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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 conп¬Ѓguration

Theorem 2.2 A projective plane satisп¬Ѓes DesarguesвЂ™ Theorem if and only if it is
В pВЎ LВў -transitive for all points p and lines L.
24 2. Projective planes

Proof Let us take another look at the Desargues conп¬Ѓguration (Fig. 2.1). It is
clear that any central conп¬Ѓguration 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
п¬Ѓxed 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
speciп¬Ѓcations agree is precisely DesarguesвЂ™ Theorem.

The proof shows a little more. Once the action of the central collineation on
one point of M oВЎ L M is known, the collineation is completely determined.
C)
 1 
So, if DesarguesвЂ™ Theorem holds, then these groups of central collineations act
sharply transitively on the relevant set.
Now the additive and multiplicative structures of the п¬Ѓeld turn up as groups
of elations and homologies respectively with п¬Ѓxed 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 п¬Ѓelds are called Desar-
guesian planes.
There is much more to be said about the relationships among conп¬Ѓguration
theorems, coordinatisation, and central collineations. I refer to DembowskiвЂ™s book
for some of these. One such relation is of particular importance.
PappusвЂ™ Theorem is the assertion that, if alternate vertices of a hexagon are
collinear (that is, the п¬Ѓrst, third and п¬Ѓfth, 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 satisп¬Ѓes PappusвЂ™ Theorem if and only if it is
В
isomorphic to PG 2ВЎ F for some commutative п¬Ѓeld F.
Вў
Proof The proof involves two steps. First, a purely geometric argument shows
that PappusвЂ™ Theorem implies DesarguesвЂ™. This is shown in Fig. 2.3. This п¬Ѓg-
ure shows a potential Desargues conп¬Ѓguration, in which the required collinearity
is shown by three applications of PappusвЂ™ 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: PappusвЂ™ 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
ВҐ ВҐ ВҐ ВҐ
pusвЂ™ Theorem, applied to the hexagon osc2 b1 c1 a1 , shows that qВЎ uВЎ t are collinear;
applied to osb1 c2 b2 a2 , shows that rВЎ uВЎ v are collinear; and applied to b1tuvc2s (us-
ing the two collinearities just established), shows that pВЎ qВЎ r are collinear. The
derived collinearities are shown as dotted lines in the п¬Ѓgure. (Note that the п¬Ѓgure
shows only the generic case of DesarguesвЂ™ 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, PappusвЂ™ Theorem is equivalent to the commutativity of multiplication. (See
Exercise 3.)

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

Remark. It follows from Theorems 2.1 and 2.3 and WedderburnвЂ™s Theorem 1.1
that, in a п¬Ѓnite projective plane, DesarguesвЂ™ Theorem implies PappusвЂ™. No geo-
metric proof of this implication is known.
A similar treatment of afп¬Ѓne 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.
В  В  В  В
3. In PG 2ВЎ F , let the vertices of a hexagon be 1ВЎ 0ВЎ 0Вў , 0ВЎ 0ВЎ 1Вў , 0ВЎ 1ВЎ 0Вў ,
Вў
В  В  В  `В¦В
1ВЎ О± 1ВЎ 1Вў , 1ВЎ 1ВЎ 0Вў and ОІВЎ ОІ О± 1Вў 1Вў . Show that alternate vertices lie on the
В¦ ВЎВ
d В  bВ  В¦ В
lines deп¬Ѓned by the column vectors 0ВЎ 0ВЎ 1Вў and О± 1ВЎ 1ВЎ 0Вў . Show that bВ
a
c
a
bВ
opposite sides meet in the points О±ВЎ 0ВЎ 1Вў , 0ВЎ ОІО±ВЎ 1Вў and 1ВЎ ОІ О± 1Вў 1Вў . Show
c ВЎe В¦
fВ
that the second and third of these lie on the line ОІВЎ 1ВЎ ОІО±Вў , which also contains c
a
the п¬Ѓrst if and only if О±ОІ ОІО±. ВҐ
2.3. Projectivities 27

