<< стр. 4(всего 7)СОДЕРЖАНИЕ >>
smallest index which is not excluded, and adjoin it to . At the conclusion, any
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 п¬Ѓrst L denotes the points and lines in 3-dimensional projective space
over F.
5.2. Some special diagrams 73

Now we turn to afп¬Ѓne spaces, where similar results hold. The label A f on a
stroke will denote the class of afп¬Ѓne planes.

Theorem 5.8 A geometry with diagram
Af
  D'
 
is an afп¬Ѓne space of п¬Ѓnite dimension.

Proof It is a linear space whose planes are afп¬Ѓne (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 О 1 be the plane through L1 and L2 , and О 2 the plane through p and L3 ,
where p is a point of L1 . Then О 1 and О 2 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 О 2 contains
HВЎ
L3 and q, hence L2 , and so is equal to О 1 , 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 inп¬Ѓnite-dimensional) afп¬Ѓne space.

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

Exercises
1. Consider a geometry of rank n with diagram
L
  D)
in which all lines have the same п¬Ѓnite cardinality k, and all the projective planes
have the same п¬Ѓnite order q.
74 5. Buekenhout geometries

( ВЎ
(a) If n 4, prove that the geometry is either projective (q k 1) or afп¬Ѓne
(q k).
ВЎ Вў
(b) If n 3, prove that q k 1В© kВ© k2 or k k2 1ВЈ .
ВЎ
ВЎ
(This result is due to Doyen and Hubaut ).
2. Construct an inп¬Ѓnite вЂњfree-likeвЂќ 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
  
  
ВЈВў
(n nodes) from an ovoid in PG nВ© F (see Section 4.4).
6

Polar spaces

Now we begin on our second major theme, polar spaces. This chapter corresponds
to the п¬Ѓrst 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 п¬Ѓrst sec-
tion, on polarities of projective spaces, provides motivation for the introduction of

6.1 Dualities and polarities
В
Recall that the dual V of a п¬Ѓnite-dimensional (left) vector space V over a
skew п¬Ѓeld F can be regarded as a left vector space of the same dimension over
ВЎ
the opposite п¬Ѓeld F , and there is thus an inclusion-reversing bijection between
Вў Вў ВЎ ВЎ
the projective spaces PG nВЈ F and PG nВЈ F . If it happens that F and F are iso-
В¤ В¤Вў
morphic, then there exists a duality of PG nВЈ F , an inclusion-reversing bijection В¤
Вў
of PG nВЈ F . В¤
Вў
Conversely, if PG nВЈ F admits a duality (for n 1), then F is isomorphic to
В¤ ВҐ
ВЎ
F , as follows from the FTPG (see Section 1.3). We will examine this conclusion
and make it more detailed.
Вў
So let ПЂ be a duality of PG nВЈ F , n 1. Composing ПЂ with the natural isomor- В¤ВЎ ВҐ
Вў Вў
phism from PG nВЈ F to PG nВЈ F , we obtain an inclusion-preserving bijection Оё
ВўВ¤ В¤
Вў ВЎ
from PG nВЈ F to PG nВЈ F . According to the FTPG, Оё is induced by a semilinear
В¤ В¤ В
transformation T from V F nВ§ 1 to its dual space V , associated with an isomor-
В¦
ВЎ
phism Пѓ : F F , which can be regarded as being an anti-automorphism of F:
ВЁ

75
76 6. Polar spaces

that is,
Вў
v1 v2 T v1 T v2 T
В¤ В¦ ВЈ
О±Пѓ vT
О±vВ¤ T В¦ 
Deп¬Ѓne a function b : V V F by the rule
 ВЁ
Вў Вў Вў
b vВЈ wВ¤ vВ¤ wT

В¦ В¤
ВЈ
В
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 п¬Ѓrst argument, and semilinear as a function
of the second вЂ” this means that
Вў Вў Вў
b vВЈ w1 w2 b vВЈ w1 b vВЈ w2
В¤
and Вў Вў
О±Пѓ b vВЈ wВ¤
b vВЈ О±wВ¤ 
В¦ 

(The preп¬Ѓx вЂњsesqui-вЂќ means вЂњone-and-a-halfвЂќ.) If we need to emphasise the anti-
automorphism Пѓ, we say that b is Пѓ-sesquilinear. If Пѓ is the identity, then the form
is bilinear.
The form b is also non-degenerate, in the sense that
Вў
 ВўВў
w V b vВЈ wВ¤ 0 v 0
 В¤ В¦  В¦
and Вў
 ВўВў
v V b vВЈ wВ¤ 0 w 0
 В¤ В¦  В¦
(The second condition asserts that T is one-to-one, so that if w 0 then wT is !В¦
a non-zero functional. The п¬Ѓrst asserts that T is onto: only the zero vector is
annihilated by every functional in the dual space.)
So, we have:
Вў
Theorem 6.1 Any duality of PG nВЈ F , for n 1, is induced by a non-degenerate В¤ ВҐ
Пѓ-sesquilinear form on the underlying vector space, where Пѓ 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
Вў
U U v V : b vВЈ wВ¤ 0 for all w U
"ВЁ &В¦\$#
%  
В¦  ('
6.1. Dualities and polarities 77

Obviously, a duality applied twice is a collineation. The most important types Вў
of dualities are those whose square is the identity. A polarity of PG nВЈ F is a В¤
Вў
duality which satisп¬Ѓes U U for all п¬‚ats U of PG nВЈ F .
) В¦ ## В¤
It is a bit difп¬Ѓcult 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 вЂњprimitiveвЂќ. Вў Вў
A sesquilinear form b is reп¬‚exive if b vВЈ wВ¤ 0 implies b wВЈ vВ¤ 0.

