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Generalized quadrangles of order (s, t) with [s-t]=2

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Generalized quadrangles of order (s, t) with [s-t]=2
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Miller, Mark Anderson
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ix, 150 leaves : ; 28 cm

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Finite generalized quadrangles ( lcsh )
Finite generalized quadrangles ( fast )
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theses ( marcgt )
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Includes bibliographical references (leaves 146-150).
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Department of Mathematical and Statistical Sciences
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by Mark Anderson Miller.

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Full Text
GENERALIZED QUADRANGLES OF ORDER
(S, T) WITH \S-T\ = 2
by
Mark Anderson Miller
B.S.E., John Brown University, 1988
M.S., University of Colorado at Denver, 1993
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Doctor of Philosophy
Applied Mathematics
1999


This thesis for the Doctor of Philosophy
degree by
Mark Anderson Miller
has been approved
by
Stanley E. Payne
William E. Cherowitzo
Sylvia A. Hobart
J. Richard Lundgren
William J. Wolfe
Date


Miller, Mark Anderson (Ph.D., Applied Mathematics)
Generalized Quadrangles of Order (s,t) with \s t\ =2
Thesis directed by Professor Stanley E. Payne
ABSTRACT
The subject of this thesis is generalized quadrangles (GQ) whose pa-
rameters differ by two. The roles of regular points, lines and ovoids are ex-
amined extensively. Such regularities allow for connections between various
GQ.
Affine planes associated with GQ(q + 1 ,q 1) are shown to be iso-
morphic to those associated with GQ(q,q). Using a new idea, the grid-like
axiom, the affine planes associated with certain regular ovoids are shown to
be isomorphic. While regular ovoids can be used to obtain numerous other
ovoids, the fan containing a pivotal ovoid is shown to be unique. Moreover if a
fan contains more than one regular ovoid it is shown to contain every regular
ovoid of the GQ.
A new characterization of the GQ(q + 1, q 1) arising from a g-arc is
provided. After considering the characterizations due to De Soete and Thas and


to Payne, the standard coordinatization of the GQ is introduced. By examining
the coordinatizing held permutations, Paynes sixth characterization axiom is
shown to be equivalent to having desarguesian planes arise from the pivotal
ovoids.
Collineations of the known GQ(q + 1, new result here is that in only one case are there collineations which are not
induced by semilinear maps of the underlying vector space.
An appendix is provided as a brief introduction to projective geometry
with a list of the known hyperovals.
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
Signed
Stanley E. Payne


DEDICATION
This thesis is dedicated to my mother Janet, my sister Michelle, and my wife Adriana.


ACKNOWLEDGMENTS
An endeavor of this magnitude of course leaves many people to thank.
I would like to acknowledge my thesis committee and thank them for their
willingness to undertake this project with me. In particular I thank my advisor
and mentor Stanley Payne for his tireless efforts to aid me in becoming a
mathematician, and I thank Bill Cherowitzo who has acted as my co-mentor
here at CL-Denver.
While in progress, portions of this thesis were presented at the Dis-
crete Mathematics Seminar at the University of Colorado at Denver, the Al-
gebraic Combinatorics Seminar at Colorado State University, and an AMS
Special Session in Groups and Geometry at Kansas State University. I would
like to thank the various organizers, Richard Lundgren, Robert Liebler, and
Ernest Shult for giving me the opportunity to present and refine this material.
My appreciation is extended to the CU-Denver Mathematics Gradu-
ate Committee for the financial assistance I received here which allowed me to
pursue my research. Moreover, I gratefully acknowledge the generosity of the


Warren Bateman Family exhibited through their funding of the Bateman Teach-
ing Fellowship of which I was a recipient. This fellowship was established in
memory of Lynn Bateman whose reputation as an outstanding mathematics
educator inspires us all.
Over the past four years many friends and family members have pro-
duced an immeasurable amount of encouragement. While I cannot name them
all here, I do wish to thank a few specific individuals whose tangible support
proved to be both necessary and sufficient: C.B. and Tonny Euser, Allen and
Leanne Holder, Matthew and Kimberly Lockhart, Michael and Nancy Miller,
and my parents, Roy and Janet Miller. Finally I thank my wife Adriana whose
constant, unfailing love has helped in ways I cannot express. I am forever in
your debt.
M. A. Miller
Denver, Colorado


CONTENTS
Chapter
1. Introduction and review ........................................... 1
1.1 Preface............................................................... 1
1.2 Historical background ................................................ 4
1.3 Definitions, examples, and observations............................... 5
2. Known examples ................................................... 16
2.1 Two classical examples: W(q) and Q(4, q)............................. 16
2.2 A construction by Tits............................................... 17
2.3 Constructions by Ahrens and Szekeres ................................ 20
2.4 A construction by Payne.............................................. 23
3. Regular ovoids in S............................................... 29
3.1 Affine planes and ovoids............................................. 29
3.2 Intersection of traces with ovoids of M.............................. 32
3.3 Constricting about a regular ovoid .................................. 34
3.4 Regularity in Soo and planar isomorphisms............................ 35
3.5 Grid-like Fans....................................................... 41
viii


4. Regularity in S..................................................... 50
4.1 A theorem of Thas and van Maldeghem .............................. 50
4.2 Regular pairs..................................................... 60
5. Characterizations................................................... 69
5.1 (0,2)-sets........................................................ 69
5.2 Axioms of Payne................................................... 91
5.3 Coordinatizing the GQ............................................ 104
5.4 Relating Paynes Axioms to a and /?.............................. Ill
6. Collineations...................................................... 118
6.1 Collineations of 6.2 Collineations of P(W(q), x) .................................. 124
7. Other constructions................................................ 126
7.1 Constructions using groups and cosets ........................ 126
7.2 Constructions using groups and designs........................ 132
Appendix
A. Projective planes and 3-space..................................... 136
B. Arcs and ovals.................................................... 141
References............................................................ 146
IX


