
\bbe_i^*\colon \R^{n+1}\to\R\quad\hbox{defined by}\quad \bbe_i^*(\bbe_j)=\de_{ij}=
\begin{cases}
1&\hbox{if $i=j$;}\\
0&\hbox{if $i\ne j$.}
\end{cases}
\]

Further, there is a natural one-to-one correspondence between subspaces of
$\R^{n+1}$ and its dual: a subspace $V\subset\R^{n+1}$ corresponds to its
annihilator (perpendicular) subspace $V^\perp$, that is, the set of linear
forms $\al\colon\R^{n+1}\to\R$ vanishing on $V$. By elementary linear
algebra, $\dim V+\dim V^\perp=n+1$. Hence we obtain the following
correspondence between elements of the geometry of projective linear
subspaces of $\PP^n=\PP(\R^{n+1})$ and those of $(\PP^n)^*=\PP(\R^{n+1})^*$:
\newcommand{\dualcorr}{\quad\longleftrightarrow\quad}
\begin{align*}
E= \PP(V)=\PP^d\subset\PP^n&\dualcorr E^\perp = \PP^{n-d-1}=\PP(V^\perp)\subset(\PP^n)^* \\
\hbox{point } P=\PP^0\in\PP^n&\dualcorr \hbox{hyperplane }\PP^{n-1}=\Pi\subset(\PP^n)^*\\
\hbox{subspace }E_1\subset E_2&\dualcorr \hbox{supspace }E_1^\perp\supset E_2^\perp \\
\hbox{intersection $E_1\cap E_2$}&\dualcorr \hbox{span
$\Span{E^\perp_1,E^\perp_2}$}\\
\hbox{span $\Span{E_1,E_2}$}&\dualcorr \hbox{intersection
$E^\perp_1\cap E^\perp_2$}
\end{align*}

The case of $\PP^2$ is special and particularly illustrative: hyperplanes in
$\PP^2$ are simply lines $L=\PP^1\subset\PP^2$; points are dual to lines,
and the line through two points is dual to the intersection of two lines.

\begin{prop}[\rm Principle of duality for $\PP^2$] Every theorem
concerning points and lines in $\PP^2$ has a dual theorem, obtained from the
original one via the following substitutions:
\begin{align*}
\hbox{points $P$}&\dualcorr \hbox{lines $L$}\\
\hbox{lines $L$}&\dualcorr \hbox{points $P$}\\
\hbox{line $P_1P_2$ (= the span
$\Span{P_1,P_2}$)}&\dualcorr \hbox{point of intersection $L_1\cap
L_2$}\\
\hbox{intersection $L_1\cap L_2$}&\dualcorr \hbox{line
$P_1P_2$}
\end{align*}
\end{prop}

This means that given a theorem and its proof about points and lines
in $\PP^2$, you get a new theorem and its proof by replacing points by
lines etc., in a completely automatic way. For example, the dual of
Desargues' theorem in $\PP^2$ is its converse (which is why I omitted
the proof in \ref{Desargues}). For the dual of Pappus' theorem, see
\exref{ex!Pap}. \index{duality|)}

\section{Axiomatic projective geometry}
\label{Axiomatic projective geometry}
\index{axiomatic projective geometry|(}
\markright{\hfill\thesection\ Axiomatic projective geometry\quad}
An {\em axiomatic projective plane} $\Pi$ consists of two sets
\[
{\rm Points}(\Pi)\hbox{ and }{\rm Lines}(\Pi)
\]
and a relation
\[
\operatorname{Incidence}(\Pi)\subset\operatorname{Points}
(\Pi)\times\operatorname{Lines}(\Pi),
\]
usually called an {\em incidence relation}. If
$(P,L)\in\operatorname{Incidence}(\Pi)$, we say
that ``point $P$ is on line $L$'' or ``line $L$ passes through point
$P$''; because this is an axiomatic system, we might as well say
with David Hilbert ``beer mug $P$ is on table $L$''.

This data is subject to the following axioms:
\begin{enumerate}

\item Every line has at least $3$ points.

\item Every point has at least $3$ lines through it.

\item Through any two distinct points there is a unique line.

