<<

. 2
( 7)



>>

333
5
A A@ 9 9 8 333
5
8 A6 6 A
333
@9 5
8 4
76 7 6 7 6 5 4444
6
p q r
8
7 6 7 6 6 68 6 6 6 5 5 4444 a2
646464 4 4 4
7 7 88 b2
7 44 4 4
c2



Figure 2.1: The Desargues con¬guration



Theorem 2.2 A projective plane satis¬es Desargues™ Theorem if and only if it is
 p¡ L¢ -transitive for all points p and lines L.
24 2. Projective planes

Proof Let us take another look at the Desargues con¬guration (Fig. 2.1). It is
clear that any central con¬guration with centre o and axis L which carries a1 to a2
is completely determined at every point b1 not on M. (The line a1 a2 meets L at a
¬xed point r and is mapped to b1 b2 ; so b2 is the intersection of ra2 and ob1 .) Now,
if we replace M with another line MB through o, we get another determination of
the action of the collineation. It is easy to see that the condition that these two
speci¬cations agree is precisely Desargues™ Theorem.

The proof shows a little more. Once the action of the central collineation on
one point of M o¡ L M is known, the collineation is completely determined.
C)
 1 
So, if Desargues™ Theorem holds, then these groups of central collineations act
sharply transitively on the relevant set.
Now the additive and multiplicative structures of the ¬eld turn up as groups
of elations and homologies respectively with ¬xed centre and axis. We see im-
mediately that these structures are both groups. More of the axioms are easily
deduced too. For example, let L be a line, and consider all elations with axis L
(and arbitrary centre on L). This set is a group G. For each point p on L, the
elations with centre p form a normal subgroup. These normal subgroups partition
the non-identity elements of G, since a non-identity elation has at most one cen-
tre. But a group having such a partition is abelian (see Exercise 2). So addition is
commutative.
In view of this theorem, projective planes over skew ¬elds are called Desar-
guesian planes.
There is much more to be said about the relationships among con¬guration
theorems, coordinatisation, and central collineations. I refer to Dembowski™s book
for some of these. One such relation is of particular importance.
Pappus™ Theorem is the assertion that, if alternate vertices of a hexagon are
collinear (that is, the ¬rst, third and ¬fth, and also the second, fourth and sixth),
then also the three points of intersection of opposite edges are collinear. See
Fig. 2.2.

Theorem 2.3 A projective plane satis¬es Pappus™ Theorem if and only if it is
 
isomorphic to PG 2¡ F for some commutative ¬eld F.
¢
Proof The proof involves two steps. First, a purely geometric argument shows
that Pappus™ Theorem implies Desargues™. This is shown in Fig. 2.3. This ¬g-
ure shows a potential Desargues con¬guration, in which the required collinearity
is shown by three applications of Pappus™ Theorem. The proof requires four new
2.2. Desarguesian and Pappian planes 25

G G EP PE P P P
G E E P PI I P D D P P P P
E I I D H @ H P@H
G
G
G I E E DH D H @
EH H @D @
GI
IGG H E @ D
I GH @ E EDD
I H HG
I H G @@ ED I
HH DE
G@ IH
G @F F F F F F F I
FFFFFF I HF F H F F

Figure 2.2: Pappus™ Theorem

points, s a1 b1 a2 c2 , t b1 c1 os, u b1 c2 oa1 , and v b2 c2 os. Now Pap-
1 1 1 1
¥ ¥ ¥ ¥
pus™ Theorem, applied to the hexagon osc2 b1 c1 a1 , shows that q¡ u¡ t are collinear;
applied to osb1 c2 b2 a2 , shows that r¡ u¡ v are collinear; and applied to b1tuvc2s (us-
ing the two collinearities just established), shows that p¡ q¡ r are collinear. The
derived collinearities are shown as dotted lines in the ¬gure. (Note that the ¬gure
shows only the generic case of Desargues™ Theorem; it is necessary to take care
of the possible degeneracies as well.)
The second step involves the use of coordinates to show that, in a Desarguesian
plane, Pappus™ Theorem is equivalent to the commutativity of multiplication. (See
Exercise 3.)

In view of this, projective planes over commutative ¬elds are called Pappian
planes.

Remark. It follows from Theorems 2.1 and 2.3 and Wedderburn™s Theorem 1.1
that, in a ¬nite projective plane, Desargues™ Theorem implies Pappus™. No geo-
metric proof of this implication is known.
A similar treatment of af¬ne planes is possible.

Exercises
1. (a) Show that a collineation which has a centre has an axis, and vice versa.
26 2. Projective planes


99 u
9
9
9
9
9
9
9
9
WW V V S S o
W VV S
3W W 3 3X V Q S S t
c1
W X V VY Y Q Y Q S Q Q Q b1
W
WY X Y X Y V V S S Q Q Q Q Q Q a1
Y X WX W QQQ
S
V QQQ
Y7 7Y X W S
V QQQ
S
YT T U U W s
V R
S r
RRRR
T T TUT T V
p q
U U T T TV T T T S S R a2
RRRR
T RT RT R R R
U U VV
b2 R
4 4U R R R
4 c2
4444
4444
44 4 v



Figure 2.3: Pappus implies Desargues



(b) Show that a collineation cannot have more than one centre.

2. The group G has a family of proper normal subgroups which partition the
non-identity elements of G. Prove that G is abelian.
       
