Geometric Algebra (Interscience Tracts in Pure & Applied Mathematics)

-**

restriction

W

W + U, the cosets A + U W + U by cosets of the subspace

union of these cosets forms the space are, therefore, the stratification of

W+

U of

U. This shows

f(W) = (U

W

+

TP)/C7.

What

is

the kernel

we have \f/(A) =
For

all

with the image

(U

+

W)/U.

H

Since everything was canonical

W

we have

+

are subspaces of V then (U and W)/U If are canonically isomorphic. = In the special case we find that V/U and W/(U TF) = W/0 = TF are canonically isomorphic. Suppose now that

THEOREM

and

W/(U

1.2.

C\

U

W]

V

U

W

n

GEOMETRIC ALGEBRA

8

U

only the subspace such that V =

V W?

of

U

is

Does there

given.

Such a subspace

exist a subspace

W

be called supple-

shall

mentary to U. Let S be a basis of U and complete S to a basis S VJ T = 7 of V where S and T are disjoint. Put T7 = (T), then f7 + = and obviously V 7 TT. This construction involves choices and is far from being canonical.

W

U

THEOREM 1.3. To every subspace canonical way) supplementary spaces

of

V

one can find (in a non-

W for which V = U W. Each of however, canonically supplementary subspaces W W = U W then W, the morphic space V/U.IfV = U W canonically isomorphic V W an isomorphism into then the image f(S) of a / these

2

is

2

l

to

If

iso-

is,

to

.

is

:

V will at least be independent. One concludes the inequality V < dim W. Should / be also onto then equality holds. In our construction of W we also saw that dim V = dim U +

basis of

dim

W and since W ~ V/U one obtains

dim

V = dim U + dim V = / W, that

dim hence

also,

whenever

dim Let

now

t/i

C

C/2

7=

C

dim

17

F/C7

+ dim F.

C^ be subspaces of V. Find subspaces T7 2

and TF3 such that

and, therefore,

We have dim U /Ui = dim W = dim TF dim U /U, = dim(TT U then if *7 C C ^2 proved: 2

2

2

3

3)

3

l

dim

(1.1)

Let U,

=

t/a/f/!

now

U

and

0, C72

=

C7,

dim(C7

,

dim

W + dim W 2

3

.

W

3 and Thus we have

,

= dim ^/EA

W be

= dim

r/3 /C72

+ dim U /U 3

two given subspaces

= U + W. We obtain + W) = dim U + dim(C7

2

of V.

.

Use

C/3

= dim

C7

+ dim

+

W)/U

TF/(C7

H W).

(1.1) for

CHAPTER

we add on both sides dim(C7 we get dim(U W) = dim If

W

H

dim(C7

Next we use

+

F)

(1.1) for

dim V/(U

If

H

we add dim V/(U

dim V/(U

+

9

I

H W) and use dim W/(U C\ W) +

+ dim(C7 H

W) = dim U

+ dim TF.

= U Ci W, U2 = W, U3 = V: WO = dim TF/(C7 n WO + dim F/TF = dim (17 + WO/17 + dim 7/TF. CA

+

W) and

W)

use

+ dim (U + W)/U =

dim

V/U

we obtain dim V/(U

+W) + dim

V/(U

H WO

= dim

F/J7

+ dim

F/TP.

dimension of V is finite all subspaces of V have finite dimenthen our interest will be concentrated however, dim V = on two types of subspaces U. Those whose dimension is finite and, on the other hand, those which are extremely large, namely those which have a finite dimensional supplement. For spaces of the second If the

sion. If,

,

but dim V/U is finite; dim U tells us very little type dim U about U, but dim V/U gives us the amount by which U differs from the whole space V. We give, therefore, to dim V/U a formal status by

DEFINITION

1.3.

The dimension

of the space

V/U

is called

the

codimension of U:

codim

The

various results

THEOREM

1.4.

U=

dim V/U.

we have obtained

The following

are expressed in

rules hold between dimensions

and

codimensions of subspaces:

dim

(1.2) (1.3)

(1.4)

U+

codim

U

==

dim 7,

+ WO + dim(f7 H W) = dim U + dim W, codim(C7 + WO + codim([7 H W) = codim U + codim W. dim(f7

These rules are of little value unless the terms on one side are could not (then those on the other side are also) since an

finite

be transposed to the other side by subtraction.

<

GEOMETRIC ALGEBRA

10

Spaces of dimension one are called lines, of dimension two planes of codimension one are called hyperplanes.

and spaces

More

3.

homomorphisms

V be right vector spaces over a

V and

Let

detailed structure of

field

k and denote by

Hom(7, V) the set of all homomorphisms of V into 7'. We shall make Horn (7, V) into an abelian additive group by defining an addition: e Hom(7, V), let / + g be the map which sends X e V onto the vector f(X) + g(X) of 7'; in other words, = f(X) + g(X). (/ + g)(X) That / + g is a homomorphism and that the addition is associative

If

/ and g are

the vector

and commutative

is

easily checked.

X e V onto the

V

The map which sends every

element of is obviously the 7') and shall also be denoted by 0. If / e Hom(7, 7'), then the map onto / which sends (f(X)) is a homomorphism

vector

vector of

Hom(7,

X

+ (/)=

and indeed /

a)

V

=

is

established.

possible to give more structure to are going to investigate some of the possibilities.

In special situations

Hom(7, 70 and we

The group property

0.

is

it

7.

Hom(7, 7) maps 7 into 7; one also calls it an endomorphism Hom(7, 7), then it is possible to combine them to a map gf V -A 7 -* 7 as we did in 1 gf(X) = g(f(X)). One sees immediately that gf is also a homomorphism of 7 > 7.

An

element of

of 7. If /, g E

:

:

Since

= 01/00 and 0CA

+

/0(*)

=

rffiGK)

=

(0/1

+

see that both distributive laws hold; Hom(7, 7) now becomes a ring. This ring has a unit element, namely the identity map. The maps / which are a one-to-one correspondence lead to an

we

inverse

map

l

/"

which

is

also in

Horn (7, 7). These maps / form

CHAPTER

11

I

therefore a group under multiplication. All of Chapter is finite. to the study of this group if dim

IV

is

devoted

V

Let us

now

investigate

some elementary properties

of

Hom(F, V)

if dim V = n is finite. Let / e Hom(F, F) and let U be the kernel of /. Then V/U c~ /(F) so that the dimension of the image /(F)

dim U. This shows that / is an onto map if and only if dim [7 0, i.e., if and only if / is an isomorphism into. A n be a basis of F and set f(A<) = B If A2 Let A

is

n

=

l

X

= A&i

,

,

+

+

/(X)

(1.5)

t

,

A 2 x2

+A = B lXl

n

xn

e

.

F then

+ B x> + 2

+ B xn n

.

Conversely choose any n vectors B { e F and define a map / by (1.5). sees easily that / e Hom(F, F) and that f(Ai) = B< Consequently / is completely determined by the images B { of the basis elements A i and the J? can be any system of n vectors of F. If we

One

.

t

express each

B

t

by the

basis

A,

=

B,

,

n

f(A,)

=

53 A,a, i9

j

=

1, 2,

,

n,

F-l

then we see that / is described by an n-by-n matrix (a,,-) where i is the index of the rows and j the index of the columns. Let g t Hom(F, F) be given by the matrix (6,-,) which means that

Then (/

+

g)(A t )

=

E

x-l

and

4\ z -l F

A.(a.,

+

GEOMETRIC ALGEBRA

12

We

+

/ g is described by the matrix (a tl + &<,) and fg is the reason for defining addition and multia This by (2"-i A/)see that

-

plication of matrices

Under

by

and multiplication the correbecomes an isomorphism between Hom(F, V)

this definition of addition * (a,-/)

spondence /

and the ring

of all n-by-n matrices. This isomorphism is far from canonical since

choice of the basis

A

-

t

it

depends on the

for V.

Let g be another element of Hom(F, F), but suppose that g is one-to-one. Let (&,,-) be the matrix associated with the element l

gfg~

of

Hom(F,

F).

The meaning 0/flf '(A,)

=

of the matrix

EA

9

b

(&,-,-)

is

that

.

-l

If

we apply

1

g'

to this equation

it

becomes

any one-to-one onto map of F the vectors g~ (A w ) are another basis of F, and g can be chosen in such a way that gT^A,) is any given basis of F. Looking at the equation from this point of view we see that the matrix (&,-/) is the one which would describe / l if we had chosen g~ (A ) as basis of F. Therefore: The matrix describing / in terms of the new basis is the same as 1 the one describing gfg' in terms of the old basis A, In this statement g was the map which carries the "new basis" g~ (A 9 ) into the Since 0f

l

l

is

9

.

l

old one, l

g(g- (A 9))

This g

now

is,

therefore, a fixed

that /

>

A, g

>

map

= A,

.

once the new basis

D are the descriptions

of

is

given. Suppose in terms of

/ and g

> DAD" the original basis. Then gfg' The attitude should be that g is fixed, determined by the old and the new basis, and that / ranges over Hom(F, F). We can state 1

1

.

CHAPTER

13

I

THEOREM

The ring Hom(F, F) is isomorphic to the ring of 1.5. matrices with elements in k. The isomorphism depends on n-by-n the choice of a basis. Let g be an element of Hom(F, F) which carries a all

selected

new

basis into the old one

and suppose

D describes g

that g

A is the description of any f in terms in terms of the old basis. If f 1 of the old basis, then DAD' is the description of this same f in terms >

of the

new

basis.

Mathematical education is still suffering from the enthusiams which the discovery of this isomorphism has aroused. The result has been that geometry was eliminated and replaced by computations. Instead of the intuitive maps of a space preserving addition and multiplication by scalars (these maps have an immediate geometric meaning), matrices have been introduced. From the innumerable absurdities from a pedagogical point of view let me point out one example and contrast it with the direct description. Matrix method: A product of a matrix A and a vector (which is then an n-tuple of numbers) is defined; it is also a vector. Now the poor student has to swallow the following definition: A vector is called an eigen vector if a number X exists such that

X

X

AX =

\X.

Going through the formalism, the characteristic equation, one then ends up with theorems like: If a matrix A has n distinct eigen values, then a matrix D can be found such that DAD~ is a diagonal l

matrix.

The student if

will of course learn all this since

he does not. Instead one should argue

of the space

F into itself.

Given a

like this:

Does there

he

will fail the course

linear transformation

exist a line

which

/

kept fixed one include the value should then In order to eigen modify by /? the question by asking whether a line is mapped into itself. This

means

is

of course for a vector spanning the line that

f(X)

=

\X.

Having thus motivated the problem, the matrix

A

describing /

moment for the actual computation of X. It again. Then one proves all the customary theorems

will enter only for

a

should disappear without ever talking of matrices and asks the question: Suppose we can find a basis of F which consists of eigen vectors; what does

GEOMETRIC ALGEBRA

14

this' imply for the geometric description of /? Well, the space is stretched in the various directions of the basis by factors which are

the eigen values. Only then does one ask

what

this

description of / by a matrix in terms of this

basis.

We have obviously

the diagonal form. I should of course soften lately

for the

reproach since books have appeared

my

stress this point of view so that

which

means

improvements are to

be expected.

my experience that proofs involving matrices can be shortened 50% if one throws the matrices out. Sometimes it can not be

It is

by

done; a determinant may have to be computed. Talking of determinants we assume that the reader

with them. In Chapter IV the non-commutative case; case. If

k

is

a commutative

we

use a

new

basis,

A

familiar

we

give a definition which works even in let us right now stick to the commutative field

and /

the determinant of / as follows: Let put det / = det A. If

is

A

Hom(F, F), then we define be a matrix describing / and

c

has to be replaced by

DAD~

l ,

the deter-

1

minant becomes det D- det A- (det D)" by the multiplication theorem of determinants; since k is commutative det D cancels and

we

see that the

map

=

/ -> det / is

well defined

and canonical.

If g

fg corresponds to the matrix shows det fg = det /-det g.

THEOREM

det

A

corresponds to the matrix B, then and the multiplication theorem

AB

There exists a well defined

1.6.

Hom(F, F) called the determinant of

>

an endomorphism

det(/0)

=

map

k

(if

k

is

commutative)

/. // satisfies

det /-det

g.

In view of this fact it should be possible to describe det / in an intrinsic manner. The reader will find such a description in Bourbaki, Algfcbre, 6)

V

Chapter is

III.

a two-sided space.

V

is both a right and a left Suppose that vector space over k and that we have the additional rule (aA)b =

DEFINITION

1.4.

CHAPTER a(Ab) for over k.

all

a,btk and A

t

15

T

V. Then we

call

V

THEOREM 1.7. // V is a right and a then Hom(F, V) can be made into a left

over

7c,

V a two-sided space two-sided vector space

vector space over k in

a

canonical way.

To this effect we have to define a product af for a e / Hom(F, V). We mean by af the function which sends the X e V onto the vector a-f(X) in V. Since e

(af)(X

+

7)

=

a-f(X

+

7)

=

k and vector

a

and (a/)(X6)

=

=

a-f(Xb)

a

a/Hom(F,

7').

The equations (a(/

((a

+

+

6)/)(X)

0))(X)

=

(a

=

a((f

=

(af)(X)

+

+

6)-/(Z)

g)(X))

+ =

=

a(f(X)

(ag)(X)

a/(X)

+

=

(af

+

6/(X)

=

ajr)(X),

(a/

+

show that IIom(F, F') is a left space over k. in a natural The question arises: Can one make any right space way into a two-sided space by defining a product aX from the left? One thinks of course of the definition aX = Xa. But then a(bX) = (bX)a = (Xb)a = X(ba) = (ba)X, whereas we should have obtained (ab)X. This "natural" definition works only if the field k is com-

V

mutative.

THEOREM field k,

then

V and V are vector spaces over a Hom(F, V) can be made in a natural way 1.7'.

//

commutative into

a

vector

space over k. If k is again any field, then the most important example of a two-sided vector space is the field k itself if one defines addition

V

and multiplication as they are defined in the field. this case in the next paragraph.

We shall investigate

GEOMETRIC ALGEBRA

16

4. Duality

DEFINITION set

The elements

V

If

1.5.

V = Hom(F,

k) is

a

a right vector space over fc, then the vector space over k called the dual of V.

is

left

of t^ are called functionals of V.

^, ^,

the definition of a functional

=

J5)

The

and pairings


a

it is

map V

+

A

fc

To

repeat

such that

2)

operations between functionals are fe

+

is

a

3)

F

If

=

*)(A)

left

+

*(A)

*(A),

vector space over

/c

4)

we

also define the dual space

= Hom(F, fc). In order to obtain complete symmetry we write, however, (AV instead of (A). V is a right vector space over k. t^

The notation

will

become

still


if

simpler

if

we

write
V is a left space).

Rewriting

1)

and

3)

these rules take on the form of "distributive laws" and 2) and 4) become "associative laws". In other words let us regard
kind of "product" of an element that
e V*


and an element

A

e

V

such

E k.

This change of notation suggests a generalisation where a right are given (W not necessarily the dual space V and a left space of V), together with a product AB e k for A e W, B E V.

W

DEFINITION

If

1.6.

we say that a

AB e

fc

is

W a and V a right vector space over W and V into F given, provided a product

pairing of defined for all

left

is

fc,

is

A

TF and

E

all

B e F such that the following

rules hold: 1)

A(B

3)

(A t

We

+B = + A )B = 2)

l

2

notice that V"

Our task

is

and

AB, A!JS

F

to study V"

+ AB + A B, 2

2

,

2)

A(Bb)

=

4)

(aA)J5

=

are naturally paired into fc. closely and to investigate general

more

pairings.

Let {Ai} be a basis of the right space F where the indices i range over some set / which we have used for indexing. Let

X

X= since

X

A -z

where x< 9* holds only for & finite number of must be & finite linear combination of {A,-}. Then

]T)f

t

-

t

indices,

CHAPTER X

2
=

This shows that

t

all

t

Xi

17

I

=

X)

fl.*

the a< are known. Select conversely by any condition and define

restricted

The sum on the right side makes sense since x j holds only for a finite number of the subscripts i. The two equations
+

(X)

v $ vi

i

THEOREM c

9

a.

Thus we have

a basis of V, then for randomly given one and only one

// {A {

k (one for each

Let which


= 2* M,t =

get also
a

=
and

>(F)

A = X" A

i

is

}

i) there is

-

.

words

=

and

=

for i 7* j. In other finite linear combination of the

Ai

1

>

t

A,-

.

v

-

t

t

W

.

This means that


alone already determines the

as values of

6,-

2

(p

&> i s therefore, on the basis vectors. The linear combination Q unique, the

W

t

Since there are as

WQ

many ^>

as there are A<

t

,

we

get

dim

W

= dim

.

V.

can be described as the set of those functional

then

we

W

Q

get

since

contains obviously

dim V = dim

W C Q

have dim


If

,

V' <

and and,

and our notion

=

<

.

WQ

is

that only the zero functional vanishes on

analogue: If

WQ = W =

of dimension is very crude, we also then certainly not the whole space V. If ^ aud -4. 5^ 0, then at least one a,- 5^ 0. For this j a ^ 0. By the definition of a functional we know trivially

V'

A = 2* ^ we get A

?

t^.

functional of V, i.e., dim V = then dim

all

A

e

V, and

Let us now state our

if


=0 for all

results.

all of
t

V.

Now we

V, then

see

4=0.

an

GEOMETRIC ALGEBRA

18

THEOREM

We

1.9.

V =

always have dim

dim V. If we know

0/or all
=


0/or

A, then

all


=

0; if

we know
=

t \

.

% }

W

DEFINITION

1.7.

A is orthogonal to B.

W

we say that

BtV

and

If

.

is

F

If

A

If

W

W, B

t

is

V and AB =

e

a subspace of

we

W and V

shall say that a subspace of V,

Q

f or A t orthogonal to VQ provided AB = is a subspace of V, one sees easily that the set

W which are orthogonal to V V H C W. Similarly each given subspace W of subspace W% of V. We have trivially V C (V*)*-

Q

V%

a subspace of rise to a

of all vectors in

is

W

W gives

:

W

Abbreviating we

F?* instead of (F?)*. Of special importance is the subspace F* of which are orthogonal to all consisting of all vectors of vectors of F. We shall call F* the left kernel of our pairing. Similarly write

W

W

the subspace W* of Notice that Theorem 1.9

we

call

both kernels are

F

the right kernel of our pairing. tells us that in the pairing of F and

0.

Suppose that in our pairing of

Each

A

W gives us a function

t

F

W and F into

on

F with

the

fc

left

values in

kernel

fc,

is 0.

namely

^(JQ = AX. The

function

A

is

easily seen to

be a homomorphism

F

fc,

V.

i.e.,
This suggests studying the onto
map

W

>

V

which maps the vector

.

*

MX = (A+ E)X = AX + BX = (^

and <poA

X=

(aA)X = a(AX) =

The two equations ^+/? =

map

W

>

of those

Xt

V.

is zero.

A

? c

is


+


a homomorphism.

TF f or which

is

and

The

^> aA

a A

=

X.

&PA

mean

kernel of this

the zero function:

map

AX =

that our consists for all

However we had assumed that the left kernel of our pairing > V Hence A = 0. Our map is, therefore, an isomorphism

into (not always onto).

W

CHAPTER Similarly,

V

the right kernel

if

is 0,

19

I

one obtains an isomorphism into:

-+TF.

W

be a subspace of left kernel is 0. Let V the of is which We can W$ Q subspace orthogonal to find in a natural way a new pairing of the space and the space Q e and of a V/W* into k by defining as product of a vector

Suppose again that the

W and

W

W

.

X W

+ W*

Y

coset

X-(Y This pairing

Y - Y

XY =

is

^F!

and That

.

+

TT?)

well defined. Indeed

pairing is obvious. right kernel if

X

since

e W*<>

l

XY

F/TF? the element

in

this

W

JTF.

Y + W% = Y + W*Q implies we have X(Y - FO = or l

new

What

XY =

in

is

=

of k:

multiplication satisfies the axioms of a the right kernel? Y W* will be in the

is

X

+

W

This means Y e W* and, the zero element of the space V/W* W*<> = W*Q therefore, Y The right kernel of our new pairing is 0. We can use the previously for all

+

e

.

.

,

method to construct an isomorphism into

established

(canonical)

:

V/W What

W

Q

associated with

X- (A If

of

F

is

+

Q

X-(Y+

W*,)

= XA

V/W*Q

;

the func-

(X ranging over TF

).

As before

Fo)

setting

= XY,

XtV*

Q

left kernel;

XY =

Yt

V.

But

since

of our original pairing

is 0,

this

for all

is 0.

We

X

obtain, therefore,

in full clarity.

e

F/F

satisfies

.

the axioms.

will lie in the left kernel if

an isomorphism

into:

.

urged not to give up but to go over until he sees everything formulate our results:

inexperienced reader

all definitions

and

F

we have assumed that the left kernel means that the left kernel of our new

VI - F/Fo

THEOREM

Y+

,

this pairing is well defined

This time we ask for the

The

of

it is

a given subspace of V, we can also define a natural pairing

VI and V/VQ by

pairing

+ W*

map? Given an element A

is this

tional of

is

and mappings again and again

We

1.10.

Let

W

and

V

be paired into k

and assume

that

GEOMETRIC ALGEBRA

20

F*

the left kernel

F

W

W and

Q be a subspace of of our pairing is 0. Let exist natural isomorphisms into:

a subspace of F. There

(1.6)

F->F/F

(1.7)

O

Since these

maps

.

are isomorphisms into, the dimensions of the

spaces on the left are < to the dimensions of the spaces on the right. Under the symbol dim the ~ on top of a space can be dropped because of Theorem 1.9. We obtain dim V/W*Q < dim TF and dim V*Q <

dim F/Fo

,

< dim

codim TF$ If

the

which we can also write in the form

inequalities

We put F = first, we get

W*Q

W

dim F?

and

<

in the second inequality

<

dim Fo**

W?

codim

codim

.

and combine

W that W C

<

F

dim

with

it

.

And now we remember the trivial fact dim W < dim W** Therefore we get equality

W* *

Q

so that

.

dim TF ** = codim

(1.8)

The main significance is finite. Since dim

dim

W

.

when

of this formula appears in the case

W C W$*

W

W*Q = ,

we

see that simply

TF* =

W

.

In the isomorphism (1 .6) both spaces have the same finite dimension, our map is, therefore, onto and thus V/W*Q may be regarded naturally The map (1.7) should be used for F = W% and as the dual of Q

W

.

W

into V/W*Q This map is onto again becomes an isomorphism of and V/W% is naturally and we see now that each of the spaces .

W

the dual of the other.

In

on

(1.8)

no

restriction

Wo = W. We

We if

Q is

we can use the formula

necessary;

W* = dim W.

should mention that our results are true for

dim

W

all

subspaces of

itself is finite.

What can we do if now you

pairing (by

and

W

codim

(1.9)

W

on

obtain

the

left

kernel

F*

is

not 0?

We make

are used to this procedure) between

F by defining (X

+

F*)F = XY,

X+

F*

e

TF/F*,

7c

F.

a

new

W/V*

CHAPTER

The element This means

X

21

I

X+ e

V* lies in the left kernel if XY = for all Y e V. F*, and hence X + F* = F*, the zero element of kernel is now zero. The right kernel is obviously

F/F*. The left the old W*. Equation (1.9) tells us that the codimension of W* in F is the same as the dimension of our left factor which is W/V*.

We

obtain, therefore,

dim V/W*

(1.10)

= dim W/V*.

W

Suppose now that both kernels F* and W* are zero and that dim (1.10) shows that dim F is also finite and equals dim W. In this case we can use all our results unrestrictedly on subspaces = W, then W*Q = and F. If we choose for instance of both Q and F is naturally and F/WO = F; we see that each of the spaces

is finite.

W

W

W

The

reader should again visualize the map: =
the dual of the other.

W

X

AX

W W* of a subspace W C W and the subspace W\ C F. Any we have merely to put obtainable from a TF subspace F C F = of W V*Q And distinct subspaces W give distinct images since W*Q = W\ implies (by starring) W = W It therefore a one-to-one an inclusion W C Wi lattice inverting, correspondence which -

Q

Q

is

Q

;

.

l

is

implies

W* D W*

is

.

Q

i.e.,

(a strict inclusion

becomes

strict again).

Again let

us collect our results.

W

Assume and V paired into k: 1.11. dim W/V* = dim V/W*', in particular, if one

THEOREM a)

W/V*

and

V/W*

dimensions are equal. b) // the

left

dim

(1.11)

If

dim WQ

W

Q

is finite, then

V/W%

W

and

kernel is

of the spaces and the

dimensional, the other one is also,

is finite

=

W C W, then

codim

W** =

W*Q =

W

Q

dim

and each

W*.*. of the spaces

W

Q

and

is naturally the dual of the other. is finite, then each of the spaces and dim // both kernels are <-> W% and V is naturally the dual of the other. The correspondence

W

c)

W

W

and the subspaces of V. It between the subspaces of reverses any inclusion relation (in the strict sense). In case our pairing is the one between tf and F we can strengthen

is one-to-one

the results.

We know already that both kernels are 0.

Let

F

be any

GEOMETRIC ALGEBRA

22

V

subspace of

and consider the

Theorem 1.10 F* > V/VQ It is an isomorphism iito show that it is onto. Let

map

(1.7) of

and we

We

(not necessarily of finite dimension) :

.

shall

-

Then

MX +

F)

=

+ Y + Fo) = X(X + Fo) + (Y + XX + Fo) + v(Y + Fo) = XX) +
and

F )a) = This implies that

XX) =

X* +

a functional of F.

is

Fo)

= X^o) = Xaero

+ F )a Suppose X e F
F/F

element of

,

)

then

=

0.

and, therefore, belongs to F? We contend has the given y as image under the isomorphism (1.7) and that this will show the ontoness of our map. What is its image? We had


vanishes on

all

F

of

.

make the new pairing of FH and F/F by defining Y (X + F ) = YX (Y t V*Q X e F); for Y = $ t V% this means (X + F ) = &X and this function & (X + F ) on F/ F is the image. But &X was by definition y(X + F ) so that this function on F/F is indeed ^?. Let A $ F The space F? is now naturally the dual of F/F The coset A + V is not the zero element of F/F so that a functional to

,

.


e

can be found such that
F/F

The corresponding &

F =

)

7*

.

(Theorem

1.9).

+F

) 5^ 0. This
A

F

F

F

* proved F* = F Finally, we set in formula (1.11) of Theorem 1.11: * = we Since Wl = F dim TF codim W% for Q the space F * = codim F which supplements the analogue of (1. 11), obtain dim F dim F = codim F * which is true in all pairings. Thus we see in our Q <-> Wf> special case that there is a one-to-one correspondence of finite dimension and subspaces of between subspaces Q F with finite codimension. Indeed, if codim F is finite, then dim F? = V* is the space which gives W* = F = codim Fo is finite and there is only one such space. If W^ is given and of finite dimension, .

,

W

,

W C W W

.

,

W

;

CHAPTER then (1.11) shows codim

w* Wo

W*<>

23

I

= dim

W

W

and

Q

= VI

gives

V

.

W

Consider the pairing between fi = and V. = V V* then and is F?* Q any subspace of V, If naturally the dual of F/Fo We have not only dim V = codim VH (established in

THEOREM

V

1.12.

is

.

Theorem

W

W

Q <->

1.11), but also dim F* * is one-to-one between all

=

codim

V

.

The correspondence finite dimen-

subspaces W C W of

subspaces of V with finite codimension. Similar results would not hold in general for subspaces of V. sion

and

all

Let us look especially at a hyperplane F of V. Then codim V = 1, hence dim V% = 1. Let VI = <^>) (

X

X

X

\[/

W

W

The proof of these simple facts about hyperplanes is burdened by too much theory. Let us see whether we can not get them from scratch: > k functional

Take a

This

is

a

^

plane

F

F/Fo

is

map F

Consider the canonical map F > F/F with kernel F 1-dimensional, hence isomorphic (not canonical) to k. The F/Fo > k is then a functional v? with kernel F Take .

.

.

and two functionals ^>, ^ with kernel F Then 4 l = = 6 ab~ and the functional will a 7* 5^

F

A

any vector

.

\l/

:

THEOUEM

1.13.

of solutions
any y

up

to

j

a

of

Vr

A

=

hyperplane where

gives in this

left factor 9*

of 5.

F

way a

is

o/

F

can be described as the

an element

hyperplane.


of

V

is

determined by

and

set

conversely

F

k.

Linear equations

We illustrate the theory developed thus far by applying it to the theory of linear equations. The beginner should not expect that we

GEOMETRIC ALGEBRA

24

be able to develop miracle methods for solving equations. An is still best found by the elementary method of suc-

will

actual solution

cessive elimination.

Let

+

=

a lnxn

b

l

,

be a system of ra equations with n unknown o\ whose coefficients a,, and right sides 6 are given elements of a field k. The following objects will play a role in the discussion: rows and n columns. 1) The matrix (a,,) of the coefficients with k and a fixed basis V over 2) A right n-dimensional vector space t

m

E E l

,

-

2

,

,

interpreted in

En x n 3) ,

of V.

A solution

(x l

,

z2

,

,

zn ) of

(1. 12) shall

V by the "solution vector" X = E&i + E2 x +

The dual space V' with the basis ^ 2 y? n dual to En With the f-th row of (a t; ) we associate the functional 2

,

,

,

,

.

,

-

,

X

If

=

#!#!

+

JBJ 2 o: 2

(this follows easily

4)

be

+

2

.

E

,

En

The subspace

sion r

is

+

from ^

+A

T

t

/?,-

n

=

xn

any vector

is

number

V

.

then

5 tJ ).

W of ^ spanned by ^

the maximal

of

m)

^2 ^m Its dimenindependent vectors among

of linearly

,

,

,

-

the \l/i and consequently the maximal number of left linearly independent row vectors of (a,/). We call r therefore the left row rank of (a t/ ). 5)

In

it

(a,-,)

The m-dimensional right space S m we place the n column vectors A l

and

A

also

the vector

An

B =

(bi

,

of m-tuples of elements of k. ,

A2

bz

,

,

,

,

An

6 m ).

of the matrix

The

vectors

span a subspace C7 of S OT and its dimension shall be called the right column rank of the matrix (,). In our setup the equations (1.12) can be rewritten in two ways. In the space S m they are obviously equivalent with the vector equa-

A!

,

tion

-

2

,

,

will

CHAPTER

+A

A,x,

(1.13)

But they can

+

x2

(fcZ,

+Ax X = b,

n n

also be written as

B =

(1.14)

2

25

I

\f/ t

*2 X,

,

=

or as

Xe

iMT),

,

B.

7.

Our equations need not have a bility is stated in the following

solution. The problem of solvaway: Consider the matrix (a
given and fixed. For which vectors B does a solution exist? Equation (1.13) tells us that the vectors B for which a solution exists

must

Equation following

map

V -A S m = (fcX, *

of

map /

X

vectors

is

Sm

of

B must

.

lie

in the

image space of the

:

f(X) This

us that

(1.14) tells

those vectors

U

in the subspace

lie

2

X,

,

XtV.

fJQ,

obviously a homomorphism. Its kernel consists of = 1, which ^,X = (i m), hence of the

X for

,

which are orthogonal to

$ m and, therefore, ^ which they span. The kernel is orthogonal to the space therefore W*. We see that the image space U V/W*, hence \l/i

WC

dim dim

U = codim W*. U = dim W. We left

row rank

,

\f/ 2

^

But codim W*

= dim

have therefore the

=

of (a t; )

,

,

right

and we obtain

W^,

rule:

column rank

of (a,/),

a rule which facilitates sometimes the computation of ranks. In most applications the answer we have found thus far is satisfactory: The equations have a solution if and only if B belongs to a certain r-dimensional subspace U of S m the point being that know now that r is the left row rank of (a,,). Frequently one asks when the equations have a solution for ,

B

t

S m This means U = S m .

or r

=

m, and

is

true

if

and only

if

we all

the

rows of (a M ) are left linearly independent. Suppose now that B is of such a nature that the equations have a = 6 (i 1, 2, solution and let Q be a special solution: ^vY m). Then the equations can be rewritten as

X

-

t

*.(X

and mean that

X

X

- Z = )

must belong to the subspace

solutions consist of the coset familiar rule: general solution of the

,

homogeneous equations.