2.3 Projectivities
В  В
Let О  X be a projective plane. Temporarily, let LВў be the set of points
incident with L; and let xВў be the set of lines incident with x. If x is not incident
В  В
with L, there is a natural bijection between LВў and xВў : 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
В  В  В
any two sets xВў or LВў . In particular, for any line L, we obtain a set P LВў of
В  В
projectivities from LВў to itself (or self-projectivities), and analogously a set P xВў
for any point x. В  В  В  В
The sets P LВў and P xВў are actually groups of permutations of LВў or xВў .
(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
ВўВ  ВўВ
groups are naturally isomorphic: if g is any projectivity from L1 to L2 , say,
ВўВ  ВўВ  В
then gh 1 P L1 g P L2 . So the group P LВў of self-projectivities on a line is an
ВҐ
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

В
Proposition 2.4 The permutation group P LВў is 3-transitive.

Proof It sufп¬Ѓces to show that there is a projectivity п¬Ѓxing any two points x1 2x ВЎ
L and mapping any further point y1 to any other point y2 . In general, we will use
28 2. Projective planes
В  rsВў В
the notation вЂњ(L1 to L2 via p)вЂќ for the composite of the perspectivities L1 pВў
Вў В  tВ
and pВў L2 . Let Mi be any other lines through xi (i 1ВЎ 2), u a point on M1 ,
r ВҐ
and zi M2 (i 1ВЎ 2) such that yi uzi are collinear (i 1ВЎ 2). Then the product of

ВҐ ВҐ
(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 п¬Ѓrst 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

В
Theorem 2.5 The group P LВў of projectivities on a projective plane О  is sharply
3-transitive if and only if О  is pappian.

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

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

g (L1 to L2 via x) (L2 to L3 via y)
u
ВҐ
п¬Ѓxes 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 xВЎ y and L1 L3 are collinear.
1
Now the point z is determined, and PappusвЂ™ Theorem is equivalent to the assertion
that it maps a random point p of L1 correctly. (Fig. 2.5 is just PappusвЂ™ 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
В
Pappian plane is isomorphic to PG 2ВЎ F for some commutative п¬Ѓeld F. Now it is
Вў
easily checked that any self-projectivity on a line is induced by a linear fractional
В
transformation (an element of PGL 2ВЎ F ; and this group is sharply 3-transitive.
Вў
In the п¬Ѓnite 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 п¬Ѓnite planes is a single Boolean invariant.
In the inп¬Ѓnite case, however, more interesting possibilities arise. If the plane
has order О±, then the group of projectivities has О± generators, and so has order О±;
so it can never be the symmetric group (which has order 2О± ). Barlotti  gave an
example in which the stabiliser of any six points is the identity, and the stabiliser
of any п¬Ѓve points is a free group. On the other hand, Schleiermacher  showed
that, if the stabiliser of any п¬Ѓve 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-
п¬Ѓguration theorems, and central collineations; the deп¬Ѓnition 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 afп¬Ѓne) spaces. The
principal results are due to Veblen and Young.

В
3.1 The GF 2ВЎ case
In the last section, we saw an axiomatic characterisation of the geometries
PGВў 2ВЈ F (as projective planes satisfying DesarguesвЂ™ Theorem). We turn now to
В¤
the characterisation of projective spaces of arbitrary dimension, due to Veblen and
Young. Since the points and the subspaces of any п¬Ѓxed 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 п¬Ѓrst of several times in these notes, we will give a detailed and self-
contained argument for the case of GFВў 2В¤ , 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:
ВҐ
(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 п¬Ѓnite dimen-
ВҐ
sional) projective space over GFВў 2В¤ .

31
32 3. Coordinatisation of projective spaces

htb y
ВЁВ¦
В¦ ВЁ
В¦ ВЁ
В¦ ВЁ
В§В§ В¦ В¦ ВЁ
ВЁ
В§В§ В¦
ВЁ В§
ВЁВ§ В¦
ВЁВ§ В§ В¦ z
В§ВЁ
В§ В¦
В§В§ ВЁ В¦
В§
В§ В¦
ВЁ
В§В§ В¦
ВЁ
В§
 
  
Figure 3.1: VeblenвЂ™s Axiom

Remark We will see later that, more or less, conditions (a) and (b) charac-
terise arbitrary projective spaces. Condition (c) obviously speciп¬Ѓes that the п¬Ѓeld
is GFВў 2В¤ . The phrase вЂњnot necessarily п¬Ѓnite dimensionalвЂќ should be interpreted
as meaning that X and can be identiп¬Ѓed with the subspaces of rank 1 and 2
ВҐ
respectively of a vector space over GFВў 2В¤ , not necessarily of п¬Ѓnite rank.

