<<

. 5
( 7)



>>

This proof works in both the ¬nite and the in¬nite case. If G is ¬nite, we let
s and t be the orders of G; that is, any line has s 1 points and any point lies on ¥
t 1 lines, so that the diagram is
¥
  ¢ 
¡
s t

For the classical polar spaces over GF¦ q§ , we have s q and t q1© µ , where ¨ ¨
µ is given in Table 6.5.1.
From now on, “generalised quadrangle” will be abbreviated to GQ.
The next result summarises some properties of ¬nite GQs.

Theorem 7.2 Let G be a ¬nite GQ with orders s t.
(a) G has s 1§ st 1§ points and t 1§ st 1§ lines.
¥¦ 
¦ ¥ ¥¦ 
¦ ¥
(b) s t divides st s 1§ t 1§ ;
¥ ¥¦ ¥
¦
s2 ;
(c) if s 1, then t
 
7.1. Generalised quadrangles 99

t 2.
(d) if t 1, then s
 
Proof (a) is proved by elementary counting, like that in Section 6.5. (b) is shown
by an argument involving eigenvalues of matrices, in the spirit of the proof of
the Friendship Theorem outlined in Exercise 2.2.4. (c) is proved by elementary
counting (see Exercise 3), and (d) is dual to (c).
In particular, if s 2, then t 4; and the case t 3 is excluded by (b) above.
¨  ¨
So t 1 2 or 4. These three values are realised by the three orthogonal rank 2
¨
polar spaces over GF¦ 2§ . We will see, as a special case of a later result, that these
are the only GQs with s 2. However, this result is suf¬ciently interesting to be
¨
worth another proof which generalises it in a different direction.
Theorem 7.3 Let G be a GQ with orders s 2 and t. Then t 1 2 or 4; and
¨ ¨
there is a unique geometry for each value of t.
Note the generalisation: t is not assumed to be ¬nite!
£
Proof Take a point and call it ∞; let Li : i I be the set of lines containing  ¤
∞. Number the points other than ∞ on Li as pi0 and pi1 . Now, for any point q £
not collinear with p, there is a function fq : I 0 1¤ de¬ned by the rule that 
the unique point of Li collinear with q is pi fq i . We use the function fq as a label 
for q. Let X be the set of points not collinear with ∞. We consider the possible
relationships of points in X . Write q r if q and r are collinear. 
1. If q r X satisfy q r, then fq and fr agree in just one position, viz., the
 
unique index i for which the line Li through ∞ meets the line qr.
2. If q r are not collinear but some point of X is collinear with both, then fq
and fr agree in all but two positions; for all but two values are changed twice, the
remaining two being changed just once.
3. Otherwise, fq fr ; for all the common neighbours of q and r are adjacent
¨
to ∞.
Note too that, for any i I and q X , there is a point r q for which fq and
  
fr agree only in i, viz., the last point of the line through q meeting Li .
Now suppose (as we may) that I 2, and choose distinct i j k I. Given
!
  
q X , choose r s t X such that fq and fr agree only in i, fr and fs only in j,
 
and fs and ft only on k. Then clearly fq and ft agree in precisely the three points
i j k, since these values are changed twice and all others three times. By the case

analysis, it follows that I 3 or I 2 3, as required. "
¨ $
# ¨
The uniqueness also follows from this analysis, with a little more work: we
know enough about the structure of X that the entire geometry can be recon-
structed.
100 7. Axioms for polar spaces

Problem. Can there exist a GQ with s ¬nite (s 1) and t in¬nite? 
The proof above shows that there is no such GQ with s 2. It is also known ¨
that there is no GQ with s 3 or s 4 and t in¬nite, though the proofs are much
¨ ¨
harder. (This is due to Kantor and Brouwer for s 3, and Cherlin for s 4.) ¨ ¨
Beyond this, nothing is known, though Cherlin™s argument could in principle be
extended to larger values of s.
The GQs with s 2 and t 1 2 have simple descriptions. For s 1 we have
¨ ¨ ¨
the 3 3 grid. For t 2, take the points to be all the 2-element subsets of a set
% ¨
of cardinality 6, and the lines to be all partitions of the 6-set into three disjoint
2-subsets. The GQ with order 2 4§ is a little harder to describe. The implicit
¦
construction in Theorem 7.3 is one of the simplest ” the functions fq are all those
which take the value 1 an even number of times, each such function representing
a unique point. This GQ also arises in classical algebraic geometry, as the Schl¨ ¬‚i
a
con¬guration of 27 lines in a general cubic surface, lying three at a time in 45
planes.
In the classical polar spaces, the orders s and t are both powers of the same
prime. There are examples where this is not the case ” see Exercise 5.

Exercises
1. Prove that the dual of a GQ is a GQ.
2. Prove that a GQ with two points on any line is a complete bipartite graph.
3. Let G be a ¬nite GQ with orders s t, where s 1. Let p1 and p2 be non- 
adjacent points, and let xn be the number of points p3 adjacent to neither p1 nor
p2 for which there are exactly n common neighbours of p1 p2 and p3 . Show that 

‘ xn s2 t st s t
¨ # # ¥ 
‘ nxn s¦ t 1§ t 1§
¨ ¥ # ¦ &

‘ n¦ n ¡
1§ xn t 1§ t t 1§
# ¨ ¥¦ #¦

Hence prove that t s2 , with equality if and only if any three pairwise non-

collinear points have exactly s 1 common neighbours.
¥
4. In this exercise, we use the terminology of coding theory (as in Section
3.2). Consider the space V of words of length 6 with even weight. This is a vector
space of rank 5 over GF 2§ . The “standard inner product” on V is a bilinear form
¦
which is alternating (by the even weight condition); its radical V is spanned by '
the unique word of weight 6. Thus, V V is a vector space of rank 4 carrying a
( '
non-degenerate alternating bilinear form. The 15 non-zero vectors of this space
7.2. Diagrams for polar spaces 101

are cosets of V containing a word of weight 2 and the complementary word
'
of weight 4, and so can be identi¬ed with the 2-subsets of a 6-set. Extend this
identi¬cation to an isomorphism between the combinatorial description of the GQ
with orders (2, 2) and the rank 2 symplectic polar space over GF¦ 2§ .
5. Let q be an even prime power, and let C be a hyperoval in Π PG¦ 2 q§ ,
¨
a set of q 2 points meeting every line in 0 or 2 points (see Section 4.3). Now
¥
take Π to be the hyperplane at in¬nity of AG¦ 3 q§ . Let G be the geometry whose
points are all the points of AG¦ 3 q§ , and whose lines are all the lines of AG¦ 3 q§
which meet Π in a point of C. Prove that G is a GQ with orders q 1 q 1§ .
#¦ ¥
6. Construct “free” GQs.