X + =

M7 *. We

W*

of V.

The

have, therefore, the

special solution

+

general solution

GEOMETRIC ALGEBRA

26

When

the solution unique? This means W* = and conse= t^; therefore, r = n. Let us review the extreme cases

is

quently W for r:

r r

= n

= m

means

solvability for all

means uniqueness

of the solution,

B; if it exists.

m =

n (the frequent case of as many equations as there are unknowns), then r = n means solvability for all B as well as uniqueness of the solution, and is also equivalent with W* = Should

which implies that the homogeneous equations have only the

trivial

solution. This fact is used frequently in applications.

X +

TF* which we have given for the solutions Q a geometric language. F shall be called a Let VQ be any subspace of V. A coset A F is merely given as a set of vectors, linear variety. If the coset A then one can get back the subspace F by subtracting from each F a special one, say A. The subspace F shall be vector of A

The

description

calls for

+

+

+

called the direction of the linear variety.

(The intuitive picture

is,

of course, to think of vectors as "arrows" from the origin and to as consisting of the endpoints of think of the linear variety can now say that the solutions of "arrow" in .) every

A + F

We

A + F

our equations form a linear variety which passes through a special F is a linear variety, solution and has the direction W*. If A

+

then dim

F

be called the dimension of the linear variety. For the solutions of our equations the dimension is dim W* = codim shall

W

and, therefore,

n

r.

One may also ask for the converse. Given a linear variety A + F What can be said about all equations \I/X = b ($ e V") which are satisfied by all X t A + F ? They must be satisfied by A, which = $(A), and thus they have the form $(X or A) = gives b = e Ft gives really ^(l^o) 0; $ must, therefore, belong to F? Any ^ an equation: set b = \f/A\ since \f/(X A) = 0, if X e A + F we = dim V% and let ^ ^ 2 be a get \[/(X) = b. If we put r = X ^ A (i = 1, basis of F? then r) is a set of linear equations .

.

,

,

,

,

and

\1/ {

t

their solution satisfies

X A z F?* = codim Ft = n

Fo

;

A

+

,

\[/ r

,

$i(X Fo

is

A)

=

(i

=

-

1,

,

the set of solutions, and

r)

so that

dim

F =

r.

The elementary geometry

of linear varieties is very simple. Let

CHAPTER

27

I

= A +

Vi and L 2 = B + F2 be two linear varieties with direcL2 not and F2 They need not intersect. When is L X V X t If and a a F if t and vector vector 2 2 exist empty? only = B + X 2 orA-B = (--YO + X 2 This means such that A J3 e V l + F2 If L! r\ Z/2 is not empty and C a common vector, LI

tions Vi

.

l

A+X

H

t

l

.

l

.

then

we can

+ (F! n

C

= C

write LI

7 ).

The

a

Its

dimension

We

must

l

t

B

+ +

B) B)

L2

,

L

also consider the "join"

L

(A (A and

+ F + F

Vi FI

L

2

2

The

.

But

.

t

+F

C

nL

t

non-empty L! C\

linear variety containing both and all differences of

A

2

,

direction of

for

(still

+V L = o

L2

V

l

Z/ 2 =*

HF OF

2

.

that of FI 2 of Lj and L 2 the smallest Its direction must contain is

.

,

and L2 V and F2 .

hence LX r\

,

therefore,

2 is,

L2 )

2

l

hence

,

vectors of

all

B +

join contains B, hence certainly this is a linear variety and contains

L

l

hence L! o

Its direction is

-

L2 = 5

+

(A

+

F!

+ F We notice that there is a case

B)

(A

2

+

5)

F,

+ F

a

.

.

distinction: 1)

Lj P\

L! o

We

L2

is

L2 = B

+ V

V,

2

,

l

dim(L,

and

L 2 ) = dim(F,

o

+F

2 ).

obtain

dim(Li 2)

+

B t V + F2

A

not empty, then

L! Pi

La

n L + dim(Z/!

is

2)

o

L 2 ) = dim

dim L2

.

A - B 4 V + V2 and = 1 + dim(F + F2 ).

empty. Then o

+

L,

La)

1

r

Let us illustrate this in a few special cases. then L! = A, L 2 = B, F! = F2 = 0; if L v

If

dim L

L2

?

,

= dim L 2 =

0,

then they do not

meet and dim(L! o Lj o

L2

is

L 2) =

the unique "line" through

A

1

;

and B.

Suppose n = dim F = 2, dim L^ = dim L 2 empty we have dim(Li o L 2 ) = 1 + dim(F!

dim(F! If

+F =

Lj C\

2)

L2

is

1,

V = F l

2

=

1.

+ F

(parallel lines).

not empty, then

dim(L 1

n L + dim^ o L a)

a)

=

2.

If 2)

L

<

l

C\ 2,

L2

is

hence

GEOMETfcIC ALGEBRA

28 If

L!

L!

C\L 2

5*

L2

,

L2 ) >

then dim(Lj o

The

L2 ) =

hence din^Lj C\

0,

a "point". 6. Suggestions for

2-4

1,

an exercise

come to a clearer understanding of the content of he works out by himself the analogue for ordinary additively

reader will

if

written abelian groups: Let R be the additive group of real numbers and Z the subgroup of the ordinary integers. We must first familiarize ourselves with the

R/Z. The elements of R/Z are the cosets a a + Z and b + Z, are equal if and only

factor group

a

e

R.

Two cosets, To

+Z if

a

with b is

be understood that we Z by merely giving a (where a is only defined up to an describe a integer). Such a coset may be multiplied by an integer (this can be done in any additive group) but a product of two cosets is not well

an

integer.

simplify notations

it shall

+

defined.

One

main properties of R/Z is that it is "divisible": R/Z and an integer n > one can find an element

of the

To an

element a t b such that nb =

a. However, this & is not uniquely defined. Aside from the coset given by a/n one has also the cosets a/n + i/n for

0
b.

now go

over our whole exposition on vector the word vector space by abelian group, spaces, replacing everywhere and dimension by order of the group. Whenever subspace by subgroup a plus sign occurs between dimensions it has to be replaced by a reader should

We

reference to the field k has to be dropped. keep V now means additive group, its order. have no difficulty with the symbol (S) but

Any

product sign.

the notations for easier comparison:

dim

We

V

disregard the notion of independence of a set. abelian group we mean elements A l A 2

V

orders are e 1

that

mA + l

1

,

e2

,

,

+

er e

Z

a basis of a

,

,

m2A 2

By

,

A

r

such that the A< generate

+mA =

of

V

finite

V

whose and such

and only if each r r (w< E Z) The reader may consult any book on group -

if

a multiple of e theory for the proof that a finite abelian group has a basis. The notion of a basis does not work any longer for infinite groups and has to be abandoned. Consequently we can not prove Theorem 1.3

nti is

t

.

and prove directly that (1.1), namely

dim

for

C/a/C/i

U C U C U 1

2

3

the analogue of equation

= dim U./^dim U,/U2

,

CHAPTER

We

can now prove Theorem 1.4. In 3 we need only the fact Hom(F, V) can be made into an abelian group; then we immedi-

holds.

that

29

I

ately go over to 4. It is in the analogue of Definition 1.5 that the special group R/Z comes into play. define F = Horn (7, R/Z) and talk in Definition

We

W

a pairing of abelian groups and V into R/Z. If V is finite with the basis A (e the order of A ), let

by letting

{

t

4

t

{

%

{

t

t

t

t

t

l

.

,-

t

{

.

X

X



X

+

YtU,mtZ. He should try the definition
map

well defined? In

with an a

matics.

^

0?

Now

The reader

and other topics

will

7.

Let ofG.

G

meet

it

again in the theory of Banach spaces

of analysis.

Notions oj group theory

be a group, not necessarily commutative, and S any subset

DEFINITION 1.8. The set of all x t G for which the set xS is the same as the set Sx is called the normalizer N s of the set S. The set which xs = sx for all s t S is called the centralizer Z s of S. The set of all x e G which commute with all elements of G, in other words the centralizer Z of G is called the center of G. of all

x

c

G for

30

Ns

GEOMETRIC ALGEBRA

and Z R are subgroups

G and

of

invariant (normal) subgroup of

the center

is

a commutative

(?.

Let / be a homomorphism of G into some other group. When is the image group f(G) commutative? /(a)/(b) = /(&)/() is equivalent 1 is the kernel of / it moans aba'V e with /(aba'V ) = 1. If

K

The element l

is

bab~ a~

l

abcT'b"

1

is

H

commutators. Conversely, let all commutators. If x t G and h hence

.

commutator of a and b. Its inverse b and a. The kernel K contains all be any subgroup of G which contains

called the

the commutator of

,

K

1

l

xH C Hx and similarly Hx

invariant and since the kernel

H, then xh = xhx~ h~ -h-x e Hx, xH. Such an is automatically

e

H

C

H

l

of the canonical

map G

>

G/H

commutators, the factor group G/H will be abelian. The smallest subgroup (?' of G which we can form in this way consists of all products of commutators and (?' is another way of contains

all

CK

expressing the condition

we

found.

DEFINITION 1.9. The set of all products of commutators of G called the commutator subgroup G' of G. The factor group G/G' is abelian and the image of G under a homomorphism / will be an abelian group if and only if G' is contained in the kernel K of /. A subgroup of G which contains 0' is necessarily invariant and G/H commutative. Let e be any element of G and consider the following map is

H

v e :G-+G:

=


We

have

(xy)

=

l

cxc"

cxyc~

1

exc"

1

.

1

eye"

=

c

(x)

so that

is

a homomorphism. c

(d(x))

= =

c(dzd"

V

1

=

l

cdx(cd)~

=

e

d(x),

the identity. If x t G is ^(x) ^(^-'(2)) which shows that

words

c
d

=

.

^

The map

is

=



fl

G onto (?,

words an automorphism of G. This particular type of automorphism

is

an isomorphism

of

^

in other


.

.

and

is

a homomorphism. The element

c will

be in the kernel

if

=

1,

CHAPTER cxc~*

=

x for

all

x

e

G.

The

31

I

kernel of this

map

is

the center of

(?.

We see that IQ

of

Let us call two subsets S under some
U =

~ G/Z

S and T ?C

(S).

of

are equivalent, [7 and S are equivalent.

G

Then

^(T), then

How many

.

equivalent

/S

=

T

if

is

*v-.(T); and

=

U =

the image

if

T and

C7

^

C (S) so that vWe(>S) sets are equivalent to a given S?

=
We

have to decide:

S =

when

is

means S = aSaT

(S)

1



,

or

.

l



.

,

:

.

,

N

s ) of the normaliser of S. equal to the index (G Of special importance is the case where S consists of one element = {a}. Then each
ranges over

(?, is

:

.

The elements

G are,

therefore, partitioned into equivalence classes: l a is equivalent to all cac~ The number of elements in the equivalence class of a is a ). Which of these equivalence classes contain of

.

(G:N

one element only? For such an a we must have a = cac~ for all c t G. They are, therefore, the elements a of the center Z G of G. l

Denoting by #(S) the number of elements in a set S we have, therefore, #(#
Counting the number of elements in

G by

equivalence classes leads

to a formula,

#(G)

(1.15)

=

#(Z

)

+

E (G

:

#),

a

where ]T) means only rather vaguely "sum over certain a" but where each (G N a ) > 1 since those cases where (G Na ) = 1 have already been counted in #(Z ). :

:

Although the following application is not needed, we can not resist the temptation of mentioning it. r Let G be a group whose order #((?) = p (r > 1) is a power of a is a power of p since (G a ) divides prime p. Then each term of r is p Since each term is > 1, the whole ]^ a divisible by p. Therefore

2

.

:

N

GEOMETRIC ALGEBRA

32

#(Z

=

)

J^ a

#(?

that the order of If

is divisible

Z

is

>

Z

1,

by

p.

We

have the famous theorem

not just identity.

is

multiplicative group G we may adjoin to it a zero = a = for any a in G and also for a = 0, such that a

we have a

element

and obtain a new

set

which

is

not a group but shall be called rather

inaccurately a group with 0-element.

element" we shall mean a DEFINITION 1.10. By a 'group with = a= union of an ordinary group and an element such that a for all a in the set. We have now to describe what we mean by an ordered group: It shall be a group G with an additional binary relation a < b satisfying all laws one would like to have true. Denote by S the set of all a > 1. We would like to conclude from a > 1, 6 > 1 that ab > 1. S will have to be closed under multiplication. One also would like to conclude that a > 1 is equivalent with a~ < 1 and certainly exclude a = 1. Furthermore, we would like to have each element either > I or = 1 or < 1. This shows that we should postulate 1 S where /S" 1 denotes the that G be the disjoint union S' {1} set of inverses of S. Finally a > 1 should imply ba > 6; multiplying 1 on the right by b' we would like to get bob' > 1. Let us try this *

l

W

W

1

as a definition.

A group G is said to be ordered if a set S is 1.11. the with out properties: singled l S, {1} 1) G is the disjoint union S~ S (closed under multiplication), 2) S-S 1 S. 3) for any 6 t G we have bSb' 1 1 bSb' and, therefore, Sor S b~*Sb fe" we get Applying 3) for DEFINITION

U

U

C

C C

bSb-

1

We or a&~

or ab" it

=

C

S.

1

e 1

t

>

l

l

l

a E S. This gives b>(b" a)b 1 whether one says b~ a irrelevant S. It is, therefore, S. Notice that a > 1 really means a e S as we would

define a

b as

meaning

b

e e

S S

like

to be.

Given any pair a, b. If b~ a e S, then a > 6; if b~ a = 1, then l = then a~ b e $, hence b > a and these possibilities a 5; if 6~ a e S" " exclude each other. Any two elements are comparable' \ 1 l Suppose a > b and b > c, then ab~ e S and be" e S, hence l aV l -bc~ = ac~ e S and we get a > c, the transitivity of >. Suppose a > b and let c be any element in (?; then b" a c S and l

1

l

1

,

l

1

CHAPTER ab~

l

33

I

S and consequently (cb)~ (ca) = b~ a e S and S which means that a > b implies ca > cb as well l

e 1

alT e

l

l

ac(bc)~ as ac

>

= be.

One can multiply an If

ac

a

> If

a

> bd.

>

b

and

c

inequality by c. d then ac > ad and ad

>

One can multiply two b then b~

l

>

a"

1 ,

>

bd,

hence by transitivity

inequalities.

since this

means ab~

l

e S.

All intuitive laws for inequalities are satisfied.

DEFINITION if

if

A

1.12.

"group with 0"

U

=

G

is

called ordered

the ordinary group G is ordered in the sense of Definition 1.11 and is defined to be < than any element of G.

Again

laws are true.

all intuitive

8. Notions of field theory

We have remarked repeatedly that a field k need not have commutative multiplication but the addition is of course commutative. The set of non-zero elements of k forms a multiplicative group which shall be denoted by k*. As in any additive group we can multiply an element a t k by an ordinary integer n t Z (Z shall denote the set of ordinary integers) and get an element na t k. In any additive commutative group = na ma, n(ma) = (nw)a, n(a b) = one has the rules (n in) a na nb. In k one sees easily the additional rule

+

+

+

+

(nd)(mb)

For instance

if

n and (

and upon expansion

a

ra are

>

+a+

.

.

=

(nra)(ab).

the

left side is

.)( b

+b+

.

.

.)

of this product one gets the right side.

The question whether na =

does not depend on a l with na-a~ = or ne = where e is the unit element of k. This equation does certainly not hold f or n = 1. > If we map the ring Z of integers into k by n ne, then this map is a n whenever 9^ 0, then it is an isoring homomorphism. If ne 5^

Let a

7* 0.

since this is equivalent

mprphism; k contains in this case an isomorphic replica of Z and, since is a field, an isomorphic replica of the field Q of rational numbers. A field fc of this type is called a field of characteristic 0. Suppose, on the other hand, that the map n>ne has a non-zero kernel H, a subgroup consists of all multiples of the additive group Z. Such a subgroup 5* k

GEOMETRIC ALGEBRA

34

pv of the smallest positive integer p in H. As remarked earlier p j* 1. If p were not a prime number, then p = db with positive a, & which

<

are

p. Since

a

true since k

is

field of this

type and

The only

<

v

<

=

p-e field.

we would This number p 0,

a prime number.

is

Z

distinct images of

p

The p

1.

which is not get (ae)(be) = is called the characteristic of a

in

fc

are the p elements ve with among them are

non-zero elements

1

closed under multiplication and, since in a field the cancellation law holds, they form a group. The p elements ve form, therefore, a subfield Q 9 of fc (which is isomorphic to the field of residue classes Z/pZ of

integers modulo p). From now on we will denote the unit element of fc by 1, the elements vl simply by v with the understanding that v is to be read modulo the characteristic of fc. If fc is a subfield of a field F, we may regard F as a left vector space over fc, taking as definition of the vector space operations the ones we already have in F. This space F has a dimension (over fc) which is called the left degree of F over fc and denoted by [F fc]. :

One could of course also define a right degree; it is an unsolved problem to decide whether any connection exists between these two degrees. We shall stick consistently to the left degree. By a left fc-basis {a,} of F one means, therefore, a set of elements in F such that any

F is a unique finite left linear combination of these basis elements with coefficients in fc. We may write ft = 2, x a i where can hold only for a finite number of indices i. Xi E fc and where x j& If F is a subfield of E and fc a subfield of F, let { T, be a left

element of

*

{

}

E

F-basis of

we can x lf

t

and

a

{a,-}

A = ^/

write

k where only a

left fc-basis of F.

0/F/

finite

w ^h

Pt e

number

F

Given an element A xu a an d /

e

E

2

of the x,, are j& 0.

We

obtain

i.i

If,

A =

conversely,

2,-

(2<

we have a sum

#t/.)r/

;

this

A

2./ x a a i^i

and

call it

determines the coefficients

4, then

2< #<*

and each ]), #,,, determines the x it uniquely. This shows that the set {a r/} is a fc-basis of E and we have proved of the F/ uniquely

t

the formula

[E If

a

t

fc,

normalizer

then we

N

a

call

of a. If x t

:

fc]

=

[E

the set of

N

a

and

:

F][F

ysNa

,

k].

k for which xa = ax the then a: y e 2V. and XT/ e JV.

x

all

:

t

CHAPTER

35

I

Should x 7* 0, then xa = ax implies oaT = x~ a, hence x~ t m This proves that a is a subfield of fc. Af* is the group theoretical normalizer of a in the group k*. The set of all x t k such that xy = yx for aH y t k forms also a field Zk and again we have that Z\ is the center of the group k*. We remark that trivially Z k m . A certain geometric problem is connected with fields k which contain only a finite number s of elements. The characteristic of such a field must be a prime p > 0. If k is a subfield of F and [F fc] = r, then each element of F is obtained uniquely in the form 1

l

N

l

.

N

CN

:

where the

a,

form a

fc-basis of F.

Hence we

see that

F

contains

r

s

elements.

Denote by Zk the center of k and let q be the number of elements Zk Call n = [k Zk ] and let a be any element of k. Then ZA C Na Ck and we introduce the degrees: da = [JV Zk ] and ea = [k a ]. The relation n = d ea shows that da is a divisor of n. We had called g the number of elements in Z k qn and gd * are, therefore, the number of elements in fc, N a respectively, and the num-

in

:

.

:

:

N

;

,

ber of elements in

Z\ N*a

d*

n

are q 1, q 1, q apply now formula (1.15) of 7 to the group G a formula which looks like this: ,

,

fc*

We

f-l-(q-l)+

(1.16)

1,

=

respectively.

fc*

and obtain

2;

where y^ d means vaguely a sum of terms, each one of the form d n 1) and the same term possibly repeated. Indeed, (q l)/(q n a (0:N*) = (q - l)/(^ - 1). The d should always be a divisor of n and be < n since each (G N*) should be > 1. Our aim is to show that a formula like (1.16) can not exist if n > 1. If we succeed, we will have proved n = 1 and, therefore, This would mean that fc itself is commutative. To give the fc = Z k proof we first need some facts about the so-called cyclotomic poly:

.

nomials. It is well

the

field of

(LI?)

known that the polynomial x* complex numbers: *-

-

1

-

n

(*

-

1

)

can be factored in

36

GEOMETRIC ALGEBRA

ranges over the n-th roots of unity: c = 1. If d is the precise order of e, then e is called a primalive d-th root of unity. Every d-th n

wherfe

root of unity will appear

the n-th roots of unity

among

if

d

is

a

divisor of n. Let us define

=

*<(*)

where

II

-

(

)

shall range only over the primitive d-th roots of unity.

e

Grouping the factors of

(1.17)

-

x-

(i.i8)

1

we

find

n *x*>

=

d|n

where d ranges over

all divisors of n. These polynomials 3> n (z) are the cyclotomic polynomials. The $ n (x) have obviously highest coefficient 1. We contend now that all coefficients of 3> n (x) are integers.

=

Since &i(x)

assume

it is

x

1,

this contention is true for

proved for all

j(x)

<

with d

n.

n

1

and we may

We know

then that

(1.18) has the form

z

n

-

1

=

*(*)/(*)

where f(x) is the product of the factors ^j(x) with d < n and d\n. Therefore f(x) has integral coefficients and its highest coefficient is 1. The desired polynomial $ n (x) can, therefore, be obtained by dividing x

n

f(x). Remembering the way this quotient is computed fact that the highest coefficient of f(x) is 1, we see that

by

1

and the

&n (x)

has integral coefficients. Let now d be a divisor of n but d

s

-

-

i

=

<

n

n.

Then

*,(*);

aid

each term $a(z) since d\n.

nomial

But

3> n (z)

divides x

n

d

appear as a factor on the right side of (1.18)

will

will still I

as well as (x

d

n

the quotient (x be one of the factors.

T* n] in

n

l)/(x

Thus we

1)

the poly-

see that

3> n (x)

d

l)/(z

1),

.

x

and both f(x) and x

=

we

q,

n

g(x) will

1

have integral

coefficients.

If

we n

set

1 see that the integer $(
and (q and can conclude that the integer

$n (q) must

divide q

1.

CHAPTER

37

I

Now we estimate the size of the integer <((/). It is a product of terms q The absolute value of q 6 is the distance of the point c on the unit circle arid the point q > 2 on the real axis. Each factor is certainly > 1 even > q 1 in absolute value and can be equal = 1 only for to q 1. This case e = 1 does not occur if n > 1 since c is a primitive n-th root of unity. Thus certainly !$((?) > q 1 if n > 1. We see clearly that $(
,

,

I

THEOREM

1.14.

Every

with a finite number of elements is

field,

commutative.

DEFINITION 1.13. Let / be a map of a field k into some field F which is one-to-one into and is a homomorphism for addition. If

/

satisfies

(1.19)

=

f(ab)

for all a, b e

fc

we

call

f(d)f(V)

/ an isomorphism

(1.20)

of k into F. If

/

satisfies

=

f(ab)

we call / an antiisomorphism of k into F. Let us remark that it suffices to assume that / is a homomorphism for addition, does not map all of k onto 0, and satisfies either (1.19) or (1.20). Indeed, if /(a) = for a single a 9* 0, then it would already follow that /(afc) 0. Since fc is a field, afc = fc, which contradicts our assumption. Hua has discovered a beautiful theorem which has a nice geometric for all a, b e k

application:

THEOREM

1.15.

//

o-

is

a

map

of a field k into some field

F

which

satisfies the following conditions:

a

1)

is

a homomorphism for addition 1 a 7* we have 0(a~ ) = (o-(a))" j

l

2) for

the inverse of

maps

3) er(l)

then


=

(x

=

=

1

in 2),

l)(a? 1.

If

+

;

i.e.,

we assume

thai


the inverse of the image,

1;

is either

REMARK. a

an element onto

an isomorphism or an antiisomorphism of k

Suppose a then x = 1)

=

0.

satisfies er(l)

Thus,

we put r(a) =

only conditions 1) and 2 1 x = of that is x

satisfies

if

tr(a),

o-

into F.

,

does not satisfy condition

then T

Set

2). 1

=

3),
will satisfy all three conditions

38

GEOMETRIC ALGEBRA

which means that
into.

We

a, & t k; a, 1

1

(b' 1

(IT

-

1

is

a)"" 1

to the right:

a)"

a'

an identity: 1 l 6 ^ and a"" 7* b. Then the expression a~ + well defined. Let us factor out a' to the left and

establish

first

Assume

+ (IT 1

1

=

1

a)"

l

a~ ((b~

-

l

+ a)(b~ -

= a-V'OT 1

We

1

l

a)

a)-

1

a)"

.

can, therefore, take the inverse: 1

(a"

+

-

l

(b~

1

1

a)" )'

=

l

(b~

-

d)ba

=

a

-

aba.

Thus we have

-

a

+

l

(a~

-

1

(6"

1

1

a)' )-

=

aba.

we apply a

to the left side, conditions 1) and 2) allow us to a each time with the operation of taking the inverse interchange we will end and up with an expression like the one we started with, 9 but a replaced by a and 6 replaced by 6*. The left side becomes a'&'a' and we have shown that If

(abaY

(1.21) (1.21) is also true

a or b are But a'Va*

if

side of (1.21) is a*. (1.21) is true for all a, 6 e k. 1 in (1.21); Set 6

= a'&V. 1

0. If a""

=

=

a*a~ a*

=

(aT-

ff

6,

=

then ab a" which

=

1,

the

=

2

(a )'

(1.22)

+

b:

6')'

=

Replace a in (1.22) by a (a*

If

+ ab +

we use

(1.23)

(1.22)

ba

+

and condition (06)'

+

(ba)'

(a*)'

1)

=

we

+ a'b' +

b'a'

finally obtain

a'b'

+

b'a'.

+

left

shows that

(&*)'.

CHAPTER

Now

comes the main

(1.24)

((ab)'

trick.

-

Let

39

I

and consider

a, b 5^

-

a'bO(abr'((ab)'

bV).

Multiplying out (1.24) becomes equal to (ab)'

(1.25)

Use

- bV -

a'b'

+

a'b'(ab)~'b'a'.

=

on b'(ab)"'b' (b(abr'b)', write a'(b'(ab)"b')a' a'(b(dbr bya' and use (1.21) again. We obtain (1.21)

-

l

l

(a-b(abr b-ay

Thus

=

(ba)'.

(1.24) is equal to

- bV -

(ab)'

which

is

zero

by

+

a'b'

(ba)'

Thus the product

(1.23).

(1.24) is zero

and one

of

=

0,

the factors must vanish.

We now know

that

or b'a' Sa'b'

which

is

much

nearer to what

we want

to show. For a or b

(1.26) is trivially true.

We

ask whether the following situation could occur in a, b, c, d exist in k such that

k:

can four

elements

(ab)'

=

a'b'

^ bV,

(1.27)

We

shall derive a contradiction from this assumption. Let x be any element in k and use (1.26) on a and b

+

+

(a'(b

+ x)) 9

(1.28)

(a(b

<

left side is (ab

+

ax)' 9

=

a'b'

+ aV

or

((b

The

9

x)

x:

+

*)

V=

= a'b' = a'x'.

+

b'a'

+ V.

(ax). If the

first

case of

If the second case of (1.28) happens we get (ax) 9 9 9 happens remember that a'b' 5^ b'a' so that certainly (ax) 7* x a 9 This means, by (1.26), that (ax) = a'x'. We have, therefore, always

(1.28)

.

GEOMETRIC ALGEBRA

40

=

(ax)'

+

(c(d

a'x'.

The same method is used on

and

x))

((c

+

x)d)'. Collecting

(1.29)

(oar)'

=

a'x',

(1.30)

(xbY

=

x'b',

(1.31)

(ex)'

=

x'c',

(1.32)

(xd)'

=

d'x'.

=

Set x

=

and x

d in

and x

(1.29)

b in (1.31).

We

a'd'

(1.33)

=

a in

+

the expressions ((a x)b)', four cases the results are:

all

(1.32); set also

=

x

c in (1.30)

obtain

=

and

d'a'

c'6"

= 6V.

Finally:

f (a

(1.34) ((a

+ c)(b +

=

d)Y

\ab)

~\~

\ad)

~\~

=

a'b'

+ a'd' + c'b' + c'f

or

+

d)'(o

computation of the

direct

d)'

< ((b

A

+ c)'(b +

(co)

b'a'

+ d'a' + b'c' + dV.

left side gives:

(ca)

~\~

+ c)' = =

a b

~{~

ad

~f*

c b

~\~

d

c

.

on the right of (1.34) would give c'd = dV which contradicts (1.27). Using (1.33) for the second possibility we = b a* and this contradicts (1.27) again. Our theorem is get

The

ff

first possibility

aV

ff

finally established; the fact that (1.27)

means that

in (1.26)

we

either

have the

or else the second one for all a, 6 e

9.

DEFINITION

1.14.

A

k

is

for all a, b e

fc

k.

Ordered

field

can not happen obviously

first possibility

fields

said to be ordered,

if,

first

of

all,

ordered as an additive group. Rewriting Definition 1.11 in the additive notation (the third condition in 1.11 is not necessary since addition is commutative) this means that a set P of so-called "posi-

it is

tive" elements 1)

2)

is

singled out such that

=

-P VJ {0} VJ P (disjoint), P + P C P (P is closed under addition). k

To

these

3)

P-P

two additive conditions we add a multiplicative one:

CP

(a

product of positive elements

is positive).

CHAPTER

We

a>6bya

41

I

mention again that a > is can use all of and, therefore, consequences Definition 1.11 of course in the additive form among them for b > a. Let us now derive multipliexample that a > b implies define

now

equivalent with a

6eP,

P

t

cative consequences. a) If c e

and

(a

6) e

>

and a

6)c e

c(a

ifO

> 6, then a - b e P. Therefore, c(a - b) c P hence ca > cb and ac > be. If c e P, then P, cb ca and be hence ac are reversed > > P, (inequalities

P

c).

> 0, then a- a > 0. If > a multiplication by a reverses the inequality, hence also in this case a a > 0. Squares of non-zero 2 elements are positive and especially 1 = I > 0. Therefore > 1. b) If a

-

We can write c" > c implies > 1

2

1

c-(c~ l c~

)

and

>

see that c

implies c"

1

>

0,

.

1

and on the right by c" keeps Multiplying on the left by c j 1 l the inequality: a > b implies cac" > cbc~ The special cases 6 = l and show that cPc~ P and, if S is the set of elements l a > 1, that cSc~ S. The set P is an invariant multiplicative l subgroup of A;*; a > b is equivalent with ab~ > 1, hence with .

6=1

C

C

ab~

l

e S. If

a

E

1

a" gives a"

by The group

P and neither > > 1 or a" e S.

1

not

=

then

1,

1

>

a.

Multiplying

1

1

P is the disjoint union

l

VJ S and our ordering 1 an ordering of the multiplicative group P. c) Since 1 e P, any sum of terms 1 is in P which shows that the characteristic of k must be 0. The ordinary positive integers are also positive in k. The ordering of A; induces, therefore, on the sub field Q of rational numbers the usual ordering. This proves also that the field of rational numbers can only be ordered in the usual

S~ ^J

{

}

of k induces

way. In some geometric problems the ordering of a field arises in a quite different manner. We are led to another definition.

