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0 0 0

have the same colour);

p q if p q or p q;
f
1 2
0 0 0

p q if neither p q nor q p (this holds only if p and q have the same
f
e 0 0

colour).

Step 2 There do not exist points pВЎ q with p 2 q and q 2 p. 0 0

For, if so, then (with p1 p, q1 q) there are points p2 q2 such that В¤ В¤ ВЎ

p1 p2 q1 q2 p1
1 1 1 2
0 0 0 0 Q

Let ci be the colour of pi and qi , i 1ВЎ 2. By Step 1, the colouring of the plane В¤

p1 p2 q1 is determined; and every point of this plane off the line p1 p2 . In partic-
ular, if x1 p1 q1 , x1 p1 q1 , then every point of x1 p2 except p2 has colour c1 .
ВЁ 3В¤ ВЎ

Similarly, every point of x1 q2 except q2 has colour c1 ; and then every point of x1 x2
except x2 has colour c1 , where x2 p2 q2 , x2 p2 q2 . ВЁ 3В¤ ВЎ

But, by the same argument, every point of x1 x2 except x1 has colour c2 , giving

Step 3 is a partial order. 0

The antisymmetry follows by deп¬Ѓnition for 1 and by Step 2 for 2 ; we must 0 0
prove transitivity. So suppose that p q r, and consider cases. If p 1 q 1 r, 0 0 0 0
then p 2 r by deп¬Ѓnition. If p 1 q 2 r or p 2 q 1 r, then p and r have
0 0 0 0 0

different colours and so are comparable; and r 1 p contradicts Step 2. Finally, if 0
p 2 q 2 r, then p 1 s 1 q for some s; then s 1 r, so that p 2 r.
0 0 0 0 0 0

Step 4 If p q r or p q r, then p r; and if p q r, then p r.
0 e e 0 0 e e e

Suppose that p q r. If p 1 q, then p and r have different colours and so are
0 e 0
comparable; and r 1 p would imply r q by Step 3, so p 1 r. The next case is
0 0 0
similar. The last assertion is a simple consequence of the other two.
4.3. SegreвЂ™s Theorem 51

Step 5 If p q, then p r for all points r of pq except p; and the points of pq
0 0

other than p are pairwise incomparable.
This holds by assumption if p 1 q, and by the proof of Step 2 if p 2 q. 0 0
В  В
Now let S pВў q: p q , and T pВў q:q p.
g
В¤ 30 h
В¤ 0
E E
G G

В  В  В  В  В
Step 6 S pВў and T pВў are subspaces, with p S pВў T pВў . Moreover, T pВў is ВЁ `
В  В
the union of the spaces S qВў for q p, and is spanned by the points of S pВў with 0
colour different from that of p; and we have
В  В
p q implies S pВў S qВў ;
e g
В¤

В  В
q p implies S qВў T pВў .
0 p
i

All of this follows by straightforward argument from the preceding steps. В
Now the proof of the Theorem follows: we set S pВў : p X , and let qa
В¤ ВЁ
E G
В  В
f S pВў be the colour of p. The conclusions of the Theorem follow from the
Вў8

assertions in Step 6.

Remark. The only place in the above argument where the hypothesis F 2  r

В
was used was in Step 1. Now PG 2ВЎ 2Вў has seven points; so, up to complementa-
В
tion, a subset of PG 2ВЎ 2Вў is empty, a point, a line with a point removed, a line, or
a triangle. Only the last case fails to satisfy the conclusion of the Theorem. So we
have the following result:

Theorem 4.10 The conclusions of Theorems 4.8 and 4.9 remain true in the case
В
F GF 2Вў provided that we add the extra hypothesis that no colour class inter-
В¤

sects a plane in a triangle (or, in 4.8, that no plane meets S in a triangle or the
complement of one).

Exercise
1. Prove that Theorem 4.8 holds in any projective plane of order greater than 2
(not necessarily Desarguesian).

4.3 SegreвЂ™s Theorem
For projective geometries over п¬Ѓnite п¬Ѓelds, it is very natural to ask for char-
acterisations of interesting sets of points by hypotheses on their intersections with
52 4. Various topics

lines. Very much п¬Ѓner discriminations are possible with п¬Ѓnite than with inп¬Ѓnite
cardinal numbers; for example, all inп¬Ѓnite subsets of a countably inп¬Ѓnite set whose
complements are also inп¬Ѓnite are alike.
It is not my intention to survey even a small part of this vast literature. But I
will describe one of the earliest and most celebrated results of this kind. I begin
with some generalities about algebraic curves. Assume that F is a commutative
п¬Ѓeld.
If a polynomial f in x1 xnY 1 is homogeneous, that is, a sum of terms all of
ВЎ Qs8Q ВЎ
Q
0 implies f О±vВў 0 for all О± F. So, it f vanishes
В  В
the same degree, then f vВў g
В¤ g
В¤ ВЁ
В
at a non-zero vector, then it vanishes at the rank 1 subspace (the point of PG nВЎ F ) Вў
it spans. The algebraic variety deп¬Ѓned by f is the set of points spanned by zeros
of f . We are concerned here only with the case n 2, in which case (assuming В¤

that f does not vanish identically) this set is called an algebraic curve. В
Now consider the case where f has degree 2, and F GF qВў , where q is an В¤

odd prime power. The curve it deп¬Ѓnes may be a single point, or a line, or two lines; В
but, if none of these occurs, then it is equivalent (under the group PGL 3ВЎ qВў ) to
the curve deп¬Ѓned by the equation x2 x2 x2 0 (see Exercise 1). Any curve
! ! В¤
1 2 3
equivalent to this one is called a conic (or irreducible conic).
It can be shown (see Exercise 2) that a conic has q 1 points, no three of !

which are collinear. The converse assertion is the content of SegreвЂ™s Theorem:
В
Theorem 4.11 (SegreвЂ™s Theorem) For q odd, a set of q 1 points in PG 2ВЎ qВў ,
!

with no three collinear, is a conic.
Proof Let be an oval. We begin with some combinatorial analysis which ap-
t

plies in any plane of odd order; then we introduce coordinates.

Step 1 Any point not on lies on 0 or 2 tangents.
t

Proof Let p be a point not on . Since q 1 is even, and an even number
t В¤wvu
t !

of points lie on secants through p, an even number must lie on tangents also. Let
xi be the number of points outside which lie on i tangents. Now we have
t

в€‘ xi q2
В¤ ВЎ

в€‘ ixi
В
q 1Вў qВЎ
В¤ !

в€‘i i
В  В
1Вў xi q 1Вў q

(These are all obtained by double counting. The п¬Ѓrst holds because there are q2
points outside ; the second because there are q 1 tangents (one at each point
t !
4.3. SegreвЂ™s Theorem 53

of ), each containing q points not on ; and the third because any two tangents
t t

intersect at a unique point outside .) t
В
From these equations, we see that в€‘ i i 2Вў xi 0. But the term i 1 in the
sum vanishes (any point lies on an even number of tangents); the terms i 0 and В¤
В
i 2 clearly vanish, and i i 2Вў 0 for any other value of i. So xi 0 for all i 0
 В¤ 3В¤

or 2, proving the assertion.

