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\begin{document}

\title{Projective Geometry}
\author{Richard Earl \\
%EndAName
Based on notes by Andrew Dancer}
\date{Trinity Term 2018}
\maketitle

\noindent \underline{\textbf{LECTURE\ LAYOUT}}\bigskip

\noindent 1-2: Projective Spaces (as $\mathbb{P}(V)$ of a vector space $V$).
Homogeneous Co-ordinates. Linear Subspaces.

\noindent 3-4: Projective Transformations. General Position. Desargues
Theorem. Cross-ratio.

\noindent 5: Dual Spaces. Duality.

\noindent 6-7: Symmetric Bilinear Forms. Conics. Singular conics, singular
points. Projective equivalence of non-singular conics.

\noindent 7-8: Correspondence between $\mathbb{P}^{1}$ and a non-singular
conic. Simple applications to Diophantine Equations.\bigskip

\noindent \underline{\textbf{SUGGESTED READING}}

\noindent 1) N.J. Hitchin, Maths Institute notes on Projective Geometry,
found at

\begin{center}
http://people.maths.ox.ac.uk/hitchin/hitchinnotes/hitchinnotes.html
\end{center}

\noindent 2) M. Reid and B. Szendr\H{o}i, Geometry and topology, Cambridge
University Press, 2005 (Chapter 5).\bigskip

\noindent 3) R. Casse, Projective Geometry, An Introduction, Oxford
University Press (2006)\bigskip

\textbf{Further Reading:}

\noindent 4) A.\ Beutelspachter and U. Rosenbaum, Projective Geometry,
Cambridge (1998)\bigskip

\chapter{Introduction}

We will start by considering a theorem that is not actually part of this
course -- namely \textit{Bezout's Theorem} -- but a clean and general result
which readily motivates the worth of projective geometry. (For those
interested, the theorem is part of the Part B course Algebraic Curves.)

Bezout's Theorem is a first significant result in \textit{algebraic geometry}%
. There as many different types of geometry as there are different
properties of interest to geometers. Differential geometry, for example, is
interested in smooth objects (spaces and maps) to which calculus can be
applied. Riemannian geometry is focused on metric properties. Algebraic
geometry is unsurprisingly interested in objects that can be described using
the language of algebraic, and proved using the theorem's of algebra.

So, for example, curves defined by polynomials such as $x^{2}+xy+y^{2}=1$
would be of interest to an algebraic geometer whereas the curve with
equation $y=e^{x}$ would not be. B\'{e}zout's Theorem addresses a natural
first question: how many times do two curves, defined by polynomials of
degrees $m$ and $n$, intersect?

If we begin with $m=n=1$ then we are talking about two lines. These
typically meet in a point but we recognize that this wouldn't be the case if
the lines are parallel. If $m=1$ and $n=2,$ so that we're considering a line
and a conic, then there can be as many as two intersections. We appreciate
that there may be no intersections -- with $y=0$ and $y=x^{2}+1$ -- but that
can be circumvented by working with complex numbers, and we can see that the
answer might be just one -- with $y=0$ and $y=x^{2}$ -- but we could think
of this as a double contact or repeated root in some sense. But we are still
left with cases like $y=x$ and $(y-x)^{2}=1$ which appear to have no
intersection, or $y=0$ and $y^{2}=x$ which has one "single contact"
intersection. Think about the $m=n=2$ case and you'll find the number of
intersections can be $0,1,2,3,4$.

Perhaps, then, the best we can do is to say that the two curves meet in at
most $mn$ points. Even the use of complex numbers and appreciation of
multiple contacts cannot completely resolve the issue. It turns out, though,
that all we are missing is the notion of \textit{points at infinity.} Once
we properly introduce the notion of parallel lines meeting at a point at
infinity then B\'{e}zout's Theorem states that the two curves have $mn$
intersections, counting multiple contacts, using complex numbers, and
including points at infinity.

So given two parallel lines, we will agree that they meet at some idealized
point at infinity. As lines should only meet once this point at infinity
lies in both directions. Given a third parallel line, it will meet each of
these two lines in a point at infinity, and so in fact at the same point at
infinity. So to each family of parallel lines there is a single point at
infinity. Put another way there is a point at infinity for each gradient $m$%
, that is the lines $y=mx+c$ all meet in the same point at infinity. And we
need to remember to allow $m=\infty $ as a possible gradient, relating to
the family of parallel vertical lines. \ These points at infinity make the 
\textit{line at infinity.}

\begin{equation*}
%TCIMACRO{%
%\FRAME{itbpFU}{2.0185in}{2.0358in}{0in}{\Qcb{Parallel lines with the same point at infinity $P_{\infty }$}}{}{lines.jpg}{%
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%BeginExpansion
\ifcase\msipdfoutput
\FRAME{itbpFU}{2.0185in}{2.0358in}{0in}{\Qcb{Parallel lines with the same
point at infinity $P_{\infty }$}}{}{lines.jpg}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 2.0185in;height 2.0358in;depth
0in;original-width 5.0004in;original-height 5.0419in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename 'lines.jpg';file-properties
"XNPEU";}}%
\else
\parbox[b]{2.0185in}{\begin{center}
\includegraphics[
natheight=5.0419in, natwidth=5.0004in, height=2.0358in, width=2.0185in]
{C:/Users/Richard/Work/Lectures/Projective Geometry/2018/graphics/lines__1.jpg}%
\\
Parallel lines with the same point at infinity $P_{\infty }$%
\end{center}}%
\fi
%EndExpansion
\end{equation*}

Note though that these `points at infinity' aren't special in any way, or
rather we've only made them special by our choice of where to put our affine 
$xy$-axes. The family of parallel lines passing through a point at infinity,
properly judged from infinity, would not look any different from the family
of lines passing through the origin.

If we return to our earlier examples when B\'{e}zout appeared not to hold:

\begin{itemize}
\item $y=0$, $y^{2}=x.$ The parabola and line meet a second time at the
point at infinity at the "end" of the $x$-axis

\item $y=x,$ $(y-x)^{2}=1$. The two lines $y=x\pm 1$ both meet $y=x$ at a
point at infinity in the same way that $y=0$ and $y^{2}=x^{2}$ meet at the
origin.
\end{itemize}

\section{The Projective Plane}

We need, then, a rigorous, formal way of introducing these points at
infinity if we are to prove geometric results involving them. For fixed $m$
the lines $y=mx+c$ all meet at a point at infinity.This point at infinity is
where the points $(x,mx)$ move to as $x\rightarrow \pm \infty .$ So it's the
ratio of $x$ and $y$ that is important here. Somehow we want to include all
the points $(x,y)$ of the standard affine plane $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}$ and a line at infinity including the points $\left( \infty ,m\infty
\right) $ where $m\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\cup \left\{ \infty \right\} $.

We cannot make easy meaning of $(\infty ,m\infty )$ but if we recognize this 
$\infty $ as the consequence of some erroneous division by zero, then we can
describe our "extended" plane using \textbf{homogeneous co-ordinates, }first
introduced by Mobius in 1827.

\begin{definition}
Given real $x_{0},x_{1},x_{2},$ not all zero, then we write $[x_{0}\colon
x_{1}\colon x_{2}]$ for the equivalence class of $(x_{0},x_{1},x_{2})\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}\backslash \{\mathbf{0}\}$ under the equivalence relation%
\begin{equation*}
(x_{0},x_{1},x_{2})\sim (\lambda x_{0},\lambda x_{1},\lambda x_{2})\text{%
\qquad where }\lambda \neq 0.
\end{equation*}
\end{definition}

How does this help us with the previous discussion? Well if $x_{0}\neq 0$
then we may divide by $x_{0}$ (i.e. set $\lambda =1/x_{0}$) to see that such
equivalence classes can be represented as $[1\colon x\colon y]$ where $%
x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}.$ These are "most" of the equivalence
classes and $[1\colon x\colon y]$ can be identified with the point $\left(
x,y\right) \in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}.$ And the remaining equivalence classes, when $x_{0}=0$ are $[0\colon
1\colon m]$ when $x_{1}\neq 0$ which corresponds to the point at infinity $%
(\infty ,m\infty ),$ and finally $[0\colon 0\colon 1]$ which corresponds to "%
$m=\infty $" the point at infinity of the vertical lines.

If we return to the earlier "problematic" examples we see now that

\begin{itemize}
\item $y=0$, $y^{2}=x.$ The parabola and line meet a second time at $%
[0\colon 1\colon 0]$.

\item $y=x,$ $(y-x)^{2}=1$. The two lines $y=x\pm 1$ both meet $y=x$ at $%
[0\colon 1\colon 1].$
\end{itemize}

Whilst here, and remembering that $x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}$, we
can see that the affine lines $y=mx+c$ would become%
\begin{equation*}
x_{2}=mx_{1}+cx_{0}
\end{equation*}%
and that each passes through the point at infinity $[0\colon 1\colon m].$
Further the parabola $y^{2}=x$ would become $x_{2}^{2}=x_{0}x_{1}$ and we
see this does indeed pass through the point $[0\colon 1\colon 0].$

The variables $x_{1}/x_{0}$ and $x_{2}/x_{0}$ are known as \textbf{%
inhomogeneous co-ordinates.}\bigskip

\chapter{Projective Spaces\label{Projective spaces chapter}}

In this chapter we define the basic objects of study in this course,
projective spaces. Looking to generalize the earlier discussion we consider
projective spaces over any field ${\mathbb{F}}$, not just $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ or $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.$ Further we will consider finite-dimensional spaces over ${\mathbb{F}}$ so
that we can also better appreciate the effect on a projective space of a
change of basis.

So throughout $V$ will denote a finite-dimensional vector space over a field 
${\mathbb{F}}$. We shall denote by ${\mathbb{F}}^{\ast }$ the multiplicative
group of nonzero elements of ${\mathbb{F}}$.

\begin{definition}
\label{defproj}The \textbf{projective space} ${\mathbb{P}}(V)$ consists of
the equivalence classes $[v],$ where $v\in V,v\neq 0,$ under the equivalence
relation $v\sim \lambda v$ for all $\lambda \in {\mathbb{F}}^{\ast }.$

Note that ${\mathbb{P}}(V)$ can equally be identified as the set of $1$%
-dimensional subspaces of $V$.

Or, in the language of group actions, we might write%
\begin{equation*}
{\mathbb{P}}(V)=\frac{V\backslash \{0\}}{{\mathbb{F}}^{\ast }}
\end{equation*}%
where ${\mathbb{F}}^{\ast }$ acts by scalar multiplication.
\end{definition}

\begin{remark}
Note that, currently, a projective space is simply a set. It has no
particular algebraic structure. We will in due course discuss the topology
of some projective spaces.
\end{remark}

\begin{definition}
We define the \textbf{dimension} of ${\mathbb{P}}(V)$ as $\dim V-1$. This
reflects our intuition that factoring out the ${\mathbb{F}}^{\ast }$ action
has lowered the dimension by one.

We will use the notation ${\mathbb{P}}^{d}$ to denote a projective space of
dimension $d$ and will write ${\mathbb{FP}}^{d}$ if we wish to highlight the
field.