1. Introduction and review
1.1 Preface
While countless articles have been published on the topic of general-
ized quadrangles, relatively few self-contained references in this area are avail-
able. One notable exception is Finite generalized quadrangles by Payne and
Thas which was published in 1984. Since then there has been somewhat of an
explosion of interest in GQ. It now appears that due to the great proliferation
of work in this held no suitable sequel would be fathomable. So it seems that
a better task might be collecting much of what is known about a particular
type of GQ and providing a sort of specialized reference manual. This is the
motivation for what follows with the focus on generalized quadrangles whose
parameters differ by 2. The purpose of this thesis is to gather together in
one place much of the relevant work to date involving these GQ, including
work which originated with this author. Portions of the material here were
previously reported with Payne in [PM98].
Chapter one provides an introduction to generalized quadrangles with
1


some basic definitions and small examples. In chapter two the known exam-
ples are presented as are properties of spreads/ovoids in these examples. In
particular the hyperbolic lines in the pivotal ovoids are seen to be related to
affine lines and planar arcs.
Chapter three deals primarily with regular ovoids of GQ(q + 1, q 1)
and the corresponding affine planes. In addition to the known result that
pivotal ovoids correspond to regular points, the following material is new:
(1) The planes corresponding to ovoids are shown to be isomorphic, and
sometimes identical, to planes associated with GQ of order q in theo-
rems 3.4.4 and 3.4.5.
(2) The Grid-like Axiom is introduced in section 3.5.
(3) Every regular ovoid is shown to be pivotal for exactly one fan in theo-
rem 3.5.4.
(4) Planes associated with ovoids of a grid-like fan are shown to be iso-
morphic in theorem 3.5.5.
Chapter four begins with an elaboration of the proof by Thas and
van Maldeghem showing that no non-classical GQ(q + 1 ,q 1) can have all
points regular. Next regular pairs of points are examined leading to a new
classification of regular ovoids.
In chapter five, a new characterization of the GQ(q + 1, q 1) arising
2


from a q-arc is provided. After considering the characterizations due to De
Soete and Thas and to Payne, the standard coordinatization of the GQ is
introduced. By examining the coordinating held permutations, Paynes sixth
characterization axiom is shown to be equivalent to having desarguesian planes
arise from the pivotal ovoids.
Collineations of the known GQ(q+1, q1) are discussed in chapter six.
The main new result here is that in only one case are there collineations of the
quadrangle which are not collineations of projective three-space. The chapter
concludes with a clarification of remarks made in [PM98] regarding findings
of [GJS94], In Chapter Seven, two previously known alternative constructions
are provided: one due to Payne and the other to De Bruyn.
An appendix is provided as a brief introduction to projective geometry
with a list of the known hyperovals.
While no work can be completely self-contained, an attempt has been
made to keep the reader from having to consult an unwieldy number of refer-
ences to understand the material here. Most terms not defined in this thesis can
be found in either [Dem68] or [PT84], A nice overview of generalized quad-
rangles is given in [Tha95] as well as [Pay96], the former of which concerns
generalized n-gons with n not necessarily equal to 4.
3


1.2 Historical background
The study of finite generalized quadrangles (GQ) is a relatively young
branch of discrete mathematics. J. Tits first introduced the notion of a gener-
alized polygon in 1959 [Tit59]. For the next decade some progress was made in
this field, particularly in the study of what are now termed the classical gener-
alized quadrangles (see for example [Dem68] and [FH64]); then from the late
1960s through the early 1980s many geometers and algebraists began looking
more deeply at this subject. A plethora of new results and a good number of
new examples were discovered.
As the focus of this thesis is on GQ(s,t) where s and t differ by
two, some background on these GQ may be in order. (The roles of s and t
are described in the next section.) In the late 1960s and early 1970s new
constructions for GQ(s,s + 2) were given by Ahrens and Szekeres in [AS69]
and independently by Hall in [Hal71]. This inspired Paynes work in [Pay71b]
which gave new GQ(s, s 2).
With these constructions in hand, two questions naturally arise: What
are the collineations of these GQ, and what are the defining characteristics of
these GQ? Much work has been done in an attempt to answer these questions.
These two questions serve as a motivation for much of this thesis.
4


1.3 Definitions, examples, and observations
We begin with some basic definitions and then look at a few small
examples. Let V and B be two non-empty sets, called points and lines, with
an incidence relation X such that there are positive integers s and t satisfying
Gl) Each point is incident with f+1 lines; any two points are mutually
incident with at most one line.
G2) Each line is incident with s +1 points; any two lines are mutually
incident with at most one point.
G3) Given a line L and a point x not incident with L there is a unique
point y and a unique line M such that x X M X y X L.
Such a collection S = (V, B.X) is called a generalized quadrangle
of order (s,t) written GQ(s,t); when s = t the GQ is said to have order s.
The dual of a GQ(s,t) is the GQ(t,s) obtained by interchanging the roles of
points and lines. Any theorem or definition given for a GQ can be dualized by
interchanging points and lines. It will be assumed that whenever a definition
or theorem is given, its dual has also been given.
Two points incident with a common line are collinear and two lines
incident with a common point are concurrent, x ~ y means that x and y are
either collinear if x and y are points or concurrent if x and y are lines.
If X is a set of points (respectively, lines) of S, then X1- denotes the
5


set of all points collinear (resp., lines concurrent) with everything in A"; X1- is
called the trace of X. If X = {a;} is a singleton set, it is common to write X1-
as xL. The span of X. written Xx, is the set of all points collinear (resp.,
lines concurrent) with all of X. By convention, x x. It is worth noting
that (X^)1- = XL.
Given any two noncollinear points x and y, {agt/}-11- is called the
hyperbolic line through x and //. Dually the hyperbolic point on two
nonconcurrent lines L and M is the span {/,. .1/} .
Often GQ under examination here will be related to projective geome-
tries. To avoid confusion, the symbol < x, y > will be used for the projective
space spanned by x and y whereas xy will indicate the line of the GQ con-
taining x and y. Sometimes the term S.
Straightforward arguments demonstrate the following:
For x t- 'P. |re-11 = st + s + 1.
For L E B, = st + t + 1.
\P\ = (s + l)(st + 1).
\B\ = (t + l)(st + 1).
For two noncollinear points x, y, |{rc, t/}| = t + 1
and 2 < |{rc, t/}J"L| 6