7.2 Diagrams for polar spaces
The inductive properties of polar spaces are exactly what is needed to show
that they are diagram geometries.

Proposition 7.4 A classical polar space of rank n belongs to the diagram
    3210 
  )))    

with n nodes.

Proof Given a variety U of rank d, the varieties contained in it form a projective
space of dimension d 1, while the varieties containing it are those of the polar
#
space U U of rank n d; moreover, any variety contained in U is incident with
(' #
any variety containing U . Since a rank 2 polar space is a generalised quadrangle,
it follows by induction that residues of varieties are correctly described by the
diagram.

This diagram is commonly referred to as Cn .
By analogy with Section 5.2, it might be thought that any geometry with di-
agram Cn for n 3 is a classical polar space. This is false for several reasons,
4
which we will see at various points. But ¬rst, here is one example of a geometry
with diagram C3 which is nothing like a polar space, even though it is highly sym-
metrical. This geometry was discovered by Arnold Neumaier, and is referred to
as Neumaier™s geometry or the A7 -geometry.
Let X be a set of seven points. The structure of a projective plane PG 2 2§ can
¦
be imposed on X in 30 different ways ” this number is the index of PGL¦ 3 2§
102 7. Axioms for polar spaces

in the symmetric group S7 . Since PGL¦ 3 2§ contains no odd permutations, it is
contained in the alternating group A7 with index 15, and so the 30 planes fall into
two orbits of length 15 under A7 . Now we take the points, lines, and planes of
the geometry to be respectively the elements of X , the 3-element subsets of X ,
and one orbit of A7 on PG¦ 2 2§ s. Incidence between points and lines, or between
lines and planes, is de¬ned by membership; and every point is incident with every
plane.
It is clear that the residue of a plane is a projective plane PG¦ 2 2§ , while
the residue of a line is a digon. Consider the residue of a point x. The lines
incident with x can be identi¬ed with the 2-element subsets of the 6-element set
£
Y X x¤ . Given a plane, its three lines containing x partition Y into three
¨ 5
2-sets. It is easy to check that, given such a triple of lines, there are just two
ways to draw the remaining four lines to complete PG¦ 2 2§ , and that these two
are related by an odd permutation of X . So our chosen orbit of planes has exactly
one member inducing the given partition of Y , and the planes incident with x can
be identi¬ed with all the partitions of Y into three 2-sets. As we saw in Section
7.1, this incidence structure is a generalised quadrangle with order 2 2.
We conclude that the geometry has the diagram
    ¢ 
¡
2 2 2

This example shows that, even in a geometry with such a simple diagram,
a variety is not necessarily determined by its point-shadow (all planes have the
same point-shadow!); the intersection of point-shadows of varieties need not be
the point-shadow of a variety, and the points and lines need not form a partial
L    
linear space. So the special properties of linear diagrams with all strokes
do not extend. However, classical polar spaces do have these nice properties.
A C3 -geometry in which every point and every plane are incident is called ¬‚at.
Neumaier™s geometry is the only known ¬nite example of such a geometry. Some
in¬nite examples were constructed by Sarah Rees; we now describe these. First,
a re-interpretation of Neumaier™s geometry.
Consider the rank 6 vector space V of all binary words of length 7 having even
weight. On V , we can de¬ne a quadratic form by the rule
1 ¡
f v§ 2 wt¦ v§ mod 2§

¨ 7
¦
The bilinear form obtained by polarising f is just the usual dot product, since
¡ ¡
wt¦ v w§ wt¦ v§ wt¦ w§ 2v w
¥ 6
¨ ¥8 9
#
7.2. Diagrams for polar spaces 103

It follows that f is non-singular: the only vector orthogonal to V is the all-1 word,
which is not in V . Now the points of X , which index the coordinates, are in one-
one correspondence with the seven words of weight 6, which are non-singular
vectors. The lines correspond to the vectors of weight 4, which comprise all the
singular vectors.
We saw in Section 6.4 that the planes on the quadric fall into two families,
such that two planes of the same family meet in a subspace of even codimension
(necessarily a point), while planes of different families meet in a subspace of odd
codimension (the empty set or a line). Now a plane on the quadric contains seven
non-zero singular vectors (of weight 4), any two of which are orthogonal, and so
meet in an even number of points, necessarily 2. The complements of these 4-sets
form seven 3-sets, any two meeting in one point, so forming a projective plane
PG¦ 2 2§ . It is readily checked that the two classes of planes correspond exactly
to the two orbits of A7 we described earlier. So the points, lines and planes of
Neumaier™s geometry can be identi¬ed with a special set of seven non-singular
points, the singular points, and one family of planes on the quadric. Incidence
between the non-singular and the singular points is de¬ned by orthogonality.
Now we reverse the procedure. We start with a hyperbolic quadric Q in
PG¦ 5 F , that is, a quadric of rank 3 with germ zero. A set S of non-singular
§
points is called an exterior set if it has the property that, given any line L of
Q, a unique point of S is orthogonal to L. Now consider the geometry G whose
POINTS, LINES and PLANES are the points of S, the points of Q, and one family
of planes on Q; incidence between POINTS and LINES is de¬ned by orthogonal-
ity, that between LINES and PLANES is incidence in the polar space, and every
POINT is incident with every PLANE.
Such a geometry belongs to the diagram C3 . For the residue of a PLANE Π is
a projective plane, naturally the dual of Π. (The correspondence between points
of S and lines of Π is bijective; for, given x S, x' cannot contain Π, since a polar

space in PG¦ 4 q§ cannot have rank 3, and so it meets Π in a line.) The residue of
a POINT x is the polar space x' , which as we™ve seen is rank 2, and so a GQ. And
of course the POINTS and PLANES incident with a LINE form a digon.
Thas showed that no further ¬nite examples can be constructed in this way:

Theorem 7.5 There is no exterior set for the hyperbolic quadric in PG 5 q§ for
¦
q 2.


However, Rees (who ¬rst described this construction) observed that there are
in¬nite examples. Consider the case where F ; let the form be x1 x2 x3 x4

@ ¥ ¥
104 7. Axioms for polar spaces

x5 x6 . Now the space of rank 3 spanned by 1 1 0 0 0 0§ , 0 0 1 1 0 0§ and
¦ ¦
0 0 0 0 1 1§ is positive de¬nite, and so is disjoint from the quadric; the points
¦
spanned by vectors in this space form an exterior set.
Now we turn to hyperbolic quadrics in general. As we saw in Section 6.4, the
maximal t.s. subspaces on such a quadric Q of rank n can be partitioned into two
families, so that a ¬‚at of dimension n 2 lies in a unique member of each family.
#
We construct a new geometry by letting these ¬‚ats be varieties of different types.
Now there is no need to retain the ¬‚ats of dimension n 2, since such a ¬‚at is the
#
intersection of the two maximal ¬‚ats containing it.