If $\dim {\mathbb{P}}(V)=1$ then ${\mathbb{P}}(V)$ is called a \textbf{%
projective line}, and if $\dim {\mathbb{P}}(V)=2$ then ${\mathbb{P}}(V)$ is
called a \textbf{projective plane}.
\end{definition}

\begin{example}
(a) The real projective line $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\mathbb{P}^{1}$is%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}-\{(0,0)\}}{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{\ast }}.
\end{equation*}%
As a quotient topological space this is the circle. It is straightforwardly
homeomorphic to $S^{1}/\left\{ \pm 1\right\} $ which in turn is homeomorphic
to%
\begin{equation*}
\frac{\text{upper semi-circle}}{-1\sim 1}\cong S^{1}.
\end{equation*}%
(b) The complex projective line $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\mathbb{P}^{1}$is%
\begin{equation*}
\frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{2}-\{(0,0)\}}{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{\ast }}.
\end{equation*}%
As a quotient topological space this is the sphere, the Riemann sphere most
naturally, as we can identify $z\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $[1\colon z]$ and $\infty $ as $\left[ 0\colon 1\right] .$

(c) The real projective plane $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\mathbb{P}^{2}$ was the subject of the discussion in the previous chapter.
\end{example}

\begin{definition}
If $U$ is a subspace of $V$, then ${\mathbb{P}}(U)$ is a subset of ${\mathbb{%
P}}(V)$ called a \textbf{linear subspace}. In particular, if $\dim U=2$, we
obtain a projective line (usually just referred to as a line) in ${\mathbb{P}%
}(V)$. If $\dim U=\dim V-1$, then we call ${\mathbb{P}}(U)$ a \textbf{%
hyperplane.}
\end{definition}

\begin{proposition}
\label{linejoin} Through any 2 distinct points in ${\mathbb{P}}(V)$, there
is a unique projective line.
\end{proposition}

\begin{proof}
Let $[u]\neq \lbrack v]$ in ${\mathbb{P}}(V)$, so $u,v$ are linearly
independent. The unique line containing $[u],[v]$ is now ${\mathbb{P}}%
(\langle u,v\rangle ).$ Certainly this is a projective line and if ${\mathbb{%
P}}(U)$ were another such line with $\dim U=2$ then we'd have $u,v\in U$ and
so $U=\langle u,v\rangle .$\bigskip 
\end{proof}

We can immediately see that intersection properties are nicer in projective
space than in a vector space.

\begin{proposition}
\label{linesmeet} In the projective plane, any 2 distinct projective lines
meet in a unique point.
\end{proposition}

\begin{proof}
We can write the projective plane as ${\mathbb{P}}(V)$ for a 3-dimensional
vector space $V$ and the projective lines as ${\mathbb{P}}(U_1), {\mathbb{P}}%
(U_2)$ for two distinct 2-dimensional subspaces $U_1$ and $U_2$ of $V$.

Now recall the formula 
\begin{equation*}
\dim (U_{1}+U_{2})+\dim (U_{1}\cap U_{2})=\dim (U_{1})+\dim (U_{2}).
\end{equation*}%
As $U_{1},U_{2}$ are distinct 2-dimensional subspaces, the sum $U_{1}+U_{2}$
strictly contains $U_{1}$ and hence is of dimension greater than $2$, so is
the full 3-dimensional space $V$. Hence the formula shows $U_{1}\cap U_{2}$
is $1$-dimensional, and this represents the unique point in projective space
where ${\mathbb{P}}(U_{1})$ meets ${\mathbb{P}}(U_{2})$.
\end{proof}

\section{Homogeneous Co-ordinates.}

Say that $\dim V=n+1$ and $v_{0},\ldots ,v_{n}$ is a basis for $V$. Then any 
$v\in V$ can be uniquely written%
\begin{equation*}
v=x_{0}v_{0}+x_{1}v_{1}+\cdots +x_{n}v_{n}
\end{equation*}%
and so $[v]$ can be given the \textbf{homogeneous co-ordinates} 
\begin{equation*}
\lbrack v]=[x_{0}\colon x_{1}\colon \cdots \colon x_{n}]
\end{equation*}%
which are unique up to multiplication by a non-zero scalar. Recall that not
all of $x_{0},\ldots ,x_{n}$ can be zero.

\begin{example}
Let ${\mathbb{P}}(V)$ be a projective line and $P_{0},P_{1},P_{2}$ be three
distinct points in ${\mathbb{P}}(V)$. Then there is a basis for $V$ with
respect to which%
\begin{equation*}
P_{0}=[1\colon 0],\qquad P_{1}=[1\colon 1],\qquad P_{2}=[0\colon 1].
\end{equation*}%
Note that the inhomogeneous co-ordinates of $P_{0},P_{1},P_{2}$ are $%
0,1,\infty $ respectively.
\end{example}

\begin{proof}
Say that $P_{0}=[v_{0}]$ and $P_{1}=[v_{1}].$ As $P_{0}\neq P_{1}$ then $%
v_{0},v_{1}$ are linearly independent and hence a basis of $V.$ So, if $%
P_{2}=[v_{2}]$ we may write%
\begin{equation*}
v_{2}=\alpha _{0}v_{0}+\alpha _{1}v_{1}
\end{equation*}%
and as $P_{0},P_{1},P_{2}$ are distinct then neither $\alpha _{0}$ nor $%
\alpha _{1}$ are zero. So with respect to the basis $\left\{ \alpha
_{0}v_{0},\alpha _{1}v_{1}\right\} $ for $V$ it's the case that $%
P_{0},P_{1},P_{2}$ have the homogeneous co-ordinates above.
\end{proof}

\begin{remark}
This result may seem familiar from the study of M\"{o}bius transformations,
where we saw three distinct points of the extended complex plane can be
mapped to $0,1,\infty $ by a unique M\"{o}bius transformation. We shall see
in due course that the M\"{o}bius transformations are the projective
transformations of $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\mathbb{P}^{1}.$
\end{remark}

\begin{example}
Consider the line in affine space ${\mathbb{R}}^{2}$ with equation $y=2x$.
We can complete this to a projective line in ${\mathbb{R}}{\mathbb{P}}^{2}$
by embedding ${\mathbb{R}}^{2}$ in ${\mathbb{R}}{\mathbb{P}}^{2}$ via $%
(x,y)\mapsto \lbrack 1,x,y]$. Recall that $x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}$
so that in terms of homogeneous coordinates $[x_{0},x_{1},x_{2}]$, the
projective line has equation $x_{2}-2x_{1}=0$.
\end{example}

\begin{example}
Write down the equation of the line connecting the points $[1\colon 2\colon
0]$ and $\left[ 0\colon 1\colon 1\right] $ in $\mathbb{P}(\mathbb{F}^{3}).$
What is its equation in inhomogeneous co-ordinates? What is its point at
infinity?
\end{example}

\begin{solution}
Let $x_{0},x_{1},x_{2}$ be the homogeneous coordinates in $\mathbb{F}^{3}.$
A (projective) line has the form $\mathbb{P}(U)$ where $U$ isq a $2$%
-dimensional subspace of $\mathbb{F}^{3}$ and such subspaces have an
equation of the form%
\begin{equation*}
a_{0}x_{0}+a_{1}x_{1}+a_{2}x_{2}=0.
\end{equation*}%
So we have $a_{0}+2a_{1}=0$ and $a_{1}+a_{2}=0$ so that the (projective)
line has equation%
\begin{equation*}
-2a_{1}x_{0}+a_{1}x_{1}-a_{1}x_{2}=0
\end{equation*}%
or, as such equations are unique only up to multiplication by a non-zero
scalar, then we can write this as%
\begin{equation*}
2xx_{0}-x_{1}+x_{2}=0.
\end{equation*}%
In terms of inhomogeneous co-ordinates $x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}$
this is the line $2-x+y=0$ and its point at infinity (when $x_{0}=0$) is $%
[0\colon 1\colon 1].$
\end{solution}

\begin{example}
What projective conic does the hyperbola $xy=1$ correspond to in ${\mathbb{P}%
}^{2}$? What are its points at infinity? What are the points at infinity of
the parabola $y^{2}=x$?
\end{example}

\begin{solution}
Recalling that $x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}$ then the projectivized
version of the hyperbola has equation $x_{1}x_{2}=x_{0}^{2}.$ Note that its
two points at infinity are $[0\colon 1\colon 0]$ and $[0\colon 0\colon 1]$,
respectively the points at infinity of the $x$-axis and $y$-axis. The
parbola $y^{2}=x$ has projectivized equation $x_{2}^{2}=x_{1}x_{0}$ which
has point at infinity $[0\colon 1\colon 0].$
\end{solution}

Generally, we may decompose $n$-dimensional projective space ${\mathbb{F}}{%
\mathbb{P}}^{n}={\mathbb{P(F}}^{n+1})$ as the union of 2 sets 
\begin{equation*}
S_{\infty }=\{[x_{0},\ldots ,x_{n}]:x_{0}=0:x_{i}\;\mathrm{not\;all}\;0\}
\end{equation*}%
\begin{equation*}
S_{aff}=\{[x_{0},\ldots ,x_{n}]:x_{0}\neq 0\}
\end{equation*}%
Clearly $S_{\infty }$ may be identified with projective space ${\mathbb{F}}{%
\mathbb{P}}^{n-1}$ of dimension one lower. In $S_{\mathrm{aff}}$, every
point may be written as $[1,t_{1},\ldots ,t_{n}]$ where $t_{i}=x_{i}/x_{0}$,
and this sets up an identification of $S_{\mathrm{aff}}$ with ${\mathbb{F}}%
^{n}$. So we have a disjoint union 
\begin{equation}
{\mathbb{F}}{\mathbb{P}}^{n}={\mathbb{F}}^{n}\cup {\mathbb{F}}{\mathbb{P}}%
^{n-1}.  \label{decomp}
\end{equation}%
Intuitively, we are adding some points at infinity -- in fact a \textbf{%
hyperplane at infinity} ${\mathbb{F}}{\mathbb{P}}^{n-1}$-- to the \textbf{%
affine space} ${\mathbb{F}}^{n}$ to obtain the projective space ${\mathbb{F}}%
{\mathbb{P}}^{n}$. As we mentioned in the introduction, this ensures that
projective space has nicer properties than affine space, especially as
regards intersection theory. For example, going back to Proposition \ref%
{linesmeet}, we can see that parallel lines in affine space ${\mathbb{F}}%
^{2} $ generate projective lines in ${\mathbb{F}}{\mathbb{P}}^{2}$ that meet
in the ${\mathbb{F}}{\mathbb{P}}^{1}$ at infinity.

It is important to realise that the decomposition (\ref{decomp}) is not
canonical. There is nothing special or different about the points at
infinity. They are deemed such only by our choice of where to place an
affine space ${\mathbb{F}}^{n}$ within the projective space ${\mathbb{F}}{%
\mathbb{P}}^{n}$. We could , for example, choose any other coordinate $x_{i}$
and decompose projective space according to whether $x_{i}$ is zero or
nonzero.

In fact, it is often useful to consider the subsets ${\mathcal{U}}_{i}$ of ${%
\mathbb{F}}{\mathbb{P}}^{n}$ given by 
\begin{equation*}
{\mathcal{U}}_{i}=\{[x_{0},\ldots ,x_{n}]:x_{i}\neq 0\}
\end{equation*}%
The ${\mathcal{U}}_{i}$ cover ${\mathbb{F}}{\mathbb{P}}^{n}$, as every point
in ${\mathbb{F}}{\mathbb{P}}^{n}$ has \textit{some} coordinate $x_{i}$
nonzero. As above, each ${\mathcal{U}}_{i}$ may be identified with ${\mathbb{%
F}}^{n}$. So we have covered projective space by open sets each with an
identification with affine space. If ${\mathbb{F}}={\mathbb{R}}$ or ${%
\mathbb{C}}$ this endows ${\mathbb{R}}{\mathbb{P}}^{n}$ and ${\mathbb{C}}{%
\mathbb{P}}^{n}$ with the structure of an $n$-dimensional \textbf{manifold}
and $n$-dimensional \textbf{complex manifold} respectively. (More on this in
the Part A course Introduction to Manifolds.)

\begin{remark}
(Off-syllabus) If we take the field ${\mathbb{F}}$ to be ${\mathbb{R}}$ or ${%
\mathbb{C}}$ then in fact we can put a topology on projective space, related
to the Euclidean topology on ${\mathbb{R}}^{n}$ or ${\mathbb{C}}^{n}$.