For two nonconcurrent lines L, M, \{L, M}\ = s + 1
and 2 < \{L, M}J-\ < s + 1.
Two noncollinear points form a regular pair provided their span
attains the upper bound, t + 1; two collinear points are dehned to form a
regular pair. An individual point is a regular point provided it forms a
regular pair with every other point. Dually, { /.. M} is a regular pair of lines
provided either { L. M}J- = s + 1 or L and M are concurrent; a regular line
is one which forms a regular pair with every other line. A point is coregular
provided all lines incident with it are regular.
The following three observations will be useful when working with
regularity.
Observation 1.3.1 {o^o^} is regular if and only if whenever {2/1,2/2} ^
Ui-x-,} then {yi,y2} = {xi.x2}
Proof: If Xi ~ x2, then clearly the observation holds as {xi,x2} =
{.r 1. Now assume x\ and x2 are not collinear. Suppose {a; 1,0:2} is
regular and let {.rt} = {0:1,.... xt. 1}. If {2/1,2/2} Q { c 1. } then each
Xi is in {yi,y2}L. Because \{yi,y2}L\ = t + 1, {2/1, y2}L = {xi, x2}L.
Now suppose {2/1,2/2}^ = { c 1. } whenever {2/1,2/2} Q
This forces |{a;i,a;2}| = t + 1 in which case {x\,x2} is regular.
7


Observation 1.3.2 If {xi,x2} is regular then zi,z2 E {.r if and only
if {zi,z2} = {./i. } .
Proof: The proof is similar to that of observation 1.3.1 and is left
to the reader.
Observation 1.3.3 A point X\ is regular if and only if every pair of points in
is regular.
Proof: Assume x\ is regular. Let yi,y2 E xf, x2 E {yi,
By Observation 1.3.1, {yi,y2)'L = Hence
Therefore {yi,y2} is regular.
Conversely, assume every pair of points in Xi is regular. Let x2 be
some point other than x\ and let {yi,y2} C {xi^x^^. By Observation 1.3.1,
as {x\,x2} C {yi,y2}, is regular. Therefore x\ is regular.
In the future, the proofs of numbered observations will be left to the
reader.
An ovoid is a collection of st + 1 pairwise noncollinear points. For
k < st + 1, a k-cap is a set of k pairwise non-collinear points. A spread is a
collection of st + 1 pairwise nonconcurrent lines.
A set of ovoids which partitions the point-set is a fan. A set of spreads
8


which partitions the line-set is called a packing. An ovoid (resp. spread) is
said to be regular if the span of any pair of its points (resp. lines) is of
maximum size and is contained in the ovoid (resp. spread).
A GQ(s,t) with s or t equal to 1 is called thin; otherwise the GQ is
thick. A smallest thick GQ would have parameters s = t = 2. Such a GQ is
constructed below. For the most part, the GQ examined in this thesis will be
thick. However, it is sometimes instructive to begin with small, thin examples.
Example 1.3.4 The 4x4 Grid:
Let B be the lines of a 4 x 4 grid and let V be the points of intersection on the
grid. Observe that this gives a GQ(3,1).
The next example is the dual of the first.
Example 1.3.5 I\ u (The Complete Balanced Bipartite Graph on 8 Nodes):
Let V be the node set of /\ u and let B be the edge set. This gives a GQ( 1,3).
In general for n > 1, any n x n grid is a GQ(n 1,1) whose dual is
Kn,n (sometimes called a dual grid) which is a GQ(l,n 1). However the
case n = 4 is of particular interest here. To see why, first consider the smallest
thick GQ.
Example 1.3.6 GQ(2,2):
Let S = {1,2,..., 6} and let V = { {i,j}\i,j E S,i ^ j} be the set of duads,
i.e. subsets of size 2, from S A syntheme is a triple of disjoint members
9


of V. For example {{1, 2}, {3, 4}, {5, 6}} is a syntheme. Let B be the set of
all synthemes formed from the duads in V. Straightforward verification shows
that S = (V, B, e) is a GQ of order 2.
Sometimes it is helpful to view a particularly GQ from more than one
point of view.
The GQ(2, 2) may also be viewed differently using algebra. Let Z2 be
the group of order 2 and set G = Z2 x Z2 x Z2. For i e {1, 2, 3} let e* be the
element with a 1 in the ith position and zeros elsewhere, let e be the identity
element, and let j = J2ei- Construct the following subgroups of G: A* =
{e, e*}, A* = {e, Let the points of S be the elements of G, cosets of subgroups in JF*,
and the set T. Let the lines of S be the subgroups in F together with their
cosets. If incidence is given by containment and inclusion, then S is isomorphic
to example 1.3.6.
Similarly, examples 1.3.4 and 1.3.5 may be viewed from an algebraic
point of view. Before this is done a relationship between the three examples
is established. At first glance examples 1.3.4 and 1.3.5 may seem unrelated to
1.3.6. However, there is a very elegant connection. Let S be a GQ(2,2) with
a point x, and let Px be the points of S which are not in x. Let Bi be the
lines of S which do not contain x (note that each of these lines has a point
10


