Foundations of Projective Geometry [2009 ed.] 4871878376, 9784871878371

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Foundations of Projective Geometry

Robin Hartshorne

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FOUNDATIONS OF PROJECTIVE

GEOMETRY

Robin Hartshorne

ISHI PRESS INTERNATIONAL

Foundations of Projective Geometry

by Robin Hartshorne First published as “Foundations of Projective Geometry Lecture Notes at Harvard University” by W. A. Benjamin, Inc, 1967

Second printing, 1976 Third printing, 1978 Fourth printing, 1980 This new edition with an added appendix published in 2009 by The Ishi Press in New York and Tokyo

Copyright © 1996, 2009 by Robin Hartshorne Robin Hartshorne Department of Mathematics University of California Berkeley, CA 94720-3840 All rights reserved according to International Law. No part of this book may be reproduced or otherwise copied without the written permission of the publisher.

ISBN 4-87187-837-6

978-4-87187-837-1 Ishi Press International 1664 Davidson Avenue, Suite 1B Bronx NY 10453 917-507-7226 Printed in the United States of America

PREFACE TO THE NEW EDITION

This little book first appeared more than forty years ago. In spare language, it develops the basic ideas of projective geometry in the plane, starting from the incidence axioms, and culminating in the introduction of coordinates. This last result shows that any projective plane satisfying Desargues’ axiom is isomorphic to the Cartesian plane over a division ring. Furthermore the division ring is commutative, 1.e. is a field, if and only if the plane satisfies Pappus’ axiom. I have decided to reissue the book as is, only correcting typographic errors and adding a new Appendix on the simple group of order 168. This Appendix, which requires knowledge of elementary group theory up through the Sylow theorems, shows how the projective plane of seven points is revealed naturally in the internal structure of the subgroups of the simple group of order 168.

My own encounter with projective geometry began in my senior year of high school, when I was an exchange student in Germany. My math teacher, perhaps annoyed with my incessant questions, gave me a book with the implied message: Here, go away and read this. The book was Theodore Reye’s Geometrie der Lage. Now Reye was a disciple of K. G. C. von Staudt, whose book of the same title in 1847 was the first to develop projective geometry without using measurement. He took the ideas of projective correspondences between geometrical figures that had been developed earlier

ill

using numbers as coordinates by Monge, Poncelet, Steiner, and others, and showed that they could be basic a _ few from purely synthetically derived assumptions. His tour de force, which was not really understood until much later, was the recovery of a field of coefficients out of the internal structure of the geometry. When I came to Harvard as a freshman the following year, I was ready for Oscar Zariski’s course on projective geometry. He used some notes of Emil Artin, which were later published in his book Geometric Algebra (1957). It is von Staudt’s idea that is at the core of the theorem of introduction of coordinates presented in Chapter 7 of this book, though the proof here is based on Artin’s method, via Zariski’s lectures.

When I first came to Berkeley, I taught the course on projective geometry several times. Then for many years I focused on my research activities in algebraic geometry. It was only later that I came back to teaching elementary geometry, and then it was a course on Euclidean and non-Euclidean geometry. This allowed me to go back and study the tradition of Euclid’s Elements, the long struggle over the parallel postulate, the development of non-Euclidean geometry in the early 19" century, and Hilbert’s masterful reformulation of the axiomatic basis of Euclidean and non-Euclidean geometry in his Festschrift Grundlagen der Geometrie (1899). After teaching the course several times and accumulating notes along the way, I wrote down what I had learned in the book Geometry: Euclid and Beyond (2000). 1V

If I were to rewrite this book on projective geometry now, it would certainly come out very different. I would want to trace the historical development of the ideas before they crystallized into the axiomatic form presented here. The first stirrings of projective geometry in the early 19" century were an outgrowth of Euclidean geometry, obtained by adding points at infinity and studying properties of figures preserved under projection (cf. Poncelet’s Propriétés projectives des figures, 1822). The foundations were intuitive, with something like what we now call the real numbers in the background, not explicitly acknowledged, and free use of measurement of lengths and angles whenever that was convenient. For example, the preservation of the cross-ratio under projection goes all the way back to Pappus. Steiner made a serious attempt to bring the subject into a coherent whole in his Systematische Entwickelung... (1832). And von Staudt made a great step towards abstract synthetic projective geometry in his Geometrie der Lage (1847). Von Staudt was criticized for his proof of the fundamental theorem (FT in this book) because it used some unexplained ideas on continuity. But the critics, Felix Klein in particular, were equally confused about continuity, and it was not

until the end of the 19" century with Pasch and Hilbert that the meaning and role of continuity were really understood. The possibility of developing projective geometry axiomatically without continuity, which would apply to finite geometries as well as the standard model over the real numbers, emerged only at the beginning of

the 20" century.

There are also several further topics that I would want to include, and that I regret not writing down earlier, but which would take considerable space to develop in a satisfactory manner. One is the projective theory of conics. Here the point would be first to explain the equivalence of the geometric definition of Apollonius and the analytic definition by quadratic equations with the projective definitions of Steiner as the locus of intersections of lines in two projectively related pencils, and of von Staudt as the set of self-conjugate points of a projective polarity. Then one can easily prove the beautiful theorems of Pascal and Brianchon about the “mystic hexagram” associated with six points on a conic. To include some more modern mathematics, one might even go so far as to derive the group law on the set of points of a plane cubic curve. Another subject I would want to treat is the circle of ideas arising out of Klein’s Erlanger Programm of 1872 that “all geometry is projective geometry.” His point is to recognize Euclidean and non-Euclidean plane geometries as special cases of projective geometry, by leaving out a line, or by looking only at the set of points inside a conic, and characterizing these geometries by their associated groups of transformations. A modern extension of these ideas appears in the book of F. Bachmann, Aufbau der Geometrie aus dem Spiegelungsbegriff (1959). As far as I know, his ideas have not yet received a satisfactory treatment in English.

Vi

If you like the purely synthetic approach to geometry, you may enjoy von Staudt’s treatment of the quadric surfaces in projective three-space, extending even to a discussion of the twisted cubic curve. But | would be the first to admit that in higher dimensions, and for varieties of higher degree the synthetic methods become overly cumbersome and that modern algebraic methods are more convenient. If you follow that path, you may find yourself, like me, becoming an algebraic geometer.

Robin Hartshorne

Berkeley, CA December 2009

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CHAPTER INTRODUCTION

1. : AFFINE

Projective Properties

rotation

which

geometry are

Thus

of distance

However,

under

of the most

available

of Euclidean

what

is true

and what

Affine

geometry

Let us

start with

plane geometry,

dinary

with properties

stretchinj,,

of incidence.

translation,

devclopment

important

plane,

to see

PLANES.

or

of the theory,

and angle will play no part.

one

those

PROJECTIVE

in the axiomatic

is the real projective

(e.g.

AND

is concerned

invariant

of the plane.

the notions

PLANES

examples

of the theory

and there we will use all the techniques

geometry

and analytic geometry)

is not true.

some

of the most

elementary

facts

of or-

for our

which we will take as axioms

synthetic

development. Definition,

An

called

points,

anda

lowing

three

axioms,

on

1",

of the line

l.

plane

set of subsets,

'! 1 passes

or

affine

Al

- A3.

through

We

is a set,

whose

called lines, will

use

P '' to mean

the

elements

satisfying terminology

the point

P

are

the fol'' P lies

is an element

A 1,Given

two distinct points

one line containing both We they have

no points

A 3.

all.

Q,

there

is one and only

Q. if they are

equal,

line

m,

There

which

1 anda

point

P, noton

is parallel to

1,

exist three non-collinear

is said to be collinear

if there

1,

there

and which points.

exists

a line

P il

is not equal to

Peiiesvon

1 containing

Q.

ae

1M m

the intersection

of

1a

1 is parallel to

m.

em

¥

for all.

4

there

>

implies.

2

if and only if.

exists.

1 and

through

(A set of points

Notation, Ee

is one

passes

)

ip)

or if

in common.

Givena

and only one line

them

and

and

say that two lines are parallel

A2.

Pi ac Po

P

P

m.

P.

Example geometry,

: The

satifies

ordinary

plane,

the axioms

Al

known

to us from

- A 3, and therefore

Euclidean is an affine

plane.

A convenient way

Cartesian

coordinates,

is represented

of representing

this plane is by introducing

as in analytic geometry.

as a pair

(x,y)

Thus

of real numbers.

a point

( We write

P

x,y €R.)

if

Proposition

1.1

Definition,

A relation

the following three

Parallelism

properties

1. Reflexive

:

~

is an equivalence

is an equivalence

relation if it has

: a~wa

2. Symmetric

:

a~b>b~a

3.

:

a~b

and

b~c

F®

We

must

check

the three

Transitive

Proof

of Proposition:

1. Any line is parallel to itself,

2.1 ||

mm

relation.

|| 1 by definition

awc.

by definition.

properties

Some

ae | m,

If l=n, PRG

m

|n

, we wish

there is nothing to prove.

iineethenwm

is impossible, empty

and

set),

anette sOth

by axiom

and so

| m,

A 2.

to prove

1 || n.

If 14n,

and there is a point

and

through

pass

We conclude that

P,

which

1Nn=@

(the

l | is

Proposition

1.2

Two

distinct

lines

have

at most

one

point

in cOmmon.

For

if

1, m

both pass

nein loxy axai@ren

AN I,

i = in.

Example: an affine

An affine

plane with

Indeed,

non-collinear

Hosea

IP (Ol I,

there

P,

joining

is a line Now

would

ZR

|

m

two

distinct points

at least four

points.

P,Q,

There

is

points.

points.

ia;

1 through

the line QR

plane has

by A 3 there

three

a line

four

through

are

Call

qWeyeiees iS

parallel Q,

I PQ,

and

to R, which

passing

is not parallelto

exists

through m

by

Al.

Similarly,

R.

(1 { m).

For

if it were,

then we

have

PQ || m || | aR and hence because

PQ PQ

|| QR

4QR,

by Proposition 1.1. and both

contain

Q,

This is impossible,

however,

Hence m,

which

1 must

is parallel sons

to

ievon

PQ?

indeed

a fourth point.

Now

they meet On

meet PQ,

in some

This

hand,

a Q.

proves

the lines

(for example

point

and different

4 PaeandsS

consider

the other

m

S.

Since

from

Similarly

PQ,

S S

lies on

does

not

S 4 Ren Sm om

Ss

the first assertion.

PR

and

QS.

It may

in the real projective

it is consistent with

happen

that

plane they will (proof

the axioms

to assume

?)).

that

they do not meet.

In that case we P,Q,R,S,

and

have an affine plane

six lines

easily that the axioms

affine

passing

through

to some

line

A1-

A3

A pencil

some

1.

are

Y

to each

+x, 2 Tx,

P,

verified.

and

This

one

can

verify

is the smallest

second

x

4Tx,,

T:

and

a) the set of all lines

b)

the set of all lines parallel

case

we

speak

X

Vy €Y,

of a pencil of parallel

correspondence

X——> Y

of the set

is either

or

A one-to-one

is amapping element

of lines

point

In the

Definition.

x,

PQ,PR,PS,QR,QS,RS,

of four points

plane. Definition,

and

consisting

(i.e. arule

an element

between

T,

two sets

X

which associates

T(x) = y € Y)

4x GexXeesuch

lines.

such

thatw(x

ee -

eS

Jee

06

ee

4}

eli —

Aer es & 7

are. —

——-

+

2

4a

21

CHAPTER

Definition operaton,

3.

DIGRESSION

: A group

is a set

called multiplication,

Gl.

ON GROUPS

(Associativity)

AND

AUTOMORPHISMS

G, together with a binary

written For

ab,

all

such that

a,b,c

€G,

(ab)c = a(bc) Gace

& Se

There

exists

ACS

ILo ASA

For

each

an element

1 €G _ such

that

for all a.

a €QG,

there

exists

an element

awe G

such that aa The

The

element

element

at

Note

ba.

However,

is called inverse

we

ineve alll

permutations

If

of the set

8)» B> € G

to be the permutation

(GaGa

Hee oS

2.

morphisms

@iii,

1. Let

S

of

QG

or unit element.

a.

the product

say the group

Examples.

S.

1 is called the identity,

that in general

G 4.

onto

= aye = le

ab

may

is abelian,

CG,

be different from

or commutative

if

Als S le.

be any set,

and let

S. A permutation

is a

are two permutations, obtained by performing

G

be the set of

1-1

mapping

we define first

of

S

88> IG,

Bo: then

8):

Sp,

Let

of

(g,8,) (x) = g,(g,()).

)

C

and let

bea

C, i.e.

configuration,

G

be the set of auto-

the set of those permutations

of

C

which

jay

send lines onto lines. morphisms

is written

8)» 8>> by performing

, and

first

85

of two auto-

£8

Again we define the product

then

8):

This

group

Aut C.

Definition: to another,

A homomorphism

is a mapping

of the set

@ : G, =

G, to the set

G,

of one group

G,

such that

(ab) = p(ayp(b) for each

a,b

€ G)-

An isomorphism which

is

1-1

and

subset

Note

S,

let

subgroup

of

G

Let

x €S,

bea

such

condition

Example.

set

Let

HCG,

this

group with another,

is

a homomorphism

onto.

Definition. empty

of one

G=

group.

A subgroup

that for any

implies Perm

a,b

€H,

of ab

G

is a non-

€H,

and

“bs Seale

hk

1€H.

S,

the group

and let

H=

{ this map

is onto.

gH

So suppose

by

h }|—> gh. By definition of gH,

hy»h, € H

have

the same

image.

Then

gh, = gh..

Multiplying on the left by cae we deduce Corollary

subgroup

3.2

Let

Proof

of elements

: Indeed,

as

finite

and

1€H.

Finally,

note

that two

suppose

group,

and let

H

bea

we

of

H

have

g€G,

of

H).

the same

then

gH,

g'H

and g'H

are

either

have an element

number

g € gH,

equal,

in common,

of

since H.

or disjoint.

namely

have

has ,» we have

y = gh" = g'h'h the opposite

result follows

g=

Sania

th" Gigi pe

inclusion,

€ asl Ogric piri:

so they are

equal.

immediately.

ee

a Example. of the group

Perm

subgroup

G

of

x.

ghi= othe.

y € gH,

By symmetry

The

and

If

of left cosets

is the union of the left cosets

cosets

on the right by

for any

(number

all the left cosets

Thus G

gH

X= Multiplying

#(H)-

H, by the proposition.

g=g'l,

Hence

bea

. Then

#(G) =

Indeed,

G

h, =h, .

Let S

leaving

S

bea

finite set,

of permutations x

H={ g€G

of

fixed:

| g(x) = x}.

andlet S.

Let

G

be a subgroup

x €S,

and let

H

be the

24

Let

for

g' = gh

Indeed,

Then for any

g(x) =y.

and suppose

g €G,

2 (x)e=-y-

g' €’gH;

h @Heyso

some

g'(x) = gh(x) = g(x) = ysuch that

element

be some

g'' € G

, let

Conversely

Then

g(x) = y.

gg" (x) = ey) = x, so

gg" €H, and =]

GaP

ee

W

eaters

Thus

gH = {g'€G | g'(x) = y}. It follows

that the number

of left cosets

of point in the orbitof x

under

y €S

for some

such that

y = g(x) sag (ey i

Definition: is transitive

element

#(G)

=n!

3.3

Let

orbit of

So we

(orbit

element

is all of

example,

#(G)= Corollary

*

is equal to the number x

is the set of points

conclude

x)

GC Perm S of permutations

if the orbit of some

So in the above

The

H

g €G.

H(H)

A group

that the orbit of every

Then

G.

of

if

#(H)S

bea

G

is the whole

of

of a set S.

S

It follows

S.

is transitive,

#(S). set with

n elements,

and

let

G=

Perm;

25

Proof:

permutation, let G

x €S.

By induction

so

on

#(G)=1.

Let

H

is transitive,

n.

Solet

be the subgroup

since

one

If

n=

S

1, there

have

n+

is only the identity

1

of permutations

can permute

x

elements, leaving

with any

other

x

and fixed.

element

of S.

Hence

#(G)= But

of

H

S,

H#(H)*

is just the group

so

#(H)

=n!

#(S)=(n+1)°

of permutations

#(H).

of the remaining

by the induction hypothesis.

n

elements

Hence

#(G) = (n+ 1)! q.end.

Later of

in the course,

automorphisms

In particular,

is equivalent

"large

we

of a projective will

show

in a sense

will content

simple

plane,

and

that the axiom

to the statement

enough",

moment, of a few

we

we will have much

that the group

certain

configurations.

with

of its subgroups.

P 5 (''Desargues

theorem'"')

of automorphisms

which will be made

ourselves

to do with the group

precise

calculating

later.

is

For

the

the automorphisms

26

Automorphisms

of the

Call the plane its seven

points

(this suggests obtained

7.

as

Plane

of Seven

Points.

Name

A,B,C,D,P,Q,R

how it could be

by completing

plane of four points.) are

Projective

the affine

Then its lines

shown.

Proposition

3.4

G = Aut 7 is transitive. Proof:

We will write

down

some

elements

of

G

explicitly.

a = (AC)(BD) for example. change

This notation

B and D''.

More

means

generally

"' interchange

A and C, and inter-

a symbol

(A)A,,-. ee) 1e

means

''send

A) to A 2?

Multiplication on

the

of two

right first,

TN > ay AN 37° Sr RAG

such

then

the

symbols next

on

r=]

to Aw

is defined the

right,

and

A. to A) ai :

by performing and

so

the one

a

b = (AB)(CD) Thus

we

see

already

that

A

canbe

sent to

ab =(AC)(BD)(AB)(CD) ba = (AB)(CD)(AC)(BD) Thus

we

can

also

send

A

to

Ds

B

or to

C.

= (AD)(BC) = (AD)(BC)

= ab.

We

calculate

Gin

Another

automorphism

is

c = (BQ)(DR) Since

the orbit

contains

Q

of

and

A

already

R.

Finally

contains

B,C,D,

we

see

that it also

d = (PA)(BQ) shows

that the

orbit

of

Proposition

of

m

leaving

P

Proof

of

A

3.5

H

is

Theorem

3.6

non-collinear

of

; which

elements

G

is

Proof:

Let of

KCH K

leave

on the set of

K

easily.

Given points

and

carry

Q

{A,B,C,D},

Hence We

K

is transitive.

be the subgroup

of automorphisms

is transitive on the set above

are

all in

H,

mw -{ P}.

so that the orbit

=q7-{P}.

two sets

of

G

A), A.A,

m7, there

is one

and

A,'A,'A,!

and only one

of

automorphism

. The number

of

7:°6-4 = 168,

We

is uniquely

l,a,b,ab.

a,b,c

the above

be the subgroup P

H

so

A) to A) 1 ; A, to A, ', and A. 3 toA.'3

sends

in

HCG

Then

that

7,

{A,B,C,D,Q,R}

three m

is all of

Let

fixed.