3. In PG 2¡ F , let the vertices of a hexagon be 1¡ 0¡ 0¢ , 0¡ 0¡ 1¢ , 0¡ 1¡ 0¢ ,
¢
      `¦ 
1¡ ± 1¡ 1¢ , 1¡ 1¡ 0¢ and β¡ β ± 1¢ 1¢ . Show that alternate vertices lie on the
¦ ¡ 
d   b  ¦  
lines de¬ned by the column vectors 0¡ 0¡ 1¢ and ± 1¡ 1¡ 0¢ . Show that b 
a
c
a

opposite sides meet in the points ±¡ 0¡ 1¢ , 0¡ β±¡ 1¢ and 1¡ β ± 1¢ 1¢ . Show
c ¡e ¦

that the second and third of these lie on the line β¡ 1¡ β±¢ , which also contains c
a
the ¬rst if and only if ±β β±. ¥
2.3. Projectivities 27

2.3 Projectivities
   
Let Π X be a projective plane. Temporarily, let L¢ be the set of points
¢g%©¡ ¥  
incident with L; and let x¢ be the set of lines incident with x. If x is not incident
   
with L, there is a natural bijection between L¢ and x¢ : each point on L lies
on a unique line through x. This bijection is called a perspectivity. By iterating
perspectivities and their inverses, we get a bijection (called a projectivity) between
     
any two sets x¢ or L¢ . In particular, for any line L, we obtain a set P L¢ of
   
projectivities from L¢ to itself (or self-projectivities), and analogously a set P x¢
for any point x.        
The sets P L¢ and P x¢ are actually groups of permutations of L¢ or x¢ .
(Any self-projectivity is the composition of a chain of perspectivities; the product
of two self-projectivities corresponds to the concatenation of the chains, while the
inverse corresponds to the chain in reverse order.) Moreover, these permutation
¢  ¢ 
groups are naturally isomorphic: if g is any projectivity from L1 to L2 , say,
¢  ¢   
then gh 1 P L1 g P L2 . So the group P L¢ of self-projectivities on a line is an
¥
invariant of the projective plane. It turns out that the structure of this group carries
information about the plane which is closely related to concepts we have already
seen.

z2
pp 5 5
5 p
M2
5 p
qq 5
z1 p
pq 5
qp ii i
5 u
i 5i5 qp
qp
5i i qp
5 ii qp
5i p
q
M1
ii
5 p
q
5 p
q
i L
x1 x2 y2 y1


Figure 2.4: 3-transitivity

 
Proposition 2.4 The permutation group P L¢ is 3-transitive.

Proof It suf¬ces to show that there is a projectivity ¬xing any two points x1 2x ¡
L and mapping any further point y1 to any other point y2 . In general, we will use
28 2. Projective planes
  rs¢  
the notation “(L1 to L2 via p)” for the composite of the perspectivities L1 p¢
¢   t 
and p¢ L2 . Let Mi be any other lines through xi (i 1¡ 2), u a point on M1 ,
r ¥
and zi M2 (i 1¡ 2) such that yi uzi are collinear (i 1¡ 2). Then the product of

¥ ¥
(L to M1 via z1 ) and (M1 to L via z2 ) is the required projectivity (Fig. 2.4.)

A permutation group G is sharply t-transitive if, given any two t-tuples of
distinct points, there is a unique element of G carrying the ¬rst to the second (in
order). The main result about groups of projectivities is the following theorem:

G G EP PE P P P
G E E P PI I P D D P P P P
L2
E I I D H @ H P@H
G
G
G I E E DH D H @
EH H D@ @
GI z
G
I
p
I G G H H E@ @E D D
I H
I H H G G @ @ E DE D
L1 L3 I
HH DE
G@ IH
G @F F F F F F F
x
I
FFFFFF I HF F H F F
y



Figure 2.5: Composition of projectivities

 
Theorem 2.5 The group P L¢ of projectivities on a projective plane Π is sharply
3-transitive if and only if Π is pappian.

Proof We sketch the proof. The crucial step is the equivalence of Pappus™ Theo-
rem to the following assertion:

Let L1 2 3 be non-concurrent lines, and x and y two points such
L L
¡¡
that the projectivity

g (L1 to L2 via x) (L2 to L3 via y)
u
¥
¬xes L1 L3 . Then there is a point z such that the projectivity g is
1
equal to (L1 to L3 via z).
2.3. Projectivities 29

The hypothesis is equivalent to the assertion that x¡ y and L1 L3 are collinear.
1
Now the point z is determined, and Pappus™ Theorem is equivalent to the assertion
that it maps a random point p of L1 correctly. (Fig. 2.5 is just Pappus™ Theorem.)
Now this assertion allows long chains of projectivities to be shortened, so that
their action can be controlled.

The converse can be seen another way. By Theorem 2.3, we know that a
 
Pappian plane is isomorphic to PG 2¡ F for some commutative ¬eld F. Now it is
¢
easily checked that any self-projectivity on a line is induced by a linear fractional
 
transformation (an element of PGL 2¡ F ; and this group is sharply 3-transitive.
¢
In the ¬nite case, there are very few 3-transitive groups apart from the sym-
metric and alternating groups; and, for all known non-Pappian planes, the group of
projectivities is indeed symmetric or alternating (though it is not known whether
this is necessarily so). Both possibilities occur; so, at present, all that this provides
us for non-Pappian ¬nite planes is a single Boolean invariant.
In the in¬nite case, however, more interesting possibilities arise. If the plane
has order ±, then the group of projectivities has ± generators, and so has order ±;
so it can never be the symmetric group (which has order 2± ). Barlotti [1] gave an
example in which the stabiliser of any six points is the identity, and the stabiliser
of any ¬ve points is a free group. On the other hand, Schleiermacher [25] showed
that, if the stabiliser of any ¬ve points is trivial, then the stabiliser of any three
points is trivial (and the plane is Pappian).
Further developments involve deeper relationships between projectivities, con-
¬guration theorems, and central collineations; the de¬nition and study of projec-
tivities in other incidence structures; and so on.
3

Coordinatisation of projective
spaces

In this chapter, we describe axiom systems for projective (and af¬ne) spaces. The
principal results are due to Veblen and Young.

 
3.1 The GF 2¡ case
In the last section, we saw an axiomatic characterisation of the geometries
PG¢ 2£ F (as projective planes satisfying Desargues™ Theorem). We turn now to
¤
the characterisation of projective spaces of arbitrary dimension, due to Veblen and
Young. Since the points and the subspaces of any ¬xed dimension determine the
geometry, we expect an axiomatisation in terms of these. Obviously the case of
points and lines will be the simplest.
For the ¬rst of several times in these notes, we will give a detailed and self-
contained argument for the case of GF¢ 2¤ , and treat the general case in rather less
detail.
Theorem 3.1 Let X be a set of points, a set of subsets of X called lines. Assume:
¥
(a) any two points lie on a unique line;
(b) a line meeting two sides of a triangle, not at a vertex, meets the third side;
(c) a line contains exactly three points.
Then X and are the sets of points and lines in a (not necessarily ¬nite dimen-
¥
sional) projective space over GF¢ 2¤ .