\item Any two distinct lines meet in a unique point.
\end{enumerate}
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}a}
\centering \input{proj_cupsandsaucers.pstex_t}
% \leavevmode\hbox{\epsfysize=3.5 true cm\epsffile{proj_cups_saucers.eps}}
\caption{Axiomatic projective plane}
\label{axiomatic}
\end{figure}
Note that these axioms are obviously dual: you can replace the beer mugs on
the tables throughout, and vice versa, and the axioms continue to hold.

More generally an {\em axiomatic projective space} has a lattice of
projective linear subspaces, the incidence relation $\subset$, intersection
and linear span, and suitable axioms. It is best not to insist a priori that
the dimension of the space or its projective linear subspaces is specified.
The most important case is the infinite dimensional case, which von Neumann
used to give axiomatic foundations to quantum mechanics, when dimensions of
projective linear subspaces can take values in $\R$ or the value $\infty$.

\paragraph{Introducing coordinates in axiomatic projective planes} The real
projective plane $\PP^2=\PP^2_\R$ discussed thus far is certainly not the
only axiomatic projective plane: given any field $k$, you can take
$\PP^2_k=\bigl(k^3\setminus\{0\}\bigr)/{\sim}$ where $(x_0,x_1,x_2)\sim(\la
x_0,\la x_1,\la x_2)$ for $0\ne\la\in k$. It is an easy exercise to show
that axioms 1 to 4 continue to hold in $\PP^2_k$. For example, if $k=\FF_2$
you get an axiomatic projective plane with 7 points and 7 lines.
(See~\exref{ex!no of points}.)

For this purpose, $k$ has to be a {\em division ring}, meaning that $ax=b$
has a solution for every $a,b\in A$ with $a\ne0$, but it is not necessary
that $k$ is commutative: you just have to take care that in the equivalence
relation $(x_0,x_1,x_2)\sim(\la x_0,\la x_1,\la x_2)$ only {\em left}
multiplication by $\la\in k^*$ is allowed, and the linear subspaces of $k^3$
used to define lines are {\em right} $k$-subspaces. Indeed, even the
associative law on $k$ can be weakened, although some kind of associativity
is required in order that $(x_0,x_1,x_2)\sim(\la x_0,\la x_1,\la x_2)$ is an
equivalence relation. For a nontrivial example, do
Ex.~\ref{ch!lie}.\ref{ex!octonions}. In this course, I do not have time for
a detailed discussion of the following result, one of the most beautiful
contributions of geometry to pure algebra; for details, consult
Hartshorne~\cite{Ha}.

\begin{thm}[\rm Hilbert's construction]
An axiomatic projective plane $\Pi$ gives rise to a division ring $A$
such that $\Pi=\PP^2_A$. Moreover,
\begin{align*}
\hbox{$A$ is an associative ring}& \iff\hbox{Desargues' theorem holds in $\Pi$};\\
\hbox{$A$ is a commutative ring}& \iff\hbox{Pappus' theorem holds in $\Pi$}.
\end{align*}
\index{Desargues' theorem} \index{Pappus' theorem}
\end{thm} \unskip

\paragraph{Flavour of proof}
We must make a number of choices in $\Pi$. First pick a line $L_\infty$ to
serve as the line at infinity,\index{hyperplane!at infinity} three points
$P_\infty$, $Q_\infty$ and $R_\infty$ on it, and another two lines $L,L_1$
distinct from $L_\infty$. The elements of the division algebra $A$ are the
\begin{figure}[ht]
\renewcommand{\thefigure}{\arabic{chapter}.\arabic{section}b}
\centering
\leavevmode\hbox{\epsfysize=5.8 true cm\epsffile{proj_addition.eps}}
\caption{Geometric construction of addition}
\label{add}
\end{figure}
points of $L$ except for $\infty=L\cap L_\infty$. Now pick 2 points of
$L\setminus\infty$, and call them~0 and~1. The algebraic operation $+$ is
constructed in terms of parallels (since we have fixed $L_\infty$, two lines
of $\Pi$ are {\em parallel} if their intersection is on $L_\infty$) and
$\times$ in terms of similarity. For example, addition is defined as in
Figure~\ref{add}.
\index{axiomatic projective geometry|)}