: Note

under

A

leaving

fixed,

a,b€K. by where

is just the group

conclude

from

HIG) = #(H) =

Q

one

fixed.

they also leave

since

determined

analysis

step farther

Therefore R

fixed.

consisting

the previous

K

the point

elements

is transitive

an element A,

as One

of the four elements

discussion

#(H)* #(n) #(K)>

since

On the other hand, it sends

as follows.

#(n - {P}),

that

sees

28

whence #(G)

The

SO A

first statement

statements,

LO Ge

of the theorem

but it is a little tricky.

1) Since

G

is transitive, =

2)

eee

Again

We

we

follows

from

do it in three

can find

g €G

the previous

steps.

such that

!

since

G

is transitive,

we

can find

8) € G such that

g,(P) = A): Then =

We have

supposed

that

'

A 4 A,, and

=I

8 (A,) and are distinct from element

h €H_

P.

such

But H

Ay 4 A,!

=!

checks

then

is transitive

7 - {P},

so there is an

a.) rs

that

al

the property '

=e

!

gi(A,) = A, '

3) any

on

that

La

has

;

(gg,) - (A,')

hg, Ei (A,)) gevtan bc= (eg) 22) One

a. AMatove

Thus

part 2)

two distinct points.

ss

i=

shows

that any

Changing

two

notation,

distinct points we

write

g

can be sent into

instead

of

g',

so

29

we

may

assume

g(A,) = A, aa

1

g(A,) ak = A, ! ;

Choose

8) € G

such that Ai

8, (P) 8, (Q)

by part are

2).

Then

=

Ay:

since

non-collinear,

we

8,

-1

ALA

deduce

A, as

are non-collinear,

that

(A3), (gg)

P,Q

-1

A

MOIS t

a"

of the points

(A,')

are non-collinear.

In other words,

{A,B,C,D}

there is an element

. Thus

, and each

and

these

last two points are

k € K

in the set

such that

k(g, (A) = (gg) (A, One

check easily that =

8 = gg)kg, is the required

element

of

G:

g'(A)) = Ay!l '

=

'

g ' (A) -= A, For

the uniqueness

of non-collinear

second

triples.

in

of

in

7.

G

of

G

let us count the number

The first can

and the last in

Since the order

transformations

é

of this element,

points

6 ways,

'

is

4 ways.

168,

be chosen

Thus

in

there are

of triples

7 ways,

168

the

such

there must be exactly one

sending a given triple into another

such triple. q.e.d.

30

A Similar

analysis

Plane

Affine

ofthe

Automorphisms

shows that the group of automorphisms 432,

points can be taken into any three non-collinear

SS,

>

>

t

non-collinear

and any three

SO

xX,ed

9: 8-6 =

has order

points.

9

of

9 points

the affine plane of

of

points by a unique

element of the group.

show

that there

is a unique

automorphism

of non-collinear

A,,A,,A, ae aes S

given triple

do this for each of the triples the inverse becomes

Here we of a projective

is,

that

(x) +53)

plane.

a triple

X€IR.

P,Q,A

For

points.

and A), A,

with the second.

of the Real Projective

study another Recall

important

to

intoa

then one

can

A! , and compose The

proof thus

and

A

Plane

example

(Ax, AX,» AX3)

not all zero.

mere

oaee

Ge Chea

Smo

not all zero,

re present

SKS)

plane is defined

coordinates

line is the set of points which

1

of the automorphisms

that the real projective

of real numbers,

aes

as ER,

sending

: A point is given by homogeneous

That

be sufficient

it would

simpler.

Automorphisms

nN e o,

A), Aj,A3,

of the first automorphism

much

as follows

3.6,

In proof of theorem

:

Note

and with

the

same

(x)+X5%3) the

.

convention

point,

for any

satisfy an equation

of the fc

31

Brief

review

is a collection

each

of which

Ae

Bayar

of

of matrices

n

may

real

take

numbers,

values

21

21

{EYAY

the first subscript

is defined

Aga C=

indices

, say

i,j,

Hence

grees pare } . The matrix

is

MW

a

[In

a

Zz.

apy

determines

Ge

2

2

ann

the row,

and the second

subscript

the column.

The product

where

n.

ij

nl

n)

1 to

by two

of real numbers

in a square:

es

order

nXnmatrix

indexed

from

ll

determines

An

ees rAapr Agger eAA

usually written

Here

.

i

of two matrices

A = fa;

Amys)

jah =

to be

Deo

G

and n

it

peetartoas

by;

eae

j

AN

ein bj ave Dies

in

nj

7a

ter)

(both of

B24

There

matrices

to

is also

a function

IR _, which

Da.

determinant,

is characterized

If A,B

are

two

the

= det

A’ det B a

°

Note

det(C(a))

incidentally

a multiplicative Ne

(AB)

i

is associative.

Ze

One Gi=

can prove 7A

A

)

ool

has

I = C(l) be haves

the following facts

(BG)

(In general

A matrix

Oo

= a.

that the identity matrix

identity.

nXn

by the following two properties

For each a € IR, let G(a) = ( ass Then

matrices

set of

matrices,

det(A:B)

D2.

from

ise.

as

:

multiipliicationrmon

it is not commutative. )

a multiplicative

inverse

As

if and

only if det A 4 o. Hence group

under

the set of

n Xn

multiplication,

3.

Let

of simultaneous

A=

linear

matrices

denoted

(a;)

eV) IA

Tee

es

Iipaoanld5 1 n

For

Waa

proofs

det

and

A4o

formsa

.

consider

the set

has a solution

se 9|s)

nn

det A 4 o , then this set of equations

if this set of equations

det

equations

pee aig maole ei

If

with

GL(n,R)

be a matrix,

Bore Gr Gk ees SP Goo J War72 mall I

by

A

n

has

n

5

a solution.

for all possible

Conversely, choices

of

A 4 OF

on these

statements,

We will take them for granted,

tefer

to any book

and use them without

on algebra.

comment

in the

35

rest of the course. and

2.

Because

linear equations

(One

can prove x peo

to say

is the same

x,

easily that are

Now ™

let

(a)

We

define

goes

64

eee)

b

n

be a

be the real projective

of that system

of

b

A

A=

1

by

x

x

a solution

from

as saying that

x

A

3 follows

n

3 X 3 matrix

plane,

a transformation

)

of real numbers,

with homogeneous

Ty

of

m

as follows:

and let

coordinates

Xs X42Xq-

The point (x),X513)

into the point T

4 () »

X52

3)

-_ a

i] (x)

i] aay)

' ere

)

where

1

eS!

Soy

1

A

1

Proposition with plane

detA

4 o,

Siew

GEES) I ta PR ho! Tobyes pies

ape a)Eat

3.7

then

ieee?

If

TA

yrer

A

oncete

isa

3X3

matrix

is an automorphism

of real numbers

of the real projective

7.

Proof 1)

If we

(x,', 5's! defined.

: We must observe

We

replace

(x, »X5 »X3)

) is replaced must

also

several by

by (Ax,', Ax,"

check

that

things. (Ax), AX, AX)

, then

A%3') , so the mapping

X)',X5' 9X3"

are

is well-

not all zero.

Indeed,

34

in a matrix

notation,

1 A

1 -

x,

=

X5

3 ee

where

(x5

)stands

"3

But

since

the left by

for

=

fo)

x5

fe)

fo)

Gia

(@)

(e)

A

has aninverse

A

, we

1

the matrix

fe)

1

det A4o, A

as

!

a7

_,

aa and so multiplying

on

have

1

(x) = A (x') (where

are

(x)

stands

all zero,

the

is a well-defined

for the column

x,

are also all zero,

map

of

.

2) inverse

into

to

we

must

(x')

TA

check

shows

must

that

that

T,71

be one-to-one

TS

takes

lines

ne and surjective.

into lines.

let

be the equation (x),x553)

+

C5X>5 =e

of a line.

We

Let

A

=]

= (b..) .

J

C3%3

must

satisfy this equation

line.

TA

-]

(x) =A _

Ty » hence

Finally,

a

(= etc. ) So if the x," x which is impossible. Thus

7.

;

Cx)

the new

m

expression

mapping 3)

Indeed,

The

vector

=o

(*)

find a new

(*), Then

its image we

have

line,

such

(x)', x

(2

that whenever

',x,')

lies on

55

for each

i.

Thus

if

(x). x

2

»X3)

satisfy

(*) . Then

(x)', 5's x3")

will satisfy the equation @ if APP oyBj se Uy) ) cs

which

is

(Zc.b.)x, This

te 0 co (& b 25%) ') ) + c,(% D3 5%; UN } =

is the equation

three

+ (£ cib,)x,

!

of the required

t

earn + (Z cb. .)x, \ =O.

line.

We

have

only to check that the

coefficients Se De= )) any

for

j=1,2,3,

the argument

are in

ste ste

(**)

not all zero.

1) above

: The

fomorsca)

uAue

:

But this argument equations

is analogous

(**) re present

to

the fact that

(cls estncs! |

1 ae

t

'

'

where

(c),¢5+¢3)

Multiplying terms

by

A

which

1%

A4o

are

equal

era.

and

fo)

fe)

fo)

re)

oO

fe)

since

that the

Cc." were

3.8

det A'+ 0.

if and only

Let

A

c,

can be expressed

all zero,

and

of

ce rai; for alin (5%

is

c. would

be all

A'

7

be two

3x 3

Then the automorphisms

if there

the

in

(*) is a line.

is an automorphism

Proposition

det

if the

is impossible

Hence

ed a |

on the right shows

of the c.! . Hence

zero,

=

sh

a real

number

T,

ji : Oo,

matrices

A

and

such

with

Tas

that

of

w

A” = XA,

36

the x, ! will just be changed by

of

Ty

(o,0,1)

on four

Tat

and

respectively

‘

7,

of

specific points

. Now Cl

@

fle

=;

:

»P, m, Ehavel

2

(©.

411

°

=

a5)

°

points of

P,P

points

(onlyvole

as)

T(P,) = A's

two

(l,0,0),

aay

fo)

these

namely

ll

T ,(P)) ef

Now

study the action

We will then

fir

. Let us call these

(1,1,1)

, and

becuase

).

TA ay Ah

suppose

Conversely,

T, A =T AY

X,

is sucha

if there

Clearly

Proof:

sets

of coordinates

7 , so there must 1

a's)

are

supposed

exista

AXECR,A

Tat

to the points

to represent

4 ©,

such

the same

that

=

hee Se '

=

!

=

a'o1 = §y@21 Say Similarly,

the numbers

applying

Ea

TA

X, €IR i

and

and

X,¢€R,

x

a'i2 = 2412

aie

cis

223 = 4423

=

2133

to the point

Q.

mn

!

oA,

s

'

Ty

!

such that

_

2°32 = +2432 apply

t

40,

'

222 = 2422

Now

both

P 2? and

43433

We find

P, » we

find

37

: A

ied 2s

1

=

ao) + a> + a53

1

Similarly

number

for

Tat EA

pf 40

az) + az> + a33

vaing

such that

T ,(Q) = T,1(Q)

T 1(Q)

=p-

,

so there

T , (Q) .

Now

is a real

using all our

, we find

equations

aj,0,> #) + a0, - #) + 2,0, - #) = 0 ao,( - H+ 2,00, - #) + a,4( 4-H) = 0 BU) el aes In other words,

Hence

the point (dj “Hs A5- Hs A3-H)

Ay = A, = A3 =u.

all zero,

cannot

Is sent into (0,0,0)

(We saw this before:

be sent into (0,0,0)

by

A.

.

a triple of numbers, Hence

7H

ENO

Au

and we

are

done.

group,

written

not =—Or

and A37H =) SOMA

NAM

Definition: PGL(2,IR)

of

Hence matrix

A,

A'

A=

KR,

is the group

T, for some

3X

3 matrix

of

m.

Then

PGL(2,IR)

of real numbers,

represent the same

of four points,

general linear

over

fa;,)

Theorem

plane

2

an element

a real number

Ay = A, = r, =p,

The projective

order

n of the form

EW Lie Lemna

element

) 40 such that 3.9

no three

Let

A with

of

det A4io.

is represented

with

by a

3 X 3

det A4d o , and two matrices

of the group

if and only if there

is

A! = QA.

A,B,C,D

of which

of all automorphisms

are

and

A',B',C',D'

collinear,

there is a unique automorphism

be two sets

in the real projective T € PGL(2,IR)

such

38

that

-T(A)rarA)..

T(B)ieeBo

Proof : Let

P),P,,P3,2

(o, 0,1) and (1,1,1) prove

the theorem

suppose @~

we

can

sneditto.

io | sends

considered in the case

A,B,C,D,

. Then

A,B,C,D=

P

1D)

points

senditto

(1,0,0)

(0,1, 0)

it will be sufficient to P,P,

P,, Pi»P3,2

~

=D

Q.

Indeed,

into any other.

A',B',C',D'.

Let

Then

into A',B',C',D', have

homogeneous

coordinates

(a),a,,2,)

A (c), Cos c3) and (d},d,, 43) » respectively.

find a matrix such

above

andlet

A,B,C,D

gard

be the four

send the quadruple

Let A,B,C,D,

(b),b,,b,)

ar (C) iC

(e;5) , with determinant

Then we

4 0, and real numbers

F

must

A,U,V,p

that

T(P,) = Avani) Ce

Aa, See

hese 7S

T(P,) = 13) ad. Oe

Ub. Es t.5

mS ha Pa. 3

T(P,) = C ive.

vere

Pedy

E.(@e) t= Dias

joysl. =

ut ti2+

Clearly it will be sufficient to take

p =1,

43

2e3

gs Ib.

and find

,u,v 4 0 such that

Aa, + ub, + vey=d

Aa,

Lemma (a),a,,a,)

collinear

I

Au 2

ra, + ub, Tyce |

jon3

3.10

Ub, tuc,

+

Let

‘ (b),b,,b,)

if and only if

A,B,C

be three points in

/ (c),¢55¢3)

respectively.

m7, with

Then

coordinates

A,B,C,

are

ded

1

2

3

b,

b,

b,

acon Proof of lemma. if there

is a line,

leleede

h. not all zero, of 4 o

A,B,C.

The points

hes?

that

seen

equations

non-trivial

=o

(#3.

i.e.

such a set of equations.

collinear

if and only

aeers «3am is satisfied by the coordinates

that the determinant set of numbers

on p.

if and only

solution,

are

say

such that this equations

We have

fear!

A,B,C

Ceahiex

22

if and only if for each

of linear

boo

with equation

hex

ato

30)

a unique

solution.

Now

they exist

our

h, are

set

It follows

b. = 0, the set of equations

not all zero.

Therefore

(a;;)is

(b.), the corresponding

have

if for

of a matrix

has a

solutions

of

# the determinant

above

case,

non-

is zero. Proof collinear

of theorem

, hence

rX,U,V,

we

can

solve

are all

equations

. In our

A,B,C

are

by the lemma,

det}

Hence

, continued

+ o.

say that

1

1

1

a,

b,

cy,

te

Acryl

the equations

Indeed

above

, suppose,

+0 :

for

say

low). (see note below)

)\,yw,v . NowI

X=0.

claim

Then our

40

and

hence

b

det

which

is impossible Note:

1

c

1

d

b,

cy

We

eee

1

d,

, by the lemma,

We must

use

=o

since

B,€,D

of a matrix

obtained by reflecting One

can see

A=

the entries

of the trans-

of the matrix itself.

(a;;) tobe

of the matrix

A

We

= (a; ) aeltiis

in the main

diagonal.

easily that

(AGaB); Now

not collinear.

the fact that the determinant

pose of a matrix is equal to the determinant define the transpose

are

consider

eB eeAt

the function from

the set of matrices

to the real numbers

given by

A—— Then

this function

therefore

det (A).

satisfies

it is the same det(A)

define

conditions

as the determinant

Dl,

function

D2

on

p.

30,

. Hence

= det(AT)

So we have found We

the two

),p,U¥

all 4 o

which

satisfy the equations

above.

ty by the equations Aaa 1

i

eis wey AE Then

A,B,C

(t,5) is a matrix

, with determinant

are non-collinear

!) , so

4 oO (again by the lemma,

T » given by the matrix

(t..) , qJ

since

41

is an element For

of

of

PGL(2,IR)

the uniqueness,

PGL(2,IR)

3.8 , the matrices

differ by a scalar

sends

suppose

which accomplish

Proposition

PGL(2

which

multiple,

PP

that

T

2

»P3,2

and

T'

LOMASDa Gre

are

two

elements

our task.

Then by the proof of

(t;5) and

(el)

and hence

defining

give the same

al

element

meleee

of

,R) la Go Clo Our

next main

to be a subgroup projective

of

plane,

theorem

Aut,

will be that

the group

is actually

PGL(2,IR)

, whichwe know

of automorphisms

of the real

equal to it:

PGL(2)RR) =Autr The

preliminary

statement

,

will follow

which

A

have

field is a set

some

F, together with two operations

the following properties.

ipale

ey ae is) = |e)ar A

V2 igi) Ia

Bee

(a+b)

c)

Fe.

=

F 4,

¥Va€F,-—]

In other words,

F

470 21a. + (b+

o€F

NE Alp jog (o © ID

SuchsthatmansOl=sOLtsde= ia.

-a€F

is an abelian

YVacrF

such that a+ (-a)=0.

group

under

addition.

IP By,

ab = ba

Vieapibacebie

If (),

a(bc) = (ab)c

Vilas bac 16 by

Fie

after

results.

Definition + , *

and proof of this theorem

aglio.a uch thatwe ea) Grasa

¥Vae€F

42

gtiohta cor see

F 8.

V

ip Me

a(bEtnc ssa barca Cc

normal to assume

also

Definition: 1-1

mapping

o

ofl.

If of

F

F

Via

a group

form

elements

So the non-zero

under

a‘

=1.

a

ee

Decree

(it is

multiplication

).

is afield, onto

such that

an automorphism

F, written

a——>

aoe

of

F

isa

such that

(a+ b) 7 = a7 + b? (ab)? = a’b”

fori

DCE

Bua Galt followeltha

Proposition projective

3.11

plane which

P, = (0,0,1)

and

leaves

»

(Note we

)Then there

such

be any automorphism

fixed the points

Q = (1,1,1) .

be given by a matrix. of real numbers,

Let

choke som eels conte)

is

of the real

P, =1(1 50,0) 40h

do not assume

an automorphism

that og

2 ©

(ose

ole

can

of the field

that

(m1 p43) = (79 XQ”) for each point

(x, x5.)