DEFINITION

properties hold.

however, 1)

A

1.15.

<

binary relation a

fixed

a

e

2)

it,

others

For any

is

fixed

a

weakly ordered if a such that the usual ordering

said to be

to field operations presupposes,

axioms: k the ordering of k shall either be preserved x a. (Notice that some a may

map x may reverse

or reversed under the

preserve

k

The connection

much weaker

For any

field

b is defined in k

c ft*

+

it.)

the ordering of k shall either be preserved

GEOMETRIC ALGEBRA

42

map x

or reversed under the

x

* xa.

is

(Nothing

said about a

map

ax.) If

we say elements

that either a

<

l

a2

are in the arrangement a^ < < an or a x a 2

,

o2

>

>

-

,

,

>

a

.

a n we mean The axioms

say that the maps preserve the arrangements. There is one freak case: the field with two elements 0, 1 has obviously only one arrangement so that any one-to-one map preserves it; the field can be weakly ordered. Suppose from now on that k has at least three elements. We prove certain facts about such a weakly ordered field. 1)

k can not have the characteristic

distinct elements of k

Let indeed

2.

and suppose that

A;

is

0, a, 6 be three weakly ordered and has

characteristic 2. a) Our elements are arranged as 0, a, 6. If we add a we get the b and if we add b we get fc, a + &, 0. These arrangement a, 0, a two arrangements combined give a, 0, a 6, 6 and this contradicts

+

+

the original one. b)

is a, 0, b. Adding a and 6 we get 0, a, a whose combination gives either 0, a, b, a +

The arrangement

and a 0, 6, a,

+

fe,

a

+

6,

The arrangement

is

is

0,

a/2 and a/2 we a which, together

a/2 and a/2 Adding Combined with the original one

a/2, a/2.

a/2,

a,

get a, 0,

a,

a/2, a/2. (The case 0, a/2,

change in

and consider the

fc*

0, a/2,

and a/2,

gives

we

The arrangement

and

a/2,

e

Adding

a/2, 0, a/2.

get the arrangements a, with the original one give

b)

6

b contradicting the assumption.

These two types exhaust the possibilities. 2) fc has now a characteristic 7* 2. Let a a/2. possible arrangements of 0, a/2 and a)

+

b or

0, a.

a/2

is

just a sign

a.)

Both these hypotheses lead to the arrangement a, 0,

which must, therefore, hold in any Adding a we also get 0, a, 2a. 3)

a

field

k for any non-zero element.

Suppose two elements are in the arrangement

CHAPTER

43

I

0, a, b.

Adding a we get a, 2a, a + 6. Combining it with 0, a, 2a we see that also have 0, a, a + 6. Thus we have the following rule: If two elements a and b are on the same side of 0, then their sum is also on this same side. Denote now by P the set of all elements which are on the same

we

as the element

side of

1.

Then we know

-PU and that field.

{0}

that k

is

the disjoint union

UP

P+PC

It is

P. These are the first two axioms of an ordered worth while noticing that we used only condition 1) of

Definition 1.15.

as 1. Multiplying a, b t P. Then a is on the same side of we get that ab is on the same side of as 6. But this means ab*P. The field k is ordered. We see that the maps x > ax (for a j& 0)

Suppose b

by

also either preserve or reverse the ordering.

THEOREM 1.16. // a field k is weakly ordered and has more than two elements, then it is an ordered field. Hilbert has constructed an ordered non-commutative field and we

shall present his example.

F be any field and a an automorphism of F. We denote by 9 a the image of an element a t F under <7 and by a* its image under the automorphism or*. We construct an extension field k of F consisting of all formal Let

power

series

which contain only a finite number of non-zero coi; in other words we assume that for each can be found such that a, = for i < N. power series an integer The addition in k is defined as usual but the multiplication is not the ordinary one. To multiply two power series one first multiplies

in a variable efficients

t

with a negative

N

termwise:

GEOMETRIC ALGEBRA

44

But now one does not assume that f can be interchanged with

6,

;

instead of an ordinary interchange one uses the rule t*b

= 6'V

which for our product produces the

Z

series

*,&;'"'.

Finally one collects terms with the will, therefore be

same power

of

/.

The

coefficient

c r of

a good exercise to make all this rigorous. We indicate the and leave the details to the reader. Since a power series is steps the coefficients we may just as well say that the elements given by of k are functions f(i) on the integers with value in F, thinking of It is

the value f(i) as the coefficient of t\ We consider only functions / whenever i < whose value f(i) = (some integer depending on /). The addition of functions is defined as usual but the product

N

fg shall be the function whose value on

One has number i

<

to verify that the of non-zero terms

i is

given by

sum on and

the right side has only a finite also that (fg)(i) = whenever

N + M- N is the integer belonging to / and M the one belonging

tog. It is easy to

show that the two we wish

associative law hold. Since

we have power

distributive laws

and also the

to consider k as extension of

F

map F

isomorphically into k. The naive description by series suggests how to do it. An element a e F is mapped onto to

the function a which satisfies d(0) = a and d(i) = for i 0. One > a is an a F that of into k. The the isomorphism ring verifies^ F 1 the unit of of element turns unit of out to be the k. If image

^

/ e k, then af is the function whose value at has at i the value f(i)-a' m

i is

a-f(f)

)

whereas fa

%

.

One wishes also to get the power t To this effect one denotes by t m the function which has value 1 at m and j/alue at all other .

integers.

Now

one proves tmtn

=

/ m+n

and

/<,

=

I,

the unit element.

CHAPTER Therefore /7 /

= r. Now one

fc,

tm

/!

by

/

we

see that

we can

write

to drop the bar on top of

then//

w

is

(/O(0 =

We

and abbreviating

can prove

One begins If / e

=

45

I

really

-

f(i

wish to show that k

a, identifying a with a. a shift by m. One easily verifies the formulas

and

m) a

is

(r/)(i)

field, i.e.,

=

(/(f

- m)Y\

that a non-zero element has

an inverse. If / ?* 0, then by taking a suitable f* and a suitable m a e F, the new function g = a// will satisfy 0(0) = 1 and g(i) = for i < (its power series has no negative terms and begins with 1). If one can show that g has an inverse, then / will have the inverse m l = a~ gt~ m ). One tries a hypothetical inverse h of t g~ a (since / the same type, &(0) = 1, h(i) = for i < 0. For /i(l), A(2), one l

-

finds equations (starting with A(0)

=

1)

+ h(l)g(iy + + h(l)g(oy

=

0,

h(G)g(2)

=

0,

h(G)g(2)

=

0,

fe(0) ff (l)

The inverse which allow successively to compute /i(l), /i(2), l be a left inverse: f~ f = 1. But mere group theory shows that l then ff~ = 1. Our ring fc is an extension field of F. At this point one can safely return to the naive power series = 2^ a t e fc and if / ^ 0, then there exists description of fc. If / if i < but a# ^ 0. We shall call such that a,- = an integer this a N the lowest term of /. If a N is the lowest term of / and b M the lowest term of g, then a N b M will be the lowest term of fg. It is now time to talk of the ordering of fc. We assume that F is an ordered field and try to define a set P of elements of fc which shall whose lowest induce an ordering of fc. Let P consist of those / 9 a F: 0. The of condition term a^ is positive in the ordering N > .

will

l

-

t

N

N

fc= the

PUJOJUP sum

of

v

(disjoint) is obviously satisfied; the rule that

two elements

in

P lies also in P is not difficult

to verify.

There remains the product rule. Here we obviously have to assume of our automorphism that a > implies a* > 0. Suppose that a has

GEOMETRIC ALGEBRA

46

imply that a*

<

0. It follows

1

ff

is

property.

aN

will also

implied by a > which shows that or" has our property. 1 needed o-" in order to be sure that all powers of o- have the same

>

that a

We

<

then a

this property;

Now we

respectively,

,

can prove the product

bM

of /,

respectively,

lowest term aN b'M of fg will be positive.

We

rule. If

the lowest terms

g are positive, then the see that fc is ordered.

We

=

a*t. As soon as o- is not the be non-commutative. This reduces the construction problem a little. We must find a field F (commutative or not) which is ordered and which has an automorphism
Is

fc

non-commutative?

identity

automorphism our

have

field

fc

ta

will

variable x:

]C

<*&*>

but this time with the ordinary multiplication of power series (hence with identity as automorphism). Let us order F by the lowest term; possible since the positivity in Q trivially.

this

is

We

F which

automorphism preserves the an automorphism
identity

must

preserves positivity. It

is

find

precisely:

/= To show That

that

o-

is

a,*'-r- Z<2V.

an automorphism

of

F is a rather

preserves the sign of the lowest term This finishes the construction of fc. it

THEOREM

1.17.

is

trivial exercise.

completely obvious.

There do exist ordered non-commutative

fields.

An ordered field has characteristic field Q of rational numbers. Since the

field

the usual way, the ordering of

in agreement with the usual

fc

is

and contains,

Q

therefore, the can be ordered only in

ordering of Q.

DEFINITION

An

ordered field fc is said to be archimedean one can find an integer n such that a < n. Let fc be archimedean and a e fc. There are integers n, m such that m < a < n. Every element of fc lies a < n and a < ra; then between integers. Consider now the set C of all rational numbers > a and the set D of all rational numbers < a. Neither C nor D

if

for

any a

e

1.16.

fc

CHAPTER

47

I

be empty and these sets form a Dedekind cut in the rational numbers. Denote by /(a) the real number defined by this cut. We have a map / k R where R is the field of real numbers and will show that / is one-to-one into. Let a 7* b, say 6 > a; we shall construct a rational number which lies strictly between a and 6 and 1 a)" thereby prove /(a) 7* f(b). There is an integer m > (b Since b a > ma > 1. Among the integers n we obtain mb which satisfy n > ma there is a smallest one. For this n we have will

:

.

ma > n The

>

1

n

= n

ma)

(mb

mb

+ ma and get mb >

n.

inequality

mb > n > ma shows

m >

and

b

> n/m >

a.

We

conclude f(b)

>

/(a).

From

the properties of Dedekind cuts one deduces that / is an isomorphism of k into R; f preserves the ordering as our previous discussion shows.

THEOREM

An

1.18.

of fact, isomorphic

to

archimedean field is commutative, as a matter a subfield of the real numbers in its natural

ordering.

We saw that Q admits only one ordering. The field R of real numbers admits also only one ordering. Indeed squares of non-zero elements are positive in any ordering. Let a c R be positive in some ordering in the of R. Then a can not be a square which means that a > then a is in the ordinary sense ordinary sense. Conversely if a > a square and, therefore, positive in the given ordering. Let now cr be an automorphism of R] a carries squares into squares and preserves, therefore, the positivity in JR. If a < 6, then a* < V. The automorphism a- induces on Q the identity. Let a be a given

element of 72. It is uniquely described by the Dedekind cut which it produces among the rational numbers. Since or preserves inequalities 9 and leaves all rational numbers fixed, a will have the same Dedekind cut as

a.

But

this

means

a"

=

o.

THEOREM 1.19. The field R of real numbers can be ordered only in the usual way. Every automorphism of R is the identity. 10. Valuations 1.17. Let fc be a group with 0-element. We

DEFINITION tive

field

shall

and G an ordered multiplicaassume both k and G to be

GEOMETRIC ALGEBRA

48

A map of k into (?,

commutative.

denoted by a

lute value of a") is called a valuation of conditions:

= = |ob|

1)

2)

+

3) |a

and only

if

|a|

if it

|a|

0,

|a|.|b|,

< Max

b|

REMARK.

(|a|, |6|).

may be

Condition 3)

<

1

+

1

3a) |a| implies Proof: Conditions 1)

1

Suppose 3a)

is

<

replaced

a|

<

2)

imply

and

by the

rule

1.

=

|1|

1

so that 3) implies 3a). if a = b = 0. and 3a) implies

true. Condition 3) certainly holds

Otherwise assume that

+

=

a

if

fc

(read as "absosatisfies the following >

<

|a|

|6|.

Then

\a/b\

<

1

Multiplying both sides by |6| we get 3). All we shall really need are some examples of valuations. They are easily constructed by means of another notion. |1

a/b\

1.

DEFINITION 1.18. Let k be a commutative field. A subring R said to be a valuation ring if for every non-zero element of k 1 either a or a" lies in R. 1 The elements of k for which both a and a" are in R form a multiplicative group C7, called the group of units. Those non-zero elements l for which a~ but not a is in R form a set S, and those for which a, but not a" is in R form the set S~ consisting of the inverses of

is

l

1

,

the elements in S. Clearly,

=

k

and

this

union

The

is disjoint.

R = If a, 6 e

and b~

l

1

fiT

1

-*?-

The

R and

ab

=

a(a~ b~

C

S' and

l

1

U

cosets of

l

t C7,

this

union

)

1

form the {0}

R

ring

cfV

S'

is: 1

1

e

U

1 ;

U.

R, hence a'

1

=

6- (a

V

1

)

t

R

U and

b E U. This proves Sand SC7 S. 88 consequently

shows a

t

t

C

C

set

U S^/UV

[U/U]

U

S/C7

a group with 0-element, S/U a may, therefore, consider k/U as an ordered group

is disjoint.

semi-group. We with 0-element.

A

then

UV S,

S-'U

{0} VJ

R which S' U C S"

= and

U

{0}

now

k/U

is

> |a| shall easily constructed: the map a merely be the canonical map of k onto k/U, in other words

valuation

is

CHAPTER |a|

Conditions

means

1)

and

49

I

= aU. The

2) are trivially satisfied.

aU is either in

S~

l

/U or in U/U =

inequality

|a|

<

1

shows that |a| < 1 + a e R and conse-

this

1;

equivalent with a e R and implies trivially 1 quently |1 a| < 1. The case R = k is not excluded. $ is then the empty set and and U/ U = 1 We call this valuation k/ U has only two elements, the trivial valuation. Now our examples: = Q the field of rational numbers and p a given 1) Let k fixed prime. Let R be the set of those elements of Q which can be written with a denominator relatively prime to p. R is obviously a valuation ring. U consists of those fractions whose numerator and denominator are prime to p. If a 5^ 0, then |a| = aU = p*U where p is the contribution of p to a if we factor a into prime powers; < 1 means a e R, hence i > 0. If |a| = p*U and |6| = p'U, then |a| < |&| implies |a/6| = p*~' U < 1 or i j > 0, i > j. An element |a| of Q is "small" if it is divisible by a high power of p. This valuation of Q is called the p-adic valuation. 2) Let k be an ordered field. An element of k is said to be finite if it lies between two integers; if it does not, we may call it infinite. One sees easily that the set of all finite elements forms a valuation ring R. The corresponding set S consists of the infinite elements of k, l the set S~ may be called the set of infinitely small elements; |a| < 1 means now that a is finite. This valuation will be trivial if and only if k is archimedean. If k a gives a classification of the infinitely is not archimedean, then large or small elements "according to size".

is

+

.

%

|

\

CHAPTER

II

Affine and Projective Geometry

1.

Introduction and the

first

three axioms

We

are all familiar with analytic geometry where a point in a described by a pair (x, y) of real numbers, a straight line by plane a linear, a conic by a quadratic equation. Analytic geometry enables is

us to reduce any elementary geometric problem to a mere algebraic one. The intersection of a straight line and a circle suggests, however, enlarging the system by introducing a new plane whose points are

An obvious generalisation of this procedure the following. Let k be a given field; construct a plane whose "points" are the pairs (x, y) of elements of k and define lines by linear equations. It is of course comparatively easy to work out the geometric theorems which would hold in such a geometry.

pairs of complex numbers. is

A much more fascinating problem is, however, the converse. Given a plane geometry whose objects are the elements of two sets, the set of points and the set of lines; assume that certain axioms of geometric nature are true. Is it possible to find a field k such that the points of our geometry can be described by coordinates from A; arid the lines by linear equations? We shall first state the needed axioms in a loose fashion. The first two are the incidence axioms: there is exactly one line through two distinct points and exactly one parallel to a given line through

a given point. The third axiom eliminates freak cases of geometries which are easily enumerated and merely assures us of the existence of enough points and lines. For a while we shall work only with these three axioms and prepare the ground for the fourth (and last) which is

the most interesting one.

One introduces certain symmetries of the geometry called dilatations. They are (except for freaks) one-to-one maps of the plane onto itself which move all points of a line into points of a parallel line. The identity map is such a dilatation and with only our three 51

GEOMETRIC ALGEBRA

52

axioms at ones disposal, one should not expect the geometry to have any others. This is where the fourth axiom comes into play. It is split into two parts, 4a and 4b. Axiom 4a postulates the existence of a translation (a special kind of dilatation) which moves a given point into any other given point. With this axiom 4a alone one can already construct a certain field fc. Axiom 4b shall now ensure the existence of enough dilatations of the remaining kind (which resemble the transformations of similarity in the euclidean plane). Using

axiom 4b one can show that the points can be described by coordinates from k and the lines by linear equations.

One

can, therefore, conclude that one can coordinatize

any geo-

metry which contains enough symmetries.

We shall now start with a precise formulation of the axioms. We are given two sets: a set of "points" and a set of "lines". We are also given a basic binary relation between a given point P and a given line I: "P lies on V (which may or may not be true for the given pair P and I). All the axioms can be expressed in terms of this one basic relation. However it is clear that the language would become intolerably clumsy if one would actually do this. We shall, therefore, use obvious synonyms for the binary relation as "P is on I" or "I contains P" or "I goes through P". If P lies on both I and 9

I and m meet in P and if P is the only point on and m, then we say that "Z and m intersect in P" or that the intersection of I and m". At the beginning we shall be a

m we may say that both

"P

is

I

little stricter

in our language until the reader has learned to replace by the basic relation, but we will soon relax.

the synonyms

For a similar reason we did not draw of this chapter.

The

figures in the first

paragraphs

will, nevertheless, get a better underhe makes simple sketches for himself.

reader

standing of the proofs if Our first task is now to define what

is

meant by "parallelism".

DEFINITION 2.1. If Z and m are two lines such that either I = m or that no point P lies on both I and m, then we call I and m parallel and write Z m. If Z and m are not parallel, we write Z Jf m. If Z Jf m, then there is at least one point P which lies on both ||

Z

and m.

Given two distinct points P and Q, there exists a unique P lies on Z and Q lies on 1. We write Z = P + Q. line If Z Jf m, then there is exactly one point P which lies on both I and m.

AXIOM Z

1.

such that

CHAPTER

hence

I

line

by axiom

points, then,

1,

=

I

m,

m.

\\

AXIOM one

two such

there were

if

Indeed,

53

II

Given a point

2.

P

and a

line

I,

there exists one and only

m such that P lies on m and such that m

THEOREM

\\

I.

"Parallel" is an equivalence relation.

2.1.

obviously reflexive and symmetric. To prove the If there is no point on both J3 transitivity assume l^ \\12 and 1 2 then then li If there is a point P on both ^ and 1 3 1A li and Z 3 Proof: It

is

.

||

.

,

by axiom 2

/t

(since

DEFINITION

THEOREM ir 2

,

and

number of any line.

2.2.

lines

I

and

12

13

\\

12 )

we have

=

lv

13

hence again

,

equivalence class of parallel lines

is

called

lines.

that

Suppose

there

and

this

be a line of

ir l

there exists exactly one point

,

three

exist

Then any number is equal

of parallel lines.

7T 3

Proof: Let

||

An

2.2.

a pencil of parallel

wi

,

\\

m a line of

pencil to the

ir 2

.

distinct

pencils,

w contains the same number of points on

We have

/ Jf

m so that

P of Z which also lies on m. On the other

be any point of I. There is exactly one line m' m, hence such that Q lies on m We have, therefore, a one-to-one correspondence between the points of I and the lines of 7r 2 the number

hand one

Q

let

line of

\\

;

7r 2

.

:

of points of

/

have shown:

is

the same as the

number

of lines of

w2

.

Thus

far

we

two

distinct pencils are given, then each line of one as contains pencil many points as there are lines in the other pencil. If TT is any pencil, then it is certainly distinct from at least two of If

our three pencils, say IT j KI and TT j ir2 The number of points on a line of TTI is equal to the number of lines in ?r and equal to the

number

of lines in

AXIOM does not

3. lie

ir 2

.

The theorem

follows easily.

There exist three distinct points A, B, C such that C on the line A + #. We also say that there exist three non-

collinear points.

+ B and A + C are not parallel: otherwise (containing For the point A) they would be equal and C would lie on A + B + C and A + C B + C. There do the same reason A + B The lines A

JB.

Jf

exist

rem

now

Jf

at least three distinct pencils of parallel lines

2.2 applies.

and Theo-

GEOMETRIC ALGEBRA

54

EXERCISE

Enumerate

1.

carefully (not being prejudiced at all

about existence of points or lines or points on lines) all cases where the first two axioms hold but where axiom 3 is violated.

EXERCISE of points in

2. Find the geometry with the least possible number which all three axioms hold. Show that its structure is

unique.

2. Dilatations

and

translations

In modern mathematics the investigation of the symmetries of a given mathematical structure has always yielded the most powerful results.

Symmetries are maps which preserve certain properties.

In our case they

DEFINITION tion

if it

will

be maps preserving "direction".

A map

2.3.

P

Let two distinct points given. If onZ'.

We

o-

of points into points

I' is

the line

||

P+

1

can give two examples. all points onto one given point.

and any other The identity map.

generate dilatation 2)

We

called a dilata-

and Q and their images P and Q' be Q which passes through P', then Q' lies

Map

1)

is

has the following property:

We

shall call this a de-

dilatation non-degenerate.

can now immediately prove

THEOREM

2.3.

A

dilatation a is uniquely determined by the images

and Q. Should P' = Q', then a is degenare points mapped into P'. Should P' 7^ Q', then a is one-to-one and onto (every point is an image).

P

P', Q' of two distinct points

erate

and

all

Proof: 1) Let R be any point which is not onP Let V be the line through P which is 1 dilatation R' must lie on I Similarly, R' 1

||

.

+ Q. Then R + P R + Q. R + P. By definition of a Jf

will lie

on the

line I"

which

R + Q and passes through Q'. The lines V and I" are not parallel and contain R We see that R is uniquely determined. 2) Let now R be a point of P + Q, say R j* P. Select any point S not on P + Q. The point R is not on P + S; otherwise R and P would lie on P + Q and P + S, hence P + Q = P + S, contradicting

is

||

f

f

.

the choice of S.

The image

S' of

S is determined by

1);

we know

the

CHAPTER

55

II

S. By 1) the images P and S' of P and /S, and R is not on P image of R is determined. 1 = Q'. The degenerate map r which maps every 3) Suppose P into P' has the same effect as
+

f

1

f

.

+

f

||

||

.

+

.

+

+

.

+

||

+

r

1

+

This finishes the proof of the theorem. As an immediate consequence we have the

COROLLARY. er

DEFINITION

Any line

THEOREM a

o-

and the

Let a be a non-degenerate dilatation,

2.4.

is

P and vP shall be called a trace of P. unique and = P +
We

Let

2.4.

trace of P. If

Proof:

Q


lies

can assume

+

line
aQ

a non-degenerate dilatation, then
be

on

1,

Q I \\

P so that = P +

7*

Z

by

The

P

a point and

Q.

Then aQ But

definition of a dilatation.

+

COROLLARY.

P a point. P ^ o-P,

If

I,

vQ have the point
=

has two fixed points, then a

containing

then the trace

I

If the dilatation

the identity map.

is

intersection of

^ aP I

and

are, therefore, equal.

two non-parallel

traces

is

a

fixed point.

The

possibilities for the traces of

All lines, if and only if or 2) All lines through a point 1)

=

a non-singular dilatation a are:

1.

P if

Indeed, any line through trace since


j& 1

P

is

cr 7* 1 and if P is a fixed point of o-. a trace and no other line could be a

can not have more than one fixed point.

56

GEOMETRIC ALGEBRA

A certain pencil of parallel lines if

3)

This exhausts

DEFINITION

all possibilities

A

2.5.

or has no fixed point. and suggests a definition.

non-singular dilatation T shall be called a 1 or if r has no fixed point. If r is a trans-

=

translation

if

lation j&

then the traces of r form a pencil w of parallel lines which

we

1,

either r

shall call the "direction" of

THEOREM

A

2.5.

T.

translation r is uniquely determined by the image

of one point P. (Notice that we do not claim the existence of translations with pre-assigned image point.)

Proof: Let

be a trace of r which contains P. Any line parallel V l,l'j*l and Q on I'. Then rQ must lie on I'. Put m = P Q and let ra' be parallel to m and contain rP. By definition of a dilatation rQ must also lie on m'. rQ lies on both V and m' and I Jf m' since I Jf m. Thus rQ is uniquely determined. But a dilation is uniquely determined by the image of two points. We shall assume from now on that dilatations are non-singular unless we explicitly state the opposite. The definition of a dilatation can then be simplified to: Whenever P ^ Q we have
/

is

/

also a trace of r; let

\\

+

1

+

\\

,

.

P+Q The map

o-i
map which l


P

7*

v~

will l

also be

Q and

l


we can form the inverse

one-to-one and onto.

P +

l

(T'Q

||


P)

+

If

P

l

a(
Q)

9*

Q

then

= P + Q

1

which shows o-' to be a dilatation. The dilatations form, therefore, a group. 1 If r is a translation, then r" is also a translation. Suppose indeed l l that r~ has a fixed point P. Then r~ P = P. Applying r to it we get P = rP which is only possible for the translation r = 1 in which 1 case r"" = 1 is also a translation. Assume r Y and r2 are translations and suppose that r^r 2 has a fixed point:

r^P =

r^r 2

lation in

= any

1

=

l

P. Thus r 2 and

I

have which means that they are equal: r 2 = ri" which is a translation and we see that r^ 2 is a trans-

the same effect on

Then

P. Then r 2 P

case.

ri

TI

P

The

1

.

translations form, therefore, a group.

CHAPTER

57

II

Let a be a dilatation and r a translation. We wish to show that I = TI is again a translation. If r l has a fixed point P, then 1 l l l
Suppose now that T l

P +

1

P

and that the pencil l P and is,

7* I

TT

is

the direction of

T.

therefore, a line of TT. l l I = Since er is a dilatation, cr" P TcT'P cr(er~ P) CT(T
T0~

the r-trace of a~

is

+

+

+

||

+

these results:

THEOREM 2.6. The dilatations form a group D and the translations form an invariant sub-group T of D. If a- is a dilatation and T 5* 1 a 1 translation, then T and oro-" have the same direction.

THEOREM TT

2.7.

Identity

and

of parallel lines as direction

then and a retrace of rP. Proof: If T 5*

1,

the translations having

a given pencil

form a group.

P + rP =

rP

+

T

-I

(rP)

is

a r-trace of

P

If TI 7^ 1 and T 2 T^ 1 have the direction ?r, then P + T 2 P lies in v and contains T2 P. It contains, therefore, also TiT 2 P, by Theorem 2.4. If TiT 2

TiT 2

P

P =

THEOREM

T

is

=

P, then T!T 2

which

1; if

T^P ^

P, then

P +

T2

P = P +

a trace of

is

// translations with different directions

2.8.

then

exist,

a commutative group.

Proof: 1)

Suppose

and T2 have

TI

different directions.

direction as

By Theorem

T^T^ has the same direction as T hence the If T^T^T^ 9* 1, then T^^rVi" has the Ti"

the translation

2

,

1

1

.

direction as T 2 l

.

But l

so that TiT 2 Ti T*

TI

and

2.6

same same

1

have the same direction as TI This is a contradiction TI which shows TiT 2 = T2 Ti

TaTrVi"

has also the direction

.

and consequently r^r 2 ri r^ = 1 2) Suppose TI and T 2 have the same direction. By assumption there exists a translation T S whose direction is different from that of Therefore T S TI = TiT 3 The translation T 2 T3 must also TI and T 2 have a direction different from that of TI otherwise T^-T^ = T3 would have the same direction. We know, therefore, also that l

1

.

.

.

;

From

TiT a T 3

Ti'(T 2 T3)

=

=

we

T 2 TiT3

(T2 T 3)-Tx

get

now

GEOMETRIC ALGEBRA

58

REMARK.

It

is

conceivable that the geometry contains only

IT (aside from identity). It is an undecided question whether T has to be commutative in this case. Probably there do exist counter-examples, but none is known.

translations with a single direction

3. Construction of the field

We can hope for a "good" geometry only enough symmetries. We postulate, therefore, AXIOM

Given any two points lation TPQ which moves P into Q: 4a.

T PQ (P)

P

=

if

the geometry has

and Q, there

exists a trans-

Q.

REMARK. Since the image of one point determines a translation, TPQ is unique; especially rPP = 1. Since we now obviously have translations with different directions, our group T of translations be commutative. The geometric meaning of axiom 4a another geometric interpretation shall be given later. We shall define new maps which will send T into T. will

A map

a

homomorphism" homomorphism

if:

DEFINITION preserving 1) It is

2.6.

a

:

T

of T,

>

T

shall

meaning by

is

obvious,

be called a "trace

this that

(

use the symbol r

(We

a

for the

image of

T

under the

map

a.

for

convenience.) 2) It preserves traces, or, more precisely: the traces of r are among a a This means that either r = 1, or that r and T" the traces of r .

have the same

direction.

Important examples of trace-preserving homomorphisms are the following. a)

r t

= 1 for all every translation T of T onto identity: r T. Clearly both conditions are satisfied. We denote this very

special

b)

a

Map

map by

since r

=

1 is

very suggestive. 1 be denoted by 1. Thus r

The

identity which

will

Map

each r onto

inverse r"

=

r for all

map

of course

TtT. c)

by

T~

I .

We

have

its

1 .

We

denote this

CHAPTER

)- =

T

( TlT2

59

tt

JV

=

rrV

"*

T

arid r have the same traces. commutative; r 1 he a fixed dilatation. Map each r onto orcr" We know already that r and error" have the same traces. Furthermore

since

is

d) Let

<7

.

1

-i

\

/


which shows condition

The by

1.

=

-i

-i

We do not give a special name to this map.

set of all trace-preserving

shall

be denoted

We may

construct a

homomorphisms

ft.

DEFINITION

2.7.

Let

new map which sends

a,

be elements of

ft

r into r

" -

T

P

and

T+' =

We may

also send r into (r^)

THEOREM Under

// a and

2.9.

ft

map a

+

ft.

Thus

V.

T

and

call this

a

belong to k, then

this definition the set

element

a

fc.

shall call this

a- ft. Therefore

map

+

ft

and

aft

belong to k.

k becomes an associative ring with unit

1.

Proof: ( Tl

1)

r2 )

af/3

= (r^r (r,r 2 Y (by = r\r 2T?r? (since a, =

r"r?r "r?

(since

T

definition of

a

+

ft

e k)

is

commutative).

ft)

Hence (T

If r

=

If

a

then r T j 1, then 1,

= / = its

1,

hence r

direction

TT

a

^ =

occurs

1.

among tt

r" and, therefore, also

a

+

ft

e

among

the traces of r -r

the traces of r ft

=

T

a+ft .

a

and

This shows

fc.

2)

The (/)"

traces of T are

=

T''.

among

those of r

f ,

hence among those of

GEOMETRIC ALGEBRA

60 3) If a,

j9,

y

E


we show

fc,

__ "

+i

T

T

(a

_ Til ^T

T

r

==

r r

=

T

^

=

=

13

T T

(since

=

T

.

r

=

a

r):

+ I

Tis commutative)

=

This shows that k

T -. T is

T

a:

a .

+ (-l)a0:

a T

=

l. r

6)

.+c-i>

""""

7

+

ft+a

+a = T

T

(P

+ *:

/J

5) 0+

+

3F).

+

a

4)

<*

--.0 + T

__ ~~~*

'

(we used the associativity of

+/?