31
32 3. Coordinatisation of projective spaces

htb y
¨¦
¦ ¨
¦ ¨
¦ ¨
§§ ¦ ¦ ¨
x© y §
¨
§§ ¦
¨ §
¨§ ¦
¨§ § ¦ z
§¨
§ ¦
§§ ¨ ¦
§
§ ¦
¨
§§ ¦
¨
§
x y© z x© y© z x© y z
 
  
©
Figure 3.1: Veblen™s Axiom

Remark We will see later that, more or less, conditions (a) and (b) charac-
terise arbitrary projective spaces. Condition (c) obviously speci¬es that the ¬eld
is GF¢ 2¤ . The phrase “not necessarily ¬nite dimensional” should be interpreted
as meaning that X and can be identi¬ed with the subspaces of rank 1 and 2
¥
respectively of a vector space over GF¢ 2¤ , not necessarily of ¬nite rank.

Proof Since 1 is the only non-zero scalar in GF¢ 2¤ , the points of projective space
can be identi¬ed with the non-zero vectors; lines are then triples of non-zero vec-
tors with sum 0. Our job is to reconstruct this space.
Let 0 be an element not in X , and set V X 0 . Now de¬ne an addition in  

V as follows:
for all v V, 0 v v 0 v and v v 0;
! "  "  " 
for all x£ y X with x y, x y z, where z is the third point of the line
! # " 
containing x and y.
We claim that V£ is an abelian group. Commutativity is clear; 0 is the
%$ ¢
¤"
identity, and each element is its own inverse. Only the associative law is non-
trivial; and the only non-trivial case, when x£ y£ z are distinct non-collinear points,
follows immediately from Veblen™s axiom (b) (see Fig. 3.1.1).
Next, we de¬ne scalar multiplication over GF¢ 2¤ , in the only possible way:
0 v 0, 1 v v for all v V . The only non-trivial vector space axiom is 1
& & ! "¢
1¤ v 1 v 1 v, and this follows from v v 0.
'
& & "& " 
Finally, 0£ x£ y£ z is a rank 2 subspace if and only if x y z.
 " 
There is a different but even simpler characterisation in terms of hyperplanes,
which foreshadows some later developments.
3.1. The GF¢ 2¤ case 33

Let be any family of subsets of X . The subset Y of X is called a subspace
¥
if any member of which contains two points of Y is wholly contained within Y .
¥
(Thus, the empty set, the whole of X , and any singleton are trivially subspaces.)
The subspace Y is called a hyperplane if it intersects every member of (neces- ¥
sarily in one or all of its points).

Theorem 3.2 Let be a collection of subsets of X . Suppose that
¥
(a) every set in has cardinality 3;
¥
(b) any two points of X lie in at least one member of ; ¥
(c) every point of X lies outside some hyperplane.

Then X and are the point and line sets of a projective geometry over GF¢ 2¤ , not
¥
necessarily ¬nite dimensional.

Proof Let be the set of hyperplanes. For each point x X , we de¬ne a function
( !
px : GF¢ 2¤ by the rule
( )
0 if x H;
!
px H
20¤ ¢
1 1 if x H.
#!
By condition (c), px is non-zero for all x X . !
Let P px : x X . We claim that P 0 is a subspace of the vector space
3
 !  

GF¢ 2¤ of functions from to GF¢ 2¤ . Take x£ y X , and let x£ y£ z be any set
5
4 ( ! 
in containing x and y. Then a hyperplane contains z if and only if it contains
¥
both or neither of x and y; so pz px py . The claim follows.  "
Now the map x px is 1“1, since if px py then px py 0, contradicting
6)  " 
the preceding paragraph. Clearly this map takes members of to lines. The ¥
theorem is proved.

Remark The fact that two points lie in a unique line turns out to be a conse-
quence of the other assumptions.

Exercises
1. Suppose that conditions (a) and (c) of Theorem 3.1.2 hold. Prove that two
points of X lie in at most one member of . ¥
34 3. Coordinatisation of projective spaces

2. Let X satisfy conditions (a) and (c) of Theorem 3.1.1. Let Y be the
87£ ¢
¤¥
set of points x of X with the following property: for any two lines x£ y1 y2 and
 £ 
x£ z1 z2 containing x, the lines y1 z1 and y2 z2 intersect. Prove that Y is a subspace
 £
of X .
3. Let X be a set of points, a collection of subsets of X called lines. Assume
9
that any two points lie in at least one line, and that every point lies outside some
hyperplane. Show that, if the line size is not restricted to be 3, then we cannot
conclude that X and are the point and line sets of a projective space, even if
9
any two points lie on exactly one line. [Hint: In a projective plane, any line is
a hyperplane. Select three lines L1 L2 L3 forming a triangle. Show that it is
£ £
possible to delete some points, and to add some lines, so that L1 L2 L3 remain
£ £
hyperplanes.]
4. Let X satisfy the hypotheses of the previous question. Assume addi-
@95£ ¢
¤
tionally that any two points lie on a unique line, and that some hyperplane is a line
and is ¬nite. Prove that there is a number n such that any hyperplane contains n 1 "
points, any point lies on n 1 lines, and the total number of lines is n2 n 1.
" " "

3.2 An application
I now give a brief application to coding theory. This application is a bit spuri-
ous, since a more general result can be proved by a different but equally simple ar-
gument; but it demonstrates an important link between these ¬elds. Additionally,
the procedure can be reversed, to give characterisations of other combinatorial
designs using theorems about codes.
The problem tackled by the theory of error-correcting codes is to send a mes-
sage over a noisy channel in which some distortion may occur, so that the errors
can be corrected but the price paid in loss of speed is not too great. This is not the
place to discuss coding theory in detail. We simplify by assuming that a message
transmitted over the channel is a sequence of blocks, each block being an n-tuple
of bits (zeros or ones). We also assume that we can be con¬dent that, during the
transmission of a single block, no more than e bits are transmitted incorrectly (a
zero changed to a one or vice versa). The Hamming distance between two blocks
is the number of coordinates in which they differ; that is, the number of errors re-
quired to change one into the other. A code is just a set of “codewords” or blocks
(n-tuples of bits), containing more than one codeword. It is e-error correcting if
the Hamming distance between two codewords is at least 2e 1. (The reason for
"
the name is that, by the triangle inequality, an arbitrary word cannot lie at distance
3.2. An application 35

e or less from more than one codeword. By our assumption, the received word
lies at distance e or less from the transmitted codeword; so this codeword can be
recovered.)
To maximise the transmission rate, we need as many codewords as possible.
The optimum is obtained when every word lies within distance e of a (unique)
codeword. In other words, the closed balls of radius e centred at the codewords
¬ll the space of all words without any overlap! A code with this property is called
perfect e-error-correcting.
Encoding and decoding are made much easier if the code is linear, that is, it is
a GF 2¤ -subspace of the vector space GF¢ 2¤ n of all words.
¢