В¦ 
В¦

Proposition 6.2 A duality is a polarity if and only if the sesquilinear form deп¬Ѓning
it is reп¬‚exive.

Proof b is reп¬‚exive if and only if

v w2 w v2
4#3 1
0
 53 1
#0
Hence, if b is reп¬‚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,
В¦ 7# 8 В¦# В¦
# 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 classiп¬Ѓcation of reп¬‚exive forms. For convenience, from
now on F will always be assumed to be commutative. (Note that, if the anti-
automorphism Пѓ is an automorphism, and in particular if Пѓ is the identity, then F
is automatically commutative.) Вў Вў
b vВЈ wВ¤ Пѓ for all vВЈ w V .
The form b is said to be Пѓ-Hermitian if b wВЈ vВ¤ B
В¦ 
Вў
This implies that, for any v, b vВЈ vВ¤ lies in the п¬Ѓxed п¬Ѓeld of Пѓ. If Пѓ is the identity,
such a form (which is bilinear) is called symmetric. Вў
A bilinear form b is called alternating if b vВЈ vВ¤ 0 for all v V . This implies
Вў7
В¦ 
Вў Вў
that b wВЈ vВ¤ b vВЈ wВ¤ for all vВЈ w V . (Expand b v wВЈ v wВ¤ 0, and note
DC
8В¦  E
В¦
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 reп¬‚exive. Conversely, we have the
following:
78 6. Polar spaces

Theorem 6.3 A non-degenerate reп¬‚exive Пѓ-sesquilinear form is either alternat-
ing, or a scalar multiple of a Пѓ-Hermitian form. In the latter case, if Пѓ 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
Пѓ2 1, and then the proof of the theorem is given in the case of a bilinear form
В¦
(that is, when Пѓ 1). В¦

Proposition 6.4 If b is a non-zero reп¬‚exive Пѓ-sesquilinear form, then Пѓ2 is the
identity.

Proof Note п¬Ѓrst that a form is Пѓ-sesquilinear if and only if it is additive in each
variable and satisп¬Ѓes
Вў Вў Вў Вў
b vВЈ wВ¤ ОІПѓ
b О±vВЈ wВ¤ О±b vВЈ wВ¤ b vВЈ ОІwВ¤

В¦ ВЈF 
В¦ 

Вў
Step 1 If b is alternating, then Пѓ 1. For we can choose v and w with b vВЈ wВ¤
В¦ G
В¦
Вў
1. Then for any О± F, we have
b wВЈ vВ¤
8 
В¦ 
Вў
О± О±b vВЈ wВ¤
В¦
Вў
b О±vВЈ wВ¤
В¦
Вў
b wВЈ О±vВ¤
В¦ 8
Вў
b wВЈ vВ¤ О±Пѓ
В¦ 8
О±Пѓ
В¦ 

(Note that this step does not require non-degeneracy, merely that b is not identi-
cally zero.)
Вў
So we can assume that there exists v with b vВЈ vВ¤ 0. Multiplying b by a non- H
!В¦ Вў
zero scalar (this does not affect the hypotheses), we may assume that b vВЈ vВ¤ 1. 
В¦