Theorem 7.6 Let Q be a hyperbolic quadric of rank n 3. Let G be the geometry
4
whose ¬‚ats are the t.s. subspaces of dimension different from n 2, where the #
two families of ¬‚ats of dimension n 1 are assigned different types. Incidence
#
between ¬‚ats, at least one of which has dimension less than n 1, is as usual; #
while n 1§ -¬‚ats of different types are incident if they intersect in an n 2§ -¬‚at.
#¦ #¦
Then the geometry has diagram
    3110 
  )))    

 

(n nodes).

Proof We need only check the residue of a ¬‚at of dimension n 3: the rest #
follows by induction, as in Proposition 7.4. Such a ¬‚at cannot be the intersection
of two n 1§ -¬‚ats of different types; so any two such ¬‚ats of different types

containing it are incident.

This diagram is denoted by Dn . The result holds also for n 2, provided that¨
we interpret D2 as two unconnected nodes ” the quadric has two families of lines,
each line of one family meeting each line of the other.

Exercises
1. Prove that the line joining two points of an exterior set to the quadric Q is
disjoint from Q.
2. Prove that an exterior set to a quadric in PG¦ 5 q§ must have q2 q 1 ¥ ¥
points.
3. Show that the plane constructed in Rees™ example is an exterior set.
7.3. Tits and Buekenhout“Shult 105

7.3 Tits and Buekenhout“Shult
We now begin working towards the axiomatisation of polar spaces. This major
result of Tits (building on earlier work of Veldkamp) will not be proved completely
here, but the next four sections should give some impression of how the proof
works.
Tits™ theorem characterises a class of spaces which almost coincides with the
classical polar spaces of rank at least 3. There are a few additional examples
of rank 3, some of which will be described later. I will use the term abstract
polar space for a geometry satisfying the axioms. In fact, Tits™ axioms describe
all subspaces of arbitrary dimension; an alternative axiom system, proposed by
Buekenhout and Shult, involves only points and lines (in the spirit of the Veblen“
Young axioms for projective spaces). In this section, I show the equivalence of
these axiom systems.
Temporarily, then, an abstract polar space of type T is a geometry satisfying
the conditions (P1)“(P4) of Section 6.4, repeated here for convenience.

(P1) Any ¬‚at, together with the ¬‚ats it contains, is a projective space of dimen-
sion at most r 1.
#
(P2) The intersection of any family of ¬‚ats is a ¬‚at.

(P3) If U is a ¬‚at of dimension r 1 and p a point not in U , then the union of the
#
lines joining p to points of U is a ¬‚at W of dimension r 1; and U W is a
# B
hyperplane in both U and W .

(P4) There exist two disjoint ¬‚ats of dimension r 1.
#
An abstract polar space of type BS is a geometry of points and lines satisfying
the following conditions. In these axioms, a subspace is a set S of points with the
property that if a line L contains two points of S, then L S; a singular subspace
C
is a subspace, any two of whose points are collinear.

(BS1) Any line contains at least three points.

(BS2) No point is collinear with all others.

(BS3) Any chain of singular subspaces is of ¬nite length.

(BS4) If the point p is not on the line L, then p is collinear with one or all points
of L.
106 7. Axioms for polar spaces

(Note that (BS4) is our earlier (BS), and is the key condition here.)

Theorem 7.7 (a) The points and lines of an abstract polar space of type T form
an abstract polar space of type BS.
(b) The singular subspaces of an abstract polar space of type BS form an abstract
polar space of type T.

Proof (a) It is an easy deduction from (P1)“(P4) that any subspace is contained
in a subspace of dimension n 1. For let U be a subspace, and W a subspace
#
of dimension n 1 for which U W has dimension as large as possible; if p
# B 
U W , then (P3) gives a subspace of dimension n 1 containing p and U W ,
5 # B
contradicting maximality.
Now, if L is a line and p a point not on L, let W be a subspace of dimension
n 1 containing L. If p W , then p is collinear with every point of L; otherwise,
# 
the neighbours of p in W form a hyperplane, meeting L in one or all of its points.
Thus, (BS4) holds. The other conditions are clear.
(b) Now let G be an abstract polar space of type BS. Call two points adjacent
if they are collinear; this gives the point set a graph structure. Every maximal
clique in the graph is a subspace. For let S be a maximal clique, and p q S; let

L be a line containing p and q. Any point of S L is collinear with p and q, and so
5
with every point of L; thus S L is a clique, and by maximality, L S.
D C
If p S (where S is a maximal clique), then the set of neighbours of p in S is a
E
hyperplane. Every point q S lies outside such a hyperplane; for, by (BS2), there

is a point p not adjacent to q. As we saw in Section 3.1, if every line has size 3,
then this implies that S is a projective space; but this deduction cannot be made in
general. However, in the present situation, Buekenhout and Shult are able to show
that S is indeed a projective space. (In particular, this implies that two points lie on
at most one line. For the union of two lines through two common points is a clique
by (BS4), and so would be contained in a maximal clique. However, Buekenhout
and Shult have to show that two points lie on at most one line before they know
that the subspaces are projective spaces; the proof is surprisingly tricky.)
Any singular subspace lies in some maximal clique, and so is itself a projective
space. Thus (P1) holds; and the remaining axioms can now be veri¬ed.

We will now simplify the terminology by using the term “abstract polar space”
equally for either type.
The induction principles we used in classical polar spaces work in almost the
same way in abstract polar spaces.
7.4. Recognising hyperbolic quadrics 107

Proposition 7.8 Let U be a d 1§ -dimensional subspace of an abstract polar
¦ #
space of rank n. Then the subspaces containing U form an abstract polar space
of rank n d.
#

Exercise
1. Show directly that, in an abstract polar space of type BS having three points
on any line, any two points lie on at most one line, and singular subspaces are
projective.