Proof:

of

f.

We note that

©

must leave the line since it contains

xX, =0 fixed

Ps

and

P.-

We will take this line as

the line

at infinity,

the affine

plane

and consider

x, 40.

R eS

A=T-

{x, = o} +

R= (0,1,0)

F,= (9,0,1)

43

Our

automorphism

automorphism

@

then sends

of the affine plane.

A

into itself,

We will use

affine

and so is an

coordinates

rw \

= %/%,

vas x,/x,. Since

@~

leaves

horizontal leaves

lines,

fixed

hence

an

asc

Pada nex

Let

P. and

vertical

P, = (0,0)

Q = (1,1),

and

lines.

lines

Besides

into

that,

Y

4

it

AS

y

(a”, at)

bea Then

is also on the X-axis

we

into vertical

1 A an ,

(0,0)

(a7, 0)

so it can be written-as

Thus

lines

send horizontal

siand thease ot)

X-axis.

for a suitable

P, , it will

it leaves

(a,0o)

point on the @(a,o)

fixed

element

ery)

(\,¢)

Tee 0)

Xx

aT ER.

a mapping

define

(ope Ast SS and we

see

immediately

The line fixed.

Vertical

x =y lines (a,a)

that

o7

=0

and

is sent into itself, go into vertical = (line x = yf)

lines.

vee I

because Hence

(line x = a)

is sent into

(avira) = (line x = y) /f (line x = a)

P,

and

the point

Q

are

44

goes

into itself,

go into horizontal

lines

horizontal

Similarly,

so we

deduce

lines,

and the

Y-axis

that

g(o,a) = (o,a). if

Finally,

andi

is any point

(a,b)

, we

deduce

x=a

the lines

by drawing

ve—s beatae

Glarb) =e(age balm Hence

the action

o : R—~>

by the mapping

By the way, the

X-axis

of

onto

since

itself

ina

Now we will show a,b

€R,,

construct

and

consider

the point

@

|

on the affine

which we

plane

have

1-1manner,

that

so

g

of

the points

(a,0)

- to - one

of

, (b,o0) onthe

(a+ b,0) geometrically

as follows

it must

A,

is One

is an automorphism

go

R.

:

(0,0) (a%o)

(bo)

(LF0) (a+b,0)

(orb?0} = ( (arb) o )

send and

onto.

Let

X-axis.

(0,1)

(a,0)

expressed

constructed.

is an automorphism

@

is completely

We

can

45

Now

@

Ie

Draw

(des

Draw

She

Get

4,

Draw

the line joining

Gye

Draw

the line parallel to

6,

Intersect

sends

and it sends

the line x=a (a,l)

the line (b,o0)

:

and parallelism.

y=1

into

Hence

by intersection

5.

y=1

@

1,2.

(0,1) and

with the

4. through

it sends

. It preserves

also sends

(b,o).

(a,1).

X-axis.

intoitself;

ib? 0)

of

x=a

joins d

(a+ b,0) into

into

x=

and intersections, Oo

(a +b

0

,o).

Therefore

(ab) Ooj='a By another

geometrically

(0,0)

bs

construction,

from

the points

(1,0)

(4,9)

we

can obtain the point

(a,0) and

(b, o/

ae

(ab, 0)

(b, 0)

(ba, 0)

46

Le

Draw

x=a

ae

Intersect

Se

oings

4,

Draw a line parallel

ye

Intersect

x = y

tom

(ob. Oo).

(Lele

with

4.

(a,a).

to obtain

we

(a,a).

through

3.

to

the

fixed,

(1,1)

@~ leaves

Since

with

X-axis. similarly

see

construction

by this

that

tabhaciacb oe Hence

g

is an automorphism

Now of ©

ona

we

return

point with

GaseGul

of the line

with

x3 = o

(x),x,,)).

(x). x5) . Oo

are we

ule

Hence

(x, nes 2

oO

to the projective homogeneous

x3 = 0,

(which

Now ©

of the field of

this

we

this

is left fixed by

of itis

(x) +53)

point as

@)

point is in

4)

,l1) . Therefore

m , and study the effect

plane

coordinates

write

latter

real numbers.

intersection

and the line joining

A,

, whose

by intersecting

the

:

and has

affine

homogeneous

the transformed

(0,0, 1)

coordinates

coordinates

lines,

find

iy

a: Pee

P(x), x5, 0°) = (x) »X5 20). Gasere. affine

x, + o

the point

(x),

»x.,)

ees

is in

A-,

and has

coordinates

y = So

. Then

(x,y)

oO

/x.. ey

= (x wy) = (9/257, x, Cy

) .

.

This

last equations

because

47

g is an automorphism , so takes quotients ~(x,y)

has

homogeneous

coordinates

into quotients.

(se x5

Go

pcs)

Therefore

and we

are

done.

q.e.d.

Proposition numbers

proceed

Let

can prove

that

go

rational

4)

n?=n

for any positive integer

x = we

Ons O

Tomson)

€R,

Then

xo#

be a sequence

Then the sequence

converges

Indeed,

Bele

,

40> sO;

DEES)

Conversely

because

x7 >o

is an

the inverse

. Therefore

:

be a real number.

la -a| < €. n

n.

fixed.

(a/b)? = at be . Hence

is also an automorphism.

to

, by induction,

atom

Ab 2 so) xm (Gc)°e 1 El(b “1.2ye)

5)

We

fixed. If x

en ose

also

(ait b)? mao 4+ b° 4s Hence

MEE Js) 4 Oy,

numbers

of the real numbers.

ia.e

all the integers

leaves

such that

be an automorphism

T(t

3)

g

of the field of real

steps.

ihe =

2)

of

@

in several

1)

Hence

only automorphism

is the identity automorphism. Proof:

we

3.12 The

this

Using the previous

says

of real numbers,

{a} VE>o,

results,

converges = N

to such

and let

a that

this is equivalent to

a

{ae} n>N»=

48

n

inthe

€>o two

a? ,

to

converges

a

Thus

g

which

qe €@,

rational numbers

we

is any real number,

a€IR

If

6).

So the

number.

equivalent.

are

conditions

is a rational

€

if

co _ €

so

definition,

rational

to consider

it is sufficient

Furthermore,

a°| < ee, .

Ja i=

we

a” Sed

Then

a.

to

of the limit.

a” , by the uniqueness

a=

so

and

converges

of

can find a sequence

is the identity. Gaceds

PGL(2,IR).

y(Q)=D

T(P,)=C,

(Possible by theorem

T(Q)=D. m1

which

P

leaves

an automorphism

of

Proposition

it can be written

3.10

that

such

T € PGL(2,RR)

a

Choose

.

o(P,) = AS x

AP) =) Ney o(P,) aa

Let

» € Auta.

Let

is already

o € Aut 7

that any

Itis sufficient to show

Proof: in

IR) = Aut 7. PGL(2,

3.13

Theorem

1

T(P)) As

3.9).

»P,P3,

T(P,) as

fixed.

is

tly

Then

Hence

by

(5553) —> (4472737) for some

automorphism

the identity,

so

Teo

o of

IR.

But by the last Proposition

is the identity,

so

g

is

» = T € PGL(2,R). qread.

Note

that specific properties

into Proposition over

an arbitrary

in Chapter

6.

3.11. field.

The

of the real numbers

rest of the argument

In fact,

we

will

study

would

this more

entered

have

only

been valid

general

situation

49

CHAPTER

We

4.

will now

can deduce

from

ELEMENTARY

SYNTHETIC

study the properties

the axioms

PROJECTIVE

of a projective

P1-P4

GEOMETRY

plane which

(and occasionally

P 5,

we

P6, P7

to be defined)

Proposition set of lines

in

7,

Let

m

bea projective

and define a line*

in

7*

plane.

Let

m*

Then

f*

is a projective

m . Furthermore, Proof

will call them alee)

ID Gy a3y

: We

this

7

verify

P1*,...

P 4*

are

P*,Q*

If

of 7:

7,

it says,

statement

into

P 2%

says

point *

at least one

eS

there

fixed

called the dual projective plane

P 5 ,sodoes

the asioms

7.

P1l-P4

to distinguish

and

If 1*

two

for

them from

m*,

and we

P1-P4.

containing if

are three

two

distinct

from

m*

are two lines* In

three

77,

this

which

P 1 and

in

in

lines of ty iy

Nabeet

points*

7.

( We

7, levine

1D Zhe

m*,

from

follows

lines

translate

they have at

that two pencils

says

non-collinear

non-concurrent

Wesel,

containing

follows

line in common,

are

and

, then

If we

Q*.

P*

are

1,m

m*

of

points*

This

in common.

There

distinct

of lines

pencil

is a unique

a unique point in common.

least one

some

BE? -

is a unique line*

then there

have

satisfies

must

P 1*,

there

if

plane,

be the

to be a pencil of lines

7.( A pencil of lines is the set of all lines passing through

point.) of

in

4.1

of lines

Pl.

in

m*.

say three

This

or more

50

lines are are

concurrent

contained

linear are

if they all pass

in a pencil

points

A,B,C.

This

Then

says

Every

one

By

sees

that every

1 be some

soya

JP,

easily

i.e.

are three

that the lines

in

m*

pencil in

@

has at least three lines.

if they

non-colAB,AC,

BC

Hence

IP Zh,

points

A,B,C.

has at least three

Cs

Let the

line not passing

the pencil of lines

y= IP,

has at least three points*

P, and

Aheesia lay

at least three

Il joe

through

lines

a = PA,

IC.

Now

we

will assume

P5,

Desargues

axiom,

and we

wish

to

.

prove P 5*

points* collinear,

of

Let

m* , such

and

points *

are

P 3 there

point,

line*

pencil be centered at

P

of lines.)

some

not concurrent,

P 4%

let

through

collinear,

O*,

A¥, B¥,

that

A*, B¥, Cx ;

C*,

O*A*A'*

AX',

B*'

Cx'

| OXB*BIX

'*, B'*, C'*

are

not

be

seven

, O*FC*C'X collinear.

distinct

are Then

the

51

Translated

Let o,b,b'; not

into

#7,

this says the following

0,a,b,c,a',b',c'

0,c,c'

concurrent.

be seven

concurrent,

are Then

the

lines,

and such that

such that a,b,c;

0,a,a'; a',b',c'

lines

p = (a.b) U (a'.b')

q = (a.c) U (a’.c’) eS (where

U denotes

intersection

(1d, QU

(se)

the line joining two points,

of two lines

) are

concurrent.

and

.

denotes

the

are

52

in such a way

of the diagram

we will label the points

statement,

this

To prove

So let

as to be able to apply P 5.

@l-somaras IN

Ox 10). \o

AO

CRECe

1B) S Bigle

Bi=saac Geaa Sb:

Gs

O, A;B;C;A°,

Then we

conlude

ale B,C’

satisty

ot)

the hypothesis

FS,

90

that

Pe=sA BAL

Beep ac

Ol

Gasp eac.

A GrAl

ReaD GarbuC ll——p Cd] are

collinear.

But

PQ =r,

so this says

that

p,q,r

are

concurrent.

q.eé.d. Corollary

4.2

about a projective

PAS

Pea

(Principle

plane

(respectively

obtained from

S

of Duality)

fm , which

can

mis —elon>)

ae

by interchanging

. Let

be proved

Then

the

S from

'dual''

be any

statement

the axioms

statement

the words

point


eeLine

lies on




passes

collinear




concurrent

intersection


join

through

etc.

S*,

53

can also be proved from Proof:

Indeed,

dual projective

S*

plane

5),

as we

have

Remarks: by sending

a point

isa point of

projective 2. m7 .

I believe

order

There

P

of

9(10

one

Definition: points

and

three

of which

m

P1*

see

the plane

ona

t—~>

into the pencil of lines

can

line)

- P 4% ( resp.

through

plane

7**

P , which

finite projective

, drawing

planes

sides

of

of this.

points

all six lines

of opposite

to the plane

is the configuration

by taking four

of the

.

will give an example

and then taking the intersection

, obtained

we, need not be isomorphic

obtained

collinear

7**

easily that this is an isomorphism

A complete quadrangle

six lines

are

applied to the

P 1 - P 4 ( respectively

map

of the non-Desarguesian

points

seven

is a natural

with the projective

However,

from

in turn follow from

1.

m

it follows

S

P 1 - P 5).

just shown.

7** , One

plane

P 1 - P 4 ( respectively

is just the statement

1* , hence

P 1l* - P 5*) . But these P1-P

the axioms

of

A,B,C,D,

connecting

them

no ,

:

1D & IN3} 5 (1D)

The

points

seven

points

CeebD

Rea

Dea

P,Q,R

It may quadrangle

Ors

are

happen

are

).

G

called diagonal points of the complete

that the diagonal

collinear

However,

points

(as for example

this never

P,Q,R

of a complete

in the projective

happens

quadrangle.

plane of

in the real projective

plane

54

(as we will see below) pathological

, and in general

phenomenon,

, it is to be regarded as a

hence we will make

an axiom

saying this

should not happen.

P7. quadrangle

Then T

(Fano's Axiom). are never

4.3

Proof:

A,B,C,D

no three

of them

(0,0,1) Hence

The

are

of the real projective

points

plane

it will be

sufficient to show that the diagonal points of this com-

collinear.

(tT 0)",

They

are not are

(l,o0,1)

sands (osleaieec

see if they are

collinear,

real projective

,

plane

be the vertices

collinear

(1,0, 0) (0,1,0)

plete quadrangle

of a complete

collinear.

Proposition Let

The diagonal points

7

, so we

which

, (1,1,1)

of

satisfies

a complete

P 7. quadrangle.

can find an automorphism

carries

A,B,C,D

respectively

into the

( by theorem

3.9)

.

BHD)

we

apply lemma

3.10

det

Since

2 4 Oo,

1

oO

1

1

1

fo)

fo)

1

iL

we

m* , hence

of

that the points

P7

ina

P 7*,

The

three

of which

three

lines

translated

statement

Definition: lines

and

projective

not collinear.

plane

#7

of duality also applies

diagonal lines

This

seven

are

implies

P 7*

in regard to

P 7.

Proof:

concurrent.

4.4

determinant

= fe

the principle

consequences

following:

calculate

conclude

Proposition in

, and

of a complete

requires

A complete

six points, are

into the language

some

concurrent,

explanation

says

the

are never

:

is the configuration

by taking four

their

7,

quadrilateral

quadrilateral

obtained

of

six points

lines

of

a,b,c,d,

of intersection

no

, and the

p = (a.b) U (c.d) ' (a.c) U (b. q = d)

r

joining

diagonal

opposite

lines

this would

(a, d) U (b. c)

pairs

of points.

of the complete

To prove

and suppose

'

P 7*

, let

that the three

show

These

lines

p,q,r

are

called the

quadrilateral. a,b,c,d

diagonal

be a complete

lines

p,q,r

quadrilateral,

were

that the diagonal points of the complete

concurrent,

quadrangle

Then

56

ABCD

, where

Ags bad Bi-acad

G@i=Faab Diawae Cue

were

collinear,

which

Remark:

contradicts

The astute

nition of a complete

P7.

reader

quadrilateral

*K.

will have

P 7*

noticed

is true.

that the defi-

is the ''dual'' of the definition

complete quadrangle.

In general,I expect from

construct

the duals

for himself

Hence

now

of all definitions,

of a

on that the reader

theorems,

will

and proofs.

Bye

Harmonic

Points.

Definition

. An ordered quadruple

on a line is called a harmonic X,Y,Z,W

such

quadrangle

(say

and

C,D

that

A

and

A=

XY.

B

XS

quadruple B

are

of distinct points

A,B,C,D

if there is a complete quadrangle

diagonal

points

of the complete

ZW Zain)

lie on the remaining

two

sides

of the quadrangle

(say

CEGXWaeran deel) eG) YoZe)ig

A

c

In symbols

, we write

B

>

H(AB,CD)

if

harmonic

forma

A,B,C,D

quadruple.

Note A,B,C,D

that if

are

ABCD

distinct

quadrangle

XYZW

points does

not make

is a harmonic

implies

that the diagonal

are not collinear. much

quadruple,

sense

unless

then the fact that

points

of a defining

In fact , the notion of

4

P 7

is

Fano's

Axiom

harmonic satisfied,

58

we

hence

Proposition

A

B

and

roles.

coincide

with the definition

Furthermore

with

to

Proof:

the line

ABC

and

Draw

points.

of harmonic

of

from

and

C

and

X,Y,Z,W

H(BA,CD)

D

distinct points

P 5) , this point

D

is unique.

or

the notation

etc.

such that

A,B,C,

since

play symmetrical

to make

be three

A,B,C

the definition,

D

point of A

speak

, there is a point

(assuming

harmonic

respect

Let

4.6

P 7)

{assuming

fourth

roles,

could permute

once

Proposition

Then

immediately

play symmetrical

In fact,

we

H(AB, CD) = H(BA, CD) = H(AB, DC) =H(BA, DC).

4.5:

This follows

Proof:

this when

assume

always

will

the harmonic

on a line.

H(AB,CD). D

conjugate

is called the of

B.

two lines

. Draw a line

n

1, m

through

through

C,

A,

different from

different from

k

ABC.

C

a9

Then join

B

to

respectively. intersect

A,B,C

l.n,

Then join

ABC

at

r.m

D.

B

to

and

Then

by

by construction we

Now we

assume

P 5,

point.

Given

is another

point such

is a complete

quadrangle

m.n.

s.l

P7

. Hence

fourth harmonic

D'

and join

toformaline we

have

A,B,C

B

XZ. YW

D

t.

r PS

Let

t

is distinct from

H(AB, CD). the uniqueness

construct

H(AB,CD').

KYZW _ such

XY.ZW

seethat

and will prove

that

A=

Call these lines

D

of the

as above.

Suppose

Then by definition,

there

that

GC ¢€ Xw DIRCRYeZ Call

1I' = AX

, m!'=

construction,

Thus independent showing

ae

AZ,

applied to

l1',m',n',

of the choice vary

If we

Let

be defined

D

quadrangle

X'Y'Z'wW'

m,

.

and

of

one

Step 1.

tinct from

n' = CX

replace by

KYZW.

that our

1,m,n. of

. Then we

will give

it is sufficient to show

that if we

complete

and

We

1,m,n,

Let

W = m.n

D

1', we

above,

D

and label

is

steps,

remains

by

the same.

D.

the resulting

line through

abtained from

belongs

of

get the same

be another

label the quadrangle

(Note the point

construction

the point

as

1'

D'.

do this in three

1 bya line 1,m,n

see that the above

1',m,n

A, dis.

to both quadrangles. )

(Y'Z') and

. (ABC)

sides meet

IN,

BregX Hence

and

by

Y'Z'

,

proof in this

of

respectively

1 and Step 3.

iH]

Clue

on

AB,

If we replace is identical

which

m

by

with

sides,

namely

is what we wanted m!',

we

YZ

to prove.

get the same

D.