Theorem 3.3 A linear perfect 1-error-correcting code has length 2d A 1 for some
d 1; there is a unique such code of any length having this form.
B

Remark These unique codes are called Hamming codes. Their relation to pro-
jective spaces will be made clear by the proof below.

Proof Let C be such a code, of length n. Obviously it contains 0. We de¬ne the
weight wt¢ v¤ of any word v to be its Hamming distance from 0. The weight of
any non-zero codeword is at least 3. Now let X be the set of coordinate places,
and the set of triples of points of X which support codewords (i.e., for which a
¥
codeword has 1s in just those positions).
We verify the hypotheses of Theorem 3.1. Condition (c) is clear.
Let x and y be coordinate positions, and let w be the word with entries 1
in positions x and y and 0 elsewhere. w is not a codeword, so there must be a
unique codeword c at distance 1 from w; then c must have weight 3 and support
containing x and y. So (a) holds.
Let x£ y£ r x£ z£ q y£ z£ p be the supports of codewords u£ v£ w. By linear-
 DC
£ DC

ity, u v w is a codeword, and its support is p£ q£ r . So (b) holds.
" " 
A
Thus X and are the points and lines of a projective space PG¢ d 1£ 2¤ for
¥
some d 1; the number of points is n 2d 1. Moreover, it™s easy to see that C
A
B 
is spanned by its words of weight 3 (see Exercise 1), so it is uniquely determined
by d.

Note, incidentally, that the automorphism group of the Hamming code is the
same as that of the projective space, viz. PGL¢ d 2¤ . £
36 3. Coordinatisation of projective spaces

Exercise
1. Prove that a perfect linear code is spanned by its words of minimum weight.
(Use induction on the weight. If w is any non-zero codeword, there is a codeword
u whose support contains e 1 points of the support of w; then u w has smaller
" "
weight than w.)
2. Prove that if a perfect e-error-correcting code of length n exists, then
e
n
‘ i
FE
i 0

is a power of 2. Deduce that, if e 3, then n 7 or 23. (Hint: the cubic polynomial
 
in n factorises.)

Remark. The case n 7 is trivial. For n 23, there is a unique code (up to
 
isometry), the so-called binary Golay code.
3. Verify the following decoding scheme for the Hamming code Hd of length
2d 1. Let Md be the 2d 1 d matrix over GF¢ 2¤ whose rows are the base
A A G
2d 1. Show that the null space of the
A
2 representations of the integers 1£ 2£ 5III
£HHH
matrix Md is precisely Hd . Now let w be received when a codeword is transmitted,
and assume that at most one error has occurred. Prove that
if wHd 0, then w is correct;

if wHd is the ith row of Hd , then the ith position is incorrect.


3.3 The general case
The general coordinatisation theorem is the same as Theorem 3.1, with the
hypothesis “three points per line” weakened to “at least three points per line”.
Accordingly we consider geometries with point set X and line set (where is ¥ ¥
a set of subsets of X ) satisfying:
(LS1) Any line contains at least two points.

(LS2) Two points lie in a unique line.
Such a geometry is called a linear space. Recall that a subspace is a set of points
which contains the (unique) line through any two of its points. In a linear space,
in addition to the trivial subspaces (the empty set, singletons, and X ), any line is
3.3. The general case 37

a subspace. Any subspace, equipped with the lines it contains, is a linear space in
its own right.
A linear space is called thick if it satis¬es:

(LS1+) Any line contains at least three points.

Finally, we will impose Veblen™s Axiom:

(V) A line meeting two sides of a triangle, not at a vertex, meets the third side
also.

Theorem 3.4 (Veblen“Young Theorem) Let X be a linear space, which is
8¥7£ ¢
¤
thick and satis¬es Veblen™s Axiom (V). Then one of the following holds:

/
0;
(a) X 3P

/
0;
(b) X 1,
RQ Q 3¥

(c) X, X 3;
UQ Q  S3¥

T
(d) X is a projective plane;
87£ ¢
¤¥
(e) X is a projective space over a skew ¬eld, not necessarily of ¬nite dimen-
87£ ¢
¤¥
sion.


Remark It is common to restrict to ¬nite-dimensional projective spaces by adding
the additional hypothesis that any chain of subspaces has ¬nite length.

Proof (outline) The key observation provides us with lots of subspaces.

Lemma 3.5 Let X be a linear space satisfying Veblen™s axiom. Let Y be a
87£ ¢
¤¥
subspace, and p a point not in Y ; let Z be the union of the lines joining p to points
of Y . Then Z is a subspace, and Y is a hyperplane in Z.

Proof Let q and r be points of Z. There are several cases, of which the generic
case is that where q£ r Y and the lines pq and pr meet Y in distinct points s, t.
#!
By (V), the lines qr and st meet at a point u of Y . If v is another point of qr, then
by (V) again, the line pv meets st at a point of Y ; so v Z. !
38 3. Coordinatisation of projective spaces

We write this subspace as Y pW .
£V
Now, if L is a line and p a point not in L, then L£ pW is a projective plane. (It
V
is a subspace in which L is a hyperplane; all that has to be shown is that every
line is a hyperplane, which follows once we show that L£ pW contains no proper
V
subspace properly containing a line.)
The theorem is clearly true if there do not exist four non-coplanar points; so
we may suppose that such points do exist.
We claim that Desargues™ Theorem holds. To see this examine the geometric
proof of Desargues™ Theorem in Section 1.2; it is obvious for any non-planar con-
¬guration, and the planar case follows by several applications of the non-planar
case. Now the same argument applies here.
It follows from Theorem 2.1 that every plane in our space can be coordinatised
by a skew ¬eld.
To complete the proof, we have to show that the coordinatisation can be ex-
tended consistently to the whole space. For this, ¬rst one shows that the skew
¬elds coordinatising all planes are the same: this can be proved for planes within
a 3-dimensional subspace by means of central collineations, and the result extends
by connectedness to all pairs of planes. The remainder of the argument involves
careful book-keeping.
From this, we can ¬nd a classi¬cation of not necessarily thick linear spaces
satisfying Veblen™s axiom. The sum of a family of linear spaces is de¬ned as
follows. The point set is the disjoint union of the point sets of the constituent
spaces. Lines are of two types:

(a) all lines of the constituent spaces;

(b) all pairs of points from different constituents.