7.4 Recognising hyperbolic quadrics
There are two special cases where the proof of the characterisation of po-
lar spaces is substantially easier, namely, hyperbolic quadrics and quadrics over
GF¦ 2§ ; they will be treated in this section and the next.
In the case of a hyperbolic quadric, we bypass the need to reconstruct the
quadric by simply showing that there is a unique example of each rank over any
¬eld. First, we observe that the partition of the maximal subspaces into two types
follows directly from the axioms; properties of the actual model are not required.
We begin with a general result on abstract polar spaces.
An abstract polar space G can be regarded as a point-line geometry, as we™ve
seen. Sometimes it is useful to consider a “dual” situation, de¬ning a geometry GF
whose POINTs are the maximal subspaces of G and whose LINEs are the next-
to-maximal subspaces, incidence being reversed inclusion. We call this geometry
a dual polar space. In a dual polar space, we de¬ne the distance between two
POINTs to be the number of LINEs on a shortest path joining them.

Proposition 7.9 Let GF be a dual polar space.

(a) The distance between two POINTs is the codimension of their intersection.

(b) Given a POINT p and a LINE L, there is a unique POINT of L nearest to p.


Proof Let U1 , U2 be maximal subspaces. By the inductive principle (Proposi-
/
tion 7.8), we may assume that U1 U2 0. (It is clear that any path from U1 to
B ¨
U2 , in which not all terms contain U1 U2 , must have length strictly greater than
B
the codimension of U1 U2 ; so, once the result is proved in the quotient, no such
B
path can be minimal.)
108 7. Axioms for polar spaces

Now each point of U1 is collinear (in G) with a hyperplane in U2 , and vice
versa; so, given any hyperplane H in U2 , there is a unique point of U1 adjacent to
H, and hence (by (P3)) a unique maximal subspace containing H and meeting U1 .
The result follows.
(b) Let U be a maximal subspace and W a subspace of rank one less than
/
maximal. As before, we may assume that U W 0. Now there is a unique point
B ¨
p U collinear with all points of W . Then W pH is the uniue POINT on the LINE
 G 
W nearest to the POINT U .

Proposition 7.10 Let G be an abstract polar space of rank n, in which any n #¦
2§ -dimensional subspace is contained in exactly two maximal subspaces. Then
the maximal subspaces fall into two families, the intersection of two subspaces
having even codimension in each if and only if the subspaces belong to the same
family.

Proof The associated dual polar space is a graph. By Proposition 7.9(b), the
graph is bipartite, since if an odd circuit exists, then there is one of minimal length,
and both vertices on any edge are then equidistant from the opposite vertex in the
cycle.

Now, in any abstract polar space of rank n 4, in which lines contain at least
4
three points, any maximal subspace is isomorphic to PG n 1 F for some skew #¦ §
¬eld F. Now an easy connectedness argument shows that the same ¬eld F coor-
dinatises every maximal subspace.

Theorem 7.11 Let G be an abstract polar space of rank n 4, in which each next- 4
to-maximal subspace is contained in exactly two maximal subspaces. Assume that
some maximal subspace is isomorphic to PG n 1 F . Then F is commutative,
#¦ §
and G is isomorphic to the hyperbolic quadric of rank n over F.

Proof It is enough to show that F is commutative and that n and F uniquely deter-
mine the geometry, since the hyperbolic quadric clearly has the required property.
Rather than prove F commutative, I will show merely that it is isomorphic
to its opposite. It suf¬ces to show this when n 4. Take two maximal sub- ¨
spaces meeting in a plane Π, and a point p Π. By the FTPG, both maximal

subspaces are isomorphic to PG¦ 3 F . Now consider the residue of p. This is a
§
projective space, in which there is a plane isomorphic to PG¦ 2 F , and a point PI
§
residue isomorphic to PG¦ 2 F . Hence F F . The stronger statement that F is

§ I
commutative is shown by Tits. He observes that the quotient of p has a polarity
7.5. Recognising quadrics over GF 2§ 109
¦

interchanging a point and a plane incident with it, and ¬xing every line incident
with both; and this can only happen in a projective 3-space over a commutative
¬eld.
Let U1 and U2 be disjoint maximal subspaces. Note that they have the same
type if n is even, opposite types if n is odd. Let p be any point in neither subspace.
Then for i 1 2, there is a unique maximal subspace Wi containing p and meeting
¨
Ui in a hyperplane. Then Wi has the opposite type to Ui , so W1 and W2 have the
same type if n is even, opposite types if n is odd. Thus, their intersection has
codimension congruent to n mod 2. Since p W1 W2 , the intersection is at least
 B
a line. But their distance in the dual polar space is at least n 2, since U1 and U2
#
have distance n; so W1 W2 is a line L. Clearly L meets both U1 and U2 .
B
Each point of U1 is adjacent to a hyperplane of U2, and vice versa; so U1
and U2 are naturally duals. Now the lines joining points of U1 and U2 are easily
described, and it is not hard to show that the whole geometry is determined.


7.5 Recognising quadrics over GF 2R Q

In this section, we determine the abstract polar spaces with three points on
every line. Since we are given information only about points and lines, the BS
approach is the natural one. The result here was ¬rst found by Shult (assuming
a constant number of lines per point) and Seidel (in general), and was a crucial
precursor of the Buekenhout“Shult Theorem (Theorem 7.7). Shult and Seidel
proved the theorem by induction on the rank: a rank 2 polar space is a generalised
quadrangle, and the classi¬cation in this case is Theorem 7.3. The elegant direct
argument given here is due to Jonathan Hall.
Let G be an abstract polar space with three points per line. We have already
seen that the facts that two points lie on at most one line, and that maximal singular
subspaces are projective spaces, are proved more easily under this hypothesis than
in general. But here is a direct proof of the ¬rst assertion. Suppose that the points
£ £
a and b lie on two lines a b x¤ and a b y¤ . Then y is collinear with a and b, and
£
so also with x; so there is a line x y z¤ for some z, and both a and b are joined
to z. Any further point is joined to both or neither x and y, and so is joined to z,
contradicting (BS2).
De¬ne a graph “ whose vertices are the points, two vertices being adjacent if
they are collinear. The graph has the following property:
£ £
(T) every edge x y¤ lies in a triangle x y z¤ with the property that any further
£
point is joined to one or all of x y z¤ .
110 7. Axioms for polar spaces

This is called the triangle property. Shult and Seidel phrased their result as the
determination of ¬nite graphs with the triangle property. (The argument just given
shows that, in a graph with the triangle property in which no vertex is adjacent to
all others, there is a unique triangle with the property speci¬ed by (T) containing
any edge. Thus, the graph and the polar space determine each other.) The proof
given below is not the original argument of Shult and Seidel, which used induc-
tion, but is a direct argument due to Jonathan Hall (having the added feature that
it works equally well for in¬nite-dimensional spaces).
Theorem 7.12 An abstract polar space in which each line contains three points
is a quadric over GF¦ 2§ .
Proof As noted above, we may assume instead that we have a graph “ with the
triangle property (T), having at least one edge, and having no vertex adjacent to
all others. Let X be the vertex set of the graph “, and let F GF¦ 2§ . We begin ¨
ˆ
with the vector space V of all functions from X to F which are zero everywhere
except on a ¬nite set, with pointwise operations. (If X is ¬nite, then V is just the
ˆˆ
space F X of all functions from X to F.) Let x V be the characteristic function of