Proof Since 1 is the only non-zero scalar in GFВў 2В¤ , the points of projective space
can be identiп¬Ѓed 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 deп¬Ѓne an addition in  

V as follows:
for all v V, 0 v v 0 v and v v 0;
! "  "  " 
for all xВЈ y X with x y, x y z, where z is the third point of the line
! # " 
containing x and y.
We claim that VВЈ is an abelian group. Commutativity is clear; 0 is the
%\$ Вў
В¤"
identity, and each element is its own inverse. Only the associative law is non-
trivial; and the only non-trivial case, when xВЈ yВЈ z are distinct non-collinear points,
follows immediately from VeblenвЂ™s axiom (b) (see Fig. 3.1.1).
Next, we deп¬Ѓne scalar multiplication over GFВў 2В¤ , in the only possible way:
0 v 0, 1 v v for all v V . The only non-trivial vector space axiom is 1
& & ! "Вў
1В¤ v 1 v 1 v, and this follows from v v 0.
'
& & "& " 
Finally, 0ВЈ xВЈ yВЈ 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 GFВў 2В¤ case 33

Let be any family of subsets of X . The subset Y of X is called a subspace
ВҐ
if any member of which contains two points of Y is wholly contained within Y .
ВҐ
(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- ВҐ
sarily in one or all of its points).

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

Then X and are the point and line sets of a projective geometry over GFВў 2В¤ , not
ВҐ
necessarily п¬Ѓnite dimensional.

Proof Let be the set of hyperplanes. For each point x X , we deп¬Ѓne a function
( !
px : GFВў 2В¤ by the rule
( )
0 if x H;
!
px H
20В¤ Вў
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
 !  

GFВў 2В¤ of functions from to GFВў 2В¤ . Take xВЈ y X , and let xВЈ yВЈ z be any set
5
4 ( ! 
in containing x and y. Then a hyperplane contains z if and only if it contains
ВҐ
both or neither of x and y; so pz px py . The claim follows.  "
Now the map x px is 1вЂ“1, since if px py then px py 0, contradicting
6)  " 
the preceding paragraph. Clearly this map takes members of to lines. The ВҐ
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 . ВҐ
34 3. Coordinatisation of projective spaces

2. Let X satisfy conditions (a) and (c) of Theorem 3.1.1. Let Y be the
87ВЈ Вў
В¤ВҐ
set of points x of X with the following property: for any two lines xВЈ y1 y2 and
 ВЈ 
xВЈ z1 z2 containing x, the lines y1 z1 and y2 z2 intersect. Prove that Y is a subspace
 ВЈ
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
ВЈ ВЈ
possible to delete some points, and to add some lines, so that L1 L2 L3 remain
ВЈ ВЈ
hyperplanes.]
4. Let X satisfy the hypotheses of the previous question. Assume addi-
@95ВЈ Вў
В¤
tionally that any two points lie on a unique line, and that some hyperplane is a line
and is п¬Ѓnite. 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 п¬Ѓelds. 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 conп¬Ѓdent 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 вЂњcodewordsвЂќ 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
п¬Ѓll 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 2В¤ -subspace of the vector space GFВў 2В¤ n of all words.
Вў