For ${\mathbb{R}}^{n+1}$, this proceeds by observing that Definition \ref%
{defproj} is equivalent to saying that 
\begin{equation*}
{\mathbb{R}}{\mathbb{P}}^{n}=\frac{\{v\in S^{n}\subset {\mathbb{R}}^{n+1}\}}{%
v\sim -v}
\end{equation*}%
(where $S^{n}$ is the unit sphere in ${\mathbb{R}}^{n+1}$), because every
nonzero vector $v=(x_{1},\ldots ,x_{n})\in {\mathbb{R}}^{n+1}$ may be scaled
by ${\mathbb{R}}^{\ast }$ to an element of norm one, which is unique up to
replacing $v$ by $-v$. So we have exhibited real projective space as the
quotient of the sphere by a ${\mathbb{Z}}_{2}$ action. We can thus endow
real projective space with the quotient topology, which is compact (as the
sphere is a compact subset of Euclidean space) and Hausdorff (as it is the
quotient of a Hausdorff space by the action of a finite (and hence compact)
group). Similar ideas may be used to topologize complex projective space.

Real and complex projective spaces may thus be viewed as compactifications
of the corresponding affine spaces. In particular, the projective lines over
these fields are the one-point compactifications of ${\mathbb{R}}$ and ${%
\mathbb{C}}$ respectively. In the complex case, as we alluded to above, we
may view the projective line as the Riemann sphere and in the real case we
obtain the circle.

For general fields ${\mathbb{F}}$, we do not have an analogue of the
Euclidean topology on ${\mathbb{F}}^{n}$, so these ideas are not applicable.
In algebraic geometry there is a standard topology for projective spaces
over general fields, the \textbf{Zariski topology}, but it has very
different properties-- in particular it has fewer open sets and is not
Hausdorff. (This is covered more in the Part C Algebraic\ Geometry course.)
\end{remark}

More generally we can see that subsets of projective space can be defined by
the zero sets of homogeneous polynomials. We say a polynomial $P(x)$ is 
\textbf{homogeneous} of degree $j$ if there exists a positive integer $j$
such that $P(\lambda x)=\lambda ^{j}P(x)$ for all $x$. (Equivalently, all
the terms in $P(x)$ are of total degree $j$). For example, $%
x_{0}x_{1}^{2}+x_{0}x_{1}x_{2}+x_{2}^{3}$ is homogeneous of degree 3, but $%
x_{0}+x_{1}x_{2}$ is not homogeneous.

Homogeneity is the condition that ensures that the equation $P(x)=0$ is
well-defined on projective space. A \textbf{projective algebraic variety} is
a subset of projective space defined by a system of homogeneous polynomial
equations. If the equations are all of degree 1 we define the linear
subspaces. We shall later investigate the case of \textit{quadrics}, which
are defined by a single homogeneous quadratic polynomial.

\section{Axiomatization of Projective Planes (Off-syllabus)}

Alternatively an axiomatic approach can be taken to introducing projective
planes. In this approach, a projective plane consists of collections $%
\mathcal{P}$ of points and $\mathcal{L}$ of lines satisfying:

\begin{itemize}
\item given two distinct points, there is a unique line containing them.

\item any two lines have at least one point in common.

\item any line contains at least three points.

\item there are at least two lines.
\end{itemize}

The above actually provides a somewhat more general notion of what it is to
be a projective plane. The smallest projective plane is the order 2 \textit{%
Fano plane} (named after the Italian mathematician Gino Fano (1871-1952))
which is often represented as below:%
\begin{equation*}
%TCIMACRO{%
%\FRAME{itbpFU}{1.8628in}{1.6328in}{0in}{\Qcb{Fano Plane}}{}{fano.png}{%
%\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.8628in;height 1.6328in;depth 0in;original-width 7.1044in;original-height 6.2189in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'fano.png';file-properties "XNPEU";}}}%
%BeginExpansion
\ifcase\msipdfoutput
\FRAME{itbpFU}{1.8628in}{1.6328in}{0in}{\Qcb{Fano Plane}}{}{fano.png}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "USEDEF";valid_file "F";width 1.8628in;height 1.6328in;depth
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"XNPEU";}}%
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\parbox[b]{1.8628in}{\begin{center}
\includegraphics[
natheight=6.2189in, natwidth=7.1044in, height=1.6328in, width=1.8628in]
{C:/Users/Richard/Work/Lectures/Projective Geometry/2018/graphics/fano__2.png}%
\\
Fano Plane
\end{center}}%
\fi
%EndExpansion
\end{equation*}%
where the dots are points and the line segments/circle are lines. This is
the same as $\mathbb{F}_{2}\mathbb{P}^{2}.$ The \textit{order} of a
projective plane is one less than the number of points in any projective
line.

However there are four non-isomorphic projective planes of order $9,$ the
usual $\mathbb{F}_{9}\mathbb{P}^{2}$ and three further planes in which
Desargues' Theorem does not hold (see later). The only \textit{known} finite
projective planes have an order which is the power of a prime. There is no
projective plane of order 10, but this is only known to be true using
lengthy computer elimination and it is still an open problem as to whether
there is a projective plane of order 12.

\chapter{Projective transformations\label{Proj Trans Chapter}}

Whenever we introduce a class of mathematical objects, we are always
interested in the transformations between them. For example with groups we
consider homomorphisms, with rings we consider ring homomorphisms, with
topological spaces we consider continuous maps, and with vector spaces
linear maps.

We have defined projective spaces in terms of quotients of vector spaces. It
is therefore natural to consider maps of projective spaces induced by linear
maps of vector spaces. The obvious definition is 
\begin{equation*}
\tau :[v]\mapsto \lbrack Tv]
\end{equation*}%
where $[v]$ is the point of projective space represented by $v\in
V\backslash \{0\}$.

There are two potential problems we must consider first. One, as always with
defining maps on quotient spaces, is to check that the map is well-defined.
That is, we must check that if $[v]=[w]$ then $[Tv]=[Tw]$. In our situation
this is clear from the linearity of $T$, and the fact that $[v]=[w]$ if and
only if $v$ is a nonzero scalar multiple of $w$. The second problem is that
only nonzero vectors represent points of projective space, so we need $Tv$
to be nonzero whenever $v$ is, that is, we need $T$ to be injective.

\begin{definition}
If $T:V\rightarrow W$ is an injective linear transformation, we define the
associated \textbf{projective linear transformation} by%
\begin{equation*}
\tau :v\mapsto \lbrack Tv]
\end{equation*}%
We will generally be interested in the case when $V=W$ so that $T$ is
invertible.
\end{definition}

Note that any nonzero scalar multiple of $T$ represents the same projective
transformation as does $T$. In fact the assignment $T\mapsto \tau $ defines
a homomorphism from $GL(V)$, the group of invertible linear transformations
of $V$, onto the group of projective linear transformations of ${\mathbb{P}}%
(V)$. The kernel of this map is the (normal) subgroup of scalar invertible
linear transformations, that is, nonzero scalar multiples of the identity.
Therefore, using the first isomorphism theorem for groups, we can make the
definition.

\begin{definition}
The group of projective linear transformations of ${\mathbb{P}}(V)$ is 
\begin{equation*}
PGL(V)=\frac{GL(V)}{\{\lambda I:\lambda \in {\mathbb{F}}^{\ast }\}}.
\end{equation*}
\end{definition}

More concretely, if we identify $V$ with ${\mathbb{F}}^{n+1}$ by choosing a
basis, then we write the group $PGL(V)$ as $PGL(n+1,{\mathbb{F}})$, the
quotient of the group of $(n+1)\times (n+1)$ invertible matrices over ${%
\mathbb{F}}$ by the subgroup of nonzero scalar matrices.

Of course we can write projective transformations in terms of homogeneous
coordinates. We illustrate this in the case of the projective line.

\begin{example}
Consider an invertible linear map $T:{\mathbb{F}}^{2}\rightarrow {\mathbb{F}}%
^{2}$ given by $T:(x,y)^{T}\mapsto (ax+by,cx+dy)^{T}$ with $ad-bc\neq 0$.

Working on an affine patch $y\neq 0$, we can rewrite the associated
projective linear transformation of ${\mathbb{F}}{\mathbb{P}}^{1}$ as 
\begin{equation*}
\left[ \frac{x}{y},1\right] \mapsto \left[ \frac{ax+by}{cx+dy},1\right] ,
\end{equation*}%
so in terms of the inhomogeneous coordinate $t=\frac{x}{y}$ as 
\begin{equation*}
t\mapsto \frac{at+b}{ct+d}.
\end{equation*}%
In the case ${\mathbb{F}}={\mathbb{C}}$, we have encountered these
transformations before: they are the \textbf{M\"{o}bius transformations} of
the Riemann sphere ${\mathbb{C}}{\mathbb{P}}^{1}$. (The point at infinity $%
\infty $ in the Riemann sphere is here identified with $[1,0]$).
\end{example}

\begin{example}
Find the order of the group $PGL(2,{\mathbb{F}})$ where ${\mathbb{F}}$ is a
finite field with $q$ elements.
\end{example}

\begin{solution}
A representative of an element of $PGL(2,{\mathbb{F}})$ is an invertible
matrix and so its columns form a basis for ${\mathbb{F}}^{2}.$ The first
vector can therefore be any vector in ${\mathbb{F}}^{2}\backslash \{0\}$,
and there are $q^{2}-1$ such vectors. Now the second vector need to be
independent of the first column. As ${\mathbb{F}}^{\ast }={\mathbb{F}}%
\backslash \{0\}$ then there are $q-1$ vectors that are dependent on the
first vector and so $q^{2}-q$ independent vectors. The number of invertible
matrices over ${\mathbb{F}}$ is then $(q^{2}-1)(q^{2}-q).$ However each
transformation in $PGL(2,{\mathbb{F}})$ is represented by $q-1$ invertible
matrices (which are scalar multiples of one another). Finally we have%
\begin{equation*}
\left\vert PGL(2,{\mathbb{F}})\right\vert =\frac{(q^{2}-1)(q^{2}-q)}{(q-1)}%
=(q-1)q(q+1).
\end{equation*}
\end{solution}

We may recall from complex analysis the result that given an ordered triple
of distinct points in the Riemann sphere, there is a unique M\"obius
transformation sending the triple to $(0,1, \infty)$. Hence the group $%
PGL(2, {\mathbb{C}})$ of M\"obius transformations acts transitively on the
set of ordered triples of distinct points in the projective line.

What does the condition that the points are distinct mean in terms of
projective geometry? Well, two points in projective space are equal if and
only if their representative vectors are proportional, which for two vectors
is equivalent to saying they are dependent. This motivates the following
definition.

\begin{definition}
In an $n$-dimensional projective space ${\mathbb{F}}{\mathbb{P}}^{n}$, we
say that $n+2$ points are \emph{in }\textbf{general position} if each subset
of $n+1$ of these points has linearly independent representative vectors
\end{definition}

So in the $n=1$ case, this just says we have a triple of distinct points.
And four points are in general position in ${\mathbb{P}}^{2}$ if no three
are collinear.

\medskip We now prove the following theorem, which generalises the above
result about M\"{o}bius transformations.

\begin{theorem}
(\textbf{General Position Theorem}) Let $X_{0},\ldots ,X_{n+1}$ and $%
Y_{0},\ldots ,Y_{n+1}$ be two $(n+2)$-tuples of points in $n$-dimensional
projective space, such that each $(n+2)$-tuple is in general position. Then
there exists a unique projective linear transformation $\tau $ such that $%
\tau (X_{i})=Y_{i}$ for each $i$.
\end{theorem}

\begin{proof}
Let $X_{i}=[v_{i}]$ for $i=0,\ldots ,n+1$, that is, $v_{i}$ are
representative vectors for $X_{i}$. The general position hypothesis implies
that $v_{0},\ldots ,v_{n}$ form a basis for the vector space. For the last
point $X_{n+1}$, we have 
\begin{equation*}
v_{n+1}=\sum_{i=0}^{n}\lambda _{i}v_{i}
\end{equation*}%
for some scalars $\lambda _{i}$.

Now, all $\lambda _{i}$ are nonzero, again using the general position
hypothesis (for if one is zero then we get a dependency relation between $%
v_{n+1}$ and $n$ of the other $v_{i}$). So we may in fact replace $v_{i}$ by 
$\lambda _{i}v_{i}$ and take 
\begin{equation*}
v_{n+1}=\sum_{i=0}^{n}v_{i}
\end{equation*}%
Again using general position, this representation of $v_{n+1}$ is unique.