Remark Points not on are called exterior points or interior points according
t

as they lie on 2 or 0 tangents, by analogy with the real case. But the analogy goes
no further. In the real case, every line through an interior point is a secant; this is
false for п¬Ѓnite planes.

В
Step 2 The product of all the non-zero elements of GF qВў is equal to 1.

Proof The solutions of the quadratic x2 1 are x 1 and x 1; these are the
В¤ В¤ В¤
only elements equal to their multiplicative inverses. So, in the product of all the
non-zero elements, everything except 1 and 1 pairs off with its inverse, leaving
these two elements unpaired.

For the next two steps, note that we can choose the coordinate system so that
the sides of a given triangle have equations x 0, y 0 and z 0 (and the opposite
В¤ В¤ В¤

vertices are 1ВЎ 0ВЎ 0y , 0ВЎ 1ВЎ 0y , and 0ВЎ 0ВЎ 1y respectively). WeвЂ™ll call this the triangle
x x x

of reference.

Step 3 Suppose that concurrent lines through the vertices of the triangle of ref-
erence meet the opposite sides in the points 0ВЎ 1ВЎ ay , bВЎ 0ВЎ 1y , and 1ВЎ cВЎ 0y . Then
x x x

abc 1. В¤

Proof The equations of the concurrent lines are z ay, x bz and y cx respec- В¤ В¤ В¤

tively; the point of concurrency must satisfy all three equations, whence abc 1.) В¤

Remark This result is equivalent to the classical Theorem of Menelaus.

Step 4 Let the vertices of the triangle of reference be chosen to be three points
of , and let the tangents at these points have equations z ay, x bz and y cx
t В¤ В¤ В¤
respectively. Then abc 1. В¤
54 4. Various topics

Proof There are q 2 further points of , say p1 pqВЂ 2 . Consider the point
QQ
1ВЎ 0ВЎ 0y . It lies on the tangent z ay, meeting the opposite side in 0ВЎ 1ВЎ ay ; two se-
x В¤ x

cants which are sides of the triangle; and q 2 further secants, through p1 pqВЂ 2 . В© ВЎ 88Q ВЎ
QQ
qВЂ 2
Let the secant through pi meet the opposite side in 0ВЎ 1ВЎ ai . Then a в€ЏiВЃ 1 ai 1, x y В©TВ¤
qВЂ 2 qВЂ 2
by Step 2. If bi ci are similarly deп¬Ѓned, we have also b в€ЏiВЃ 1 bi c в€ЏiВЃ 1 ci 1.
ВЎ В¤ В‚В¤

Thus
qВЂ 2
abc в€Џ ai bi ci
В
1
gВў
В¤
iВЃ 1
But, by Step 3, ai bi ci 1 for i 1ВЎ q 2; so abc 1.
В¤ В¤ ВЎ s8Q В© В¤
QQ

Step 5 Given any three points pВЎ qВЎ r of , there is a conic passing through
t a

pВЎ qВЎ r and having the same tangents at these points as does . t

Proof Choosing coordinates as in Step 4, the conic with equation

yz czx caxy 0

can be checked to have the required property. (For example, 1ВЎ 0ВЎ 0y lies on this x

conic; and, putting z ay, we obtain ay2 0, so 1ВЎ 0ВЎ 0y is the unique point of the
В¤ В¤ x

conic on this line.)

Step 6 Now we are п¬Ѓnished if we can show that the conic of Step 5 passes
a

through an arbitrary further point s of . t

Proof Let and be the conics passing through pВЎ qВЎ s and pВЎ rВЎ s respectively
a a
 %

and having the correct tangents there. Let the conics , and have equations a a a
 Вѓ


f 0, f 0, f 0 respectively. (These equations are determined up to a
В¤ В¤ %

В¤
constant factor.) Let L p Lq Lr Ls be the tangents to at pВЎ qВЎ rВЎ s respectively.
ВЎ ВЎ ВЎ t
Since all three conics are tangent to L p at p, we can choose the normalisation so
that f f f agree identically on L p .
ВЎ ВЎ Вѓ

Now consider the restrictions of f and f to Ls . Both are quadratic functions  %


having a double zero at s, and the values at the point Ls L p coincide; so the A
two functions agree identically on Ls . Similarly, f and f agree on Lq , and f 

and f agree on Lr . But then f , f and f all agree at the point Lq Lr . So the
%
  %
 A
quadratic functions f and f agree on L p , Ls , and Lq Lr , which forces them to
 %
 A
be equal. So the three conics coincide, and our claim is proved (and with it SegreвЂ™s
Theorem).
4.3. SegreвЂ™s Theorem 55

The argument in the last part of the proof can be generalised to give the follow-
ing result (of which it forms the case n q 1, m 2, with L1 Ln the tangents
В¤ ! В¤ ВЎ 88Q ВЎ
QQ
to the oval, and pi1 pi2 the point of tangency of Li taken with multiplicity 2).
ВЎ
E G

В
Proposition 4.12 Let L1 Ln be lines in PG 2ВЎ qВў , no three concurrent. Let
ВЎ 88Q ВЎ
QQ
pi1 pim be points of Li , not necessarily distinct, but lying on none of the other
ВЎ Q88Q ВЎ
Q
LiВ„ . Suppose that, for any three of the lines, there is an algebraic curve of degree
m whose intersections with those lines are precisely the speciп¬Ѓed points (counted
with the appropriate multiplicity). Then there is a curve of degree m, meeting each
line in just the speciп¬Ѓed points.

Proposition 4.12 has been generalised  to arbitrary sets of lines (without
the assumption that no three are concurrent).
В
Proposition 4.13 Let L1 Ln be lines in PG 2ВЎ qВў . Let pi1 pim be points of
ВЎ s8Q ВЎ ВЎ 88Q ВЎ
QQ QQ
Li , not necessarily distinct, but lying on none of the other LiВ„ . Suppose that, for any
three of the lines which form a triangle, and for the set of all lines passing through
any point of the plane (whenever there are at least three such lines), there is an
algebraic curve of degree m whose intersections with those lines are precisely the
speciп¬Ѓed points (counted with the appropriate multiplicity). Then there is a curve
of degree m, meeting each line in just the speciп¬Ѓed points.
В
The analogue of SegreвЂ™s Theorem over GF qВў with even q is false. In this case,
the tangents to an oval S all pass through a single point n, the nucleus of the oval
(Exercise 4); and, for any p S, the set S n p is also an oval. But, if q 4,
ВЁ 
В…D
E В†` G
E G
then at most one of these ovals can be a conic (see Exercise 5: these ovals have
q common points). For sufп¬Ѓciently large q (viz., q 64), and also for q 16, В‡ В¤

there are other ovals, not arising from this construction. We refer to  or  for
up-to-date information on ovals in planes of even order.
We saw that there are ovals in inп¬Ѓnite projective planes which are not conics.
However, there is a remarkable characterisation of conics due to Buekenhout. A
hexagon is said to be Pascalian if the three points of intersection of opposite sides
are collinear. In this terminology, PappusвЂ™ Theorem asserts that a hexagon whose
vertices lie alternately on two lines is Pascalian. Since a pair of lines forms a
вЂњdegenerate conicвЂќ, this theorem is generalised by PascalвЂ™s Theorem:

Theorem 4.14 (PascalвЂ™s Theorem) In a Pappian projective plane, a hexagon in-
scribed in a conic is Pascalian.
56 4. Various topics

We know from Theorem 2.3 that a projective plane satisfying PappusвЂ™ Theo-
В
rem is isomorphic to PG 2ВЎ F for a commutative п¬Ѓeld F. The theorem of Bueken-
Вў

hout completes this circle of ideas. Its proof is group-theoretic, using a character-
В
isation of PGL 2ВЎ F as sharply 3-transitive group due to Tits.
Вў

Theorem 4.15 (BuekenhoutвЂ™s Theorem) Let S be an oval in a projective plane
О . Suppose that every hexagon with vertices in S is Pascalian. Then О  is isomor-
phic to PG 2ВЎ F for some commutative п¬Ѓeld F, and S is a conic in О .
В
Вў

Exercises

1. (a) By completing the square, prove that any homogeneous polynomial of
degree 2 in n variables, over a commutative п¬Ѓeld F with characteristic different
from 2, is equivalent (by non-singular linear transformation) to the polynomial

О±1 x2 О±n x2
1 n
! ! 88Q
QQ Q

(b) Prove that multiplication of any О±i in the above form by a square in F gives
an equivalent form.
(c) Now let F GF qВў and n 3; let О· be a п¬Ѓxed nonsquare in F. Show that
В
В¤ В¤

the curves deп¬Ѓned by x2 , x2 x2 and x2 О·x2 are respectively a line, two lines,

and a point. Show that there exists О± such that О· 1 О±2 . Observing that В¤ !

О±yВў О±x О· x2
В  В  В
2 2
y2
x yВў
! ! В© В¤ ! wВў
ВЎ

prove that the forms x2 x2 x2 and x2 О·x2 О·x2 are equivalent. Deduce the
1 2 3 1 2 3
! ! ! !
В
classiп¬Ѓcation of curves of degree 2 over GF qВў given in the text.
2. Count the number of secants through an exterior point and through an
interior point of an oval in a projective plane of odd order q. Also, count the
number of points of each type.
3. Prove that a curve of degree 2 over any commutative п¬Ѓeld is empty, a point,
a line, a pair of lines, or an oval. Prove also that a curve of degree 2 over a п¬Ѓnite
п¬Ѓeld is non-empty.
4. Prove that, if q is even, then the tangents to an oval in a projective plane
of order q are concurrent. Deduce that there is a set of q 2 points with no three !
collinear, having no tangents (i.e., meeting every line in 0 or 2 points). (Removing
any one of these points then gives an oval.)
4.4. Ovoids and inversive planes 57

Remark. A set of n 2 points in a plane of order n, no three collinear, is called
!

a hyperoval.
5. Prove that, in any inп¬Ѓnite projective plane, for any integer k 1, there is a 

set of points meeting every line in exactly k points.
В
6. Prove that п¬Ѓve points of PG 2ВЎ F , with no three collinear, are contained in Вў
В  В
a unique conic. (Take four of the points to be the standard set 1ВЎ 0ВЎ 0Вў , 0ВЎ 1ВЎ 0Вў ,
0ВЎ 0ВЎ 1Вў and 1ВЎ 1ВЎ 1Вў ; the п¬Ѓfth is 1ВЎ О±ВЎ ОІВў , where О± and ОІ are distinct from one
В  В  В

another and from 0 and 1.)

4.4 Ovoids and inversive planes
Ovoids are 3-dimensional analogues of ovals. They have added importance
because of their connection with inversive planes, which are one-point extensions
of afп¬Ѓne planes. (The traditional example is the relation between the Riemann
sphere and the вЂњextended complex planeвЂќ.)
Fields in this section are commutative.
В
An ovoid in PG 3ВЎ F is a set of points with the properties
Вў t

(O1) no three points of are collinear;
t

(O2) the tangents to through a point of form a plane pencil.
t t

(If a set of points satisп¬Ѓes (O1), a line is called a secant, tangent or passant if it
meets the set in 2, 1 or 0 points respectively. The plane containing the tangents to
an ovoid at a point x is called the tangent plane at x.)
The classical examples of ovoids are the elliptic quadrics. Let О±x2 ОІx Оі ! !

be an irreducible quadratic over the п¬Ѓeld F. The elliptic quadric consists of the
В  В
points of PG 3ВЎ F whose coordinates x1 x2 x3 x4 satisfy
Вў ВЎ ВЎ ВЎ Вў

О±x2 ОІx3 x4 Оіx2
x1 x2 0
3 4
! ! ! В¤ Q

The proof that these points do form an ovoid is left as an exercise.
Over п¬Ѓnite п¬Ѓelds, ovoids are rare. Barlotti and Panella showed the following
analogue of SegreвЂ™s theorem on ovals:
В
Theorem 4.16 Any ovoid in PG 3ВЎ qВў , for q an odd prime power, is an elliptic
58 4. Various topics

For even q, just one further family is known, the SuzukiвЂ“Tits ovoids, which we
will construct in Section 8.4.
An inversive plane is, as said above, a one-point extension of an afп¬Ѓne plane.
В
That is, it is a pair X , where X is a set of points, and a collection of subsets
В‰В€ВЎ
Вўa a

of X called circles, satisfying
(I1) any three points lie in a unique circle;

(I2) if xВЎ y are points and C a circle with x C and y C, then there is a unique
ВЁ 7ВЁ

circle C satisfying y C and C C x;
ВЁ U
В¤ E
A G

(I3) there exist four non-concircular points.
It is readily checked that, for x X , the points different from x and circles con-
ВЁ

taining x form an afп¬Ѓne plane. The order of the inversive plane is the (common)
order of its derived afп¬Ѓne planes.

Proposition 4.17 The points and non-trivial plane sections of an ovoid form an
inversive plane.