Вў
Step 2 Assume for a contradiction that Пѓ2 1. For any vector w, if b wВЈ vВ¤ 0, !В¦ I
!В¦
Вў
then we can replace w by its product with a non-zero scalar to assume b wВЈ vВ¤ 1. G
В¦
Вў Вў Вў
Then b w vВЈ vВ¤ 0, and so b vВЈ w vВ¤ 0, whence b vВЈ wВ¤ 1. We claim that
8 В¦ 8 
В¦ 
В¦
Вў
b wВЈ wВ¤ 1. 
В¦
6.1. Dualities and polarities 79
Вў Вў
Proof Suppose that О± b wВЈ wВ¤ 1. Note п¬Ѓrst that b w О±vВЈ vВ¤ 0, and В¦ P
!В¦ 8 E
В¦
Вў
0, whence О± О±Пѓ . Take any element О» F with О» 1, and
so b wВЈ w О±vВ¤ 8 C
В¦ В¦  !В¦
Вў Вў
choose Вµ F such that ВµПѓ 1 О»В¤ 1 О± О»В¤ . Since О± 1, we have Вµ 1; and
 В¦ 8 R
Q 8 !В¦ !В¦
ВµПѓ О»ВµПѓ О± О»
8 В¦ 8
Вў Вў
ВµПѓ 1 ВµПѓ
This implies, п¬Ѓrst, that О» О± 1, and second that
В¦ 8 В¤ 8 RВ¤
Q
Вў
ВµПѓ О»ВµПѓ
О»vВЈ w О± О»
bw ВµvВ¤ 0
8 8 В¦ 8 8 В¦
Вў
О»vВ¤
Hence b w ВµvВЈ w 0, and we obtain
8 8 
В¦
О»Пѓ ВµО»Пѓ
О± Вµ 0
8 8 В¦
Applying Пѓ to this equation and using the fact that О±Пѓ О±, we obtain
В¦
ВµПѓ О»Пѓ О» Пѓ ВµПѓ
2 2
О± 0ВЈ
8 8 В¦
whence Вў Вў
О»Пѓ ВµПѓ 1 ВµПѓ
2
О± О»
1
В¦ QВ¤
В¦ 8 В¤ 8
But О» was an arbitrary element different from 1. Since clearly 1Пѓ 1, we have
В¦
Пѓ2 1, contrary to assumption.
В¦
Вў
Step 3 Let W v . Then V v2 W , and rk W 1. For any x W , we
ВўВ¦ TU SВ¦
0 WВ¤
V 
#
Вў Вў
have b vВЈ vВ¤ b v xВЈ vВ¤ 1, and so by Step 2, we have b v xВЈ v xВ¤ 1.
H
В¦ В¦X X
В¦
Вў
Thus b xВЈ xВ¤ 2. Putting x 0, we see that F must have characteristic 2, and
8YВ¦C В¦
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
Вў Вў Вў Вў
b uВЈ vВ¤ b uВЈ wВ¤ b uВЈ wВ¤ b uВЈ vВ¤ 0
G
8 7
В¦
by commutativity; that is, using bilinearity,
Вў Вў Вў
b uВЈ b uВЈ vВ¤ w b uВЈ wВ¤ vВ¤ 0
8 В¦
By reп¬‚exivity, ВўВў Вў
b b uВЈ vВ¤ w b uВЈ wВ¤ vВЈ uВ¤ 0ВЈ
8 В¦
80 6. Polar spaces

whence bilinearity again gives
Вў Вў Вў Вў
b uВЈ vВ¤ b wВЈ uВ¤ b uВЈ wВ¤ b vВЈ uВ¤ (6.1)
В¦ 

Вў Вў
Call a vector u good if b uВЈ vВ¤ b vВЈ uВ¤ 0 for some v. By (6.1), if u is good,
В¦ !I
В¦
Вў Вў Вў
then b uВЈ wВ¤ b wВЈ uВ¤ for all w. Also, if u is good and b uВЈ vВ¤ 0, then v is good.
7
В¦ !X
В¦ Вў
But, given any two non-zero vectors u1 u2 , there exists v with b ui vВ¤ 0 for ВЈ aВЈ
!В¦
Вў
i 1ВЈ 2. (For there exist v1 v2 with b ui vi 0 for i 1ВЈ 2, by non-degeneracy;
В¦ ВЈ PВ¤ ВЈ
!В¦ В¦
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
В¦
Вў ВўВў Вў
b uВЈ uВ¤ b uВЈ vВ¤ b vВЈ uВ¤ 0
G
8 В¦3
В¤
Вў
for all uВЈ v. So, if u is not good, then b uВЈ uВ¤ 0; and, if no vector is good, then b

В¦
is alternating.

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

Exercises
1. Let b be a sesquilinear form on V . Deп¬Ѓne the left and right radicals of b to
be the subsets bВў
 Вў
v V : w V b vВЈ wВ¤ 0'
%   В¤ В¦
and bВў
 Вў
v V: w V b wВЈ vВ¤ 0'
%   В¤ В¦
respectively. Prove that the left and right radicals are subspaces of the same rank
(if V has п¬Ѓnite rank).
(Note: If the left and right radicals are equal, this subspace is called the radical
of b. This holds if b is reп¬‚exive.)
2. Give an example of a bilinear form on an inп¬Ѓnite-rank vector space whose
3. Let Пѓ be a (non-identity) automorphism of F of order 2. Let E be the
Вў
subп¬Ѓeld Fix ПѓВ¤ .
(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 О» Вў
Вў
F E. Then О» is quadratic over E, so E О»В¤ has degree 2 over E. Now E О»В¤
c
contains an element П‰ such that П‰Пѓ П‰ (if the characteristic is not 2) or П‰Пѓ
8YВ¦ В¦
6.2. Hermitian and quadratic forms 81

П‰ 1 (if the characteristic is 2). Now, given two such elements, their quotient or
difference respectively is п¬Ѓxed by Пѓ, so lies in E.]
(b) Prove that
F : О»О»Пѓ Оµe ОµПѓ : Оµ
О» 1' F
%  В¦ &d
%В¦  ('

