9

3-dimensional subspace is contained in a unique member of . Then the points,

9

lines, planes, and subspaces in form a geometry with the diagram

L L

©

where the ¬rst L denotes the points and lines in 3-dimensional projective space

over F.

5.2. Some special diagrams 73

Now we turn to af¬ne spaces, where similar results hold. The label A f on a

stroke will denote the class of af¬ne planes.

Theorem 5.8 A geometry with diagram

Af

D'

is an af¬ne space of ¬nite dimension.

Proof It is a linear space whose planes are af¬ne (that is, satisfying condition

(AS1) of Section 11). We must show that parallelism is transitive. So suppose

that L1 L2 L3 , but L1 L3 . Then all three lines lie in a subspace of dimension

GG IGH

3; so it is enough to deduce a contradiction in the case of geometries of rank 3.

Note that, for a geometry with diagram A f , two planes which have a

common point must meet in a line.

Let Π1 be the plane through L1 and L2 , and Π2 the plane through p and L3 ,

where p is a point of L1 . Then Π1 and Π2 both contain p, so they meet in a line

M L1 . Then M is not parallel to L2 , so meets it in a point q, But then Π2 contains

H¡

L3 and q, hence L2 , and so is equal to Π1 , a contradiction.

The fact that all linear subspaces are shadows of varieties is proved as in The-

orem 5.6.

Theorem 5.9 A geometry with diagram

Af L

in which some line has more than three points, consists of the points, lines and

planes of a (possibly in¬nite-dimensional) af¬ne space.

The proof is as for Theorem 5.7, using Buekenhout™s Theorem 3.10.

Exercises

1. Consider a geometry of rank n with diagram

L

D)

©

in which all lines have the same ¬nite cardinality k, and all the projective planes

have the same ¬nite order q.

74 5. Buekenhout geometries

( ¡

(a) If n 4, prove that the geometry is either projective (q k 1) or af¬ne

(q k).

¡ ¢

(b) If n 3, prove that q k 1© k© k2 or k k2 1£ .

¡

¡

(This result is due to Doyen and Hubaut [16]).

2. Construct an in¬nite “free-like” geometry with diagram

c

(Ensure that three points lie in a unique plane, while two planes meet in two

points.)

c Af .

3. (a) Show that an inversive plane belongs to the diagram

What are the varieties?

(b) Show how to construct a geometry with diagram

c Af

£¢

(n nodes) from an ovoid in PG n© F (see Section 4.4).

6

Polar spaces

Now we begin on our second major theme, polar spaces. This chapter corresponds

to the ¬rst half of Chapter 1, and gives the algebraic description of polar spaces.

The algebraic background required is more elaborate (vector spaces with forms,

rather than just vector spaces), accounting for the increased length. The ¬rst sec-

tion, on polarities of projective spaces, provides motivation for the introduction of

the (Hermitian and quadratic) forms.

6.1 Dualities and polarities

Recall that the dual V of a ¬nite-dimensional (left) vector space V over a

skew ¬eld F can be regarded as a left vector space of the same dimension over

¡

the opposite ¬eld F , and there is thus an inclusion-reversing bijection between

¢ ¢ ¡ ¡

the projective spaces PG n£ F and PG n£ F . If it happens that F and F are iso-

¤ ¤¢

morphic, then there exists a duality of PG n£ F , an inclusion-reversing bijection ¤

¢

of PG n£ F . ¤

¢

Conversely, if PG n£ F admits a duality (for n 1), then F is isomorphic to

¤ ¥

¡

F , as follows from the FTPG (see Section 1.3). We will examine this conclusion

and make it more detailed.

¢

So let π be a duality of PG n£ F , n 1. Composing π with the natural isomor- ¤¡ ¥

¢ ¢

phism from PG n£ F to PG n£ F , we obtain an inclusion-preserving bijection θ

¢¤ ¤

¢ ¡

from PG n£ F to PG n£ F . According to the FTPG, θ is induced by a semilinear

¤ ¤

transformation T from V F n§ 1 to its dual space V , associated with an isomor-

¦

¡

phism σ : F F , which can be regarded as being an anti-automorphism of F:

¨

75

76 6. Polar spaces

that is,

¢

v1 v2 T v1 T v2 T

¤ ¦ £

¢© ©

±σ vT

±v¤ T ¦

De¬ne a function b : V V F by the rule

¨

¢ ¢ ¢

b v£ w¤ v¤ wT

¦ ¤

£

that is, the result of applying the element wT of V to v. Then b is a sesquilinear

form: it is linear as a function of the ¬rst argument, and semilinear as a function

of the second ” this means that

¢ ¢ ¢

b v£ w1 w2 b v£ w1 b v£ w2

¤

¦ ©¤ ¤

©

and ¢ ¢

±σ b v£ w¤

b v£ ±w¤

¦

(The pre¬x “sesqui-” means “one-and-a-half”.) If we need to emphasise the anti-

automorphism σ, we say that b is σ-sesquilinear. If σ is the identity, then the form

is bilinear.

The form b is also non-degenerate, in the sense that

¢

¢¢

w V b v£ w¤ 0 v 0

¤ ¦ ¦

and ¢

¢¢

v V b v£ w¤ 0 w 0

¤ ¦ ¦

(The second condition asserts that T is one-to-one, so that if w 0 then wT is !¦

a non-zero functional. The ¬rst asserts that T is onto: only the zero vector is

annihilated by every functional in the dual space.)

So, we have:

¢

Theorem 6.1 Any duality of PG n£ F , for n 1, is induced by a non-degenerate ¤ ¥

σ-sesquilinear form on the underlying vector space, where σ is an anti-automorphism

of F.

Conversely, any non-degenerate sesquilinear form on V induces a duality. We

can short-circuit the passage to the dual space, and write the duality as

¢

U U v V : b v£ w¤ 0 for all w U

"¨ &¦$#

%

¦ ('

6.1. Dualities and polarities 77

Obviously, a duality applied twice is a collineation. The most important types ¢

of dualities are those whose square is the identity. A polarity of PG n£ F is a ¤

¢

duality which satis¬es U U for all ¬‚ats U of PG n£ F .

) ¦ ## ¤

It is a bit dif¬cult to motivate the detailed study of polarities at this stage; but

it will turn out that they give rise to a class of geometries (the polar spaces) with

properties similar to those of projective spaces. To put it somewhat vaguely, we

are trying to add some extra structure to a projective space; if a duality is not

a polarity, then its square is a non-identity collineation, and some of the extra

structure arises from this collineation. Only in the case of a polarity is the extra

structure “primitive”. ¢ ¢

A sesquilinear form b is re¬‚exive if b v£ w¤ 0 implies b w£ v¤ 0.