£
ˆ
the singleton set x¤ . The functions x, for x X , form a basis for V . We de¬ne a
ˆ 
ˆ ˆ
bilinear form b on V by setting
0 if x y or x is joined to y,
¨
ˆˆˆ
b¦ x y§T6

1 otherwise,

and extending linearly, and a quadratic form fˆ by setting fˆ x§
ˆ 0 for all x X

¨ 
ˆ
and extending to V by the rule
ˆ
fˆ v fˆ v§ fˆ w§ ¡
w§ b¦ v w§
¥¦ 3
¨ 8¦
¥ ¢¦
¥
ˆ
Note that both b and fˆ are well-de¬ned.
Let R be the radical of fˆ; that is, R is the subspace
£ ˆ
V : fˆ v§
ˆ ˆ
v 0 b¦ v w§ 0 for all w V
 V¦
¨ V
¨  W¤
ˆ
and set V V R. Then b and fˆ induce bilinear and quadratic forms b f on V : for
ˆ
¨ (
fˆ v§ (and this is well-de¬ned, that is, independent
example, we have f v R§ ¥¦ 6
¨ ¦
of the choice of coset representative). Now let x x R V .
¯ˆ ¨ ¥ 
We claim that the embedding x x has the required properties; in other words,
¯ X
it is one-to-one; its image is the quadric de¬ned by f ; and two vertices are adjacent
if and only if the corresponding points of the quadric are orthogonal. We proceed
in a series of steps.
7.5. Recognising quadrics over GF 2§ 111
¦
£
Step 1 Let x y z¤ be a special triangle, as in the statement of the triangle prop-
erty (T). Then x y z 0.
¯¯¯ ¥ ¥ ¨
It is required to show that r x y z R. We have
ˆˆˆ ¨ ¥ ¥ 

ˆˆ ˆˆˆ ˆˆˆ ˆˆˆ
b¦ r v§ b¦ x v§ b¦ y v§ b¦ z v§ 0
6
¨ ¢
¥ ¢
¥ 6
¨

for all v X , by the triangle property; and


ˆˆˆ ˆ ˆˆ ˆˆˆ
fˆ r§ fˆ x§ fˆ y§ fˆ z
ˆ ˆ ˆ b¦ x y§ b¦ y z b¦ z x§ 0
¨6 ¦ 8¦
¥ 8¦
¥ ¢§ ¦
¥ ¢
¥ ¢§
¥ V
¨

by de¬nition.


Step 2 The map x x is one-to-one on X .
¯ X
ˆˆˆ
Suppose that x y. Then r x y R. Hence b¦ x y§
¯¯ ˆˆ 0, and so x is joined
¨ ¨ ¥  3
¨
to y. Let z be the third vertex of the special triangle containing x and y. Then
z x y R by Step 1, and so z is joined to all other points of X , contrary to
ˆˆˆ ¨ ¥ 
assumption.


Step 3 Any quadrangle is contained in a 3 3 grid. %
£
Let x y z w¤ be a quadrangle. Letting x y x y, etc., we see that x y
¯¯ ¥ ¨ ¥ ¥
is not joined to z or w, and hence is joined to z w. Similarly, y z is joined ¥ ¥
to w x; and the third point in the special triangle through each of these pairs is
¥
x y z w, completing the grid. (See Fig. 7.1.)
¥ ¥ ¥

Step 4 For any v V , write v ‘iY I xi , where xi
¯ X , and the number m I
 ¨  `¨
of summands is minimal (for the given v). Then

(a) m 3;


(b) the points xi are pairwise non-adjacent.

This is the crucial step, and needs four sub-stages.


Substep 4.1 Assertion (b) is true.
If xi x j , we could replace xi x j by the third point xk of the special triangle,
 ¥
and obtain a shorter expression.
112 7. Axioms for polar spaces


w z© w z




x© y© z© w
w© x y© z




x x© y y



Figure 7.1: A grid

Substep 4.2 If L is a line on x1 , and y a point of L which is adjacent to x2 ,
then y xi for all i I.
 £
If not, let L x1 y z¤ , and suppose that xi z. Then xi is joined to the third
¨  
point w of the line x2 y. Let u be the third point on xi w. Then z p1 u xi x2 ,
¯¯ ¯¯ ¯ ¥ ¥ ¥ ¨
and we can replace x1 x2 xi by the shorter expression z u.
¯ ¯ ¯ ¯¯
¥ ¥ ¥

Substep 4.3 There are two points y z joined to all xi .
Each line through x1 contains a point with this property, by Substep 4.2. It is
easily seen that if x1 lies on a unique line, then one of the points on this line is
adjacent to all others, contrary to assumption.

Substep 4.4 m 3.  £ £
Suppose not. Considering the quadrangles x1 y x2 z¤ and x3 y x4 z¤ , we    
¬nd (by Step 3) points a and b with

¯ ¡
x1
¯ y
¯ x2
¯ z
¯ a
¯ x3
¯ y
¯ x4
¯ z
¯ b
¥ ¥ ¥ ¨ ¥ ¥ ¥ ¨
¯
But then x1
¯ x2
¯ x3
¯ x4
¯ a
¯ b, a shorter expression.
¥ ¥ ¥ ¨ ¥

Step 5 If v V, v 0, and f v§ 0, then v x for some x
¯ X.
 E¨ 6¦
¨ ¨ 
7.6. The general case 113

If not then, by Step 4, either v x y, or v x
¯¯ ¯ y
¯ z, where points x y (and
¯
¨ ¥ ¨ ¥ ¥
z) are (pairwise) non-adjacent. In the second case,
¡
f v§ f x§
¯ f y§
¯ fz
¯ b¦ x y§
¯¯ b¦ y z
¯¯ b¦ z x§
¯¯ 0 0 0 1 1 1 1

¨ ¥8 ¦ a¦
¥ a§ ¦
¥ ¥a 8§
¥ V
¨ ¥ ¥ ¥ ¥ ¥ ¨

The other case is similar but easier.

Step 6 x y if and only if b¦ x y§
¯¯ 0.
 6
¨
This is true by de¬nition.


7.6 The general case
A weak form of the general classi¬cation of polar spaces, by Veldkamp and
Tits, can be stated as follows.

