. 3
( 7)


0 0 0

have the same colour);

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

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


Step 2 There do not exist points p¡ 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 ¤ ¤ ¡

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

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

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 .
¨ 3¤ ¡

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 . ¨ 3¤ ¡

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

Step 3 is a partial order. 0

The antisymmetry follows by de¬nition 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 de¬nition. 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. Segre™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
Now let S p¢ q: p q , and T p¢ q:q p.
¤ 30 h
¤ 0

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

q p implies S q¢ T p¢ .
0 p

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

assertions in Step 6.

Remark. The only place in the above argument where the hypothesis F 2  r

was used was in Step 1. Now PG 2¡ 2¢ has seven points; so, up to complementa-
tion, a subset of PG 2¡ 2¢ 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
F GF 2¢ provided that we add the extra hypothesis that no colour class inter-

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

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

4.3 Segre™s Theorem
For projective geometries over ¬nite ¬elds, 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 ¬ner discriminations are possible with ¬nite than with in¬nite
cardinal numbers; for example, all in¬nite subsets of a countably in¬nite set whose
complements are also in¬nite 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
If a polynomial f in x1 xnY 1 is homogeneous, that is, a sum of terms all of
¡ Qs8Q ¡
0 implies f ±v¢ 0 for all ± F. So, it f vanishes
the same degree, then f v¢ g
¤ g
¤ ¨
at a non-zero vector, then it vanishes at the rank 1 subspace (the point of PG n¡ F ) ¢
it spans. The algebraic variety de¬ned 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 ¤

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

odd prime power. The curve it de¬nes may be a single point, or a line, or two lines;  
but, if none of these occurs, then it is equivalent (under the group PGL 3¡ q¢ ) to
the curve de¬ned by the equation x2 x2 x2 0 (see Exercise 1). Any curve
! ! ¤
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 Segre™s Theorem:
Theorem 4.11 (Segre™s Theorem) For q odd, a set of q 1 points in PG 2¡ q¢ ,

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

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

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

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

‘ xi q2
¤ ¡

‘ ixi
q 1¢ q¡
¤ !

‘i i
1¢ xi q 1¢ q
© ¤ ! Q

(These are all obtained by double counting. The ¬rst holds because there are q2
points outside ; the second because there are q 1 tangents (one at each point
t !
4.3. Segre™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
From these equations, we see that ‘ i i 2¢ 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 ¤
i 2 clearly vanish, and i i 2¢ 0 for any other value of i. So xi 0 for all i 0
¤ © h
 ¤ 3¤

or 2, proving the assertion.

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

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 ¬nite planes.

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

Proof The solutions of the quadratic x2 1 are x 1 and x 1; these are the
¤ ¤ ¤
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
¤ ¤ ¤

vertices are 1¡ 0¡ 0y , 0¡ 1¡ 0y , and 0¡ 0¡ 1y respectively). We™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 0¡ 1¡ ay , b¡ 0¡ 1y , and 1¡ c¡ 0y . Then
x x x

abc 1. ¤

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

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

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 ¤ ¤ ¤
respectively. Then abc 1. ¤
54 4. Various topics

Proof There are q 2 further points of , say p1 pq€ 2 . Consider the point
© t ¡ 88Q ¡
1¡ 0¡ 0y . It lies on the tangent z ay, meeting the opposite side in 0¡ 1¡ ay ; two se-
x ¤ x

cants which are sides of the triangle; and q 2 further secants, through p1 pq€ 2 . © ¡ 88Q ¡
q€ 2
Let the secant through pi meet the opposite side in 0¡ 1¡ ai . Then a ∏i 1 ai 1, x y ©T¤
q€ 2 q€ 2
by Step 2. If bi ci are similarly de¬ned, we have also b ∏i 1 bi c ∏i 1 ci 1.
¡ ¤ ‚¤

q€ 2
abc ∏ ai bi ci
© Q
i 1
But, by Step 3, ai bi ci 1 for i 1¡ q 2; so abc 1.
¤ ¤ ¡ s8Q © ¤

Step 5 Given any three points p¡ q¡ r of , there is a conic passing through
t a

p¡ q¡ 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, 1¡ 0¡ 0y lies on this x

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

conic on this line.)