¦

¦

Proposition 6.2 A duality is a polarity if and only if the sesquilinear form de¬ning

it is re¬‚exive.

Proof b is re¬‚exive if and only if

v w2 w v2

4#3 1

0

53 1

#0

Hence, if b is re¬‚exive, then U U for all subspaces U . But by non-degeneracy,

6 ##

dimU dimV dimU dimU ; and so U U for all U . Conversely,

¦ 7# 8 ¦# ¦

# 7#

#

given a polarity , if w v2 , then v v2 w2 (since inclusions are

) # 9

0 0@ A6 7#

0 # #

reversed).

We now turn to the classi¬cation of re¬‚exive forms. For convenience, from

now on F will always be assumed to be commutative. (Note that, if the anti-

automorphism σ is an automorphism, and in particular if σ is the identity, then F

is automatically commutative.) ¢ ¢

b v£ w¤ σ for all v£ w V .

The form b is said to be σ-Hermitian if b w£ v¤ B

¦

¢

This implies that, for any v, b v£ v¤ lies in the ¬xed ¬eld of σ. If σ is the identity,

such a form (which is bilinear) is called symmetric. ¢

A bilinear form b is called alternating if b v£ v¤ 0 for all v V . This implies

¢7

¦

¢ ¢

that b w£ v¤ b v£ w¤ for all v£ w V . (Expand b v w£ v w¤ 0, and note

DC

8¦ E

¦

© ©

that two of the four terms are zero.) Hence, if the characteristic is 2, then any

alternating form is symmetric (but not conversely); but, in characteristic different

from 2, only the zero form is both symmetric and alternating.

Clearly, an alternating or Hermitian form is re¬‚exive. Conversely, we have the

following:

78 6. Polar spaces

Theorem 6.3 A non-degenerate re¬‚exive σ-sesquilinear form is either alternat-

ing, or a scalar multiple of a σ-Hermitian form. In the latter case, if σ is the

identity, then the scalar can be taken to be 1.

I will not give the complete proof of this theorem. The next result shows that

σ2 1, and then the proof of the theorem is given in the case of a bilinear form

¦

(that is, when σ 1). ¦

Proposition 6.4 If b is a non-zero re¬‚exive σ-sesquilinear form, then σ2 is the

identity.

Proof Note ¬rst that a form is σ-sesquilinear if and only if it is additive in each

variable and satis¬es

¢ ¢ ¢ ¢

b v£ w¤ βσ

b ±v£ w¤ ±b v£ w¤ b v£ βw¤

¦ £F

¦

¢

Step 1 If b is alternating, then σ 1. For we can choose v and w with b v£ w¤

¦ G

¦

¢

1. Then for any ± F, we have

b w£ v¤

8

¦

¢

± ±b v£ w¤

¦

¢

b ±v£ w¤

¦

¢

b w£ ±v¤

¦ 8

¢

b w£ v¤ ±σ

¦ 8

±σ

¦

(Note that this step does not require non-degeneracy, merely that b is not identi-

cally zero.)

¢

So we can assume that there exists v with b v£ v¤ 0. Multiplying b by a non- H

!¦ ¢

zero scalar (this does not affect the hypotheses), we may assume that b v£ v¤ 1.

¦

¢

Step 2 Assume for a contradiction that σ2 1. For any vector w, if b w£ v¤ 0, !¦ I

!¦

¢

then we can replace w by its product with a non-zero scalar to assume b w£ v¤ 1. G

¦

¢ ¢ ¢

Then b w v£ v¤ 0, and so b v£ w v¤ 0, whence b v£ w¤ 1. We claim that

8 ¦ 8

¦

¦

¢

b w£ w¤ 1.

¦

6.1. Dualities and polarities 79

¢ ¢

Proof Suppose that ± b w£ w¤ 1. Note ¬rst that b w ±v£ v¤ 0, and ¦ P

!¦ 8 E

¦

¢

0, whence ± ±σ . Take any element » F with » 1, and

so b w£ w ±v¤ 8 C

¦ ¦ !¦

¢ ¢

choose µ F such that µσ 1 »¤ 1 ± »¤ . Since ± 1, we have µ 1; and

¦ 8 R

Q 8 !¦ !¦

µσ »µσ ± »

8 ¦ 8

¢ ¢

µσ 1 µσ

This implies, ¬rst, that » ± 1, and second that

¦ 8 ¤ 8 R¤

Q

¢

µσ »µσ

»v£ w ± »

bw µv¤ 0

8 8 ¦ 8 8 ¦

©

¢

»v¤

Hence b w µv£ w 0, and we obtain

8 8

¦

»σ µ»σ

± µ 0

8 8 ¦

©

Applying σ to this equation and using the fact that ±σ ±, we obtain

¦

µσ »σ » σ µσ

2 2

± 0£

8 8 ¦

©

whence ¢ ¢

»σ µσ 1 µσ

2

± »

1

¦ Q¤

¦ 8 ¤ 8

But » was an arbitrary element different from 1. Since clearly 1σ 1, we have

¦

σ2 1, contrary to assumption.

¦

¢

Step 3 Let W v . Then V v2 W , and rk W 1. For any x W , we

¢¦ TU S¦

0 W¤

V

#

¢ ¢

have b v£ v¤ b v x£ v¤ 1, and so by Step 2, we have b v x£ v x¤ 1.

H

¦ ¦X X

¦

© © ©

¢

Thus b x£ x¤ 2. Putting x 0, we see that F must have characteristic 2, and

8Y¦C ¦

that b W is alternating. But then Step 1 shows that b W is identically zero, whence

` `

W is contained in the radical of b, contrary to the assumed non-degeneracy.

Proof of Theorem 6.3 We have

¢ ¢ ¢ ¢

b u£ v¤ b u£ w¤ b u£ w¤ b u£ v¤ 0

G

8 7

¦

by commutativity; that is, using bilinearity,

¢ ¢ ¢

b u£ b u£ v¤ w b u£ w¤ v¤ 0

8 ¦

By re¬‚exivity, ¢¢ ¢

b b u£ v¤ w b u£ w¤ v£ u¤ 0£

8 ¦

80 6. Polar spaces

whence bilinearity again gives

¢ ¢ ¢ ¢

b u£ v¤ b w£ u¤ b u£ w¤ b v£ u¤ (6.1)

¦

¢ ¢

Call a vector u good if b u£ v¤ b v£ u¤ 0 for some v. By (6.1), if u is good,

¦ !I

¦

¢ ¢ ¢

then b u£ w¤ b w£ u¤ for all w. Also, if u is good and b u£ v¤ 0, then v is good.