removed from it). For each y E Px, let Ly be a new line which joins y to the
unique remaining point of {aqi/jT-1, and let B2 be the set of all such Ly. Let
Bx = Bi U B2. Then Px and Bx form the point and line sets (respectively) of
a GQ( 1,3).
Now return to S and let L be any line of S. We create a similarly
derived structure as follows. Let BL be the lines of S which are not in IT; let
Pi be the points of S which are not on L. For each M E BL, let xm be a new
point incident with the unique remaining line of {L. M} Define P2 to be
the set of all such with PL = Pi U P2. The resulting structure with point
set Pl and line set BL is a GQ(3,1).
This relationship between the three GQ is not coincidental; in fact
it lies at the heart of this thesis. For this reason this process is explained in
more detail here. Let S = (V,B,X) be any GQ(q,q) with a regular point x.
Let Vx = V xL. Let Bx = Blx U Bj. where Blx is the set of lines which are
not incident with x, and B~ = { {x/yj^ly E V x }. i.e. B2X is the set of
hyperbolic lines through x. If the incidence Xx is given by that of X and by
containment, it is easy to verify that the resulting structure Sx = (Vx, Bx,Xx)
has q points on each line, and each point of Sx is on q + 2 points. That G3
holds is shown below in cases.
(1) Let L eB\ and let p be a point of Vx not on L. In this case p and L
11


are also elements of S. There is a unique line M E B incident with p
and a point z of L.
(a) If z E Vx, then M is the unique line of B\ collinear in Sx with
p and a point of L. If some hyperbolic line of B2 confined p and
a point tv of L, then tv and p would both be collinear in S with
a point of xx which would give a triangle in S. Hence M is the
unique line of Bx through p and a point of L.
(b) If z 0 Vx, then z E {x,p}. As x is regular, every line through
z contains a unique point of {.r.p) Specifically, L contains a
unique point tv of {.r.p) The hyperbolic line {.r.p} is the
unique line of Bx containing p and a point of L.
(2) Let L E B2 and let p be a point of Vx not on L. Counting the number of
points in Sx gives \PX\ = \P\ = (q + l)(q2 +1) (q + l)q 1 = q3.
In S there are q points in L\{a;}. Each of these points is on q +1 lines.
As no point of Vx can be on more than one such line, these lines cover
q(q + l)(q 1) = q3 points of VX\L\ i.e. these lines partition the points
of VX\L. Hence there is a unique line M E Blx which is incident with
p and a point of L.
Because p 0 L, the regularity of x implies {.r.p) has no point of
Vx in common with L. Therefore M is the unique line of Bx which is
12


incident with p and a point of L.
This shows that Sx is a GQ(q 1 ,q + 1). Most GQ constructions
presented here are similarly straightforward to verify. For this reason it will be
often left to the reader to check that a given incidence structure satisfies Gl,
G2, and G3.
The process of forming Sx from S is called expanding about the
regular point x. Sx is sometimes written P(S,x).
Dually, if L is a regular line of S, replacing x with L and interchanging
the roles of points and lines gives a GQ(q + 1, P(S, L). Naturally this process is called expanding about the regular line
L.
If these ideas are interpreted algebraicly, observe that the 4x4 dual
grid can be constructed using groups. Let P+ = T U {At} where A = {g j}-
The points of the dual grid can be viewed as points of G\ the lines can be
viewed as cosets of subgroups in P+.
The 4x4 grid can be constructed similarly. Let T' = {Mi, A, AA, A A}.
The points of the grid are the elements of G and the cosets of subgroups in A.
Section 7.1 of this thesis explores this interpretation further.
In some respects expansion about regular points and lines is the GQ-
analogue to the relationship between projective and affine planes. If 7r is a
13


projective plane, a line L and all of the points on L can be removed from n to
form the corresponding affine plane. Similarly, a point p and all of the lines
through p can be removed from tt to form the corresponding dual affine plane.
We know that to every projective plane of order n there corresponds
an affine plane of order n. Every known GQ(q, q) has at least one regular
point or one regular line, and hence gives constructions of GQ(q 1,^ + 1)
or GQ(q + 1 ,q 1). Moreover, every known GQ whose parameters differ by
2 arise by the method of expansion. However, expanding about two different
regular points may produce to two non-isomorphic GQ.
Recall that every known projective plane has prime power order. Sim-
ilarly in every known example of GQ with parameters (s, s), (s 1, s + 1), or
(s + 1, s 1), s is a power of a prime. This may seem unusual at first reading,
however further examination shows a strong relationship between projective
geometries and GQ.
We close this chapter with some special constructions of planes relat-
ing to GQ. Readers unfamiliar with projective and affine planes are directed to
the appendix. Let S = (V, £>,Z) be a GQ(q, q) with a regular point p incident
with a regular line L.
Construction 1.3.7 The projective plane ir*(p) = ( V*(p),C*(p) ):
Let T*{p) = p. Let C*(p) = {{x,y}\x,y e ir1}, i.e. C*(p) consists of the
14


given by containment. Clearly \V*(p)\ = q2 + q + 1 and each member of C*(p)
contains q + 1 points. Because any pair of points in V*(p) determines a unique
member of £*(p), ir*(p) is a projective plane of order q.
Construction 1.3.8 The affine plane ir(p) = ( T(p),C(p) ):
Let C(p) = £*(p)\{L}, and let T{p) be the points of T{p) with the points of
L removed. This makes 7r(p) an affine plane of order q.
Construction 1.3.9 The projective plane it*(L) = ( T*(L),C*(L) ):
Let V*{L) = { { M. iV}-L|M, N e L } and £1 (I.) = L. Incidence is given by
containment. This is a projective plane of order q.
Construction 1.3.10 The affine plane 7r*{L) = ( £>*(£),£*(£) ):
Remove L and all of its points to form the affine plane 7r* (L) = ( V* (L), £* (L) ).
The projective planes constructed in 1.3.7 and 1.3.9 are well known.
The related affine planes of constructions 1.3.8 and 1.3.10 will be revisited in
section 3.4.
15