+ 0) + 7 =

(-n.

=

.( T

T

=

)-

=

1

r.

a commutative group under addition.

+ 7)a

(0

7)

=

+ T:

Pot

Notice that only definitions of addition and multiplication are used. 8)

o(j8

Here we had to use that a

+ 7) =

is

o0

+ 07:

a homomorphism.

9)

=

l-o

10)

T

1

-

(T

a

1

)

-

T*

a

and

and

=

! 1

T"'

=

a:

1

(r )"

=

This establishes all axioms for a ring. In our next theorem we use axiom 4a to the

THEOREM exists

2.10.

a unique

Let

a

dilatation

e a-

fc,

a

^

which has

and

P

P

be

r".

full extent.

a given point. There

as fixed point and such that

CHAPTER -1

61

II

r c T.

for all

Proof:

Suppose such a

1)

the translation rPQ

.


a

TPQ

P

this to the point

Apply

TP (P)

Let

exists.

Q

be any point and select for T

Then

=

=

1

(TTpgCr

.

remembering that

=

^^(P)

leaves

cr

C7T PO (P)

=

P

fixed:


Thus
(2.1)

=

r* PQ (P).

how to compute the image of a given point Q under 0: takes rPO computes TPQ and applies this to P. This proves not only the uniqueness but shows us also which map
One

,

prove our theorem. a t k and allow even a = 0. Define a map a on the Let now 2) formula (2.1). We contend that a is a dilatation, but points Q by tried in order to

possibly degenerate.

Q and R

Let

P

R

into

be two distinct points. The translation T QR TPQ moves

since r QR TPQ (P)

=

r QR (Q)

=

72.

Therefore

TQRTPQ If

we apply a

to this equation,

we

Applying both sides to the point

by formula

hence,

get

P we

obtain

(2.1),

(2.2)

be a line through (P) = 1 "(P) = satisfy to be a dilatation. By (2.1) we have
Let

i

of TO*

1(P)

.

= P

so that a has

every point T PQ

=

is

mapped

1 for all

Q.

Any

P

into

;

as fixed point. If

P

and


(2.1) gives

is

degenerate, then

P =

r?o(P); hence

translation r has the form rPQ

,

namely

for

GEOMETRIC ALGEBRA

62

Q = a

,

r (p)

T* 0,

Therefore r

m

then

Let now a the form

cr

is

=

a

1 f or all

T

r e

which means a

=

0. If

not degenerate.

we know

?& 0. Since


=

we can

P,

write (2.1) in

or

Q =

l


This means that the translation

and

is

consequently rPO

.

>

~1

o

1

1

P

moves

(rp )((r" )~

into

Hence -1

a


or

We have already mentioned that r PQ can be any translation and have, therefore, T*

But

=

1

r-

T e T.

for all


r,

this is the contention of our theorem.

REMARK.

Example

shows the converse. preserving

If

d) of the trace-preserving homomorphisms 1 * arcr" is a dilatation, then r is a trace-

a-

homomorphism

a; it

can not be

since

=

1

orcr""

1

holds

=

1 and not for all r. We see, therefore, that example d) only f or T has already exhausted all possibilities for a trace-preserving homowe can find a
fixed point for

THEOREM


k

2.11.

is

afield (possibly not commutative under multi-

plication).

Proof: Let

atk and a

Find a dilatation
5^ 0.

The map which sends r into o^Vcr = 1 of k which we shall call a" Thus

l

l

l

(r~ r(
is

)~

also

.

a

r

=

1

ffTff"

and

r

a

=

*

l


Now, er'a""

r

1

=

/

(T

a~'

I

\a )

=

/

0(0

1

\

T
1

=

r

=

1

r

a

r

=

1

oro-""

.

an element

CHAPTER

63

II

and a~ a l

r

which shows aa" inverse

is

=

af^a

// r

2.12.

or r = or a =

1

1

)

= 1

If r

1.

a

a

= = r

1 ft

and

=

r

1

r

establishes the existence of

holds for one a and one holds for particular a, ft,

=

1

and a

)- =

1

1

(T

Suppose now r

=

a

=

then r

5* 0,

=

~1\ I/ (
an

T,

then either

r,

then either

ft.

Proof: Suppose r"

a

<7

established).

THEOREM a = T =

=

1

axa"

/

(T

uniqueness follows from group theory, provided the

(its

existence

=

T**.

I"'

7* 0.

=

Apply a~ or

l

r

Multiplying by r~

ft

=

l

:

1.

we have

r

a ~ ft

=

1.

If

1.

(The element

is

of course

(

1) ft

and

also

/3(

l),

since

holds in (-1)) = ft-Q = 0. The rule 0-0 = every distributive ring but is also checked directly: r"' = (r / = ft

+ ft-(-l) =

0(1

+

4. Introduction of coordinates

Our geometry is still not "symmetric" enough. We shall need one more axiom which we shall put into two forms; the equivalence of the two forms will be a separate theorem.

AXIOM if T!

r2

=

7^

r

4b.

1,

If T!

and

r 2 T^ 1, T!

r 2 are translations with the

3^ r 2

,

same

then there exists an a in

fc

traces

and

such that

a i

.

= rl if r 2 = 1, then r 2 = T? but we wanted to state the axiom in 3^ needed, 2.12 shows Theorem that the a is unique. its weakest form. form the other of the axiom. Now we state REMARK. Thus only TI

If 73

=

7!, then r 2

;

.

1 is

AXIOM 4b P (for a given point P). Given two points Q and R such that P, Q, R are distinct but lie on a line; then there exists a dilatation o- which has P as fixed point and moves Q into R. 2.13. // axiom 4b P holds for one point P, then axiom 4b and axiom 4b implies axiom 4b P for all points P.

THEOREM is true,

GEOMETRIC ALGEBRA

64 Proof:

P

true for a particular P. Let TI and r 2 be 1) Assume axiom 4b translations satisfying the hypothesis of axiom 4b and set ?J? = Q and T 2 = R. Since r l and r 2 have the same traces, the points P, Q,

P

R

and are

are collinear by the assumptions on TI and r 2 fixed
.

>

r2 Let a e k such that r = ow" Then r 2 = r" 2) Assume that axiom 4b is true and let P, Q, 72 be three distinct collinear points. Put r = rPO r 2 = rPR Since P, Q, J? are collinear, T! and r 2 have the same traces, TI 3^ 1, r 2 ^ 1 and TX 5^ r 2 By axiom 4b there exists an a 9* such that r 2 = r" By Theorem 2.10 1

=

a

1

.

.

.

.

,

.

.

there exists a dilatation with fixed point for all r e T, especially

=

T2 If

we apply

P we


this to

=

T2 (T

,

=

get r 2 (P)

P

such that r

(TTi

a

=

1

orcr"

.

ori(P) or

R =

er(Q); i.e.,

o-

is

the desired dilatation.

The geometric meaning shall

be given

THEOREM directions.

of

P

axiom 4b

clear

is

and another one

later.

2.14.

To any

Let

7* 1, r 2 7* 1 be translations with different

TI

translation T e

T

there exist

unique elements

a,

ft

in k such that T

Proof:

=

The commutativity

=

r irj

is clear

rjr

i

since

.

T

commutative. Q. Let li be a retrace

is

P be any point and suppose r(P) = 1) P and 2 a retrace through Q. Then through and 2 a point R on Let

Z

/

/i

Zt

Jf / 2

so that there

is

.

The translation rPR The translation r RQ is

is

either 1 or has the

also either

1

same direction as TI same direction as r 2 elements a and ft in k .

or has the

By axiom 4b and the remark there do exist such that TPR = r" and T RQ = r 2 Then r 2 r1(P) = and hence r/ = r 2 r" But rPO is our r. = r \r\ then rf" 7 = rj" If r?" 7 5^ 2) If T T{ .

.

T RQ rPR (P)

= Q

.

1

side of Ti~

y

=

r2

" /s

has the direction of r l

which is not possible. Thus = 6. shows a = 7 and of T 2

.

,

7 rj*"

=

1

1,

then the

left

the right side the direction and r 2 "^ = 1 Theorem 2.12

,

.

CHAPTER

65

II

We

are now ready to introduce coordinates. In ordinary geometry one describes them by selecting an origin 0, by drawing two coordinate axis and by marking "unit points" on each of the axis. We shall do precisely the same. We select a point as "origin" and two translations TI ^ 1 and r2 j* 1 with different traces. The

naive meaning of TI and r 2 and the r2 -trace through axis

is

The retrace through be thought of as the coordinate

the following:

shall

and the two points ri(0) and r 2 (0) as the "unit points". now P be any point. Write the translation r p in the form

Let

with the unique

,

If conversely (,

17

77)

k and assign to P the pair ( 77) as coordinates. a given pair, then T\T\ is a translation, it takes

1

,

is

=

P

so that TOP into a certain point have the given coordinates (, 77).

We

For

we get TOO =

P=

1

=

rirj

write

,

and

P =

this point

P

will

(f, 77).

T\T^ so that the origin

has coordinates

(0, 0).

For

(1,

we have

0)

"unit point" on the the second axis.

rlrj

=

first axis

=

TI

TOP so that r^O) is indeed the r 2 (0) the unit point on

and similarly

Let now I be any line, P = (a, 0) a point on I and r = r \T\ a translation 7* 1 which has / as trace. Let Q = (, 77) be any point on I. Then TPQ is either 1 or has the direction of r. By axiom 4b and the remark we have

=

TPQ

with some

t

e

T

fc.

Z will appear among the traces of T' / (P) = Q which shows that every r* is some TPQ with Q on I. We also know that t is uniquely determined by TPQ (Theorem 2.12). To get the coordinates of Q we must compute TOO We have

Conversely, for any /,

is

some point on

;

i

*\<

T T 2; (Y i

Thus Q = ranges over a, 0, 7, 5

and

(fy fc.

+

a, J5

+

_

a $

T i T2

ty + a tt+ft T2 T!

0) will range over the points of

How much freedom can we have

still

be certain that'Q

=

points of a line? Since P = T*IT* can be any point, a,

(ty

/}

+

a, t8

I

if t

in the choices of

+ 0) describes

are arbitrary; T

the

= T>J had

GEOMETRIC ALGEBRA

66 to be a translation j* If this is the case,

which means that y and line through P which

1

then the

8

are not both 0.

is

a r-trace

the

is

formula describes.

line that our

We may abbreviate in vectorial notation: We put A = (7, d) and P = (a, 0). Then

t

ranging over

fc,

gives the points of a line.

A

j

is

the only restric-

tion.

The reader may be more familiar with another form

of the equation

of a straight line:

Put Q

We a)

7

1

ay~

d.

=

(x, y).

Then x =

ty

+

=

y

a,

/

+

ft.

consider two cases:

Then / = xy~

7^ 0.

This

is

l

l

ay~ and consequently y

=

1

xy"

8

+

ft

the form i

y Converse,

if

y

= xm

+

b

"parametric form" with y

y

= xm +

put x

=

1, 8

= =

/,

6.

2/

m, a

= tm = 0,

+

b

=

and we have a b.

so that y is arbitrary and x = a. 7 = 0. Then 5 5^ We have certainly recovered the usual description of points and b)

lines

by

coordinates.

5. Ajfine

We field

geometry based on a given

field

now at the other end. Suppose that k is a given us build up an affine geometry based on k. define a point P to be an ordered pair (, 77) of elements e k. rj shall start

and

let

We We shall identify P also with the vector and use vector notation. A line will be defined as the point set P + tA where P is a given point, A a given non-zero vector and where / ranges over k. The ,

(

,

rj)

Z

relation that a point Q lies on a line I shall merely mean that Q is an element of the point set we called L Our problem is now to prove that this geometry satisfies all our axioms. Let us first consider the possible intersections of a line P tA

+

and Q + uB where A and B are non-zero vectors and where u range over k. For a common point we must have

P

+

tA

= Q

+ uB

/

and

CHAPTER

and we must shall

now

find the

and u

/

67

II

satisfying this equation

which we

write in the form

- uB = Q -

tA

(2.3)

B are left linearly independent an equation always implies that x = y = 0. Then any vector can be expressed uniquely in the form xA + yB and (2.3) merely requires P be expressed in this form. We see that there is exactly one that Q / and u solving (2.3) and this means that the intersection is exactly one point. Case 2. A and B are left linearly dependent elements a, ft in k Case

xA

+

1.

A

P.

and

yB =

which are not both zero such that aA + &B = 0. Then a j 0; pB = 0, hence /3 = 0, since we have assumed B 9* 0. = ?A where 7 = -/T * ^ 0. Similarly ft j* 0. Then We can then simplify the line Q + uB = Q + uyA =* Q + vA with y = uy-j if w ranges over &, so will v. We may assume, therefore, B = A and (2.3) becomes exist

otherwise

for,

1

(2.4)

If

(t

P =

+

a

= Q -

P.

of A, we can not possibly solve this the two lines do not intersect, they are parallel. If d A is a left multiple of A, then Q = P 8 A and the line

+

uA becomes P

and we if

t*)A

P is not a left multiple

Q

equation

Q Q

-

see that the

+

+

(u

two

$)A. If

u ranges over

A*,

so does

u

+

d

lines are equal and, therefore, parallel again.

We see now: two lines P + tA and Q + uB are parallel if and only B = yA. A pencil of parallel lines consists of the lines P + tA for fixed A if we let P range over all points. Our

analysis has also

containing two

Then Q

points.

we

get P,

shown that there can be at most one line now P ^ Q be two distinct

distinct points. Let

- P ^

so that

f or t

=

tA

a given line and

1

we

get

Q

P+

t(Q

as points on

-

P)

a

is

this line.

line.

For

t

Thus axiom

= 1

is

is

a

verified. If

P +

parallel line

is

and contains is axiom

uniqueness. This

The

points

the line

A +

A = t(B

(0, 0),

A) =

Q

Q

(for

a given point, then

u =

0).

Our

Q

+

uA

discussion shows the

2.

B = (/,

(1, 0),

C =

0) contains

(0, 1)

are not collinear:

A and B but not C. Axiom 3

holds.

Consider next the following

map

a: select

a

e

k and any vector C.

GEOMETRIC ALGEBRA

68

Put '

aX =

(2.5)

+

ctX

C.

a = 0, then every point is mapped onto C and a is a dilatation. Let a 7* 0. The image of the line P + tA is a(P + tA) + C = aP + C + atA = (aP + C) + uA since u = at will range over fc If

if J

does.

P+

+

+

uA which is tA is the line (aP C) a tA. Thus a is a dilatation and non-degenerate one, goes into a line.

The image

of the line

parallel to P +

since a line

The

fixed points of

X

=

a-

must

ctX

+

satisfy

C,

(1

- a)X =

C.

We

have two cases: a 5* 1; then (1 a)~ C is the only fixed point. a = 1; if C T* 0, there is no fixed point, if C = 0, then l

1)

2)

the identity. We see that the translations

X

X +

among our maps are by T C

C. Let us denote this translation


of the

=

1,

form

Obviously = 4a and T Q - P (P) = P which shows verifies axiom Q P) (Q that the translations r c are all translations of our geometry. To verify axiom 4b P let P, Q, B be three collinear points, P 7* Q, .

+

7* R. Since J? lies on the line P + t(Q P) which contains and Q, we can determine a in such a way that

P

This a

will

be

?*

P + a(Q - P) = /?. since R j& P. With this a we v(X) = aX + P - aP.

P

form the dilatation

Clearly
geometry.

Next we determine all trace-preserving homomorphisms of the translation group T. By Theorem 2.10 we may proceed as follows. C select as the point the origin (0, 0). If v(X) = has (0, 0) as fixed point then (0, 0) = C. The dilatations with

aX +

P

We

P

as fixed point are of the form
= aXj

a

^

0.

CHAPTER

69

II 1

The a

is

all

and different dilatations homomorphisms j* l l give different homomorphisms. We have
or

uniquely determined by


The maps

r

ora"

>

are then

trace-preserving

will

the translation r c

we

get as image

=


=

We

l


X)

aCcT'X

a

see that for a given

+

=

cKcT'X

C)

=

the

7*

homomorphism. To

+

C)

X + a<7 =

map r c > we have

r aC

.

r a(7 is the desired

add the 0-map which maps each r c into 1 = T O This also can be written in the form * r aC rc namely for a = 0. r aC for a t k by a and let A; be the Denote now the map r c field of all trace-preserving homomorphisms. We have just seen that a > a is a one-to-one correspondence between the field k and the field k. We contend that it is an isomorphism and have to this effect = a + J5 and that aft = aft. ft merely to show that a trace-preserving

it

to

.

j

+

The map a sends

r c into r aC

,

ft

sends

into r ftc

it

+

and a

ft

(by

definition of addition in k) into T aC 'TftC

But r aC (X

a

+

j8

+

ftO

=

X+

sends, therefore r c into r (a+/5 )c and a ft sends r c intone and aft sends

The map

under a which

is

Ta$c

.

aft

+

+ aC

ftC

ft

does the same.

into the image of r0c does the same. The fields fc and k are it

> a. indeed isomorphic under the map a Now we introduce coordinates based on the

field k.

We

select our

for the coordinates in k, select TI = r c origin (0, 0) also as origin = (0, 1). Let (f, ?;) be a given point P. T 2 = T D where C = (1, 0) and ,

D

Then TI

=

r f (7T, D (0, 0)

The

TC

=

=

r f (7

and

ric(riD)

=

rj 17!)

=

TO

=

+^=

(f,

translation rfrf will send the origin into P. rj) to the point P.

was to assign the coordinates (, that

Tr,

P =

receives merely the coordinates

(, 77) coordinates on K.

D r;)

;

=

The

P. prescription

We see,

(|, ?/) if

therefore,

we base

the

GEOMETRIC ALGEBRA

70

This means that the coordinates are "the same" apart from our isomorphism and shows that on the whole we have recovered our original field k. Notice that the isomorphism k -> k is canonical.

theorem

6. Desargues'

Assume now of our geometry only that the first three axioms hold. The following theorem may or may not hold in our geometry (see Figure

1).

THEOREM

2.15.

Let

li

,

12

,

J3

be distinct lines

a point P. Let Q, Q' be points on on 1 2 and S, S' points on 1 3 which are distinct from P We assume

parallel or meet in

Q and contend

+ R\\Q' + R'

and

Q

+

+

S'.

S

\\

Q'

which are

either

R, R points if our lines meet. 1

li

+

,

S'

that then

R + S This theorem

is

||

R'

called Desargues' theorem.

o

4

z

Figure

REMARK. is

There are certain

obviously true:

1

trivial cases in

which

this

theorem

CHAPTER

71

II

Q = Q', then Q + R = Q' + R' and since 12 and Q + R common but not Q it follows that R = R' and S = S'. Thus we may assume, if we wish, that the points are distinct. 1)

have

If

R

in

2) If Q, R,

S

statement is If the lines

Zi

are collinear, then Q', true again. ,

12

are parallel

13

,

we

shall call our

they meet in a point P, then we shall

If

S' are collinear

72',

call it

theorem

D

a.

DP.

D

THEOREM

and the

P

2.16. Axiom 4a implies a and axiom 4b implies DP. Proof: Let a be the translation which sends Q into Q' (if we wish

to prove

Da

or the dilatation with

)

into Q' (if we wish to prove DP). the traces of 0. Since

Q we

f

as fixed point which sends lines


=

Q'

similarly
=

S'.

+R

= R and

obtain
P

The


||

+

li

+

,

12

,

J3

will

Q

be among


Since

R + S our theorem

THEOREM

||

R

+


=

R'

+

S'

is true.

Da

2.17.

implies axiom 4a and

DP

axiom 4b P.

Proof: line contains only two points. Then there are two a given pencil of parallel lines, hence only four points A, B, each one C, B, A C, D in our geometry. There are six lines A is it Therefore the which two containing only expressed. points by 1)

Suppose each

lines in

+

+

-

-

geometry must have a unique structure. On the other hand there field k with only two elements 0, 1 and the rule 1 + 1=0. The geometry based on this field has only two points on a line and is therefore the four point geometry. This proves that all axioms hold in this four point geometry, especially axioms 4a and 4b. We may, therefore, assume that each line contains at least three points and it follows that to two lines one still can find a point not on these lines. this is

a

2)

Assume

Da

to associate a

not on the Let I Q

.

pair Q, Q' of distinct points we are going which will be defined only for points

With any Q>Q

map T line Q + Q

'

R

r .

+ Q' and contain R; then j* Q + Q'. Let m Q + R and m contain Q'; then m 7* Q + R. ||

||

/

I

GEOMETRIC ALGEBRA

72

+ Q'>ffQ + #we have m. Let R' be the point of interm and (Figure 2). R' lies on hence not on Q + Q', it lies on m, hence not on Q + R. C

Since

I jf

section of

I

I,

This shows R'

t*

Q' and R'

7*

R. Q+Q

S

S

1

1

Figure 2

The ||

Q

construction

+

Q' and

image R' of

R

+ R

Q and

T

+

R described in simpler terms: R R'. These statements describe the Q'

may be \\

+

Q Q '

'.

R R

we may now construct the map T will obviously send Q into Q'. Let now S be a point not on R Q>Q/ We have: R or Q + Q' and S' the image of S under T With

r

'

this pair R, R'

It

'.

+ +

R' R' are S' and these three distinct. Furthermore $ lines + + Q' Q S Q' + S'. Since we assume D a we Q + R Q' + #' and Q R' S + S' S R' S'. But the statements R can imply R S R + S' mean now that S' is also the image of S under and R R R R R Q 9 and r T \ The two maps r agree wherever both are defined. Since we assume that a point S outside Q + Q' and R + R' does St8 and know of the three maps T Q Q ', exist we can also form r R Rf S 8 T T that they agree wherever two or all three of the maps The desired map r is now the combination of the three defined. are for any point T the image shall be the image under one of the maps: three maps (for whichever it is defined). This map r has the property that r(Q) = Q' as we have remarked already when we introduced R Rf T If we can show that r is a dilatation, then it is clear that r will be a translation since all traces have turned out to be parallel and the proof will be complete (the case Q = Q' is trivial identity .

II

II

+

||

+

+

\\

\\

+

+

\\

f

\\

'

'

'

'

'

'

'

'

'

'

,

'

.

is

the map).

Let U,

S

+

V be two distinct points. One of the lines Q + Q', R + R', U and V and we may assume that Q + Q

S' will not contain

1

CHAPTER is

this line so that

U+ V Q +

Q',

||

we may use

T

73

II

Q Q' '

for

computing images.

If

U + V contains also U' and V. We may, U + V Jf Q + Q'. Then U and V are in the

then

assume that same position as our previous therefore,

72 and S for which we have proved R' S'. This finishes the proof. 3) If we assume DP, we can prove 4b P by a method completely analoguous to the previous case using the lines through P instead of lines parallel to Q Q'. We can leave the details to the reader. Because of Theorems 2.16 and 2.17 our geometry is frequently called Desarguian geometry, our plane a Desarguian plane. Desargues' theorem is the other geometric interpretation of the axioms 4a and 4b which we have promised. A third geometric interpretation shall be discussed later.

that

72

+

S

+

||

+

7.

9 Pappus theorem and the commutative law

One is is

of the simple but fascinating results in foundations of geometry the fact that one can find a simple geometric configuration which equivalent with the commutative law for multiplication of our

field k.

Select

an arbitrary point P. An element a ^ of k can be obtained 2.10) by one and only one dilatation o- with P as fixed the form

(Theorem point in

T* j

_

l

r
then T** aft

=

= (V)

l

(T a

a

rcr~

=

.

l

a

(T

Ta~ a~

l

= 00

=

r(a

l
)~

cr a is oon one hand and r uniquely a/5 ro-aj on the other. Since and this means that the determined by a. we have o- a/s =
,

s^

GEOMETRIC ALGEBRA

74

multiplicative group of non-zero elements of k is isomorphic to the group of dilatations which have P as fixed point, fc will be commutative

and only

if

if

this

group of dilatations

is

commutative.

and m through P and let Q be any point 7* P on I. If <TI is a dilatation which has P as fixed point, then / is a o-j-trace, 0iQ = Q will be 9* P and may be any point on I by axiom 4b P. We describe, therefore,
now two

distinct lines

I

f

l

f

+ R' R + Q'\\ Q

To have
<TiQ or

||

(T 1

<7i(r 2

+ +

*iQ
a 2 = o^! it is Q = T. Since

+R and we see that Q + # Q

*Jt'

=

Q'

*2Q'

=

R'

Q

f

+

S,

necessary and sufficient to have

||

+ wJl =

Q

Q

+

lies

on

(7^2/e

I

it is

= cr^Q

determined by

+

S

+

T S is the condition for commutativity. can now forget <TI and a 2 and look at the configuration. T on I and S on m are determined by ||

We

+ R'\\Q'+S and Q' + R R' + T and now we must have Q + R T + S. Six lines Q + R', R' + T, 72 + Q occur in this configuration which T + S, S + Q', Q' + m of lines as a be viewed may hexagon "inscribed" in the pair with our six points as "vertices". Q + R' and Q' + S, Q' + R and R' + Tj R + Q and T + S are pairs of opposite sides of the hexagon. Q

\\

\\

72,

Z,

The is

configuration states that the third.

This configuration

is

if

known

two

of these pairs are parallel, so

as Pappus' theorem

and we now

may state THEOREM

2.18.

The

field

k

is

commutative if and only

if

Pappus

9

theorem holds.

EXERCISE. I

and

m

Show

that Pappus' theorem holds

if

the two lines

are parallel and that for this case the commutativity of k

CHAPTER is

not needed.

Do

75

II

not use a computation with coordinates.

Make

use of translations.

a geometry contains only a

If

T

group

is

clearly finite

and so

finite

will

Theorem

familiarized himself with

number

be the

of points, then the

field k. If

1.14 of

the reader has

Chapter

I,

8 he will

see the following geometric application.

THEOREM

2.19.

In a Desarguian plane with only a Pappus holds.

finite

number

of points the theorem of

No

purely geometric proof 8.

is

known

for

Theorem

2.19.

Ordered geometry

In some geometries, for instance in the case of the euclidean plane, the points of each line appear ordered. This ordering can, however, not be distinguished from the reversed ordering. An invariant way to describe such a situation is by means of a ternary

P

"between" Q and R. Hilbert has axiomaand the reader may consult his book "Grundlagen der Geometric". We shall just follow the more naive way, considering a linear ordering and the reversed ordering as

relation: the point

lies

tized this ternary relation

equivalent.

Theorem

The plane.

We

presuppose a knowledge of Chapter

I,

9,

at least

1.16.

ordering must, however, be related to the geometry of the following definition describes this relation.

The

A

DEFINITION

2.8. plane is said to be ordered if the set of points on each line is linearly ordered; 2) parallel projection of the points of one line onto the points of another line either preserves or reverses the ordering. 1)

The main 9,

result will be the following

theorem

(see

Chapter

I,

Definition 1.15).

THEOREM a weak

2.20.

An ordering of a plane geometry induces canonically

ordering of the field k

an ordering

and a weak ordering

of k induces canonically

of the geometry.

Proof: 1)

Assume the geometry

is

ordered.

We

order the

field

k by the

following procedure: Select a point P and a translation T ** 1; a then, as a ranges over fc, the point r (P) will range over all points of the r-trace / of P. We have, therefore, a one-to-one correspondence

GEOMETRIC ALGEBRA

76

a

between the elements a of fc and the points r (P) of of k shall be the one induced by the ordering of I.

We

must

first

the selected

P

show that

of all

and

I.

The

this ordering does not

ordering

depend on

r.

Let us replace r by a translation TI 5^ 1 whose direction is different from that of r and let li be the retrace of P. We use again Theorem 2.10 stating that we can find for each non-zero element a t k a unique dilatation a a which has P as fixed point and which satisfies TO =
r-(P)

=


and

r(P)

a

r(P) and r (P) are different from

P and

are also different from r(P)

+

r,(P)

||

(r a

.

lie

on

Hence To(P)

li

and .

l


T Q (r^ (P)

=

= cr^P).

r^P)

P

=

lie

on

J,

n(P) and r"(P)

We have

+ ^(P) = ra (P) + a r (P) + T(P) are all

r(P)

r
This means that the lines that parallel a T*(P) is obtained from r (P) by parallel projection. This is, of = since this gives the point P which is the course, also true if a intersection of I and h By our assumption the ordering of k induced by P and r is either the same or the reverse of the one induced by .

P

and

TI

.

and T { have the same direction, replace P, r first by some T where r 2 has a different direction and then P, T 2 by P, TI. 2 P, This shows the independence from T. Let us now replace P by a point Q and assume first that Q does not lie on I. Call li the T-trace of Q; the lines / and Z, are parallel but distinct. We have If r

P+Q

||

T*(P) a

+

a

T (Q)

which shows again that the points T (Q) are obtained from the points a T (P) by parallel projection. The induced ordering is the same. If Q lies on I, replace P first by a point Qi not on I and then Q! by Q. Our ordering of k does not depend on P and T in the weak sense that it may possibly be reversed. Now we must show that k is weakly ordered: Let 8 be a fixed element of k and order k first by P and T, then by T*(P) and T. The two orderings will either be the same or reversed. In the first case

CHAPTER

77

II

a

the a will be ordered as the points r (P), in the second case as the a a+ - x 8 points T (r\P)) = r '(P). This shows that the map x

+

either preserves or reverses the ordering. Order also, if d 5* 0, k first by and T and then by and r*. In the first case, the a are (x e

/c)

P

P

a

ordered as the points r (P), in the second case as the points (/) "(P) = r a *(P). This shows that the map x * xd either preserves or reverses the ordering, k

indeed weakly ordered.

is

Assume now that k is weakly ordered. Let I be a given line. Select a point P on I and a translation r ^ \ which has Z as trace. a r (P) will range over the points of I if a ranges over k. Order the points on I as the corresponding a are ordered. We have to show 2)

again that the ordering does not depend on the choices of P and Another point Q of I has the form Q = /(P) for some fixed d t a+5

T.

k.

d is the same or the r (Q) = r (P). The ordering of the a reverse of the ordering of the a and this shows the independence on a

Then

+

the choice of Q. Let T V be another translation 7* trace I

r

a5

(P).

Then n = r with some d 7* 0. The a and the ad are ordered 5

with the same

1

We find r(P) = either in the

a

(r')

same or

(P)

=

in the

reversed way. This shows the independence of our prescription. Finally we must show that parallel projection either preserves be distinct or reverses the ordering of the points on lines. Let I and

m

lines

and suppose we are given a

points of a)

I

I

and

onto the points of m.

m

lines. Select

are parallel.

P

on

I

We have

Then we may

but select

Q on

mapping the two cases: the same r for both

parallel projection

to distinguish select

m to be the image of P under the

given parallel projection. The points of I are ordered according to a a r (P), those of according to r (Q). But

m

P+Q

r"(P)

+

r-(0) a

a

which means that r (Q)

II

is

the image of r (P) under our parallel

projection.

b) I and m intersect in P. Let r be any translation 7* 1 with trace and Q the image of P under T. Q will be a point ^ P on the line I. Under our parallel projection it will have an image Q' on m which is 9* P. There is a translation TI which carries P into Q'. Its trace I

m

P

be m. Order the points of by r"(P). As in 1) we have

through

r"(P)

will

=

er a

r(P)

=


and


a

I

by

=

r (P), the points of

ir.r.CP)

=


(Q')

GEOMETRIC ALGEBRA

78 if

a

Then

T* 0.

Q

+

Q'\\ *a(Q)

which shows that r"(P)

+

*.(Q')

= T(P) a

the image of T (P) by our parallel prothis is true if a = 0. We see that the ordering of jection; trivially to the corresponds ordering of I under our parallel projection. that the Notice 3) way to go from a weakly ordered field to an is

m

ordered geometry and the reverse procedure are consistent with

each other.