It is clearly a linear space.

Theorem 3.6 A linear space satisfying Veblen™s axiom is the sum of linear spaces
of types (b)“(e) in the conclusion of Theorem 3.4.

Proof Let X be such a space. De¬ne a relation on X by the rule that x y
87£ ¢
¤¥ X X
if either x y, or the line containing x and y is thick (has at least three points).

We claim ¬rst that is an equivalence relation. Re¬‚exivity and symmetry are
X
clear; so assume that x y and y z, where we may assume that x£ y and z are all
X X
distinct. If these points are collinear, then x z; so suppose not; let x1 and z1 be
X
3.4. Lattices 39

further points on the lines xy and yz respectively. By (V), the line x1 z1 contains a
point of xz different from x and z, as required.
So X is the disjoint union of equivalence classes. We show next that any
equivalence class is a subspace. So let x y. Then x z for every point z of the
X X
line xy; so this line is contained in the equivalence class of x.
So each equivalence class is a non-empty thick linear space, and hence a point,
line, projective plane, or projective space over a skew ¬eld, by Theorem 3.4. It is
clear that the whole space is the sum of its components.

A geometry satisfying the conclusion of Theorem 3.6 is called a generalised
projective space. Its ¬‚ats are its (linear) subspaces; these are precisely the sums
of ¬‚ats of the components. The term “projective space” is sometimes extended to
mean “thick generalised projective space” (i.e., to include single points, lines with
at least three points, and not necessarily Desarguesian projective planes).


3.4 Lattices
Another point of view is to regard the ¬‚ats of a projective space as forming a
lattice. We discuss this in the present section.
A lattice is a set L with two binary operations and (called join and meet),
Y `
and two constants 0 and 1, satisfying the following axioms:

(L1) and are idempotent, commutative, and associative;
Y `
(L2) x x y¤ x and x x y¤ x;
` aY
¢ b
 Y a`
¢ 

(L3) x 0 x, x 1 x.
`  Y 
It follows from these axioms that x y x holds if and only if x y y holds.
`  Y 
We write x y if these equivalent conditions hold. Then L£ is a partially
c @I ¢
¤c
ordered set with greatest element 1 and least element 0; x y and x y are the Y `
least upper bound and greatest lower bound of x and y respectively. Conversely,
any partially ordered set in which least upper bounds and greatest lower bounds
of all pairs of elements exist, and there is a least element and a greatest element,
gives rise to a lattice.
In a lattice, an atom is a non-zero element a such that a x 0 or a for any ` 
x; in other words, an element greater than zero but minimal subject to this. The
lattice is called atomic if every element is a join of atoms.
A lattice is modular if it satis¬es:
40 3. Coordinatisation of projective spaces

(M) If x z, then x y z¤ x y¤ z for all y.
c Y d0 ` aY
¢
¢ e
`
(Note that, if x z, then x y z¤ x y¤ z in any lattice.)
c Y gf ` aY
¢
¢c `h

Theorem 3.7 A lattice is a generalised projective space of ¬nite dimension if and
only if it is atomic and modular.

Proof The forward implication is an exercise. Suppose that the lattice L is atomic
and modular. Let X be the set of atoms. Identify every element z of the lattice with
the set x X : x z . (This map is 1“1; it translates meets to intersections, and
! c
the lattice order to the inclusion order.)
Let x£ y£ z be atoms, and suppose that z x y. Then trivially x z x y.
c Y Y c Y
Suppose that these two elements are unequal. Then y x z. Since y is an atom,
#c Y
y x z¤ 0, and so x y x z¤ x. But x y¤ x z¤ x z, contradicting
' Y 8`
¢
 'I Y ¢8` 8Y
¢
¤ p Y 8i Y ¢
¢`
 Y
modularity. So x z x y. Hence, if we de¬ne lines to be joins of pairs of atoms,
Y Y
it follows that two points lie in a unique line.
Now we demonstrate Veblen™s axiom. Let u£ v be points on x y, x z respec- Y Y
tively, where xyz is a triangle. Suppose that y z¤ u v¤ 0. Then y u v z,
p Y rq Y ¢
¢`
 Y Y T
so y u v y z; in other words, y u v¤ y z¤ y z. On the other hand,
TYY Y Y v Y ut Y sY
¢
¢ `¤

y u v¤ y z¤ y 0 y, contradicting modularity. So the lines y z and
t Y a`h Y taY
¢¢
¢
¤ Y  Y
u v meet. Y
By Theorem 3.6, the linear subspace is a generalised projective geometry.
Clearly the geometry has ¬nite dimension. We leave it as an exercise to show
that every ¬‚at of the geometry is an element of the lattice.

Exercises
1. Complete the proof of Theorem 3.7.
2. Show that an atomic lattice satisfying the distributive laws is modular, and
deduce that it is isomorphic to the lattice of subsets of a ¬nite set.