Proof A plane section of the ovoid is non-trivial if it contains more than one
t

point. Any three points of are non-collinear, and so deп¬Ѓne a unique plane
t

section. Given x, the points of different from x and the circles containing x
t

correspond to the lines through x not in the tangent plane Tx and the planes through
x different from Tx ; these are the points of the quotient space not incident with the
line Tx x and the lines different from Tx x, which form an afп¬Ѓne plane.
7 7

An inversive plane arising from an ovoid in this way is called egglike. Dem-
bowski proved:

Theorem 4.18 Any inversive plane of even order is egglike (and so its order is a
power of 2).

This is not known to hold for odd order, but no counterexamples are known.
There are conп¬Ѓguration theorems (the bundle theorem and MiquelвЂ™s theorem
respectively) which characterise egglike inversive planes and вЂњclassicalвЂќ inversive
planes (coming from the elliptic quadric) respectively. В
Higher-dimensional objects can also be deп¬Ѓned. A set of points of PG nВЎ F t Вў

is an ovoid if
(O1) no three points of are collinear;
t
4.5. Projective lines 59

(O2вЂ™) the tangents to through a point x of are all the lines through x in a
t t
В
hyperplane of PG nВЎ F . Вў

В
Proposition 4.19 If F is п¬Ѓnite and n 4, then PG nВЎ F contains no ovoid.
В‡ Вў

However, there can exist such ovoids over inп¬Ѓnite п¬Ѓelds (Exercise 3).

Exercises
1. Prove Proposition 4.19. [Hint: it sufп¬Ѓces to prove it for n 4.] В¤
В
2. Prove that, for q odd, a set of points in PG 3ВЎ qВў which satisп¬Ѓes (O1) has
cardinality at most q2 1, with equality if and only if it is an ovoid.
!

(This is true for q even, q 2 also, though the proof is much harder. For q 2,
 В¤
В
the complement of a hyperplane is a set of 8 points in PG 3ВЎ 2Вў satisfying (O1).) В
3. Show that the set of points of PG nВЎ whose coordinates satisfy ВўВ’В‘
Вђ

x2 x2
x1 x2 0
3 n
! ! ! 88Q В¤
QQ

is an ovoid.

4.5 Projective lines
A projective line over a п¬Ѓeld F has no non-trivial structure as an incidence
geometry. From the Kleinian point of view, though, it does have geometric struc- В
ture, derived from the fact that the group PGL 2ВЎ F operates on it. As we saw Вў

earlier, the action of this group is 3-transitive (sharply so if F is commutative),
and can even be 4-transitive for special skew п¬Ѓelds of characteristic 2. However,
we assume in this section that the п¬Ѓeld is commutative.
It is conventional to label the points of the projective line over F with elements
в€ћ , as follows: the point 1ВЎ О±Вў is labelled by О±, and the point 0ВЎ 1Вў
В  В
of F  
В“D
E  
G
by в€ћ. (If we regard points of PG 2ВЎ F as lines in the afп¬Ѓne plane AG 2ВЎ F , then
В  В
Вў Вў

the label of a point is the slope of the corresponding line.)
В
Since PGL 2ВЎ F is sharply 3-transitive, distinguishing three points must give
Вў

unique descriptions to all the others. This is conveniently done by means of the
cross ratio, the function from 4-tuples of distinct points to F 0ВЎ 1 , deп¬Ѓned by EВ”` G
В  В
x1 x3 x4 x2
f x1 x2 x3 x4 В  В
ВЎ ВЎ ВЎ gВў
В¤ Q
x1 x4 x3 x2
60 4. Various topics

In calculating cross ratio, we use the same conventions for dealing with в€ћ as
В
when elements of PGL 2ВЎ F are represented by linear fractional transformations;
Вў

for example, в€ћ О± в€ћ, and О±в€ћ7 ОІв€ћ О±7 ОІ. Slightly differing forms of the cross

ratio are often used; the one given here has the property that f в€ћВЎ 0ВЎ 1ВЎ О±Вў О±.
В
В•
В¤

В
Proposition 4.20 The group of permutations of PG 1ВЎ F preserving the cross ra- Вў
В
tio is PGL 2ВЎ F . Вў

Proof Calculation establishes that linear fractional transformations do preserve
cross ratio. Also, the cross ratio as a function of its fourth argument, with the п¬Ѓrst
three п¬Ѓxed, is one-to-one, so a permutation which preserves cross ratio and п¬Ѓxes
three points is the identity. The result follows from these two assertions.

The cross ratio of four points is unaltered if the arguments are permuted in
В  В
two cycles of length 2: for example, f x3 x4 x1 x2 f x1 x2 x3 x4 . These per-
ВЎ ВЎ ВЎ gВў
В¤ ВЎ ВЎ ВЎ Вў

mutations, together with the identity, form a normal subgroup of index six in the
symmetric group S4 . Thus, in general, six different values are obtained by permut-
ing the arguments. If О± is one of these values, the others are 1 О±, 17 О±, О± 1Вў О±,
В
7

17 1 О±Вў , and О±7 О± 1Вў . There are two special cases where the number of values
В  В

1ВЎ 2ВЎ 1 ,
is smaller, that is, where two of the six coincide. The relevant sets are В©E
2 G
П‰ВЎ П‰ 2 , where П‰ is a primitive cube root of unity. A quadruple of points
is called harmonic if its cross ratios belong to the п¬Ѓrst set, equianharmonic if they
belong to the second. The п¬Ѓrst type occurs over any п¬Ѓeld of characteristic differ-
ent from 2, while the second occurs only if F contains primitive cube roots of 1.
(But note that, if F has characteristic 3, then the two types effectively coincide:
1 2 1 , and the cross ratio of a harmonic quadruple is invariant under all
2
permutations of its arguments!)
In the arguments below, we regard a вЂњquadrupleвЂќ as being an equivalence class
of ordered quadruples (all having the same cross-ratio). So, for example, a har-
monic quadruple (in characteristic different from 3) is a 4-set with a distinguished
partition into two 2-sets.

Proposition 4.21 Suppose that the characteristic of F is not equal to 2. Then the В
group of permutations which preserve the set of harmonic quadruples is PО“L 2ВЎ F . Вў

В
Proof Again, any element of PО“L 2ВЎ F preserves the set of harmonic quadru- Вў
В
ples. To see the converse, note that PGL 2ВЎ F contains a unique conjugacy class Вў
4.5. Projective lines 61

of involutions having two п¬Ѓxed points, and that, if x1 x2 are п¬Ѓxed points and ВЎ
В
x3 x4 a 2-cycle of such an involution, then x1 x2 x3 x4 is harmonic (and the
ВЎ Вў ВЎ ВЎ ВЎ
E G
disthinguished partition is x1 x2 x3 x4 ). Thus, these involutions can be re-
ВЎ EВЎ G ВЎ
rE
E rG
G
constructed from the set of harmonic quadruples. So any permutation preserving
the harmonic quadruples normalises the group G generated by these involutions.
В
We see below that G is PSL 2ВЎ F if F contains square roots of 1, or contains this
В
group as a subgroup of index 2 otherwise. The normaliser of G is thus PО“L 2ВЎ F , Вў
as required.
В
(PSL nВЎ F is the group induced on the projective space by the invertible linear
Вў

transformations with determinant 1.)
We look further at the claim about G in the above proof. A transvection is a В
linear transformation g with all eigenvalues equal to 1, for which ker g 1Вў has В©

codimension 1. In our present case, any 2 2 upper unitriangular matrix different В–

from the identity is a transvection. The collineation of projective space induced
by a transvection is called an elation. An elation is characterised by the fact that
its п¬Ѓxed points form a hyperplane, known as the axis of the elation. Dually, an
elation п¬Ѓxes every line through a point, called the centre of the elation, which is
incident with the axis. In the present case n 2, the centre and axis of an elation В¤
coincide.
В  В
Proposition 4.22 The elations in PGL 2ВЎ F generate PSL 2ВЎ F . Вў Вў