[The left-hand set clearly contains the right. For the reverse inclusion, separate
into cases according as the characteristic is 2 or not. Вў
If the characteristic is not 2, then we can take F E П‰В¤ , where П‰2 О± E В¦ В¦ 
and П‰Пѓ П‰. If О» 1, then take Оµ 1; otherwise, if О» a bП‰, take Оµ
fВ¦ Вў
8 В¦ В¦ В¦ В¦
bО± a 1В¤ П‰. 8 В© Вў
If the characteristic is 2, show that we can take F E П‰В¤ , where П‰2 П‰ О± В¦ В¦
0, О± E, and П‰Пѓ П‰ 1. Again, if О» 1, set Оµ 1; else, if О» a bП‰, take
Вў В¦ В¦ В¦ В¦
Оµ a 1В¤ bП‰.]
4. Use the result of Exercise 3 to complete the proof of Theorem 6.3 in general.
Вў
[If b uВЈ uВ¤ 0 for all u, the form b is alternating and bilinear. If not, suppose
g
В¦
Вў Вў Вў
0 and let b uВЈ uВ¤ Пѓ О»b uВЈ uВ¤ . Choosing Оµ as in Exercise 2 and re-
that b uВЈ uВ¤ h
!В¦ В¦
normalising b, show that we may assume that О» 1, and (with this choice) that b В¦
is Hermitian.]

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 Пѓ be an automorphism of the commutative п¬Ѓeld F, of order dividing 2. Let
Вў Вў
О» F : О»Пѓ О»' be the п¬Ѓxed п¬Ѓeld of Пѓ, and Tr ПѓВ¤ О» О»Пѓ : О» F
Fix ПѓВ¤ iC
%В¦  В¦ В© Вў ig Вў
%В¦  '
the trace of Пѓ. Since Пѓ2 is the identity, it is clear that Fix ПѓВ¤ Tr ПѓВ¤ . Moreover, B
p
Вў
if Пѓ is the identity, then Fix ПѓВ¤ F, and 
В¦
0 if F has characteristic 2,
Вў
Tr ПѓВ¤ r
qВ¦
F otherwise.
Вў
Let b be a Пѓ-Hermitian form. We observed in the last section that b vВЈ vВ¤ I

Вў Вў Вў
Fix ПѓВ¤ for all v V . We call the form b trace-valued if b vВЈ vВ¤ Tr ПѓВ¤ for all
 I

v V. 
Вў Вў
Proposition 6.5 We have Tr ПѓВ¤ Fix ПѓВ¤ unless the characteristic of F is 2 and
C
В¦
Пѓ is the identity.
82 6. Polar spaces
Вў Вў
Proof E Fix ПѓВ¤ is a п¬Ѓeld, and K Tr ПѓВ¤ is an E-vector space contained in E
В¦ В¦
(Exercise 1). So, if K E, then K 0, and Пѓ is the map x x. But, since Пѓ is
!В¦ В¦ "ВЁ 8
a п¬Ѓeld automorphism, this implies that the characteristic is 2 and Пѓ 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 deп¬Ѓned 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 deп¬Ѓned by so-called pseudoquadratic forms. These
will be discussed brieп¬‚y in Section 7.6.
Let V be a vector space over F. A quadratic form on V is a function f : V F ВЁ
satisfying
s
Вў Вў
f О»vВ¤ О»2 f vВ¤ for all О» F, v V;

В¦  
s
Вў Вў Вў Вў
fv wВ¤ f vВ¤ f wВ¤ b vВЈ wВ¤ , where b is bilinear.

В¦
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
Вў Вў Вў Вў
b vВЈ wВ¤ fv wВ¤ f vВ¤ f wВ¤

В¦ G
8 G
8
and Вў Вў
1
f vВ¤ 2b vВЈ vВ¤

В¦ ВЈF
the latter equation coming from the substitution v w in the second deп¬Ѓning В¦
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; weвЂ™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 п¬Ѓeld of characteristic 2, which
Вў Вў
is not alternating. Set f vВ¤ b vВЈ vВ¤ . Then we have 
В¦
Вў Вў
f О»vВ¤ О»2 f vВ¤

В¦
and Вў Вў Вў
fv wВ¤ f vВ¤ f wВ¤

В¦ F
ВЈ
6.2. Hermitian and quadratic forms 83
Вў Вў
since b vВЈ wВ¤ b wВЈ vВ¤ 0. Thus f is вЂњalmostвЂќ a semilinear form; the map О» О»2
t
В¦ "ВЁ
is a homomorphism of the п¬Ѓeld 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 вЂњprimitiveвЂќ. 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 вЂњtypesвЂќ of
forms to study: alternating bilinear forms; Пѓ-Hermitian forms where Пѓ is not the
We have to deп¬Ѓne 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
ВўВў Вў
 Вў
f vВ¤ 0& w V b vВЈ wВ¤ 0В¤ v 0

В¦  В¤ 
В¦  В¦
where b is the associated bilinear form; that is, if the form f is non-zero on every
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 satisп¬Ѓes
Вў Вў Вў
f v wВ¤ f vВ¤ f wВ¤
u
В¦ 
ВЈ
f О»vВ¤ О»2 f vВ¤
v
В¦ F
As above, f is very nearly semilinear. The п¬Ѓeld 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 п¬Ѓeld F.
(a) If the characteristic of F is not 2, then b is non-degenerate.
(b) If F is a perfect п¬Ѓeld of characteristic 2, then the radical of b has rank at
most 1.
84 6. Polar spaces

Exercises
1. Let Пѓ be an automorphism of a commutative п¬Ѓeld F such that Пѓ2 is the
identity. Вў
(a) Prove that Fix ПѓВ¤ is a subп¬Ѓeld of F.
Вў
(b) Prove that Tr ПѓВ¤ is closed under addition, and under multiplication by
Вў
elements of Fix ПѓВ¤ .
2. Let b be an alternating bilinear form on a vector space V over a п¬Ѓeld F of
Вў
characteristic 2. Let vi : i I be a basis for V , and q any function from I to F.
 В¤ Вў Вў
Show that there is a unique quadratic form with the properties that f vi q iВ¤ 5В¤
В¦
for every i I, and f polarises to b.