2. Known examples
In this chapter a number of GQ will be constructed from objects in
projective 3-space. For these constructions some notation for the projective
objects is first established. See the appendix for more on projective planes,
projective 3-space, ovals, and oval permutations. Let F = GF(q) and let a
be an oval permutation of F. Let A = (0,1, 0, 0),B = (0, 0,1, 0), Q = {ws =
(l,s,s,0)|s e F} U {A}; thus fi is an oval in the plane 11^ = [0,0,0,1]T
embedded in PG(3,q) = Q. Let fi+ = fi U {!?} and let fi- = If q is
even then B is the nucleus of fi.
2.1 Two classical examples: W(q) and Q(4, q)
For two classical examples we visit polar spaces. Let v be a symplectic
polarity of Q. Hence for a point .r f Q. .r1' is a plane containing x. For example
let v be given by the alternating form g{u,v) = uMvT where M is the skew-
symmetric matrix:
0 0 0 1
0 0 10
The lines of Q which are fixed by v
0 -1 0 0
1 0 0 0
are called totally isotropic lines. The points of Q together with the totally
16


isotropic lines of Q under v form a GQ(q, q) written W(q) in which every point
is regular. (The polarity u is suppressed in the notation because all such GQ
are isomorphic.)
Now let f : F5 ^ F be some non-degenerate quadratic form and let
Q be the associated quadric. This means b(x,y) = f(x + y) f(x) f(y)
is a symmetric bilinear form, and () = {.r e- PG(A, q)\f(x) = 0}. That /
is non-degenerate means that for each x there is at least one y such that
f(x + y) ^ f(x) + f(y). The points of Q together with the projective lines
whose points are all in Q also form a GQ(q,q), written Q(4, q). The two
GQ, W(q) and Q(4, q), are point-line duals of one another. They are self-dual
exactly when q is even [PT84],
2.2 A construction by Tits
The first non-classical example we will examine is one due to Tits.
It was first reported in [Dem68]. In [Pay85a] Payne modified Tits description
slightly. It is this modified description which is presented here.
Define an incidence structure S = where points are of the
following two types:
i) Points of and
ii) Planes of Q which contain either exactly 1 or q + 1 points of fi.
17


The lines of S are the lines of Q which meet 0 in exactly one point.
Incidence is given by that of Q. Let wq, ... ,wq be the points of fi with the
respective tangent lines labeled LQ,... ,Lq. Verification that Gl, G2, and G3
hold follows from well known counts from projective geometry. These are
outlined below.
If p is a type i) point of V, then the lines of B through p are exactly
the q + 1 projective lines < p,wk >,0 < k < q + 1. If p is the plane 11 v. then
the lines of B incident with p are the q + 1 tangents to fi. Finally if p is any
other type ii) point, then in Q, p contains exactly one point of fi, say Wj. The
lines of B incident with p are the q + 1 projective lines through wj in the plane
p.
If L = Lk is a tangent to fi in 11 v. then the points of V incident with
IIoo are the q + l projective planes through Lk. If L is a line which meets
in the point Wj, then the type i) points incident with L are the q projective
points of L\{wj}, and the unique type ii) point incident with L is the plane
through L and Lj.
Let L be a line of B, let pi be a type i) point not incident with L,
and let P2 be a type ii) point not incident with L. Projectively, let L meet fi
at wj, and let M =< p\,Wj >. As M and L meet in S at the type ii) point
< L, M >, M is the unique line of B incident with pi and a point of L. If L is
18


a tangent line to fi, let M be the line of p2 which is also tangent to fi; in this
case L and M are both incident in S with the point Iloo. If L is not a tangent
line then in Q, L meets Iloo at a point ir r If p2 = 11 v then let M = Lj:
otherwise if p2 fi fi = Lk, let y be the projective point common to L and p2
and let M =< y,wk > In either case M is the unique line of B incident with
p2 and a point of L.
With Gl, G2, and G3 satisfied, this construction is shown to yield a
GQ(q, q) which is often written T2(fi) or does not provide new quadrangles; however when q even and with
T(tt) is different from W(q) and Q(4, q). Assume for the remainder of this
section that q is even. In this case B is the nucleus of fi.
Proposition 2.2.1 Iloo is a regular point of T2(Q).
Proof: Let x be a point of T2(Q) not case x is a projective point. Observe that {Iloo, a;}-1 is the set of planes through
< x,B >, and {11 v../ } is the set of q projective points on < x, B > other
than B together with Iloo. Thus {11 v../ } is as large as possible.
Proposition 2.2.2 The lines of Iloo which are tangent to fi are regular lines
of r2(n).
Proof: Let L a line of Iloo tangent to fi at p and let M be a line
19


not coincident in T2(fi) with L; this means M is a projective line meeting 0
at some point r ^ p. Observe that {L, .1/} is the set of q lines spanned by p
and a point of M other than r together with the line < B,r >; and {L, M}J-
is the q lines through r in < M,p > other than < r,p > together with L.
2.3 Constructions by Ahrens and Szekeres
Ahrens and Szekeres provided two examples of non-classical GQ which
we describe here. For q odd the original description of AS(q) is as follows:
Points of AS(q) are the points of affine three-space AG(3, q). Lines of AS(q)
are the following curves of AG(3, q):
(i) x = a, y = a, z = b,
(ii) x = a, y = a, z = b,
(iii) x = ca2 ba + a, y = ^2ca + b, z = a.
Payne has given the following alternative construction of this GQ(q
1,5 + 1): Choose any point x of W(q) (recall x is regular in W(q)). Let Vx be
the points of W(q) which are not contained in a totally isotropic line through
x. Let £>,. = B\ U £>2 where B\ is the set of totally isotropic lines of W(q) and
£>2 = {{+ y}x\y E Vx}. Incidence is given by containment. Notice that this
is the process of expanding W(q) about x which gives the GQ(q 1,5 + 1)
20