Step 6 Now we are ¬nished if we can show that the conic of Step 5 passes

through an arbitrary further point s of . t

Proof Let and be the conics passing through p¡ q¡ s and p¡ r¡ s respectively
a a

and having the correct tangents there. Let the conics , and have equations a a a

f 0, f 0, f 0 respectively. (These equations are determined up to a
¤ ¤ %

constant factor.) Let L p Lq Lr Ls be the tangents to at p¡ q¡ r¡ s respectively.
¡ ¡ ¡ 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 .
¡ ¡ ƒ

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
quadratic functions f and f agree on L p , Ls , and Lq Lr , which forces them to
be equal. So the three conics coincide, and our claim is proved (and with it Segre™s
4.3. Segre™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
¤ ! ¤ ¡ 88Q ¡
to the oval, and pi1 pi2 the point of tangency of Li taken with multiplicity 2).

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

Proposition 4.12 has been generalised [32] to arbitrary sets of lines (without
the assumption that no three are concurrent).
Proposition 4.13 Let L1 Ln be lines in PG 2¡ q¢ . Let pi1 pim be points of
¡ s8Q ¡ ¡ 88Q ¡
Li , not necessarily distinct, but lying on none of the other Li„ . 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
speci¬ed points (counted with the appropriate multiplicity). Then there is a curve
of degree m, meeting each line in just the speci¬ed points.
The analogue of Segre™s Theorem over GF q¢ 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,
E †` G
then at most one of these ovals can be a conic (see Exercise 5: these ovals have
q common points). For suf¬ciently large q (viz., q 64), and also for q 16, ‡ ¤

there are other ovals, not arising from this construction. We refer to [3] or [14] for
up-to-date information on ovals in planes of even order.
We saw that there are ovals in in¬nite 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, Pappus™ Theorem asserts that a hexagon whose
vertices lie alternately on two lines is Pascalian. Since a pair of lines forms a
“degenerate conic”, this theorem is generalised by Pascal™s Theorem:

Theorem 4.14 (Pascal™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 Pappus™ Theo-
rem is isomorphic to PG 2¡ F for a commutative ¬eld F. The theorem of Bueken-

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

Theorem 4.15 (Buekenhout™s Theorem) Let S be an oval in a projective plane
Π. Suppose that every hexagon with vertices in S is Pascalian. Then Π is isomor-
phic to PG 2¡ F for some commutative ¬eld F, and S is a conic in Π.


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

±1 x2 ±n x2
1 n
! ! 88Q

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

the curves de¬ned by x2 , x2 x2 and x2 ·x2 are respectively a line, two lines,
11 2 1 2 © ©

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

±y¢ ±x · x2
2 2
x y¢
! ! © ¤ ! w¢

prove that the forms x2 x2 x2 and x2 ·x2 ·x2 are equivalent. Deduce the
1 2 3 1 2 3
! ! ! !
classi¬cation of curves of degree 2 over GF q¢ 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 ¬eld is empty, a point,
a line, a pair of lines, or an oval. Prove also that a curve of degree 2 over a ¬nite
¬eld 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 in¬nite projective plane, for any integer k 1, there is a 

set of points meeting every line in exactly k points.
6. Prove that ¬ve points of PG 2¡ F , with no three collinear, are contained in ¢
a unique conic. (Take four of the points to be the standard set 1¡ 0¡ 0¢ , 0¡ 1¡ 0¢ ,
0¡ 0¡ 1¢ and 1¡ 1¡ 1¢ ; the ¬fth is 1¡ ±¡ β¢ , where ± and β are distinct from one

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 af¬ne planes. (The traditional example is the relation between the Riemann
sphere and the “extended complex plane”.)
Fields in this section are commutative.
An ovoid in PG 3¡ F is a set of points with the properties
¢ t

(O1) no three points of are collinear;

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

(If a set of points satis¬es (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 ±x2 βx γ ! !

be an irreducible quadratic over the ¬eld F. The elliptic quadric consists of the
points of PG 3¡ F whose coordinates x1 x2 x3 x4 satisfy
¢ ¡ ¡ ¡ ¢

±x2 βx3 x4 γx2
x1 x2 0
3 4
! ! ! ¤ Q

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

For even q, just one further family is known, the Suzuki“Tits ovoids, which we
will construct in Section 8.4.
An inversive plane is, as said above, a one-point extension of an af¬ne plane.
That is, it is a pair X , where X is a set of points, and a collection of subsets
¢a a