7

¦ !X

¦ ¢

But, given any two non-zero vectors u1 u2 , there exists v with b ui v¤ 0 for £ a£

!¦

¢

i 1£ 2. (For there exist v1 v2 with b ui vi 0 for i 1£ 2, by non-degeneracy;

¦ £ P¤ £

!¦ ¦

and at least one of v1 v2 v1 v2 has the required property.) So, if some vector is

£ £ ©

good, then every non-zero vector is good, and b is symmetric.

But, putting u w in (6.1) gives

¦

¢ ¢¢ ¢

b u£ u¤ b u£ v¤ b v£ u¤ 0

G

8 ¦3

¤

¢

for all u£ v. So, if u is not good, then b u£ u¤ 0; and, if no vector is good, then b

¦

is alternating.

In the next few sections, we develop this theme further.

Exercises

1. Let b be a sesquilinear form on V . De¬ne the left and right radicals of b to

be the subsets b¢

¢

v V : w V b v£ w¤ 0'

% ¤ ¦

and b¢

¢

v V: w V b w£ v¤ 0'

% ¤ ¦

respectively. Prove that the left and right radicals are subspaces of the same rank

(if V has ¬nite rank).

(Note: If the left and right radicals are equal, this subspace is called the radical

of b. This holds if b is re¬‚exive.)

2. Give an example of a bilinear form on an in¬nite-rank vector space whose

left radical is zero and whose right radical is non-zero.

3. Let σ be a (non-identity) automorphism of F of order 2. Let E be the

¢

sub¬eld Fix σ¤ .

(a) Prove that F is of degree 2 over E, i.e., a rank 2 E-vector space.

[See any textbook on Galois theory. Alternately, argue as follows: Take » ¢

¢

F E. Then » is quadratic over E, so E »¤ has degree 2 over E. Now E »¤

c

contains an element ω such that ωσ ω (if the characteristic is not 2) or ωσ

8Y¦ ¦

6.2. Hermitian and quadratic forms 81

ω 1 (if the characteristic is 2). Now, given two such elements, their quotient or

©

difference respectively is ¬xed by σ, so lies in E.]

(b) Prove that

F : »»σ µe µσ : µ

» 1' F

% ¦ &d

%¦ ('

[The left-hand set clearly contains the right. For the reverse inclusion, separate

into cases according as the characteristic is 2 or not. ¢

If the characteristic is not 2, then we can take F E ω¤ , where ω2 ± E ¦ ¦

and ωσ ω. If » 1, then take µ 1; otherwise, if » a bω, take µ

f¦ ¢

8 ¦ ¦ ¦ ¦

©

b± a 1¤ ω. 8 © ¢

If the characteristic is 2, show that we can take F E ω¤ , where ω2 ω ± ¦ ¦

© ©

0, ± E, and ωσ ω 1. Again, if » 1, set µ 1; else, if » a bω, take

¢ ¦ ¦ ¦ ¦

© ©

µ a 1¤ bω.]

¦ © ©

4. Use the result of Exercise 3 to complete the proof of Theorem 6.3 in general.

¢

[If b u£ u¤ 0 for all u, the form b is alternating and bilinear. If not, suppose

g

¦

¢ ¢ ¢

0 and let b u£ u¤ σ »b u£ u¤ . Choosing µ as in Exercise 2 and re-

that b u£ u¤ h

!¦ ¦

normalising b, show that we may assume that » 1, and (with this choice) that b ¦

is Hermitian.]

6.2 Hermitian and quadratic forms

We now change ground slightly from the last section. On the one hand, we

restrict things by excluding some bilinear forms from the discussion; on the other,

we introduce quadratic forms. The loss and gain exactly balance if the character-

istic is not 2; but, in characteristic 2, we make a net gain.

Let σ be an automorphism of the commutative ¬eld F, of order dividing 2. Let

¢ ¢

» F : »σ »' be the ¬xed ¬eld of σ, and Tr σ¤ » »σ : » F

Fix σ¤ iC

%¦ ¦ © ¢ ig ¢

%¦ '

the trace of σ. Since σ2 is the identity, it is clear that Fix σ¤ Tr σ¤ . Moreover, B

p

¢

if σ is the identity, then Fix σ¤ F, and

¦

0 if F has characteristic 2,

¢

Tr σ¤ r

q¦

F otherwise.

¢

Let b be a σ-Hermitian form. We observed in the last section that b v£ v¤ I

¢ ¢ ¢

Fix σ¤ for all v V . We call the form b trace-valued if b v£ v¤ Tr σ¤ for all

I

v V.

¢ ¢

Proposition 6.5 We have Tr σ¤ Fix σ¤ unless the characteristic of F is 2 and

C

¦

σ is the identity.

82 6. Polar spaces

¢ ¢

Proof E Fix σ¤ is a ¬eld, and K Tr σ¤ is an E-vector space contained in E

¦ ¦

(Exercise 1). So, if K E, then K 0, and σ is the map x x. But, since σ is

!¦ ¦ "¨ 8

a ¬eld automorphism, this implies that the characteristic is 2 and σ is the identity.

Thus, in characteristic 2, symmetric bilinear forms which are not alternat-

ing are not trace-valued; but this is the only obstruction. We introduce quadratic

forms to repair this damage. But, of course, quadratic forms can be de¬ned in any

characteristic. However, we note at this point that Proposition 6.5 depends in a

crucial way on the commutativity of F; this leaves open the possibility of addi-

tional types of polar spaces de¬ned by so-called pseudoquadratic forms. These

will be discussed brie¬‚y in Section 7.6.

Let V be a vector space over F. A quadratic form on V is a function f : V F ¨

satisfying

s

¢ ¢

f »v¤ »2 f v¤ for all » F, v V;

¦

s

¢ ¢ ¢ ¢

fv w¤ f v¤ f w¤ b v£ w¤ , where b is bilinear.

¦

© © ©

Now, if the characteristic of F is not 2, then b is a symmetric bilinear form.

Each of f and b determines the other, by

¢ ¢ ¢ ¢

b v£ w¤ fv w¤ f v¤ f w¤

¦ G

8 G

8

©

and ¢ ¢

1

f v¤ 2b v£ v¤

¦ £F

the latter equation coming from the substitution v w in the second de¬ning ¦

condition. So nothing new is obtained.