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 deп¬Ѓne the
weight wtВў vВ¤ 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
ВҐ
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 xВЈ yВЈ r xВЈ zВЈ q yВЈ zВЈ p be the supports of codewords uВЈ vВЈ w. By linear-
 DC
ВЈ DC
ВЈ
ity, u v w is a codeword, and its support is pВЈ qВЈ r . So (b) holds.
" " 
A
Thus X and are the points and lines of a projective space PGВў d 1ВЈ 2В¤ for
ВҐ
some d 1; the number of points is n 2d 1. Moreover, itвЂ™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. PGLВў d 2В¤ . ВЈ
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
в€‘ 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 GFВў 2В¤ whose rows are the base
A A G
2d 1. Show that the null space of the
A
2 representations of the integers 1ВЈ 2ВЈ 5III
ВЈHHH
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 вЂњthree points per lineвЂќ weakened to вЂњat least three points per lineвЂќ.
Accordingly we consider geometries with point set X and line set (where is ВҐ ВҐ
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 satisп¬Ѓes:

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

Finally, we will impose VeblenвЂ™s Axiom:

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

Theorem 3.4 (VeblenвЂ“Young Theorem) Let X be a linear space, which is
8ВҐ7ВЈ Вў
В¤
thick and satisп¬Ѓes VeblenвЂ™s Axiom (V). Then one of the following holds:

/
0;
(a) X 3P
ВҐ
/
0;
(b) X 1,
RQ Q 3ВҐ

(c) X, X 3;
UQ Q  S3ВҐ

T
(d) X is a projective plane;
87ВЈ Вў
В¤ВҐ
(e) X is a projective space over a skew п¬Ѓeld, not necessarily of п¬Ѓnite dimen-
87ВЈ Вў
В¤ВҐ
sion.

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

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

Lemma 3.5 Let X be a linear space satisfying VeblenвЂ™s axiom. Let Y be a
87ВЈ Вў
В¤ВҐ
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 qВЈ 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 .
ВЈV
Now, if L is a line and p a point not in L, then LВЈ 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 LВЈ 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 DesarguesвЂ™ Theorem holds. To see this examine the geometric
proof of DesarguesвЂ™ Theorem in Section 1.2; it is obvious for any non-planar con-
п¬Ѓguration, 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 п¬Ѓeld.
To complete the proof, we have to show that the coordinatisation can be ex-
tended consistently to the whole space. For this, п¬Ѓrst one shows that the skew
п¬Ѓelds 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 п¬Ѓnd a classiп¬Ѓcation of not necessarily thick linear spaces
satisfying VeblenвЂ™s axiom. The sum of a family of linear spaces is deп¬Ѓned 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 VeblenвЂ™s axiom is the sum of linear spaces
of types (b)вЂ“(e) in the conclusion of Theorem 3.4.

Proof Let X be such a space. Deп¬Ѓne a relation on X by the rule that x y
87ВЈ Вў
В¤ВҐ X X
if either x y, or the line containing x and y is thick (has at least three points).

We claim п¬Ѓrst that is an equivalence relation. Reп¬‚exivity and symmetry are
X
clear; so assume that x y and y z, where we may assume that xВЈ 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 п¬Ѓeld, 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 п¬‚ats are its (linear) subspaces; these are precisely the sums
of п¬‚ats of the components. The term вЂњprojective spaceвЂќ is sometimes extended to
mean вЂњthick generalised projective spaceвЂќ (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 п¬‚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 yВ¤ x and x x yВ¤ x;
` aY
Вў b
 Y a`
Вў 

(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 LВЈ is a partially
c @I Вў
В¤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 satisп¬Ѓes:
40 3. Coordinatisation of projective spaces

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

Theorem 3.7 A lattice is a generalised projective space of п¬Ѓnite 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 1вЂ“1; it translates meets to intersections, and
! c
the lattice order to the inclusion order.)
Let xВЈ yВЈ 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 zВ¤ 0, and so x y x zВ¤ x. But x yВ¤ x zВ¤ x z, contradicting
' Y 8`
Вў
 'I Y Вў8` 8Y
Вў
В¤ p Y 8i Y Вў
Вў`
 Y
modularity. So x z x y. Hence, if we deп¬Ѓne lines to be joins of pairs of atoms,
Y Y
it follows that two points lie in a unique line.
Now we demonstrate VeblenвЂ™s axiom. Let uВЈ v be points on x y, x z respec- Y Y
tively, where xyz is a triangle. Suppose that y zВ¤ u vВ¤ 0. Then y u v z,
p Y rq Y Вў
Вў`
 Y Y T
so y u v y z; in other words, y u vВ¤ y zВ¤ y z. On the other hand,
TYY Y Y v Y ut Y sY
Вў
Вў `В¤