The geometry with only four points is a freak case. Each line contains only two points, it is ordered trivially since only one ordering and its reverse are possible. This geometry corresponds to the field with two elements which

Chapter

I,

9,

is

Theorem

a freak case of a weakly ordered 1.16 implies

field.

now

THEOREM

2.21. Aside from the plane with only 4 points and the with only two elements the following is true: Any ordering of a Desarguian plane induces canonically an ordering of its field k and conversely each ordering of k induces an ordering of the plane.

field

We

shall leave aside the freak case.

DEFINITION 2.9. An ordering of a Desarguian plane archimedean if it has the following property:

is

called

same direction and if between r^P) and r 2 (P), then there exists an such that r 2 (P) lies between P and r?(P). integer n > Let us translate what this means for the field. We can write r 2 = r" The three points r?(P) = P, r^P) and r 2 (P) = r"(P) are ordered as are 0, 1, a. The assumption is that does not lie between 1 and a so that a > 0. We wish to find an integer n such that r"(P) lies between r?(P) and r?(P). This means < a < n. By definition If TI 5^

a point

P

1,

r 2 7* 1 are translations with the

does not

lie

.

1.16 of

1.18 of

Chapter I, 9 this means that Chapter I, 9 we see

fc

is

archimedean. Using Theorem

THEOREM 2.22. The necessary and sufficient condition for an ordered geometry to come from a field k which is isomorphic to a subfield of the field of real numbers (in its natural ordering) is the archimedean postulate. Notice, therefore, that in

of

Pappus

holds

an archimedean plane

the theorem

the field is necessarily commutative.

Chapter I, 9, Theorem 1.17 shows that one can not in general prove Pappus' theorem in an ordered geometry.

CHAPTER

THEOREM

79

II

There do exist ordered Desarguian planes in which Pappus does not hold. These geometries are, of course,

2.23.

the theorem of

by necessity non-archimedean.

EXERCISE. A consequence of the ordering axioms is the fact ax (a 7* 0) either preserves or reverses the ordering. that a map x Give a geometric interpretation. 9.

Let

II

Harmonic points

We

be a Desarguian plane.

figuration (Figure 4)

shall describe a certain con-

:

A

c

B

D

Figure 4

B, C on a line I select three parallel which go respectively through A, B, C and are not C. The arbitrary elements are, parallel to I. Select on 1 3 a point P 12 1 3 but not I therefore, the pencil TT of parallel lines containing ^ and the point P 9* C on 1 B The line P + A is not 12 otherwise it would be the line Z 3 and A does not lie on 13 The lines P + A and 12 intersect in a point R which is not on otherwise it would have to be the point B, and B does not lie on P + A since P is not on I. Similarly P + B will meet the line Z in a point S which is not on /. The lines / and R + S are certainly distinct. If they are not parallel, they will meet in a

To three given distinct points A,

lines

li

,

12

,

I*

~

,

,

.

;

||

.

I',

t

point D;

D =

if

they are

The

contention

is

and on the point P. of

D

parallel,

we

shall indicate this fact

by putting

oo.

since

we

shall

that

D

will

not depend on the selected pencil

IT

We shall prove this by computing the coordinates need this computation for a later purpose.

GEOMETRIC ALGEBRA

80

The and

/

TI

on origin of our coordinate system shall be a given point a given translation 7* 1 with I as trace. This alone determines

X on

already the coordinates of any point

=

I

:

X

=

(

,

0)

where

is

We

X.

have, therefore, the freedom to select T 2 as we please (without changing the coordinates of I) provided T2 9* 1, has not the trace I. We select r a as the translation which

determined by r\(O)

C into

the point P. be the coordinates of A, B, C. The map r\ and r^r\ carries into P. The point P has, therefore,

carries the point

Let

(a, 0), (6, 0), (c, 0)

into

carries

(7

the coordinates

(c,

1).

The elements

are distinct,

a, 6, c

li

point on ^ will have coordinates (a, = rJ(A); a point on Z 2 will have coordinates (6, r5r!(0) A is equation of the line P

are traces of r2

Any

.

,

12

,

Is

since

77)

77').

The

+

=

y since

we

it is satisfied

by

(c,

1)

and

=

(b

a)(c

(a, 0).

-

1

a)'

R =

(&,

17)

lies

on

this line

and

find r?

Thus we have

S =

=

T/

6)(c

= xm

e since

=

72

-

(a, (a

Let x

-

(x

+

1

a)"

a)(c 1

a)"

a)(c

(6, (6

-

.

and we

)

find,

by symmetry,

1

fr)" ).

n be the equation for R + S (it is not of the form not ii). The points JK and S lie on it and we

R + S is

||

infer: /

(&

2 6)

-

a)(c

-

1

)~ l

(a If

b)~

b)(c

m=

We

stop for a

(c

+n + n.

>

a appears on both sides and we obtain

we subtract, the factor b

(2.7)

= bm = am

-

1

a)'

+

(c

-

1

6)-

moment and ask whether

m

.

can be

0.

This would

imply (c

or 2c

=

a

+

=

%(a

b.

-

a)

=

-(c

If the characteristic of

k

6)

is 5^

2 this happens

if

and

+ 6), C is the "midpoint" of A and B. If the charac= a + &ora = 6 which is against our assumpteristic is two, then = and only the characteristic of k is j& 2 tion; therefore m and if c = (a + 6). For m = the equation of R + S is y = n; only

if c

if

if

if

CHAPTER

81

II

D=

+

R S are parallel, hence does not depend on TT and P. 7* 0, D = (d, 0). Since D Suppose that we obtain

it

means that I and

where

D =

m

=

d

we multiply

If

(a

m(c

-

first

6)(a

-

1

nm"

we

case

-

on y

= xm

+

n

1 .

- br m~ = l

a

-

d.

of the left side:

1

6)"

=

= = = Hence, from

lies

get l

6)(c

compute the inverse

=

-

- &)-% - b)(a - &)- a)-'(c - 6)(a - 6)' + (a - b)~ (c - a)-'((c - a) + (a - 6))(a - 6)" + (c - 6)-' + (c - a)' + (a - 6)(a - b)" + (c - a)" 2(a ((c

+

1

a)'

(e

1

l

1

1

(a

-

1

ft)'

1

1

1

.

(2.8)

a

and

m""

by

(2.6)

(2.8)

Let us

Thus the

.

oo

:

-

d

=

(2(a

-

1

6)"

+

(c

-

1

1

a)" )'

finally

d

(2.9)

=

a

-

(2(a

-

1

6)"

+

(c

-

1

a)" )"

1 .

This result proves, of course, more than we have contended; notice that the expression for d uses only addition, subtraction and inverting. Suppose that the points of our geometry are the ordered pairs (f 77) from a given field k. We have written the equation of a straight ,

c = with the coefficients on the right. yb can be retraced; it originates from the fact that all our mappings were written on the left of the object we wanted to map. Suppose that we had done it the other way around. Then we would also have obtained a field k connected with the geometry. This field would now not be the the same as the field fc, it would be an anti-isomorphic replica of fc, the multiplication reversed. If, on the other hand, we start with a given field and call the ordered pairs points, we may introduce two geometries which in general line in the

form xa

The reason

for this

+

+

82

GEOMETRIC ALGEBRA

will

xa

not be the same. In one of them lines are given by equations c = yb (let us call them "right lines"), in the other by

+

+

equations ax the o>axis: y

+

+c=

6/y

=

most

0,

0.

Only a few

lines will coincide for instance

of the others will not (unless the field

is

commutative). If we are given the three points (a, 0), we can construct D in the "left" geometry

over our fixed

field k. It is

with the same value is

on the x-axis

0)

(6, 0), (c,

or in the "right" geometry

we would end up d since multiplication of two elements

pretty obvious that

(2.9) for

not involved in the formula.

shall call A, 5, C, D harmonic points and D the fourth harmonic point to A, B, C. Then D is the same in the left and in the right geometry. This is the first reason why we have written d in the apparently clumsy form (2.9). Is D distinct from A, B, C? D = A would mean that R + S contains A. Since S 9* A, R + S = li which is false. By symmetry D ^ B. How about D = C? If the characteristic of k is 2, then (2.9) shows

We

=

d

((c

a)" )"

and we have indeed D = C. Let k have a characteristic get from (2.9)

c(c

2(c

-

a 1

a)' 1

a)'

a

c

which

1

1

a

=

5^ 2. If

a

(c

=

a)

we had d = 1

1

-

= -2(a ft)' = a +6,

then we would

c,

= -(2(a- 6)- + (c - a)' = -2(a - 6)- - (c - a)'

c

1

1

)'

,

1

,

1 ,

c

=

6,

is false.

THEOREM

2.24.

harmonic point

D

// the characteristic of k is 2, then the fourth coincides with C and this configuration characterizes

the fact that k has characteristic 2. // k does not have characteristic 2 (for instance in an ordered geometry with more than four points) the will be distinct. four harmonic points A, B, C, Assume now that the characteristic of A; is 5^ 2.

D

The

construction

not change

if

A

itself

and

B

exhibits a

symmetry in A and B. D does But there are more sym-

are interchanged.

CHAPTER

83

II

can be changed into an equivalent form by transposing the term a and taking an inverse: metries

(2.9)

(d

which

is

-

= -2(a -

1

a)'

1

b)'

-

(c

-

l

a)~

equivalent with

-

(d

(2.10)

Interchanging C, Consider now

x~

l

+

1

a)'

D

+

(a

-

=

1

b)-

(a

-

1

c)'

+

(b

-

1

a)'

.

the condition remains the same.

l

y~

= x~\x

+

= y'\x

l

y)y~

+

y)x~\

Using this identity on both sides of (2.10) in two different ways 1 1 1 and taking care that the factors (b a)" and (a 6)" = (fe a)" are factored out on both sides of the equation either to the left or to the right, we obtain after a cancellation two more equivalent forms: (2.11)

(d

(2.12)

(d

-

6)(d

-

l

a)~ (d

1

a)"

-

= -(b -

c)(a

= -(a -

cT*(b

6)

If we interchange A and C and we obtain

(6

-

just shifting terms (d

which

it

-

-

1

c)"

1

c)'

-

same time

= -(d -

a)(c

-

,

c).

B and D in

(2.11)

1

a)"

;

becomes

d)~\b

-

d)

=

-(c

-

a)-\b

-

c)

only in sign from (2.12) and is, therefore, correct. that one should talk of two harmonic pairs A, B and C, have established all the symmetries. We have, of course,

differs

We see since

d)(b

at the

-

we

assumed

D

D

7*

oo 9

a restriction, which will be easily removed in the

next paragraphs. If our geometry

is ordered, the two harmonic pairs A, B and C, D each Assume this were not the case. Placing the other. separate axis coordinate appropriately and making use of the symmetries we could assume that a < b < c < d. But then the left side of (2.11)

would be positive and the right

side

would be negative which

is

a

contradiction.

Assume still that the characteristic of k is ^ 2. The notion of two harmonic pairs has introduced some structure in as poor a

GEOMETRIC ALGEBRA

84

geometry as that of a one dimensional line of a plane. It is now very natural to ask how "rigid" this structure is, in other words, what symmetries it has. Let a be a one-to-one map of the line I onto itself which carries harmonic points (finite points) into harmonic points. First of all 1 o-" will have the same property. To show this, let A', B', C', D' be 1 harmonic points and A, J5, C, D their images under cr" If the fourth harmonic point D! of A, B, C is finite, then A', B', C", i) are harmonic and, therefore,
1

+

<

.

whenever a

5^ 6 (considering

the triplet

a, b, (a

the equality is trivial. Make at first the special assumption that

For 6

=

0, (2.13)


+

6)/2). If

=

and

a

o-(l)

= =

6

1.

becomes

2

This can be used on (2.13): (2.14)


which means that a

we

get a(

Let a

1)

=

+

-1

is

(c

=

6)

*(a)

+

(K&)

an isomorphism under addition. Since


=

1

1.

6=

+1 and harmonic point

1,

for the fourth

Since

=

+

+

=

1

I)"

c 7* 0, 1,

(-(c

+

1)

+

1.

l)(c

From

+

1

I)-

we

(2.9)

=

-c(c

obtain

+

1

I)'

we have d

= -1

In other words 1


+

(c

+

*

l)
1,

1, c, c""

= -1

+ 1 + c' = 1

c"

1 .

are harmonic. Their images

are again harmonic showing

1,

1,

CHAPTER

-

a(f')

(2.15)

85

II 1

We))'

,

1 so that only c j* is trivially true for c = needed. Conversely, if a map satisfies (2.14) and (2.15), then one glance at formula (2.9) tells us that if d is the fourth harmonic

an equation which is

then a(d) will be the fourth harmonic point of o-(a), since only sums, differences and inverses arc involved in
point of

a, 6, c,

such maps. We turn

now to Chapter I, 8, Theorem 1.15 and see that a is an automorphism or an antiautomorphism of k. or o-(l) = 1 may be To settle the general case where
THEOREM

2.25.

The one-to-one onto maps & of a (finite) pairs are of the form

line onto itself

which preserve harmonic

=


^

where a

morphism

and where r

ax

is either

r

+

6

an automorphism

or

an antiauto-

of k.

REMARK.

only automorphism of k (k must be would be inner

If identity is the

commutative in such a

case, since otherwise there

automorphisms), for instance, if k is the field of real numbers, then = ax b. This result was first obtained by von Staudt. a(x)

+

10. If

V is

over a

onto

The fundamental theorem of a

left

V

a vector space over a field k and if there exists an isomorphism

k and

left

vector space

which maps k then we can generalize the notion of a homomorphism in an

field

A;',

f

projective geometry

/*

obvious fashion.

DEFINITION

2.10.

A map

respect to the isomorphism M

X

:

if

V

>

V

is

called semi-linear with

GEOMETRIC ALGEBRA

86

and

+

\(X

1)

all

a

=

Y)

\(X)

+ X(F), 2)

\(aX)

=
for all

X,

YtV

e k.

The

description of all semi-linear maps follows the same pattern A2 A n is a basis of F, if as that of homomorphisms. If A B B2 # are arbitrary vectors in (thought of as images of l

{

,

,

,

,

the basis vectors

A

t

and

)

= 2*

-Y

if

,

V

,

x \^\

i

s

an arbitrary vector of

Fthen

Z o^

=

X(X)

t-i

a semi-linear

is

map

of

V

if

V

and one-to-one and onto

V

and every semi-linear map is of and only if the B are independent, if and only if the B are a basis of V.

into

this form. It is one-to-one

t

t

We can leave these rather trivial details to the reader. With each left vector space V over k we form a new

object, the

corresponding projective space V. Its elements are no longer the single vectors of F, they are the subspaces U of F. To each subspace U of F we assign a projective dimension: dim,, U = dim

171,

smaller than the ordinary dimension. take over the terminology "points' and "lines" for subspaces with projective dimension 0, respectively 1. Thus the lines of F

by

just

1

We

'

of F and the planes of F become the "lines" of F of ordinary dimension n gets (as element of F) 1. The 0-subspace of F should be the projective dimension n of F and has projective dimension as element" of the "empty thought

become the "points"

The whole space

F.

An

1.

incidence relation

introduced in F;

is

it

shall

merely be the

U C U between the subspaces of F. Intersection Ui C\ U and sum U + U can then be explained in terms of the incidence relation of F: Ui C\ U is the "largest" element conU + U is the "smallest" element which tained in Ui and U contains both U and U inclusion relation

2

l

2

2

l

2

2

2

l

DEFINITION

U C U

3)

l

2

if

implies

As an example linear


=

map Xf7,

X

:

F

then

.

A map

V

V

of the elements^ of a proa onto the elements of projective spaceF' is called = dim F', 2) a is one-to-one and onto, 1) dim F

2.11.

F

jective space a collineation

2

l

,

o-



:

C *U

2

.

of a collineation suppose that there exists a semi>

is

V

which

is

one-to-one and onto._If

obviously a collineation of

V

>

we

define

V and we say

CHAPTER that a

induced by

is

87

II

The main purpose

X.

prove that every collineation

is

of this paragraph induced by some semi-linear

to

is

map

F > 3._ Let V V be a collineation, U a subspace of V and dim V = n. We can find a sequence = U C U\ C U C C Un = V of = dim U is U such that one of the terms. where subspaces, j,

if

dim


'

>

:

2

f

C

C

C

By


.

Since

C7 t

5^

-

C7,-

+1

and

The dimen^ = n with dim i. But strictly This implies dim at/, = i and
.

V

puts an upper bound n on dim at/, consequently dim C7 = dim erf/. A collineation preserves the dimen.

sion.

Let [/i and U2 be subspaces of V and suppose that
;

C

dim(U

W

HW

+ dim W = dim + dim aW = dimCaC/! +
W) = dim

+

1

+

W H

f/j

o-C/,

2

hen co

On

the other hand

W are subspaces of

as well as

f/i

+ aW C

+

+ W and

C/!

C +

+

TF) or cri7 2 ^(C/! IF). Since
+

U

.

1

preserves the inclusion relation and since intersection can be explained in terms of the inclusion relation we obtain

as well as

and sum

U

o-"

also that


H

U = 2)

rUi

H *U

Suppose that we know of (they are U = L!

the

+

+

+

O-L!


lines

+

L2 -

-

of

+ +


2


F). r

and

+ U =

a(Ui

2)

at/!

+

(7f/ 2

.

only the effect on the "points" of If

U

is

any subspace

where each

and we

see that

L or

-

t

is

is

a

line,

F

of

then

completely

F

and

aU =

known

if

GEOMETRIC ALGEBRA

88

on the

its effect

lines of

+

(7/> 2

(T//3

.

We

can

V

is

known.

Whenever L

the following property:

on

Its effect

C

l

L

+

2

lines will

L3

now formulate and prove

have

C

then
the fundamental

theorem of projective geometry:

THEOREM

2.26.

theorem

(Fundamental

of projective geometry). n 3 over fields vectorjspaces of equal dimension k respectively fc', and the corresponding projective spaces. Let a be a one-to-one (onto) correspondence of the "points" of and the

Let

V and V

1

be

>

left

V

V

"points" of V which has

V

the following property:

Whenever

three distinct

C

+

L2 L3 "points" LH L 2 L 3 (they are lines of V) are collinear L l
\

,

C

of course be extended in at most one

+

way

to

,

.

a collineation but we contend r

more. There exists an isomorphism /* of k onto k and a semi-linear map X of V onto (with respect to p) such that the collineation which X induces on V agrees with on the points of V. If Xi is another semi1 linear map with respect to an isomorphism MI of k onto k which also

V

induces this collineation^ then \\(X) = \(&X) for some fixed a 7* = (axa' 1 )". For any a 7* of k and the isomorphism PI is given by x^ the map \(aX) will be semi-linear and induce the same collineation as X.

The isomorphism n morphisms of k.

determined by

is t therefore,


up

to

inner auto-

REMARK 1. The assumption that LI L 2 L 3 are distinct is L3 will imply trivially that vL^
,

C

+

irrele-

+


n = 2, then dimp V = 1. There is only one line, and a completely random one-to-one correspondence of the points of would be a collineation. The statement would V and the points of be false for any field with at least 5 elements; only for F 2 F 3 F 4

REMARK

If

2.

V

,

,

would

it still

hold.

Proof:

We

1)

L

7*

L

r

show by induction on

C

then
o^i then L

+L_ r

+

o-L r _i

o"L

^

2)

t

(A

is


r: if

+

+

0L

r

L .

spanned by a vector

7* 0) arid

B

e

L

r

.

C

LI This is

&L r

+ L + 2

+ L L = L If A e L + L2 + r

trivial if

A + B where

Then

r

,

.

l

C erLi + o-L + L C (A) + L we


by induction hypothesis and from

<7\A) ~p

We

+

2

r

get

.

shall use frequently the following reasoning.

Let

C

and

D

CHAPTER

89

II

be independent vectors and L (C) (D) but L ^ (D). Then L a a is spanned by vector a(7 ^ 0, which may be replaced bD, with l a~ bD = C dD. The element d is then uniquely deterby C

C

+

+

+

+

mined by L. Let

3)

L

A

be a basis of 7, L<

t

+L +

+

-

2

t

C

o-L

quently

V

line of

any

Ln

L

if

;

+


is

=

+

have the form

will

o-L n

We have V = L C 7 and conse-

(A<>.

V then

line of

any

+


=

(A,-), o-L,

Since our

.

which shows

o-L

map


is

+

onto,

o-L 2

+

=

^' so that the AJ span V. But dim F' = n implies that the A( are a basis of V. The line (A l A.) will be (A,) (A,-) A -} = (A( and distinct from (A -) if i > 2; hence tr(A l 6,-A<); since (A t A,-} ?* (A ) we will have 6, 5^ 0. Let us replace At by

+


+

+

+

C

+

t

+

t

x

the equivalent vector 6 A< t

=

.

Then

=


(A() for

i

>

1

and

A,) AJ) for i > 2.
+

iKAt

+ +

C

+

+

^

+

(A(

+ x'Ag) and, since (A! + xA

x' 7* y'.

The map

j

2)

+

(A!

given by x

> fc'

fc

#'

yA 2 )

if

x

9* y,

we have

therefore, one-to-one

is,

and 1' = 1 by 3). but only into, thus far. Clearly 0' = In a similar way we could find a map x > x" such that xA 3 ) = (Al x"Ai). We contend now that x' = x" for C^A! all x e fc. We can assume x 9* 0. #A 3 ) lies in (A 2 ) + (A 3 ) on one hand and in The line (a:A 2 #A 3 ) on the other. Its image is, therefore, xA 2 ) (A! (A! a of vector (A 2 ) spanned by (A 3) and also by a vector of The x'A 2 x"A$. only possibility for the image is ) (A( (A( A 3 ) whose image zA 3 ) = (A 2 But the line (z'A 2 2 a/'Ag). (xA = A 3 ). This shows is by the same reasoning (l'A 2 2 l"Aa) (A

+

+

+

+

+

+

+

+

+

r

(x

A2 -

We

- Ai) and x" = x' follows. (A 2 could have taken any vector A, with

=

x"A'*}

Instead of

A we 3

conclude that there ff

5)

Suppose a(A l

+

x'-iA'-i)

is

lies

in (Ai

+ A +

Its

image

is,

+ +

(A

+

is

but one

+

l

o; 2

A2

xA<)

+

proved. The

map

= -

(Al

+

x

+

>

x'

i

> >

3.

2,

x'Aft.

Xr^Ar^) = (Al

line (A!

i

such that, for

+

z2 A 2

+

+

x 2 Ai

+

+

xrA r)

+

^r-iA,.-.!) + (A r ) and is distinct from (A r ). spanned by a vector of the form Al + x 2 A 2 uA'r Our line is also in (A! + x r A r ) + (A 2 ) x'r-iA'r -i (A r _i) which implies that the image is in (A I + x'r A'r ) 2

2

therefore,

+

.

+ + +

90

GEOMETRIC ALGEBRA

(A 2)

We

+ (A'_i). Since the image must use A{ we get now u =

H;

x'r

.

know, therefore,

+xA +

+

=

+ Zs'Ai + + xiA$. + x A is in (A + 6) The image of the line (x A + + (A' It also lies in (A + x A + + x A + (A,), its image, + zA> + (AJ>. Clearly
-

2

2

xn A n) 2

2

2

2

-


n)

n

2)

-

n ).

v

n

n)

f

2

2

2

2

A given line of F' which has the form (A + yA

( 2 ) is not the image (A 2) + (A n ) as 6) has shown. Hence it must be the image of a line (Ai x2 A 2 xn A n) which implies x 2 = y * and shows that the map k fc' is onto.

7)

+ +

of a line in

+

(KA,

8)

(x

+

+

+

y)A 2

A

+

=

3>

y

+

(A[

+

(x

+A

y)'AJ

3').

But

+

(A,

+

(x

+A C

y)A 2

3)

(A,

+ xA + 2)


+

A 3)

A2

+

AJ).

so that

(A(

We

+

+

(x

y)'A'2

deduce easily (x ff

9)

(A

+

l

+

yY = x

+

xyA 2

C

Ai)

+

(A(

+

7

xA 3 ) =

+

+

x'A'2 )

f

(y

2/'.

(A(

+

(xyYA 2

+

x'AJ)

and (A

+

1

xyA 2

+

xA 3 }

C

(xyYA 2

+

x'AS)

(A,)

+

(yA 2

+A

3>

hence (A{

+

1

which implies /z of fc onto A;'.

The

line (x^A^

this is clear

(A!

+ +

(xy)

l

Xi

if

+

=

x'y'.

n

or Xi

+

+

x2 A 2

Our map x

+zA

^ =

=

C (A[) +

n) 1.


>

+

+

+ zA) as image;

has (x?A( If x v 9^0 our line

x^XnAn) and

its

Ai)

an isomorphism

is

image (A(

also given

+

x^xZAi

by

+

+ nAn)x^XnAn) is the same line as (x^A( + 10) Let \ be the semi-linear map of F onto (with respect to For any line L of F the isomorphism /x) which sends A onto AJ we have shown o-L = XL. This proves the first part of the theorem.

V

t

The

.

rest is

11)

very easy. Suppose \i is another semi-linear

map

of

F

onto F' which

CHAPTER

91

II

has the same effect on the lines of V. Then

V

which keeps every

V

X'%

a

is

map

of

V

onto

map every non-zero of V onto some non-zero multiple of X. Let vector and Y be Y will go of vectors V. The three vectors X, F, independent l = into \' \^(X) But Y). Y) respectively aX, 0F, y(X X^X^X line of

fixed. It will

X

X

X+

+

+

= aX + 0Y. A comparison with yX + yY shows a = 0. If F are dependent but ?^ and Z a third independent X and F take on the same factor, under X~% then both vector, as Z. It follows now that \~ \i(X) = aX with the same a for all X (the case that X = is trivial) and consequently X^-Y) = \(aX). and define the map X by \i(X) = X(ofX). For any 12) Let a 5^ e we have X^X) = X(oflX) = X^a^-aX) = (aflO"^^).

+

1

X" Xi(F)

-XT,

,

l

t

fc

ft

The map \i is, therefore, semi-linear and its isomorphism ^ is given ^ M X(A) which by x = (axa~*Y- K A 7*0, then X (A) = X(A) = = and X induce the same shows that X^A) collineation. X(A) so that \i The proof of our theorem is complete. t

For any semi-linear map ^> V" Suppose now V ^>

V

We

see that X 2 X!

Very interesting

is

shall

of

V

is

semi-linear again

and that X 2 Xi

=

X2X t

.

results are obtained in the case of collineations

of a protective space itself

us denote its isomorphism by X. a succession of semi-linear maps. Then

X, let

V

onto

itself.

The

semi-linear

maps

of

V

onto

form a group 8; the collineations of V form also a group which be denoted by PS. Each element X of S induces a collineation onto itself. The map S -A PS which associates with each X e S

the collineation j(\) induced by it (j(\) is merely the effect X has on the projective space V) is clearly a homomorphism of S onto PS;

an onto map by the fundamental theorem of projective geometry. kernel of j consists of those X e S which keep every line of V fixed and we have seen that there is one such map X for each a 7^ of k, hence for each a 1 7c*, namely the map \ a (X) = aX. We may then map fc* into S by a -^ \ a since \ a \ = \ aft this map i is an isomorphism of k* into S and the image of i will be the kernel of j. it is

The

;

It is

,

ft

customary to express such a situation by a diagram 1

->

fc*

A S -4 PiS ->

1

where two entirely trivial maps are added: every element of PS is mapped onto 1, and 1 > ft* maps 1 onto the unit element of A;*.

GEOMETRIC ALGEBRA

92

In this succession the image of each map is the kernel of the next map; if this situation occurs one calls the sequence of maps an exact sequence. Thus the statement that our sequence is exact all the previous ones, i.e., that i is an isomorphism, that image is the kernel of j and that j is onto. Denote now by A (k) the automorphisms of k. We can map S -A A(k), where p denotes the map X > X which associates with each semi-linear map X the automorphism X. The map p is a homomorphism as we have seen and it is an onto map since every auto-

contains

its

is possible in some semi-linear map. kernel GL (general linear group) of the map p consists of those elements of S for which the automorphism is identity. They

morphism

The

are merely our old homomorphisms of another exact sequence 1

->

GL

V onto

^ S -A A(k) ->

V.

We have,

therefore,

1

the mere injection map of GL into S. GL induces on V a special collineation called ve transformation of V. These transformations form a subproject!

where

inj indicates

Each element

of

PS denoted by PGL (projective general linear group), and GL onto PGL is just the restriction of j to GL. We shall map

group of the

of

(rather sloppily) call

it

j again.

The

kernel consists of those X a

which are homomorphisms. They must satisfy \ a (xX) = x\ a (X) or ax = xa for all x e k. This means that a comes from the center Z of k, hence from Z*, since a ? 0. We get the exact sequence

Denote by I(k) the inner automorphisms

of k.

An

element of

PS determines an automorphism of k only up to an inner one so that we get only an element of the factor group A(k)/I(k) associated with each element of PS. Let us denote by p' this onto map PS -^ A(k)/I(k). Its kernel consists of the collineations which are induced

by elements

of

S

belonging to an inner automorphism.

since one can change the automorphism by any inner one we can assume that the element of S belongs to the identity automorphism,

But

that

it is

in

GL. This shows that

exact sequence

may

PGL is the kernel

of p'

and another

be written down.

The situation will become less confusing if the reader studies the following diagram which incorporates all the statements we have

CHAPTER

93

II

made and explains in full detail the consequences of the fundamental theorem of projective geometry in the case of collineations of the

V

space

onto

itself:

1

I

PGL

->1

PS

->1

I*' -

1

I 1

The symbol

"inj" indicates a mere injection map, "can" the onto the factor group. The map p associates with 1 > axcT its kernel is clearly Z*. the inner automorphism x

canonical

a

t fc*

map

;

Every row and every column is an exact sequence. Every square of the diagram is "commutative"; this means for instance for the square whose left upper corner is S that the map j followed by = can p. p' gives the same result as p followed by can, i.e., p'j In later chapters we shall study subgroups F of GL. The effect which F has on the projective space V, therefore the image of F under j, will be denoted by PF. The kernel of this onto map F > PF consists of those \ a which belong to F. In all cases which we shall consider this kernel will be the center of F, and this will allow us to define PF as the factor group of F modulo its center. There is a geometric construction called projection (or perspectivity) which dominated the classical projective geometry and which will give

us

still

a better understanding of the fundamental theorem

of projective geometry.

V

are subspaces of a space ft Assume that both spaces V and N > dim V = dim = n > 2; notice that we allow

V

of dimension

n =

2.

T V:

Let

well as to

a =

v

be a subspace of

T = v

9

r,

ft

which

is

supplementary to

vnT =

vnT =

o.

V

as

94

GEOMETRIC ALGEBRA

W

Let ,W be a subspace of Q which contains T and intersect F = C7. Clearly 7 T7 TF but the reverse inclusion with F: is also true. Any vector of 12 = V T has the form A = B C =* 7? e with F and C e T. If A e W, then B A C lies in TF and F,

Wn

+

C

+

U + T and we obtain W = C7 + T. U of F and set W = C7 + T, then W H C TF, we can write A = J5 + C where C J5 e 7 and C t T. The vector C = A - B lies in F as well as in T and, since V C\ T = 0, we get C = 0, A = B t U. For subspaces C7 of F and spaces W which contain T the two consequently A If we start with any subspace TF Pi 7; if A t [7 F therefore in

e

C7;

equations

U = WC\V

TF=(7 + T

and

are, therefore, equivalent.