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

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

circle C satisfying y C and C C x;
¨ U
¤ E

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

taining x form an af¬ne plane. The order of the inversive plane is the (common)
order of its derived af¬ne 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

point. Any three points of are non-collinear, and so de¬ne a unique plane

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

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 af¬ne 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 con¬guration theorems (the bundle theorem and Miquel™s theorem
respectively) which characterise egglike inversive planes and “classical” inversive
planes (coming from the elliptic quadric) respectively.  
Higher-dimensional objects can also be de¬ned. A set of points of PG n¡ F t ¢

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

(O2™) the tangents to through a point x of are all the lines through x in a
t t
hyperplane of PG n¡ F . ¢

Proposition 4.19 If F is ¬nite and n 4, then PG n¡ F contains no ovoid.
‡ ¢

However, there can exist such ovoids over in¬nite ¬elds (Exercise 3).

1. Prove Proposition 4.19. [Hint: it suf¬ces to prove it for n 4.] ¤
2. Prove that, for q odd, a set of points in PG 3¡ q¢ which satis¬es (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,
the complement of a hyperplane is a set of 8 points in PG 3¡ 2¢ satisfying (O1).)  
3. Show that the set of points of PG n¡ whose coordinates satisfy ¢’‘

x2 x2
x1 x2 0
3 n
! ! ! 88Q ¤

is an ovoid.

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

earlier, the action of this group is 3-transitive (sharply so if F is commutative),
and can even be 4-transitive for special skew ¬elds of characteristic 2. However,
we assume in this section that the ¬eld is commutative.
It is conventional to label the points of the projective line over F with elements
∞ , as follows: the point 1¡ ±¢ is labelled by ±, and the point 0¡ 1¢
of F  
by ∞. (If we regard points of PG 2¡ F as lines in the af¬ne plane AG 2¡ F , then
¢ ¢

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

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 0¡ 1 , de¬ned by E”` G
x1 x3 x4 x2
  © ¢ © ¢
f x1 x2 x3 x4    
¡ ¡ ¡ g¢
¤ Q
x1 x4 x3 x2
© ¢ © ¢
60 4. Various topics

In calculating cross ratio, we use the same conventions for dealing with ∞ as
when elements of PGL 2¡ F are represented by linear fractional transformations;

for example, ∞ ± ∞, and ±∞7 β∞ ±7 β. Slightly differing forms of the cross
© ¤ ¤

ratio are often used; the one given here has the property that f ∞¡ 0¡ 1¡ ±¢ ±.


Proposition 4.20 The group of permutations of PG 1¡ F preserving the cross ra- ¢
tio is PGL 2¡ F . ¢

Proof Calculation establishes that linear fractional transformations do preserve
cross ratio. Also, the cross ratio as a function of its fourth argument, with the ¬rst
three ¬xed, is one-to-one, so a permutation which preserves cross ratio and ¬xes
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
two cycles of length 2: for example, f x3 x4 x1 x2 f x1 x2 x3 x4 . These per-
¡ ¡ ¡ g¢
¤ ¡ ¡ ¡ ¢

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 ± is one of these values, the others are 1 ±, 17 ±, ± 1¢ ±,
© © 8

17 1 ±¢ , and ±7 ± 1¢ . There are two special cases where the number of values
© ©

1¡ 2¡ 1 ,
is smaller, that is, where two of the six coincide. The relevant sets are ©E
2 G
ω¡ ω 2 , where ω is a primitive cube root of unity. A quadruple of points
and ©E 8
© G
is called harmonic if its cross ratios belong to the ¬rst set, equianharmonic if they
belong to the second. The ¬rst type occurs over any ¬eld 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
© ¤ ¤
permutations of its arguments!)
In the arguments below, we regard a “quadruple” 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  
group of permutations which preserve the set of harmonic quadruples is P“L 2¡ F . ¢

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

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

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

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

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 ¬xed points form a hyperplane, known as the axis of the elation. Dually, an
elation ¬xes 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 ¤
Proposition 4.22 The elations in PGL 2¡ F generate PSL 2¡ F . ¢ ¢

Proof The elations ¬xing a speci¬ed 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 ± 1 17 β and γ ±7 β, then ‚¤
© © ¤

— — — — —
1β 17 β
11 10 10 0
±1 γ1 β
¤ ¡
01 01 0
˜ ˜ ˜ ˜ © ˜

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

Now, if two distinct involutions have a common ¬xed 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
contains PSL 2¡ F . ¢
62 4. Various topics