The

that of

Step 1.

, interchanging

n!' we

get the

same

the

m. If we replace

The proof in this case of the corresponding the quadrangle

of corresponding

MOE IONE

must meet

case

XYZ

:

P 5, the third pair of corresponding

Step 2.

roles

B

and

A

in

pairs

Two

W.

from

are perspective

X'Y'Z'

that

i.e.

D,

that the two triangles

observe

Indeed,

= D.

through

passes

Y'Z'

that the line

show

We must

complete

formed

by

n

by

is more

difficult

quadrangle

1,m,n,

which

, since all four points

change. defines

D.

So let D,

Let

XYZW X'Y'Z'W'

be

61

be the quadrangle

formed

also meets

at

ABC

Consider Corresponding

hence

by

point

O.

1,m,n'.

sides

meet

in

XYW

In other words,

Similarly, (in that order)

and

A,B,C,

P 5* , the two triangles

ina point

We

must

show

that

Y'Z'

D.

the triangles

OW all meet

by

must

W'Z'X'

(in that order)

respectively,

which

be perspective

are

from

.

collinear,

some

the lines

NEFG) 5 ehatal

\NP,6

O.

by considering the triangles

, and applying

P 5*

once more,

ZWX

and

we deduce

Y'X'W' that the

62

lines

WNC).

are

and

concurrent

XW'

WB

. Since

+ X'W

, we

and

from

W'Z'Y'

meet

in

A

conclude

andmeZi

O, in that order

B,

fram

respectively.

“leeInustimeectionuther

the three

that their point of intersection

, the quadrangles

are perspective and

d.A\i7-

two of these lines are among

In other words perspective

5 eiavel

lines

.

XYZW

and

In particular, O.

Two

Hence

pairs

above,

is also

W'Z'Y'X'

are

the triangles

DCE

XYZ

of corresponding

the third pair of sides,

Absalom

O.

sides

YZ

zy.

Guends Proposition

(assuming P 5)

4.7

also

Let

A,B,C,D

CD,AB

be four

harmonic

are four harmonic

Combining with proposition

4.5

points.

Then

points.

, we find therefore

H(AB, CD) © H(BA, CD) = H(AB, DC) « H(BA, DC) H(CD,

AB) #

Proof: and let

H(DC,

AB) = H(CD

(see diagram

XYZW

be a complete

BA) # H(DC, BA).

on next page) quadrangle

.

We assume

H(AB,CD),

as in the definition

of harmonic

quadruple. Draw

Then

XTUZ

points;

Passes

B

DX

and

CZ,

is a complete lies

through

on

XZ,

A.

For

and let them

quadrangle

meet

with

C,D

so it will be sufficient

then we will have

in

U.

Let

XW.YZ

= 2 ee

as two of its diagonal to prove

H(CD,AB).

that

TU

63

Consider ponding

Hence

the two triangles

sides meet

by

in

D,B,C

concurrent,

is what we

1.

are four points

harmonic

quadruple,

any three prove

For points,

this later

field of three

wanted

In the projective of any line.

which

Their are

corres-

collinear.

vertices,

namely

These

plane of thirteen points, four points

always

it will be sufficient to show that

then there will always and it must : The

to prove.

forma

in any order.

To prove this, this plane.

respectively,

YTW.

AO) 5 AS

which

Examples: there

and

P 5* , the lines joining corresponding

DONE are

KUZ

which

be a fourth harmonic

be the fourth point on the line.

plane of 13 points

elements,

P 7

is the projective

is of characteristic

3.

holds

in

point to We will

plane over

But

P7

the

holds

64

quadruple

harmonic

(see problem

INE

18D,

BC

AD

of distances

7 Mee

l'

and Projectivities

A perspectivity

antother

line

obtained

in the following way

Formeach map

forma

ABCD

# 20).

Definition:

Then

points

four

if and only if the product

Perspectivities

ive

plane,

Euclidean

In the real

2.

4 Ze

any field of characteristic

plane over

in the projective

(both considered

pointmAl

A—~>A'.

:

is a mappingof

as

Let

sets O

OA™

of points

bea

one

line

1

) , which

can be

point not on either

Gils

sdraw

andier

This

is a perspectivity.

into

OAumect

sl

In symbols

we

1

or

sinwAl write

O i

which

says

or

which

"1

is mappedinto AD

says

gl ae

nN

Geer

''the points

>NO

1'

by a perspectivity

l') '"'.

center

at

AL BUGware.

A,B,C

(of the line

perspectivity with center O into the points (of the line

with

1)

are mapped

A',B',C'

via a

, respectively

O',

65

Note

that a perspectivity

that its inverse

X (as a point of

One tivities

can

is always

is also a perspectivity.

1) is sent into itself,

see

easily that

Note

For

pointof

AU

7'O

BitGciuy"

,

, then

l').

perspec-

in the diagram

example,

and

O - A'B'C'Y'

and

X =1.1'

of two or more

have

ABCY

and onto,

also thatif

X (asa

a composition

need not be a perspectivity.

above,we

one-to-one

66

another

LEIS

TTLOMEACET Duet Gueien

Note

of

lye De CLiv.e

nes

that a projectivity

Proposition

4.8

a projectivity, tivities

identity map

of

since we

Naturally, we would

cannot

we would how

like to know

in the following be four

times

. The

like to study many

times

inverse

of perspecThe

this

group,

transitive

that is is three

,

of a projectivity

of perspectivities.

the chain

only reverse

two propositions

Proposition

be two triples

need

chain

is

(in fact a perspectivity)

1 into itself is a projectivity

PJ(l)

one

of perspectivities.

is still a chain

and acts as the identity element in is projectivity,

PJ(1).

of two projectivities

the result of performing

by another

and onto,

the set of projectivities

. Then

that the composition

because

followed

line

which we will call

a group,

: Notice

one-to-one

also is always

1 bea

Let

1 into itself forms Proof

A,B,C,...

points

if the projectivity takes

7 A'B'C'...

55

as

write

1Z anid.

Wewrite

of perspectivities.

a composition

be expressed

can

1) , which

be equal to

l' (which may

1 into

line

of one

a mapping

is

A projectivity

:

Definition

re

the following

make

we

Therefore

Y.

which is different from

it

COM

goes

Y

However,

into itself.

to send 1.1'' = Y

would have

a perspectivity,

were

l''

1! to

from

map,

if the composed

Now

and in particular,

it is.

times

We

will see

transitive,

but

transitive. 4.9

of three

Let

1 bea

distinct points

line,

andilet

each.

Then

A,B,C,

there

and

A‘, B.C!

is a projectivity

67

1 into itself which Proof:

Let

not pass through projectmn,

sends 1'

A

B..u

A,B,C

bea

or

A'B'C!

A',B',C'.

line different from

A'.

from

into

Let

pe tagl! =

O

1, and which

be any point not on

4 giving.

A‘), BB, Cl

does

1,1',

so we

and

have

AVUBYUG"

A

and

A ¢1',

from

lto

A" 41.

1' , taking

Now it is sufficient ABC

forget the original points problem

into

A''B'C"

A',B',C'

tO construct a projectivity . Drop double primes

€1.Thus

, and

we have the following

:

4"

A

Let points

on

1, l,

1'

B’

be two distinct lines;

andlet

furthermore

that

1 to l'

carries

which

A',B',C'

be three

A ¢ eeancerAs A,B,C

into

¢ 1.

Ce

let

A,B,C

be three

distinct points

To construct

A',B',C'

on

distinct

l', assume

projectivity from

, respectively.

68

(AIS

Draw

, JNVI8) eS 1B)”

A Gi

A.

Ge Gamre

B'

and

C'',

joining

1"

, and let

AA',AB',AC',A'B,A'C

Draw

ate Aw ee Linen

AA

and let it meet

sends i] ABC

S

ALBiGtt

AL

BIG

>>

Thus

found the required

have

we

projectivity

as

a composition

of

two perspectivities.

takes

A projectivity

4.10

Proposition

harmonic

quadruples.

into harmonic Proof

: Since

a projectivity

is

a composition

it will be sufficient to show that a perspectivity druples

into harmonic So suppose

Let)

A’) B'y CD!

ey

= l'

eS

mapping,

separately.

of the two lines

is one

we may

itis

assume Let

such

that

1

vand

H(ABy

obe theirtimages.

A

O . 1

H(AB, CD)

of perspectivities,

takes

harmonic

qua-

Ay Ss, Gy

Drak:

quadruple.

12 lI!

te. is the same

quadruples

A

Let

CD)

where

I =sABY


AS

PILI Gu

7y ¥

.

=

o8s

‘6

:

H®

6

te

e

é

*

9

‘ee

.4

a —

owes

»

7

:

>

-

7 3

—

:

ie ow?

5

aw

Ceher

a

ht

>27p

(60)

> oo"

j

ow

epi

Wr8

:

, OATipe

Rave: 7

a

8

7

~*~.

c~

7

~

i

Py

”

i

a

_

1

“hh

: —

a

~~

_

\

ce ‘eh i , >

;

=

—

‘

-.

7

i Cesed

a

a

~s @

lhe

af

@

io Ws

OF

|

.

ST

¥

ne '

,

—

4

»

7 ‘

:

bdigp Vf -

wi

,

.

or

-

»S

:

a

——

ry

;

:

-

74

i

|

ee

te

m

i)

Bn

he =i wes

4) CD jena (ye

Re

ieew,

ae ti

CHAPTER THEOREM

FOR

In this

states

5.

PAPPUS'

PROJECTIVITIES

chapter,

that there

we

come

three

turns

Out this theorem

does

P 7,

so weintroduce

P 6,

points,

theorem,

i.e.

PJ(l)

to the ''fundamental

Pappus'

three

times

which

transitive.

the axioms

axiom.

Then we

the fundamental

theorem

theorem"',

sending three points into any

is exactly

and conversely,

. Fundamental

1 bealine.

distinct points into!

FUNDAMENTAL

P1-

and Pappus'

P 5 and

can prove

theorem

It

the funda-

implies

axiom,

P 6.

and then give

afterwards. FT

Let

THE

ON A LINE.

not follow from

We will state the Fundamental proofs

, AND

is a unique projectivity

other

mental

AXIOM

on

le esuch that 6.

Let

A,B,C

1. Then ABC

Pappus'

theorem and

there

= A'B'C! Axiom

.

(for projectivities A',B',C',

is one ;

on a line) .

be two triples

of three

and only one projectivity

of

1

Ue

Let

1 and

P=

distinct

bpesthncemarcn

Define

X.

1', different from

tinct point on

be three

Dun Gum

eteAls

X = (See

1, different from

on

points

A,B,C

Let

be two distinct lines.

1'

AB'.A'B

ORNATE

Roe Then

P,Q,

and

Rare)

Proposition and so the principle Proposition

Proof: construct

IAC! IG

Let

P,Q,R.

collinear:

5.1

P 6

implies

the dual of Pappus'

of duality extends. 5.2

P 6

is true

(problem

We

take

1

P 6%,

21) plane.

be as in the statement,

to be the line at infinity,

statement

Axiom,

in the real projective

1,1',A,B,C,A',B',C'

to proving the following Page)

-

in Euclidean

and thus

geometry

and reduce

(see following

:

Let be three

1'

be a line in the affine

distinct point on

different from and define

We

1'

1' . Then draw

P,Q,R

as

shown.

sill study various

Cutting with lines of direction

.

Let

Euclidean A,B,C

lines through Prove

ratios

that

plane.

be three

A' P,Q,R

Let

distinct

in diections are

: Cutting with lines

A, we

have

A',B',C', directions,

B,C,

collinear.

in direction

ts

ACB Bic’,

ee |

ee Ps

Therefore

TR RC'!

ae psi

TRegERC. A'P PS But

A TQC'~

AA'QS

ae ice This proves

that

PQ,

QR

are

one .

pTRa RCY A'P+PS

(similar

triangles)

ear, A'S , so

toy AO

ATQR~ JUNO)

so

-

parallel,

AA'QP.

Hence

ZZ INNO)

hence

equal lines. -e.

(See problem

22

for another

proof of this proposition.)

74

Pid

FT

5.3

Proposition

P 6 (in the presence

inplies

of

P 1-

Of course.)

Proof

elete

la liwAN DaGaA

P 6. We will assume

295.2 Cmmberd

the fundamental

theorem,

Salnutheased lenvenisgn and will prove

that

2 = ANB) 4 AAS

OpeaFA GLsALG

R= are

BG. BC

(not shown in diagram)

collinear.

Drawn

be the line 4.9 , we follows

Ab

PQ,

AUB

sand.

and let

e Drawer

1'' meet

AA'

can construct a projectivity

Glan

in

A".

sending

Gunman

Then, ABC

to

Can)

ae

Lae

as in Proposition A'B'C',

as

: AY ]

Let

Y=1.1'

On points

A

, andlet

as follows:

yur

jt

F

>>

Y' =1'.1"

. Then

these

two perspectivities

act

ABCY Now

We

let

consider

>>

B'C

meet

the chain

INE SONGL 1"

in

>>

R',

AB UG

and let

Nales BR'

meet

l'

in

C'",

of perspectivities

! il

B

}"!

B

A This

]!

,

A

takes '

ABCY

2

PBURwY.

Bowe have

two projectivities

ABY

A'B'Y'

into

they are

. We

the same.

on the same

that there

from

conclude

(Note that

line,

from

1' , each

of which

the Fundamental

by composing

sending

takes

Theorem

is stated for two triples

is a unique projectivity

Therefore

1 to

FT

but it follows

lie on different lines.

2 yorsse wag

that

of points

with any perspectivity

ABC A A'B'C'

also if they

)

the images

PLOJectivitiess..1.6-m

Cu——GCllne

of

C

must

there1ores

be the same Rua

under botu

Re,esOm

rt, hk ware

collinear. q.e.d,. Now P1-P6.

we We

Lemma

Then

1

come must

5.4

a)

1,m,n

b)

O,P

to the proof of the Fundamental prove

Let

are and

l1.n

1 Q

n

of subisidiary

1

n,

concurrent,

is perspective

perspectivity

a number

are

to

gives

n,

with

1 4n,

Theorem results

from

first.

and suppose

either

or

collinear.

i.e.

the same

there

map

is a point

as

Q,

such

the projectivity

that the

aK n above.

76

Proof : (Problems

Lemma

5,5

let

23,

1

24,

m

n,

m'

a)nor

4cand points

b)

of the previous

O" €n,

and)

!

m'

A

Proof) On

E

that

P'S

lemma

holds.

Then

there

is a line

Ll. such that

n

A

projectivity

from

Letylim.nsO.e

1 to

n.

be given.

Let

A,A'

be two

points

euancalet

AA' RectOPemcetenwinw

collinear,

Oa

14n, and suppose

!

1 2

gives the same

with

>I

'0 2

neither

and 25).

O'! 4 X,

Gael)

aL

Oo BB'

Cele

A

>I

Ouest

cence

assumed

1, Draw

O'A,

so

O'¢

ae respectively.

O,P,l.n=X

O'A',

are

and let them

not

mect

Now meet

in

correspondig

O,P,O'

sides of the triangles

, respectively,

the lines joining corresponding

the line joining Thus

varies

m)

D,D',

passes

vertices

m,

1

are

collinear,

are

. Thus

by

D

and

and

Y,

A'B'D'

hence

concurrent.

through the point

is determined

along the line

which

ABD

by

Thus

P 5*, m)>

Y=l.m. soas

A' varies,

our original projectivity

D'

is equal to

the projectivity

Performing

P20

Pal

the same

esand find a new il o

m'

=

argument

line n

fA

gives

the original projectivity.

mis,

again,

so that

we

can move

P

to

78

Lemma projectivity

5.6.

17 1'

Let

1 and

1!' be two distinct lines.

can be expressed

as the

composition

Then any

of two perspec-

tivities. Proof:

arbitrary

chain of perspectivities.

by induction

of length to prove

A projectivity was

defined

Thus

, that a chain of length

n-

1 . Looking

that a chain

of

at one

as

a composition

it will be sufficient

n>

2

can be reduced

end of the chain,

3 perspectivities

can

of an

to show,

to a chain

it will be sufficient

be reduced

to a composition

of two perspectivities. The

line

m

argument

of the previous

can be moved

lemma

actually

so as to avoid any given point.

easily (details left to reader

) that it is sufficient

shows

that the

Thus

to prove

one

can

see

the following

let

Ply

be a chain of three perspectivities, projectivity

170

can be expressed

with

1 +0.

Then the resulting

as a product

of at most

two per-

spectivities. IMB 5 Wi

are

reduced

we may 5.5,

can

be moved

are

On

we

0,m

GP

NSM

Of

We

AG

trivially to two perspectivities,

assume

and

te Sl

have

l,m,n,o either

are m

all distinct.

; O

so that the centers

respectively.

, in which

msi

oF.

using lemma Second,

case

we

n=0,

5.4a.

using lemma are

of the perspectivities

done,

or

ane nand

we

So 5. 4b n

n=o0

:

79

So we

Within Zz=n.o,

we

could

pt and

have

have

Oma lilac stinccm draw

moved

ORGCeXe72—sheemiaTOVect

h=

On

XZ

m,

ie

ClOmmand

.Wemay

by lemma

sluic rnin

assume

5.5

that

to make

hepmatics le taeDae

le

tmoXe= ml a rrles

X q o (indeed,

X ¢ o).

ee elee

Therefore

80

meet

meet

in a point

N.

In other

vary

D',H'

see that as

and we

by

is determined

N

, the line

always

D'H'

sides

corresponding

the remaining

P 5

Thus

QR.

on

N

by

hence

Q,R,

in

sides

Corresponding

Z.

from

are perspective

C'D'H'

and

CDH

Now,

DH

alone,

through

passes

words,

ruZ

X,

point

M € PQ.

again,

P 5

sousing

from

and

ABH

the triangles

Similarly,

find that

we

are

A'B'H' and

AH

prespective meetina

A'H'

Hence M

h.

oe

A

the original

have

So we

projectivity

as the

represented

composition

of two perspectivities

ruz

Theorem Proof

A',B',C', ate)

5.6

fe)

P1l-P6

: Given

, we

N A

must

a line

show

imply

1, and

two

the Fundamental triples

Theorem.

of distinct

there is a unique projectivity

points

sending

A,B,C,

ABC

AIBC! Choose

a few

special

Call them

a line cases

A',B',C',

AM

l' , not passing to the reader),

still.