We start now with any U C F, form TF = ?7 + T and intersect TF with V: U = TF H F'. This gives a one-to-one correspondence between the subspaces 7 of F and the subspaces U' of V which is 7

f

completely described by the one equation

U+T = U C U

Since

l

C

implies U{

2

a collineation of V' with center !T.

is

V

U'

t^2

+

T.

this one-to-one correspondence

onto F' which one

calls projection of

V

onto

We allow also n = 2 and can, therefore, not make use of the fundamental theorem of projective geometry. We proceed directly. A2 A n be a basis of F and denote by (AJ) the image Let A T and this leads to a of the line (A.). Then (A,> + T = (A<> B t where B e T7 since A { $ T the element formula A = a -AJ l

,

,

,

+

+

-

t

ai is 7^ 0.

t

t

We may replace AJ by the equivalent vector a.AJ

simplifying the formula to

+

;

xn A n

=

+

XiA{

-

A = t

+

,

thereby

This leads to x^A + AJ + JB x n A'n + C with C t T. If we denote -

t

.

l

by X the homomorphism (with respect to the identity automorphism X tT and obtain of k) which sends A onto A( we can write \(X) t

,

(X(X)>

+

T =

<X>

+

T.

This formula shows that the linear map X induces on F our prois not an arbitrary homomorphism. The map X F e F Pi T = (X) T (in different interShould then (X)

jection.

:

X

V

V

+

+

CHAPTER

95

II

X

pretations) shows that (X) is kept fixed by the projection; X(X) and V C\ T = is, therefore, not only in T but also in implies

V

H V fixed.

X(X) = X, our map X keeps every vector of V Let us start conversely with a linear map X onto and which keeps every vector of V C\ TO the set of all vectors of the form X(X)

:

V

V

over V. It

V

is

obvious that

TQ

is

- X

e X(X) X(X) e F' we have X(X) e F showing that our vector X(X)

TQ

is 0.

v

r\

We X(X)

Indeed,

=

if

H

is

Denote by where X ranges

fixed.

X

its intersection

X+ V

=

with

7; since

*

7' and consequently, X(X) = = 0. Similarly we can prove

X

o.

have

!F

C

+ 7

7'

any vector

is

a subspace of Q; 7, then X(X) e

7' which

>

7 +

and, therefore,

7'

V+

C

7. If

in 7', then

X(X)

=

(X(X)

-

X)

+X

e

To

+

7.

C 2 o + 7, and, therefore, 7 + 7' = T + V. = 7 + 7'; since 7 H T we can 7' H 3T = Similarly V + TQ T = 7' TQ Let ^ be a subspace of fl write 7 + V = 7 T Then which is supplementary to 7 + 7'. Set T = T 7+

This shows

7'

.

.

t

7

To

T,

We may project from T

+

(X(X)}

T, and

this

.

= 7

T = Q =

7'

T.

-X

e T C T we get <X> + T = Since X(X) shows that our X will induce the projection

from T.

X

X e 7 n_7'

condition that X(X) = for all in the geometric meaning projective spaces

The

has no direct

7 and V where individual

vectors of 12 do not appear. Let us see if we can replace it by a condition which has a geometric meaning. The projection will leave every C\ line of fixed. Suppose now 7', i.e., every J^point" of

7

n

7

V

V

7

given which leaves every "point" of fixed. Does it come from a projection? Let us assume 7 C\ 7') > 2. Should n = 2, then this is only possible if 7 = 7'; dim(7 our or must be identity and is clearly induced by the identity map X, > 3, we may use the i.e., a comes from a projection trivially. If n fundamental theorem of projective geometry. The collineation
V


:

is

H

nV

V

X

V

GEOMETRIC ALGEBRA

96

V C\V

(whose dimension is > 2). We may now change 1 the map X by the factor of and the new map X will still induce a e V for all but satisfy \(X) = Denote by /z the autoof V we have ?* morphism of k to which X belongs. For an \(aX) = a"\(X) = tfX on one hand and X(aX) = aX on the other. This proves that /z is the identity automorphism, i.e., that X is an ordinary homomorphism of V onto V. Since it keeps every vector of V C\ fixed, our collineation is indeed a projection. We postpone for a while the discussion of the cases where dim(F C\ V) < 1. In classical synthetic projective geometry a collineation of two of the space

X

X

HV

.

HV

X

V

V

of Q is called a projectivity if it can be obtained subspaces V and by a sequence of projections. Each of these projections (possibly onto other subspaces) is induced_by a linear map and consequently also the projectivity of V onto V. Let us consider first the special

V

case of a projectivity of onto V. It is induced by a linear map, therefore, by an element of GL(V) and is consequently an element of PGL(V). The question arises whether every element of PGL(V)

H

N

a projectivity. Let be any hyperplane of V. Since > n we can find a subspace of 12 which intersects V in the hyperplane H. and We can find a linear map of V onto which is identity on carries a given vector A t V which is outside onto a given vector B of which is outside H. This map will come from a projection. We can then project back onto V and move B onto any given vector A' of V which is outside H. The combined map is a projectivity > V such that X is of V onto V which is induced by a map X V 1 = // A on and that We such X(A) identity can, for instance, achieve a "stretching" X(A) = aA of the vector A or we can also C where C is any vector of H. In the early achieve X(A) = A IV we are of part Chapter going to see that every element of GL(V) can be obtained by a succession of maps of this type (with varying hyperplanes H). If we anticipate this result then we can say that the projectivities of V onto V are merely the elements of PGL(V). in If V and are different spaces we may first project V onto is

V

V

H

H

V

:

.

+

V

V

V

onto V, hence some fashion and follow it by ajyojectivity of by an arbitrary element of PGL(V). It is clear nowjbhat any linear

map

of

V onto

V induces a projectivity of V onto V. This explains

the dominant role played

by the group PGL(V)

in the synthetic

approach. We return to the question which collineations of a space

V

onto

CHAPTER

97

II

V are projections from a suitable center. One has certainly to assume that every "point" of V C\ V remains fixed and we have

a space

>

dim(F C\ V)

2. In order to dispose to strengthen the assumption. shall assume that the given collineation cr is a projectivity, i.e., that

seen that this

enough

is

of the remaining cases

if

We

we have

obtained by a sequence of projections. This implies the simplifiis induced by a linear map X of V onto and, as we have seen, is equivalent with this statement.

it is

V

cation that a

V

V C\

If

=

T

the center

then no condition on X was necessary to construct is, therefore, settled in the

0,

of the projection. This case

affirmative.

_There remains

dim(F C\ V) =

the case where

when

1,

V

and

= (A) we C\ V. Since V intersect in a point. Let (A) = aA with a a linear map must have A (A) conversely, X of V onto V for which A (A) = a A (and a can have any non-zero then X a suitable induce a projectivity of V value in k Once a onto V which keeps (A) given, one has only a

= V


9* 0. If,

for

will

X)

fixed.

freedom

left in

the choice of

A

k

commutative.

replace \(X)

by \i(X)

=

map. But this implies that ft is now see whether we can achieve

for a suitable choice of

=

little

We may

X.

We

\i(A)

a-

is

\((3X) provided Xi is again a linear must taken from the center of k.

fta.

is

ft.

This leads to the equation

and we see that such a ft can only be found if a is in the center of k. Since a. might take on any value one has to assume that is

1

We gather our results in the language of projective

geometry.

THEOREM

Let

2.27.

&

a

be

leaves all points of 1)

cr

n

suitable center

2) //

then

(V

If dinip

from a

V

V

is

3) //

and

or

H V fixed.

a

We

collineation of

can

1,

T

a perspectivity)

(
then

V have an empty

cr

is

and

V

onto

a projection of

intersection

again a projection from a suitable

V

and V^jproper

V

V which

state:

V) > is

V

projective space,

subspaces of equal dimension and

V

onto

V

.

and

if

cr

is

a

projectivity,

center.

a point and if & is a projectivity, then the field k is commutative. If k is not com-

intersect in

always a projection if mutative, then this will not always be true. __ The most important special case of this theorem, when J2 is a distinct lines leads to thejfchird of our alternatives. plane and 7,

it is

V

A

projectivity of the line

V

onto the line

V which keeps the point

GEOMETRIC ALGEBRA

98

of intersection fixed is in general a projection only if the field is commutative. It depends on the configuration of Pappus. collineation of a space V onto itself a projectivity has a quite different answer. One must assume dim p V > 2 and now the problem reduces to PGL = PS; this is equivalent with A(fc) = I(k) by our diagram, every automorphism of k must be inner. In the classical case fc = R, the field of real numbers, both conditions are satisfied. The field R is commutative and identity is

The question whether every

is

the only automorphism of R.

Our

theorem in

last

this

paragraph concerns the characterisation

of the identity.

THEOREM

2.28.

V

Let

be a protective space over a commutative

+

I points a be a projectivity of V onto V which leaves n V) and assume no n of these points lie on a hyperplane. assert that a is the identity. This theorem is false if k is not commuta-

field k. Let

= dim

fixed (n

We tive.

Proof: Let (A ), (A^, (A n) be the n + 1 points. The vectors A n are a basis of V and if A = 2<-i a all fr, are X(A ) = )?-i faA-i that \(X) = f$X equal. From X(A ) = 0A we deduce for any vector since k is commutative. This was our contention. If k is not commutative select a, ft e fc such that aft j (la. Let X be the linear map which sends A into A, (i = 1, n). The vector A = )?-i ^- i g mapped onto A The collineation a which X induces leaves the points ,

t

,

,

,

t

t

t

,

X

-

t

t

-

t

,

.

The point (X) = (A! 0A 2 ) is moved into (aA l a~ ($aA 2 ) which is different from (X).

+

(A,) fixed.

(A!

+

11.

The

l

+ /taA = 2)

projectile plane

Consider an affine Desarguian plane and let k be its field. Let V a plane of V. be a three dimensional left vector space over fc and We shall construct a new Desarguian plane Tl w as follows: The "points" of n^ shall be those lines L of V which are not

W

contained in

W.

CHAPTER

The "lines" The incidence

of tt w shall be those planes

relation shall merely be (A 2 ^ 3 ) and V = (A, A 2 spanned by a unique vector ^! %A 2

W

Let

,

ry)

A

L

=

,

,

+

(

99

II

+

U

C A3

,

of

V

which are

^

W

.

U. >.

rjA 3

If

L

C

and we

W, then L

is

shall associate

as coordinates to L.

U will have an intersection (aA 2 0A 3 ) with W. U can dA 3 be spanned by <xA 2 #A 3 and by some vector ^Li yA 2 Since L ( TF, a "point" L on U is spanned by a unique vector of the form (Ai dA 3 ) yA 2 f}A 3 ). The coordinates of L t(aA 2

+

"line"

+

+

+

+

+

+

.

+

are, therefore, (ta

+ y,tp+

S)

= *(,$

+

(7, 5).

the equation of a line in parametric form and the only restricthat (a, /5) ~ (0, 0). We see that II ^ is coordinatized by the

This

is

tion

is

field

k and that

it

has, therefore, the

same structure as our

original

Desarguian plane. obtained from the projective plane V by deleting the "points" on W. thus have the following picture: If we take a projective plane

The plane the "line"

II

^

is

W and

all

We V and eliminate one line and all the points on it, we get a Desarguian plane. All the Desarguian planes which can be obtained from V by this process have the "same" type of affine geometry. If two lines of V meet on a point of the deleted line W, then they do not meet in n^ and are, therefore, parallel.

Each deleted point

Figure 5

is

characterized

by a

GEOMETRIC ALGEBRA

100 pencil

Uw

,of

and

From

parallel lines of

W

itself

Uw

We

.

the infinite line of

the affine configurations

call

II ^

we can

them

"infinite points" of

.

get project! ve configurations

and prove them by deleting appropriate

lines.

1

indicates a project! ve configuration. If Figure 5 the line X it becomes the affine Desargues: The lines l v lines

w m x

,

2

become

parallel so that the lines HI

,

n 2 are

we 12

,

delete

and the

also parallel;

they must, therefore, meet in a point on the line X. This configuration is known as the projective Desargues. It contains of course the affine Desargues as a special case if X is the "infinite" line. o

c

A

B

D

Figure 6

Figure 6 is another such configuration. Draw a line X through the point Q which does not go through A, B, C. If we delete the line X it becomes the construction of the fourth harmonic point D.

The

fact that the construction is independent in its affine form means that the point Q could be moved to any other point on X, hence to any point Q' in V such that the line through Q and Q'

does not go through A, B, or

C

(provided of course

it is

outside

I).

By two successive moves of this type we can bring it into any position. Thus the point D is always the same no matter how P and Q are chosen. This fact

is

known

as the theorem of the complete quadri-

lateral.

The

reader

Any two

may

find for himself the projective Pappus. meet in a unique point and through

distinct lines of

distinct points of

V

V

there goes exactly one

line.

Each

line of

V

contains at least three points. figure 5 is meant to be in a plane. For a it could be interpreted in space.

that

later

purpose

it is

drawn

in

such a

way

CHAPTER It is

101

II

now clear on what axioms one would

base projective geometry

in a plane:

Two Two

1)

distinct points determine a unique line

on which they

lie.

distinct lines intersect in exactly one point. 3) Each line contains at least 3 points and there are 3 non-collinear

2)

points.

The projective Desargues.

4)

V

is a projective plane of this kind, then one deletes a line and the points on it. One obtains an affine plane in which the affine Desargues holds. This affine plane can be coordinatized by a field fc.

If

all

With

this field

k one can construct a projective plane of the same is to be remarked, however, that the

structure as the given one. It field is

not canonically constructed, that

it

involves the choice of

A

canonical field associated with the plane would have to be explained as an equivalence class of fields (one for each

the deleted

line.

deleted line).

We

have never mentioned the axioms

spaces of a dimension

n

>

2. It is

for affine or projective

approximately clear

how

these

axioms would look. In the projective case the incidence axioms might be the rules connecting dimensions of intersection and sum, in the affine case the rules connecting the dimensions of intersection and join. We had given these rules in Chapter I. If n > 2 no Desargues is necessary, one can prove it from the incidence axioms. One look at Figure 5 suggests viewing the configuration as the projection of a three-dimensional configuration onto a plane. The three-dimensional configuration is easily proved. The reader should work out the details of such a proof. The theorem of Desargues is, therefore, the necessary and sufficient condition for a plane geometry to be extendable to a (or more) dimensions. Indeed, one can prove Desargues 2 on the other hand, coordinatize the plane if Desargues > and, such a plane can clearly be imbedded in a higher dimenholds; but This is again another geometric interpretation for sional space.

geometry in 3

if

n

Desargues. If one replaces Desargues' theorem by Pappus, then, as Hessenberg has shown, Desargues can be implied from Pappus. It would be nice to have a proof of this fact in the style of our presentation. One may also think of replacing Desargues' theorem by the theo-

rem

of the complete quadrilateral. Very interesting results have been obtained which the reader may look up in the literature.

GEOMETRIC ALGEBRA

102

EXERCISES. We can of course also associate a projective space with a right vector space. be two vector spaces over fields k respectively Let now V and k' (V can be right and either left or right, or V can be left and either left or right). Assume they have both the same dimension. Let U > U r be a one-to-one correspondence (onto) of the subspaces

V

V

V

V

of

V

and the subspaces

of

V. We

V a correlation of V and V

a collineation of

1)

Let

F

be a

V over some

left

field k'

V

and

shall call it

if C/i

Ui

if

C U C U

implies U(

2

implies U(

2

C U D Ui

2

,

.

vector space over k. Find a right vector space such that there exists a collineation of V" onto

T'. 2)

To each left vector space V there exists a canonically constructed

V

over the right vector space > V. correlation

same

field

k together with a canonical

V

3) Use 1) and 2) to describe all collineations and correlations, making also use of the fundamental theorem of projective geometry. > V 4) Suppose there exists a correlation V (the same space). What is the necessary and sufficient condition on the field k of 7? > V Consider the group of all correlations and collineations of V and investigate its structure. Draw diagrams. Let T be an antiautomorphism of k (if k is commutative every automorphism is also an antiautomorphism). Call a "product" k a generalized pairing of V and V into k if the two disV X V tributive laws hold and if in case V is a right vector space the

following conditions are satisfied:

X(Ya)

=

of the dual V of such a pairing. governing 5) Suppose a correlation V

Making use

(Xd)Y

(XY)a,

V

=

r

a (XY).

one can easily describe the laws

V

Prove that there exists a generalized pairing of V and V into k which allows us to describe the correlation algebraically. Find all pairings describing a given >

exists.

correlation. 6)

Consider a correlation between two projective planes. Give

the images of the configurations of Desargues, Pappus and the complete quadrilateral. Does this constitute a proof of the image configurations?

CHAPTER Let

7)

II

and

II'

be two Desarguian

k' the fields of trace-preserving

groups n onto

T II'

affine planes,

homomorphisms

respectively T'. Let X

which preserves

103

II

lines.

:

>

II

k respectively

of the translation

n' be a one-to-one

map

of

Prove:

XTX" = T'. For each a t k the map 1

a)

b)

a trace-preserving homomorphism of of &' which we may denote by a*

is

T

7 ,

hence a certain element

.

The map X k

>

:

c)

onto

k' given

by a

a*

is

an isomorphism

of k

k'.

d) If coordinates are given in n then the coordinate system of II' can be placed in such a position that the image of the point (x, y) in II

the point (a;*, y*) of II'. Describe all such maps X by formulas.

is

e)

Using Figure 6 give a sequence of three projections which thereby produce a projectivity a of the line I through A, R onto itself with the following properties: A and B are fixed points; if C is distinct from A and B then its image under a is the fourth harmonic point D. 8)

What

the order of the projectivity
V onto itself which is of order 2 and has at least one Prove that among the linear maps of V onto itself which induce
projectivity of fixed "point".

W

The

projectivity

a-

is

completely described by the pair

17,

W

of

W equivalent to

subspaces (U, TT, U). 10) Let k be commutative and of characteristic 7*

a projectivity of a point

is

by

line

necessity a

I

which

map

of

type.

line

Show

that

of order 2

computation. Returning to 9) metric constructions. Assume that k

Give a projectivity of a

2.

and which has a fixed the type described in 8). Avoid any describe the projectivity or by geois

which

is

is of

the

field

of quaternions. is not of this

order 2 and

CHAPTER

III

Symplectic and Orthogonal Geometry 1.

Metric structures on vector spaces

This chapter presupposes a knowledge of Chapter I, 2-4. We with vector spaces V of finite dimension n over a commutative field k. A left space V shall be made canonically into a two-sided space by the definition will deal exclusively

aX

Xa Since

V

has become a

consider pairings of e k is defined for

XY

V

left

and

X,

all

a

,

Xe

e k,

V.

space as well as a right space

we can

V into k. In such a pairing a product Y e V and will now satisfy the rules XY,

= a(XY)

X(aY) (the last rule since

aY = Fa and

since

fc

commutative).

is

X XY

,

XX

= as something reader should think intuitively of like "length" of the vector and of as something related to the "angle" between and Y. We shall say that such a product defines a "metric structure" on V and investigate first how such a ^l n of V. structure can be described in terms of a basis A^ A 2 The

X

2

X

,

,

,

We set AtA =

(3.2)

i,j

0,,-e/c,

t

=

1,2,

-,

n.

Let (3.3)

X

=

Z aU,

F =

,

F-l

be any two vectors. The rules (3.4)

XY

Z

ffrA,

M-l

(3.1)

=

allow us to express their product:

E

g^y,

,

r.M-1

which shows that we know

XY

if

the g if are known.

105

GEOMETRIC ALGEBRA

106

Let us select conversely arbitrary elements g fi in A; and define a product and F given by (3.3)) by means of (3.4). (of vectors This product satisfies obviously the rules (3.1) and is, therefore, a pairing. If, in (3.4), we specialise to A< and F to A f we again get

X

XY

X

(3.2).

A function of the variables #,

and y^ of the type (3.4) is called of bilinear functions is equivalent with a

Hence the study

bilinear.

study of metric structures on V. For a given pairing the g if will depend on the choice of the basis. Let us see how a change of the basis will affect the g ti Let B l B 2 B n be a new basis of V\ then B f = ]C"-i A,a 9i with certain an c fc. The BJ will form a basis of V if and only if the determinant of the matrix (a,-,-) is 7* 0. With the new basis we construct the .

,

,

,

new

(/7,:

In matrix form this can be written as (3.5)

=

(<7^)

where

(<*i<)

07
the transpose of the matrix

(a/,) is

(a.,-)-

Our pairing will have a left kernel UQ and a right kernel F They are now of course subspaces of the same space V. We know dim V/Uo = dim V/V which implies that UQ and V have the same .

U

dimension. Let us compute

X

=

;_,

X>A *

=

certainly XA,-

j

=

only

1, 2,

wil1

n we

,

be in

(j

=

get

U

in terms of our basis

if

and only

1, 2,

,

X- (2J-i

XY =

if

n). Conversely,

Z/M^M)

=

0.

A

9

.

A

vector

Then

for all F. if

=

XA,-

X will be in

7

if

for

and

if

(3.6)

0,,*,

=

j-1,2,

0,

F-l

-.-.n.

We are especially interested in the question: when will U be merely the subspace of F? Since V has the same dimension this also implies that V Q = 0. The equations (3.6) should only have the trivial solution

matrix

and

(
this is the case

if

and only

from

0.

We

is different

DEFINITION non-singular

the determinant of the

a certain terminology.

We call a vector space V with a metric structure the kernels of the pairing are 0. The determinant

3.1.

if

if

shall use

CHAPTER

G = and

det(0,,) shall

107

III

be called the discriminant of V. The necessary V to be non-singular is G 7* 0.

sufficient condition for

Let us return to

(3.5)

and denote by

G

the determinant of (g^)

B

(hence the discriminant of V as defined by the basis determinants in (3.5) we obtain
(3.7)

=

w ).

Taking

(

The geometric of

V only up

set-up alone determines, therefore, the discriminant to a square factor.

W

DEFINITION 3.2. Let V and be vector spaces over k, and be an isomorphism of V into W. Suppose that metric structures are defined on both V and W. We shall call a an isometry of

W V into W

V

if

a "preserves products":

XY

=

(
The most important 1

V

also such

is

o-"

so

is

X,YtV.

that of isometrics of

an isometry and

if

and


V

onto V. Then

T are isometrics of

V

onto

is or.

DEFINITION

The

case

for all

Let V be a space with a metric structure. V onto F form a group which we will call the group

3.3.

isometrics of

of V.

be an endomorphism of V into V and A, a basis of V. Put = B if is an isometry, then we must have B S, = A A, aA, = 1, 2, ri). Suppose, conversely, that these equations hold. (i, j = X If E?! x,A and F = E"-i U^^ then

Let

o-

t

;

o-

t

t

,

v

and

We have

The map

cr

will

n

n

r.M-1

*,M-1

be an isometry

that the kernel of


is 0.

If

if

we can show

V

is

(or assume if necessary) non-singular this follows auto-

GEOMETRIC ALGEBRA

108 matically. Indeed, e V. But
Y

that X

is


if

=

in the left kernel

U

aX-aY =

then certainly

0,

XY so that XY =

=

Ye

for all

and therefore

0, if

for all

V. This

we assume

means

V to be

non-singular.

THEOREM 3.1. Let = 5, of V and aA t

for all

i,

~

j

;

an endomorphism V -* V, A t a basis an isometry if and only if AiA = #,-#, n and if the kernel of

will be


1, 2,

,

f

unnecessary if V is non-singular. we write B, = ]>3?-i A,a vi then the matrix (a,-,-) is the one describing the endomorphism a in the sense of Chapter I, 3. Since

is

If

,

BB = t

f

~g7j

=

QH

an isometry, we get from

for

(3.5)

and

(3.7)

(3.8)

G =

(3.9)

Should

F

THEOREM V, then det det

=

o-

-(?

be non-singular, then 3.2.
1,

//

=

/Aen

=

2

(det(a,,.))

V

G

2

-(?.

(det a)

and we get det

7*


=

1.

non-singular and o- an isometry of V onto o- = det // +1, then is called a rotation; if caM a a reflexion. The rotations form an invariant is

1.

w

subgroup of the group of V whose index is at most 2. The last part follows from the fact that the map a > det
and whose image reflections or If

we form

if

is

either

1

+1

or just

the characteristic of

ft

(in case there are

no

is 2).

X = X-X we find from 2

X = 2

(3.10)

E

(3.4)

gvXM

.

<,'-!

This

is

an expression which depends quadratically on the

therefore, called a quadratic form. the coefficient of x^x,- is g if g fi if i

is,

+

select the g if in

such a

way

that

X

2

The

^

j.

x,

and

coefficient of x? is g i{

We

notice that

,

we can

becomes a given quadratic

even possible in several ways. define a quadratic form intrinsically (without reference to a basis) as a map Q of V into k (but not a linear one) which satisfies

form;

it is

One can

the two conditions:

Q(aX)

1)

2)

The

=

2

a Q(X),

a

function of two variables X,

XeV.

t k,

Y

e

V

given by

CHAPTER

Q(X which we

+

Y)

-

109

III

Q(X)

-

Q(7),

X

o Y, is a pairing of V and F into k. denote by start with an arbitrary pairing and define o Y one finds then this Q(X) satisfies condition 1). For

shall

XY X (X + 7) - X - F = XY + YX. It will satisfy condition 2). We remark, however, that X o if

Indeed,

=

Q(X)

we

X XoY= 2

,

2

2

2

F

is

not

the original pairing. If

Suppose now, conversely, that Q(X) satisfies the two conditions. and Y = we put X = l + 2 + + r in condition 2)

X

X

we

Q(X

If

X

X^

find

we use induction on

+ Let Aj

,

A

2

X,

,

we

+ X _0

+X + 2

r

get

+X =

+

,

r

l

r)

A n be a basis

of

V and

let

X

Then Q(X)

= i

This shows that Q(X) depends indeed quadratically on the

The

z<

pairing

(3.11)

XoY=

Q(X

+

Y)

-

Q(X)

-

Q(F),

which we get from a quadratic form, is very special. If we put in (3.11) and notice that Q(2X) = 4Q(Z) we obtain

XoX

(3.12)

Equation (3.13)

We a)

.

(3.11)

=

Y =

X

2Q(X).

shows also that

X o Y = Y o X.

distinguish now two cases: The characteristic of k is ^ 2. Then we can write Q(X) = o X) and see that the quadratic form differs inessentially from

GEOMETRIC ALGEBRA

110

X

X. The pairing symmetric pairing XY We would then get o

Q(X

+

which

-

7)

Q(X)

X

o Y is symmetric. Can there be another = YX for which we also have Q(X) = $X-X1

-

Q(Y)

= j(x + Y)(X shows X-F = X oY.

The quadratic form Q(X)

+

-

7)

determines,

jo:

- jyy = x- Y

therefore,

uniquely a

symmetric pairing such that

X-X =

2Q(X).

is the study of quadratic forms we may just as well 2 symmetric pairing and call X the corresponding quadratic form since it differs from Q(X) by the inessential constant factor 2.

If

our aim

start with a

b)

The

Equation

situation (3.12)

XY =

pairing succeed. Indeed,

since the if

if

X

two terms

the characteristic

term

x&2 To .

very different

is

the characteristic of k

if

=

]C<-i

^^

A ,4 ,,,

and

is

>

then

A A J

One can

two.

start with

t

x x J

l

is

for instance never obtain a

is not a desirable not uniquely determined by the

2. Definitions of symplectic

and orthogonal geometry

Suppose again that an arbitrary pairing given.

As

vector

B if AB =

any pairing

(for i 7* j) will cancel

an unsymmetric pairing

procedure since such a pairing quadratic form.

in

we

of

shall call a vector

V and V into A orthogonal

is

V

k

is

to a

This raises immediately the problem: In which AB = imply that BA = 0? Let us suppose has this property and let A, B, C t V. Remembering that k 0.

metric structures does

that

is 2.

becomes X o X = 0. If we start with a symmetric YX and hope to obtain Q(X) from X2 we do not

commutative we get

A((AC)B

-

(AB)C)

=

(AC)(AB)

-

Hence ((AC)B

-

(AE)C)A

=

(AB)(AC)

=

0.

CHAPTER

111

III

or

C = A we

For then

=

(AC)(BA)

(3.14)

obtain

A

(CA)(AB).

= A 2 -(AJ3).

2

-(BA)

we can conclude JBA = AB. In

Should

A

2

5*

other words:

BA then A 2 = and similarly B 2 = 0. Suppose now that V contains two special vectors A and B such 2 for any vector C. that AB j BA. We intend to show that C = This would certainly be true if AC ^ CA so that we may assume AC = CA. But (3.14) is then consistent with AB 9* BA only if AC = CA = 0. We may interchange A and B and see that we can also assume BC = CB = 0. But now If

AB

7*

(A

+ C)B = AB ^ BA = B(A + Q. + C) = 0; since A = (from AB 2

Consequently (A

AC = CA =

2

we get indeed C = 2

7*

BA) and

We see that there are two types

0.

of metric structures with our property: 1)

The

symplectic geometry.

Here we postulate

X = for all X V. Replacing the vector X by X + Y we get X + XY + YX + 2

e

(3.15)

2

Y2 =

0,

hence

XY = -YX,

(3.16)

X,Y*

V.

= implies indeed BA = 0. (3.16) shows that condition (3.16) one can not get (3.15) as the special case = X, since k may have characteristic 2. For the g tj we obtain in a symplectic geometry

AB

Equation

From

V

(3.17)

A

bilinear

g^

=

g tj

0,

form 5^?./.i g ti x y j i

for

= -g

which

X =

ti

.

(3.17) holds is called

skew

2

== which symmetric. ]C?.f-i 9n x x i is shows that (3.17) the necessary and sufficient condition for a and that such a geometry is equivalent with symplectic geometry the study of skew symmetric bilinear forms.

If (3.17) is satisfied,

2)

*

The orthogonal geometry.

V is not symplectic but if it has our property, then, by necessity, XY - YX for all X, Yt V. (3.18) If

GEOMETRIC ALGEBRA

112

This is a symmetric pairing which we have also met a while ago in connection with quadratic forms. In this case AB = implies again

BA =

0.

^

If the characteristic of k is

2, this geometry is entirely satisfactory since the symmetric pairings are in one-to-one correspondence with

X

2

is the quadratic quadratic forms and one may simply say that form connected with our pairing. If, however, k has characteristic 2, then the symmetric pairings are not general enough. In this case one starts with any quadratic

form Q(X) and defines a pairing by (3.11). But notation and write this pairing as XY. Hence

XY =

(3.19)

Equation

(3.12)

Q(X

+

F)

-

Q(X)

-

let

us simplify the

Q(F).

shows that

X = 2

(3.20)

The underlying teristic is 2,

pairing

then there

is

is,

0.

therefore, symplectic

no difference between

In intuitive geometric language:

V

(if

the charac-

and

(3.18)

(3.16)).

has a symplectic geometry

by an additional quadratic form (measuring lengths if you is related to the symplectic geometry by (3.19). The V is called in both cases an orthogonal geometry. of geometry refined

wish) which

To

X + 2

repeat the connections with the quadratic form: (X Y2 2XY if the characteristic F) = 2, Q(X

Q(Y)

+ + XY

+

^

for characteristic 2.

One may

call

these

+

2

F)

=

+

Q(X) two formulas

the Pythagorean theorem or also the "law of cosine'

'

which they

become in special cases. But we have to restrict

we T^

ourselves. For the sake of simplicity assume tacitly that the field has a characteristic 2 if we deal with an orthogonal geometry. In the case of a symplectic shall always

geometry we

The

shall not

make any

restrictions.

consult the quoted literature for more details on the case of an orthogonal geometry when the characteristic interested reader

may

is 2.