3.5 Af¬ne spaces
Veblen™s axiom in a linear space is equivalent to the assertion that three non-
collinear points lie in a subspace which is a projective plane. It might be hoped that
replacing “projective plane” by “af¬ne plane” here would give an axiomatisation
of af¬ne spaces. We will see that this is almost true.
3.5. Af¬ne spaces 41

Recall from Section 2.1 the de¬nition of an af¬ne plane, and the fact that par-
allelism is an equivalence relation in an af¬ne plane, where two lines are parallel
if they are equal or disjoint.
Now suppose that X is a linear space satisfying the following condition:
w7£ ¢
¤¥
(AS1) There is a collection of subspaces with the properties that each member
x
of is an af¬ne plane, and that any three non-collinear points are contained
x
in a unique member of . x
First, a few remarks about such spaces.
1. All lines have the same cardinality. For two intersecting lines lie in an af¬ne
plane, and so are equicardinal; and, given two disjoint lines, there is a line meeting
both.
2. It would be simpler to say “any three points generate an af¬ne plane”, where
the subspace generated by a set is the intersection of all subspaces containing it.
This formulation is equivalent if the cardinality of a line is not 2. (Af¬ne spaces
of order greater than 2 have no non-trivial proper subspaces.) But, if lines have
cardinality 2, then any pair of points is a line, and so any three points form a sub-
space which is a generalised projective plane. However, we do want a formulation
which includes this case.
3. In a linear space satisfying (AS1), two lines are said to be parallel if either
they are equal, or they are disjoint and contained in a member of (and hence
x
parallel there). Now Playfair™s Axiom holds: given a line L and point p, there is a
unique line parallel to L and containing p. Moreover, parallelism is re¬‚exive and
symmetric, but not necessarily transitive. We will impose the further condition:

(AS2) Parallelism is transitive.

Theorem 3.8 A linear space satisfying (AS1) and (AS2) is empty, a single point,
a single line, an af¬ne plane, or the con¬guration of points and lines in a (not
necessarily ¬nite-dimensional) af¬ne space.

Proof Let X be the linear space. We may assume that it is not empty, a
87£ ¢
¤¥
point, a line, or an af¬ne plane (i.e., that there exist four non-coplanar points).

Step 1. De¬ne a solid to be the union of all the lines in a parallel class C which
meet a plane Π , where Π contains no line of C. Then any four non-coplanar
8!
x
points lie in a unique solid, and any solid is a subspace.
42 3. Coordinatisation of projective spaces

That a solid is a subspace is shown by considering cases, of which the generic
one runs as follows. Let p£ q be points such that the lines of C containing p and q
meet Π in distinct points x and y. Then x£ y£ p£ q lie in an af¬ne plane; so the line
of C through a point r of pq meets Π in a point x of xy.
Now the fact that the solid is determined by any four non-coplanar points
follows by showing that it has no non-trivial proper subspaces except planes (if
the cardinality of a line is not 2) or by counting (otherwise).
In a solid, if a plane Π contains no parallel to a line L, then Π meets L in a
single point. Hence any two planes in a solid are disjoint or meet in a line.

Step 2. If two planes Π and Πy contain lines from two different parallel classes,
then every line of Π is parallel to a line of Πy .
Suppose not, and let L£ M N be lines of Π, concurrent at p, and py a point of
£
Πy such that the lines Ly My through py parallel to L and M lie in Πy , but the line
£
Ny parallel to N does not. The whole con¬guration lies in a solid; so the planes
NNy and Πy , with a common point py , meet in a line K. Now K is coplanar with N
but not parallel to it, so K N is a point q. Then Π and Πy meet in q, and hence in
€
a line J. But then J is parallel to both L and M, a contradiction.
We call two such planes parallel.

Step 3. We build the embedding projective space. Here I will use a typographic
convention to distinguish the two related spaces: elements of the space we are
building will be written in CAPITALS. The POINTS are the points of X and the
parallel classes of lines of . The LINES are the lines of and the parallel classes
x ¥
of planes in . Incidence is hopefully obvious: as in the old space, together with
x
incidence between any line and its parallel class, as well as between a parallel
class C of lines and a parallel class of planes if a plane in contains a line in C.
 
By Step 2, this is a linear space; and clearly every LINE contains at least three
POINTS. We call the new POINTS and LINES (i.e., the parallel classes) “ideal”.

Step 4. We verify Veblen™s Axiom. Any three points which are not all “ideal” lie
in an af¬ne plane with its points at in¬nity adjoined, i.e., a projective plane. So let
pqr be a triangle of “ideal” POINTS, s and t POINTS on pq and pr respectively,
and o a point of X . Let P£ Q£ R£ S£ T be the lines through o in the parallel classes
p£ q£ r£ s£ t respectively. Then these ¬ve lines lie in a solid, so the planes QR and
ST (having the point o in common) meet in a line u. The parallel class U of u is
the required POINT on qr and st.
3.6. Transitivity of parallelism 43

By Theorem 3.4, the extended geometry is a projective space. The points at
in¬nity obviously form a hyperplane, and so the original points and lines form an
af¬ne space.

We spell the result out in the case where lines have cardinality 2, but referring
only to parallelism, not to the planes.

Corollary 3.9 Suppose that the 2-element subsets of a set X are partitioned into
“parallel classes” so that each class partitions X . Suppose that, for any four
points p£ q£ r£ s X , if pq rs, then pr qs. Then the points and parallelism are
! ‚ ‚
those of an af¬ne space over GF 2¤ .
¢
Here, we have used the notation to mean “belong to the same parallel class

as”. The result follows immediately from the theorem, on de¬ning to be the set
x
of 4-element subsets which are the union of two parallel 2-subsets.

Exercises
1. Give a direct proof of the Corollary, in the spirit of Section 3.1.


3.6 Transitivity of parallelism
A remarkable theorem of Buekenhout [6] shows that it is not necessary to
assume axiom (AS2) (the transitivity of parallelism) in Theorem 3.8, provided
that the cardinality of a line is at least 4. Examples due to Hall [19] show that the
condition really is needed if lines have cardinality 3.