So we

through

any

and project

have

reduced

of the points

A',B',C',

It will be suffident ay NEW

ABCZ

Cc

EL

onto

l'.

to the problem

aGmelmel

all different frOm A

(I leave

1,1'

Bue Guwaraels

to show

that there

is a unique

projectivity

sending

81

We Hence

already

know

one

it will be sufficient

such

projectivity,

to show

from

that any Other

Proposition

such

4. 9.

projectivity

is

equal to his one.

Case

Let

l.

i

Suppose

21"

the other

send

ABC

P=

AB

FAS

Ota

C

ALG

and let 1'' be the line joining I claim

that

the two triangles Their

C

1"

and

sides

P

and

A

’

X.

A'BC , which in

P,Q,X

determined

the perspectivity

Te Oe jes

.

is actually a perspectivity.

Consider

Q.

through

meet

is already

varies,

= A'B'C!

passes

AB'C'

corresponding Hence

that as

l''

projectivity

by

Indeed,

we

apply

are perspective respectively P

and

X.

P 5 to

from

O.

. This

shows,

82

and

the projectivity

eS aeAA coincide. Case

2.

Then by lemma (exactly)

their

Suppose

5.6,

the other

it can be expressed

two perspectivities,

centers

diagram

lie on

1'

and

as the composition

and by lemma

5.5,

1, respectively.

we

Thus

se 105

ABR

i] 1 ee; ls

and

PaaS A

1".

Similarly by Qe

ol!"

construct

can

we

of

assume

have

that

the following

4

P 6 applied to

hes>

>>

Now

if

D' = RD".1'.

D€1

is anarbitrary

Then consider AD

are collinear,

i.e.

ie ALD

are

define

P 6 applied to ee Ae

AD'.A'D

also by the projectivity

point,

ADR

D'' = R'D.1'"',

and

and

It says

A'D'R'.

aD

€ 1", which means

of Proposition

4.9

. Hence

that

D

goes into

D'

the two projectivities

equal. q.e.d.

Proposition Proof

5.7

:( see

the hypothesis

P 6implies

diagram

of Desargues

4

We will make

onp.

P 5. 81)

Theorem

three applications

of

Let

O,A,B,C,A',B',C'

(p 5) , and construct

P 6

to prove

that

satisfy

P,Q,R.

P,Q,R,

are

collinear.

Step 1. to the

and

Extend

A'C'

to meet

AB

at

S.

Then we apply

P 6

lines

conclude

O

Cc

C'

B

S

A

that

1 = OS;BC U=

OA]

BC!

Q are collinear. all distinct

( Note to apply

, and

of the two lines.

P 6, we

O,C,C',B,S,A

But

P 6

are

should check that

all different from

is trivial if not.

)

B,S,A

are

the intersection

85

Step 2.

and

conclude

are

collinear.

Weapply

O

B

Cree

Ale

a second

time,

to the two triples

Bi eS

that

Step 3.

We

apply

|S conclude

P 6 a third time

al me

(Vv and

P 6

T

, to the two triples

) S

that R

Pee

are

Dorm

VeEDVeOcepEce

(lavoro

LU Sby otep J;

collinear. quend.

Corollary ar NY

syanoy

corresponds

1 4 l'

5.8 ( of Fundamental

Theorem)

is a perspectivity

the intersection

to itself.

©

.

A projectivity point

X =1.1'

ni)

0@6igumsD

Tl

\

e

:

“y

Js

fl

i

=

6‘)

-

:

a

*')

~

a Gh

4

———

.

rs

A

Sg

=

_.

eter iad

Die) >

14

{

‘3

i

f%

a

C3 € F , and are not all zero.

on the left,

then

satisfying

of the form aa a

here

is the set of all points

as

there

the plane

defined

by

of distinct planes.

x, = 0

by an earlier

are

mand

in this projective

result, (theorem

2. 1)

92

Now we will study the group our

projective

of automorphisms

of

plane.

Definition:

A matrix

if there

is a matrix

matrix.

(Note that in general

division

ring.

xed

However,

just the matrices

with

Proposition

of elements

Aut(P,.)

of

F.

A=

(a; ) of elements

PESuUchethat

Kare

determinants

if we

are

working

determinant

6.2.

)Let

Then

= ray.

of

F

=I,

is

the

invertible identity

do not make

sense

over

F,

a field

over

these

a

are

4 Ors)

A=

Sore

be an invertible

3 X 3

matrix

the equations

3

Re eS Gy, BS il

define

j=l

an automorphism

Proof

Then

Th

T.

. Analogous

Proposition and

Let

have

the

Analogous

Proposition

Then

center

of

F

‘

A,A' same

be two

3.7

invertible

effect on the four

Q=(1,1,1)

@

q.v.

thereisa

matrices.

points

P S(tco}

REF,

X40

>

Au = Aye.

Proof:

AL,

P

to proof of Proposition

P, = (0,1, '0) | Po > (0,0,1), such) thatw

of

A

6.3.

Ta

1 eS

ay

aT F.

(6.4,

to Propositions Let)

€ F&F | A

3, 8

Oo , and

is the identity transformation Otherwise, »X

ee WhelenGmnismtine

aN

»x,)

—>

automorphism Bt

ur

q.v.

of

consider

P

is the automorphism oO

(x,

of

F

oO

10)

given

by

Deis

# Ais given

the matrix

in by

the

»

ne)

(Such an automorfhism Proof:

In general,

(Ax), Ax,, Ax) .

Re o

This

Wee

Ty

of

6.5.

Ons

of

of

We

of the form

6.6.

However,

and Corollary

to the point

coordinates

But

for all x, i.e.

by

such that if

X

matrices,

Then

“Al= A).

bres ap

then by

A'’

A'

PGL(2,F)

peraLee Ly 1? TON

Let

A,B,C,D

and

Since

A

Then

, 1(By=

to Theorem

matrix

GL(3,F)

of

1(A)=

the group

invertible

in the center

8°

es

F

A.

( Thus

of invertible

A',B',C',D' there

5,

matrices,

3.9

be two

is an element

1(C)=

CY,

£(D)

=D”,

q.v. T

it will be unique,

is its own

of automorphism

F.)

the transformation

is commutative,

6.5,

be invertible

(\L)em SOM

of the group

Analogous

F

A'

for some

that in general

if

and

no 3 collinear.

TeePGr2Fy such that

Note

F«, \x = x)

Conversely,

denote

TA

of points,

Proof:

F.)

the second assertion.

F >} 40%

by scalars

Proposition quadruples

(x), x 273)

of

F.

is the quotient

by multiplication

A

is clear.

\€ center

PGL(2,F)

of

smmACm— mA AG eAUcm

Definition.

P.

Let

center

Proof : 2 Ee SE

tS.

pe

ay FW

& all : =the+s ere

ie). 445

il

Oe wasnt

|

yeh yet b= Bt oss

ta

sami

101

CHAPTER

7.

INTRODUCTION PROJECTIVE

7

In this

chapter

isomorphic

to

ring

F

? Or

we

ask

a projective

alternatively,

ring

x, €

F,

to points

A

necessary

condition

Desargues'

axiom,

satisfies axiom

P

is We

F,

will

be

and

the

begin

take

was

a

and

associative,

The

one

a

for

5,

And

PP. ,

are

to be

we

given

we

some

can

we

division

find

a

(x), x5, x),

equations?

is that

should

seen will

plane

by linear

possible

have

in fact

for

plane 7,

coordinates

lines

since

is a projective

form

a projective

this

3.11

proofs

would

that

would

set

ring,

the

that Pe always

see

that

1 be

of points

was

some

the

1,

on

Desargues'

that

addition

messy

using

these

was and

and

(0, o) and

define

would

associative

rather

plane

and

problem

in A,

points

through

one

introduction

to this

construction

Then

prove

multiplication

coordinatize

line

the

the

approach

non-collinear

40-45).

i.e.

namely

naive

geometrical

(pp.

involve

A

three Let

the

in F be

problem,

A.

plane

(0,1).

be

to

F

simpler

Choose

(0,0),

division

F

finally

of the

that

P

affine

of Proposition

that

etc.

when

homogeneous

6.1).

following:

multiplication

proof

PLANE.

plane

such

with

in an

(1,0),

Now

(1,0).

of ™ ,

IN A

sufficient.

them

call

assign

5 (Theorem

of coordinates

would

and

COORDINATES

question,

given

division

satisfy

the

OF

addition

given have

in the to verify

commutative distributive,

diagrams.

Then

coordinates

on

l,

102

and prove

that lines were

the approach which tive Geometry",

given by linear

is used in Seidenberg's

chapter

However, on the principle

we

will use

a slightly more

uses

will be less work

to be done.

a study of certain

automorphisms

automorphism

PQ

|| P'Q',

lines

©

of

where

Let A,

plane

leaves

a stretching

Hence

m=AU

high-powered

in Projec-

of an affine

such

that for any

Or,

if we

i

, then

=Q'. think of ©

, 1,

metiod,

techniques,

we will first address

there

ourselves

plane.

A dilatation

two distinct

In other A

as

, pointwise

is an

points

words, contained

is an automorphism

In the real affine plane

in the ratio.

this is

''Lectures

sophisticated

A be an affine plane.

the line at infinity

Examples:

more

g(P) = P' , @Q)

into parallel lines.

projective

which

:

book,

In fact,

3.

that if one

Definition

equations.

P,Q,

@

takes

ina of

ff

’

fixed.

Ine =(ix.y) (sey GR)

k, given by equations

to

103

kx < ee*

is a dilatation.

ky

i'

Indeed,

let

O

be the point e

(0,0)

U

O= (c,o)

Then

©

stretches

two points,

points

clearly

Another

PQ

example

away

from

O

k-times,

and

if

P,Q

are

any

|| P'Q' , by similar triangles. of a dilatation

of

An

is given by a translation

ee Sede A Se

See Is

(4b)

Invthis

so

PQ

case,

any point.

P

|| P'Q'

again,

for any

Without

asking

for the moment

dilatations

in a given

affine

is translated

by the vector

from

O

to

(a,b),

P,Q.

plane

A,

whether let us

there

study

are any

some

non-trivial

of their

properties.

104

Proposition dilatations, of

A,

Aut

Dil

(A)

Let

forms

A

be an affine

a subgroup

plane.

of the

Then

group

the

set of

of all automorphisms

A.

Proof:

Indeed,

is a dilatation,

follows

7.1.

and

we

must

that the

immediately

from

see

inverse

that the product

of a dilatation

of two dilatations

is a dilatation.

the fact that parallelism

This

is an equivalence

relation, Proposition

7.2.

A dilatation

which

leaves

two

distinct points

fixed is the identity. Proof:

Let

Sar aes

p

be any point not on

bea

PQ.

PR

dilatation,

Let

let

o(R)=R'.

P,Q

be fixed,

,

Then we

and

let

R

have

|| PR!

and

QR || QR' since

@

since

R.¢ PQ...

is a dilatation.

also fixed. Same fixed,

But

argument so

@

PR.

R

an arbitrary

was P

and

QR={R}

R,

we

and

R'€ OR.

, andse

point not on

see

that every

But

R=R"', PQ.

PR +QR

i.e,

R

Applying

point of

PQ

is

the

is also

is the identity.

i.e.

7.3.

Proof; see that

A dilatation

any two dilatations

on two distinct points

So we

R'€ PR

Hence

to

Corollary two points,

Hence

Indeed,

P,Q, »)

a dilatation

-l

are ©

is determined ©,

, which

by the images behave

the same

of way

equal.

leaves

different

P,Q from

fixed, the

so is the identity.

identity

can

have

at

105

most

one

fixed point.

no fixed points

the identity

pomt

7.4.

these

two

©.

PP'

with

P,Q

different

,wehave

PP'

from

| QQ'

the

, where

But the fact that

PP'

into itself,

QQ'

into itself.

PRI

PP)

so

}{ QQ'.

intersect

line

that

@

ina

©

is a

sends

and

©

RG

the

sends

(Forexample

Proposition

, let R € PP'.

PP')) Hence 7.5.

(A)

of the group

a normal

subgroup

O-= Dil(A)

dilatations

is a dilatation with no fixed points,

is atranslation,

two points

lines

implies

is

If ©

Suppose

dilatation

Tran

for those

o(Q)=Q'.

Proof: Then

name

.

then for any

y(P) = P',

a special

A translation

Proposition

identity,

have

:

Definition. Or

We

The

(©)

=O,

translations

of dilatations of

Dil(A)

of

, i.e.

A.

Then

so’ of

© A

PR

|| P'R',

is a fixed point form

xX

a subgroup

Furthermore,

for any

but

Tran(A)

T € Tran(A)

, and

, ora} € Tran(A).

Proof: is a translation,

T.,T, We 2

First we and

be translations,

fixed point

P.

Then

must

the

check

inverse

then

TT, Wz

T(P) = 1h

that the product

of a translation

is a dilatation.

T (P') =P.

of two translations

is a translation.

Suppose

If Q

Let

it has a

is any point not

106

, then let

on

PP'

We

have

e'

?

Q' = T,(Q) P

L

Pro-

by the previous

POLL Oates Hence

ioe

Q

position

PR aOW @ as

is determined

Q'

P!' , and the line

through

a similar

For Applying

the same

77, Sid

ience

m

is a translation.

is a translation,

translation

so

point is fixed, of a

inverse

the

Clearly

form

so the translations

_alsoitixed:

Os

Hence

find every

we

Q,

to

Q.

through

7,(Q') = Ou

,

l | PQ

of the line

intersection

|| PP'

reason,

reasoning TT,

the

of

a subgroup

Dil(A). Now

T € Tran(A)

let

If it has no fixed points,

a dilatation.

a fixed point

P,

fhaspastizced

pOlntyem

then

G,

we

say

H

is anormal

if

G

example,

in an abelian

Now

we

to the question

come

of Desargues'

two existence axiom.

This

roe

is a group, of

G

and

if

is one

, so

is a subgroup and

¥gCG,

is normal.

of translations

Desargues' are

H

subgroup

of existence

If it has

OK 7

Vh€H

every

problems

ok.

Bye oor

= idee

oral:

group,

for this we will need

will find that these

maim

subgroup

For

dilatations,and

forms

Cus

(i=nl

Definition . In general, of

it is a translation,

ora | (P) = P implies tlenCem

is certainly

as

Then

.

. o € Dil(A)

axiom

. In fact,

equivalent

of those

cases

and we

to two affine

where

an

axiom

107

about

some

configuration

of the space.

Here

Our

has

geometry

become

clear

is equivalent

Desargues' '

"'enough'"'

form

the

axiom

1,m,n

be

three

parallel

C,C'

all distinct points.

2 and

Boe

©

CG: Note

plane

So

P 5

O

A5a on

says

the

Bens

5C

ii) fGansiation

;

plane

follows line

form

A P 5

at infinity

1g

Qe

Cl.

ACHRA'G!

is contained in

@w.

. Our

in a projective

Indeed, hypotheses

1,m,n state

meet that

that

BbOP

Glog,

| BC.

statements i)

A

wecEran

ela,

The

are axiom

Given 7

Let

7.6.

Theorem two

A

Assume

affine

NW Reape

mew

axiom.)

PS ABA's '!

will

i A'C' | ‘Then

that if our

7 , then

in a point

sche

AC

which

lines

setmrAos

ABA’

in a sense

that

theorems.

(distinct)

€n,

of the geometry

is equivalent to saying

automorphisms

A 5 a (Small Desargues' Let

to a property

equivalent A5a

bean

A

that

Then the following

:

holds

any two points

such

affine plane.

in

A

P,P'€

T(P) = P's

A,

there

exists

a unique

108

We

assume

taking

P

(i) = (ii)

Proof:

identity is a translation

to

P +P’.

aS

Nee

sending

T

construct a translation

P

g

P

we will set out to

Now

so

P' , and it is the only one,

So suppose

there is nothing to prove.

then the

Pp"

P=

if

A5a.

/ i>

Step 1

formation

We

Topp!

define

of

we

for

on

P,P',Q,

Q q ly

and we

©Q*eas

set

then tor any

Top: (Q) = ©!

corner

the fourth

Tpp! ‘(@)r eo er ¢ PP'

-

, and

R¢ OOF

have

Indeed,

define

R!

- Tppr

Then bysA5a. so we

have

OR)

defined defined.

O'R,

pretRm

Step 3. define

(R)

also

RY SG

now

where

follows:

, as

of the parallelogram Stepraouit

a trans-

A-1,

PP'

is the line

1

.

.

Ip?.

fo

T

Starting with to be

T pp

at a given point,

or since

P,P',Q, TQ0! we

taking

Q' = Top: (Q) , we

,» whichever

saw

they agree

one

happens

where

can

to be

they are

both

109

Step 4. Um ae

Note

that if

whenever Step

point,

let

5.

R

they are

Clearly

X', Y',Z'

7

is any point,

both defined.

is

1-1

and

be their images.

and

T(R) = R',

This

follows

onto . If

X;Y,Z

then

as above. are

collinear

Then

TiS) eaten) and

Tie):

T(Z) So it follows are

immediately

collinear.

from

the

Hence

construction

iseastmanslation

translation (ii)

statement Then since

assume

A5a,

A.

ONE

One

DO Me a

sees

immediately

with no fixed points,

hence

tone):

of

T

follows

the existence

given

and let

hypotheses,

from

the fact that a

of translations,

1,m,n,A,A',B,B',C,C',

T

be atranslation

T(B) = B'

and

taking

T(C) = C'.

and must as

A Hence

in the

into

A',

BC

|| B'C'

is a dilatation. Proposition

Proof: Gase

bela pointes;

7.7.

Let

ll, T

vet

f,T'

and)

Try

(Assuming

f

A5a)

be translations. translatesin

=P.

T(Q) = Tr'(P) and

=

of

ere

point is the identity.

Suppose

of

by our T

a fixed

> (i) . We

A5a.

that it is a dilatation

, the uniqueness

with

the definition of

is an automorphism

es sandmatitakes

Finally

deduce

Tf

from

7'(P') = TT(P)

T'(P) =O.

Tran(A)

We

different

Then

must

is an

show

directions,

abelian

group.

TT' = T'T. Wet

P

110

are

hence

are

so

equal,

Gaser2.) translation axiom

A

f

= T'T

TT!

andes,

3

to ensure

in that direction.)

there

T

and

TT*

are

T

and

T*

not used

direction,

(here we

use

is another

Wet

Tem

Theorem

direction,

7.6

and

AS5a).

bera and

a translation

are

= (rtreyrre

in different

T'TT#T® Since

have

P,P',Q,

on

Then

qr! = teers) since

the same

direction

that

so far we

. (Note

Jaren

in a different

of the parallelogram

vertex

fourth

the

as

found

both

directions.This

equals

= TT,

in different

directions. qoe€aq-

Definition. We

This

say

G

Let

bea

is the semi-direct

1)

H

2)

Hig) Koes l}

3)

H

is anormal

implies

a \POcmew

G

and

K

=

ll,

Definition. be the subgroup

of

product

subgroup

together

that every

group,

and let of

of

be subgroups.

and K

if

G

generate

element

H

H,K

G.

g ©G

can

be written

uniquely

as

Int il, kh E i.