On the other hand, if the characteristic of F is 2, then b is an alternating bilinear

form, and f cannot be recovered from b. Indeed, many different quadratic forms

correspond to the same bilinear form. (Note that the quadratic form does give

extra structure to the vector space; we™ll see that this structure is geometrically

similar to that provided by an alternating or Hermitian form.)

We say that the bilinear form is obtained by polarisation of f .

Now let b be a symmetric bilinear form over a ¬eld of characteristic 2, which

¢ ¢

is not alternating. Set f v¤ b v£ v¤ . Then we have

¦

¢ ¢

f »v¤ »2 f v¤

¦

and ¢ ¢ ¢

fv w¤ f v¤ f w¤

¦ F

£

© ©

6.2. Hermitian and quadratic forms 83

¢ ¢

since b v£ w¤ b w£ v¤ 0. Thus f is “almost” a semilinear form; the map » »2

t

¦ "¨

©

is a homomorphism of the ¬eld F with kernel 0, but it may fail to be an automor-

phism. But in any case, the kernel of f is a subspace of V , and the restriction of

b to this subspace is an alternating bilinear form. So again, in the spirit of the

vague comment motivating the study of polarities in the last section, the structure

provided by the form b is not “primitive”. For this reason, we do not consider

symmetric bilinear forms in characteristic 2 at all. However, as indicated above,

we will consider quadratic forms in characteristic 2.

Now, in characteristic different from 2, we can take either quadratic forms or

symmetric bilinear forms, since the structural content is the same. For consistency,

we will take quadratic forms in this case too. This leaves us with three “types” of

forms to study: alternating bilinear forms; σ-Hermitian forms where σ is not the

identity; and quadratic forms.

We have to de¬ne the analogue of non-degeneracy for quadratic forms. Of

course, we could require that the bilinear form obtained by polarisation is non-

degenerate; but this is too restrictive. We say that a quadratic form f is non-

singular if

¢¢ ¢

¢

f v¤ 0& w V b v£ w¤ 0¤ v 0

¦ ¤

¦ ¦

where b is the associated bilinear form; that is, if the form f is non-zero on every

non-zero vector of the radical.

If the characteristic is not 2, then non-singularity is equivalent to non-degeneracy

of the bilinear form.

Now suppose that the characteristic is 2, and let W be the radical. Then b is

identically zero on W ; so the restriction of f to W satis¬es

¢ ¢ ¢

f v w¤ f v¤ f w¤

u

¦

£

¢© ©¢

f »v¤ »2 f v¤

v

¦ F

As above, f is very nearly semilinear. The ¬eld F is called perfect if every element

is a square. In this case, f is indeed semilinear, and its kernel is a hyperplane of

W . We conclude:

Theorem 6.6 Let f be a non-singular quadratic form, which polarises to b, over

a ¬eld F.

(a) If the characteristic of F is not 2, then b is non-degenerate.

(b) If F is a perfect ¬eld of characteristic 2, then the radical of b has rank at

most 1.

84 6. Polar spaces

Exercises

1. Let σ be an automorphism of a commutative ¬eld F such that σ2 is the

identity. ¢

(a) Prove that Fix σ¤ is a sub¬eld of F.

¢

(b) Prove that Tr σ¤ is closed under addition, and under multiplication by

¢

elements of Fix σ¤ .

2. Let b be an alternating bilinear form on a vector space V over a ¬eld F of

¢

characteristic 2. Let vi : i I be a basis for V , and q any function from I to F.

¤ ¢ ¢

Show that there is a unique quadratic form with the properties that f vi q i¤ 5¤

¦

for every i I, and f polarises to b.

3. (a) Construct an imperfect ¬eld of characteristic 2.

(b) Construct a non-singular quadratic form with the property that the radical

of the associated bilinear form has rank greater than 1.

4. Show that ¬nite ¬elds of characteristic 2 are perfect. (Hint: the multiplica-

tive group is cyclic of odd order.)

6.3 Classi¬cation of forms

As explained in the last section, we now consider a vector space V of ¬nite

rank equipped with a form of one of the following types: a non-degenerate alter-

nating bilinear form b; a non-degenerate σ-Hermitian form b, where σ is not the

identity; or a non-singular quadratic form f . In the third case, we let b be the bi-

linear form obtained by polarising f ; then b is alternating or symmetric according

as the characteristic is or is not 2, but b may be degenerate. In the other two cases,

¢ ¢

we de¬ne a function f : V F de¬ned by f v¤ b v£ v¤ ” this is identically

¨ H

¦

zero if b is alternating. See Exercise 1 for the Hermitian case.

¢

We say that V is anisotropic if f v¤ 0 for all v 0. Also, V is a hyperbolic

w

!¦ ¢ !¦

¢ ¢

line if it is spanned by vectors v and w with f v¤ f w¤ 0 and b v£ w¤ 1.

5

¦ 5

¦ E

¦

(The vectors v and w are linearly independent, so V has rank 2; so, projectively, it

is a “line”.)

Theorem 6.7 A space carrying a form of one of the above types is the direct sum

of a number r of hyperbolic lines and an anisotropic space U . The number r and

the isomorphism type of U are invariants of V .

Proof If V is anisotropic, then there is nothing to prove. (V cannot contain a ¢

hyperbolic line.) So suppose that V contains a vector v 0 with f v¤ 0. !¦

¦

6.3. Classi¬cation of forms 85

¢

We claim that there is a vector w with b v£ w¤ 0. In the alternating and x

!¦

Hermitian cases, this follows immediately from the non-degeneracy of the form.

In the quadratic case, if no such vector exists, then v is in the radical of b; but v is

a singular vector, contradicting the non-singularity of f .

¢

Multiplying w by a non-zero constant, we may assume that b v£ w¤ 1.

¦

¢

Now, for any value of », we have b v£ w »v¤ 1. We wish to choose » so

8

¦

¢

that f w »v¤ 0; then v and w will span a hyperbolic line. Now we distinguish

8

¦

cases. If b is alternating, then any value of » works. If b is Hermitian, we have

¢ ¢ ¢ ¢ ¢

»b v£ w¤ »σb w£ v¤ »»σ f v¤

»v¤

fw f w¤

8 v

¦ y

8 y

8 ©

¢

¢

» »σ ;

f w¤

¦ 8y ¤

©

¢ ¢

and, since b is trace-valued, there exists » with Tr »¤ f w¤ . Finally, if f is

H

¦

quadratic, we have

¢ ¢ ¢ ¢

»v¤ »b w£ v¤ »2 f v¤

fw f w¤

8 v ¦ y

8 ©

¢

»£

f w¤

¦ 8y

¢

so we choose » f w¤ .