Proof The elations п¬Ѓxing a speciп¬Ѓed point, together with the identity, form a
group which acts sharply transitively on the remaining points. Hence the group
generated by the elations is 2-transitive. If О± 1 17 ОІ and Оі О±7 ОІ, then В‚В¤

В— В— В— В— В—
1ОІ 17 ОІ
11 10 10 0
О±1 Оі1 ОІ
В¤ ВЎ
01 01 0
В˜ В˜ В˜ В˜ В© В˜

so the two-point stabiliser in the group generated by all the elations contains that
В
in PSL 2ВЎ F . But elations have determinant 1, and so the group they generate is
Вў
В
a subgroup of PSL 2ВЎ F . So we have equality. Вў

Now, if two distinct involutions have a common п¬Ѓxed point, then their prod-
uct is a elation. Since all elations are conjugate, all can be realised in this way.
Thus the group G in the proof of Proposition 4.21 contains all elations, and hence
В
contains PSL 2ВЎ F . Вў
62 4. Various topics

We conclude with a different way of giving structure to the projective line.
Suppose that E is a subп¬Ѓeld of F. Then в€ћ E is a subset of the projective line
E D
G
в€ћ F having the structure (in any of the senses previously deп¬Ѓned) of projec-
E D
G В
tive line over E. We call any image of this set under an element of PGL 2ВЎ F a Вў

circle. Then any three points lie in a unique circle. The points and circles form an
incidence structure which is an extension of the point-line structure of afп¬Ѓne space
В
AG nВЎ E , where n is the degree of F over E. (For consider the blocks containing
Вў
в€ћ. On removing the point в€ћ, we can regard F as an E-vector space of rank n; E
itself is an afп¬Ѓne line, and the elements of PGL 2ВЎ F п¬Ѓxing в€ћ are afп¬Ѓne transfor-
В
Вў

mations; so, for any circle C containing в€ћ, C в€ћ is an afп¬Ѓne line. Since three EВ’` G
points lie in a unique circle, every afп¬Ѓne line arises in this way.)
Sometimes, as we will see, this geometry can be represented as the points and
plane sections of a quadric over E. the most familiar example is the Riemann
sphere, which is the projective line over , and can be identiп¬Ѓed with a sphere in
В™

real 3-space so that the вЂњcirclesвЂќ are plane sections.

4.6 Generation and simplicity
В
In this section, we extend to arbitrary rank the statement that PSL nВЎ F is Вў

generated by elations, and show that this group is simple, except in two special
cases.
As before, F is a commutative п¬Ѓeld.
В
Theorem 4.23 For any n 2, the group PSL nВЎ F is generated by all elations.
В‡ Вў

Proof We use induction on n, the case n 2 having been settled by Proposi- В¤
tion 4.22. The induction is based on the fact that, if W is a subspace of the axis of
an elation g, then g induces an elation on the quotient projective space modulo W .
В
Given g PSL nВЎ F , with g 1, we have to express g as a product of elations.
ВЁ Вў 3В¤

We may suppose that g п¬Ѓxes a point x. (For, if xg y x, and h is any elation В¤ 3В¤

mapping x to y, then gh 1 п¬Ѓxes x, and gh 1 is a product of elations if and only if
ВЂ ВЂ

g is.
By induction, we may multiply g by a product of elations (whose axes contain
x) to obtain an element п¬Ѓxing every line through x; so we may assume that g
itself does so. Considering a matrix representing g, and using the fact that g ВЁ
В
PSL nВЎ F , we see that g is an elation.
Вў

Theorem 4.24 Suppose that either n 3, or n 2 and F 3. Then any non-
В‡ В¤  U
В  В
trivial normal subgroup of PGL nВЎ F contains PSL nВЎ F . Вў Вў
4.6. Generation and simplicity 63

Proof We begin with an observation вЂ” if N is a normal subgroup of G, and
g 1 g1 1 gg 1 is the commutator
g N, g1 G, then gВЎ g1 N, where gВЎ g1 ВЂ
ВЂ
ВЁ ВЁ x dy
ВЁ x ey

of g and g1 вЂ” and a lemma:
В
Lemma 4.25 Under the hypotheses of Theorem 4.24, if g PGL nВЎ F maps the ВЁ Вў
В  В
point p1 of PG n 1ВЎ F to the point p2 , then there exists g1 PGL nВЎ F which

п¬Ѓxes p1 and p2 and doesnвЂ™t commute with g.

Proof Case 1: p2 g p3 p2 . We can choose g1 to п¬Ѓx p1 and p2 and move p3 . В¤ 3В¤

(If p1 p2 p3 are not collinear, this is clear. If they are collinear, use the fact that
ВЎ ВЎ
В
PGL 2ВЎ F is 3-transitive on the projective line, which has more than three points.
Вў

Case 2: p2 g p1 . Then g п¬Ѓxes the line p1 p2 , and we can choose coordinates
В¤

on this line so that p1 в€ћ, p2 0. Now g acts as x О±7 x for some О± F. Let g1 В¤ В¤ fb ВЁ
induce x ОІx on this line; then gВЎ g1 induces x ОІ2 x. So choose ОІ 0ВЎ 1ВЎ 1,
fb x y fb 3В¤ 8

as we may since F 3.  