3. (a) Construct an imperfect п¬Ѓeld 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 п¬Ѓnite п¬Ѓelds of characteristic 2 are perfect. (Hint: the multiplica-
tive group is cyclic of odd order.)

6.3 Classiп¬Ѓcation of forms
As explained in the last section, we now consider a vector space V of п¬Ѓnite
rank equipped with a form of one of the following types: a non-degenerate alter-
nating bilinear form b; a non-degenerate Пѓ-Hermitian form b, where Пѓ 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,
Вў Вў
we deп¬Ѓne a function f : V F deп¬Ѓned by f vВ¤ b vВЈ vВ¤ вЂ” this is identically
ВЁ H
В¦
zero if b is alternating. See Exercise 1 for the Hermitian case.
Вў
We say that V is anisotropic if f vВ¤ 0 for all v 0. Also, V is a hyperbolic
w
!В¦ Вў !В¦
Вў Вў
line if it is spanned by vectors v and w with f vВ¤ f wВ¤ 0 and b vВЈ wВ¤ 1.
5
В¦ 5
В¦ E
В¦
(The vectors v and w are linearly independent, so V has rank 2; so, projectively, it
is a вЂњlineвЂќ.)

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 Вў
hyperbolic line.) So suppose that V contains a vector v 0 with f vВ¤ 0. !В¦ 
В¦
6.3. Classiп¬Ѓcation of forms 85
Вў
We claim that there is a vector w with b vВЈ wВ¤ 0. In the alternating and x
!В¦
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 .
Вў
Multiplying w by a non-zero constant, we may assume that b vВЈ wВ¤ 1. 
В¦
Вў
Now, for any value of О», we have b vВЈ w О»vВ¤ 1. We wish to choose О» so
8 
В¦
Вў
that f w О»vВ¤ 0; then v and w will span a hyperbolic line. Now we distinguish
8 
В¦
cases. If b is alternating, then any value of О» works. If b is Hermitian, we have
Вў Вў Вў Вў Вў
О»b vВЈ wВ¤ О»Пѓb wВЈ vВ¤ О»О»Пѓ f vВ¤
О»vВ¤
fw f wВ¤
8 v
В¦ y
8 y
Вў
Вў
О» О»Пѓ ;
f wВ¤
В¦ 8y В¤
Вў Вў
and, since b is trace-valued, there exists О» with Tr О»В¤ f wВ¤ . Finally, if f is
H
В¦
Вў Вў Вў Вў
О»vВ¤ О»b wВЈ vВ¤ О»2 f vВ¤
fw f wВ¤
8 v В¦ y
Вў
О»ВЈ
f wВ¤
В¦ 8y
Вў
so we choose О» f wВ¤ .
В¦
Now let W1 be the hyperbolic line vВЈ w О»v2 , and let V1 W1 , where orthog-
0 8 В¦ #
onality is deп¬Ѓned with respect to the form b. It is easily checked that V V1 W1 , В¦ 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 classiп¬Ѓcation of forms over a given п¬Ѓeld, it is necessary to
determine all the anisotropic spaces. In general, this is not possible; for exam-
ple, the study of positive deп¬Ѓnite 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 п¬Ѓnite п¬Ѓelds.
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, SylvesterвЂ™s theorem asserts that any quadratic form in
n variables is equivalent to the form

x2 x2 x2 x2
 8 RВЂВЂ18
8 ВЈ
1 r rВ§ 1 rВ§ s

for some rВЈ s with r s n. If the form is non-singular, then r s n. If both r
ВЃ В¦
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 deп¬Ѓnite or negative deп¬Ѓnite; 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. ВЃ
(b) If Пѓ denotes complex conjugation, the situation for Пѓ-Hermitian forms is
the same as for quadratic forms over the reals: anisotropic forms are positive or
negative deп¬Ѓnite, and there is a unique form of each type, one the negative of the
other.
For п¬Ѓnite п¬Ѓelds, the position is as follows.
Вў
Theorem 6.10 (a) An anisotropic quadratic form in n variables over GF qВ¤ exists
if and only if n 2. There is a unique form for each n except when n 1 and q is
ВЃ В¦
odd, in which case there are two forms, one a non-square multiple of the other.
(b) Let q r2 and let Пѓ be the п¬Ѓeld automorphism О± О±r . Then there is
В¦ "ВЁ
an anisotropic Пѓ-Hermitian form in n variables if and only if n 1. The form is ВЃ
unique in each case.