Similarly we can take $Y_i = [w_i]$ for $i=0, \ldots, n+1$, with $w_{n+1} =
\sum_{i=0}^{n} w_i$, where $w_0, \ldots, w_n$ is a basis.

Now there exists a linear transformation $T$ with $T: v_i \mapsto w_i$ for $%
i=0, \ldots, n$. Linearity and the formulae for $v_{n+1},w_{n+1}$ also imply 
$T : v_{n+1} \mapsto w_{n+1}$, as required.

If $S$ is another linear transformation inducing a projective transformation
with the required property, then $Sv_{i}=\mu _{i}w_{i}$ for $i=0,\ldots ,n+1$%
, where $\mu _{i}$ are nonzero scalars. Now 
\begin{equation*}
\mu _{n+1}w_{n+1}=Sv_{n+1}=\sum_{i=0}^{n}Sv_{i}=\sum_{i=0}^{n}\mu _{i}w_{i},
\end{equation*}%
so $w_{n+1}=\sum_{i=0}^{n}(\mu _{i}/\mu _{n+1})w_{i}$ and by uniqueness of
this representation we see all the $\mu _{i}$ are equal. Hence $S=\mu T$ and
they induce the same projective map.
\end{proof}

\begin{remark}
The general position theorem might also be appreciated as follows: given $%
n+2 $ points $X_{0},\ldots ,X_{n+1}$ in general position in an $n$%
-dimensional projective space ${\mathbb{P}}(V)$ then there is a basis for $V$
with respect to which the $n+2$ points have co-ordinates%
\begin{equation*}
\left[ 1\colon 0\colon \cdots \colon 0\right] ,\quad \left[ 0\colon 1\colon
\cdots \colon 0\right] ,\quad \ldots \quad \left[ 0\colon 0\colon \cdots
\colon 1\right] ,\quad \left[ 1\colon 1\colon \cdots \colon 1\right] .
\end{equation*}%
This follows as, given any basis, the points with the above co-ordinates are
in general position, and so the projective map found in the previous theorem
might instead be viewed as a change of basis map.
\end{remark}

As an application of the General Position Theorem, we prove a celebrated
classical result of projective geometry, Desargues's Theorem. (Girard
Desargues 1591-1661 was a French mathematician.)

\begin{theorem}
(\textbf{Desargues}) Let $O,A,A^{\prime },B,B^{\prime },C,C^{\prime }$ be
seven distinct points in a projective space such that the lines $AA^{\prime
},BB^{\prime }$ and $CC^{\prime }$ are distinct and concurrent at $P$. Then
the points of intersection 
\begin{equation*}
P=BC\cap B^{\prime }C^{\prime },\qquad Q=CA\cap C^{\prime }A^{\prime
},\qquad R=AB\cap A^{\prime }B^{\prime },
\end{equation*}
are collinear.%
\begin{equation*}
%TCIMACRO{%
%\FRAME{itbpF}{5.1941in}{4.0776in}{0in}{}{}{desarguestheorem.png}{%
%\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 5.1941in;height 4.0776in;depth 0in;original-width 5.1387in;original-height 4.0274in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'DesarguesTheorem.png';file-properties "XNPEU";}}}%
%BeginExpansion
%EndExpansion
\end{equation*}
\end{theorem}

\begin{proof}
Fix a representative vector $x$ for $O.$ As $O,A,A^{\prime }$ are distinct,
then we can choose representative vectors $a,a^{\prime }$ such that%
\begin{equation*}
x=a+a^{\prime }
\end{equation*}%
and likewise we can choose representative vectors $b,b^{\prime },c,c^{\prime
}$ such that $x=b+b^{\prime }$ and $x=c+c^{\prime }.$ These equations imply $%
a-b=b^{\prime }-a^{\prime }=r$. As $a-b\in \langle a,b\rangle $ is a
representative vector for a point on the line $AB$ and $a^{\prime
}-b^{\prime }$ similarly represents a point on the line $A^{\prime
}B^{\prime }$ then $r$ is a representative vector of $R=AB\cap A^{\prime
}B^{\prime }$. Similarly $b-c$ and $c-a$ are representative vectors for $%
P=BC\cap B^{\prime }C^{\prime }$ and $Q=CA\cap C^{\prime }A^{\prime }$
respectively. But 
\begin{equation*}
p+q+r=(b-c)+(c-a)+(a-b)=0
\end{equation*}
so that these three representative vectors are linearly dependent, hence the
points they represent are collinear.
\end{proof}

\begin{example}
In the projective plane 4 points are in general position if and only if no 3
are collinear. So we see that any two such quadruples in the plane are
projectively equivalent.
\end{example}

The Theorem of Pappus is another classical result that may be proved using
general position arguments, in this case by assigning the points $%
A,B,C^{\prime },B^{\prime }$ (say) the co-ordinates $[1\colon 0\colon
0],[0\colon 1\colon 0],[0\colon 0\colon 1],[1\colon 1\colon 1],$ and then
explicitly calculating the co-ordinates of the remaining points.

\begin{theorem}
(\textbf{Pappus}) Let $A,B,C$ and $A^{\prime },B^{\prime },C^{\prime }$ be
two collinear triples of distinct points in the projective plane. Then the
three points 
\begin{equation*}
P=AB^{\prime }\cap A^{\prime }B,\qquad Q=CA^{\prime }\cap C^{\prime }A\qquad
R=BC^{\prime }\cap B^{\prime }C,
\end{equation*}
are collinear.%
\begin{equation*}
%TCIMACRO{%
%\FRAME{itbpF}{3.1704in}{1.8922in}{0in}{}{}{pappus.gif}{%
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%BeginExpansion
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\FRAME{itbpF}{3.1704in}{1.8922in}{0in}{}{}{pappus.gif}{%
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\else
\includegraphics[
natheight=1.8542in, natwidth=3.1254in, height=1.8922in, width=3.1704in]
{C:/Users/Richard/Work/Lectures/Projective Geometry/2018/graphics/pappus__3.pdf}%
\fi
%EndExpansion
\end{equation*}
\end{theorem}

\begin{proof}
Exercise 8 on problem sheet 1.
\end{proof}

\begin{remark}
(Off-syllabus) Recall earlier that we gave a list of four axioms for what it
is to be a projective plane. In a similar fashion projective spaces of
higher dimensions than two can be axiomatized. It turns out that any
projective space of space of dimension at least three is Desarguesian in
that Desargues' Theorem necessarily holds in them. Hilbert was the first to
appreciate that (axiomatized) projective planes need not be Desarguesian. It
turns out that Desargues theorem holds in a projective plane $P$ if and only
if $P=\mathbb{P}(V)$ where $V$ is a "vector space" over a division ring -- a
division ring satisfies all the axioms of a field other than the
commutativity of multiplication. In turn a Desarguesian projective plane can
be expressed as a projective space of a vector space over a field if and
only if Pappus' Theoem holds.
\end{remark}

\chapter{Cross-ratio}

Let us return to the case of the projective line. We know that any two
triples of distinct points are equivalent under the action of the projective
linear group. What can we say about quadruples? It turns out that there is a
single numerical invariant which distinguishes orbits of quadruples of
distinct points in the projective line under the projective group.

\begin{definition}
Let $x_{i}=[\xi _{i},\eta _{i}],$ where $i=0,\ldots ,3,$ be four distinct
points in the projective line ${\mathbb{F}}{\mathbb{P}}^{1}$.

The \textbf{cross-ratio} of the ordered quadruple is 
\begin{equation*}
(x_{0}x_{1}:x_{2}x_{3})=\frac{(\xi _{0}\eta _{2}-\xi _{2}\eta _{0})(\xi
_{1}\eta _{3}-\xi _{3}\eta _{1})}{(\xi _{0}\eta _{3}-\xi _{3}\eta _{0})(\xi
_{1}\eta _{2}-\xi _{2}\eta _{1})}
\end{equation*}
\end{definition}

We can observe that if we scale any pair $(\xi _{i},\eta _{i})$ then the
numerator and denominator both scale by the same factor, so the quotient on
the right hand side is unchanged. The cross-ratio is therefore well-defined.
Moreover, under projective transformations 
\begin{equation*}
\left( 
\begin{array}{cc}
\xi _{i} & \xi _{j} \\ 
\eta _{i} & \eta _{j}%
\end{array}%
\right) \mapsto \left( 
\begin{array}{cc}
a & b \\ 
c & d%
\end{array}%
\right) \left( 
\begin{array}{cc}
\xi _{i} & \xi _{j} \\ 
\eta _{i} & \eta _{j}%
\end{array}%
\right)
\end{equation*}%
by the determinant product rule we see that each bracket $\xi _{i}\eta
_{j}-\xi _{j}\eta _{i}$ scales by the (nonzero!) determinant $ad-bc$, and
hence the cross-ratio is invariant under a projective transformation. Thus
we have shown

\begin{proposition}
Cross-ratio is a projective invariant.
\end{proposition}

So any 2 quadruples that are projectively equivalent must have the same
cross-ratio. What about the converse? We would like to show that any two
quadruples with the same cross-ratio are projectively equivalent. Stated
like that, it seems like an involved calculation, but we can greatly
simplify it using the general position theorem. As a result of that theorem
we may projectively map $x_{0},x_{1},x_{2}$ to the points $[1,0],[0,1],[1,1]$
and we now know that doing this will not change the cross-ratio. Then 
\begin{equation*}
(x_{0}x_{1}:x_{2}x_{3})=\frac{(1.1-0.1)(0.\eta _{3}-\xi _{3}.1)}{(1.\eta
_{3}-0.\xi _{3})(0.1-1.1)}=\frac{\xi _{3}}{\eta _{3}}
\end{equation*}%
As the points are distinct, we may write $x_{3}=[\lambda ,1]$ for $\lambda
\neq 0,1$, and now the cross-ratio is $\lambda $. So any quadruple of
distinct points is projectively equivalent to the quadruple $%
[1,0],[0,1],[1,1],[\lambda ,1]$, where $\lambda $ is the cross-ratio. We
have proved the following result.

\begin{theorem}
Two quadruples of distinct points in the projective line are projectively
equivalent if and only if their cross-ratios are equal.
\end{theorem}

We might instead have defined the cross-ratio in terms of inhomogeneous
co-ordinates by setting%
\begin{equation}
(x_{0}x_{1}:x_{2}x_{3})=\frac{(x_{2}-x_{0})(x_{3}-x_{1})}{%
(x_{3}-x_{0})(x_{2}-x_{1})}  \label{moduli}
\end{equation}%
where we are now identifying $x_{i}$ with the inhomogeneous co-ordinate $%
\eta _{i}/\xi _{i}.$ Some care needs to be taken in properly understanding
what happens when one of these inhomogeneous

\begin{remark}
(Off-syllabus) Notice that the cross-ratio in (\ref{moduli}) does not take
the values $0$, $1$ or $\infty $. The cross-ratio thus sets up a bijection
between

\begin{itemize}
\item quadruples of distinct points in ${\mathbb{F}}{\mathbb{P}}^{1}$ modulo
the action of the projective linear group $PGL(2,{\mathbb{F}})$, and

\item the projective line ${\mathbb{F}}{\mathbb{P}}^{1}$ with points $%
0,1,\infty $ removed.
\end{itemize}

Note that as $x_{2}$ coincides with $x_{0}$ or $x_{3}$ coincides with $x_{1}$
then the cross-ratio approaches $0.$ Likewise as $x_{3}$ coincides with $%
x_{0}$ or $x_{2}$ coincides with $x_{1}$ then the cross-ratio approaches $%
\infty .$ Finally as $x_{3}$ coincides with $x_{2}$ or $x_{1}$ coincides
with $x_{0}$ the cross-ratio approaches $1$. Thus, in some sense all of ${%
\mathbb{F}}{\mathbb{P}}^{1}$ can be used to represent orbits of quadruples,
though we have found $0,1,\infty $ each correspond to two orbits and no
point of ${\mathbb{F}}{\mathbb{P}}^{1}$ corresponds to an orbit with more
than two points coinciding.