В
So let N be a non-trivial subgroup of PGL nВЎ F . Suppose that g N maps the Вў ВЁ
hyperplane H1 to H2 H1 . By the dual form of the Lemma, there exists g1 п¬Ѓxing 3В¤

H1 and H2 and not commuting with g; then gВЎ g1 п¬Ѓxes H2 . So we may assume x y

that g N п¬Ѓxes a hyperplane H.
ВЁ

Next, suppose that g doesnвЂ™t п¬Ѓx H pointwise. The group of elations with
axis H is isomorphic to the additive group of a vector space whose associated
projective space is H; so there is a transvection g1 with axis H not commuting with
g. Then gВЎ g1 п¬Ѓxes H pointwise. So we may assume that g п¬Ѓxes H pointwise.
x y

If g is not an elation, then it is a homology (induced by a diagonalisable linear
map with two eigenvalues, one having multiplicity n 1; equivalently, its п¬Ѓxed В©

points form a hyperplane and one additional point). Now if g1 is an elation with
axis H, then gВЎ g1 is a non-identity elation.
x y

We conclude that N contains an elation. But then N contains all elations (since В
they are conjugate), whence N contains PSL nВЎ F . Вў
В  В  В
For small n and small п¬Ѓnite п¬Ѓelds F GF qВў , the group PSL nВЎ qВў PSL nВЎ F В¤ g
В¤ Вў

is familiar in other guises. For n 2, recall that it is sharply 3-transitive of degree В¤
В  В  В
q 1. Hence we have PSL 2ВЎ 2Вў S3 , PSL 2ВЎ 3Вў A4 , and PSL 2ВЎ 4Вў A5 (the
! i
hВ¤ i
hВ¤ p
hВ¤
alternating groups of degrees 4 and 5 вЂ” the former is not simple, the latter is the В
unique simple group of order 60). Less obviously, PSL 2ВЎ 5Вў A5 , since it is also hВ¤
В  В
simple of order 60. Furthermore, PSL 2ВЎ 7Вў PSL 3ВЎ 2Вў (the unique simple group j
hВ¤
В  В
of order 168), PSL 2ВЎ 9Вў A6 , and PSL 4ВЎ 2Вў A8 (for reasons we will see later). В•
hВ¤ hВ¤В•
64 4. Various topics

There has been a lot of work, much of it with a very geometric п¬‚avour, con-
cerning groups generated by subsets of the set of elations. For example, McLaugh-
lin [22, 23] found all irreducible groups generated by вЂњfull elation subgroupsвЂќ (all
elations with given centre and axis). This result was put in a wider context by
Cameron and Hall . (In particular, they extended the result to spaces of inп¬Ѓ-
nite dimension.) Note that an important ingredient in the arguments of Cameron
and Hall is Theorem 4.9: under slight additional hypotheses, the set of all elation
centres satisп¬Ѓes the conditions on a colour class in that theorem. The result of
Theorem 4.9, together with the irreducibility of the group, then implies that every
point is an elation centre.

Exercises
1. (a) Prove that the non-negative integer m is the number of п¬Ѓxed points of
В
an element of PGL nВЎ qВў if and only if, when written in the base q, its digits are
non-decreasing and have sum not exceeding n.
(b) (Harder) Prove that the non-negative integer m is the number of п¬Ѓxed
В
points of an element of PО“L nВЎ F if and only if there exists r such that q is a
Вў

power of r and, when m is written in the base r, its digits are non-decreasing and
have sum at most n.
2. Prove that a simple group of order 60 possesses п¬Ѓve Sylow 2-subgroups,
which it permutes by conjugation; deduce that such a group is isomorphic to A5 .
3. Modify the proof of Theorem 4.6.2 to show that, under the same hypotheses,
В
PSL nВЎ F is simple. [It is only necessary to show that the various g1 s can be
Вў
В
chosen to lie in PSL nВЎ F . The only case where this fails is Case 2 of the Lemma
Вў
В
when n 2, F GF 5Вў .]
В¤ В¤

4. (a) Let О  be a projective plane of order 4 containing a hyperoval X (six
points, no three collinear). Prove that there are natural bijections between the set
of lines meeting X in two points and the set of 2-subsets of X ; and between the
set of points outside X and the set of partitions of X into three 2-subsets. Find a
similar description of a set bijective with the set of lines disjoint from X . Hence
show that О  is unique (up to isomorphism).
(b) Let О  be a projective plane of order 4. Prove that any four points, no
three collinear, are contained in a hyperoval. Hence show that there is a unique
projective plane of order 4 (up to isomorphism).
(See Cameron and Van Lint [F] for more on the underlying combinatorial
principle.)
5

Buekenhout geometries

Francis Buekenhout introduced an approach to geometry which has the advan-
tages of being both general, and local (a geometry is studied via its residues of
small rank). In this chapter, we introduce BuekenhoutвЂ™s geometries, and illus-
trate with projective spaces and related objects. Further examples will occur later
(polar spaces).

5.1 Buekenhout geometries

So far, nothing has been said in general about what a вЂњgeometryвЂќ is. Projective
and afп¬Ѓne geometries have been deп¬Ѓned as collections of subspaces, but even the
structure carried by the set of subspaces was left a bit vague (except in Section 3.4,
where we used the inclusion partial order to characterise generalised projective
spaces as lattices). In this section, I will follow an approach due to Buekenhout
(inspired by the early work of Tits on buildings).
Before giving the formal deп¬Ѓnition, let us remark that the subspaces or п¬‚ats
of a projective geometry are of various types (i.e., of various dimensions); may
or may not be incident (two subspaces are incident if one contains the other); and
are partially ordered by inclusion. To allow for duality, we do not want to take the
partial order as basic; and, as we will see, the betweenness relation derived from
it can be deduced from the type and incidence relations. So we regard type and
incidence as basic.
A geometry, or Buekenhout geometry, then, has the following ingredients: a
set X of varieties, a symmetric incidence relation I on X , a п¬Ѓnite set в€† of types,
В
and a type map П„ : X в€†. We require the following axiom:

65
66 5. Buekenhout geometries

(B1) Two varieties of the same type are incident if and only if they are equal.
In other words, a geometry is a multipartite graph, where we have names for the
multipartite blocks (вЂњtypesвЂќ) of the graph. We mostly use familiar geometric lan-
guage for incidence; but sometimes, graph-theoretic terms like diameter and girth
will be useful. But one graph-theoretic concept is vital; a geometry is connected
if the graph of varieties and incidence is connected.
The rank of a geometry is the number of types.
A п¬‚ag is a set of pairwise incident varieties. It follows from (B1) that the
members of a п¬‚ag have different types. A geometry satisп¬Ѓes the transversality
condition if the following strengthening of (B1) holds:
(B2) (a) Every п¬‚ag is contained in a maximal п¬‚ag.

(b) Every maximal п¬‚ag contains one variety of each type.
All geometries here will satisfy transversality.
ВЈВў ВЎ
Let F be a п¬‚ag in a geometry G. The residue GF R F of F is deп¬Ѓned as
follows: the set of varieties is

В§ В¦ В¤ВҐВЎ ВЁВ¦
XF x X F : xIy for all y F;

ВЈВўВ§ ВЎ
the set of types is в€†F в€† П„ F ; and incidence and the type map are the restric-
tions of those in G. It satisп¬Ѓes (B1) (resp. (B2)) if G does. The type of a п¬‚ag
or residue is its image under the type map, and the cotype is the complement of
the type in в€†; so the type of GF is the cotype of F. The rank and corank are the
cardinalities of the type and cotype.
A transversal geometry is called thick (resp. п¬Ѓrm thin) if every п¬‚ag of corank
1 is contained in at least three (resp. at least two, exactly two) maximal п¬‚ags.
A property holds residually in a geometry if it holds in every residue of rank at
least 2. (Residues of rank 1 are sets without structure.) In particular, all geometries
of interest are residually connected; in effect, we assume residual connectedness
as an axiom:
(B3) All residues of rank at least 2 are connected.
The next result illustrates this concept.