Proof (a) Consider п¬Ѓrst the case where the characteristic is not 2. The multiplica-
Вў
tive group of GF qВ¤ is cyclic of even order q 1; so the squares form a subgroup
8
of index 2, and if О· is a п¬Ѓxed non-square, then every non-square has the form О·О±2
for some О±. It follows easily that any quadratic form in one variable is equivalent
to either x2 or О·x2 .
6.3. Classiп¬Ѓcation 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 О·y2 , О·x2 О·y2 . В© В© В©
Вў
Suppose п¬Ѓrst that q 1 mod 4В¤ . Then 1 is a square, say 1 ОІ2 . (In
В‚ 8 8 В¦
the multiplicative group, 1 has order 2, so lies in the subgroup of even order
8
Вў Вў Вў
x ОІyВ¤ x ОІyВ¤ , and the п¬Ѓrst and
1 2 y2
2 q 1В¤ consisting of squares.) Thus x
8 В¦ 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. Вў
Now consider the other case, q 1 mod 4В¤ . Then 1 is a non-square fВ‚
8 8 Вў Вў
(since the group of squares has odd order), so the second form is x yВ¤ x yВ¤ , 8
and is not anisotropic. Moreover, the set of squares is not closed under addition Вў
1
(else it would be a subgroup of the additive group, but 2 q 1В¤ doesnвЂ™t divide q); В©
so there exist two squares whose sum is a non-square. Multiplying by a suitable
square, there exist ОІВЈ Оі with ОІ2 Оі2 1. Then ВѓВ¦
8
Вў Вў Вў
ОІx ОіyВ¤ Оіx ОІyВ¤
x2 y2 2 2
8 CВ¤
В¦ 8 ВЈ
and the п¬Ѓrst 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 В© В©
Вў
vanishes at the vector ОІВЈ ОіВЈ 1В¤ ; hence there is no anisotropic form in three or more
variables.
The characteristic 2 case is an exercise (see Exercise 3). Вў
(b) Now consider Hermitian forms. If Пѓ is an automorphism of GF qВ¤ of order
2, then q is a square, say q r2 , and О±Пѓ О±r . We need the fact that every element
В¦ В¦
Вў Вў
GF rВ¤ has the form О±О±Пѓ (see Exercise 1 of Section 6.2).
of Fix ПѓВ¤ 
В¦ Вў Вў
ВµxxПѓ for some non-zero Вµ Fix ПѓВ¤ ; writing
In one variable, we have f xВ¤ В¦E 
Вµ О±О±Пѓ and replacing x by О±x, we can assume that Вµ 1.
В¦ В¦
In two variables, we can similarly take the form to be xxПѓ yyПѓ . Now 1 8 
Вў Вў
Fix ПѓВ¤ , so 1 О»О»Пѓ; then the form vanishes at 1ВЈ О»В¤ . It follows that there is no
8 В¦
anisotropic form in any larger number of variables either.

Exercises
1. Let b be a Пѓ-Hermitian form on a vector space V over F, where Пѓ is not the
Вў Вў Вў
b vВЈ vВ¤ . Let E Fix ПѓВ¤ , and let V be V regarded as an E-
identity. Set f vВ¤ B
В¦ В¦ В„
vector space by restricting scalars. Prove that f is a quadratic form on V , which ВўВ„
Вў Вў Вў Вў
b vВЈ wВ¤ b vВЈ wВ¤ Пѓ.
polarises to the bilinear form Tr bВ¤ deп¬Ѓned by Tr bВ¤ vВЈ wВ¤ 
Вў
Show further that Tr bВ¤ 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 inп¬Ѓnite dimension (a direct sum of
countably many isotropic lines).
3. Let F be a п¬Ѓnite п¬Ѓeld 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 О±x2 xy ОІy2 or
О±x2 xy ОІy2 Оіz2 respectively. Prove that forms of the п¬Ѓrst shape (with О±ВЈ ОІ 0) !В¦
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 reп¬‚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
Вў Вў
вЂ” either f is deп¬Ѓned by f vВ¤ b vВЈ vВ¤ , or b is obtained by polarising f вЂ” and
g
В¦
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 п¬‚ats are the t.i. or t.s. subspaces (in
the above sense). (Concerning the terminology: the term вЂњpolar spaceвЂќ 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 conп¬‚ict with our earlier deп¬Ѓnition 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 deп¬Ѓnitions 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 п¬‚at, together with the п¬‚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 п¬‚ats is a п¬‚at.
(P3) If U is a п¬‚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 п¬‚at W of dimension r 1; and U W is a 8 В…
hyperplane in both U and W .
Вў
Proof Let p w2 . The function v b vВЈ wВ¤ on the vector space U is linear; let
В†В¦
0 "ВЁ
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.
ВЈ 0iВ¦ В… В¦
(P4) There exist two disjoint п¬‚ats of dimension r 1.
8
Proof Use the hyperbolic-anisotropic decomposition again. If L1 Lr are the В‡ВЂВЂВЂВЈ
ВЈ
hyperbolic lines, and vi wi are the distinguished spanning vectors in Li , then the
ВЈ
required п¬‚ats are v1 vr and w1 wr .
0 ВЈВ‡ВЂВЂВЂВЈ
 2 0 3ВЂВЂ3ВЈ
ВЈ 2
Next, we specialise to the case r 2. (A polar space of rank 1 is just an
В¦
unstructured collection of points.) A polar space of rank 2 consists of points and
lines, and has the following properties. (The п¬Ѓrst 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 rЛ† 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 вЂњsmallerвЂќ 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 xВЈ 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 п¬‚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 deп¬Ѓning this space
can be taken to be x1 x2 x3 x4 x2r 1 x2r .В© ВЂВЂ В©