Let Dil(A)

O

bea

point in

consisting

A,

of those

and define dilatations

Dil (A) ©

such

to that

o(O)= O. Proposition

an

d

Dj Dil (A) :

7.8

Dil(A)

is the

semi-direct

product

of

Tran(A)

ll

Proof: of

1) we

seen

that

Tran(A)

is a normal

subgroup

Dil(A).

By

We @) & UBeevaVN)) (7 Dil (A)

being a translation 3) Let

have

T

Let

bea

and

existence

hop

distinct

©

but

Let

g{O)=Q.

such

eeena a Dil (A)

@ has a fixed point,

be the identity.

@ € Dil(A).

translation

tine tay, ©) a Tran(A)

it must

, then

yo € Dil (A) 5 SOQ

generate

Dil{A)

= rr!

. Note

here

gy we

shews have

that used

of translations.

a (Bis Desargues.

points

Axiom)

in the affine plane

A,

Let

and assume

OFA, AS

are

collinear

(0), 1B), let

are

collinear

Once Ge

are

collinear

AB

|| A'B'

AGeATCS

BCE OC

O, AB,

C, A’, BY, C'!

that

be

the

112

Note

that this

in a projective

plane

Theorem the affine

follows

The

following

two

The

O,P,FP'

axiom

are

A5b

collinear,

suche thaitmo(©)s=—©Omandag) Proof 7.6,

holds

A

in

so the details

ay

there

(>)

: The proof

(i) = (ii)

yee

P 5,

if

A

is embedded

statements

are

equivalent,

in

A.

(ii) Given any three points and

from

7.

7.9.

plane

(i)

statement

O,P,P', exists

with

a unique

analogous

will be left to the reader.

for points

O,P,P'

Q

dilatation

O,

o

of

A,

ieee

is entirely

. Given

P40, P'¢

Here

, as above,

not OnNthe line

0

1

P

to the proof of theorem is an outline

define

:

a transformation

containing

O,P,P'

je

R

Q Q’

as follows

.

7)

O,P,P'

the intersection

agrees Hence

of the line

OQ

with

Now

if 0, p, p' (2) = Q'

with

0,9,Q'

one

can

define

the line

, one

proves

(Q) = Q' through

using

(defined

similarly)

whenever

go

either

and

to be

one,

, where

Q'

:

P'

A5b

, parallel

that

both are

0(O)=O.

j

hy

6

to

pp

defined.

Then

go

PQ.

113

is defined

everywhere.

C= 0, R,R! and onto. takes for

whenever

Gorollarys

that if

is defined. results,

A

is a dilatation.

of

. Let A5b.

Let

A'.

Then

into

Remark:

A5b = Ada

go

o

Using

, although

Indeed,

The

Let

bea

dilatation

P

that

P,P'

+P'. to

into A5a

Q

Let

P' (which

P',Q',Q'"

fixed

show

and that

1 - 1

easily that it PQ

| o(P)o(Q)

follows

from

which

is a dilatation, the

theorems

O

o(B) = B',

BC

7. 6

leaves

and

g(C) = C'.

|| B'C'

and

from

fixed and

7.9,

we

can

show

the geometrical

that

statements.

assume

P

p’

be two points. a translation

P' , which holds,

bea

of the parallelogram

P

can

g is

since

Ox

will P,

arbithary.

Let

and

clearly

, then

be given satisfying the

by the hypotheses,

this is not obvious

let us

We will construct

Pueeatre

R 4 O

uniqueness

0

show

Now

one

O,A,B,C,A',B',C'

from the fact that

sending

=R',

(23.

hypotheses

A5b.

o(R)

so is an automorphism,

so o

(ii) =(i)

So

the latter

into lines,

P,Q,

sends

show

But using the previous

lines

any

Next

, md

are

point noton on

P,P',Q.

PP',

and let

Q'

O

point on

Let

bea

0, be a dilatation which leaves exists

by theorem

collinear,

sending

Q'

so there

to*Q'.

7.9) exists

. Let

be the fourth

O

fixed,

7 (Q) = (©)

a dilatation

a,

Pew:

vertex 4 P,

and sends e er leaving

P'

114

Now

One

itself a dilatation. Now

X

: but if

P4P'

translation into

P=

P'

we

could have

sending

P

to

P).

Hence

7.6,

A5a

P' , soby

Now plane

A.

has no fixed points

T

, so

| QQ'

Ie)

theorem

we

come

In fact,

things,

while we

objects

:

1) We will define a division 2)

is in

We

will

assign

1-1 correspondence

3) in terms

of coordinates

is a

sending

P

program

ring

coordinates

in the affine

to construct

a few more

is to construct

the following

F.

to the points

with the set of ordered

of

A,

pairs

We will find the equation of an arbitrary

sothat

A

of elements

translation

of

of F.

A,

of the coordinates

4)

We will find the equation of an arbitrary

5)

Finally,

equations,

and

we will show

this will prove

In the course details

is a translation

we will find it convenient

So our

which

assuming

holds.

to the construction

are at it.

implicitly

(we are

taken the identiy, fT

Q).

, similar for

collinear

if

(because

QQ'

on

and

T(Q)=Q'.

and

T(P) = P'

PP'

lie on

must

|| XP' = X,P,P'

XP

is a fixed point

isis

T

any fixed point of

easily that

sees

it is

of dilatations,

a product

Being

T = 0,0 Zleg

consider

to verify,

indications,

of these

so we

and leave

that the lines

that

A

A

is isomorphic

constructions,

will not attempt

the trivial

in

dilatation. are

given by linear

to the affine

plane

Ne

there will be about a thousand

to do them

verifications

all,

but will

to the reader.

gve

115

1) Definition

of

Piscaelin

Now

let

F

omen

be the

tia

and

let

sends

{@eabe

Oe 1

Awan

set of points

€ Paice

translation which a 4-0,

F.

2

takes

of

OspOlnts OnslweGallathenm

o

into

Onl

1.

avis a point of

be the unique into

dhnoctw

1).

a (here we

dilatation

of

let

use A

Th be the unique

Ada).

which

If

a€F

leaves

o

and

fixed

,

a.

Now

we

define

cer.

define

addition

and multiplication

in

F

as follows.

a 3p le) = Footy 2) = 7 (b). Since

the translations

addition

and

how

is associative

that

ditive

form

at (bec)

Bucy 1s

Dian;

:

Thus

simpler

Note

also

we

and that

is an abelian

group

these

verifications

are

from

T

element,

we

see

immediately

that

:

F

plan suggested on pp.

Now

i

is the identity

inverse.

group,

commutative

(at b)etecr=

:

o

much

and

an abelian

Ta

=a

:

(0) Nemoame

smtbena de: '

under

addition.

than if we

{Notice

had followed

the

96 - 97.) our

-T

definition

T

define multiplication

of addition

that

we

have

tows allied), be Gar

as follows

:

o times

anything

is

116

we

Dace tar, bs 0,

fama

define

= 0, (a) == o 9, i)(1) ab =

Now

a group,

form

since the dilatations

see

we

that

immediately

(Ab)ict-na (bc) 2

Lee

(oj ai a

=

Therefore

the non-zero

plication.

Furthermore,

It remains

reason

, one

ifese full =)

ie

is a multiplicative

to establish

of them

the

is much

harder

our definition of multiplication

than

above,

laws

the

in

other,

is asymmetric.

If c=0 , (a+b) 0o= act be ok,

b 4 o)

(for

distributive

First

multi-

under

a group

the formulae

have

we

form

F

of

elements

inverse.

F.

For

perhaps

some

because

consider

(a+ b)c

If es o',!weuse the formulae

and find Veeye

eune

Now

applying

both

(Qa-taD) Before a lemma.

For

translations

proving any

line

=

= ie erg A

ends

Clama

in the direction

= SL

-l

OLTAG:

-1

E (hssSssek ;

of this equality.

to the point

o, we

have

clelsbGe

the other m

-l

= ogee ee sae Be

in

of

distributivity

A,

m,

let

i.e.

Tran

those

law, (A)

we

must

establish

be the group

translations

of

T € Tran(A)

117

such that either

ete

Lemma

7.10

Tran

(A)

Tu

from

T= id.

the identiy,

or

Let

m,n

airael

gpl)

and let (oy 3 AMEN

as follows:

For

PP!

each

be lines

bea

(UE) =

o

fixed,

A

point of

ewer

n

may

be the

such

A.

We

same).

different

define

a mapping

(WY)

(A) ne tue s id. and

(which

T(P) = P').

be fixed translations,

fixed

T € Tran

one Dil (A) , leaving

in

Tran, (A)

o m

|| m for all P ( Where

, there

exists

a unique

dilatation

that

patel T=oToO (Indeed,

take

@

such

that

p(T) =o0T"'a mW

(with that

,

©

Proof:

m,n

Let

7*

for all

1.

case

parallel

if necessary,

assume

i.e.

|/n. Replacing

by lines

them,

of groups,

o(t,T,) = 9(T, WT.) .

Case

treat the

m

) Define

aa

is a homomorphism

7,7, € Tran_ (A).

where

= T(0).

o).

Then

First we

g(T'(0))

that

we

may

n

pass

and

m

to

be the unique

through

Let

O.

takes

translation which

T(o0) =P',

P'

into

T''(o) = P".

P".

Then

ee ATs If

TT,

© Gnreaie

m

(eY),

ier

sbe the

90,

One CCI eae

corresponding

e=te C/T Tem gt

= 7) + 97%,

dilatations.

0; larg O00)lg 60,7ree a2

SA mens

Then

118

where

we

define

1

il

1

Similarly,

p(T.) = 757%

»

where ste

Ts"

ale

va O51

=

795

?

and

p(T)T,) SAT where

tA

Ty Ts A Bias!

to

corresponds

0,

Toe atge

g,7*0,

have

So we

Gon ee

Sty

p(T, 7,)

O(t Wit.) = TT, + | *T2* Now

9(T)T,)

ca

and

and

Tete

only happen

p(T are

MAT)

are

both

translations

both

translations T*

in the

in the

m

direction.

direction. But this

can

if

and

O(1,7,) = o(t, ko(T,), which

is what we

consider two

anted

the points

translations

TT ,(0),Q,R

Q

above.

to prove. and

R,

Then

collinear which

( To make which

we

have

implies

are

this argument

the images

O,Q,R

Q = R.)

of

collinear,

O

more

explicit

under and

also

the

119

Casicn

come

Litas

not parallelatomm

| mM

oy Sp og ae

wand: taken

by: Tran using

T'

and

i

using

T

Panel

tee

hens sO

@

z

f''' , and

Tran

(A) .

(A) .

Tran

Take

another

line

o

Define

(A)

define

3 Wieswe

(VQ) = fo)

Pee

(NY)

m

“Ge

Or

vv,

, and

Wp ¥,

are

homomorphism

by Case

l,

is a homomorphism,

(Note

the analogy

of this proof with

the proof

of Proposition

7. 7)

Cla Go Gl.

Now Consider jt T

a

rN into

oro

-1

ara

PremtOr

we

can prove

(Ac es

b) Me lnetheslemimna

LneneOmis 7

=f

the map

_ tony znoky

Aa

ra

the other

.

Now

anya,

eh, the

of

stakes

oF tells

a

= Gah us

©

elrT,) = o(T yolr,) or

6

= o(r do(r,)

|

7A

Hence

Py (ah) Applying

both

sides

to

oeNb ak kak 0, we

have

INExrls)) = Wee ar Wie


W. 2)

Project

Thus

W

back

(il)

exelaxs

acs o,

transformation,

too,

is a product of two perspectivities,

1) Project the vertical

onto

the transformation

and so is a projectivity.

This

the point

(x,o0)

direction

in

onto

1 from

x'=x+a

the point

(a,l).

isa product

This

gives x+

a.

of two perspectivit

133

the line

x= 2)

(1,1)

and

y,

getting the point

Project

(a,o0)

Y

back

Y.

onto

1, in the direction

to obtain the point

(ax,o)

of the line joining

.

(iii) x' = i x

This

transformations 1) Project

1

foe)

P oint

is a product

(x,0)

from

of three

the point

perspectivities.

(1,1)

onto

the line at infinity,

, getting W. the point

(1,0)

onto the line

2)

Project

W

from

3)

Project

Z

in the vertical direction

ee scent

o) °

back onto

x=y,

getting

1, getting the

Gaemas

Z.

134

Theorem

Let

F

Gin

1

of projectivities

of

1 into

m = P

linear

of fractional

PGL(1)

itself is equal to the group

let

let

PJ(1)

Zz

Fr”?

bea field,

the group

Then

Ce

a

the line

8.4.

be

transformations

tbe Proof

tions

: We

of three

conclude

have

special

that every

seen

that

types,

each

fractional

PGL(l)

is generated

of which

linear

by transforma-

is a projectivity.

transformation

So we

is a projectivity,

i.e.

PGL(l) © PJ(1). Now

let

Let

~

© € PJ(1) take

the points

Proposition o,l,o@

be an arbitrary

8.1,

into

there

A,B,C,

by the Fundamental there

are

is only one

equal,

i.e.

o,l,o

into

A,B,C

is a fractional

and of course

Theorem

taking

is a fractional

PGE)

this

of

1

into

itself.

respectively.

linear

Then

transformation

0,l,a

linear

on a line

into

However,

(Theorem

A,B,C.

transformation

by

taking

is also a projectivity.

for projectivities

projectivity

@

projectivity

So the

, and

5. 6)

two

so

Rwed(h) qa. ds

Remarks: of our

synthetic

Pprojectivities

1. Notice theory

on a line,

( in the form which

And that is not surprising rather

remarkable

have actually

fact.

converged,

that we have

was

the full strengh

of the Fundamental a hard

, because It says

had to use

what we

that our

and that we

theorem)

two

have

have

to prove

proved

entirely

arrived

Theorem this

result.

is really

different

in each

for

case

a

approaches

at the

5

Same

group 2.

of transformations One

may

which

occurs

More

precisely,

wonder

what

in the statement if

1'

are

let

be any point not on

isomorphic,

perspectivity

1

about

of the theorem. line,

as abstract

1' .

A

is special

is any other

PJ(l') P

of the line into itself.

1 or

Then

groups.

each

Nothing

5

is special about it.

then the groups

PJ(l)

To get such

1', and let

for

the line

an isomorphism,

»:1——~1'

a € PJ(l)

, we

and

be the

have

one

© IP").

and the mapping

a > pap! is an isomorphism One

will note,

of

P.

In fact,

which

is better

there

found

projective

way

ways.

we

have

of the line

1

1 asa

of

these

1 which

presently)

(Details

to make

So we

line in

The

P

other

left to the reader

depends

PJ(l)

say

and

PJ(l)

!).

on the choice PJ(l')

and

. Finally

F

isomorphic,

PJ(l')

a certain

namely

and using

>

is by using

Now

are

PJ(l)

F

incidence

the algebraic

group

as

the group

( this notion

is the field

will

of

= PGL(1l)

One

and

is by

properties sturcture

we will give a third way

namely

cross-ratio

, in case

examining

it in two different ways.

transformations,

preserve

been

into itself,

can describe

given by its coordinatization. terizing

.

isomorphic

that we

plane.

PJ(1')

that this isomorphism

all other

recapitulate,

considering

onto

is no one

than

transformations

have

PJ(l)

however,

non-canonically To

of

on

of the 1]

of charac-

of all permutations be explained

WL of complex

numbers,

136

sphere

of the Riemann

maps

orientation-preserving

all conformal,

of

the group

as

of this group,

interpretation

we will give a fourth

onto itself. field,

as

above,

At=a0

be four

sayy. Cd, Cake tid 200 Le

fe)

by

of the four points,

define the cross-ratio_

Then we

a,b,c,d

and let

bea 1

on the line

points

distinct

F

Let

Definition.

b-d

oh i = rernrae ice : rorr pShea BNAA)

(In case

one

: precise,

of

a,b, c,d

: if

e.g.

the projective

8.5.

over

linear

transformations

of

1 which

of

1 onto

1,

and

the set

A)

such

= A',

F

that

whenever

etc.

then

formation

does preserve

generated

by transformations

let

the group the

i.e.

b-d qe

1, as

PGL(1) group

are

more

.)

above,

coordinate

be

x

which

of fractional

of permutations

one-to-one

A,B,C,D

see

that every

the cross-ratio

four

8.2

them

preserves

cross-ratio,

with

coordinates

, it will be

Then

special

sufficient

So let

fractional

. Since

of the three

of Proposition

a,b,c,d.

definition

mappings

©

distinct

points

linear

trans-

of

= R, (A', Bt, C', D’) .

(iii)

the

and

is precisely

must

the

non-homogeneous

the cross-ratio,

: First we

make

: cross-ratio

the

. Then

R, (A, B,C, D) Proof

must

bea field,

with

{a} on

one

get for F

F,

FU

preserve 1,

Let

line over

varies

o,

we

0,

a=

Theorem

is

the group types

to see

A, B,C,D

PGL(l)

is

(i) , (ii) ,

that each be four

one

of

points

of

1,

aac A,B,C,D) =--—-. Ry, ( ) a Oo‘el

(i)

If we apply a transformation

our new points d+2

A',B',C',D'

, respectively.

Ry which

(ii)

have coordinates

(A Bc

x'=

at+A,

(atA) - (ctr) (atA) - (d+)

=

to be equal to the original

If we apply a transformation

x+XA,AEF

b+A,

‘

ct X,

(b+A) - (d+A) (DEA) R= (eX)

:

cross-ratio.

of the form

x' = \x,

AE F,

we have Age

a which

of the type

Hence

is easily seen

4 40,

b-d

-— b-c

again is clearly (iii)

If we

Aa - Ac

pe)

Ab - Ad Ab - Ac

ee Aa - Ad

equal to the first cross-ratio.

apply the transformation

x' = a x

, we have

eee

Now

Ree CAMS Bi Cin x

DL)

above

by

multiplying

and below

cross-ratio

again.

(One must

a,b,c,d

0

©

is

Thus preserves

or

we

shown

which

respectively,

R. (omlioon) or

Now

(x)

we

obtain

the special

fractional

conversely,

Then we

a Re (apioeceocn)

linear

Let

have

one

of

.

transformation

let us suppose

cross-ratio.

= x'.

the original

case when

- left to the reader)

that every

preserves

and let

abcd,

consider

separately

the cross-ratio.

a transformation

a,b,c

have

avenc

»

send

that

»

0,1,

is into

138

or

= Vives

ey

is given by the expression

@

we find that

x',

Solwing for

cx

+ a

oe

>

bac

which

is indeed a fractional

linear

transformation. q.e.d.