We conclude with a different way of giving structure to the projective line.
Suppose that E is a sub¬eld of F. Then ∞ E is a subset of the projective line
∞ F having the structure (in any of the senses previously de¬ned) of projec-
tive line over E. We call any image of this set under an element of PGL 2¡ F a ¢

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 af¬ne space
AG n¡ E , where n is the degree of F over E. (For consider the blocks containing
∞. On removing the point ∞, we can regard F as an E-vector space of rank n; E
itself is an af¬ne line, and the elements of PGL 2¡ F ¬xing ∞ are af¬ne transfor-

mations; so, for any circle C containing ∞, C ∞ is an af¬ne line. Since three E’` G
points lie in a unique circle, every af¬ne 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 identi¬ed with a sphere in

real 3-space so that the “circles” are plane sections.

4.6 Generation and simplicity
In this section, we extend to arbitrary rank the statement that PSL n¡ F is ¢

generated by elations, and show that this group is simple, except in two special
As before, F is a commutative ¬eld.
Theorem 4.23 For any n 2, the group PSL n¡ F is generated by all elations.
‡ ¢

Proof We use induction on n, the case n 2 having been settled by Proposi- ¤
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 .
Given g PSL n¡ F , with g 1, we have to express g as a product of elations.
¨ ¢ 3¤

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

mapping x to y, then gh 1 ¬xes x, and gh 1 is a product of elations if and only if
€ €

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

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

Proof We begin with an observation ” if N is a normal subgroup of G, and
g 1 g1 1 gg 1 is the commutator
g N, g1 G, then g¡ g1 N, where g¡ g1 €
¨ ¨ x dy
¨ x ey
¤ ©

of g and g1 ” and a lemma:
Lemma 4.25 Under the hypotheses of Theorem 4.24, if g PGL n¡ F maps the ¨ ¢
point p1 of PG n 1¡ F to the point p2 , then there exists g1 PGL n¡ F which
© ¢ ¨ ¢

¬xes p1 and p2 and doesn™t commute with g.

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

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

Case 2: p2 g p1 . Then g ¬xes the line p1 p2 , and we can choose coordinates

on this line so that p1 ∞, p2 0. Now g acts as x ±7 x for some ± F. Let g1 ¤ ¤ fb ¨
induce x βx on this line; then g¡ g1 induces x β2 x. So choose β 0¡ 1¡ 1,
fb x y fb 3¤ 8

as we may since F 3.  

So let N be a non-trivial subgroup of PGL n¡ F . Suppose that g N maps the ¢ ¨
hyperplane H1 to H2 H1 . By the dual form of the Lemma, there exists g1 ¬xing 3¤

H1 and H2 and not commuting with g; then g¡ g1 ¬xes H2 . So we may assume x y

that g N ¬xes a hyperplane H.

Next, suppose that g doesn™t ¬x 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 g¡ g1 ¬xes H pointwise. So we may assume that g ¬xes 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 ¬xed ©

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

We conclude that N contains an elation. But then N contains all elations (since  
they are conjugate), whence N contains PSL n¡ F . ¢
For small n and small ¬nite ¬elds F GF q¢ , the group PSL n¡ q¢ PSL n¡ F ¤ g
¤ ¢

is familiar in other guises. For n 2, recall that it is sharply 3-transitive of degree ¤
q 1. Hence we have PSL 2¡ 2¢ S3 , PSL 2¡ 3¢ A4 , and PSL 2¡ 4¢ A5 (the
! i
h¤ i
h¤ p

alternating groups of degrees 4 and 5 ” the former is not simple, the latter is the  
unique simple group of order 60). Less obviously, PSL 2¡ 5¢ A5 , since it is also h¤
simple of order 60. Furthermore, PSL 2¡ 7¢ PSL 3¡ 2¢ (the unique simple group j

of order 168), PSL 2¡ 9¢ A6 , and PSL 4¡ 2¢ A8 (for reasons we will see later). •
h¤ h¤•
64 4. Various topics

There has been a lot of work, much of it with a very geometric ¬‚avour, con-
cerning groups generated by subsets of the set of elations. For example, McLaugh-
lin [22, 23] found all irreducible groups generated by “full elation subgroups” (all
elations with given centre and axis). This result was put in a wider context by
Cameron and Hall [11]. (In particular, they extended the result to spaces of in¬-
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 satis¬es 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.