Theorem 6.11 The maximal п¬‚ats of a hyperbolic quadric fall into two classes,
with the properties that the intersection of two maximal п¬‚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
В¦Вў Вў
x1 x2 and there are just two singular points, 1ВЈ 0В¤ and 0ВЈ 1В¤ . By the inductive
0 ВЂ
2 0 3
2
principle, it follows that any п¬‚at of dimension r 2 is contained in exactly two 8
maximal п¬‚ats. Вў Вў
We take the r 1В¤ -п¬‚ats and r 2В¤ -п¬‚ats as the vertices and edges of a graph О“,
Вў8 8 Вў
that is, we join two r 1В¤ -п¬‚ats if their intersection is an r 2В¤ -п¬‚at. The theorem
8 8
will follow if we show that О“ is connected and bipartite, and that the distance
between two vertices of О“ 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.
Вў Вў Вў
Let U be a r 1В¤ -п¬‚at and K a r 2В¤ -п¬‚at. We claim that the two r 1В¤ -
8 8 8
spaces W1 W2 containing K have different distances from U . Factoring out the
ВЈ
/
t.s. subspace U K and using induction, we may assume that U K 0. Then
В… В… В¦
U K is a point p, which lies in one but not the other of W1 W2 ; say p W1 . By
В… ВЈ 
#
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 О“ 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 doesnвЂ™t happen in О“.

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 вЂњruled quadricвЂќ, 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 п¬Ѓnite п¬Ѓeld
This number, depending only on the germ, carries numerical information about
the vector space rank. The signiп¬Ѓcance of the parameter Оµ will emerge shortly.
some information about them. In the table, r denotes the polar space rank, n
scalar factor obviously deп¬Ѓne 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 classiп¬Ѓcation of п¬Ѓnite classical polar spaces was achieved by Theorem 6.7.
Finite polar spaces 6.5
п¬‚ats.
8
polar spaces together with the fact that an r 1В¤ -п¬‚at lies in exactly two maximal
Вў
3. Show that Theorem 6.11 can be proved using only properties (P1)вЂ“(P4) of
# % e#
2. Prove the assertions above about x x and xВЈ y' .
Figure 6.1: A grid
В’В” В• В“ В—В‘
В•В”
В˜ В€В€ Вђ Вђ В– В– В’ В“ В— В‘ В‰В‰ В™ В€
В‘ В˜В‰ В™
В” В—В“ В‘ В€ В™ Вђ В™ Вђ В’ В•В—
В€В˜ Вђ В’ В’ В– В–В• В• В”В— В“ В‘ В™В‰
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В“В” В‘ В–
В— В–
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В“В‘ В‘ В” В‰ В” В– В€ В– Вђ В˜ В’ В˜ В• В˜ В˜ В— В” В™ В“ В‘ В‰ В— В€В• В’ Вђ В’
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В˜В™ В•
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В™ В˜
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В€ В‰ Вђ В‘ В’ В“ В•В” В— В™ В˜ В— В– В• В” В”В’ В“ Вђ В“ В‘ В€ В‰
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В”
В’В• Вђ В— В€ В‰В™ В‘ В• В’ В˜В’ Вђ В€ В€ В˜ В‰ В‰ В‘ В‘ В– В“ В”
В˜Вђ В˜
В”В— В–В•
В•В’ В— Вђ В€В€ В™ В‰ В‘ В™ В“ В“ В‰ В˜ В‘ В“ В– В” В“ В”В–
В‰ В™В‘ В• В– В’ В’ Вђ В˜ В€В€
В’В• В— В’ Вђ В™ В‰В‘В“
Вђ
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В“ В— В—В” В” В•
Вђ В™ В™ В€В€ В€ В€ В˜ В˜ В‰В‰ В˜ В˜ В‘
В‰В™ В‘В‘ Вђ
В—В“ В’В–
В•В” В˜
В™ В™ В‰В‰ В€В€
В–Вђ
В’В’
В‘ В— В“В“ В• В”
6. Polar spaces 92
6.5. Finite polar spaces 93

must be a square.