The question of how this correspondence might be completed to include the
points $0,1,\infty $ by allowing members of the quadruple to coincide
suitably is a subtle one that leads into the branch of algebraic geometry
known as Geometric Invariant Theory.
\end{remark}

We conclude by remarking that the cross-ratio has some interesting
symmetries.

\begin{theorem}
The cross-ratio obeys the following equations: 
\begin{equation*}
(x_0 x_1 : x_2 x_3) = (x_1 x_0 : x_3 x_2) = (x_2 x_3 : x_0 x_1)
\end{equation*}
\begin{equation*}
(x_0 x_1 : x_2 x_3) = (x_1 x_0 : x_2 x_3)^{-1}
\end{equation*}
\begin{equation*}
(x_0 x_2 : x_1 x_3) = 1 - (x_0 x_1 : x_2 x_3).
\end{equation*}
\end{theorem}

\begin{proof}
Each of these identities is routine. We shall prove these using
inhomogeneous co-ordinates for ease of notation.

(a) That%
\begin{equation*}
\frac{(x_{2}-x_{0})(x_{3}-x_{1})}{(x_{3}-x_{0})(x_{2}-x_{1})}=\frac{%
(x_{3}-x_{1})(x_{2}-x_{0})}{(x_{2}-x_{1})(x_{3}-x_{0})}=\frac{%
(x_{0}-x_{2})(x_{1}-x_{3})}{(x_{1}-x_{2})(x_{0}-x_{3})}
\end{equation*}%
are trivial identities.

(b) Likewise%
\begin{equation*}
\frac{(x_{2}-x_{0})(x_{3}-x_{1})}{(x_{3}-x_{0})(x_{2}-x_{1})}=\left( \frac{%
(x_{2}-x_{1})(x_{3}-x_{0})}{(x_{3}-x_{1})(x_{2}-x_{0})}\right) ^{-1}
\end{equation*}%
is an immediate identity.

(c) Finally%
\begin{eqnarray*}
(x_{0}x_{2}:x_{1}x_{3})+(x_{0}x_{1}:x_{2}x_{3})= &&\frac{%
(x_{1}-x_{0})(x_{3}-x_{2})}{(x_{3}-x_{0})(x_{1}-x_{2})}+\frac{%
(x_{2}-x_{0})(x_{3}-x_{1})}{(x_{3}-x_{0})(x_{2}-x_{1})} \\
= &&\frac{(x_{1}-x_{0})(x_{3}-x_{2})-(x_{2}-x_{0})(x_{3}-x_{1})}{%
(x_{3}-x_{0})(x_{1}-x_{2})} \\
= &&\frac{x_{1}x_{3}+x_{0}x_{2}-x_{2}x_{3}-x_{0}x_{1}}{%
(x_{3}-x_{0})(x_{1}-x_{2})}=1.
\end{eqnarray*}
\end{proof}

\chapter{Duality}

We shall now apply some more linear algebra technology to projective
geometry. We recall that for any vector space $V$ over ${\mathbb{F}}$ we can
associate the \textit{dual space} $V^{\ast }$ of linear maps (functionals) $%
f:V\rightarrow {\mathbb{F}}$. In the finite-dimensional case (which we are
always concerned with in these notes), $V$ and $V^{\ast }$ are isomorphic,
since they are of equal dimension; however this isomorphism depends on a
choice of basis and so is not canonical. However the double dual $V^{\ast
\ast }$, that is, the dual of $V^{\ast }$, \textit{is} canonically
isomorphic to $V$. Explicitly, the map 
\begin{equation*}
\phi :V\rightarrow V^{\ast \ast }
\end{equation*}%
defined by 
\begin{equation*}
\phi (v):f\mapsto f(v)\;\;\;(f\in V^{\ast })
\end{equation*}%
defines an isomorphism between $V$ and $V^{\ast \ast }$.

We then have an inclusion-reversing correspondence between subspaces of $V$
and subspaces of $V^{\ast }$, given by associating to $U\leqslant V$ its 
\textit{annihilator} 
\begin{equation*}
U^{\circ }=\{f\in V^{\ast }:f(u)=0\;\mathrm{for\;all}\;u\in U\}
\end{equation*}%
We recall the following results from Part A Linear Algebra.

\begin{proposition}
For subspaces $U,\,U_{1},\,U_{2}$ of $V$ we have

(i) if $U_{1}\leqslant U_{2}$ then $U_{2}^{\circ }\leqslant U_{1}^{\circ }$:
that is, taking the annihilator reverses inclusion.

(ii) $(U_{1}+U_{2})^{\circ }=U_{1}^{\circ }\cap U_{2}^{\circ }.$

(iii) $(U_{1}\cap U_{2})^{\circ }=U_{1}^{\circ }+U_{2}^{\circ }.$

(iv) $\dim U+\dim U^{\circ }=\dim V.$

(v) $(U^{\circ })^{\circ }=\phi (U).$
\end{proposition}

\noindent The last statement follows from the obvious fact that $\phi
(U)\leqslant (U^{\circ })^{\circ }$, and the dimension formula (iv).

We shall usually use the canonical isomorphism $\phi $ to identify spaces
with their double duals, and subspaces with their double annihilators,
without further comment.\bigskip

Turning to projective spaces, we obtain an inclusion-reversing \textbf{%
duality correspondence} between linear subspaces of ${\mathbb{P}}(V)$ and
linear subspaces of ${\mathbb{P}}(V^{\ast })$, given by associating ${%
\mathbb{P}}(U^{\circ })$ with ${\mathbb{P}}(U)$.

In particular, points of ${\mathbb{P}}(V^{\ast })$ (which represent
1-dimensional subspaces of $V^{\ast }$) correspond to hyperplanes in ${%
\mathbb{P}}(V)$, which represent $n$-dimensional subspaces of $V$ if $\dim
V=n+1$. This is of course just the assigment to $[f]$, where $f\in V^{\ast
}-\{0\}$, of the hyperplane ${\mathbb{P}}(\ker (f))$ in ${\mathbb{P}}(V)$.
In terms of homogeneous coordinates, the point $[a_{0},\ldots ,a_{n}]$ in
the dual projective space ${\mathbb{P}}(V^{\ast })$ corresponds to the
hyperplane with equation $a_{0}x_{0}+\ldots a_{n}x_{n}=0$ in ${\mathbb{P}}%
(V) $ (scaling all the $a_{i}$ does not alter the hyperplane!). Conversely,
hyperplanes in ${\mathbb{P}}(V^{\ast })$ correspond to points in ${\mathbb{P}%
}(V^{\ast \ast })$ and thus to points in ${\mathbb{P}}(V)$.

In general, if ${\mathbb{P}}(U)$ is an $m$-dimensional linear subspace of ${%
\mathbb{P}}^{n}={\mathbb{P}}(V)$, then $U$ has dimension $m+1$, so $U^{\circ
}$ has dimension $(n+1)-(m+1)=n-m$, and hence ${\mathbb{P}}(U^{\circ })$ is
a linear subspace of ${\mathbb{P}}(V^{\ast })$ of dimension $n-m-1$.\bigskip

For the projective plane, the duality interchanges points and lines. If $%
[x],[y]$ are 2 points on a line $L$ then the lines $[x^{\circ }],[y^{\circ
}] $ meet at the point $[L^{\circ }]$. More generally a set of collinear
points corresponds under duality to a set of concurrent lines. We can
interpret $[x^{\circ }]$ as the locus in the dual plane parametrising lines
through $x$ in the original plane.

To more formally present these facts: we have $L={\mathbb{P}}(\langle
x,y\rangle )={\mathbb{P}}(\langle x\rangle +\langle y\rangle ).$ This has
dual%
\begin{equation*}
{\mathbb{P}}(\langle x,y\rangle ^{\circ })={\mathbb{P}}((\langle x\rangle
+\langle y\rangle )^{\circ })={\mathbb{P}}(\langle x\rangle ^{\circ }\cap
\langle y\rangle ^{\circ }).
\end{equation*}

Notice in fact for the projective plane that Lemma \ref{linejoin} and
Proposition \ref{linesmeet} are dual to each other, in the sense that we get
one from the other via duality. In general each theorem in projective
geometry will have a dual version. Moreover once having proved a theorem in
all projective spaces ${\mathbb{P}}(V)$ then this theorem applies equally
well to dual projective spaces ${\mathbb{P}}(V^{\ast })$ and so the dual
theorem is a free biproduct of proving the original theorem. The dual of the
theorem \textit{two distinct points lie on a unique line} has (in general)
the dual theorem \textit{two distinct hyperplanes each contain a unique
linear subspace of dimension }$\dim {\mathbb{P}}(V)-2.$

\begin{example}
The dual of Desargues's Theorem in the plane is as follows.

Let $\pi ,\alpha ,\alpha ^{\prime },\beta ,\beta ^{\prime },\gamma ,\gamma
^{\prime }$ be seven distinct lines in a projective plane such that the
points $\alpha \cap \alpha ^{\prime },\beta \cap \beta ^{\prime }$ and $%
\gamma \cap \gamma ^{\prime }$ are distinct and all lie on $\pi $. Then the
lines joining $\alpha \cap \beta ,\alpha ^{\prime }\cap \beta ^{\prime }$
and $\beta \cap \gamma ,\beta ^{\prime }\cap \gamma ^{\prime }$ and $\gamma
\cap \alpha ,\gamma ^{\prime }\cap \alpha ^{\prime }$ are concurrent.
\end{example}

\begin{example}
Give a projective plane ${\mathbb{P}}^{2},$ we can define four lines to be
in general position if no three of them are concurrent. This is equivalent
to the four points they represent in $({\mathbb{P}}^{2})^{\ast }$ being in
general position.

Under duality a line $\alpha _{0}x_{0}+\alpha _{1}x_{1}+\alpha _{2}x_{2}=0$
in ${\mathbb{P}}^{2}t$corresponds to the point $\left[ \alpha _{0}\colon
\alpha _{1}\colon \alpha _{2}\right] $ in the dual space ${\mathbb{P}}%
^{2\ast }$. So by the general position theorem, four lines in ${\mathbb{P}}%
^{2}$ which are in general position can be assumed to have the equations%
\begin{equation*}
x_{0}=0,\qquad x_{1}=0,\qquad x_{2}=0,\qquad x_{0}+x_{1}+x_{2}=0
\end{equation*}%
without any loss of generality.
\end{example}

\begin{example}
(a) Now let ${\mathbb{P}}(V)$ be a projective plane. Suppose that $%
p_{1},\ldots ,p_{6}$ are distinct points in ${\mathbb{P}}(V)$ and $%
l_{1},\ldots ,l_{4}$ are distinct lines in ${\mathbb{P}}(V)$, such that each
line $l_{j}$ contains exactly 3 of the points $p_{i}$. Show that no $p_{i}$
is contained in 3 of the lines $l_{j}$, and hence (or otherwise) prove that
each point $p_{i}$ is the intersection of exactly two of $l_{1},l_{2},l_{3}$
and $l_{4}$.

(b)\ Let $q_{1},\ldots ,q_{4}$ be the points in ${\mathbb{P}}(V^{\ast })$
dual to the lines $l_{1},\ldots ,l_{4}$ in ${\mathbb{P}}(V)$. By considering
the dual statement (or otherwise), prove that $q_{1},\ldots ,q_{4}$ are in
general position in ${\mathbb{P}}(V^{\ast })$. Explain briefly how you could
use this fact to show that the group $G$ of projective transformations of ${%
\mathbb{P}}(V)$ preserving the set $\left\{ p_{1},\ldots ,p_{6}\right\} $ is
isomorphic to the group $S_{4}$.
\end{example}

\begin{solution}
(a) Without loss of generality suppose that $p_{1}$ lies on each of $%
l_{1},l_{2},l_{3}.$ By a possible relabelling we might then assume that the
other two points on $l_{1}$ are $p_{2},p_{3}.$ The other points $%
p_{4},p_{5},p_{6}$ cannot then lie on $l_{1}$ and $p_{2}$ cannot lie on
either $l_{2}$ or $l_{3}$ without two of the lines being the same. So we
might label the other two points on $l_{2}$ as being $p_{4},p_{5}$ and we
now arrive at a contradiction as we have now run out of points to be on $%
l_{3}.$

Say now that $p_{1}$ lies only on line $l_{1}$ and that $p_{2},p_{3}$ are
the other two points on $l_{1}.$ Then $l_{2}$ may only contain one of $p_{2}$
say, $l_{3}$ may only contain $p_{3}$ and $l_{4}$ may contain none of $%
p_{1},p_{2},p_{3}$ and so contains $p_{4},p_{5},p_{6}.$ But then $l_{2}$
must contain $p_{2}$ and two of $p_{4},p_{5},p_{6}$ and so is not distinct
from $l_{4}.$

Consequently each point $p_{i}$ lies on precisely two lines as required.