written P(W(q),x). This construction works regardless of the parity of q. For
the remainder of this thesis the case where q is even will be of particular
interest.
For q even this next construction also yields a GQ(q 1,5 + 1). It
is a variation on the construction of T2(0) due to Ahrens and Szekeres [AS69]
and independently to Hall [Hal71]. Let S+ = (V+.JB+.X+) be the incidence
structure in which V+ consists of the type i) points of V, and B is the set
of lines of Q which meet in a unique point of fi+. It follows that S+ is
a GQ(q 1,5 + 1) using the incidence of Q. Such a construction is written
T*(tt+) or sometimes simply S(Q+). Notice that T2(Q+) = P(T2(Q), IIoo).
The following observations are immediate.
Observation 2.3.1 For each x E fi+, the lines of through x form a
spread.
Observation 2.3.2 The set of all such spreads formed above is a packing M.
Further examination reveals
Proposition 2.3.3 If Li,L2 are in a spread of M, then {Li, L2} is regular.
Proof: Let Lx fi L2 = y. Li, L2 determine a plane n meeting in
a line L through y. This line meets some other point x ^ y of fi+. {Li,L2}
is then the set of q lines of fI\{L} through x. Hence {Li, L2}LL is the set of q
lines in L1\{L} through y.
21


This packing in fact has a stronger property than just having regular
pairs within spreads. Observe that every pair of lines from a spread S of M
has its trace contained in some other spread of M and has its span back in S.
Such a spread S is said to be pivotal for the packing M.
Proposition 2.3.4 If Li,L2 are in different spreads of M. then {Li,L2} =
{LuL2}
Proof: Assume Sj is a spread of M. containing L\ and assume S2 is a
different spread of M containing L2. Let L e {Li, L2}\{Li, L2}. First sup-
pose {Li,^}1- contained two lines Mi,M2 from some spread S in M. Then
as S is pivotal Li,L2 are contained in a spread of Ad, a contradiction. So
has one member in each spread of Ad\{Sj, A2}. As spreads contain
only non-concurrent lines, L is in either Sj or S2. But as {L\,L2} is regular,
{ /.i. } = { /.i. /.} = { /.} . Hence {Li,L2} is contained in either
Si or S2, a contradiction.
Observe that nothing in the proof above relied on the specific con-
struction of the quadrangle, rather it relied only on having a GQ with a packing
of pivotal spreads.
22


2.4 A construction by Payne
This section provides a construction of GQ(q +1, q 1) for q even due
to Payne [Pay71b, Pay72b, Pay85a] which also uses the hyperoval fi+. When
fi+ does not contain a conic, the GQ constructed here are actually different
from the duals of those constructed by Ahrens and Szekeres.
Let Oab (respectively 0Ba) be the set of planes in Q which meet A
but not B (respectively B but not A). Finally let = T+UOab^Oba and B
be the set of lines in which contain a point of Q_. Again incidence is that
of Q. This GQ is often denoted A, B >).
Any two planes of Oab meet in a projective line. Such a line cannot
be a line of Ob a is an ovoid.
If L is a line of PG(3, q) the set of planes through L is called the
dual line in PG(3,q). If one of these planes is removed, the result is called
a dual line in AG(3,q). For any two points x0,y0 E 11 x.. let Vxo,yo be the
dual line in AG(3, q) consisting of the q planes other than IIqq which contain
the line < xq, t/o > In particular observe that Tab = {lb = [1, 0, 0,i]T\i E F}.
Observe that no projective line of Lb is a line of q2 points of 23


of Observation 2.4.1 M. = {Oab, Ob a} U Va,b is a fan of ovoids.
Each fl* has q2 + q + 1 projective points, q2 of these are pairwise non-
collinear points of the quadrangle, and the remaining q + 1 projective points
are not part of the GQ. Furthermore each fl* has q2 + q +1 projective lines and
no quadrangle lines. Consequently, the fl* will be viewed both as projective
and quadrangle objects, where the context will dictate meaning.
The dual notion of a pivotal spread is a pivotal ovoid; i.e. an ovoid
O is pivotal for a fan M. provided every pair of points in O is regular, every
pair in O has its trace in some other ovoid of M. and every pair in O has its
span in O. This leads to the next proposition.
Proposition 2.4.2 The ovoids Oab,0Ba are pivotal for A4. Moreover the
trace and span of any pair within one of these ovoids are affine lines or dual
lines in AG(3, q).
Proof: Choose ws to be a projective point of Or, and let Pi,P2 £
Va,ws- As points of planes of Vb,Ws> then Q meets Pi in a projective line through tcs; i.e., as points
of S, Q and Pi are collinear by a line of S. Likewise Q and P2 are collinear
by a line of S. Thus Q E {Pi,P2}. Hence {Pi,P2} = Vb,Ws C 0Ba, and
{Pi,P2} = Va,ws C Oab-
24