Theorem 3.10 Let X be a linear space satisfying (AS1), in which some line
87£ ¢
¤¥
contains at least four points. Then parallelism is transitive (that is, (AS2) holds),
and so X is an af¬ne space.
w7£ ¢
¤¥
To discuss the counterexamples with 2 or 3 points on a line, some terminology
is helpful. A Steiner triple system is a collection of 3-subsets of a set, any two
points lying in a unique subset of the collection. In other words, it is a linear
space with all lines of cardinality 3, or (in the terminology of Section 1.4) a 2-
v£ 3£ 1¤ design for some (possibly in¬nite) v. A Steiner quadruple system is a set
¢
of 4-subsets of a set, any three points in a unique subset in the collection (that is,
a 3-¢ v£ 4£ 1¤ design.)
44 3. Coordinatisation of projective spaces

A linear space satisfying (AS1), with two points per line, is equivalent to a
Steiner quadruple system: the distinguished 4-sets are the af¬ne planes. There
are Steiner quadruple systems aplenty; most are not af¬ne spaces over GF¢ 2¤ (for
example, because the number of points is not a power of 2). Here is an example.
Let A 1£ 2£ 3£ 4£ 5£ 6 . Let X be the set of all partitions of A into two sets of size
g

3 (so that X 10). De¬ne two types of 4-subsets of X :
ƒQ Q

(a) for all a£ b A, the set of partitions for which a£ b lie in the same part;
!
(b) for all partitions of A into three 2-sets A1 A2 A3 , the set of all partitions into
£ £
two 3-sets each of which is a transversal to the three sets Ai .

This is a Steiner quadruple system with 10 points.
In the case of three points per line, we have the following result, for which we
refer to Bruck [D] and Hall [18, 19]:

Theorem 3.11 (a) In a ¬nite Steiner triple system satisfying (AS1), the number
of points is a power of 3.

(b) For every d 4, there is a Steiner triple system with 3d points which is not
T
isomorphic to AG¢ d 3¤ . £

Exercises
1. Prove that the number of points in a Steiner triple system is either 0 or
congruent to 1 or 3 (mod 6), while the number of points in a Steiner quadruple
system is 0, 1, or congruent to 2 or 4 (mod 6).
(It is known that these conditions are suf¬cient for the existence of Steiner
triple and quadruple systems.)
2. Let X be a Steiner triple system satisfying (AS1). For each point x X ,
¤87£ ¢
¥ !
let „x be the permutation of X which ¬xes x and interchanges y and z whenever
x£ y£ z is a triple. Prove that

(a) „x is an automorphism;

(b) „2 1;

x

y, „x „y 3
(c) for x 1.
# ¢ ¤
4

Various topics

This chapter collects some topics, any of which could be expanded into an entire
chapter (or even a book!): spreads and translation planes; subsets of projective
 
spaces; projective lines; and the simplicity of PSL n¡ F . ¢




4.1 Spreads and translation planes
Let V be a vector space over F, having even rank 2n. A spread is a set of £

subspaces of V of rank n, having the property that any non-zero vector of V lies
in a unique member of . A trivial example occurs when n 1 and consists of
£ ¤ £
all the rank 1 subspaces.
The importance of spreads comes from the following result, whose proof is
straightforward.

Proposition 4.1 Let be a spread in V , and the set of all cosets of members of
£ ¥
 
. Then V¡ is an af¬ne plane. The projective plane obtained by adding a line
£ §¦
¢¥
 
at in¬nity L∞ is p¡ L∞ -transitive for all p L∞ .
¢ ¨



For ¬nite planes, the converse of the last statement is also true. An af¬ne plane
 
with the property that the projective completion is p¡ L∞ -transitive for all p L∞
¢ ¨

is called a translation plane.

Example. Let K be an extension ¬eld of F with degree n. Take V to be a rank
2 vector space over K, and the set of rank 1 K-subspaces. Then, of course,
£
 
the resulting af¬ne plane is AG 2¡ K¢ . Now forget the K-structure, and regard V

45
46 4. Various topics

as an F- vector space. Such a spread is called Desarguesian, because it can be
recognised by the fact that the af¬ne plane is Desarguesian.    
Projectively, a spread is a set of n 1¢ -dimensional ¬‚ats in PG 2n 1¡ F , © © ¢

which partitions the points of F. We will examine further the case n 1, which ¤

will be considered again in section 4.5. Assume that F is commutative.
 
Lemma 4.2 Given three pairwise skew lines in PG 3¡ F , there is a unique com- ¢

mon transversal through any point on one of the lines.

Proof Let L1 L2 L3 be the lines, and p L1 . The quotient space by p is a pro-
¡ ¡ ¨

jective plane PG 2¡ F , and Π1 p¡ L2 and Π2
 
p¡ L3 are distinct lines in this
¢ ¤
 ¤

 
plane; they meet in a unique point, which corresponds to a line M containing p
and lying in Π1 and Π2 , hence meeting L2 and L3 .

Now let be the set of common transversals to the three pairwise skew lines.



The lines in are pairwise skew, by 4.2.
 


Lemma 4.3 A common transversal to three lines of is a transversal to all of



them.

For the proof, see Exercise 2, or Section 8.4.
Let be the set of all common transversals to . The set is called a
  


regulus, and (which is also a regulus) is the opposite regulus. Thus, three
 
pairwise skew lines lie in a unique regulus.
A spread is regular if it contains the regulus through any three of its lines.

Theorem 4.4 A spread is Desarguesian if and only if it is regular.

(The proof of the forward implication is given in Exercise 2.)
If we take a regular spread, and replace the lines in a regulus in this spread
by those in the opposite regulus, the result is still a spread; for a regulus and its
opposite cover the same set of points. This process is referred to as derivation. It
gives rise to non-Desarguesian translation planes:

Proposition 4.5 If F 2, then a derivation of a regular spread is not regular.
 



Proof Choose two reguli 1 , 2 with a unique line in common. If we replace 1
  
by its opposite, then the regulus 2 contains three lines of the spread but is not

contained in the spread.
4.1. Spreads and translation planes 47

It is possible to push this much further. For example, any set of pairwise dis-
joint reguli can be replaced by their opposites. I will not discuss this any further.  
The concept of a spread of lines in PG 3¡ F can be dualised. (For the rest ¢

of the section, F is not assumed commutative.) A set of pairwise skew lines is £

called a cospread if every plane contains a (unique) line of ; in other words, if £ £
 
corresponds to a spread in the dual space PG 2¡ F . Call a bispread if it is both ¢ £
a spread and a cospread.
If F is ¬nite, then every spread is a bispread. (For there are equally many, viz.
   
q 1¢ q2 1¢ , points and planes; and a set of n pairwise skew lines accounts for
! !
 
q 1¢ n points and the same number of planes.) Moreover, a Desarguesian spread
!