Example: the line

1,

Let

F = © be the field of complex

is the projective

complex uumbers

over

, plus one additional

easily represented stereographic

line

as a sphere,

projection.

(For

complex

becomes

projecting from ponds

to the

a One-to-one Now (q.v.)

the

N pole and all

manner

correspond

of the Riemann

sphere

are

i.e.

secting

curves.

other

with the points

it is proved

conformal,

the

see

in courses linear

precisely

This

sphere,

is most via the

on functions

sphere)

.

of

Then

, the point at infinity

points

of

on the origin of the

of the sphere

corres-

correspond

in

of the complex plane. on functions

of a complex

transformations

to those

preserve

o.

book

S pole of the

onto itself which

which

any

Then

the ''plane"'

called

is placed

N pole of the sphere

that the fractional

plex plane

point,

details,

A unit sphere

( which

, thatis,

called the Riemann

a complex variable.) plane

C

numbers.

of the

one-to-one

preserve

the angles

variable

extended

com

-

transformations

orientation,

between

any

and which

two inter-

W39)

Projective

Now general,

we

any

Collineations.

come

study

automorphism

collineation,

because

Definition:

of the projective and

to the

1' = g(l)

of a projective

A projective

plane

m7,

is its image

is a mapping For

lineation.

jective plane

But we

satisfying

a fundamental points, We

by showing collineations

thiateat

see

and

collineations

P 6,

: there

is precisely

Proposition

8.6

is calleda

automorphism

1

isa

line of

the restriction

of

@

7,

@

to

1,

F

l' , should be a projectivity. is a projective

there

are may

that if

m7

then the projective is a unique

into any

elations

n= iP , where

is an

transformation

that in general,

the structure

called

then

1 to the line

that it is generated ,

In

1,

collinear,

study

@

whenever

In fact we will prove

theorem

no three

will also

P5

@,

the identity

will

collineations.

that,

under

of the line

example,

plane

collineation

such

collineations,.

into lines.

lines

it sends

ol) :li—— which

of projective

other

and

sending

four

points,

no

homologies.

is a field,

collineations

of them

three

of projective

special

kinds

Finally,

then the group

pro-

is a projective

one

of the group by certain

more

col-

satisfy

any four collinear.

collineations,

of projective we

will

show

of projective

PGL(2,F).

Let

@

be an automorphism

of

7.

Then

@

is

140

a projective that

such

P

bea point

ete

dea

collineation

is a projective 9

that

conversely

1) is a projectivity. let

exists

line

some

is an automorphism

dl.) = ee . Now let

Say

not on

1 or

say

are

Since

collinear.

that

Pi, At, As!

Let

1' = (1)

that

are

»)(A) = A,

~

, any

lo will do.

whose

restriction

is

1 the

P

(where

. In other words,

'

as

saying

Aێl

and

this is the same

denotes

the action of

the transformation

‘

°

as the perspectivity

l'

1) / Gall itty. >t

In other words,

ol, %ce)

and

P,A,A,

pio! :1'—>1!

il

to

1 be any other line,

1, . Nowif

same

is an automorphism,

collinear

el,=%

So

1s:

1, be the perspectivity

A, € 1, , then

is the same

me -

is a projectivity.

ol,

Se ey If » suppose

if and only if there

collineation

00

as

saying

o ).

141

But

wel,

jectivity,

, and

a

are

fo) and hence

©

all projectivities,

is a projective

so

el,

collineation,

is also

since

a pro-

1 was

arbitrary.

Quende Be fore we must

study

some

can prove special

much

types

of collineations,

homologies.

Then we will use them

of projective

collineations.

Definition, plane

#7,

which

has no Other

affine

X

@

plane

1) at

ff - 1a

of

@.

PQ

Hence

restricted

But of

a

P

and

and

Q

I P'Q'

to

A

m

in

@

with

axis

properties

The

restrictedto

15:

, say

line of

any

and

of the group

of the projective

Sa pointwise

fixed

, and which

1, is called the axis of the elation.

7 , with

P,QE€EA,

axis

let

1, , and let

PQ

meet

A

1, at

be the

X.

Then

Q' are

under A,

so

is a dilatation.

has no fixed points

Le , sO

line

elations

we

also meets

P'

in fact a translation. of

some

For

P'Q'

X , where

the images

@

leaves

beanelation

is fixed,o

to deduce

collineations,

called

An elation is an automorphism

fixed points.

Let

about projective

outside A

is

Conversely

any translation

of

A

gives an elation

142

an affine plane form

If

q@

1, , then we

tion of the translation

a], . Indeed,

meet

O

One a group

there

.

O.

Then

if

a

is no reason

However, we have

shown

why

17

takes

can

oe

EY

fixed,

1. into

car

:

fixed. is an

»

taken together

different

axes

be an elation

something

@.

1 p emets|

about all the elations.

with a fixed axis

| Similarly

(sat

form 1, :

at all.

First

1, (including is another

the

line,

the

fo) of

Aut mw.

Let

aT (so long as

@

©

be an automorphism

satisfies

of

P 5, there will be

pap

=]

easily be seen

to be an isomorphism

of

EB

a

vee

We

one

takes

aT into

1, into

1, , so that

can

that

leave

onto

gE

fo)

that

takes

Similarly elation.

1

with

Say they

PP'||QQ'.

the mapping

for example

and

say

9

oO

Note

should

subgroups

a——> tore

elations

speak of the direc-

of the elation

that all the elations

aa,

proup,

are both

!) . Then

are

can

P,Q,

for any

center

that the elations

elations

one

@,

we

identity) forma

which

isthe

should not suppose

For

of

a group.

is an elation with axis

1, at

gE, fo)

a group

to the fact that the translations

only refer

need

: We

Proof

1, form

the elations with axis

the identity,

if one includes

correspond,

m1) SeLIenicer

of the affine plane

, to the translations

by restriction

1

m with axis

of

elations

The

8.7

Proposition

see some

pay

details

alt

1, , @leaves

ieee has

to the

1

leaves

no other

reader.

fe)

1

pointwise

1, pointwise fixed points,

This

so it

is a familiar

143

situation

in group

theory.

Definition

of

G.

Then

we

an element

.Let

say

g €G,

bea

have

group,

Ho

aS

of

have

Proposition

and

that

fe)

is an isomorphism we

G

we

and

Hy

the following

and let

are

Ho

definition.

and

conjugate

Hy

be subgroups

subgroups,

if there

is

so that the map

h

Thus

In fact,

gh ig

Ho

onto

Hy

proved

8.8

E,

-]

Let

EB

and

fe) 1 1 , respectively.

Let

denote

™

be a projective

the groups

Then

E,

plane

of elations

and

E,

of

7

satisfying

with

axes

P 5.

1

are

conjugate

subgroups

that any

conjugate

subgroup

fo)

of

1

Aut 7. Conversely, of

E,

one

is of the form

can

see

E, , for

easily some

line

1

in

fm.

Thus

the

of

Aut 7,

set of all

fo)

elations

of

7

is the union

of the subgroup

E,

together

with

ce)

its conjugates. Definition

morphism which

has

of

which

and

above

to dilatations with axis

qf

A homology

precisely

the homology, As

.

one

O

, we

of the projective

leaves

a certain

other

fixed point

plane

7

line

be pointwise

O.

1, is called

is an auto-

fixed,

and

the axis of

is called its center. note

of the affine

that the homologies

plane

1, , and the identity,

7 - 1. Hence, they form

with

axis

if one

a group,

1, correspond

adjoins

which

the elations

we will call

H, Oo

For

any other

axis

1

Hy

is a conjugate 1

subgoup

of

Aut 7 to

H, fe)

144

O

1

and center

in a Desarguesian

projective

with axis

lo , the homologies

not on

group

Hy fee

move

a tr

since

And

1

of

P, we

point

the homologies

fe)

O

H,

are the union of the subgroup

m

can

we

plane,

Hence

Hy foe

to

is conjugate

Hy Pp

that

forma

O

and

L

line

to any other

O

1, and a point

see as above

1, , and for any point

that for any line

see

we

more,

Refining some

its

with all of

Aut 7

of

°’

conjugates.

Proposition

8.9 . Elations

col-

projective

are

and homologies

lineations.

Proof

8.6,

. By Proposition

it is sufficient to note that their

Proposition Let

A,B,C,D

of which

and

8.10.

collinear.

homologies,

such that

Then,

translation

of

such

that

have

reduced

since

m

leaves

more,

since

A! fixed,

gB)

a line

Ute

and

sends a,

A

takes

es such that (cf.

into B.G.

B"',C"',D''

is an automorphism

A

Chapter

A',

i.e.

Deintom

1D) By

no three

@ of elations

= B' , o(C) = C'

of finding a product sends

satisfying

of points,

can find a product

is Desarguesian

to the problem

a,

one

Choose

f - 1, which

a, (A) SiN!

which

Then

plane

be two quadruples

»(A) = A',

Proof : Step 1 15:

7 be a projective

A',B',C',D'

are

on

Let

is a projectivity.

which

to its axis is the identity,

any elation or homology

of

But the restriction

restriction to a single line is a projectivity.

and

¢g(D) = D'.

and VII)

A'

are not

there

anelation bw

Gu

, A',B",C'',D'"

isa qa

i

of

f7

Dose hentwe

of elations

into

and

and homologies

B',C',D'

. Further-

are

points

four

145

no

three

of which

A,B,C,D,

we

assumption

that

have

Step 2. Then

choose

using

Ee

are

collinear.

reduced

A=

a,

we

are

can

Let

reduce

D,E,

There the

are are

exists

affine

whcih

sends

leaves

D

words, By of center

into

there m

Step

and center

that

1 , and

A € \ OULD

such

reduced

that

won

q 1.

a,(B) =s Due

luher

the original problem

Gmands

with axis A=

and

BD'

A',

Guat enn Otson

are

1, , because

not collinear.

1, , such B=B',

and let them

that

So again,

a,(C) = Gla

ndEso

C=C'.

meet at

E. Now

since

A,

of

BC,

fixed,

and

E, In other is a homology

with A,

from

m-

A

the additional

and

a dilatation

plane

such

A',B',C'

case

AD

collinear,

different

a,

to the

Draw

1

we have

e Lhenm

and

an elation

Step 4.

, under

as

aby) Bu.

l, =eAB

the problem

A,D,E

again,

not collinear,

choose

line

with axis

and relabeling

tomthercascmeAs—sAgsand

A,B,C

A',B'',C'',D'',

to the original:problem

another

elation

Step 3.

relabeling

A'.

Choose

an

Thus,

axis

which

on

BC

and

sends

D

Similarly,

B, which

sends

into

E,

there

is

E

into

a homology

D!.

B,

Therefore

of

m

BB,

with

axis

leaves

AC

A,B,C

146

and two homologies.

we need three elations,

P 6.

and

plane

be a projective

leaves ©

Then

collinear.

are

of which

P5

satisfying

7 , which

of

collineation

no three

A,B,C,D,

fixed four points

m

projective

bea

»

Let

Let

8.11

Proposition

that in general,

Note

the proof of the proposition.

completes

This

D'.

into

D

and sends

fixed,

is the identity. Proof:

y

sends

~

fixed,

so

be the line

and

is a projective

on a line (Chapter

with two fixed points

But

@

fixed.

D,

are

to

fixed,

be aprojectivity,

A

and

D

ol, is a projectivity of

points

so it must

C

also leaves

So

@ restricted

and

and

1 must

by the Fundamental

Now

A

B

to

fixed the three

fixed, 5.).

Since

restricted

AD.1=F

leaves

1 pointwise

»

BC.

collineation.

© mustleave

into itself which

leaves

1

1 into itslef,

since

1

Let

B,C,F.

Theorem 7-1

isa

Hence

©

for projectivities dilatation

be the identity.

Hence

is the identity.

Theorem Let

PC(7)

m

8.12.

be a projective

the group

A',B',C', D'

and

Proof:

plane

Theorem

satisfying

P 5

of projective collineations

are

there is a unique

o(C) = C’

( Fundamental

two quadruples element

of points,

» € PC(m#)

such

for Projective

and

of

Collineations

P 6, and denote

by

#7. If

A,B,C,D

and

no three

collinear,

then

= A",

= B',;

that

(A)

dB)

o(D)\= D's Since

elations

and homologies

are projective

collineations

147

(Proposition

8.9)

to

A',B',C',D'

On

the other

eo

and

since

there

(Proposition

hand,

if

is a projective

enough

, there

y is another

8.13.

8.1l)

The

projective leaves

. Hence

group

of them

certainly

such

collineation which

is the identiy (Proposition Corollary

8.10)

are

A,B,C,D

is some

such

collineation,

A,B,C,D

© =»,

PC(m)

to send

fixed,

@. then

and

so

and @ is unique.

of projective

collineations

is generated by elations and homologies. Proof:

collinear,

Let

and let

Proposition

» € PC(m)

y

8.10

send

A,B,C,Dto

p=Pp,

so

we

»

come

into

of elations

A',B',C',D',

pg is a product

Finally;

A,B,C,D

A,B,C,D_

a product

sends

, let

be four points,

A',B',C',D'

of elations

. Construct

and homologies

Then by the uniqueness

no three

which

by

also

of the theorem,

and homologies.

to the analytic interpretation

of the projective

collineations. Theorem

8.14.

projective plane over

Let

F.

F

bea

and homologies Consider

A

First we

are

is the affine plane

ae

will show

represented

an elation

x,

x, /x,

y = x,/x,

and let

7 = P*

be the

Then

PC(7) = PGL(Z,F) Proof:

field,

a@

40,

~ that certain

very

special

and

center

elations

by matrices.

with axis

with affine

x3 = o

coordinates

(l,o,0)

. If

148

i.e.

x-direction,

inthe

A

of

is a translation

a

then

it has

equations

ae (7.

pret eeee

oy,

Xo

ek

1

yal

z

es 5

ete ae

i

a€

can find

and

(l,o,0)

where

>

F.

Now we

a

if

q@'

is any

a matrix

to

O.

A,

Then

is an elation

other such

a!

Similarly, . Passing

with center

elation, that

consider

with

TA

axis

sends

special

AE A

=i

the line

type.

center

x, =o

O,

into

1

x) 4 ©, we

is

words,

a'

a €F.

8 , with axis

(0,0) , hence is a streching in some coordinates

In other

, for some

a homology

to the affine plane

equation in homogeneous

1, , and

will be of the form

of the above

; : is represented by the matrix

(l,o,0)

3

72

tes x', =X,

with

are

equations

So its homogeneous

see

re

and center

that it is a dilatation

ratio

k +0 » and its

x!

kx

1 = x37 kx,

So it is represented

We

can

by the matrix

get another

: multiplying

matrix

representing

= b=k _

by the scalar

the same

, so we find

transformation

by

B is represented

also

by the matrix

Ome As

B

of one

before,

any other

of this form,

of the form Thus

©

lineations

homology

BH, B

, for some

we

seen

have

i.e. 8.13

have

seen

element

of

into four

points,

no three

groups

must

(Chapter

PGL(2,F)

PC(m)

by some

is represented

matrix

by a matrix

bio.

elation and every

elements

above,

can

of the group

the group

and homologies,

homology

of projective

col-

so we have

© PGL(2,F).

a unique

by the subgroup

is a conjugate

§'

they are

is generated by elations

But we

§'

bEF,

that every

But by Corollary

PC(7)

the two

bree reap i150.

so any homology

be represented by a matrix, PGL(2,F).

sail

VI)

, that over

sending

collinear.

, according

four

Since

a field

points,

F,

no three

this is already

to the Fundamental

there

is

collinear,

accomplished

Theorem

be equal. q.e.d.

above,

150

Corollary matrix

times

a product of conjugates

above.

with

Let

3 X 3

Hy

M

8.15.

apa

In particular,

ey.

2 as

GAP.

|p)

1

bea

coefficients

we

MoAb HB z 2 b, with

F

field. in

F

can

of matrices can write

2

Then

every

invertible

be written

as

of the two forms

M

a scalar

E,W and

in the form

BH1 db) eS1 Ale 3

a,

»b, ACF, b»b,,AFO,

Ae 3

a,

ee @ 1

A), pA, A,,A,,B,,B

a)

see 1

2

invertible matrices. Remark:

From

this result,

one

little effort the fact that the determinant is determined

Compare

also

uniquely

Problem

by the properties

19.

can

deduce

function D1

and

on

with

comparatively

3 X 3 D2

matrices

on page

30,

151

PROBLEMS

In the following

sitions

given in class. 1.

have

Show

the same

respondence

of points

between

completing

number

and propo-

Show

can

lines

establish

that this

in an affine

plane

a one-to-one

cor-

is also the

cardinality

of

on any line.

If there

is a line with exactly

Discuss

Prove

affine plane is the possible

n

points,

show

that the number

a

systems

of points

and lines which

VS Cy.

that the projective

of four

plane

points,

If one line in a projective

of points

6.

in the projective

Let

S bea

of

7 points

, obtained

is the smallest

the restrictions

is an affine

plane.

plane

by

possible

S

not on

of lines

in

S.

also

n

points,

find the

plane.

of

Prove

S,-

plane has

projective plane,

to be the points

S5

the affine

of parallel

that one

them).

the axioms

plane..

5.

be

(i.e.

use

explicitly.

two pencils

the affine plane

projective

you may

to them

UE) Ik, AS) Yan DS) Sh. eqeke sevors

4.

Define

that any

in the whole 3.

seyeliy

Refer

cardinality

the set of points 2.

problems,

that

and let

1, and define

Prove S

1 bea line of

(using

is isomorphic

lines

P1-

in

P 4)

to the

So that

S. to S5

comp letion

of

152

7, prove

Using the axioms

the following

except what

one

statements.

is containedin

of projective careful

Refer

three-space,

not to assume

to the axioms

P, Q lie ina plane

¥

anything

explicitly

by numb

then the line

J.

b)

A plane and a line not contained

in the plane

c)

Two

d)

A line and a point not on it lie in a unique

8.

Prove

meet

in exactly

point.

plane,

the results

distinct planes

9.

Show

i.e.

satisfies

10. from

a,b,c,d

ina

two

affine planes

and

A'

A——>

A',

which

are

with

9 points

an affine plane with

Ay =afll,

etc.

pencils of parallel lines. four points, points,

e.g. e.g

thee liline

use

Write

isomorphic.

is a one-

. ( Hint

: We know

lines has four lines in it.

label the intersections

of four points

are

if there

16 points

lines,

subsets

(You may

is a

takes lines into lines.)

of parallel

other

three-space

P 4.

one pencil

choose

plane.