Now let P\. /N e Oab such that Pi fl 11 v =< A,ws P2 fl
11 -y. =. Let Pi n P_> = A i _> = {. 1. ./ i.^. Then {Pi / L} =
{ri,..., C< Aij2, B > |w* e fi-} C
Oab- Thus we see that Oab (and by a similar argument, Ob a) is a pivotal
ovoid.
One might be tempted to think that just as the packing constructed
in the previous section consisted of pivotal spreads, the fan constructed here
consists entirely of pivotal ovoids. However this is not always the case. Up until
now, no conditions on the oval permutation a were made. At this point, the
type of oval we start with will determine certain properties of the associated
GQ.
Proposition 2.4.3 If < a >= Aut(F), then all ovoids of M. are pivotal.
Moreover the trace and span of pairs within one of these ovoids are either affine
lines, dual lines in AG(3,q), or planar arcs isomorphic to fi- which complete
to hyperovals with A and B.
Proof: Oab, Ob a were considered in the previous proposition. Now
consider each of the n^. Let X\,X2 enfl6 Pab] let xq =< X\,X2 > PlHoo.
If xQ = A, then {.rj. } = {< A,x\,Wi > |w* il } C Oab and
{.ri. =< X\,X2 > \{A} C na. Similar results hold when xq = B.
25


Now consider the situation where xQ e< x\,x2 > \{.1. B}. Let x\ =
(a, b, c, 1), x2 = (a, e, f, 1). In this case xq = (0, y, 1, 0) where y = (b + e)(c +
/)-1 7^ 0 and hence b ^ e. For each ws e there is a unique other point w§
of on < Xq,ws >. Let Ai =< X\,ws >; let A2 =< x2,w§ > and observe
Ai flA2 = (a + k, b + ks, c + ksa, 1) = (a + h, e + hs, f + hsa, 1) for some h,k ^ 0.
Matching up first coordinates gives h = k. Matching second coordinates gives
k = (£> + e)(s + s)-1, and matching third coordinates gives k = (c + f)(s + s)~a.
From this we have (h+e)(s + s)-1 = (c+f)(s + s)^a implies (s + s) = [(b+e)(c+
Z)-1]13" (we use ^e fact that a generates Aut(F)). Plugging this in above we
1 a 1
get k = (h + e)(s + s)-1 = (h + e)([(h + e)(c + /)-1]T3^)-1 = (6 + e)^3T(c + /)T3^.
Now k is written in terms of the coordinates of x\ and x2, hence we
can write {.r t} = {(a+k,b+ks,c+ksa,l)\s ef} = T where A: is as given
above. It is easy to see that is a g-arc in IIa+fe (note that k is necessarily
different from zero). Moreover if we let T1 = U {A, B} we have T1 is
projectively equivalent to fi+. To see this let ip be the transformation given
a+k b c 1
0 k 0 0
by (t,u,v,w)^ = (t,u,v,w)
. Specifically = A, B' = B,
0 0 /c 0
1 0 0 0
and (1, s, sa, 0)^ = (a + k,b + ks, c + ksa, 1).
26


Now let £- = {pr = (a, b+(b+e)r, c+(c+/)r, l)|r £ F}. For a fixed
pr, consider the line formed by pr and some (a+k, b+ks, c+ksa, 1) £ {x\,X2}-
Such a line would intersect in (1, (b + e)13" (c + /) + s, k^l{c + f )ra +
sa, 0). But [(6 + e)13^(c + /)3Tr + s]a = A;_1(c + /)r + sa. This point is
on fi- and hence this line is a GQ line. In other words each point of is
collinear in the quadrangle with each point of {xi,x2}- Since has size q
we see that in fact XT = {xi.jx2}. It is easy to see that X is a q-arc in Ii
which forces IIa to be a pivotal ovoid.
Note also that X+ = X- U {.1. B\ is projectively equivalent to il .
This equivalence is given by a where
a b cl
(t, u, v, w)a = (t, U, V, tv
0 b + e 0 0
0 0 c+f 0
10 0 0
. Clearly a maps fi+ to X+ with A and B fixed.
Therefore each IT* e Va,b is pivotal. Hence M. is a fan with every
ovoid pivotal.
Examining pairs of points from different ovoids gives the following.
Proposition 2.4.4 If all ovoids of M. are pivotal then for x\,x2 in different
27


ovoids, {.r i. } = {.r i.
Proof: The proof is similar to that of proposition 2.3.4, with line
and spread replaced by point and ovoid.
As with proposition 2.3.4, note that the proof depends not on the par-
ticular construction but only on the existence of a fan with all members pivotal.
This idea of recognizing properties of GQ independent of their constructions
will be useful in their classification and characterization.
28


3. Regular ovoids in S
In this chapter regular ovoids are examined extensively. The main
theme in this chapter is the relationship between regular ovoids and affine
planes. Along the way, a method of deriving a GQ of order q from one of order
(q + 1, q 1) is presented.
Let S = (V, B.X) be a GQ(q + 1, 5 1) with q even. Let Ooo be an
ovoid of S, i.e. is a set of 1 + (q + l)(q 1) = q2 points of S, no two
on a line. This means that each line of S contains exactly one point of O^.
Also assume that is regular: for all .r. // t- {.r. y} is a regular pair and
{x^}^ is a set of q points all in O^.
3.1 Affine planes and ovoids
This section begins by noting that the points and hyperbolic lines of
the regular ovoid O^ can be viewed as the points and lines of an affine plane.
The parallel classes of this affine plane are used to construct the remaining
ovoids of a fan. Examining these ovoids from the affine point of view will help
in determining other characteristics of the GQ.
Construct an affine plane 'k{O0O) as follows. Take the q2 points of
29