\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\label{ex!perspec} Let $x,y,z$ be coordinates in $\R^3$, and
$\Pi:(z=1)$, $\Pi':(y=1)$ two hyperplanes. Write down the
perspectivity\index{perspective} $\fie\colon\Pi\to\Pi'$ from $O=(0,0,0)$ in
terms of coordinates $(x,y)$ on $\Pi$ and $(x,z)$ on $\Pi'$. Find and
describe the points of $\Pi$ where $\fie$ is not defined. Prove that
$\fie$ takes a line $L\subset\Pi$ to a line $L'=\fie(L)\subset\Pi'$
(with a single exception). Consider the pencil of parallel lines
$y=mx+c$ of $\Pi$ (for $m$ fixed and $c$ variable), and determine how
$\fie$ maps.

\item In the notation of the preceding exercise, let
$S:(x^2+y^2=1)\subset\Pi$. Understand the effect of the
perspectivity\index{perspective} $\fie$ on $S$, both geometrically
and in coordinates. Show that a circle and a hyperbola in $\R^2$
correspond to projectively equivalent curves in $\PP^2_\R$. Account
for the 4 asymptotic directions of the hyperbola in terms of $S$.

\item In $\PP^2$, write down the equation of the line joining
$P=(1:1:0)$ and $(\al:0:\be)$; write down the point of intersection
of the 2 lines $x+y+z=0$ and $\al x+\be y=0$.

\item Let $\Pi_i\subset\R^3$ be the 3 planes of
Ex.~\ref{ch!aff}.\ref{ex!3planes}. Construct $\PP^3$ by introducing a
4th coordinate $t$, write down the planes of $\PP^3$ by homogenising
the equations of $\Pi_i$, and calculate again the intersections and
spans.\index{dimension!of intersection}

\item\label{ex6} Prove that 3 lines $L,M,N$ of $\PP^n$ that
intersect in pairs are either concurrent (have a common point)
or coplanar. \begin{hint} Use dimension of intersection.\end{hint}

\item\label{ex7} Suppose that $L,M,N$ are three lines of $\PP^4$ not
all contained in any hyperplane.\index{hyperplane} Prove that there
exists a unique line meeting all three. \begin{hint} consider first
the span $\Span{L,M}=\PP^3$.\end{hint}

\item Write down all the projective linear maps $\fie$ of $\PP^2$
taking
\[
(1:0:0)\mapsto(1:2:3),\,\, (0:1:0)\mapsto(2:1:3),\,\, (0:0:1)\mapsto(3:1:2).
\]
Now write down the unique projective linear map taking the standard frame of
reference \[(1:0:0),\quad(0:1:0),\quad(0:0:1),\quad(1:1:1)\] into
\[
(1:2:3),\quad(2:1:3),\quad(3:1:2),\quad(1:2:2)
\]
respectively. \begin{hint} Reread the proof of Theorem~\ref{PGL}.\end{hint}

\item Consider the affine linear map $\fie_0\colon\aff^2\to\aff^2$
given by
\[
(x,y)\mapsto(3x-2,4y-3).
\]
Prove that $\fie_0$ has a unique fixed point in $\aff^2$. \begin{hint}
You can do this by linear algebra, or by using the contraction mapping
theorem from metric spaces.\end{hint}

Write down the projective linear map
$\fie$ of $\PP^2$ extending $\fie_0$. Find the locus of fixed points
of $\fie$ on $\PP^2$. \begin{hint} either find the fixed points ``by
observation'', or prove that $(x:y:z)$ is a fixed point of a
projective linear map $\bx\mapsto A\bx$ if and only
if $\bx=(x,y,z)$ is an eigenvector of $A$.\end{hint}

\item Repeat the previous question for the map
\[
(x,y)\mapsto(x-y+2,x+y+3).
\]

\item\label{ex!aff lin} Suppose $z=(1-\la)x+\la y$. Write
$y=(1-\la')x+\la'z$; find $\la'$ as a function of $\la$. Similarly,
determine the effect of each permutation of $x,y,z$ on the affine
ratio $\la=(z-x)/(y-x)$. Thus permuting the 3 points $x,y,z$ defines
an action of the symmetric group $S_3$ on the set of values of $\la$.