(b) $q_{1},\ldots ,q_{4}$ not being in general position in ${\mathbb{P}}%
(V^{\ast })$ means that three of the points are collinear. The dual
statement is that three of $l_{1},\ldots ,l_{4}$ are concurrent, but then
the points $p_{1},\ldots ,p_{6}$ are not distinct and this is not the case.
Given any permutation $\sigma $ of the $q_{i}$ there is a unique projective
transformation of ${\mathbb{P}}(V^{\ast })$ effecting that permutation on
the $q_{i}$ as they are in general position (and so is then any reordering
of them). Thus the group of transformations in $PGL(V^{\ast })$ preserving $%
q_{1},\ldots ,q_{4}$ is isomorphic to $S_{4}.$ Necessarily the dual of that
map preserves the set $\left\{ l_{1},l_{2},l_{3},l_{4}\right\} $ and so
preserves the set of intersections $\left\{ p_{1},\ldots ,p_{6}\right\} .$
\end{solution}

\chapter{Bilinear Forms and Quadrics}

The next piece of algebra we consider in the context of projective geometry
is the theory of bilinear forms.

\begin{definition}
A \textbf{symmetric bilinear form} on a vector space $V$ over ${\mathbb{F}}$
is a map $B\colon V\times V\rightarrow {\mathbb{F}}$ such that

(i) $B(v,w)=B(w,v)$

(ii) $B$ is linear in $v$ (and hence, by (i), also in $w$)
\end{definition}

If an addition we have the property

(iii) if $B(v,w)=0$ for all $w$ then $v=0$,

then we say the form is \textbf{nondegenerate} or \textbf{nonsingular}%
.\bigskip

More concretely, if we choose a basis $e_{0},\ldots ,e_{n}$ then a bilinear
form is given by $B(v,w)=v^{T}Mw$ for a symmetric matrix $M$ given by $%
[M]_{ij}=B(e_{i},e_{j})$. Nondegeneracy of the form is equivalent to
nonsingularity/invertibility of the matrix $M$. Symmetric matrices form a
vector space of dimension $\frac{1}{2}\dim V(\dim V+1)$, so we can form
linear combinations of bilinear forms.

\begin{remark}
In part A linear algebra we focused particularly on inner products. Over ${%
\mathbb{R}}$, these are symmetric bilinear forms which satisfy the extra
condition of positive definiteness (that is $B(v,v)>0$ for $v\neq 0$). Over $%
{\mathbb{C}}$, positive definiteness requires the form to be conjugate
symmetric, rather than symmetric, and sesquilinear, rather than bilinear, --
that is, the form is linear in one variable and conjugate linear in the
other. We shall focus instead on bilinear forms and drop the positive
definiteness property. In fact, for most purposes nondegeneracy is a good
replacement for positive definiteness. In particular, nondegeneracy is
actually equivalent to the statement that the map from $V$ to $V^{\ast }$
defined by $v\mapsto B(v,.)$ is an isomorphism.
\end{remark}

Any bilinear form is determined (if the characteristic of ${\mathbb{F}}$
does no equal $2$), by the associated \textbf{quadratic form} 
\begin{equation*}
Q(v)=B(v,v),
\end{equation*}%
for we can recover $B$ via the \textbf{polarisation identity} 
\begin{equation*}
B(v,w)=\frac{1}{4}(B(v+w,v+w)-B(v-w,v-w))=\frac{1}{4}(Q(v+w)-Q(v-w)).
\end{equation*}

Over ${\mathbb{R}}$ or ${\mathbb{C}}$ we can diagonalise quadratic forms.

\begin{theorem}
If $v \mapsto Q(v) = B(v,v)$ is a quadratic form defined on a vector space,
then

(i) if ${\mathbb{F}}={\mathbb{C}}$, there is a basis $e_{0},\ldots ,e_{n}$
with respect to which 
\begin{equation*}
Q(v)=\lambda _{0}^{2}+\cdots +\lambda _{r}^{2}
\end{equation*}%
where $v=\sum_{i=0}^{n}\lambda _{i}e_{i}$ and where $r+1$ is the rank of $B.$

(ii) if ${\mathbb{F}}={\mathbb{R}}$, there is a basis $e_{0},\ldots ,e_{n}$
with respect to which 
\begin{equation*}
Q(v)=\lambda _{0}^{2}+\cdots +\lambda _{r}^{2}-\lambda _{r+1}^{2}-\cdots
-\lambda _{r+s}^{2}
\end{equation*}%
where $v=\sum_{i=0}^{n}\lambda _{i}e_{i}$ and where $r+s+1$ is the rank of $%
B.$
\end{theorem}

\begin{proof}
Write $B(v,v)=v^{T}Xv=\sum_{i,j}X_{ij}v_{i}v_{j}$ in some basis, where $X$
is a symmetric matrix. We can assume that some $X_{ii}$ is nonzero, because
if $X_{ij}$ is nonzero we can introduce new variables $y_{i}=\frac{1}{2}%
(v_{i}+v_{j}),\;y_{j}=\frac{1}{2}(v_{i}-v_{j})$ and now $%
v_{i}v_{j}=y_{i}^{2}-y_{j}^{2}$.

Now we complete the square. 
\begin{equation*}
\frac{1}{X_{ii}}\left( \sum_{j}X_{ij}v_{j}\right)
^{2}=X_{ii}v_{i}^{2}+2\sum_{j\neq i}X_{ij}v_{j}v_{i}+\mathrm{terms\;in}%
\;v_{j}(j\neq i).
\end{equation*}%
So by introducing the new variable $\tilde{y}_{i}=\sum X_{ij}v_{j}$ we can
put $B$ into the form 
\begin{equation*}
B(v,v)=\frac{1}{X_{ii}}y_{i}^{2}+\mathrm{terms\;in}\;v_{j}(j\neq i)
\end{equation*}%
Now we repeat the process until we have diagonalised $B$ to $\mathrm{diag}%
(\beta _{0},\ldots ,\beta _{n})$: rescaling the variables appropriately by
now setting for non-zero $\beta _{i}$%
\begin{equation*}
y_{i}\mapsto \sqrt{\beta _{i}}y_{i}\quad \text{over }%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\quad \text{and}\quad y_{i}\mapsto \sqrt{\left\vert \beta _{i}\right\vert }%
y_{i}\quad \text{over }%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,
\end{equation*}%
now brings it into the desired form. Note that over ${\mathbb{R}}$ we cannot
change the sign of $y_{i}^{2}$ by rescaling.
\end{proof}

\begin{remark}
Note that over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ we might simplify this argument by beginning with the spectral theorem,
but that the spectral theorem would not help with the complex case as we are
interested in symmetric rather than conjugate symmetric (i.e. Hermitian)
forms.
\end{remark}

Notice that the form is nondegenerate exactly when $r=n$ (in the complex
case) and $r+s=n$ (in the real case).

\begin{example}
Consider the form on ${\mathbb{R}}^{3}$.%
\begin{equation*}
x_{0}x_{1}+x_{1}x_{2}+x_{2}x_{0}.
\end{equation*}%
We change variables to 
\begin{equation*}
y_{0}=\frac{1}{2}(x_{0}+x_{1}),\;\;y_{1}=\frac{1}{2}(x_{0}-x_{1}),\;%
\;y_{2}=x_{2}
\end{equation*}%
to generate some nonzero diagonal terms. The form is now 
\begin{equation*}
y_{0}^{2}-y_{1}^{2}+2y_{2}y_{0}.
\end{equation*}%
We complete the square, writing this as 
\begin{equation*}
(y_{0}+y_{2})^{2}-y_{2}^{2}-y_{1}^{2}
\end{equation*}%
so on putting $z_{0}=y_{0}+y_{2}$ and $z_{1}=y_{1},\;z_{2}=y_{2}$ we get the
required form 
\begin{equation*}
z_{0}^{2}-z_{1}^{2}-z_{2}^{2}
\end{equation*}%
over the reals. If we work over ${\mathbb{C}}$ then scaling $z_{1},z_{2}$ by 
$i$ brings us into the standard form of a nondegenerate quadratic form over $%
{\mathbb{C}}$.
\end{example}

We have seen that linear subspaces of projective space ${\mathbb{P}}(V)$ are
projectivisations of subspaces of $V$, and hence are determined by systems
of homogeneous linear equations. The next simplest subsets of projective
space defined by polynomial equations are the \textit{quadrics}, which are
defined by the vanishing of a quadratic form.

\begin{definition}
A \textbf{quadric} is the locus of points in a projective space defined by
an equation $Q(v)=0$, where $v\mapsto Q(v)=B(v,v)$ is a (not identically
zero) quadratic form.

In the case that the projective space is a projective plane then quadrics
are more typically referred to as \textbf{conics.}

We remark that this does indeed define a subset of projective space, as $%
Q(v) $ is homogeneous of degree 2 in $v$ (cf. the remarks at the end of \S %
2).
\end{definition}

\begin{example}
In $\mathbb{P}(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3})$ consider the conic defined by the quadratic form%
\begin{equation*}
Q(x_{0},x_{1},x_{2})=x_{0}^{2}+x_{1}^{2}-x_{2}^{2}.
\end{equation*}%
In inhomogeneous co-ordinates $x=x_{1}/x_{0},y=x_{2}/x_{0}$ this could be
considered as the hyperbola%
\begin{equation*}
1+x^{2}-y^{2}=0.
\end{equation*}%
This has points at infinity $\left[ 0\colon 1\colon 1\right] $ and $\left[
0\colon 1\colon -1\right] $ (on the line $x_{0}=0$).

We might instead have considered the conic using the inhomogeneous
co-ordinates $x=x_{0}/x_{2}$, $y=x_{1}/x_{2}$ and then we would be
considering the circle with equation%
\begin{equation*}
x^{2}+y^{2}-1=0
\end{equation*}%
which has no points at infinity (on the line $x_{2}=0$).
\end{example}

\begin{example}
Find a projective transformation which takes the parabola $y^{2}=x$ to the
circle $x^{2}+y^{2}=1.$ (Here we are using $x=x_{1}/x_{0},y=x_{2}/x_{0}.$)
\end{example}

\begin{solution}
In homogeneous co-ordinates these conics (call them $C_{1},C_{2}$) are
defined by the equation $x_{2}^{2}=x_{1}x_{0}$ and $%
x_{1}^{2}+x_{2}^{2}-x_{0}^{2}=0$ and so by the matrices%
\begin{equation*}
B_{1}=\left( 
\begin{array}{ccc}
0 & \frac{1}{2} & 0 \\ 
\frac{1}{2} & 0 & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) ,\qquad B_{2}=\left( 
\begin{array}{ccc}
-1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) .
\end{equation*}%
Consider the effect of a projective transformation $\left[ T\right] $ on the
first conic. Then%
\begin{eqnarray*}
\left[ \mathbf{x}\right] \in \left[ T\right] (C_{1})\quad
\Longleftrightarrow &&\quad \left[ T^{-1}\right] \left[ \mathbf{x}\right]
\in C_{1} \\
\Longleftrightarrow &&\quad (T^{-1}\mathbf{x})^{T}B_{1}T^{-1}\mathbf{x}=0. \\
\Longleftrightarrow &&\quad \mathbf{x}^{T}(T^{-1})^{T}B_{1}T^{-1}\mathbf{x}%
=0.
\end{eqnarray*}%
And if we want $T$ to map $C_{1}$ to $C_{2}$ then we need%
\begin{equation*}
\mathbf{x}^{T}(T^{-1})^{T}B_{1}T^{-1}\mathbf{x}=0\quad \Longleftrightarrow
\quad \mathbf{x}^{T}B_{2}\mathbf{x}=0
\end{equation*}%
so it is sufficient to have%
\begin{equation*}
(T^{-1})^{T}B_{1}T^{-1}=B_{2}
\end{equation*}%
or equivalently that%
\begin{equation*}
B_{1}=T^{T}B_{2}T.
\end{equation*}%
Now the matrix $B_{1}$ is symmetric and has eigenvalues $\pm 1/2,1.$ So
there is an orthogonal matrix $P$ such that%
\begin{equation*}
P^{T}B_{1}P=\mathrm{diag}(-1/2,1/2,1).
\end{equation*}%
If we now take $Q=\mathrm{diag}(\sqrt{2},\sqrt{2},1)$ then we have%
\begin{equation*}
Q^{T}P^{T}B_{1}PQ=\mathrm{diag}(-1,1,1)=B_{2}.
\end{equation*}%
So we might choose $T=(PQ)^{-1}.$ Calculating $P$ gives%
\begin{equation*}
T=(PQ)^{-1}=\left( 
\begin{array}{ccc}
1 & 1 & 0 \\ 
-1 & 1 & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) ^{-1}=\left( 
\begin{array}{ccc}
\frac{1}{2} & -\frac{1}{2} & 0 \\ 
\frac{1}{2} & \frac{1}{2} & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) .
\end{equation*}
\end{solution}

\begin{definition}
We say a quadric is \textbf{nonsingular} if the associated symmetric
bilinear form is nondegenerate.