Proposition 5.1 Let G be a residually connected transversal geometry, and let x
and y be varieties of X , and i and j distinct types. Then there is a path from x to y
in which all varieties except possibly x and y have type i or j.
5.1. Buekenhout geometries 67

Proof The proof is by induction on the rank. For rank 2, residual connectedness
is just connectedness, and the result holds by deп¬Ѓnition. So assume the result for
all geometries of smaller rank than G.
We show п¬Ѓrst that a two-step path whose middle vertex is not of type i or j
can be replaced by a path of the type required. So let xzy be a path of length 2.
Then x and y lie in the residue of z; so the assertion follows from the inductive
hypothesis.
Now this construction reduces by one the number of interior vertices not of
type i or j on a path with speciп¬Ѓed endpoints. Repeating it as often as necessary
gives the result.

The heart of BuekenhoutвЂ™s idea is that вЂњlocalвЂќ conditions on (or axiomati-
sations of) a geometry are really conditions about residues of small rank. This
motivates the following deп¬Ѓnition of a diagram.
Let в€† be a п¬Ѓnite set. Assume that, for any distinct iВ© j в€†, a class i j of
В¦ 
geometries of rank 2 is given, whose two types of varieties are called вЂњpointsвЂќ
 
and вЂњblocksвЂќ. Suppose that the geometries in ji are the duals of those in i j .
The set в€† equipped with these collections of geometries is called a diagram. It is
represented pictorially by taking a вЂњnodeвЂќ for each element of в€†, with an вЂњedgeвЂќ
between each pair of nodes, the edge from i to j being adorned or labelled with

some symbol for the class i j . We will see examples later.
Вў Вў
A geometry G belongs to the diagram в€†В© i j : iВ© j в€†ВЈ if в€† is the set of types
ВЈ В¦
of G and, for all distinct iВ© j в€†, and all residues GF in G with rank 2 and type
В¦

iВ© jВЁ , GF is isomorphic to a member of i j (where we take points and blocks in
В¤
GF to be varieties of types i and j respectively).
In order to illustrate this idea, we need to deп¬Ѓne some classes of rank 2 ge-
ometries to use in diagrams. Some of these we have met already; but the most
important is the most trivial: A digon is a rank 2 geometry (having at least two
points and at least two blocks) in which any point and block are incident; in other
words, a complete bipartite graph containing a cycle. By abuse of notation, the
вЂњlabelled edgeвЂќ used to represent digons is the absence of an edge! This is done
in part because most of the rank 2 residues of our geometries will be digons, and
this convention leads to uncluttered pictorial representations of diagrams.
A partial linear space is a rank 2 geometry in which two points lie on at most
one line (and dually, two lines meet in at most one point). It is represented by an
edge with the label О , thus:
О
 

68 5. Buekenhout geometries

We already met the concepts linear space and generalised projective plane: they
are partial linear spaces in which the п¬Ѓrst, resp. both, occurrences of вЂњat mostвЂќ
are replaced by вЂњexactlyвЂќ. They are represented by edges with label L and with-
out any label, respectively. (Conveniently, the labels for the self-dual concepts
of вЂњpartial linear spaceвЂќ and вЂњgeneralised projective planeвЂќ coincide with their
mirror-images, while the label for вЂњlinear spaceвЂќ does not.) Note that a projec-
tive plane is a thick generalised projective plane. Another specialisation of linear
spaces, a вЂњcircleвЂќ or вЂњcomplete graphвЂќ, has all lines of cardinality 2; it is denoted
by an edge with label c.
Now we can give an example:
Proposition 5.2 A projective geometry of dimension n has the diagram
  
  


Proof Transversality and residual connectivity are straightforward to check. We
verify the rank 2 residues. Take the types to be the dimensions 0В© 1В© n 1, and

let F be a п¬‚ag of cotype iВ© jВЁ , where i j. В¤
Case 1: j i 1. Then F has the form
ВЎ
 !   !  
 
U0 U1 Ui" Ui# Un"
1 2 1

Its residue consists of all subspaces of dimension i or i 1 between Ui" 1 and Ui# 2 ;
this is clearly the projective plane based on the rank 3 vector space Ui# 2 Ui" 1. \$
%
Case 2: j i 1. Now the п¬‚ag F looks like
    "       

U0 Ui" Ui# Uj U j# Un"
1 1 1 1 1

Its residue consists of all subspaces lying either between Ui" 1 and Ui# 1 , or be-
tween U j 1 and U j# 1 . Any subspace X of the п¬Ѓrst type is incident with any sub-
"
 " &
space Y of the second, since X Ui# 1 U j 1 Y . So the residue is a digon.
In diagrams, it is convenient to label the nodes with the corresponding ele-
ments of в€†. For example, in the case of a projective geometry of dimension n, we
take the labels to be the dimensions of varieties represented by the nodes, thus:
" !'
   

0 1 2 n2 n" 1
5.1. Buekenhout geometries 69

I will use the convention that labels are placed above the nodes where possible.
This reserves the space below the nodes for another use, as follows.
A transversal geometry is said to have orders, or parameters, if there are num-
bers si (for i в€†) with the property that any п¬‚ag of cotype i is contained in exactly
В¦
si 1 maximal п¬‚ags. If so, these numbers si are the orders (or parameters). Now,
if G is a geometry with orders, then G is thick/п¬Ѓrm/thin respectively if and only
%
if all orders are 1/( 1/ 1 respectively. We will write the orders beneath the
ВЎ
nodes, where appropriate. Note that a projective plane of order n (as deп¬Ѓned ear- Вў
lier) has orders nВ© n (in the present terminology). Thus, the geometry PG nВ© qВЈ has
diagram
" )
   

0 1 2 n2 n" 1
q q q q q

We conclude this section with some general results about Buekenhout geome-
tries. These results depend on our convention that a non-edge symbolises a digon.

Proposition 5.3 Let the diagram в€† be the disjoint union of в€†1 and в€†2 , with no
edges between these sets. Then a variety with type in в€†1 and one with type in в€†2
are incident.