¦

Now let W1 be the hyperbolic line v£ w »v2 , and let V1 W1 , where orthog-

0 8 ¦ #

onality is de¬ned with respect to the form b. It is easily checked that V V1 W1 , ¦ T

and the restriction of the form to V1 is still non-degenerate or non-singular, as

appropriate. Now the existence of the decomposition follows by induction.

I will omit the proof of uniqueness.

The number r of hyperbolic lines is called the polar rank or Witt index of V . I

do not know of a commonly accepted term for U ; I will call it the germ of V , for

reasons which will become clear shortly.

To complete the classi¬cation of forms over a given ¬eld, it is necessary to

determine all the anisotropic spaces. In general, this is not possible; for exam-

ple, the study of positive de¬nite quadratic forms over the rational numbers leads

quickly into deep number-theoretic waters. I will consider the cases of the real

and complex numbers and ¬nite ¬elds.

First, though, the alternating case is trivial:

Proposition 6.8 The only anisotropic space carrying an alternating bilinear form

is the zero space.

86 6. Polar spaces

In combination with Theorem 6.7, this shows that a space carrying a non-

degenerate alternating bilinear form is a direct sum of hyperbolic lines.

Over the real numbers, Sylvester™s theorem asserts that any quadratic form in

n variables is equivalent to the form

x2 x2 x2 x2

© ©

8 R18

8 £

1 r r§ 1 r§ s

for some r£ s with r s n. If the form is non-singular, then r s n. If both r

¦

© ©

and s are non-zero, there is a non-zero singular vector (with 1 in positions 1 and

r 1, 0 elsewhere). So we have:

©

Proposition 6.9 If V is a real vector space of rank n, then an anisotropic form

on V is either positive de¬nite or negative de¬nite; there is a unique form of each

type up to invertible linear transformation, one the negative of the other.

The reals have no non-identity automorphisms, so Hermitian forms do not

arise.

Over the complex numbers, the following facts are easily shown:

(a) There is a unique non-singular quadratic form (up to equivalence) in n

variables for any n. A space carrying such a form is anisotropic if and only if

n 1.

(b) If σ denotes complex conjugation, the situation for σ-Hermitian forms is

the same as for quadratic forms over the reals: anisotropic forms are positive or

negative de¬nite, and there is a unique form of each type, one the negative of the

other.

For ¬nite ¬elds, the position is as follows.

¢

Theorem 6.10 (a) An anisotropic quadratic form in n variables over GF q¤ exists

if and only if n 2. There is a unique form for each n except when n 1 and q is

¦

odd, in which case there are two forms, one a non-square multiple of the other.

(b) Let q r2 and let σ be the ¬eld automorphism ± ±r . Then there is

¦ "¨

an anisotropic σ-Hermitian form in n variables if and only if n 1. The form is

unique in each case.

Proof (a) Consider ¬rst the case where the characteristic is not 2. The multiplica-

¢

tive group of GF q¤ is cyclic of even order q 1; so the squares form a subgroup

8

of index 2, and if · is a ¬xed non-square, then every non-square has the form ·±2

for some ±. It follows easily that any quadratic form in one variable is equivalent

to either x2 or ·x2 .

6.3. Classi¬cation of forms 87

Next, consider non-singular forms in two variables. By completing the square,

such a form is equivalent to one of x2 y2 , x2 ·y2 , ·x2 ·y2 . © © ©

¢

Suppose ¬rst that q 1 mod 4¤ . Then 1 is a square, say 1 β2 . (In

‚ 8 8 ¦

the multiplicative group, 1 has order 2, so lies in the subgroup of even order

8

¢ ¢ ¢

x βy¤ x βy¤ , and the ¬rst and

1 2 y2

2 q 1¤ consisting of squares.) Thus x

8 ¦ 8

© ©

third forms are not anisotropic. Moreover, any form in 3 or more variables, when

converted to diagonal form, contains one of these two, and so is not anisotropic

either. ¢

Now consider the other case, q 1 mod 4¤ . Then 1 is a non-square f‚

8 8 ¢ ¢

(since the group of squares has odd order), so the second form is x y¤ x y¤ , 8

©

and is not anisotropic. Moreover, the set of squares is not closed under addition ¢

1

(else it would be a subgroup of the additive group, but 2 q 1¤ doesn™t divide q); ©

so there exist two squares whose sum is a non-square. Multiplying by a suitable

square, there exist β£ γ with β2 γ2 1. Then ¦

8

©

¢ ¢ ¢

βx γy¤ γx βy¤

x2 y2 2 2

8 C¤

¦ 8 £

© © ©

and the ¬rst and third forms are equivalent. Moreover, a form in three variables

is certainly not anisotropic unless it is equivalent to x2 y2 z2 , and this form © ©

¢

vanishes at the vector β£ γ£ 1¤ ; hence there is no anisotropic form in three or more

variables.

The characteristic 2 case is an exercise (see Exercise 3). ¢

(b) Now consider Hermitian forms. If σ is an automorphism of GF q¤ of order

2, then q is a square, say q r2 , and ±σ ±r . We need the fact that every element

¦ ¦

¢ ¢

GF r¤ has the form ±±σ (see Exercise 1 of Section 6.2).

of Fix σ¤

¦ ¢ ¢

µxxσ for some non-zero µ Fix σ¤ ; writing

In one variable, we have f x¤ ¦E

µ ±±σ and replacing x by ±x, we can assume that µ 1.

¦ ¦

In two variables, we can similarly take the form to be xxσ yyσ . Now 1 8

©

¢ ¢

Fix σ¤ , so 1 »»σ; then the form vanishes at 1£ »¤ . It follows that there is no

8 ¦

anisotropic form in any larger number of variables either.

Exercises

1. Let b be a σ-Hermitian form on a vector space V over F, where σ is not the

¢ ¢ ¢

b v£ v¤ . Let E Fix σ¤ , and let V be V regarded as an E-

identity. Set f v¤ B

¦ ¦ „

vector space by restricting scalars. Prove that f is a quadratic form on V , which ¢„

¢ ¢ ¢ ¢

b v£ w¤ b v£ w¤ σ.

polarises to the bilinear form Tr b¤ de¬ned by Tr b¤ v£ w¤

¦ ©

¢

Show further that Tr b¤ is non-degenerate if and only if b is.

88 6. Polar spaces

2. Prove that there is, up to equivalence, a unique non-degenerate alternating

bilinear form on a vector space of countably in¬nite dimension (a direct sum of

countably many isotropic lines).

3. Let F be a ¬nite ¬eld of characteristic 2.

(a) Prove that every element of F has a unique square root.