is a bispread; and any derivation of a bispread is a bispread (since the concept of a
regulus is self-dual). The reader may be wondering if there are any spreads which
are not bispreads! That they exist in profusion is a consequence of the next result
/
0), and gives us lots of strange translation planes.
(take #"
¤


Theorem 4.6 Let F be an in¬nite ¬eld. Let , be sets of points and planes in " $
 
PG 3¡ F , with the property that F . Then there is a set of pairwise
¢ 201)$('&§%
! "

  £
skew lines, satisfying

(a) the point p lies on a line of if and only if p ;
£ 43 ¨
"


(b) the plane Π contains a line of if and only if Π .
£ 53 ¨
$



 
Proof We use the fact that PG 2¡ F is not the union of fewer than F points and ¢  

lines. For, if S is any set of fewer than F points and lines, and L is a line not in  
S, then L is not covered by its intersections with members of S.
The proof is a simple trans¬nite induction. (Note that we are using the Axiom
of Choice here; but, in any case, the proof is valid over any ¬eld which can be
well-ordered, in particular, over any countable ¬eld.) For readers unfamiliar with
set theory, assume that F is countable, delete the word “trans¬nite”, and ignore
comments about limit ordinals in the following argument.
Let ± be the initial ordinal of cardinality F . Well-order the points of PG 3¡ F
 
  ¢

not in and the planes not in in a single sequence of order-type ±, say Xβ :
 
" $

β ±¢ . Construct a sequence β : β ±¢ by trans¬nite recursion, as follows.
 
0 £ 0
/
Set 0 0. £ ¤

Suppose that β is a successor ordinal, say β γ 1. Suppose that Xβ is a point ¤ !
(the other case is dual). If γ contains a line incident with Xβ, then set β γ.
£ £ 6¤
£
 
Suppose not. Consider the projective plane PG 3¡ F Xβ. By our initial remark, 8¢
7
48 4. Various topics

this plane is not covered by fewer than ± lines of the form L¡ Xβ Xβ (for L γ)  7 9¨
£

or Π7 Xβ (for Π with Xβ Π) and points p¡ Xβ Xβ (for p ). So we can

$ ¨  7 4¨
"

choose a point lying outside the union of these points and lines, that is, a line Lβ
/
containing Xβ so that Lβ L 0 (for L γ ), Lβ Π (for Π ), and p Lβ (for
¤ B¨
£ 3¨ C¨
$ 3¨
A
p ). Set β Lβ .
γ

" £ @¤
£ FD
E G
If β is a limit ordinal, set
γ
β
£ I¤
H £ Q
γP β

Then is the required set of lines.
±
£



Exercises
 
1. Show that, if three pairwise skew lines in PG 3¡ F are given, then it is ¢

possible to choose coordinates so that the lines have equations

x1 x2 0;
¤ ¤


x3 x4 0;
¤ ¤


x3 x1 x4 x2 .
¤ ¡ ¤


Find the common transversals to these three lines.
2. Now let F be commutative. Show that the common transversals to any three
of the lines found in the last question are the original three lines and the lines with
equations

x3 ±, x2 x4 ±
x1 ¤ ¤


for ± F, ± 0¡ 1.
¨ 3¤

Deduce that the Desarguesian spread de¬ned by a quadratic extension of F is
regular.  
3. Prove that Lemma 4.3 is valid in PG 3¡ F if and only if F is commutative. ¢

4. Use Theorem 4.6 to show that, if F is an in¬nite ¬eld, then there is a spread
 
of lines in AG 3¡ F which contains one line from each parallel class.
¢




4.2 Some subsets of projective spaces
For most of the second half of these lecture notes, we will be considering sub-
sets of projective spaces which consist of the points (and general subspaces) on
4.2. Some subsets of projective spaces 49

which certain forms vanish identically. In this section, I will describe some more
basic subsets of projective spaces, and how to recognise them by their intersec-
tions with lines. The ¬rst example is a fact we have already met.
Proposition 4.7 (a) A set S of points in a projective space is a subspace if and
only if, for any line L, S contains no point, one point, or all points of L.
(b) A set S of points in a projective space is a hyperplane if and only if, for any
line L, S contains one or all points of L.
The main theorem of this section is a generalisation of Proposition 4.7(a).
What if we make the condition symmetric, that is, ask that S contains none, one,
all but one, or all points of any line L? The result is easiest to state in the ¬nite
case:
 
Theorem 4.8 Let S be a set of points of X PG n¡ F such that, for any line L,
¤ ¢
S contains none, one, all but one, or all points of S. Suppose that F 2. Then  R


there is a chain
/
0 X0 X1 Xm X
¤ ¤
S U88TS
SQQQ
   
of subspaces of X , such that either S X2iY X2i , or S X2iY
1 2
iX 0 iX 0
VW¤ ¢ VW¤
` `
X2iY 1 . ¢
 
The hypothesis that F 2 is necessary: over the ¬eld GF 2¢ , a line has just
 R


three points, so the four possibilities listed in the hypothesis cover all subsets of
a line. This means that any subset of the projective space satis¬es the hypothesis!
(Nevertheless, see Theorem 4.10 below.)
Note that the hypothesis on S is “self-complementary”, and the conclusion
must re¬‚ect this. It is more natural to talk about a colouring of the points with two
colours such that each colour class satis¬es the hypothesis of the theorem. In this
language, the result can be stated as follows.
Theorem 4.9 Let the points of a (possibly in¬nite) projective space X over F be
coloured with two colours c1 and c2 , such that every colour class contains none,
one, all but one, or all points of any line. Suppose that F 2. Then there is a  U

chain of subspaces of X , and a function f : c1 c2 , so that
a 5a
b ¡
E G

(a) X;
ca
¤
V


(b) for Y , there exist points of Y lying in no smaller subspace in , and all

a a
 
such points have colour f Y . ¢


The proof proceeds in a number of stages.
50 4. Various topics

Step 1 The result is true for a projective plane (Exercise 1).
Now we de¬ne four relations 1, 2, , on X , as follows:
0 0 0 e


p 1 q if p is the only point of its colour on the line pq; (this relation or
f
0

its converse holds between p and q if and only if p and q have different
colours);
p 2 q if there exists q with p r and r q (this holds only if p and q
f
1 1

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