Proof We use induction on the rank. For rank 2, в€† is the diagram of a digon,
and the result is true by deп¬Ѓnition. So assume that в€† 10 0
%
2, and (without loss of
generality) that в€†120 0
% 1.
Let Xi be the set of varieties with type in в€†i , for i 1В© 2. By the inductive
ВЎВў
4Вў Вў
В¦ 3 4ВЎ
3
hypothesis, if xВ© y X1 with xIy, then R xВЈ X2 R yВЈ X2 . (Considering R xВЈ ,
5Вў
3
we see that every variety in R xВЈ X2 is incident with y, so the left-hand set is
contained in the right-hand set. Reversing the rЛ† les of x and y establishes the
o
36 Вў В¦
result.) Now by connectedness, R xВЈ X2 is independent of x X1 . (Note that
Proposition 5.1 is being used here.) But this set must be X2 , since every variety in
X2 is incident with some variety in X1 .

A diagram is linear if the вЂњnon-digonвЂќ edges form a simple path, as in the
diagram for projective spaces in Proposition 5.3 above.
Suppose that one particular type in a geometry is selected, and varieties of that
type are called points. Then the shadow, or point-shadow, of a variety x is the set
Вў Вў
Sh xВЈ of varieties incident with x. Sometimes we write Sh0 xВЈ , where 0 is the type
of a point. In a geometry with a linear diagram, the convention is that points are
varieties of the left-most type.
70 5. Buekenhout geometries

Corollary 5.4 In a linear diagram, if xIy, and the type of y is further to the right
8Вў Вў
7
than that of x, then Sh xВЈ Sh yВЈ .
Вў
Proof R xВЈ has disconnected diagram, with points and the type of y in different
Вў
components; so, by Proposition 5.2, every point in R xВЈ is incident with y.

Exercises
1. (a) Construct a geometry which is connected but not residually connected.
(b) Show that, if G has any of the following properties, then so does any residue
of G of rank at least 2: residually connected, transversal, thick, п¬Ѓrm, thin.
2. Show that any generalised projective geometry belongs to the diagram
  
  

9
3. (a) A chamber of a transversal geometry G is a maximal п¬‚ag. Let be the
9
set of chambers of the geometry G. Form a graph with vertex set by joining
two chambers which coincide in all but one variety. G is said to be chamber-
connected if this graph is connected. Prove that a residually connected geometry
is chamber-connected, and a chamber-connected geometry is connected.
ВЈВў
(b) Consider the 3-dimensional afп¬Ѓne space AG 3В© F over the п¬Ѓeld F. Take
three types of varieties: points (type 0), lines (type 1), and parallel classes of
planes (type 2). Incidence between points and lines is as usual; a line L and
a parallel class C of planes are incident if L lies in some plane of C; and any
variety of type 0 is incident with any variety of type 2. Show that this geometry is
chamber-connected but not residually connected.
(c) Let V be a six-dimensional vector space over a п¬Ѓeld F, with a basis e1 2 3 f1 f2 f3 .
e e
Let G be the additive group of V , and let H1 , H2, H3 be the additive groups of the
three subspaces 2 3 f1 , 3 1 f2 , and 1 2 f3 . Form the coset geome-
e e e e e e

try GВ© H1 2 3 : its vaarieties of type i are the cosets of Hi in G, and two
H H
varieties are incident if and only if the corresponding cosets have non-empty in-
tersection. Show that this geometry is connected but not chamber-connected.

5.2 Some special diagrams
In this section, we п¬Ѓrst consider geometries with linear diagram in which all
strokes are linear spaces; then we specialise some or all of these linear spaces to
projective or afп¬Ѓne planes. We will see that the axiomatisations of projective and
afп¬Ѓne spaces can be expressed very simply in this formalism.
5.2. Some special diagrams 71

Theorem 5.5 Let G be a geometry with diagram
0 L1 L2 " )
n2L   

n" 1

Let varieties of type 0 and 1 be points and lines.
B
(a) The points and shadows of lines form a linear space .
B
(b) The shadow of any variety is a subspace of .
8Вў Вў
7
(c) Sh0 xВЈ Sh0 yВЈ if and only if x is incident with y.
(d) If x is a variety and p a point not incident with x, then there is a unique variety
Вў ВЎ Вў
y incident with x and p such that П„ yВЈ П„ xВЈ 1.

Proof (a) We show that two points lie on at least one line by induction on the
rank. There is a path between any two points using only points and lines, by
Proposition 5.2; so it sufп¬Ѓces to show that any such path of length greater than 2
can be shortened. So assume pILIqIMIr, where pВ© qВ© r are points and LВ© M lines. Вў
By the induction hypothesis, the POINTs L and M of R qВЈ lie in a LINE О , a
plane of G incident with L and M. By Corollary 5.4, p and q are incident with О .
Since О  is a linear space, there is a line through p and q. (The convention of using
Вў
capitals for varieties in R qВЈ is used here.)
Now suppose that two lines L and M contain the two points p and q. Consid-
Вў
ering R pВЈ , we п¬Ѓnd a plane О  incident with L and M and hence with p and q. But
О  is a linear space, so L M. ВЎ Вў
В¦
(b) Let y be any variety, and pВ© q Sh0 yВЈ . Since points and lines incident
with y form a linear space by (a), there is a line incident with pВ© q and y. This must
be the unique line incident with p and q; and, by Corollary 5.4, all its points are
Вў
incident with y and so are in Sh0 yВЈ .
Вў Вў
7
(c) The reverse implication is Corollary 5.4. So suppose that Sh0 xВЈ Sh0 yВЈ .
Вў Вў CВў Вў
В¦ 7
Take p Sh0 xВЈ . Then, in R pВЈ , we have Sh1 xВЈ Sh1 yВЈ (since these shadows
are linear subspaces) , and so xIy by induction. (The base case of the induction,
where x is a line, is covered by (b).) Вў
В¦
(d) This is clear if x is a point. Otherwise, choose q Sh0 xВЈ , and apply
Вў
induction in R qВЈ (replacing p by the line pq).
Theorem 5.6 A geometry with diagram
  D!'
  
is a generalised projective space (of п¬Ѓnite dimension).
72 5. Buekenhout geometries

Proof By Theorem 5.5(d), a potential Veblen conп¬Ѓguration lies in a plane; since
planes are projective, VeblenвЂ™s axiom holds. It remains to show that every linear
subspace is the shadow of some variety; this follows easily by induction.

Theorem 5.7 A geometry with diagram
L
  
consists of the points, lines and planes of a (possibly inп¬Ѓnite-dimensional) gener-
alised projective space.

Proof VeblenвЂ™s axiom is veriп¬Ѓed as in Theorem 5.6. It is clear that every point,
line or plane corresponds to a variety.

Remark. Consider geometries with the diagram
L L
  E!'
 

By the argument for Theorem 5.7, we have all the points, lines and planes, and
some higher-dimensional varieties, of a generalised projective space. Examples

arise by taking all the п¬‚ats of dimension at most r 1, where r is the rank. How-
ever, there are other examples. A simple case, with r 4, can be constructed as ВЎ
follows.
F
Let be a projective space of countable dimension over a п¬Ѓnite п¬Ѓeld F.