Theorem 7.13 A polar space of type T having ¬nite rank n 4 is either classical, 4
or de¬ned by a pseudoquadratic form on a vector space over a division ring of
characteristic 2.

I will not attempt to outline the proof of this theorem, but merely make some
remarks, including a “de¬nition” of a pseudoquadratic form.
Let V be a vector space over a skew ¬eld F of characteristic 2, and σ an
anti-automorphism of F satisfying σ2 1. Let K0 be the additive subgroup ¨
£
x xσ of F, and K K( K0 . A function f : V K is called a pseudo-
¥ ¤ bF
¨  F
quadratic form relative to σ if there is a σ-sesquilinear form g such that f v§ ¨c ¦
g¦ v v§ mod K0 . Equivalently, f polarises to a σ-Hermitian form f satisfying
c cσ , that is, a trace-valued form. The function
v V c F f v v§

d  fg§
¦  ¦ §
¦ c
¨ ¥ §
f de¬nes a polar space, consisting of the subspaces of V on which f vanishes
(mod K0 ). If K0 is equal to the ¬xed ¬eld of σ, then the same polar space is de-
¬ned by the Hermitian form g; so we may assume that this is not the case in the
second conclusion of Theorem 7.13. For further discussion, see Tits [S].
Tits™ result is actually better than indicated: all polar spaces of rank n 3 are 4
classi¬ed. There are two types of polar spaces of rank 3 which are not covered
by Theorem 7.13. The ¬rst exists over any non-commutative ¬eld, and will be
described in the ¬rst section of Chapter 8. The other is remarkable in consisting of
the only polar spaces whose planes are non-Desarguesian. This type is constructed
by Tits from the algebraic groups of type E6 , and again I refer to Tits for the
construction, which requires detailed knowledge of these algebraic groups. The
114 7. Axioms for polar spaces

planes actually satisfy a weakening of Desargues™ theorem known as the Moufang
condition, and can be “coordinatised” by certain alternative division rings which
generalise the Cayley numbers or octonions.
Of course, the determination of polar spaces of rank 2 (GQs) is a hopeless task!
Nevertheless, it is possible to formulate the Moufang condition for generalised
quadrangles; and all GQs satisfying the Moufang condition have been determined
(by Fong and Seitz in the ¬nite case, Tits and Weiss in general.) This effectively
completes the analogy with coordinatisation theorems for projective spaces.
The other geometric achievement of Tits in the 1974 lecture notes is the ana-
logue of the Fundamental Theorem of Projective Geometry:

Theorem 7.14 Any isomorphism between classical polar spaces of rank at least
2, which are not of symplectic or orthogonal type in characteristic 2, is induced
by a semilinear transformation of the underlying vector spaces.

The reason for the exception will be seen in Section 8.4. As in Section 1.3,
this result shows that the automorphism groups of classical polar spaces consist of
semilinear transformations modulo scalars. These groups, with some exceptions
of small rank, have “large” simple subgroups, just as happened for the automor-
phism groups of projective spaces in Section 4.6. These groups are the classical
groups, and are named after their polar spaces: symplectic, orthogonal and uni-
tary groups. For details, see the classic accounts: Dickson [K], Dieudonn´ [L], e
and Artin [B], or for more recent accounts Taylor [R], Cameron [10]. In the sym-
plectic or unitary case, the classical group consists of all the linear transformations
of determinant 1 preserving the form de¬ning the geometry, modulo scalars. In
the orthogonal case, it is sometimes necessary to pass to a subgroup of index 2.
(For example, if the polar space is a hyperbolic quadric in characteristic 2, take
the subgroup ¬xing the two families of maximal t.s. subspaces.)
8

The Klein quadric and triality

Low-dimensional hyperbolic quadrics possess a remarkably rich structure; the
Klein quadric in 5-space encodes a projective 3-space, and the triality quadric in
7-space possesses an unexpected threefold symmetry. The contents of this chapter
can be predicted from the diagrams of these geometries, since D3 is isomorphic to
A3 , and D4 has an automorphism of order 3.

8.1 The Pfaf¬an
The determinant of a skew-symmetric matrix is a square. This can be seen in
small cases by direct calculation:
 
0 a12
a2
det 12
a12 0 ¤
¡ ¢ £
©¨
0 a12 a13 a14 ©
a12 0 a23 a24
¡
2 
det a12 a34 a13 a24 a14 a23
¡
¡ ¥¦¦§  
a13 a23 0 a34
¡  £ 
a14 a24 a34 0
¡ ¡ ¡

Theorem 8.1 (a) The determinant of a skew-symmetric matrix of odd size is
zero.
(b) There is a unique polynomial Pf A in the indeterminates ai j for 1 i j
  


2n, having the properties
(i) if A is a skew-symmetric 2n 2n matrix with i j entry ai j for 1 i
  


j 2n, then

Pf A 2;
det A  
 £ 

115
116 8. The Klein quadric and triality

(ii) Pf A contains the term a12 a34 a2n! with coef¬cient 1.
1 2n
  

Proof We begin by observing that, if A is a skew-symmetric matrix, then the
form B de¬ned by
Bxy xAy" ¤
 £
is an alternating bilinear form. Moreover, B is non-degenerate if and only if A is
non-singular: for xAy 0 for all y if and only if xA 0. We know that there is
"
£ £
no non-degenerate alternating bilinear form on a space of odd dimension; so (a)
is proved.
We know also that, if A is singular, then det A 0, whereas if A is non- 
 £
singular, then there exists an invertible matrix P such that

   
0 1 0 1

PAP" diag
1 0 1 0
¤ ¤ ¤
¡ ¡
¢ $¢
¢
£

det P 2 . Thus, det A is a square in either case.
so that det A !
  
 £  
Now regard ai j as being indeterminates over the ¬eld F; that is, let K F ai j :
£¡ 
1 i j 2n be the ¬eld of fractions of the polynomial ring in n 2n 1 vari-
    

ables over F. If A is the skew-symmetric matrix with entries ai j for 1 i  
j 2n, then as we have seen, det A is a square in K. It is actually the square
 

of a polynomial. (For the polynomial ring is a unique factorisation domain; if
f g 2 , where f and g are polynomials with no common factor, then
det A &
 
 %£

det A g2 f 2 , and so f 2 divides det A ; this implies that g is a unit.) Now det A
  
 £  
contains a term
a2 a2 a2 1 2n
12 34 2n! 
corresponding to the permutation
12 34 2n 1 2n
¡
   '
 ¤
  
and so by choice of sign in the square root we may assume that (ii)(b) holds.
Clearly the polynomial Pf A is uniquely determined. 