1. (a) Prove that the non-negative integer m is the number of ¬xed points of
an element of PGL n¡ q¢ 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 ¬xed
points of an element of P“L n¡ F if and only if there exists r such that q is a

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 ¬ve 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,
PSL n¡ F is simple. [It is only necessary to show that the various g1 s can be
chosen to lie in PSL n¡ F . The only case where this fails is Case 2 of the Lemma
when n 2, F GF 5¢ .]
¤ ¤

4. (a) Let Π 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 Π is unique (up to isomorphism).
(b) Let Π 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

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 Buekenhout™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 “geometry” is. Projective
and af¬ne geometries have been de¬ned 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 de¬nition, let us remark that the subspaces or ¬‚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 ¬nite set ∆ of types,
and a type map „ : X ∆. We require the following axiom:

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 (“types”) 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 ¬‚ag is a set of pairwise incident varieties. It follows from (B1) that the
members of a ¬‚ag have different types. A geometry satis¬es the transversality
condition if the following strengthening of (B1) holds:
(B2) (a) Every ¬‚ag is contained in a maximal ¬‚ag.

(b) Every maximal ¬‚ag contains one variety of each type.
All geometries here will satisfy transversality.
£¢ ¡
Let F be a ¬‚ag in a geometry G. The residue GF R F of F is de¬ned as
follows: the set of varieties is

§ ¦ ¤¥¡ ¨¦
XF x X F : xIy for all y F;

£¢§ ¡
the set of types is ∆F ∆ „ F ; and incidence and the type map are the restric-
tions of those in G. It satis¬es (B1) (resp. (B2)) if G does. The type of a ¬‚ag
or residue is its image under the type map, and the cotype is the complement of
the type in ∆; 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. ¬rm thin) if every ¬‚ag of corank
1 is contained in at least three (resp. at least two, exactly two) maximal ¬‚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 de¬nition. So assume the result for
all geometries of smaller rank than G.
We show ¬rst 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
Now this construction reduces by one the number of interior vertices not of
type i or j on a path with speci¬ed endpoints. Repeating it as often as necessary
gives the result.

The heart of Buekenhout™s idea is that “local” conditions on (or axiomati-
sations of) a geometry are really conditions about residues of small rank. This
motivates the following de¬nition of a diagram.
Let ∆ be a ¬nite set. Assume that, for any distinct i© j ∆, a class i j of
geometries of rank 2 is given, whose two types of varieties are called “points”
and “blocks”. Suppose that the geometries in ji are the duals of those in i j .
The set ∆ equipped with these collections of geometries is called a diagram. It is
represented pictorially by taking a “node” for each element of ∆, with an “edge”
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.
¢ ¢
A geometry G belongs to the diagram ∆© i j : i© j ∆£ if ∆ is the set of types
£ ¦
of G and, for all distinct i© j ∆, and all residues GF in G with rank 2 and type

i© j¨ , GF is isomorphic to a member of i j (where we take points and blocks in
GF to be varieties of types i and j respectively).
In order to illustrate this idea, we need to de¬ne 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
“labelled edge” 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 Π, thus:

68 5. Buekenhout geometries

We already met the concepts linear space and generalised projective plane: they
are partial linear spaces in which the ¬rst, resp. both, occurrences of “at most”
are replaced by “exactly”. They are represented by edges with label L and with-
out any label, respectively. (Conveniently, the labels for the self-dual concepts
of “partial linear space” and “generalised projective plane” coincide with their
mirror-images, while the label for “linear space” does not.) Note that a projec-
tive plane is a thick generalised projective plane. Another specialisation of linear
spaces, a “circle” or “complete graph”, 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 0© 1© n 1, and

let F be a ¬‚ag of cotype i© j¨ , where i j. ¤
Case 1: j i 1. Then F has the form
 !   !  
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 ¬‚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 ¬rst 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 ∆. 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 ∆) with the property that any ¬‚ag of cotype i is contained in exactly
si 1 maximal ¬‚ags. If so, these numbers si are the orders (or parameters). Now,
if G is a geometry with orders, then G is thick/¬rm/thin respectively if and only
if all orders are 1/( 1/ 1 respectively. We will write the orders beneath the
nodes, where appropriate. Note that a projective plane of order n (as de¬ned ear- ¢
lier) has orders n© n (in the present terminology). Thus, the geometry PG n© q£ has
" )

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 ∆ be the disjoint union of ∆1 and ∆2 , with no
edges between these sets. Then a variety with type in ∆1 and one with type in ∆2
are incident.