(b) By considering the bilinear form obtained by polarisation, prove that a

non-singular form in 2 or 3 variables over F is equivalent to ±x2 xy βy2 or

© ©

±x2 xy βy2 γz2 respectively. Prove that forms of the ¬rst shape (with ±£ β 0) !¦

© © ©

are all equivalent, while those of the second shape cannot be anisotropic.

6.4 Classical polar spaces

Polar spaces describe the geometry of vector spaces carrying a re¬‚exive sesquilin-

ear form or a quadratic form in much the same way as projective spaces describe

the geometry of vector spaces. We now embark on the study of these geometries;

the three preceding sections contain the prerequisite algebra.

First, some terminology. The polar spaces associated with the three types of

forms (alternating bilinear, Hermitian, and quadratic) are referred to by the same

names as the groups associated with them: symplectic, unitary, and orthogonal

respectively. Of what do these spaces consist?

Let V be a vector space carrying a form of one of our three types. Recall that

as well as a sesquilinear form b in two variables, we have a form f in one variable

¢ ¢

” either f is de¬ned by f v¤ b v£ v¤ , or b is obtained by polarising f ” and

g

¦

we make use of both forms. A subspace of V on which b vanishes identically is

called a totally isotropic subspace (or t.i. subspace), while a subspace on which f

vanishes identically is called a totally singular subspace (or t.s. subspace). Every

t.s. subspace is t.i., but the converse is false. In the case of alternating forms, every

subspace is t.s.! I frequently use the expression t.i. or t.s. subspace, to mean a t.i.

subspace (in the symplectic or unitary case) or a t.s. subspace (in the orthogonal

case).

The classical polar space (or simply the polar space) associated with a vector

space carrying a form is the geometry whose ¬‚ats are the t.i. or t.s. subspaces (in

the above sense). (Concerning the terminology: the term “polar space” is normally

reserved for a geometry satisfying the axioms of Tits, which we will meet shortly.

But every classical polar space is a polar space, so the terminology here should

cause no confusion.) Note that, if the form is anisotropic, then the only member

of the polar space is the zero subspace. The polar rank of a classical polar space is

6.4. Classical polar spaces 89

the largest vector space rank of any t.i. or t.s. subspace; it is zero if and only if the

form is anisotropic. Where there is no confusion, polar rank will be called simply

rank. (We will soon see that there is no con¬‚ict with our earlier de¬nition of polar

rank as the number of hyperbolic lines in the decomposition of the space.) We use

the terms point, line, plane, etc., just as for projective spaces.

We now proceed to derive some properties of polar spaces. Let G be a classical

polar space of polar rank r.

First, we identify the two de¬nitions of polar space rank. We use the expres-

sion for V as the direct sum of r hyperbolic lines and an anisotropic subspace

given by Theorem 6.7. Any t.i. or t.s. subspace meets each hyperbolic line in at

most a point, and meets the anisotropic germ in the zero space; so its rank is at

most r. But the span of r t.i. or t.s. points, one chosen from each hyperbolic line,

is a t.i. or t.s. subspace of rank r.

(P1) Any ¬‚at, together with the ¬‚ats it contains, is a projective space of dimen-

sion at most r 1. 8

This is clear since a subspace of a t.i. or t.s. subspace is itself t.i. or t.s. The next

property is also clear.

(P2) The intersection of any family of ¬‚ats is a ¬‚at.

(P3) If U is a ¬‚at of dimension r 1 and p a point not in U , then the union of the

8

lines joining p to points of U is a ¬‚at W of dimension r 1; and U W is a 8 …

hyperplane in both U and W .

¢

Proof Let p w2 . The function v b v£ w¤ on the vector space U is linear; let

¦

0 "¨

K be its kernel, a hyperplane in U . Then the line (of the projective space) joining

p to a point q U is t.i. or t.s. if and only if q K; and the union of all such t.i. or

t.s. lines is a t.i. or t.s. space W K v2 , such that W U K, as required.

£ 0i¦ … ¦

(P4) There exist two disjoint ¬‚ats of dimension r 1.

8

Proof Use the hyperbolic-anisotropic decomposition again. If L1 Lr are the £

£

hyperbolic lines, and vi wi are the distinguished spanning vectors in Li , then the

£

required ¬‚ats are v1 vr and w1 wr .

0 ££

2 0 33£

£ 2

Next, we specialise to the case r 2. (A polar space of rank 1 is just an

¦

unstructured collection of points.) A polar space of rank 2 consists of points and

lines, and has the following properties. (The ¬rst two are immediate consequences

of (P2) and (P3) respectively.)

90 6. Polar spaces

(Q1) Two points lie on at most one line.

(Q2) If L is a line, and p a point not on L, then there is a unique point of L

collinear with p.

(Q3) No point is collinear with all others.

For, by (P4), there exist disjoint lines; and, given any point p, at least one of

these lines does not contain p, and p fails to be collinear with some point of this

line.

A geometry satisfying (Q1), (Q2) and (Q3) is called a generalised quadran-

gle. Such geometries play much the same rˆ le in the theory of polar spaces as

o

projective planes do in the theory of projective spaces. We will return to them

later.

Note that (Q1) holds in a polar space of arbitrary rank.

Another property of polar spaces, which is proved by almost the same argu-

ment as (P3), is the following extension of (Q2):

(BS) If L is a line, and p a point not on L, then p is collinear with one or all

points of L.

In a polar space G, for any set S of points, we let S denote the set of points

#

which are perpendicular to (that is, collinear with) every point of S. It follows

from (BS) that, for any set S, the set S is a (linear) subspace of G (that is, if two

#

points of S are collinear, then the line joining them lies wholly in S ). Moreover,

# #

for any point x, x is a hyperplane of G (that is, a subspace which meets every

#

line).

Polar spaces have good inductive properties. Let G be a classical polar space.

There are two natural ways of producing a “smaller” polar space from G:

(a) Take a point x of G, and consider the quotient space x x, the space whose

e#

points, lines, . . . are the lines, planes, . . . of G containing x.

(b) Take two non-perpendicular points x and y, and consider x£ y' .

% #

In each case, the space constructed is a classical polar space, having the same

germ as G but with polar rank one less than that of G. (Note that, in (b), the

span of x and y in the vector space is a hyperbolic line.) There are more general

versions. For example, if S is a ¬‚at of dimension d 1, then S S is a polar space

8 e#

6.4. Classical polar spaces 91

of rank r d with the same germ as G. We will see below and in the next section

8

how this inductive process can be used to obtain information about polar spaces.