Appendix for readers who have worked on some in Chapter II, If r,

fc

is

exercises

11.

not necessarily commutative but has an antiautomorphism consider a generalized pairing of a right vector space

we may

CHAPTER

V

with

itself into

XY

113

III

and investigate the corresponding problem:

fc

=

Suppose implies always YX = 0. Since the antiautomorphism and the pairing are involved the discussion is a little harder. Assume dim V > 2 and consider first the case where the kernel of the pairing

is 0.

Let A, B,

C

7 such that AB

e

A(B-(AB)~

we

and

5*

-

l

AC ^ 1

C-(AC)-

)

0.

From

=

get

-

1

(B-(AB)-

C-(AC)-')A

=

which leads to

= (ACr

r

(AB)~ (BA)

r

(CA)

=

a.

We may write (BA) = (AB) a where a depends only on A; for B for which AB = BA = this equation is trivially true. One has now to show that a is independent of A. Let A and A be r

vectors

2

l

B which is orthogonal = 0. Changing B A B A B to A 2 but not to A! and 2 i.e., = = B a A we 0. There exists also by factor may assume AiBi 1, = = a vector B 2 such that A ,B 2 1. The vector B = BI + B 2 0, A B 2 = = satisfies AJZ 1. We know three equations: 1, A # BA = (A 2J5) 2 BA, = (A.BYa,

two independent vectors. There l

;

a vector

exists l

?

1

l

1

2

l

2

2

T

2

,

,

= 0. Should A and A 2 BA 2 = 2 2 give JSAj = be dependent vectors ?^ 0, then we compare them to a third independent vector and conclude that a is the same for all A 7* 0. For A = it does not matter what a we take. The result is:

^

They

,

,

(YX) = (XYYa Case case,

I.

X

=

2

for all

YX = -(XY)

1

i

X, Y

for all

T -(XY) = (XY) a. Selecting we obtain a = 1, and are left

T

over

.

r=l,

only

if

is

the

the identity. field

fc

is

V.

X e V. We deduce then, as in the symplectic and obtain

X and Y in such a way that XY = I with (XY) = (XY) But XY can range that

e

The

all of

k and

we

see

identity can be an antiautomorphism

commutative. This means that our geometry

symplectic.

Case II. There exists a special vector

X

Q

such that Xl

=

5^ 0.

GEOMETRIC ALGEBRA

114 Set

X

Y =

=s

introduce a

and we obtain a

in our equation

-X"

=

new pairing [XY]

from the given pairing

(just

pT (XY). It

by a constant

T 0[YX] = (P(XY]yp-

[YX] =

=

T

(T

Let us

0.

l

=

differs

very inessentially

We get

factor).

[XYYfi,

l

p- [XY]'p.

We have [(Ya)X]

= P~\(Ya)X) =

l

(T a'(YX)

=

l

pT

a^[YX]

which shows that the new pairing belongs to the antiautomorphism a > /TV/3 = a\ With X we can now write

[YX] Interchanging once more: [YX] the identity.

Making that r

is

this

= [ZF]\ = [FJ?] X9 which implies that X 2 =

of order at

r (YX) = (XY)

If r

=

1,

then the

field

corresponding geometry a hermitian form. If the pairing

is

has a kernel

and obtains the same

Each

result

therefore,

assume

most 2 and that

.

must be commutative and we are

case of orthogonal geometry. If r is of order 2, then k may or

EXERCISE.

we may,

replacement in our pairing

an antiautomorphism

may

called unitary

1,

in the

not be commutative. The

and the form

X

2

is

called

one has to assume dim V/V > 2 by going to the pairing induced on V/V

VQ

7* 0,

.

of our geometries with kernel

induces on the

V

a correlation of order 2 and each correlation of projective space order 2 is induced by one of our geometries. By what geometric feature is the symplectic correlation distinguished from the others?

name "symplectic" somewhat justified? Suppose the geometry either unitary or orthogonal. What does it mean for a special 2 = in terms of the correlation? This leads to vector to satisfy Is the

is

X

the classical definition of a conic in projective geometry. 3.

Common features of orthogonal and symplectic geometry

In this paragraph we

will

study spaces with either an orthogonal

or a symplectic geometry. In either case orthogonality of vectors

CHAPTER

115

III

U

is a subspace of F, or of subspaces is unambiguously defined. If then the orthogonal subspace U* has a unique meaning. The two

kernels of the pairing are the same, they are the space V*.

DEFINITION 3.4. The kernel V* and denoted by rad V.

V

of

the radical of

is called

V

If U is a, subspace of V, then the pairing of V induces by restriction a pairing of U which will be of the same type as the one of V orthogonal or sympletic. U itself has a radical consisting of those vectors of U* which are in U. In other words

U = U C\

rad

(3.21)

DEFINITION

V

If

3.5.

V=

(3.22)

of subspaces

UC

U*,

is

the direct

U,

C/2

V.

sum -

U

-

r

which are mutually orthogonal, then we and use the symbol

V is the orthogonal sum of the Ui V = U, U (3.23)

J_

2

-

-

-

_L

U

shall

say that

.

r

V be

a vector space which is a direct sum (3.22) of subspaces Ui Suppose that a geometric structure is given to each subspace. Then there is a unique way to extend these structures to one of V

Let .

such that

V becomes the

orthogonal

sum

U

of the

%

.

If

are vectors of

V and A

(3.24)

XY

The

=

-

t

,

B

A,B,

-

t

e

Ui

;

,

then we have obviously to define

+AB + 2

+

-

2

A

T

B,

.

reader will have no difficulty in proving that (3.24) defines a V and that V will have a symplectic, respectively, ortho-

pairing on

gonal geometry

The geometry Ui and

U

f

if all

of

V

the Ui are symplectic, respectively, orthogonal. induces on each C7 t its original geometry and -

are orthogonal

if i

9* j.

+ Ur U< orthogonal to U, if Suppose V = t/i + U* + = ]T]7-i A, i j& j but do not assume that the sum is direct. Let e rad V. Then we must have XBi = for all Ai * Ui and assume or A t e rad Ui Bi e Ui which gives A t Bi = Conversely, if each ,

X

9

,

X

.

,

GEOMETRIC ALGEBRA

116

Ai

t

rad

C7 t

X t rad

then

,

V. In other words,

if

the

are mutually

(7,

orthogonal, then,

V =

raxl

(3.25)

U

Should each

V =

rad

Indeed

A

t

t

if

rad

V

0,

U =

0.

t

we

We

obtain

+ rad U

-

2

rad

i.e.,

But

non-singular.

^ =

]C"-i

U +

rad

l

be non-singular,

t

is

U +

rad

U =

sum

in this case our

AiB =

for

t

any

can, therefore, write (3.23)

.

then we obtain

0,

t

r

B

each

if

is direct.

e [/,

%

.

Hence

C7, is

non-

singular.

Consider the subspace rad

V

of

V

rad

V

and

U

let

be a supplementary

subspace

V= rad

V

is

orthogonal to

V

and, therefore, to

V =

(3.26)

U;

rad

V

E7

and we get

U.

We deduce rad

V =

Since this last

rad(rad 7) JL rad C7

sum is

direct

=

rad

V

U=

we must have rad

rad U.

J_

0.

U is, therefore,

non-singular.

V

does not in general induce a geometry on a The geometry on factor space. It does so, however, for the space F/rad V. It is natural rad F) and the coset(F rad V) to define as product of the coset(X

+

the element

+

XY: (X

(3.27)

+ rad

F)-(F

+ rad

= XY.

7)

1.3 we have mapped a supplementary space U isomorfactor space F/rad V. The map was canonical onto the phically rad V. The equation and sent the vector of U onto the coset

In Theorem

X+

X

(3.27)

means that

this

map

is

an isometry of

U onto

7/rad V. Thus

we have proved Each space U which is supplementary to rad V an orthogonal splitting (3.26). U is non-singular and canonically isometric to F/rad F.

THEOREM

3.3.

gives rise to

DEFINITION

Let

F =

U

U

V

=

U{ J- Ur be orthogonal splittings of two spaces F and Ui JL and suppose that an isometry v of C7, into U( is given for each i. 3.6.

ZA

t

JL

2

J_

-

-

JL

r

,

_L

V If

CHAPTER

X

=

]5".i At with A* of F into by

map

is

We shall of

V

=


call it

V

THEOREM isometry

E7,

V =

Let

3.4.

on/0 C7 t

.

+ A

-

_L


sum

an isometry

of

V

det

(3.31)

=

onto

a-

Ui

TAen a

is

J.

the orthogonal

+

42

(T r

JL

of the

rr

r

.

maps

=

V

t72

J_


_L

/&e orthogonal _L 02 JL

0*1

define a

be denoted by

shall

eri


we can

.

will, of course, have to check that a is an and that scalar products are preserved.

reader

into

+

*!/!,

an isometry and

(3.30)

The

=


(3.29)

which

a vector of V, then

e C7, is

V


117

III

f/ r

isomorphism

and eacA

o\

an

left

to

sum J.

cr r

and we have

det v l

det

..... det a r

o- 2

.

if T

=

TI JL T 2 JL

where the r { are also isometries of (TT

(3.32)

The proof

of this

=

J_

VI-TI

theorem

[/,

o- 2

'

'

JL T r

onto Ui then

r 2 JL

1.

o- r

rf

straight forward

is

.

and can be

the reader.

THEOREM

3.5.

subspace of V.

17**

(3.33)

Suppose

We

-

The subspace Should

U, rad

(3.34)

U

dim

U =

V

is

17

+ dim

rad C7*

non-singular and

let

U

be

any

= U

C7*

H

= dim

F,

U*.

U will be non-singular if and only if U* is non-singular.

be non-singular, then

(3.35)

Finally,

that

have always

ifV= U

J_

we have

V = U C7*. W, then U and W are non-singular and W =

C7*.

Proof: Since the kernels of our pairing are 0, we have (3.33) from the general theory of pairings. Formulas (3.33) and (3.21) imply (3.34). If

U is non-singular, then

(3.34)

shows that U*

is

non-singular

GEOMETRIC ALGEBRA

118

U+

and that the sum

V= U

(3.35). If

dim

TF,

therefore

t/*;

W

C7* is direct; since

WC

then

= U*

the dimensions

and rad

U = U

fit,

we

get

W = n - dim U -

U* and dim

H

C7*

=

0.

DEFINITION 3.7. A space is called isotropic if all products between vectors of the space are 0. The zero subspace of a space and the radical of a space are examples of isotropic subspaces. A vector A is called isotropic if A 2 = 0. In a symplectic geometry every vector

is

isotropic.

THEOREM 3.6. Let V be a space with orthogonal geometry and suppose that every vector of V is isotropic. Then V is isotropic. Proof: Under our assumption this geometry is symplectic as well

-YX

= = YX. This have, therefore, = since we assume, in case of an orthogonal geometry, implies that the characteristic of k is ?* 2.

XY

We

as orthogonal.

XY

The

following special case plays an important role in the general

theory.

We

assume that dim

V =

contains an isotropic vector If A is any vector which

V =

(AT,

We

A).

NM

such that were 0, then

N

2,

N is

V

that

is

non-singular, but that

V

?& 0.

not contained in the line (N), then

M

determine another isotropic vector = yNA. If = 1. Put = xN yA, then e rad V\ but we have assumed V to be non-singular. shall try to

M

NM

+

NA

NM

= 1. and we can determine y uniquely so that Therefore 7^ 2 = In the symplectic case is automatically satisfied so that any x is possible. If V has orthogonal geometry, x must be determined

NA

M

from

M This

is

2

=

= 2xyNA

also possible since

determined We have

2yNA

+

7*

y*A*.

and leads to a uniquely

x.

now

V =

for both geometries

N = 2

(N, M),

V=

M

2

=

0,

NM

=

1.

M) is a plane, we may impose on it a geometry and 12 = 1 (hence 21 = 1 in the orthogonal, 2 = M'2 = and NM = 1. = 1 in the symplectic case). Then N 21 and NX = If X = xN + yM e rad V, then XM = 0; hence x = = 0. V is non-singular. which gives y Conversely,

by

setting

if

n

=

g 22

(N,

=

CHAPTER

119

III

Suppose V = (Af, M) has an orthogonal geometry. X = xN + yM 2 will he isotropic if X = 2xy = 0; hence either y = 0, X = xAT or = yM. x = 0,

X

DEFINITION 3.8. A non-singular plane which contains an isotropic vector shall be called a hyperbolic plane. It can always be spanned of vectors which satisfy by a pair N,

M

N =M =

NM

2

2

0,

=

1.

M

We

shall call any such ordered pair N, a hyperbolic pair. If V a non-singular plane with orthogonal geometry and ^ is an in V such isotropic vector of V, then there exists precisely one that AT, is a hyperbolic pair. The vectors xN and z/Af are then the

N

is

M

M

only isotropic vectors of V.

DEFINITION PI

>

1*2

,

9

An

3.9.

P

shall

r

orthogonal sum of hyperbolic be called a hyperbolic space:

H =P 2r

l

JL

P

-L

2

-

-

_L

P

r

planes

.

and of course of even dimension 2r. a space irreducible if it can not be written as an of proper subspaces. Because of (3.26) we see that

It is non-singular

We may

call

orthogonal

sum

an irreducible space of

is

then

If it is isotropic,

an isotropic space

is

we

non-singular case

necessarily either non-singular or isotropic. dimension is 1 since any direct splitting

its

also

an orthogonal

splitting.

2)

rad

U =

(3.35)

=

shows U*

(AT,

A)

THEOREM sum of lines

dim

is

The A, are

V =

N

V must is

contain

non-singular

1.

be any vector of V. Since such that NA 7* 0. The plane non-singular and (3.35) shows again U* = 0, V = U.

A

3.7.

called

only if none of the

7*

V

(A,) JL (A,) JL

an orthogonal

4

t

space with orthogonal geometry is an orthogonal

V =

A

0,

Sympledic geometry. Let 0, there exists an A

7 =

discuss the

distinguish:

Orthogonal geometry. Because of Theorem 3.6, a non-isotropic vector A. The subspace U = (A) 1)

and

To

-

t

.

-

-

basis of V.

JL

V

<4 n>. is

non-singular if and

are isotropic.

non-singular symplectic space is an orthogonal

sum

of hyperbolic

GEOMETRIC ALGEBRA

120 planes, in other words

it is

a hyperbolic space.

dimension

Its

is

always

even:

THEOREM Write

U=

V

Let

3.8.

U

rad

N

N, 2 Then we can find, in V, vectors MI 2 Ni Mi is a hyperbolic pair and such let

,

U ,N

and

be non-singular

W and

_L

,

,

,

,

that

%

U =

P!

P

J_

J.

2

-

-

J.

P

basis of rad U.

such that each

r

hyperbolic planes

also orthogonal to

%

,

M

the

P = (N M ) are mutually orthogonal and V will therefore contain the non-singular space {

bea

r

,

M

any subspace of V.

W.

W

r

U

into which, in turn, contains U. Let finally v be any isometry of f some non-singular space V Then we can extend a to an isometry a .

U into

of

V.

Proof:

For r = nothing is to be proved. tion on r. The subspace

We may, therefore, use induc-

1)

U = orthogonal to

is

N

r

rad t/*


N

,

-

2

,

,

N We have N N ^).

but does not contain

=

U =

rad

Q

N

N

r

.

-

(N,

,

W

AT,.,) _L

2

-

,

,

r

The space U% must, therefore, This implies r c U* but r 4 rad U A 0. The plane (N r ^ contain a vector A such that r A) is nonin is a U* It and contained spanned by hyperbolic pair singular U * our space (/ is orthoSet P r = (N r Af r ). Since P r r r *

.

N

,

.

N

,

M

.

C

,

P

which shows that

,

contained in the non-singular has r of U dimension 1. By induction The radical Q only space P* P* on as the we whole can find hyperbolic space) hypothesis (used gonal to

r

C7

is

.

pairs

N

%

M,

,

P*

in

<

(for i

r

1)

such that the

P

%

are mutually

orthogonal and also orthogonal to W. Since they are orthogonal we have finished the to P r and since P r is also orthogonal to construction of U.

W

2)

Then

U into V, N'< = N' JL W. We can,

Let a be the isometry of
=

M

(Nl Ni such that ,

,

,

r)

M(


and

W=


therefore, find in

V

a hyperbolic pair and the planes p r = (N( M() are mutually orthogonal and orthogonal to W. The is now extended to a map 6- of U into by premap (T of U into

vectors

i

ATJ

,

is

.

,

V

scribing irMi

= Mi

V

;

&

is clearly

an isometry.

CHAPTER

121

III

V

THEOREM 3.9. (Witt's theorem). Let V and be non-singular spaces which are isometric under some isometry p. Let
U

of

V

onto

Proof 1)

in

If

V.

:_

U is

Theorem

the non-singular subspace of V which was constructed then a can be extended to an isometry of U into V.

3.8,

This reduces the problem to the case where U is non-singular, an assumption which we shall make from now on. 2) For a symplectic space V let U' be the image of U under
V

= U' _L U'*. All we have really and write V = U JL U* and to do is to show that U* and U'* are isometric. But this is clear. V and have the same dimension since the isometry p exists. 7* and U'* have, therefore, also the same dimension and they are

V

non-singular;

by Theorem

3.7 they are hyperbolic (we are in the

symplectic case) and, hence, clearly isometric. now on that our geometry is orthogonal.

We may assume from

3) Should our theorem be true for non-singular subspaces of lower dimension, then we can proceed as follows: Assume that U = t/i J_ U2 where U\ and U2 are proper subspaces of U. We shall denote by U{ the image of Ui under
U{ which
into

map

V

U\

>

which

o-

U2 (which is orthogonal to Ui) into the can mapping (by induction hypothesis) be extended X of U* onto /{* arid now we are finished since to an isometry r J_ X is the desired extension. 4) We have still to show that the theorem is true if U is non= (A) is then a line and A 2 5* 0. For the singular and irreducible. U
It will

map

the space

space C/i*; this

image C =
2

.

V

.

V

we proceed as follows. the vectors A + B and A Since (A + B)(A - B) = orthogonal. They can not both be isotropic since 2A = (A

construct this T

- B

+

B)

are

+

122

GEOMETRIC ALGEBRA

D = A +

eB (e = 1) a nonA eB By what we have seen A + cB)* = {/})* Let n be the D (/* is an orthoisometry which is identy on // and sends D into gonal sum of isometries on (D) and (Z)}*). Then not isotropic. Denote by B. isotropic one of the vectors = (A belongs to the hyperplane

(A

E)

is

H

= -A - eB, - eB. #) = A we get juA = eB e5)

Adding the two equations

(k is not of charac2 in the orthogonal case). 1 we are done. If e = +1, then pA = If e = B. Let v be the -3T. It is obviously an of V onto map which sends every vector B. This finishes the proof of our theorem. isometry of V and v^A teristic

X

Theorem

has

3.9

many important

U is naturally called

subspace

subspace of some

THEOREM

An

consequences.

maximal

isotropic isotropic subspace of V.

if

isotropic

U is not a proper

All maximal isotropic subspaces of a non-singular the same dimension r. This invariant r of V is called the

3.10.

V have

space index of V.

be maximal isotropic and suppose dim Ui < dim C7 2 We can find a linear map of f/i into U2 which is an isomorphism into. Since both f/ and U2 are isotropic such a map is trivially an isometry. By Theorem 3.9 it can be extended to an isometry a Proof: Let Ui and

U

2

.

t

and consequently Ui C
V

onto V. Thus

vU

l

C U

l

.

2

l

l

.

l

THEOREM

The dimension of a maximal hyperbolic subspace index of V. Therefore 2r < n = dim V and the maximal value %n for the index r is reached if and only if V itself is

is 2r

where

3.11.

r is the

hyperbolic (for instance in the symplectic case). 7/2* is

and only vectors. One

maximal hyperbolic

any non-zero

isotropic

if

A

hyperbolic subspace

if the space II*, does not contain

achieves,

V = H2r JL W where W may be called V determines uniquely the geometry of

W

a splitting The geometry of

therefore,

anisotropic.

(indepently of the choice

ofH 2r ). Proof: 1)

If

U is a maximal isotropic subspace of V and r =

Theorem

3.8

shows the existence

of

a hyperbolic space

dim

H

2r

of

then dimen-

[/,

CHAPTER

W

sion 2r. Let

=

7/ 2* so that

,

U

JL

H

U

Should TF

TF.

N

then would be orthogonal and not be contained in U. The

5^

0,

7

(N) would be isotropic, contradicting the fact that

was maximal 2) If

V = # 2r 1

we have

contain an isotropic vector AT hence to the subspace to 7/2r

space

123

III

U

isotropic.

of F, then

any hyperbolic subspace

2t is

H

2.

has the form

...

tf (N: Mi) and (Ni tf J is an isotropic subspace of dimension s. This shows s < r. We can = 2a JL //2
(Nl

,

M()

JL(Ni,MQ

H

,

H

,

H

H

H

H

W be

H

F onto V which carries 2a onto 2a Let 2(r - a) and images of // 2 (r-.> respectively TF under
H

see that

,

,

.

TF ;

is the space orthogonal to = then this will contain space r, isotropic vectors; if s which is isometric to W. reduces to

<

2

(

r

J_

-.)

W

H

the

2t

r,

.

If

then

THEOREM 3.12. Let V be non-singular. If Ui and U2 are isometric svbspaces of F, then U* and U* are isometric. Proof: There is an isometry of F onto F which carries Ui onto U 2 .

U* onto U*2 applications we

It carries, therefore,

For several trics of

later

shall

need a lemma about isome-

hyperbolic spaces.

THEOREM

Let

3.13.

be a hyperbolic space vector

.

N,

fixed.

V =

(N,

,

M,)

JL

<# 2

and a an isometry of

Then a

is

V

,

M

J_

2)

onto

-

JL

V which

a rotation. As a matter of fact

(N r

,

M

r)

keeps each
is of the

following form: aAT

= N.

t

,

aM, =

EN

v

a,,

+

M,

-i

where the matrix

skew symmetric Proof:

We

(a,/) is

if

have

symmetric

if the

geometry

it is

orthogonal. of course
=

JV,

.

Let

We find =

b tl

is

symplectic

and

GEOMETKIC ALGEBRA

124

we must have aN^aMj = N 1 and ft,, = for i 7* j. Thus

and, since get

=

6',,-

*M, =

Z AU,,

if

M,-

t

+

Mi

a-

an isometry, we

is

.

v-\

We

mast

=

check that rAf.-^lf,

still

+ iXa,, ,.l

tf A/ + M.YE /\M-1

MN

M,)

=

.Tf./lf,.

=

o,,

+

This leads to

0.

(M.ATX, =

0.

'

If F and we get a,- = a in this we assume case a,, a,,that the characteristic of k is ^ 2 we see that (a,-,) is skew symmetric. Obviously, in both cases, det a = +1. If

F

is

orthogonal, then

is

We The

symplectic, then

now

shall

%

is

t

t

.

,-

since

turn to isometrics of general non-singular spaces. be denoted by 1 or by l v if a reference to the

The map

needed.

satisfies (I

+

o-)X

be denoted by

1

=

X

a which sends each vector

X

=

+

0, 1

1 F if

respectively


=

and

X

onto

X

shall, therefore,

a reference to the space

V

needed.

is

Let

V= U

W. Then we can form

JL

It is obviously of interest only

otherwise

that 2

if

lu is 5^

JL l

w

2 since

lv

If

.

X

or.

Such a map

o-

satisfies

obviously

An isometry of a non-singular space V is called = I. If the characteristic of k is ^ 2 then every the form \u JL lw (resulting from a splitting V =

if a

involution is of

W}. If

=

3.14.

an involution _L

o-

the characteristic of k

U and W are characterised by 1. We shall now prove

THEOREM

the isometry

= A + B with A t U and + B. We have
would be merely

it

B e W", then aX = - A or X c T7, and o-X =

U

1

and

identity shall

space

a

=

t

=

2

the characteristic of

k

V will keep V fixed and

is 2, each involution of

every vector of a certain maximal isotropic subspace of be of the type of isometries discussed in Theorem 3.13.

Proof:

Suppose o- = 1. Then X-vY. Consequently 2

1)

(*X

-

X)(
+

The two subspaces orthogonal.

7)

U=

=

XY =

trX-o-F

(a-

aX-vY and vX-Y = SX-aY =

+

F - X-o-F - XY = 0. TF = + 1)F are, therefore,


1)F and

(o-

CHAPTER Let the characteristic of k be of by o-, a vector
2)

reversed

U C\

W=

125

III

A

9* 2.

V=

X of U is

vector
+ X W is preserved by 0. Since any vector X can be written as

cr.

Consequently

X)

%((?X

+

TT and o- = -! J. 1* 3) If the characteristic of k is 2 we are in the symplectic case. 1 and
%(vX

+

X) we

see that

C7

.

+

K

+

Then

(a

+

VC

1)

+

K. Let us write

K = dK Since

or/C

K

=

we have

,


K

Q

.

= X? and

thus

1)

(
The only vectors of K which are contained in K*Q which means that rad K is the kernel of cr 1

K%

C

^*-

are those of rad if

applied to

K

K%

.

This shows

dim Kl = dim

(a-

-

1)X?

+ dim rad K.

K

Both (0are isotropic subspaces of the non!)/* and rad singular space K*Q and have, therefore, dimensions at most J dim K% see that this value is reached, i.e., that rad must be a maximal .

K

We

K* If K denotes a maximal isotropic subspace K = rad K _L K is a maximal isotropic subspace of V. K C K shows that every vector of K is left fixed by

isotropic subspace of of

K

But

then

,

Q

.

2

l

l

cr.

2

2

DEFINITION

Suppose the characteristic of k is j& 2. If U, then we call p the type of the p involution a. We have obviously det cr = ( l) Since C7 has to be non-singular the type p must be an even number if V is symplectic; if V has orthogonal geometry, then p might be any number < n = dim V. An involution of type p = 1 shall be called a symmetry with

er=

3.10.

luJLlw and p = dim

.

respect to the hypcrplane TV.

a 180

An

involution of type 2 shall be called

rotation.

reason for the word symmetry should be clear: U = (A) a non-singular line, 17* = a non-singular hyperplane and the B (with B z W) is xA B. image of a vector xA

The

W

is

+

+

The

following theorem characterizes the isometrics

THEOREM which keeps Proof: If

3.15.

Let

all lines o-

of

V be non-singular V fixed. Then a =

keeps the line (X)

fixed,

1?

and a an isometry of

then

lr

V

.

aX = Xa

and

for

any

GEOMETRIC ALGEBRA

126

(X) we have c and on the other

Y

t

X

+

+

a(X

+

F)

=

cr(X)

+

o-(F)

= Xa

+

A

Yb.

a = c = b and we know now that we have
X

XY

.

4. Special features of orthogonal

comparison shows Xa with the same

Then JfF =

7* 0.

=

1.

geometry

In this section V will stand for a non-singular space of dimension n with an orthogonal geometry.

DEFINITION

3.11.

The group

of all isometrics of

V and

F

V

into

is

denoted by 0. The subgroup and the commutator subgroup of by 0. If it is necessary to be more specific in these symbols, then we shall put n as a subscript and inclose in a parenthesis any further information. For instance On (/c, /) denotes the orthogonal group of a space over k whose dimension is n and whose geometry is based + on the quadratic form /. is the kernel of the determinant map; O + The index of O h in is 2 the image is commutative, hence 12 called the orthogonal group of + of all rotations is denoted by O

is

C

.

since reflexions (for instance symmetries) are present state a slight improvement of Witt's theorem.

if

n

>

1.

We

U

THEOREM 3.16. Let
U

U+ V

whose determinants have opposite sign, then p will be a reflexion of V which keeps every vector of U fixed. Conversely, if p is a reflexion which keeps every vector of U fixed and TI one extension of a to F, then All we have to r 2 = TIP is another extension and det r 2 = det r do is to see whether such a reflexion exists. 1)

If T!

and

T 2 are

two extensions

of

a-

to

= rrV 2

t

.

Suppose dim U + dim rad U < n. Tho space U of Theorem of dimension dim U + dim rad f/. We can write V = _L 0* where U* ^ 0. Let X be a reflexion of U* and put p = !# _L X. 2)

3.8

is

3)

Suppose dim

U+

dim rad

U =

n.

Then F =

=

H

2r

_L

W.

CHAPTER

We

are looking for a p which

W* =

127

III

identity on

is

U and,

therefore, on

W.

on H 2r But it should keep every vector of rad (/ (which is a maximal isotropic 2r ) fixed. Theorem 3.13 shows that such a p does not subspace of It

must map

fI 2r

onto

itself

and be a

reflexion

.

H

exist.

THEOREM vector of

3.17.

Let


be

H fixed.

a hyperplane

V

an isometry

which leaves every of = 1. // is singular, then a

H is

If

H

non-singular, then a is either identity or the symmetry with respect determines the to II. The effect of an isometry on a hyperplane

H

isometry completely if II is singular, and

up

to

a symmetry with respect

to II if II is non-singular.

Proof: 1)

N

is

H

be singular. The isotropic. We have rad

Let

line

H

=

=

//

L = //* = (N) is also singular, rad H* = (N) and can write

(N)

J_

W

W

2. The plane W* is is non-singular and of dimension n where non-singular and contains N. There exists, therefore, a unique in W*. If a leaves // element- wise fixed then hyperbolic pair N,

M


=

W*. The pair

the uniqueness gives IF* as well as on W.


aM

now
is

again hyperbolic and
= N;

M which shows a to be identity on

Let // be non-singular. The line L = //* = (A) will be non= L J_ H. If a is identity on then cr = A J. 1 H singular and V and X, as isometry on L, must be zfcl/, 3) If
H

.

H

THEOREM

3.18.

Suppose

that

n

>

V contains

3 and that

If or keeps every isotropic line fixed, then a Proof: V contains a hyperbolic pair N, M.

lines.

H

l

=

1

isotropic

v

The space (N, M)*

= then *B = Be non-singular and ^ 0. Let B e (AT, M)*. If B 2 B Then 0. Assume now ^ M) (AT, (B) = (N, M, B) by assumption. is non-singular and (B) the only line in (N, M, B) which is orthogonal to (N, M). If we can prove that a- maps (N, M, B) onto itself, then we obtain again crB = Be since, by assumption,
is

GEOMETRIC ALGEBRA

128

N

2

is

M+

B or not, the vector JB Regardless of whether B = be we on isotropic. By assumption easjly seen to have, therefore, 2

one hand

= dN -

2

B -M + dB 2i

and on the other hand
I

B*-M

+ BJ = aN - |# -M + 2

a(B).

A

comparison shows first that a(B) t (N, M, B) which proves 2 = d = c. If B is selected a(B) = Be also if B 7* 0, and secondly that a 2 so that B j 0, then one sees in addition that a = d = b. This proves now that every vector of (N, M) and every vector of (AT, M}* e V and Theorem is multiplied by a. Therefore
X

3.15 shows

ff

=

lv

.

THEOREM

3.19. Suppose that & leaves every non-isotropic line but the is a hyperbolic plane over omit single case where of = a 3 Then with elements. dbl F afield Proof: Because of Theorem 3.15 we may assume that contains

V

V fixed

.

V

a hyperbolic pair N, M. 1) dim V > 3; then dim(AT, M)* > 1. Let B be a non-isotropic B are non-isotropic, hence, by vector of (N, M)*. Both B and = aB and Bb B) = (N
N+ +

+

+

+

We

use

now Theorem

V =

3.15.

(AT, M). (M) are not loft fixed, then a 2) must interchange them since they are the only isotropic lines of F. Hence
If the lines (AT),

.

is

not isotropic, hence, by assumption,

a(N

+ cM) =

d(N

+ cM) = dN +

on one hand and

a(N

+ cM)

= aM

+ a-N

dcM

CHAPTER

129

III

on the other. A comparison yields a = dc c/a = d, hence c3 = a2 a contradiction. In our next theorem, due to E. Cartari and J. Dieudonne*, we ,

try to express each isometry of

V

,

by symmetries with respect to

hyperplanes.