Proof We use induction on the rank. For rank 2, ∆ is the diagram of a digon,
and the result is true by de¬nition. So assume that ∆ 10 0
2, and (without loss of
generality) that ∆120 0
% 1.
Let Xi be the set of varieties with type in ∆i , for i 1© 2. By the inductive
4¢ ¢
¦ 3 4¡
hypothesis, if x© y X1 with xIy, then R x£ X2 R y£ X2 . (Considering R x£ ,

we see that every variety in R x£ X2 is incident with y, so the left-hand set is
contained in the right-hand set. Reversing the rˆ les of x and y establishes the
36 ¢ ¦
result.) Now by connectedness, R x£ 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 “non-digon” 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
¢ ¢
Sh x£ of varieties incident with x. Sometimes we write Sh0 x£ , 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
8¢ ¢
than that of x, then Sh x£ Sh y£ .
Proof R x£ has disconnected diagram, with points and the type of y in different
components; so, by Proposition 5.2, every point in R x£ is incident with y.

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, ¬rm, thin.
2. Show that any generalised projective geometry belongs to the diagram

3. (a) A chamber of a transversal geometry G is a maximal ¬‚ag. Let be the
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.
(b) Consider the 3-dimensional af¬ne space AG 3© F over the ¬eld 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 ¬eld 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
A©©@A©©@ ¢ ¢ A©©@
three subspaces 2 3 f1 , 3 1 f2 , and 1 2 f3 . Form the coset geome-
e e e e e e

!££ © ©
try G© H1 2 3 : its vaarieties of type i are the cosets of Hi in G, and two
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 ¬rst 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 af¬ne planes. We will see that the axiomatisations of projective and
af¬ne 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 " )

n" 1

Let varieties of type 0 and 1 be points and lines.
(a) The points and shadows of lines form a linear space .
(b) The shadow of any variety is a subspace of .
8¢ ¢
(c) Sh0 x£ Sh0 y£ 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
¢ ¡ ¢
y incident with x and p such that „ y£ „ x£ 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 suf¬ces to show that any such path of length greater than 2
can be shortened. So assume pILIqIMIr, where p© q© r are points and L© M lines. ¢
By the induction hypothesis, the POINTs L and M of R q£ lie in a LINE Π, a
plane of G incident with L and M. By Corollary 5.4, p and q are incident with Π.
Since Π is a linear space, there is a line through p and q. (The convention of using
capitals for varieties in R q£ is used here.)
Now suppose that two lines L and M contain the two points p and q. Consid-
ering R p£ , we ¬nd a plane Π incident with L and M and hence with p and q. But
Π is a linear space, so L M. ¡ ¢
(b) Let y be any variety, and p© q Sh0 y£ . Since points and lines incident
with y form a linear space by (a), there is a line incident with p© q and y. This must
be the unique line incident with p and q; and, by Corollary 5.4, all its points are
incident with y and so are in Sh0 y£ .
¢ ¢
(c) The reverse implication is Corollary 5.4. So suppose that Sh0 x£ Sh0 y£ .
¢ ¢ C¢ ¢
¦ 7
Take p Sh0 x£ . Then, in R p£ , we have Sh1 x£ Sh1 y£ (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).) ¢
(d) This is clear if x is a point. Otherwise, choose q Sh0 x£ , and apply
induction in R q£ (replacing p by the line pq).
Theorem 5.6 A geometry with diagram
is a generalised projective space (of ¬nite dimension).
72 5. Buekenhout geometries

Proof By Theorem 5.5(d), a potential Veblen con¬guration lies in a plane; since
planes are projective, Veblen™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
consists of the points, lines and planes of a (possibly in¬nite-dimensional) gener-
alised projective space.

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

Remark. Consider geometries with the diagram

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 ¬‚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 ¡
Let be a projective space of countable dimension over a ¬nite ¬eld F.
 © ©
Enumerate the 3-dimensional and 4-dimensional subspaces in lists T0 1 T and
 © © 9
F0 1 F . Now construct a set of 4-dimensional subspaces in stages as fol-
lows. At the nth stage, if Tn is already contained in a member of , do nothing. 9
Otherwise, of the in¬nitely many subspaces Fj which contain Tn , only ¬nitely

many are excluded because they contain any Tm with m n; let Fi be the one with


. 3
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