\item Let $P,Q,R=(1:0),(0:1),(1:1)$ be the standard frame of
reference of $\PP^1$.
\begin{enumerate}

\item Find the projective linear map that takes $P,Q,S$ to $Q,P,S$
(in that order); next $P,Q,S$ to $P,S,Q$. What is the effect of your
map on the affine coordinate of a point $R=(1:\la)\in\PP^1$?

\item Verify that the matrixes $\begin{pmatrix}0&1\\1&0 \end{pmatrix}$ and
$\begin{pmatrix}1&-1\\0&-1 \end{pmatrix}$ generate a group under matrix
multiplication isomorphic to the symmetric group $S_3$.

\item The cross-ratio\index{cross-ratio} of 4 points
$\bp,\bq,\br,\bs$ on a line is defined to be
\[
\{\bp,\bq;\br,\bs\}=\frac{\bp-\br}{\bp-\bs}\cdot\frac{\bq-\bs}{\bq-\br}.
\]
Explain what happens when $\bp,\bq,\br,\bs$ are permuted. Prove that
there are in general 6 values
\[
\la,\ \frac{1}{\la},\ 1-\frac{1}{\la},\ \la-\frac{1}{\la},\
\frac{1}{1-\la},\ \frac{\la}{1-\la}
\]
for the cross-ratio, and the group fixing one value is a 4-group
$V_4$.
\end{enumerate}

\item Desargues' theorem\index{Desargues' theorem} \ref{Desargues}
states that if $\triangle PQR$ and $\triangle P'Q'R'$ are two
triangles in perspective\index{perspective} from a point then the 3
points of intersection (e.g., $C=PQ\cap P'Q'$) of corresponding sides
are collinear. See Figure~\ref{fig!Des}. Give the coordinate proof.
\begin{hint} as in the proof of Theorem~\ref{Pappus}, take 4 of the
points as frame of reference,\index{frame!projective} choose
convenient notation for the 3 remaining points, find the coordinates
of $A,B,C$ and prove they are collinear.\end{hint}

\item\label{ex!Des1} Modify the argument to prove the converse of
Desargues' theorem.

\item\label{ex!Des2} State and prove the dual of Desargues' theorem.
Use the same Figure~\ref{fig!Des}.\index{duality}

\item\label{ex!Pap} State and prove\index{Pappus' theorem} the dual
of Pappus' theorem. \begin{hint} with care you can choose notation
exactly dual to that in \ref{Pappus}, e.g., $p:(x=0)$, $L=p\cap
q=(0:0:1)$, etc.\end{hint}

\item State and prove the dual of the statement of \exref{ex7}.
\begin{hint} \dots given three 2-planes of $\PP^4$ not \dots\end{hint}

\item Do the same for \exref{ex6}.

\item Let $L,L'\subset\PP^2$ be two lines. Prove that a projective
linear map $\fie\colon L\to L'$ can be written as the composite of at
most 2 perspectivities\index{perspective} $L\to M$ and $M\to L'$ from
suitably chosen points of $\PP^2$. \begin{hint} Step~1. If the point
of intersection $L\cap L'=P$ is mapped to itself by $\fie$, show that
$\fie$ is a perspectivity\index{perspective} because you can fix the
centre $O$ to deal with 3 points. Step~2. In general, choose a third
line $M$ and a centre $O$ so that $\fie$ composed with the
perspectivity\index{perspective} $\psi\colon M\to L$ is as in
Step~1.\end{hint}

\item Prove that $\PP^n$ has a decomposition as a disjoint union
of $n+1$ subsets
\[
\PP^n= \{\hbox{pt}\}\sqcup\aff^1 \sqcup\aff^2 \sqcup\cdots\sqcup\aff^n.
\]
\begin{hint} $\PP^n=\aff^n\sqcup\hbox{hyperplane at
$\infty$}$.\end{hint} \index{infinity!hyperplane at}

\item\label{ex!no of points}
If $k$ is a finite field with $q$ elements, find 2 different proofs
of
\[
\#(\PP^n_k)=1+q+q^2+\cdots+q^n.
\]
\begin{hint} the ``topological''\index{topology!of $\PP^n$} proof
uses the decomposition of the preceding exercise. The ``arithmetic''
method just counts using the definition
$\PP^n_k=\bigl(k^{n+1}\setminus0\bigr)/k^*$.\end{hint}
\end{enumerate}
\index{projective!geometry|)}