y u vВ¤ y zВ¤ y 0 y, contradicting modularity. So the lines y z and
t Y a`h Y taY
ВўВў
Вў
В¤ Y  Y
u v meet. Y
By Theorem 3.6, the linear subspace is a generalised projective geometry.
Clearly the geometry has п¬Ѓnite dimension. We leave it as an exercise to show
that every п¬‚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 п¬Ѓnite set.

3.5 Afп¬Ѓne spaces
VeblenвЂ™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 вЂњprojective planeвЂќ by вЂњafп¬Ѓne planeвЂќ here would give an axiomatisation
of afп¬Ѓne spaces. We will see that this is almost true.
3.5. Afп¬Ѓne spaces 41

Recall from Section 2.1 the deп¬Ѓnition of an afп¬Ѓne plane, and the fact that par-
allelism is an equivalence relation in an afп¬Ѓne plane, where two lines are parallel
if they are equal or disjoint.
Now suppose that X is a linear space satisfying the following condition:
w7ВЈ Вў
В¤ВҐ
(AS1) There is a collection of subspaces with the properties that each member
x
of is an afп¬Ѓne 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 afп¬Ѓne
plane, and so are equicardinal; and, given two disjoint lines, there is a line meeting
both.
2. It would be simpler to say вЂњany three points generate an afп¬Ѓne planeвЂќ, 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. (Afп¬Ѓne 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 PlayfairвЂ™s Axiom holds: given a line L and point p, there is a
unique line parallel to L and containing p. Moreover, parallelism is reп¬‚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 afп¬Ѓne plane, or the conп¬Ѓguration of points and lines in a (not
necessarily п¬Ѓnite-dimensional) afп¬Ѓne space.

Proof Let X be the linear space. We may assume that it is not empty, a
87ВЈ Вў
В¤ВҐ
point, a line, or an afп¬Ѓne plane (i.e., that there exist four non-coplanar points).

Step 1. Deп¬Ѓne a solid to be the union of all the lines in a parallel class C which
meet a plane О  , where О  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 pВЈ q be points such that the lines of C containing p and q
meet О  in distinct points x and y. Then xВЈ yВЈ pВЈ q lie in an afп¬Ѓne plane; so the line
of C through a point r of pq meets О  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 О  contains no parallel to a line L, then О  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 О  and О y contain lines from two different parallel classes,
then every line of О  is parallel to a line of О y .
Suppose not, and let LВЈ M N be lines of О , concurrent at p, and py a point of
ВЈ
О y such that the lines Ly My through py parallel to L and M lie in О y , but the line
ВЈ
Ny parallel to N does not. The whole conп¬Ѓguration lies in a solid; so the planes
NNy and О y , 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 О  and О y meet in q, and hence in
ВЂ
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 ВҐ
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.
ВЃ ВЃ
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) вЂњidealвЂќ.

Step 4. We verify VeblenвЂ™s Axiom. Any three points which are not all вЂњidealвЂќ lie
in an afп¬Ѓne plane with its points at inп¬Ѓnity adjoined, i.e., a projective plane. So let
pqr be a triangle of вЂњidealвЂќ POINTS, s and t POINTS on pq and pr respectively,
and o a point of X . Let PВЈ QВЈ RВЈ SВЈ T be the lines through o in the parallel classes
pВЈ qВЈ rВЈ sВЈ t respectively. Then these п¬Ѓve 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
inп¬Ѓnity obviously form a hyperplane, and so the original points and lines form an
afп¬Ѓne 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
вЂњparallel classesвЂќ so that each class partitions X . Suppose that, for any four
points pВЈ qВЈ rВЈ s X , if pq rs, then pr qs. Then the points and parallelism are
! В‚ В‚
those of an afп¬Ѓne space over GF 2В¤ .
Вў
Here, we have used the notation to mean вЂњbelong to the same parallel class
В‚
asвЂќ. The result follows immediately from the theorem, on deп¬Ѓning 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  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  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
87ВЈ Вў
В¤ВҐ
contains at least four points. Then parallelism is transitive (that is, (AS2) holds),
and so X is an afп¬Ѓne space.
w7ВЈ Вў
В¤ВҐ
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-
vВЈ 3ВЈ 1В¤ design for some (possibly inп¬Ѓnite) v. A Steiner quadruple system is a set
Вў
of 4-subsets of a set, any three points in a unique subset in the collection (that is,
a 3-Вў vВЈ 4ВЈ 1В¤ 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 afп¬Ѓne planes. There
are Steiner quadruple systems aplenty; most are not afп¬Ѓne spaces over GFВў 2В¤ (for
example, because the number of points is not a power of 2). Here is an example.
Let A 1ВЈ 2ВЈ 3ВЈ 4ВЈ 5ВЈ 6 . Let X be the set of all partitions of A into two sets of size
g