We investigate just one type in more detail, the so-called hyperbolic quadric

or hyperbolic orthogonal space, the orthogonal space which is a direct sum of

hyperbolic lines (that is, having germ 0). The quadratic form de¬ning this space

can be taken to be x1 x2 x3 x4 x2r 1 x2r .© ©

© Q

Theorem 6.11 The maximal ¬‚ats of a hyperbolic quadric fall into two classes,

with the properties that the intersection of two maximal ¬‚ats has even codimension

in each if and only if they belong to the same class.

Proof First, note that the result holds when r 1, since then the quadratic form is

¦¢ ¢

x1 x2 and there are just two singular points, 1£ 0¤ and 0£ 1¤ . By the inductive

0

2 0 3

2

principle, it follows that any ¬‚at of dimension r 2 is contained in exactly two 8

maximal ¬‚ats. ¢ ¢

We take the r 1¤ -¬‚ats and r 2¤ -¬‚ats as the vertices and edges of a graph “,

¢8 8 ¢

that is, we join two r 1¤ -¬‚ats if their intersection is an r 2¤ -¬‚at. The theorem

8 8

will follow if we show that “ is connected and bipartite, and that the distance

between two vertices of “ is the codimension of their intersection. Clearly the

codimension of the intersection increases by at most one with every step in the

graph, so it is at most equal to the distance. We prove equality by induction.

¢ ¢ ¢

Let U be a r 1¤ -¬‚at and K a r 2¤ -¬‚at. We claim that the two r 1¤ -

8 8 8

spaces W1 W2 containing K have different distances from U . Factoring out the

£

/

t.s. subspace U K and using induction, we may assume that U K 0. Then

… … ¦

U K is a point p, which lies in one but not the other of W1 W2 ; say p W1 . By

… £

#

induction, the distance from U to W1 is r 1; so the distance from U to W2 is at

8

most r, hence equal to r by the remark in the preceding paragraph.

This establishes the claim about the distance. The fact that “ is bipartite also

follows, since in any non-bipartite graph there exists an edge both of whose ver-

tices have the same distance from some third vertex, and the argument given shows

that this doesn™t happen in “.

In particular, the rank 2 hyperbolic quadric consists of two families of lines

forming a grid, as shown in Fig. 6.1. This is the so-called “ruled quadric”, familiar

from models such as wastepaper baskets.

Exercises

1. Prove (BS).

all spaces in the family. Note that, in the unitary case, the order of the ¬nite ¬eld

This number, depending only on the germ, carries numerical information about

the vector space rank. The signi¬cance of the parameter µ will emerge shortly.

some information about them. In the table, r denotes the polar space rank, n

scalar factor obviously de¬ne the same polar space.) The following table gives

symplectic, two unitary, and three orthogonal. (Forms which differ only by a

We subdivide these spaces into six families according to their germ, viz., one

The classi¬cation of ¬nite classical polar spaces was achieved by Theorem 6.7.

Finite polar spaces 6.5

¬‚ats.

8

polar spaces together with the fact that an r 1¤ -¬‚at lies in exactly two maximal

¢

3. Show that Theorem 6.11 can be proved using only properties (P1)“(P4) of

# % e#

2. Prove the assertions above about x x and x£ y' .

Figure 6.1: A grid

’” • “ —‘

•”

˜ – – ’ “ — ‘ ™

‘ ˜ ™

” —“ ‘ ™ ™ ’ •—

˜ ’ ’ – –• • ”— “ ‘ ™

”“ – “ ‘ ˜ ˜

” —” “ ™‘ ™ ’ — •’

“– ‘ ˜ ˜ ˜ ˜ ’ • • – ”

–— “ ™ ‘ ™ —•

“” ‘ –

— –

‘ “” – ˜ ’ ’ ˜ • ” ™ “ ‘ ™ — ’ • ’

– ™ ” “ ™ ‘ — •

™

“‘ ‘ ” ” – – ˜ ’ ˜ • ˜ ˜ — ” ™ “ ‘ — • ’ ’

™–

‘ “ “ ” – ’ ’ • ˜ • — — ˜

– ™ ™ ” “ — ‘ — • ’

˜™ •

‘ “ ” ’ – • — ˜ ™ – ” “ ‘ ’

™ ˜

‘ ‘ “ ’ ” • – – — ˜ – — — ” “ • ’ ‘

™˜ ˜™ •

˜

‘ ’ “ •” — ™ ˜ — – • ” ”’ “ “ ‘

˜™

’ ‘ • “ ”— – ™ ˜— • – ’ ‘

’ • —“ ” – ™ ˜ — ˜ • ’ – ” “ ‘

’ • ‘ — “ ” ™ – – ™ ™ ˜ — • ’ – ” “ ‘

’ • — ‘ “ ™ ” — ˜• ’˜ ˜ – ” ‘“ ”

– ‘“

’ • — ‘ ™ “ ” ™ – – — • ˜ ’ ˜ – ”“

’ • — ™ ‘ “ ™ ™ ” —–

”

’• — ™ ‘ • ’ ˜’ ˜ ‘ ‘ – “ ”

˜ ˜

”— –•

•’ — ™ ‘ ™ “ “ ˜ ‘ “ – ” “ ”–

™‘ • – ’ ’ ˜

’• — ’ ™ ‘“

–

“ — —” ” •

™ ™ ˜ ˜ ˜ ˜ ‘

™ ‘‘

—“ ’–

•” ˜

™ ™

–

’’

‘ — ““ • ”

6. Polar spaces 92

6.5. Finite polar spaces 93

must be a square.

µ

Type n

Symplectic 2r 0

1

Unitary 2r 8 2

1

Unitary 2r 1 © 2

Orthogonal 2r 1

8

Orthogonal 2r 1 0

©

Orthogonal 2r 2 1

©

Table 6.1: Finite classical polar spaces

µ

Theorem 6.12 The number of points in a ¬nite polar space of rank 1 is q1§ 1,

©

where µ is given in Table 6.1.

¢

Proof Let V be a vector space carrying a form of rank 1 over GF q¤ . Then V

is the orthogonal direct sum of a hyperbolic line L and an anisotropic germ U of

dimension k (say). Let nk be the number of points.

Suppose that k 0. If p is a point of the polar space, then p lies on the hyper-

¥

plane p ; any other hyperplane containing p is non-degenerate with polar rank 1

#

and having germ of dimension k 1. Consider a parallel class of hyperplanes in 8

the af¬ne space whose hyperplane at in¬nity is p . Each such hyperplane con- #

tains nk 1 1 points, and the hyperplane at in¬nity contains just one, viz., p. So

8Q

we have ¢

nk 1 q nk 1 1¤ 8 ¦ 8Q

£

¢

from which it follows that nk 1 n0 1¤ qk . So it is enough to prove the result

¦ 8

©

for the case k 0, that is, for a hyperbolic line.