Оµ
Type n
Symplectic 2r 0
1
Unitary 2r 8 2
1
Orthogonal 2r 1
8
Orthogonal 2r 1 0
Orthogonal 2r 2 1

Table 6.1: Finite classical polar spaces

Оµ
Theorem 6.12 The number of points in a п¬Ѓnite polar space of rank 1 is q1В§ 1,
where Оµ is given in Table 6.1.
Вў
Proof Let V be a vector space carrying a form of rank 1 over GF qВ¤ . 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-
ВҐ
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 afп¬Ѓne space whose hyperplane at inп¬Ѓnity is p . Each such hyperplane con- #
tains nk 1 1 points, and the hyperplane at inп¬Ѓnity contains just one, viz., p. So
8Q
we have Вў
nk 1 q nk 1 1В¤ 8 В¦ 8Q 
ВЈ
Вў
from which it follows that nk 1 n0 1В¤ qk . So it is enough to prove the result
В¦ 8
for the case k 0, that is, for a hyperbolic line.
В¦
In the symplectic case, each of the q 1 projective points on a line is isotropic. В©
Consider the unitary case. We can take the form to be
ВЂВў
Вў Вў
b x1 y1 x2 y2 x1 y2 y1 x2
ВЈ В¤
ВЈ ВЈ ВЂВ¤
В¦В¤ ВЈ
where x xПѓ xr , r2 q. So the isotropic points satisfy xy yx 0, that is,
В¦ В¦ В¦ В¦
Вў Вў
Tr xyВ¤ 0. How many pairs xВЈ yВ¤ satisfy this? If y 0, then x is arbitrary. If
5
В¦ В¦
y 0, then a п¬Ѓxed multiple of x is in the kernel of the trace map, a set of size q1d 2
!В¦ Вў
(since Tr is GF q1d 2 -linear). So there are
В¤
Вў Вў Вў
1В¤ q1d 2
1В¤ q1d 2
q q 1 q 1В¤
8 В¦ 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, Вў Вў
and has two singular points, 1ВЈ 0В¤ and 1ВЈ 0В¤ . 0 ВЂ
2 0 ВЂ
2
Вў Вў Вў
Оµ
Theorem 6.13 In a п¬Ѓnite polar space of rank r, there are qr 1В¤ qrВ§ 1В¤ q
8 ВЂ
e 8
1В¤ points, of which q2r 1В§ Оµ are not perpendicular to a given point.
Q
Вў Вў
Proof We let F rВ¤ be the number of points, and G rВ¤ the number not perpen- Вў
dicular to a given point. (We do not assume that G rВ¤ 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.
Вў Вў
q2 G r 1В¤ .
Step 1 G rВ¤ 
В¦ 8 Вў
Take a point x, and count pairs yВЈ zВ¤ , where y x , z x , and z y . Choos-  ! 
# # #
Вў
ing z п¬Ѓrst, there are G rВ¤ choices; then xВЈ z2 is a hyperbolic line, and y is a point in 0
Вў
xВЈ z2 , so there are F r 1В¤ choices for y. On the other hand, choosing y п¬Ѓrst, the
0 8
#
lines through y are the points of the rank r 1 polar space x x, and so there are 8 e#
Вў Вў
F r 1В¤ of them, with q points different from x on each, giving qF r 1В¤ choices
8 8
for y; then xВЈ y2 and yВЈ z2 are non-perpendicular lines in y , i.e., points of y y,
Вў0 0 e#
#
Вў
so there are G r 1В¤ choices for yВЈ z2 , and so qG r 1В¤ choices for y. thus
8 0 8
Вў Вў Вў Вў
G rВ¤ F r 1В¤ qF r 1В¤ qG r 1В¤
G
e 8 
В¦ 8 y
e 8 F
ВЈ
from which the result follows.
Вў Вў
q1В§ Оµ , it follows immediately that G rВ¤ 1В§ Оµ ,
q2r
Since G 1В¤ as required.

В¦ 
В¦ Q
Вў Вў Вў
Step 2 F rВ¤ 1 qF r 1В¤ G rВ¤ .

В¦ 8
For this, simply observe (as above) that points perpendicular to x lie on lines
of x x. e# Вў Вў Вў
Now it is just a matter of calculation that the function qr 1В¤ qrВ§ Оµ 1В¤ q 8 ВЂ
e 8
1В¤ satisп¬Ѓes the recurrence of Step 2 and correctly reduces to q1В§ Оµ 1 when r В¦
1.

Theorem 6.14 The number of maximal п¬‚ats in a п¬Ѓnite polar space of rank r is
r
в€Џ1
Вў Оµ
qiВ§ FВ¤
if 1
6.5. Finite polar spaces 95
Вў Вў
Proof Let H rВ¤ be this number. Count pairs xВЈ U , where U is a maximal п¬‚at В¤
and x U . We п¬Ѓnd that

Вў Вў Вў Вў Вў
qr
F rВ¤ H r 1В¤ H rВ¤ 1В¤ q 1В¤
G
e 8 В¦ y
e 8 ВЂ
e 8 
ВЈ
so Вў Вў Вў
Оµ
qrВ§
H rВ¤ 1 Hr 1В¤

В¦ В¤ 8 

Now the result is immediate.

It should now be clear that any reasonable counting question about п¬Ѓnite polar
spaces can be answered in terms of qВЈ rВЈ Оµ.
7

Axioms for polar spaces

The axiomatisation of polar spaces was begun by Veldkamp, completed by Tits,
and simpliп¬Ѓed 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 rЛ† le here to that of projective planes in the theory of projective
o
spaces.

We saw the deп¬Ѓnition 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:
В  ВўВ
ВЎ

The axioms (Q1)вЂ“(Q3) are self-dual; so the dual of 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
ВЈ
a singleton line pВ¤ , then every other point would be collinear with p (by (Q2)),
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.

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