3 (so that X 10). Deп¬Ѓne two types of 4-subsets of X :
ВѓQ Q

(a) for all aВЈ b A, the set of partitions for which aВЈ 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
ВЈ ВЈ
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 п¬Ѓnite 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 AGВў d 3В¤ . ВЈ

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 sufп¬Ѓcient for the existence of Steiner
triple and quadruple systems.)
2. Let X be a Steiner triple system satisfying (AS1). For each point x X ,
В¤87ВЈ Вў
ВҐ !
let П„x be the permutation of X which п¬Ѓxes x and interchanges y and z whenever
xВЈ yВЈ z is a triple. Prove that

(a) П„x is an automorphism;

(b) П„2 1;

x

y, П„x П„y 3
(c) for x 1.
# Вў В¤
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
В
spaces; projective lines; and the simplicity of PSL nВЎ F . Вў

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

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
ВЈ В¤ ВЈ
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
ВЈ ВҐ
В
. Then VВЎ is an afп¬Ѓne plane. The projective plane obtained by adding a line
ВЈ В§В¦
ВўВҐ
В
at inп¬Ѓnity Lв€ћ is pВЎ Lв€ћ -transitive for all p Lв€ћ .
Вў ВЁ

For п¬Ѓnite planes, the converse of the last statement is also true. An afп¬Ѓne plane
В
with the property that the projective completion is pВЎ Lв€ћ -transitive for all p Lв€ћ
Вў ВЁ

is called a translation plane.

Example. Let K be an extension п¬Ѓeld 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,
ВЈ
В
the resulting afп¬Ѓne plane is AG 2ВЎ KВў . 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 afп¬Ѓne plane is Desarguesian. В  В
Projectively, a spread is a set of n 1Вў -dimensional п¬‚ats in PG 2n 1ВЎ F , В© В© Вў

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

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

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-
ВЎ ВЎ ВЁ

jective plane PG 2ВЎ F , and О 1 pВЎ L2 and О 2
В
pВЎ L3 are distinct lines in this
Вў В¤
 В¤

 
plane; they meet in a unique point, which corresponds to a line M containing p
and lying in О 1 and О 2 , 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. В
The concept of a spread of lines in PG 3ВЎ F can be dualised. (For the rest Вў

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

called a cospread if every plane contains a (unique) line of ; in other words, if ВЈ ВЈ
В
corresponds to a spread in the dual space PG 2ВЎ F . Call a bispread if it is both Вў ВЈ
If F is п¬Ѓnite, then every spread is a bispread. (For there are equally many, viz.
В  В
q 1Вў q2 1Вў , points and planes; and a set of n pairwise skew lines accounts for
! !
В
q 1Вў 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 #"
В¤

Theorem 4.6 Let F be an inп¬Ѓnite п¬Ѓeld. Let , be sets of points and planes in " \$
В
PG 3ВЎ F , with the property that F . Then there is a set of pairwise
Вў 201)\$('&В§%
! "

  ВЈ
skew lines, satisfying

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

(b) the plane О  contains a line of if and only if О  .
ВЈ 53 ВЁ
\$

В
Proof We use the fact that PG 2ВЎ F is not the union of fewer than F points and Вў  