On choosing a basis, this is equivalent to the symmetric matrix defining the
form being invertible.
\end{definition}

\begin{proposition}
Non-empty, nonsingular conics in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\mathbb{P}^{2}$ are projectively equivalent.
\end{proposition}

\begin{proof}
By our earlier theorem any conic's equation can be put in the form%
\begin{equation*}
x_{0}^{2}+\cdots +x_{r}^{2}-x_{r+1}^{2}-\cdots -x_{r+s}^{2}=0.
\end{equation*}%
As $r+s$ is the rank and the conic is non-singular then $r+s+1=3$ and so we
are left considering the possibilities:%
\begin{eqnarray*}
x_{0}^{2}+x_{1}^{2}+x_{2}^{2} &=&0 \\
x_{0}^{2}+x_{1}^{2}-x_{2}^{2} &=&0 \\
x_{0}^{2}-x_{1}^{2}-x_{2}^{2} &=&0 \\
-x_{0}^{2}-x_{1}^{2}-x_{2}^{2} &=&0.
\end{eqnarray*}%
The first and last are empty conics (as we're working over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$) and the second two are clearly projectively equivalent (by permuting $%
x_{0},x_{1},x_{2}$ accordingly).
\end{proof}

\begin{remark}
We might then view an ellipse as a conic which does not intersect the line
at infinity, a parabola as a conic which is tangential to the line at
infinity, and a hyperbola as a conic which intersects the line at infinity
in two points.\pagebreak
\end{remark}

\begin{proposition}
If a quadric contains three distinct, collinear points then it contains the
entire line.
\end{proposition}

\begin{proof}
We may choose vectors $\mathbf{x}$ and $\mathbf{y}$ so that the points are $%
\left[ \mathbf{x}\right] ,\,\left[ \mathbf{y}\right] ,\,\left[ \mathbf{x}+%
\mathbf{y}\right] .$ Then we have%
\begin{equation*}
B(\mathbf{x},\mathbf{x})=0=B(\mathbf{y},\mathbf{y})
\end{equation*}%
and%
\begin{equation*}
0=B(\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y})=B(\mathbf{x},\mathbf{x})+2B(%
\mathbf{x},\mathbf{y})+B(\mathbf{y},\mathbf{y})
\end{equation*}%
so that $B(\mathbf{x},\mathbf{y})=0$ as well. Then for any point $\alpha 
\mathbf{x}+\beta \mathbf{y}$ of the line we have%
\begin{equation*}
B(\alpha \mathbf{x}+\beta \mathbf{y},\alpha \mathbf{x}+\beta \mathbf{y}%
)=\alpha ^{2}B(\mathbf{x},\mathbf{x})+2\alpha \beta B(\mathbf{x},\mathbf{y}%
)+\beta ^{2}B(\mathbf{y},\mathbf{y})=0
\end{equation*}%
as required.
\end{proof}

Over ${\mathbb{C}}$, our diagonalisation theorem tells us that a conic can
be put into one of the following three forms:

\begin{enumerate}
\item[(i)] $z_{0}^{2}+z_{1}^{2}+z_{2}^{2}=0$;

\item[(ii)] $z_{0}^{2}+z_{1}^{2}=0$;

\item[(iii)] $z_{0}^{2}=0$.
\end{enumerate}

Case (i) is the general case, when the conic is nonsingular. The remaining
two cases are the two kinds of singular conics. Case (ii) is a pair of
distinct lines -- on putting the conic in the above form $%
z_{0}^{2}+z_{1}^{2}=0$ we see the lines are $z_{0}-iz_{1}=0$ and $%
z_{0}+iz_{1}=0$, which meet at the point $[0,0,1]$ in the plane. Case (iii)
is the most degenerate -- it is a double line, i.e. a line with multiplicity
two. We can think of Cases (ii) and (iii) as singular limits or
degenerations of the generic nonsingular conics.

In fact generally (for conics, but not higher dimensional quadrics) we have:

\begin{proposition}
A conic in $\mathbb{FP}^{2}$ which contains a line is singular.
\end{proposition}

\begin{proof}
Take two points on the line in the conic and a point off the line. If we
choose representative vectors $e_{0},e_{1},e_{2}$ for these three points
then form a basis and say that $X=\left( X_{ij}\right) $ is symmetric matrix
over $\mathbb{F}$ representing the conic. Points on the line have
representative vectors $\alpha e_{0}+\beta e_{1}$ so that%
\begin{equation*}
(\alpha e_{0}+\beta e_{1})^{T}X(\alpha e_{0}+\beta e_{1})=0
\end{equation*}%
for all $\alpha ,\beta .$ But this means that $X$ has the form%
\begin{equation*}
X=\left( 
\begin{array}{ccc}
0 & 0 & X_{02} \\ 
0 & 0 & X_{12} \\ 
X_{02} & X_{12} & X_{22}%
\end{array}%
\right)
\end{equation*}%
which is a singular matrix.
\end{proof}

\begin{proposition}
Through each point $P$ of a non-singular conic $C$ in $\mathbb{FP}^{2}$
there is a unique line which meets the conic only once. This line is called
the \textbf{tangent line} to $C$ at $P.$
\end{proposition}

\begin{proof}
Say that the conic is represented by the invertible symmetric matrix $X$
with respect to some basis and say that $P=\left[ \mathbf{x}\right] .$ We
then have that $\mathbf{x}^{T}X\mathbf{x}=0$ and that $X\mathbf{x}\neq 
\mathbf{0}$ as $X$ is invertible.

A line in $\mathbb{FP}^{2}$ has the form $\mathbb{P}(\langle \mathbf{x},%
\mathbf{y}\rangle )$ where $\mathbf{y}$ is independent of $\mathbf{x.}$ This
line will meet the conic at points satisfying%
\begin{equation*}
0=(\alpha \mathbf{x}+\beta \mathbf{y})^{T}X(\alpha \mathbf{x}+\beta \mathbf{y%
})=2\alpha \beta \mathbf{y}^{T}X\mathbf{x}+\beta ^{2}\mathbf{y}^{T}X\mathbf{y%
}.
\end{equation*}%
For the line to meet the conic only at $P$ (when $\beta =0$) we must have
that%
\begin{equation*}
2\lambda \mathbf{y}^{T}X\mathbf{x}+\mathbf{y}^{T}X\mathbf{y}\neq 0\qquad
\left( \lambda =\alpha /\beta \right)
\end{equation*}%
for all scalars $\lambda \in \mathbb{F}$. This is only possible if $\mathbf{y%
}^{T}X\mathbf{x}=0$ and $\mathbf{y}^{T}X\mathbf{y}\neq 0.$ The plane of
vectors perpendicular to $X\mathbf{x}$ uniquely specifies a plane in $%
\mathbb{F}^{3}$ and so uniquely specifies a line in $\mathbb{FP}^{2}.$
\end{proof}

Pappus' Theorem might be viewed as a special case of the following theorem
as applying to singular conics.

\begin{theorem}
(\textbf{Pascal's Mystic Hexagon})

Let $A,B,C,A^{\prime },B^{\prime },C^{\prime }$ be six distinct points on a
non-singular conic. Then the intersections%
\begin{equation*}
P=AB^{\prime }\cap A^{\prime }B,\qquad Q=AC^{\prime }\cap A^{\prime
}C,\qquad R=BC^{\prime }\cap B^{\prime }C,
\end{equation*}%
are collinear.
\end{theorem}

\begin{proof}
The conic contains no line and hence the points $A,B,A^{\prime },B^{\prime }$
are in general position. So we may assume them to have co-ordinates%
\begin{equation*}
A=\left[ 1\colon 0\colon 0\right] ,\qquad B=\left[ 0\colon 1\colon 0\right]
,\qquad A^{\prime }=\left[ 0\colon 0\colon 1\right] ,\qquad B^{\prime }=%
\left[ 1\colon 1\colon 1\right] .
\end{equation*}%
Let $C=\left[ x\colon y\colon z\right] $ and $C^{\prime }=\left[ X\colon
Y\colon Z\right] $, noting that $x,y,z,X,Y,Z$ are necessarily non-zero. We
then have that%
\begin{eqnarray*}
P &=&AB^{\prime }\cap A^{\prime }B=\left\{ x_{1}=x_{2}\right\} \cap \left\{
x_{0}=0\right\} =\left[ 0\colon 1\colon 1\right] , \\
Q &=&AC^{\prime }\cap A^{\prime }C=\left\{ Zx_{1}=Yx_{2}\right\} \cap
\left\{ yx_{0}=xx_{1}\right\} =\left[ xY\colon yY\colon yZ\right] \\
R &=&BC^{\prime }\cap B^{\prime }C=\left\{ Zx_{0}=Xx_{2}\right\} \cap
\left\{ (z-y)x_{0}+(x-z)x_{1}+(y-x)x_{2}=0\right\} \\
&=&\left[ (x-z)X\colon (x-y)Z+(y-z)X\colon (x-z)Z\right] .
\end{eqnarray*}%
Now the conic is represented by a symmetric matrix%
\begin{equation*}
\left( 
\begin{array}{ccc}
0 & \alpha & \beta \\ 
\alpha & 0 & -\alpha -\beta \\ 
\beta & -\alpha -\beta & 0%
\end{array}%
\right)
\end{equation*}%
for some $\alpha ,\beta ,$ in order to pass through $A,B,A^{\prime
},B^{\prime }.$ As $C$ also lies on the conice we also have that%
\begin{equation*}
\alpha xy+\beta xz-(\alpha +\beta )yz=0
\end{equation*}%
so that%
\begin{equation*}
\frac{\alpha }{\beta }=\frac{z(y-x)}{y(x-z)}.
\end{equation*}%
As $C^{\prime }$ similarly lies on the line then%
\begin{equation*}
\frac{z(y-x)}{y(x-z)}=\frac{Z(Y-X)}{Y(X-Z)}.
\end{equation*}%
Now%
\begin{eqnarray*}
&&((x-z)X,(x-y)Z+(y-z)X,(x-z)Z)-\left( \frac{(x-z)X}{xY}\right) (xY,yY,yZ) \\
&=&\left( 0,(x-y)Z+(y-z)X-\frac{(x-z)Xy}{x},(x-z)Z-\frac{(x-z)yXZ}{xY}\right)
\\
&=&\frac{1}{xY}\left( 0,(x-y)xYZ+(y-z)xXY-(x-z)yXY,(x-z)xYZ-(x-z)yXZ\right)
\\
&=&\frac{1}{xY}\left( 0,(y-x)Y(zX-xZ),(x-z)Z\left( xY-yX\right) \right) .
\end{eqnarray*}%
If this is to be some multiple of $\left( 0,1,1\right) $ (which would show $%
P,Q,R$'s representative vectors to be dependent) we need these last two
co-ordinates to be equal. But that requirement rearranges to the condition
above connecting $x,y,z$ with $X,Y,Z.$\bigskip
\end{proof}

\begin{definition}
The \textbf{singular points} of a quadric $Q(\mathbf{v})=\mathbf{v}^{T}X%
\mathbf{v}=0$ are those points $[\mathbf{v}]$ where $X\mathbf{v}=0$.
\end{definition}

So in case (ii), where $X = \mathrm{diag}(1,1,0)$, the unique singular point
is $[0,0,1]$, the intersection point of the pair of lines. In case (iii) we
have $X = \mathrm{diag}(1,0,0)$, then the singular points are the points on
the line $z_0 =0$ : in other words every point on the conic is singular.

\begin{remark}
The conic is nonsingular if and only if $X$ is invertible, which is
equivalent to the only solution to $Xv=0$ being $v=0$. So the conic is
nonsingular if and only it has no singular points in projective space. This
justifies the terminology in the definition above.