The result for arbitrary skew-symmetric matrices is now obtained by speciali-
sation (that is, substituting values from F for the indeterminates ai j ).

Exercises

1. A one-factor on the set 1 2 2n) is a partition F of this set into n subsets ( ¤ ¤ ¤
of size 2. We represent each 2-set( i j) by the ordered pair i j with i j. The 

¤
crossing number χ F of the one-factor F is the number of pairs i j k l of

) ¤ ¤ ¤ (

 

sets in F for which i k j l.   
8.2. The Klein correspondence 117


(a) Let be the set of one-factors on the set 1 2 2n) . What is ?
n n
0 ( 021 1
¤ ¤ ¤

(b) Let A ai j be a skew-symmetric matrix of order 2n. Prove that




‘ ∏
χ7 F 8 
Pf A 1 ai j
¡
 
 £ 
F i9 j8 F
64
5 n 7 @
4

2. Show that, if A is a skew-symmetric matrix and P any invertible matrix,
then

Pf PAP" det P Pf A  A
 
 £  
det P 2 det A , and taking the square root shows that
Hint: We have det PAP "  
 £  
Pf PAP det P Pf A ; it is enough to justify the positive sign. Show that it
"  
 C£
B  
suf¬ces to consider the ˜standard™ skew-symmetric matrix
 D   
0 1 0 1

A diag
1 0 1 0
¤ ¤ ¤
¡ ¡
¢ $¢
¢
£

In this case, show that the 2n 1 2n entry in PAP contains the term p2n! 1 2n! 1 p2n 2n , ¡ "

¤

so that Pf PAP contains the diagonal entry of det P with sign 1.
"  
 
3. Show that any linear transformation of a vector space ¬xing a symplectic
form (a non-degenerate alternating bilinear form) has determinant 1.


8.2 The Klein correspondence
We begin by describing an abstract polar space which appears not to be of
classical type. Let F be a skew ¬eld, and consider the geometry de¬ned from E
PG 3 F as follows:

¤
F
the POINTs of are the lines of PG 3 F ;
E 
¤

F
the LINEs of are the plane pencils (incident point-plane pairs);
E
F
the PLANEs of are of two types: the points, and the planes.
E

A POINT and LINE are incident if the line belongs to the plane pencil (i.e., is
incident with both the point and the plane). A LINE and PLANE are incident
if the point or plane is one of the elements of the incident pair; and incidence
between a POINT and a PLANE is the usual incidence in PG 3 F . 
¤
If a PLANE is a plane Π, then the POINTs and LINEs of this PLANE corre-


spond to the lines and points of Π; so the residue of the plane is isomorphic to the
118 8. The Klein quadric and triality

dual of Π, namely, PG 2 F . On the other hand, if a PLANE is a point p, then
G
¤

the POINTs and LINEs of this PLANE are the lines and planes through p, so its
residue is the residue of p in PG 3 F , namely PG 2 F . Thus (PS1) holds. (Note
 
¤ ¤
 
that, if F is not isomorphic to its opposite, then the space contains non-isomorphic
planes, something which cannot happen in a classical polar space.)
Axiom (PS2) is clear. Consider (PS3). Suppose that the PLANE in question
is a plane Π, and the POINT not incident with it is a line L. Then L Π is a point H
p; the set of POINTs of Π collinear with L is the plane pencil de¬ned by p and Π
(which is a LINE), and the union of the LINEs joining them to L consists of all
lines through p (which is a PLANE), as required. The other case is dual.
Finally, if the point p and plane Π are non-incident, then the PLANEs they
de¬ne are disjoint, proving (PS4).
Note that any LINE is incident with just two PLANEs, one of each type; so, if
the polar space is classical, it must be a hyperbolic quadric in PG 5 F . We now 
¤

show that, if F is commutative, it is indeed this quadric in disguise! (For non-
commutative ¬elds, this is one of the exceptional rank 3 polar spaces mentioned
in Section 7.6.)
The skew-symmetric matrices of order 4 over F form a vector space of rank

6, with x12 x34 as coordinates. The Pfaf¬an is a quadratic form on this vector
¤ ¤
space, which vanishes precisely on the singular matrices. So, projectively, the
singular matrices form a quadric in PG 5 F , the so-called Klein quadric. From
I 
¤

the form of the Pfaf¬an, we see that this quadric is hyperbolic ” but in fact this
will become clear geometrically.
Any skew-symmetric matrix has even rank. In our case, a non-zero singular
skew-symmetric matrix A has rank 2, and so can be written in the form

A X v w : v" w w" v
¡

¤
£  £
for some vectors v w. Replacing these two vectors by linear combinations ±v 
¤
βw and γv δw multiplies A by a factor ±δ βγ (which is just the determinant of ¡

the transformation). So we have a map from the line of PG 3 F spanned by v and 
¤

w to the point of the Klein quadric spanned by X v w . This map is a bijection: 
¤

we have seen that it is onto, and the matrix determines the line as its row space.
This bijection has the properties predicted by our abstract treatment. Most
important,

two points of the Klein quadric are perpendicular if and only if the
corresponding lines intersect.
8.2. The Klein correspondence 119

To prove this, note that two points are perpendicular if and only if the line
joining them lies in . Now, if two lines intersect, we can take them to be u vQ
I P ¤
and u wQ ; and we have
P ¤

± u" v β u" w u" ±v βw ±v βw u
v" u w" u
¡ ¡ ¡ "
R
    ¤
  £  
so the line joining the corresponding points lies in the quadric. Conversely, if two
S 

lines are skew, then they are v1 v2 and v3 v4 , where v1 v4 is a basis;
P Q P Q ( )
¤ ¤ ¤ ¤
then the matrix
v"1 v2 v"2 v1 v"3 v4 v"4 v3 ¡ ¡

has rank 4, and is a point on the line not on . I
Hence the planes on the quadric correspond to maximal families of pairwise
intersecting lines, of which there are two types: all lines through a ¬xed point;
and all lines in a ¬xed plane. Moreover, the argument in the preceding paragraph
shows that lines on do indeed correspond to plane pencils of lines in PG 3 F .
I 
¤

This completes the identi¬cation.