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 transп¬Ѓnite induction. (Note that we are using the Axiom
of Choice here; but, in any case, the proof is valid over any п¬Ѓeld which can be
well-ordered, in particular, over any countable п¬Ѓeld.) For readers unfamiliar with
set theory, assume that F is countable, delete the word вЂњtransп¬ЃniteвЂќ, and ignore
comments about limit ordinals in the following argument.
Let О± be the initial ordinal of cardinality F . Well-order the points of PG 3ВЎ F
В
  Вў

not in and the planes not in in a single sequence of order-type О±, say XОІ :
В
" \$

ОІ О±Вў . Construct a sequence ОІ : ОІ О±Вў by transп¬Ѓnite recursion, as follows.
В
0 ВЈ 0
/
Set 0 0. ВЈ В¤

Suppose that ОІ is a successor ordinal, say ОІ Оі 1. Suppose that XОІ is a point В¤ !
(the other case is dual). If Оі contains a line incident with XОІ, then set ОІ Оі.
ВЈ ВЈ 6В¤
ВЈ
В
Suppose not. Consider the projective plane PG 3ВЎ F XОІ. By our initial remark, 8Вў
7
48 4. Various topics

this plane is not covered by fewer than О± lines of the form LВЎ XОІ XОІ (for L Оі)  7 9ВЁ
ВЈ

or О 7 XОІ (for О  with XОІ О ) and points pВЎ XОІ XОІ (for p ). So we can
@ВЁ
\$ ВЁ  7 4ВЁ
"

choose a point lying outside the union of these points and lines, that is, a line LОІ
/
containing XОІ so that LОІ L 0 (for L Оі ), LОІ О  (for О  ), and p LОІ (for
В¤ BВЁ
ВЈ 3ВЁ CВЁ
\$ 3ВЁ
A
p ). Set ОІ LОІ .
Оі
4ВЁ
" ВЈ @В¤
ВЈ FD
E G
If ОІ is a limit ordinal, set
Оі
ОІ
ВЈ IВ¤
H ВЈ Q
ОіP ОІ

Then is the required set of lines.
О±
ВЈ

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

possible to choose coordinates so that the lines have equations

x1 x2 0;
В¤ В¤

x3 x4 0;
В¤ В¤

x3 x1 x4 x2 .
В¤ ВЎ В¤

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 О±, x2 x4 О±
x1 В¤ В¤

for О± F, О± 0ВЎ 1.
ВЁ 3В¤

Deduce that the Desarguesian spread deп¬Ѓned by a quadratic extension of F is
regular. В
3. Prove that Lemma 4.3 is valid in PG 3ВЎ F if and only if F is commutative. Вў

4. Use Theorem 4.6 to show that, if F is an inп¬Ѓnite п¬Ѓeld, then there is a spread
В
of lines in AG 3ВЎ F which contains one line from each parallel class.
Вў

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 п¬Ѓrst 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 п¬Ѓnite
case:
В
Theorem 4.8 Let S be a set of points of X PG nВЎ F such that, for any line L,
В¤ Вў
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
В¤ В¤
S U88TS
SQQQ
В  В
of subspaces of X , such that either S X2iY X2i , or S X2iY
1 2
iX 0 iX 0
VWВ¤ Вў VWВ¤
` `
X2iY 1 . Вў
В
The hypothesis that F 2 is necessary: over the п¬Ѓeld GF 2Вў , 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 satisп¬Ѓes the hypothesis!
(Nevertheless, see Theorem 4.10 below.)
Note that the hypothesis on S is вЂњself-complementaryвЂќ, and the conclusion
must reп¬‚ect this. It is more natural to talk about a colouring of the points with two
colours such that each colour class satisп¬Ѓes the hypothesis of the theorem. In this
language, the result can be stated as follows.
Theorem 4.9 Let the points of a (possibly inп¬Ѓnite) 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 ВЎ
E G

(a) X;
ca
В¤
V

(b) for Y , there exist points of Y lying in no smaller subspace in , and all
dВЁ
a a
В
such points have colour f Y . Вў

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 deп¬Ѓne 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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