¦

In the symplectic case, each of the q 1 projective points on a line is isotropic. ©

Consider the unitary case. We can take the form to be

¢

¢ ¢

b x1 y1 x2 y2 x1 y2 y1 x2

£ ¤

£ £ ¤

¦¤ £

©

where x xσ xr , r2 q. So the isotropic points satisfy xy yx 0, that is,

¦ ¦ ¦ ¦

©

¢ ¢

Tr xy¤ 0. How many pairs x£ y¤ satisfy this? If y 0, then x is arbitrary. If

5

¦ ¦

y 0, then a ¬xed multiple of x is in the kernel of the trace map, a set of size q1d 2

!¦ ¢

(since Tr is GF q1d 2 -linear). So there are

¤

¢ ¢ ¢

1¤ q1d 2

1¤ q1d 2

q q 1 q 1¤

8 ¦ 8

© © ©

94 6. Polar spaces

vectors, i.e., q1d 2 1 projective points.

©

Finally, consider the orthogonal case. The quadratic form is equivalent to xy, ¢ ¢

and has two singular points, 1£ 0¤ and 1£ 0¤ . 0

2 0

2

¢ ¢ ¢

µ

Theorem 6.13 In a ¬nite polar space of rank r, there are qr 1¤ qr§ 1¤ q

8

e 8

©

1¤ points, of which q2r 1§ µ are not perpendicular to a given point.

Q

¢ ¢

Proof We let F r¤ be the number of points, and G r¤ the number not perpen- ¢

dicular to a given point. (We do not assume that G r¤ is constant; this constancy

follows from the induction that proves the theorem.) We use the two inductive

principles described at the end of the last section.

¢ ¢

q2 G r 1¤ .

Step 1 G r¤

¦ 8 ¢

Take a point x, and count pairs y£ z¤ , where y x , z x , and z y . Choos- !

# # #

¢

ing z ¬rst, there are G r¤ choices; then x£ z2 is a hyperbolic line, and y is a point in 0

¢

x£ z2 , so there are F r 1¤ choices for y. On the other hand, choosing y ¬rst, the

0 8

#

lines through y are the points of the rank r 1 polar space x x, and so there are 8 e#

¢ ¢

F r 1¤ of them, with q points different from x on each, giving qF r 1¤ choices

8 8

for y; then x£ y2 and y£ z2 are non-perpendicular lines in y , i.e., points of y y,

¢0 0 e#

#

¢

so there are G r 1¤ choices for y£ z2 , and so qG r 1¤ choices for y. thus

8 0 8

¢ ¢ ¢ ¢

G r¤ F r 1¤ qF r 1¤ qG r 1¤

G

e 8

¦ 8 y

e 8 F

£

from which the result follows.

¢ ¢

q1§ µ , it follows immediately that G r¤ 1§ µ ,

q2r

Since G 1¤ as required.

¦

¦ Q

¢ ¢ ¢

Step 2 F r¤ 1 qF r 1¤ G r¤ .

¦ 8

© ©

For this, simply observe (as above) that points perpendicular to x lie on lines

of x x. e# ¢ ¢ ¢

Now it is just a matter of calculation that the function qr 1¤ qr§ µ 1¤ q 8

e 8

©

1¤ satis¬es the recurrence of Step 2 and correctly reduces to q1§ µ 1 when r ¦

©

1.

Theorem 6.14 The number of maximal ¬‚ats in a ¬nite polar space of rank r is

r

∏1

¢ µ

qi§ F¤

©

if 1

6.5. Finite polar spaces 95

¢ ¢

Proof Let H r¤ be this number. Count pairs x£ U , where U is a maximal ¬‚at ¤

and x U . We ¬nd that

¢ ¢ ¢ ¢ ¢

qr

F r¤ H r 1¤ H r¤ 1¤ q 1¤

G

e 8 ¦ y

e 8

e 8

£

so ¢ ¢ ¢

µ

qr§

H r¤ 1 Hr 1¤

¦ ¤ 8

©

Now the result is immediate.

It should now be clear that any reasonable counting question about ¬nite polar

spaces can be answered in terms of q£ r£ µ.

7

Axioms for polar spaces

The axiomatisation of polar spaces was begun by Veldkamp, completed by Tits,

and simpli¬ed by Buekenhout, Shult, Hanssens, and others. In this chapter, the

analogue of Chapter 3, these results are discussed, and proofs given in some cases

as illustrations. We begin with a discussion of generalised quadrangles, which

play a similar rˆ le here to that of projective planes in the theory of projective

o

spaces.

7.1 Generalised quadrangles

We saw the de¬nition of a generalised quadrangle in Section 6.4: it is a rank 2

geometry satisfying the conditions

(Q1) two points lie on at most one line;

(Q2) if the point p is not on the line L, then p is collinear with exactly one point

of L;

(Q3) no point is collinear with all others.

For later use, we represent generalised quadrangles by a diagram with a double

arc, thus:

¢

¡

The axioms (Q1)“(Q3) are self-dual; so the dual of a generalised quadrangle

is also a generalised quadrangle.

Two simple classes of examples are provided by the complete bipartite graphs,

whose points fall into two disjoint sets (with at least two points in each, and whose

97

98 7. Axioms for polar spaces

lines consist of all pairs of points containing one from each set), and their duals,

the grids, some of which we met in Section 6.4. Any generalised quadrangle in

which lines have just two points is a complete bipartite graph, and dually (Exer-

cise 2). We note that any line contains at least two points, and dually: if L were

£

a singleton line p¤ , then every other point would be collinear with p (by (Q2)),

contradicting (Q3).

Apart from complete bipartite graphs and grids, all generalised quadrangles

have orders:

Theorem 7.1 Let G be a generalised quadrangle in which there is a line with at

least three points and a point on at least three lines. Then the number of points on

a line, and the number of lines through a point, are constants.

Proof First observe that, if lines L1 and L2 are disjoint, then they have the same

cardinality; for collinearity sets up a bijection between the points on L1 and those

on L2 .

Now suppose that L1 and L2 intersect. Let p be a point on neither of these

lines. Then one line through p meets L1 , and one meets L2 , so there is a line

L3 disjoint from both L1 and L2 . It follows that L1 and L2 both have the same

cardinality as L3 .

The other assertion is proved dually.