If we work over ${\mathbb{C}}$ or ${\mathbb{R}}$, then we may further
understand the notion of singular point using some ideas from part A
Multivariable Calculus. The conic is defined by the equation $f=0$ where $%
f:v\mapsto v^{T}Xv$. Now, the derivative of $f$ at $v$ in the sense of
multivariable calculus is $df_{X}:h\mapsto 2h^{T}Xv$, which has maximal rank
one unless $Xv=0$. So the singular points are the points where $df$ has less
than maximal rank, and hence where the manifold structure on the conic
breaks down. In the example above, this happens in case (ii) exactly where
the lines intersect.
\end{remark}

Nonsingular conics actually have a very nice description. If we fix a point $%
x$ on the conic, and take a projective line not containing $x$, then
projection from $x$ onto the line actually sets up a bijection between the
conic and the line. (If ${\mathbb{F}}={\mathbb{C}}$, this in fact defines a
homeomorphism between the conic and the projective line, and hence the
Riemann sphere, though note this is \textit{not} a projective equivalence).

\begin{theorem}
Let $C$ be a nonsingular conic in the projective plane, and let $x$ be a
point of $C$. Let ${\mathbb{P}}(U)$ be a projective line in the plane not
containing $x$. Then there is a bijection $\alpha \colon {\mathbb{P}}%
(U)\rightarrow C$ such that $x,y,\alpha (y)$ are collinear, for $y\in {%
\mathbb{P}}(U)$.
\end{theorem}

\begin{proof}
Let $B$ denote the nondegenerate bilinear form whose quadratic form $Q$
defines the conic $C$. Let $x = [v_0]$ be a point on $C$, so that $%
B(v_0,v_0)=0$.

For each $y \in {\mathbb{P}}(U)$, we want to see when (other than at $x$)
the projective line containing $x$ and $y$ meets the conic. We will find
that there is a unique such point and this will be $\alpha(y)$.

Let $y \in {\mathbb{P}}(U)$ have representative vector $u \in U$, so that $%
v_0, u$ are linearly independent, as we are assuming $x \notin {\mathbb{P}}%
(U)$. Consider the 2-dimensional space $W_u$ spanned by $v_0$ and $u$, so
the projective line we are considering is ${\mathbb{P}}(W_u)$. Observe that
the bilinear form $B$ cannot be identically zero on the space $W_u$ . For we
could extend to a basis $v_0, u, z$ for ${\mathbb{C}}^3$ and the orthogonal
complement of $z$ would meet $W_u$ in at least a 1-dimensional subspace,
elements of which would now be orthogonal to the whole of $V$, thus
contradicting nondegeneracy.

With respect to the basis $v_0, u$, the form $Q$ restricted to $W_u$ is 
\begin{equation*}
Q(\lambda_0 v_0 + \lambda_1 u) = 2 \lambda_0 \lambda_1 B(v_0,u) +
\lambda_1^2 B(u,u)
\end{equation*}
and $B(v_0,u), B(u,u)$ are not both zero. So the projective line ${\mathbb{P}%
}(W_u)$ meets the conic at two points. One is the basepoint $x = [v_0]$,
corresponding to $(\lambda_0,\lambda_1)=(1,0)$. The other, corresponding to $%
(\lambda_0, \lambda_1)= (B(u,u), -2B(v_0,u))$, is defined to be $\alpha(y)$.
Note that $\alpha$ is injective as given any point $z \neq x$ on the conic,
the projective line through $x,z$ meets the line ${\mathbb{P}}(U)$ in a
unique point $y$. Moreover, $\alpha(y)=x$ exactly when $B(v_0, u)=0$, which
defines a unique point in ${\mathbb{P}}(U)$.
\end{proof}

We have set up a bijection between a nonsingular conic and the projective
line. This kind of bijection is called a \textbf{rational parametrisation.}

The existence of a rational parametrisation for the conic has some nice
applications in the theory of Diophantine equations. These are polynomial
equations where we are primarily interested in rational or integral
solutions.

\begin{example}
Consider the equation 
\begin{equation*}
X^{2}-Y^{2}-Z^{2}=0
\end{equation*}%
whose solutions are Pythagorean triples. As our basepoint on the conic
defined by the above equation we may take $[1\colon 1\colon 0]$. We can take 
$X=0$ as the projective line, which does not contain the basepoint. So if $y$
is a point on the projective line with representative vector $u=(0,\lambda
_{1},\lambda _{2})$ then 
\begin{eqnarray*}
\alpha (y) &=&B(u,u)v_{0}-2B(u,v_{0})u \\
&=&-(\lambda _{1}^{2}+\lambda _{2}^{2})(1,1,0)+2\lambda _{1}(0,\lambda
_{1},\lambda _{2}) \\
&=&(-(\lambda _{1}^{2}+\lambda _{2}^{2}),\lambda _{1}^{2}-\lambda
_{2}^{2},2\lambda _{1}\lambda _{2})
\end{eqnarray*}%
It is clear that this does indeed give solutions to the equations. Replacing 
$X$ by its negative, we obtain the familiar formula for Pythagorean triples: 
\begin{equation*}
X=s^{2}+t^{2}\;\;:\;\;Y=s^{2}-t^{2}\;\;:\;\;Z=2st
\end{equation*}%
and by taking $s,t$ to be rational (respectively, integral) we get the
solution in rational numbers (respectively, integers). For example, $%
(s,t)=(2,1)$ gives $(X,Y,Z)=(5,3,4)$, while $(s,t)=(3,2)$ and $(4,3)$ give
the triples $(13,5,12)$ and $(25,7,24)$ respectively.
\end{example}

Given a non-singular conic $C$ in a projective plane $\mathbb{P}^{2}$, then
at each point there is a uniquely defined tangent line. This leads to a
well-defined map from the conic to the dual projective space $\mathbb{P}%
^{2\ast }.$ It turns out the image of $C$ is a non-singular conic, called
the \textbf{dual conic} in $\mathbb{P}^{2\ast }.$

\begin{example}
Find a rational parametrization for the dual conic in $\mathbb{P}^{2\ast }$
of $x_{1}^{2}+x_{1}x_{2}+x_{2}^{2}=x_{0}^{2}.$
\end{example}

\begin{solution}
From our earlier proof of the existence of tangent lines, we know that any
point $\mathbf{x}$ in the conic has tangent line with equation $\mathbf{y}%
^{T}X\mathbf{x}=0$ where the symmetric matrix $X$ represents the bilinear
form. Moving to inhomogenous co-ordinates $x=x_{1}/x_{0}$ and $y=x_{2}/x_{0}$
our curve has the form%
\begin{equation*}
x^{2}+xy+y^{2}=1.
\end{equation*}%
A point on the ellipse is $\left( 1,0\right) $ and so if we set $y=q(x-1)$
then we find%
\begin{equation*}
x^{2}+xq(x-1)+q^{2}(x-1)^{2}=1
\end{equation*}%
which factorizes to%
\begin{equation*}
(x-1)\left( x+1+qx+q^{2}(x-1)\right) =0.
\end{equation*}%
The second root is then%
\begin{equation*}
x=\frac{q^{2}-1}{1+q+q^{2}},
\end{equation*}%
so that%
\begin{equation*}
y=q(x-1)=\frac{-q(2+q)}{1+q+q^{2}}
\end{equation*}%
and%
\begin{equation*}
\left[ x_{0}\colon x_{1}\colon x_{2}\right] =\left[ q^{2}-1\colon
-q(2+q)\colon q^{2}+q+1\right] 
\end{equation*}%
is a rational parametrization for the conic.

Recall that we identify the line $\alpha _{0}x_{0}+\alpha _{1}x_{1}+\alpha
_{2}x_{2}=0$ with the point $\left[ \alpha _{0}\colon \alpha _{1}\colon
\alpha _{2}\right] $ in the dual space. As a symmetric matrix representing
the conic is%
\begin{equation*}
\left( 
\begin{array}{ccc}
-1 & 0 & 0 \\ 
0 & 1 & \frac{1}{2} \\ 
0 & \frac{1}{2} & 1%
\end{array}%
\right) 
\end{equation*}%
then the dual conic has a rational parametrization%
\begin{eqnarray*}
\left[ \alpha _{0}\colon \alpha _{1}\colon \alpha _{2}\right]  &=&\left[
1-q^{2}\colon \frac{1}{2}(q^{2}+q+1)-q(2+q)\colon -\frac{1}{2}%
q(2+q)+q^{2}+q+1\right]  \\
&=&\left[ 2-2q^{2}\colon 1-3q-q^{2}\colon 2+q^{2}\right] .
\end{eqnarray*}
\end{solution}

With this notion of a dual conic, we can describe the dual of Pascal's
Theorem which is in fact another classical theorem of geometry, \textit{%
Brianchon's Theorem}.

\begin{theorem}
(\textbf{Brianchon}) If a hexagon is circumscribed about a conic, the three
diagonals are concurrent.
\end{theorem}

\begin{example}
Find all the rational solutions to Pell's equation%
\begin{equation*}
x^{2}-2y^{2}=1.
\end{equation*}
\end{example}

\begin{solution}
We will work with the inhomogeneous co-ordinates $x$ and $y.$ Note that the
point $\left( 1,0\right) $ satisfies Pell's equation and consider the line $%
y=q(x-1)$ where $q$ is rational. Substituting this into Pell's equation we
find that%
\begin{equation*}
x^{2}-2q^{2}(x-1)^{2}=1.
\end{equation*}%
This factorizes easily (and in any case we know $x=1$ to be a root of the
equation) and so we have%
\begin{equation*}
(x-1)(x+1-2q^{2}(x-1))=0.
\end{equation*}%
The second root for $x$ is then%
\begin{equation*}
x=\frac{2q^{2}+1}{2q^{2}-1},
\end{equation*}%
and%
\begin{equation*}
y=q(x-1)=\frac{2q}{2q^{2}-1}.
\end{equation*}
\end{solution}

In the Part B course Algebraic Curves you will see that nonsingular curves
of higher degree in the projective plane do not admit rational
parametrisations. Indeed, over ${\mathbb{C}}$ such curves are not
homeomorphic to the Riemann sphere. The genus of a degree $d$ nonsingular
curve in the complex projective plane is $\frac{1}{2}(d-1)(d-2)$ which is
only zero for $d=1,2$ i.e. the case of lines and conics. We refer, for
example, to~\cite{K} for more on this subject.

For further reading, introducing more ideas of algebraic geometry, see:

\begin{itemize}
\item F. Kirwan, \textit{Complex algebraic curves}, LMS Student Texts, CUP,
1992.

\item K. Smith, L. Kahanp\"{a}\"{a}, P. Kek\"{a}l\"{a}inen and W. Traves, 
\textit{An invitation to algebraic geometry}, Springer Universitext, 2000.
\end{itemize}

\end{document}