THEOREM isometry of to

V

Let

3.20.

V onto V is

and dim V = n. Every most n symmetries with respect

be non-singular

a product of

at

non-singular hyperplanes. Proof: The theorem is trivial for

(7=1

and

n =

also for

use induction on n and have to distinguish 4 cases. Case 1). There exists a non-isotropic vector A,

1.

left fixed

H=

We

by

<7.

H

= /?. Let X be the restriction of a to and (A)*] then
<

-

where

r t is a

symmetry

of the space

H with respect to a hyperplane

Put Ti = IL JL Ti where L = (A). Each f leaves the hyperplane L JL H, of V fixed and is, therefore (as reflexion), a symmetry f r is \ L _L X = o- and we know in this of F. The product f^ 1 symmetries. case that
Hi

of //.

t

;

T(O-A

+

A)

=

+

o-A

A,

r(o-A

-

A)

= A -

o-A.

= 2A

which shows that TO- leaves A fixed. 2 = 1, we 1. Since r where r < n = rr a rr T 2 obtain, after multiplication by r from the left, oproduct of at most n symmetries. Case 3). n = 2. Cases 1) and 2) allow us to assume that V is contains non-zero isotropic vectors: V = (AT, M) where N,

Adding we have:

By

case

1) TO-

rcr(2A)

=

rr

r^a

1

,

M

a hyperbolic pair. a)
= aM, vM = a'W. Then

We

fixed non-isotropic vector.

means

a-

+

are in Case

aAf)

= aM

+ N

is

a

=

1

1).

= aN, aM = a~ M. We may assume a j 1 = 1. A = AT + M and
b) crN

cr(AT

since o 1

(a'

- 1)M

are not isotropic which brings us back to Case 2). can now assume that n > 3, that no non-isotropic Case 4). A is isotropic whenever vector is left fixed, and, finally, that o-A

We

A

is

not isotropic.

GEOMETRIC ALGEBRA

130

N be an isotropic vector. The space (AT)* has at least dimension

Let 2 and

radical

its

the same as that of (N)

is

it is

(i.e.,

(N)). (N)*

contains a non-isotropic vector A.

We have A 2 ^

and (A

A

assumption that a A.

+

a(A

are isotropic.

last

eAT)

=

+

cN)

The square

is,

therefore,

equation

-

-

A)(
is

We

7* 0.

conclude by our

as well as the vectors

(A

*N)

2e(trA

This

-

= A2

2

+

-

(
A)

-

2

N)

+

down

for

((rAT

=

e

+

=

2

N)

and

1

*(
0.

=

e

N)

and added.

1

We get 2(crAT - AT) = or (
2

The

W of

whether

X

aX

X is

the image I. It contains only isotropic vectors and is, of V under the map aconsequently an isotropic subspace of V. A product of any two of its

isotropic or not.

set

all

these vectors

is

vectors will be zero.

now

Let (
X t V and Y e

- X)(*Y -

But
aX-vY =

W*. Consider

=

F)

=


F =

We

is

X-
-

-

(
X e W, Y t TF* so that the last term is 0. XY since is an isornetry. Hence

X) Y. -

Furthermore,


X(Y which

-

true for

all

X

t

V.

(r7)

= Y

This means

0Y

t

rad

V =

or


see that every vector of

that a non-isotropic vector

We

is

conclude that the equality

H

is left fixed.

left fixed.

W

We

had assumed

This shows that T7*

is

W* < \n From dim W + dim W* = n we sign holds. The space V is, therefore,

know isotropic subspace. since these spaces are isotropic. an

W*

not

dim

<

\n and dim

a hyperbolic spare 2r n = 2r and IV* a maximal isotropic subspace The isometry a leaves every vector of W* fixed arid Theoof 2r rem 3.13 shows o- to be a rotation. We can conclude, for instance, that for the space V = II 2r our

H

,

.

,

theorem holds at

least for reflexions. Let

now

T

be any symmetry

V = H Then ra is a reflexion of H hence TO- = T^Z r. with s < n = 2r, but since TO- is a reflexion s must be odd, hence = rr^ < 2r = n s < 2r 1. We get a product of s + r. of

2r

2r

.


symmetries.

2

,

,

1

CHAPTER

REMARK

Let

1.

=

cr

r,r 2

respect to the hyperplane 77 t

codim

/,

+

codim

77,

where

r, .

131

III

U =H

Let

t

M = codim

l

?/,-+,

the

r< is

n 77 +

C\

2

symmetry with C\ II Then 4;

codim(/

t

+

.

7/ t4l ).

Hence codim [/"+!< codim This shows codim

<

f/ t

i,

especially

+

f7,

1.

codim (H L C\

77 2

H

-

s.

C\

77.)

<

s

or

dim(77

H

,

n 77 n

n

2

77.)

>

n

O

(^ II 2 C\ 77, is left fixed by every r, and, This shows that every vector of a subspace of a dimension at least n s is left fixed by o-. Should s < n, then aleaves some non-zero vector fixed. We conclude that an isometry 1 F ) can not be without a fixed non-zero vector (as for instance written as product of less than n symmetries. n If o- is a product of n symmetries, then det <7 = ( Thus we l)

Every vector by

therefore,

of

{

a.

.

state: If fixed,

n is even, then every reflexion keeps some non-zero vector and if n is odd, the same holds for a rotation.

REMARK

Suppose that a can not be written by less than n = symmetries, and let T be an arbitrary symmetry. Then det(ro-) = det o-, which means TOTiT 2 T, with s < n and, hence, 2.

-

a

=

for

TT (7

T. (s

I

+

<

1

n).

We see that the first factor in the product

can be an arbitrarily chosen symmetry.

Examples I.

dim 7 =

2.

Every reflexion is a symmetry and every rotation is a product of two symmetries in which the first factor can be chosen in advance = TT). Theorem 3.17 shows that a rotation o- 7* 1 can (also for 1 leave only the zero vector fixed and that any rotation is uniquely determined by the image of one non-zero vector. 1 Let r be a symmetry, a = TTI a rotation. Then rar" TIT

=

1

(rrj)'

=

l


.

We

have, therefore I

T(TT~

(3.36) If

<7j

=

rr 2

is

=

(T

1 .

another rotation, we have

=

I

T(r 2 c7T2 )T~

=

T
r~

=

(7.

=

Z

T TI T

=

GEOMETRIC ALGEBRA

132

Thus the group 02 is commutative and the interaction with the coset of symmetries is given by (3.36). Theorem 3.14 tells us that dblr are the only involutions among the rotations, if n = 2. This also allows us to answer the question when the group 2

then (3.36) shows o- = o-" = or o2 1, i.e., that there are only two rotations. 0\ has index 2 in rOa is the other coset (r a symmetry) so that there are only two symmetries in such a geometry. This means that V contains only two non-isotropic lines. If A, B is an orthogonal basis, then (A)

itself is abelian.

Should this be the

1

case,

2

,

and (B) are the only non-isotropic lines. Hence the two lines (A + B) and (A B) are isotropic and, since V can only contain two isotropic lines, we have enumerated all lines of F. The field k has only three elements. Conversely, if F is a hyperbolic plane over a field with three elements, then V contains four lines, two isotropic and two non-

and thus two rotations. have v = a' for every rotation; 2 is commutative. To summarize: If dim V = 2, but if V is not a hyperbolic plane over a field with three elements, then 2 is not commutative and 2 contains rotations 2
isotropic ones. There are two symmetries These rotations can only be dbly and we

1

dim 7 = 3. Theorem 3.17 shows that a rotation 5* 1 can not leave two independent vectors fixed. However it is a product of two symmetries and leaves, therefore, all vectors of some line (A) fixed. We call II.

(A) the axis of the rotation. Let ff be an isometry which keeps one non-zero vector A fixed and reverses some other non-zero vector B. The vectors A, B are 2 independent, o- will be a rotation which keeps every vector of 2 = 1 (but a ^ dbl). It (A, B) fixed. By Theorem 3.17 we have ofollows that

cr

is

an involution

7*

1,

hence a symmetry,

if it is

a

and a 180 rotation, if it is a rotation o- is written as product Tir 2 of symmetries, we can r 2) as product of two 180 rotations. also write a- = ( r )( Let us look finally at all rotations with given axis (A) (include We have to also
reflexion,

If

1

.

distinguish

two

cases.

= (A) (A)* and our (A) non-isotropic. We can write V group is obviously isomorphic to the group 2 of (A)*. 1)

CHAPTER 2)

(N)*

A = =

etries a

N = 2

AT,

0.

We

by the

=

have rad (N)*

= 0, (AT, J5) with NB of F for which 0N =

scribed completely

133

III

2

5* 0. 2V.

We

Then

effect it

rad (N} = (N). Hence shall determine all isom-


=

(2V)*

and


is

de-

has on the singular plane (N)* 2 yB and from B* = (er B) =

(Theorem 3.17). We set
The symmetries TX and the rotations which keep N fixed. Obviously
=

cr z

are

now

,

all

isometrics


which shows that the rotations with axis (N) form a group isomorphic to the additive group of 7c. We remark that ar x = (o- x/2 ) 2 which shows that every
Some

of these results are useful

THEOREM

3.21.

Let

dim

V >

squares of rotations consists of Proof: 1)

1^

if

n>

3.

3

The

.

centralizer of the set of

.

Suppose V contains an isotropic = (N) and we can write

line (N).

Then rad (N)* =

rad (N)

W

W

2 > 1. Let U be is non-singular and of dimension n where of dimension n 3 (possibly U = 0) a non-singular subspace of = U* _L U. This space U* is non-singular, of dimension and write V 3, and contains (N). Let cr x be a rotation of U* with axis (N);
W

GEOMETRIC ALGEBRA

134

then If we put a = (rx/2 J. l v the square of the rotation
,

.

B t U and suppose A + B is left fixed by x J_ lu Obviously A must left fixed by vs and this means A t (N). The fixed vectors are, = W a subspace whose radical is (N). therefore, those of (N) _L U
be

Q

Let

now

ro-V"

1

=

commutative with

r be 2
,

all

;

This means

squares of rotations. Then fixed so that rWQ =

rW rW is

1

rcrV" leaves every vector of that the radical r(N} of

W

(AT).

Our

.

r leaves every

= lr V does contain vectors. Let P be any not 2) Suppose isotropic X of P such a We rotation know exists of V. that there plane 2 1P Then a2 P* and let a = X that X j* 1. Write 7 = P 2 2 then we see as leaves only the vectors of P* fixed. If T
and Theorem 3.19 shows r

.

.

.

l

o-

,

A given line is an intersection of two planes > n we see and that r leaves every line of V fixed. From (since 3) Theorem 3.15 we see now r = 1^ The meaning of Theorem 3.21 will become clear a little later. Let now TI and r 2 be symmetries with respect to the non-singular = H\ and hyperplanes H and H 2 Assume again n > 3. Let (A) leaves every plane fixed.

.

.

v

(B)

= H\

.

(A) and (B) are non-singular, the space

P =

(A, B)

(of dimension < 2) not isotropic, its radical of dimension < 1. We have again rad P* = rad P, P* has dimension at least n 2 and 3. The space contains a non-singular subspace U of dimension n P\ 7/2 since U is orthogonal to both A and B. TI and r 2 U f 2 on the three dimensional are identity on U and symmetries f l can write C7*. We non-singular space

CH

i

,

,

r l r2

Let of

^

(

JL

Itf)(r 2

J_ 1^)

by a product of two 180 rotations. r,r 2 r. be any rotation of F, expressed as product s < n symmetries. Since s must be even, we can group the product pairs and replace them by 180 rotations. Thus we have

and have replaced

in

=

now

r r2 v

=

a-

THEOREM

3.22.

even number

< n

//

n

>

3,

then every rotation is a product of

an

of 180 rotations. For our next aim we need a group theoretical lemma.

LEMMA.

Let

G

following properties:

be

a group and

S a

subset of elements with the

CHAPTER HI 1)

G

2)

For x

3)

s*

Then

by 8.

is generated

=

the

e

G

135

and

8 we

s e

have xsx~~

l

e S.

r all s e S.

1

f commutator subgroup G' of

=

818981* 83*

2

(siS 2 )

,

Si e

*S,

G is generated by the commutators G' contains the squares of all elements

$

52

ofG. products of elements of the form obviously a group and property 2)

Proof: Let ff be the set of 1

with

SiS^s^s^

s lt s 2 e S.

H

all

is

shows that it is invariant in G. Let / be the canonical map

G =

G/// with kernel ff. Since = /(^/(siX hence ) 1, /(si)/( 2 ) The set /S generates G so that the image /(G) = G/# is generated by f(S). We see that /(G) is a commutative group which shows is generated that G is contained in the kernel H. Trivially, since has at most G'. of by commutators, Every generator /(G) order 2 so that every element in /(G) has at most order 2. This 2 2 proves f(x ) = 1, x c H. If o- is a symmetry (180 rotation) with respect to the hyperplane 1 (with the plane P) then rcrr" is again a symmetry (180 rotation) 2 with hyperplane rH (with plane rP) and o- = 1. The set of all

s^srW H wehave/^Sz^r ^ 1

1

1

e

>

H

1

H C

H

(the group 0*) symmetries (180 rotations) generates the group and satisfies the conditions of the lemma. 1 For n > 3 the commutator group of 0^ is generated by all (o^oa) where
The commutator group

rotations.

ftn

of

On

2

generated by all (r^a) 2 is a rotation, (r^a) is a

is

r^

where

TI and r 2 are symmetries; since square of a rotation. This shows that 12 n

is not larger than the commutator group of On The meaning of Theorem 3.21 becomes clear. We have determined the centralizer of the group Qn therefore, also + l r obviously commute with the centralizer of n and On (since every element of On ). Thus we have .

,

THEOREM

Let

3.23.

commutator group are symmetries tralizer of

,

l r is the

then

ifcl F is the

on whether

n

>

which

and contains

On Ot

then

12 n

or

On

centet of

is

Qn

all

dblp

On

or not.

On and Ot

The groups

2

0*n and

center of

lv e

3.

is generated

;

.

(r^)

The

center of l v is

the center of the factor

tin

On

a

have the same

where

squares of elements of

ofton (

In

elements have at most order 2.

by

all

On

.

and ra The cen-

TI

.Ifn is odd, reflexion). If n is even

is

groups

l

is

r

t

lv

or l v depending

O n /8n and 0*n/QH

all

GEOMETRIC ALGEBRA

136

Another important property of the group

THEOREM that

3.24.

//

> 3, U 7*

n

no proper subspace

then is

12n is

This means

irreducible.

is

12n

expressed in

into itself by all elements

mapped

Proof: Suppose that there is such a proper subspace shall derive a contradiction and distinguish two cases.

U ^

0.

We

U

= U JL W\ if = T _L p with r

non-singular. Write V and hence
is

U

The

=

involution X

implies that X

Theorem

lu

JL

lw would

in the centralizer of

is

12 n

,

is

,

then,

by

e 0(C7), p c

=

o-X

satisfy

which

Qn

e

o-

Xer

0(W). and this

a contradiction to

3.23.

U is singular. Then rad U 7* is also mapped into itself by &n and, replacing U by rad U, we may assume that U is isotropic. Let N j be a vector of U and AT, M a hyperbolic pair. Since n > 3 we can find a three-dimensional non-singular subspace W = (N, M, A) where A is a non-isotropic vector of (AT, M}*. Isotropic so that U P\ W = (N) subspaces of W have at most dimension a 1 is be a rotation of W with that U Let ^ x (remember isotropic). 2)

1

axis (M). If
M

M


N

The

N

=

and consequently


=

l

w (two vectors are kept fixed). hence into another line of

therefore, moved into a line not contained in U. Setting line

since

o-,

(N)

=

is,

2 (o-x/2 )

,

and

0-./V

W


=


_L

1

w

t

we have

o-

e

12n

^ U.

EXERCISES. +

12n if n = 2. n Investigate the irreducibility of the groups O n the ve be Consider Let the 2) space. corresponding project! 2 = 0. the of What is V defined geometric meaning by hypersurface

1)

,

,

V

X

of the

maximal

isotropic subspaces of

Vt

5. Special features of symplectic

We space

have seen in Theorem 3.7 that every non-singular symplectic is

hyperbolic:

V = H 2r = P (3.37)

geometry

=

l

(N,

JL

,

P2 M,

JL

,

-

-

N M 2

P

JL

,

r

-

2

,

,

N

r

,

M

r)

CHAPTER

=

where each P,

(N<

M

,

t-)

DEFINITION 3.12. We and in the arrangement

G

criminant

a hyperbolic plane. The

is

#,#, = M,M, =

(3.38)

137

111

0,

NMi =

#^,

1,

=

N M t

,

,

for

satisfy i/

5^

/x.

7

a basis of of the type (3.37), (3.38) (3.37) a symplectic basis of V. The dis-

call

of a symplectic basis is 1 as one easily computes.

DEFINITION

The group

3.13.

of isometrics of a non-singular

symplectic space V = H 2r of dimension n =

2r is called the symplectic

group and denoted by Spn (k). We are going to show that Spn (k) contains only rotations and shall give two proofs for this fact. Let A be a given non-zero vector of V. We ask whether Sp n (k) contains an element 0- which moves each vector X e V by some vector of the line (A):

=

X +
We should have cr(X) -

This leads to

(3.39)

Select Fo so that

We

pute
(AY

)

(3.40)

where

^

0, substitute


k

is

Y = YQ

in (3.39)

and com-

satisfies (3.39)

and con-

obtain

=

c-(AX)

a constant; (3.40) certainly

sequently

a(X)

(3.41)

=

X + c-(AX)-A.

We

verify immediately that any into V. Since V morphism of

V

metry. If

and only

c if

=

0,

AX =

plane // = (A)*. another way. Let vector of

H

=

map

of the

form

(3.41) is a

homo-

non-singular, a is indeed an isofixed if If c j 0, then a leaves

is

X to the X must hyperbelong 0, which means that

then

o-

lv

.

We

can characterize these special isometrics in be a given hyperplane. When does a leave every fixed? If t V and Y e H, then

H

X

=


=

GEOMETRIC ALGEBRA

138

X e H*. But //* is a line we suppose
a symplectic transvection in the direction A. For a fixed A our a still depends on c. If we denote it by o- c we find
A is not moved at all assume B ^ A. Any vector orthogonal to B and since A is to be changed we must have A(B A) = AB 7* 0. If A. Since we this is true, then A is moved by a multiple of B still have the c at our disposal we can indeed achieve
gonal, or

if

A =

B.

Suppose now that AB = and EC ^ 0. Indeed, if (A)*

* <>*

(A)*. If (A)* for

C = D

let

D

0.

= t

We

(A)*,

D

other non-zero

plectic transvections.

M

*

= AE ^ C and then C into

+ E we have AC

we can move A into be moved into some

C such that AC ^ C outside the hyperplane

can find a

(B)*, select

E

(B)* and

and

e *,

EG = BD

E

j (A)*;

Now

9* 0.

A non-zero vector A can vector B by at most two symB.

M

and N 2 Let now N 2 be two hyperbolic pairs. By at most two transvections we can move NI into N 2 and these transinto a pair N 2 M<3 vections will move the hyperbolic pair NI We contend that by at most two more transvections we can move into N 2 N 2 3 into N 2 2 and thus have moved NI 2 by have to show transvections. What we still at most four symplectic and N, M' are hyperbolic pairs, amounts to the following: if N, then we can find at most two transvections which keep N fixed '. and move into l

,

i

,

,

j

M

M

M

,

l

,

,

M

l

,

.

M

M

M

M

Case

1).

MM' ^

0.

vection whose direction

NM'

NM

=

1

1

M

Then we can move into M' by one transis M. Notice that N(M M) =

M

=

f

'

0; AT

remains fixed under this transvection

and we are done. Case

2).

bolic pair

MM

and

1

=

0.

M(N +

We first see that N, N +

M

)

=

MN =1^0.

M

is

also a hyper-

Therefore

we can

CHAPTER

move

M into

AT,

N + M. Now we have (N + M)M' = NM' = therefore, move N, N + M into AT, M'. This finishes

and can,

1 5^

AT,

the proof. Let U be a non-singular subspace of

V = U If r is

139

III

a transvection of

so that

7*.

J.

then 1^

[/*,

V

obviously a transvection

_L r is

of F. This shows:

NI

If

move

M

,

2V,

N

and

l

M!

,

2

M M

2

,

into AT 2

,

2

are hyperbolic pairs of [/*, then by at most four transvections of

we can

V

such

U

that each of these transvections keeps every vector of fixed. Let now G be the subgroup of Sp n (k) which is generated by

M

all

M

AT 2 t 2 symplectic transvections. Let a e Sp n (k) and A^ r r be a symplectic basis of V. Denote the images of this basis under a by N[ M{ N'r M'r There is an element 2 2

N

,

,

M

,

,

G

pairs AT,

,

M

Let

M

,

,

,

,

,

into

l

.

,

.

,

t

9

into

v

M{ NI, M'2J C7 = (2V( M( N'2

-

,

,

-

AT:

,

,

M

,

M:

-

2

9

,

.

form a symplectic basis

will

r,

,

,

M

,

N( M{ Suppose we have conG which moves the first i of our hyperbolic the corresponding N' M' The images of our first

symplectic basis under

NI

N

,

which moves NI structed an element r e TI e

,

,

,

,

ATJ

tf ,"+ i ,

,

Affi!

-

,

,

M(), then the pairs

AT r",

A

M ". r

r

J

+,

,

M'i+

,

U* (v > 1). Find a r e G which keeps U and moves the pair N"+i of fixed vector Af.'ii into the every Af element TT will the The move first i + 1 pairs J+i N(+i pair N' This shows into the that there AfJ v M, Af, corresponding pairs is an element of G which moves the whole first basis into the second and we obtain G = Sp n (k). Now we will see that det a = +1 for all elements of Sp n (k). It is indeed enough to show this if a is a transvection. Nothing is to 1 = +1. be proved if the characteristic of k is 2; for, in this case 2 =
ATJi,

M(\ belong v

,

all

to

,

.

,

{

.

,

,

THEOREM

Every element of Sp n (k)

3.25.

determinant +1). The group

Sp n (k)

is

a rotation

(i.e.,

has

is generated by the symplectic

transvections. If

o-

for all

is

a transvection

Xe

V. If r

T*T~

1

t

Sp

^

n (k),

X-X=

1

with direction A, then
then

r(a(r"X)

-

l

(r' X)) e (rA)

X

e

(A)

GEOMETRIC ALGEBRA

140

which shows that

1

ro-r"

a transvection with direction rA.

1 is

5^

Suppose now that r belongs to the center of Spn (k). Then certainly

=

I

T
The T

=

=

for our transvection and, consequently, (A) (rA). fixed and Theorem 3.14 implies element r leaves all lines of cr

V

dbl r

.

THEOREM 3.26. The center of Spn (k) consists of It is of lr order 2 if the characteristic of k is 5^ 2, of order 1 if the characteristic is 2. .

The second proof proceeds

in an entirely different manner. Let be a symplectic space, possibly singular. Take any basis of V and compute its discriminant G. If V is singular, then G = 0; if V is non-singular, then V has a symplectic basis whose discriminant is +1. By formula (3.7) we see that G is the square of an element of k. Lew now (
V

.

:

independent variables field of rational

#,-,

<

i

functions of the

and put x ti symmetric. Hence

XH

(1

to be

=

(3.42)

x it

<

3

<

n)

an(l call k the resulting

with rational coefficients. Define for i > j. The matrix (#,,-) is skew

#,,-

det(*,.,)

where / and g are polynomials in the #,/ (i < j) with integral coThe reader will be familiar with the fact that one has unique factorisation in the ring of polynomials with integral coefficients. We may assume that g is relatively prime to /. The left side of (3.42) is a polynomial with integral coefficients; it follows 2 2 1. that f is divisible by g which implies that g is a unit, g = The identity (3.42) may now be simplified to

efficients.

=

det(*, f )

2

f

and / is determined up to a sign. This is a polynomial identity which must remain true in any field and for any (skew symmetric) special values for the

z

t ,

symplectic basis, then the

.

If

we

specialize the

left side

becomes

1.

x ti to the g it of a may, therefore,

We

CHAPTER fix

the sign of /

for

a symplectic

141

III

by demanding that / should take on the value

1

basis.

THEOREM 3.27. There exists a polynomial called the Pfaffian and denoted by Pf(x ii ), which has integral coefficients and the following property: If (g if ) is skew symmetric, then detfo,,)

=

(P/W-

// the g {i come from a symplectic basis, then Pf(ga) = 1. 2 Adjoin now to Q not only the x ti but n more variables a t/ and put

It is easily seen that (?/,,-) is on both sides and extracting

JY(tf)

where

=

e

1.

when y ti = x {i

THEOREM

a square root we find

=

To determine .

skew symmetric. Taking determinants

This shows

-P/(x,,)-det(a,,) e

specialize the (a tj-) to the unit matrix

= +1.

// (g ti ) is skew symmetric and

3.28.

(a,,-)

any matrix,

then

Suppose that (a,/) is a matrix describing an element of Spn (k) which means that the symplectic space is and suppose P/(gf,-,) ^ = (jr<,) and we get det(a = +1, Then non-singular. (a/ )(5r t/ )(O t

t ,-)

another proof of Theorem 3.25. If V is a non-singular symplectic space of dimension n and any skew symmetric matrix, we can always find vectors (
,

,

,

may be proved as follows: Let F be a space with basis B n Impose on V a symplectic geometry by putting B B! 2 = BfBf g ti This space V may be singular,

This ,

,

,

.

.

7 = V

l

JL

rad

7

.

B = C + D where C< e F D, e rad 7 Then g = B.JS,- = We have g = C C,- where the vectors C come from the C.C/ Imbed now Vi in a non-singular space V non-singular space V Let

t

t

t

t

,

.

it

-

.

{i

t

l

of the right dimension.

.

t

GEOMETRIC ALGEBRA

142

We may 7,,-

=

V

take in

a symplectic basis

E.E, (then P/(y,,)

=

and

1)

,

En

,

put

set

Then

or

Theorem

3.28 shows

=

(3.43)

From (3.43) one can get we formulate as exercises.

all

det(a,,).

the properties of the Pfaffian which

one interchanges the r-th and s-th row in (0,,) and simultaneously the r-th and s-th column, then P/(
1)

t

,

C 12 =

4)

and more generally

c

r.

=

1

p/((*,.,),..,v,..:K-ir-

5) Prove, for instance, the

Pffra) 6)

For n for

7) If

Ai and

n

=

=

=

Si2C 12

<

s.

expansion

+

Xi*C l3

+

-

+

x ln C ln

.

2, 4,

is any matrix of even degree with column vectors one defines a "symplectic product" of column vectors by

(a,,-)

if

r

,

as

/

a 4l

then det(a tj ) can be computed by means of det(a,,)

=

CHAPTER 6.

Geometry over

143

III

finite fields

= Fq

be a field with q elements, F*q the multiplicative group of non-zero elements of F q Squaring the elements of F* is a homomorphism whose kernel is 1. If the characteristic is 2, then +1 = 1, the map is an isomorphism, every element is a

Let k

.

square. If the characteristic is 5^ 2, then the kernel is of order 2, the image group has order (q l)/2 and has index 2 in the whole group F* If g is a special non-square, then every non-square is of .

the form gy

2 .

V be a vector space of dimension n over F with a non-singular orthogonal geometry. We investigate first the lowest values of n. = 1, V = (A}. A may have the form a or the form go, 1) n changing A by the factor a we may assume that either A = 1 or A = g. This gives two distinct geometries: If A = 1, then the 2 of V are squares in F* if A = g, then squares of the vectors j Let

q

2

2

2

;

2

2

2

;

they are non-squares of F* = 2. A plane over F q contains q 2) n = (A, B). Attach to V a sign e if V (B), .

= +1,

put

+ =

1 lines,

V contains no isotropic line put = V is q 6 (there are just 2

if

of non-isotropic lines of e

=

+1). The number

Each

of the g

The

c

total

1.

is

xB) and

hyperbolic

The number

isotropic lines

if

c. of symmetries in 0(V) is, therefore, q + + which is in the coset of (V). (V) 0(7)

These symmetries form It follows that 0*(V) contains also q vectors.

+

(A

V

1; if

^

c

elements.

non-isotropic lines contains q of non-isotropic vectors in

number The q

1

V is,

non-zero therefore,

1 rotations produce q equivalence classes among the non-isotropic vectors of V. An equivalence class consists (by Witt's theorem) precisely of the vectors with equal

1)(<7

(<7

We

e

see that every element of vector of V.

squares.

some

).

F*q must appear as square

of

A be a vector of V with A 2 = 1, B a non-zero vector orthogonal 2 1 to A; changing B by a factor we may assume that either B = 2 2 2 2 2 = = = B x + or B y we -g. Since V (A, B) and (Ax + By) 2 2 1 and not hyperbolic if B = see that V is hyperbolic if B = g. The sign really describes the geometry completely; if c = 1, Let

then x

2

For the (2q

2

gy 1 if

is

total

the quadratic form attached to our geometry. of isotropic vectors in V we find q eq

number

+

V is hyperbolic, only the zero vector if V is not hyperbolic).

GEOMETRIC ALGEBRA

144

>

be any non-singular plane of V and B a nonisotropic vector of P*. We can find in P a vector C such that B ^ but (C C2 = -B 2 Then C Bf = 0. V must contain 3)

n

Let

3.

P

+

+

.

isotropic vectors.

According to Theorem 3.11 we can write

V = H 2r where

H

2r

W W does

_L

not contain non-zero hyperbolic and where We see that we have four possible types of geo-

is

isotropic vectors.

metries

=

(P,-

hyperbolic planes):

n odd

V = V =

(III)

(IV)

(I)

(II)

_L

P

Px

JL

P

7 =

P,

_LP a

P =

Pi

J.

P,

2

2

-

J. -

J-

P

J.

-

(n - 1)/2

(A),

1,

2

P (n - 1)/2

J_

A = A = 2

1

,

0,

n even

where TF

is

P,

P n/2

,

W,

J- PC.-,,,, -L

JL

a plane with

c

=

1.

The quadratic forms attached

to these four types are:

(I)

(II)

2x,x 2

(III)

2x x 2

(IV)

2XiX 2

t

+ + +

2x 3 x 4 2x 3 x 4 2x 3 x 4

+ + +

-

-

+ + +

1

2xn _ 3 ^- 2 2xn _ 3 x_ 2 (n "

(n - 1}/2

The

2x n _ 2 x n _

+ + +

gxl

,

2xn _!X n xl,,

-

u/

,

gxl

n/2

n/

V

It discriminants are (-J) (-l) V, (-l) (-l) n with the character worth that is while noticing together quadratic of the discriminant determines the type of geometry. It will also ,

,

where PI be important to remember that if V = PI _L Vi V of the same type. are hyperbolic plane, then V and ,

is

a

l

The

between type I and type II is inessential. If we multiply the quadratic form of type I by g it becomes equivalent to the form of type II. The difference between type III and type IV is very great; a maximal isotropic subspace is of dimension n/2 for 1 for type IV. type III and of dimension n/2 difference

CHAPTER

We

associate again a sign

e

145

III

= +1

with type III and

e

=

I

with type IV.

number of isotropic vectors of V for some n > 3 and N, is a hyperbolic pair of V, 2 and is of the same type as V; then (A/", M)* has dimension n (N, M)* contains, therefore, ^n _ 2 isotropic vectors. The space (A^) J_ (N, M}* has dimension n 1 and is orthogonal Denote now by

the