Exercise
1. This exercise gives the promised identi¬cation of PSL 4 2 with the alter- ¤

nating group A8 .
Let V be the vector space of rank 6 over GF 2 consisting of the binary words 

of length 8 having even weight modulo the subspace Z consisting of the all-zero
and all-1 words. Show that the function
1
fv Z 2 wt v mod 2
   
 £  
is well-de¬ned and is a quadratic form of rank 3 on V , whose zeros form the
Klein quadric . Show that the symmetric group S8 interchanges the two families
I
of planes on , the subgroup ¬xing the two families being the alternating group
I
A8 .
Use the Klein correspondence to show that A8 is embedded as a subgroup
of PGL 4 2 PSL 4 2 . By calculating the orders of these groups, show that
¤ ¤
 £ 
equality holds.

Remark The isomorphism between PSL 4 2 and A8 can be used to give a solu- ¤

tion to Kirkman™s schoolgirl problem. This problem asks for a schedule for ¬fteen
schoolgirls to walk in ¬ve groups of three every day for seven days, subject to the
120 8. The Klein quadric and triality

requirement that any two girls walk together in a group exactly once during the
week.
The 7 5 groups of girls are thus the blocks of a 2- 15 3 1 design. We will
 ¤
¤

take this design to consist of the points and lines of PG 3 2 . The problem is then ¤

to ¬nd a ˜parallelism™ or ˜resolution™, a partition of the lines into seven ˜parallel
classes™ each consisting of ¬ve pairwise disjoint lines.
One way to ¬nd a parallel class is to consider the underlying vector space
V 4 2 as a vector space of rank 2 over GF 4 . The ¬ve ˜points™ or rank 1
¤ 
 
subspaces over GF 4 become ¬ve pairwise disjoint lines when we restrict the


scalars to GF 2 . Scalar multiplication by a primitive element of GF 4 is an
 
 
automorphism of order 3, ¬xing all ¬ve lines, and commuting with a subgroup
SL 2 4 A5 . Moreover, if two such automorphisms oof order 3 have a com-
U ¤
T
 £
mon ¬xed line, then they generate a 2 3) -group, since the stabiliser of a line in ( ¤
GL 4 2 is a 2 3) -group. (
¤ ¤

Now, in A8 , an element of order 3 commuting with a subgroup isomorphic to
A5 is necessarily a 3-cycle. Two 3-cycles generate a 2 3) -group if and only if ( ¤
their supports intersect in 0 or 2 points. So we require a set of seven 3-subsets of
S 

1 8) , any two of which meet in one point. The lines of PG 2 2 (omitting
( ¤
¤ ¤ 
one point) have this property.


8.3 Some dualities
We have interpreted points of the Klein quadric in PG 3 F . What about the 
¤

points off the quadric?

Theorem 8.2 There is a bijection from the set of points p outside the Klein quadric
to symplectic structures on PG 3 F , with the property that a point of per-
I I

¤

pendicular to p translates under the Klein correspondence to a totally isotropic
line for the symplectic geometry.
W
Proof A point p is represented by a skew-symmetric matrix A which has
V I
non-zero Pfaf¬an (and hence is invertible), up to a scalar multiple. The matrix
de¬nes a symplectic form b, by the rule

bvw vAw"

¤ £
We must show that a line is t.i. with respect to this form if and only if the corre-
sponding point of is perpendicular to p. I
8.3. Some dualities 121

Let A be a non-singular skew-symmetric 4 4 matrix over a ¬eld F. By direct 
calculation, we show that the following assertions are equivalent, for any vectors
v w F 4:
W
¤
(a) X v w v w w v is orthogonal to A, with respect to the bilinear form ¡
" "

¤
 £
obtained by polarising the quadratic form Q X Pf X ;  
 £ 
(b) v and w are orthogonal with respect to the symplectic form with matrix A† ,
that is, vA† w 0. "
£
Here the matrices A and A† are given by

© ©¨
0 a12 a13 a14 0 a34 a24 a23
¡
© ©
a12 0 a23 a24 a34 0 a14 a13
¡ ¡ ¡
A† 
A ¡ ¥¦¦§ ¥¦
a13 a23 0 a34 a24 a14 0 a12
¤
¡ ¡
 
¦§
£ £
a14 a24 a34 0 a23 a13 a12 0
¡ ¡ ¡ ¡ ¡

Note that, if A is the matrix of the standard symplectic form, then so is A† . In
general, the map taking the point outside the quadric spanned by A to the sym-
plectic form with matrix A† is the one asserted in the theorem.
Let 1 be the symplectic GQ over F, and 2 the orthogonal GQ associated
E E
W
with the quadric vX , where is the Klein quadric and vQ . (Note that I`Y
H I V$ P I
any non-singular quadratic form of rank 2 in 5 variables is equivalent to ±x20 
x1 x2 x3 x4 for some ± 0; so any two such forms are equivalent up to scalar V

£
multiple, and de¬ne the same GQ.) We have de¬ned a map from points of 2 to E
lines of 1 . Given any point p of 1 , the lines of 1 containing p form a plane
E E E
pencil in PG 3 F , and so translate into a line of 2 . Thus we have shown: E

¤

Theorem 8.3 For any ¬eld F, the symplectic GQ in PG 3 F and the orthogonal 
¤
GQ in PG 4 F are dual. 
¤

Now let F be a ¬eld which has a Galois extension K of degree 2 and σ the
Galois automorphism of K over F. With the extension K F we can associate two &
GQs:
1 : the unitary GQ in PG 3 K , de¬ned by the Hermitian form
abE 
¤

x1 yσ x2 yσ x3 yσ x4 yσ ;
2 1 4 3
  

2: the orthogonal GQ in PG 5 F de¬ned by the quadratic form
abE 
¤

±x2 βx5 x6 γx2
x1 x2 x3 x4 5 6
    ¤
where ±x2 βx γ is an irreducible quadratic over F which splits in K.
 
122 8. The Klein quadric and triality

Theorem 8.4 The two GQs and de¬ned above are dual.
abE bE
a
1 2


Proof This is proved by “twisting the Klein correspondence”. In outline, we take
the Klein correspondence over K, and change coordinates on the quadric so that
restriction of scalars to F gives the geometry 2 , rather than the Klein quadric aE
over F; then show that the corresponding set of lines in PG 3 K are those which 
¤

are totally isotropic with respect to a Hermitian form.


Exercises
1. Prove the assertion about A and A† in the proof of Theorem 8.2.
Let be a hyperbolic quadric of rank n. If v is a non-singular vector, then the
I
quadric vX has the property
cH
I

d
F
meets every maximal subspace E of in a hyperplane of E.
I
d
We call a set satisfying this condition special. The point of the next three ex-
d
ercises is to investigate whether special sets are necessarily quadrics of the form
vX .
cH
I
2. Consider the case n 2. Let the rank 4 vector space be the space of all
£
2 2 matrices over F, and let the quadratic form be the determinant.


<<

. 5
( 7)



>>