P1!-

isomorphic

1 that each pencil of parallel be

line.

projective

the axioms

A

Construct

problem

[

one

PLANES

that any

T :

in exactly

problem.)

AFFINE

(We say that two planes

to-one mapping

meet

that any plane

of the previous

FINITE

Then

Be very

If two distinct points

them

projective

Let

S 6

is stated by the axioms.

a) joining

S1-

and let To

1,2, 3,4 be another.

construct

the plane,

to be the lines in the three

out each line explicitly,

2 = { A,,B,,C,,D,}

:

other

by naming

its

you mu

ll.

Euler

' A meeting regiments

row

of 36 Officers

must

and each

of different

of six different

be arranged

column

ranks

It has from

in 1779 posed the following problem

contains

shown

this fact that there

the

and from

six different

in such

a manner

that each

6 officers

from

£=

that this problem

is no affine

We will consider

are

ranks

different regiments

and

"' .

been

of 10 elements,

in a square

:

has no solution.

plane with

the Desargues

Deduce

36 points.

configuration,

{ 0,A,B,C,A',B',C',P,Q,R

which

is a set

}, and 10 lines,

which

subsets

O,A, A! OC, 5, p OFC, AG Bk Ps Deg ke A,C,Q

HSoAd Siar ©: By Gar Hsje Oe 52 PO. Let

G=

transitive

Aut C

be the group

12.

Show

13.

a)

on

that

Show

G

of automorphisms

is transitive

that the subgroup

a set of six letters.

on

of

of

2.

J.

G

leaving a point fixed is

154

fixed has

order

Now

planes,

the order

we

consider

results. which

call

we

Oran Aw ibe fay

2 =(tOrA

Cy A noe

3 =i

OF Be Orcs,Rey

(GASB

4 5

Cab

Oo: G==——> is an isomorphism

Pek. of

element

Perm

of groups.

of the

a permutation

induces

G

, and that the resulting mapping

{ 1,2,3,4,5}

set of five planes,

UE

Ors Ray

POA. BC.

i}

that each

Show

14,

{ 1,2, 3,4,5

Thus

G

}

is isomorphic

to the permutation

on five letters. 15.2)

Let

L,

of

subsets

further

some

the previous

from

G

of

namely

De

group

points

2.

Deduce

c)

collinear

two

leaving

G

of

that the subgroup

Show

b)

m7

et

TT, be a set of four

be the free projective

(as in class)

Show

16. projective

of the projective

that these

Prove

plane

such

pair of distinct points (Hint : First

minimal

generated

. Show that any permutation

to an automorphism b)

plane

reduce

altitude. )

are

points

A,B,C,D,

no lines.

by the configuration

of the set

plane

and

{A,B,C,D}

7

extends

7.

not the only automorphisms

of

7.

that there is no finite configuration in the real that each

lies to the

line

on a line, Euclidean

contains

at least

three

and not all the points plane,

then

choose

points,

are

every

collinear.

a triangle

with

155

17. nm

, thatis,

Let

m7 T

of

7.

( and only one)

plane.

Let

£

T

be an involution 7 sb

that

7 such

be the set of fixed point of

7.

of

> ae

Prove

that one

of the following is true : l. There

is a line

1) InAs

Case

2.

is a line

1

There

Chatha

and a point

taal

Ee

¢ 1, , such that

we

define a "line"

:

Case to be any

Let

of

Case

ire Lou ees

%

projective

be an automorphism

let

identity map

bea

3.

Y

subset

is a projective of

£,

plane,

of the form

where

(line in

#)f\Z,

which

has

in at

least two points. Prove

P 7 is not

has

that

Case

For

each

case

of a projective

the property

19.

Let

1,2,3 above,

plane

arise

only if the axiom

D1.

»

be a function from

(A: B)

@A)

'

Let

T 4 identity,

which

for i.e.

but more m

Am

the set of

2 X 2 real matrices

such that

p(A) - @B) , and

= det A,

(A similar 20.

proof a specific

of the given case.

1B) 23. o(* eas that

give without

7 , and aninvolution

fovAt= ( i e} } to the real numbers,

Prove

1. can

satisfied.

18. example

furthermore

each

a clk

{ x we) = Gycl > Joye,

involved proof would work

be the real projective (eat Ort)

sioner aulll 2, lo, fey, lS

plane,

and let

for

Usk

n Xn matrices)

.

156

BS

(15,1),

Crer(c, OF!) 1D) = {( chy @p)9)

AMG Gace leurs

ma-c R,(AB, CD) SS

is equal to

- 1 . (In general,

cross-ratio

of the four points.

geometry in the affine plane 21.

P1-P4

and

P6

Consider

projective plane,

this product

R(AB, CD)

) You may

use methods

the words

of the dual,

to prove

b-d den Pees

P 6* , of

of Euclidean

Pappus'

PQ

Pappus'axiom

Write

out this

geometry.

holds in the real projective plane) For the next three

axiom,

, etc., P 6.

make

Then

use

P 6*,

and take the line

the Euclidean plane.

of Euclidean

''point'' and "'line''

the configuration

to be the line at infinity.

is called the

x, Ey GLe

By interchanging

statement

22.

methods

are four

points if and only if the product

harmonic

a careful

CD

AB,

that

Prove

F

be four points on the "' x7 axis!

of Pappus'

( using the notation then becomes

statement,

(This

axiom

gives

in the real given in class)

a statement

and then prove a second

in

it, using

proof that

P 6

.

problems,

we

consider

the following situation

let

be a chain of two perspectivities be the resulting projectivity from

, andassume 1 to

14n.

n, and let

Let

9: 1—>n

X be the point

l-n.

:

eye

23.

1)

Prove

that if

is actually

a perspectivity

, then

@(X) =X. 2)

Now

the following

conditions are

concurrent,

b)

O,P,X

are

collinear.

With are

are

©, P, xX are

let

il

concurrent.

With

and

en (RUSE Remark

mentioned under

that one of

above

that there

, assume is a point

is the perspectivity

l>m,n

7

above, are

furthermore Q

such

1 Q roe

assume

that

‘.

( Use P5 or P5%),

also that

not concurrent,

Let

Y = l:m,

Yeo rovesthatOmisethesperspectivity:

OMmOrsErO7%)ie :

in class.

The

problems

In fact,

the initial hypotheses

equivalent

@

but that)

a—Inoneemand letmOnO©

0

Prove

the initial hypotheses

collinear,

= X , and prove

or

the initial hypotheses

collinear,

Zoe

(X)

holds

1l,m,n

1,m,n

O,P,Q

simply that

a)

24. that

assume

23,

they prove above,

24,

25

give a proof of lemma

a stonger

the following

result, three

namely,

conditions

:

(i)

y

is a perspectivity

(ii)

p(X) = X

(iii)

either

a) or

b) of

# 23 above is true.

that are

5.4

158

j isa

, andwhere

F = {a+ bjla,b€k}

Let

3,

and multiplication modulo

, with addition

be the field of 3 elements

k={0,1,2}

Let

26.

symbol.

a) Define addition and multiplication i = 2,

and prove

b) elements

Prove

of

F

27.

where

that

is cyclic of order A=F

Define

addition

in

A;

asa

using the relation

group

F* of

and denote

the

elements

and multiplication

(sy) en (xan)

the right-hand

non-zero

8.

set,

addition

(Eta

F,

is then a field.

that the multiplicative

Let

x € F.

F

in

in

(here

A

of

A

as

(x)

as follows:

the left-hand + is the

+ is the addition in F). if y is a squarein F

(xy)

(x)(y) = { (x y) if

(We say

y

is a square

Prove

a) A

b) under

in

F

if

y is not a squarein

4 z © §

is an abelian group

The non-zero

elements

such that’ under

A*

y = + Ne

+

of

A

from

a group

multiplication.

c) (0)(x) = (x)(0) = (0)

for all (x) € A

d) ( (x) + (y) ) (z) = (x)(z) + (y)(2) 28.

previous

that

F.

Let

problem

a),b),c),d)

A

bea

(i.e.

hold.).

that the left distributive

finite

A

algebra

is a finite

Note law

that

satisfying

set,

A

is missing.

for all (x) , (y),(z)

with two

would

a),b),c),d)

Operations

be a division

Prove

of the

that one

t, such

ring,

can

except

construct

€A

159

:

:

a projective

plane

I.

x, € A,

(Prove

A point

where

2

aN

over

(x) x55)

condition

fit

Verity

thevaxioms

1)

somewhere

If

A

not satisfy

Desargues

Axioms

''A,B,C some

Now

A5a,A5b,

of points)

Pl] - P 4%

You

may

equations

need

to use

define lines

!

the finiteness

of

A

and

let

are nice £

A6

of the problem

27,

ae

axiom

P 5.

projective

Thus

PS

show

that

iz an example

of

plane.

for the real affine plane

In the ordinary

down

of the form

CaGrAY

sets

is the algebra

non-Desarguesean

30.

Write

an equation

in the proof.

29.

relation

satisfying

Not all linear

2)

tsaor

)

°

determine

Two warnings:

|\nc -Aue

with

= Om

x

equations

(x) 1X55 x3)

Id, © © AN

=

cx,

or

these

LOrPany

So

+

x5

: of triples

~ ( XA, X5A, xX)

Se

or

a finite

class

A line is the set of all points ae

(Prove

as follows

is an equivalence

this is an equivalence II.

does

A

Euclidean collinear,

properties

be an

( you

plane, and

of this

abstract

define

B

this

affine

let

is between

stand for the A

and

C'"'.

relation. plane

satisfying

one-Pappus'axiom).

Al, A2, A3,

Assume

that

160

D

has a notion of betweenness

A,B,C€Z satisfies (Make

, we

have < ABC

certain axioms,

sure

given,

there were

>,

namely

enough).

i.e.

for

certain

and assume

of points

that this notion

the properties Add further

triples

you listed




earlier.

a ''completeness''

axiom,

say C : whenever and

1'' , so that no element

other

subset,

Vo Bieol

then there

VV Ge.

(Dedekind

(You may

, must

use

points,

a unique point

that your

that

(Pasch's

A €1,

subsets

elements

l1'

of the

such that

BAG oe

geometry

plane over

R

is the only

2), with

this notion

the real numbers complete

R

ordered

of .

field.)

A axiom)

:

three non-collinear

if

< BCD and

>

,

then there

two

andaG.4Ac we have, : symbols

Lee oss a)

(1234).

What

Let

Ge

S,

is the order

be the subgroup

of

G

? ( The

generated

order

by the permutation

is the number

of elements

in

161

b) (12) and

(34)

c) If so,

Let

write

HE

. What

S4 be the subgroup is the order

of

Is there an isomorphism it explicitly.

32.

The

If not,

Pappus

and 9 lines

in the diagram

group

What

is the order

of automorphism

of

?

(of abstract groups

Configuration.

of 9 points

as hown

by the permutations

) de). 3 (GS

explain why not.

is the configuration

a)

H

generated

{£,

of the

= ?

if b) Explain briefly how

you arrived at the answer 33.

a)

What

IN

a).

In the real projective

line joining the points

b)

to

(1, 0,

1) and

- x, + 2x,

In the real projective

any four points,

automorphism

2,

which

will send

no three

what

is the equation

of the lines

? plane,

any four

collinear.

we

points,

know

that there

no three

Find the coefficients

3

ean =1

which

sends

the points

of the

3) ?

with equations

ae

Gi

0

3x, tf X5 1 x, =0

automorphism

(1,

is the point of intersection x

34.

plane,

B!

i=1,2,3

is an

collinear,

=i

of an

into

IY

162

Bye (Chale 1),

(17050),

C=

(0,1, 0)

B=

(0,0; 1)

A= into

Alte (1000) = eps) = (Onl, Nite

=U

eet

pe)

hes)

respectively.

a) State the axioms

35.

b)

cc) 36.

not hold.

Outlines

"There

four

of which

are

also that.

points,

refer

no three Ov

ij, 2 2. and)

of the following

each

(Please

and

it,

to results

explain wit

sta.

state which

aes

in class,

proved

collinez

— Seand

unply) planes,

projective

are

plang

axlom

of

does

or

and give brief

of their proofs.) a) The

projective

plane

of seven

b) The real projective

plane.

c) The

plane

37.

free projective

a) Draw

a picture

IE + identity

? If so,

38.

1, l'

write

points.

by four

generated

of the projective

an automorphism

b) I there

eyes

the staement

P 5, P 6, P 7 holdin

the axioms

does

proof that they imply

Prove

For

of a projective

a complete

Give Q:

P 2, P 3, P 4

Pl,

one

T

down

of

gq

plane

such

explicitly.

points.

of seven

that

points,

7.

rv! = identity,

If not,

explain why

not.

Let

X=1-1'.

Let

C,D

Let Let

be two

a projectivity

©:

A,B

distinct

1——~>

be two distinct lines in a projective be two distinct points points

1'

on

which

1' , different

sends

AS

on

7.

l, different from

from

tO

plane

X

-

X.

Construct

me Gens

respectively,

163

Bo Let

»

on

1,

harmonic

sarily

a projectivity

the four

Let

are

(where of

on

four

m

which

send

a)

element

of

1, such

A! = y(A)

7

points

Proof

or

Pl

that for any four

, B'=9(B)

?

satisfying

= A'B'

etc.). counter

of the complete

- P 6.

points

, C'D'

Is

are

~ neces-

example.

quadrangle

on

Tes,

be a projective

A

to

Let

F

F.

plane

harmonic

1 into itself

distinct points

42. zero

of the points CD

(+1,

points

be two

dna projective

Find the diagonal points

41.

there

AB,

points

40.

B

ls beralline

be a permutation

A,B,C,D

four

let

of seven

many

automorphisms

. How

of

B ? Give

bea

Prove

plane

reasons

your

division ring, that the map

Let

of

A

7

and

are

!.,

and let

©:

points.

F —~>

\ bea F,

fixed non-

defined

by

olx) = Ax” for allx€F,

p

b)

Let

elements

has

(Recall that are

is an automorphism

defined

F=

p

be a prime

{ 0,1,...,p-1} p.)

F.

number

no automorphisms

modulo

of

. Prove

other

, where

that the field

than the identity

F

of

automorphism.

addition and multiplication

164

four in

any

order.

you

may

44,

In

wish

the

tained

in

center

be

AJA,

ordinary real

let

theorems

any

Euclidean

projective

P

bea

from

to

Prove

explicitly

class

from

which

use.

tangents

Draw

Y .

to

the

0,

the

Quote

points,

harmonic

four

are

they

that

and

A,B,C,D

points

exactly

has

£

that

Show

7.

of

line

any

be

£

let

and

j

# = IP.

let

elements,

three

with

field

the

be

F

Let

43,

meet

plane

(considered

plane),

let

point

outside

P

to

C

,

OP

at

B,

(by any method)

C

being

con-

a

circle

with

t)

,

and

let

meeting

C

at

Ay

and

OP

that

C

be

as

let

X,Y,B,P

meet

are

and

C

four

and

at

A,

t.

2

X

ar

harmonic

points.

(p -1)(p-1).

First class

prove which

that says

G = PGL(2,F). that

Then

use

the

a matrix

ple Des ee hy te

Tee of

row and

elements

is

all

C=

of

F

zeros,

has

and

(c),C5-C3,)

the Fundamental

Theorem

ts

determinant

the

points

of

m

are

for projective

# 0

A = not

if

and

only

(a,-a5.85), collinear.

collineations

of

if

B= Or 7.

you

no

(b,,b5,b,) may

use

166

BIBLIOGRAPHY This

list includes

preparing

ln

from

the present notes.

Ey

- chapter

Artin,

Artzy

abstract

a good chapter

on a plane

geometry

H.F.

University - Volume

between

1929-1940 I,

G.

chapter

I.

.

- we

to the

chapter

sketchy treatment

R.D,

of finite order''

planes,

one

has

on group

Carmichael , 1937

of which

uN. %- 0! 950-

in an affine

, Addison-Wesley different

plane,

axioms

, 1965. one

non-Desarguesian

the proof that any

given in these

- section 108 contains

.

ours.

of Geometry''

can put

planes.

, Cambridge

have

theory

of perspectivities

of Modern

to supplement

Algebra"

the very

notes.

reprint,1956

examples we

, "A survey

, ''Introduction

, Dover

chain

to a chain of length two.

Birkhoff and S. MacLane , 1941

5.

various

can be reduced

Macmillan

refer

on the various

bibliography

Interscience,

than

in

.

distinct lines

4.

directly

of coordinates

Geometry"

, ''Principles

used

for further

Algebra’,

, especially

Baker

were

to them

approach

, ''Linear

- contains

3.

which

the construction

a slightly more

R.

Refer

“Geometric

Il contains

2.

only works

to the theory

of groups

.

of finite non-Desarguesian

reproduced

in problems

projective

26-29,

167

6. Hill

H.S.M.

Coxeter

, ''The Real Projective

Plane"

, McGraw-

1949

- A good general

7.

reference

H.S.M.

- chapter

14,

Coxeter

gives

for

synthetic projective

, 'Introduction

a good brief

survey

geometry.

to Geometry",

of the basic

Wiley

topics

, 196l.

of projective

geometry.

8.

Wiley

W.T.

Fishback

, ''Projective

and Euclidean

Geometry"

,

, 1962.

- A good general

9.

D.

Hilbert and S.

Chelsea,l952

Springer - chapter

reference

(translated

, much

in the spirit of our

Cohn-Vossen

from

German

,

, ''Geometry

treatment.

and the Imagination".

''Anschauliche

Geometrie"

1932.) III on projective

configurations

is very

pleasant

reading and

quite relevant.

10.

M.

1942.

Dover

- see

chapter

Kraitchik

reprint 1953 VIL,

Nostrand - a very

A.

12

for the interpretation

, and Euler's

Seidenberg

Recreations''

, Norton

Co.,

.

section

as finite affine planes

ll.

, ''Mathematical

, ''Lectures

problem

of magic

squares

of the officers.

in Projective

Geometry"

, 1963. good general

reference

, with

emphasis

on axiomatics.

, Van

,

:

eo

‘oo

é

004 hy

a

hh

@ es ls

te

!

-

a

ie

7 a

tran

Go

io

aoe 6-4 i

a

7

:

OO

—

—_

hen

Sq?

awe

atjee3

are!

iw eh

os

is

SS

| =.

|

|

7 7

7

fis

_"ERiseciaeie

; 7

>a

7

:

6:3

Oud

:

.

:

> eq? sie

=

7

7

in

-

Tedne

_

Sere Saree

ese. i: 4

°

iL

: -

:

-

7

+ e@

@

sl,

mae

7

’

:

:

v

8

:

~

a

1 &)

cnraes om

—

a

4

Da

'

ee

ae

ve

.

ms)

Oe

—