Oao to be the points of the ^(C^). For x and y in let the hyperbolic line
{a;,?/}11 be a line of ,k{0oq). Let incidnce be given by containment. Because
000 is regular, |{rc, 2/}_L_L| = q- Furthermore by observation 1.3.2, each pair of
points in 7r(O^ is contained in a unique line of This makes
an affine plane (see the appendix).
Let /1. /b....l\, be the pairwise disjoint lines of one parallel class of
7r((9oo). The points of these lines account for all the points of the ovoid. Hence
Tl U T2 U U Tq = Ocq. The proposition below shows how a related ovoid can
be constructed from parallel classes of
Proposition 3.1.1 T1 U T\U U T'q is an ovoid of S.
Proof: Let I) = {./,. //,} Recall that T1 is the set of points
collinear with everything in Tj.
Let z\ E Tf1 = {.r 1. //1} and z2 e T^ //_.} If £2 £ Tj1 then
2/i e T2. (To see this observe that the q points of fl O^ are exactly the q
points of T2.) This is a contradiction because Tl is parallel to T2. Thus for all
1 0 j, T)1 fl T1 = 0 which implies IT1 U U T1) = q2.
If z[ E Tj1, then zi 0 z[, otherwise a triangle in S is formed. Now
suppose z\ ~ Z2; then z\ E {.r 1. ) Since z\ is collinear with each of the q
non-collinear points of Tx, 2:2 must be on one of the lines from zi to a point of
Tl, say Z2 E y\Z\. Since each of the q lines through Z2 contains exactly one
30


point of T2 = {x2,y2}, the line y\Z\ must be such a line. This means there
is a z E y\Z\ such that z 6 T2, giving two points of Ooo,yi and z, which are
collinear in S. But this cannot be as is an ovoid.
This shows that no two points of Tj1 U U are collinear in S.
Therefore Tj1 U U is an ovoid of S. m
This shows that for each parallel class of 7r(0oo) there is a corre-
sponding ovoid. These ovoids will be of primary interest. However as an aside
observe that in fact, many more ovoids can be constructed.
Proposition 3.1.2 If T' E {'/}. 1) \. 1 < % < q, then Uf=i T/ is an ovoid.
Proof: For distinct i and j, no point of Tt is collinear with a point
of Tj. Likewise by proposition 3.1.1, no point of TA is collinear with a point of
Tj-. Assume T[ = l\ and = T%- Let //i e T\,z\ E Tf1, Z2 E T%- If E l\.
then z2 ~ Z\. As shown above, this cannot happen. Hence Ti CiT,^ = 0, which
implies \T[ U U T'q\ = q2.
We need to show that no point of Ti is collinear with a point of T^.
For contradiction, suppose yi ~ z2. As above, since every line through z2 con-
tains a point of there is some z E y\z2 with z ET2. But then z and y\ are
two points of Oao which are collinear in S. This contradicts O^ is an ovoid of
S. Therefore no two points of Uf=i T' are collinear.
31


Let Eq, Ei, ... Eq he the q +1 parallel classes of lines of 7r(0oo). Then
for 0 < i < q, Oi = : T E Ej} is an ovoid as shown above, and
M = {Oooi Oo, Oi,..., Oq} is a fan of ovoids. As O^ is regular, it follows that
Ooo is pivotal for M. By construction each Oi is partitioned into hyperbolic
lines of size q whose traces partition O^, 0 < i < q. These ovoids may or may
not also be regular; in some cases they are all regular. The final theorem of
chapter 4 will help to classify the regular ovoids.
3.2 Intersection of traces with ovoids of M
In this section two propositions from [Pay85a] are reviewed. These
propositions indicate how the traces of point pairs intersect the ovoids of M.
They will be used in examing regular pairs in Chapter 4 and in characterizing
GQ in Chapter 5.
Put / = {(). I...(/}:/ = / U {oo}.
Proposition 3.2.1 Let b, d E Oj, with j E I,b ^ d. If {b, d}L fi O^ ^ 0 then
{M^cOoo.
Proof: Let b,d E Oj,b 0 d. Suppose a E {b, d}L fi O^ and
c E {b, d}1- with c 0 O^. As each line meets each ovoid of M. exactly once,
let a! = be fi CCo and let / be the unique point of ad collinear with a/. As
32


a, a' E Oao (which is pivotal) and b E {a, a'}-1 then all of {a, a'}-1 is contained
in Oj\ specifically f E Oj. But d is assumed to be in Oj, a contradiction as
d~f.
Proposition 3.2.2 Let O^, Oj, Oj be distinct ovoids of M.
i) x E Oj, y E Oj, and x ^ y implies |{:c, y}L fi 0,x\ = 1.
ii) a E Ooo, b E Oi, and a b implies |{a, C\ Oj| = 1;
Proof: For each a E O^, ax contains q(q 1) noncollinear pairs
(z,w) E Oi x Oj. As Ooo is pivotal, if a and a' are distinct points in Ooo, then
a1- and a'L can never contain the same noncollinear pair (z, w) E OjX Oj. Thus
there are (\000\)q(q 1) = q3(q 1) such pairs as a runs over the elements of
Oo-
On the other hand, counting the number of such pairs shows that
there are q2 choices for z E Oj, and q2 q choices for w E Oj with z w. This
gives q3(q 1) noncollinear pairs (z,w) E Oj x Oj, which proves part i).
For a fixed a E Cl v. .r E Oj, {a^x}1- has exactly q points. If any two
of these were in the same ovoid of M then the previous proposition would be
contradicted. This proves ii).
33


3.3 Constricting about a regular ovoid
At this point additional GQ can be constructed. Having a regular
ovoid Ooo in S allows us to form a new GQ, S^, of order q via a method
called constriction about 00c. This is done by defining the points, lines and
incidence of Points of Sqq'. Too = P\Coo together with the symbols (O0),..., (Oq)
Lines of v = B U { /' : T is a line of ~(Ox.)} U {Too}
Incidence of S^: A line of B is incident with a point of Too provided
the two were incident in S. If T is a hyperbolic line of O^, then TL is incident
with the q points it contains in S and Tx ~ (Oi) provided Oi is the (unique)
ovoid from M. containing TL. Finally is incident with each of the q + 1
points (Oi), 0 < i < q.
In chapter 1, it is was noted that the process of expanding about a
regular line to form a GQ(q +1, 5 1) from a GQ(q, q) was analogous to forming
an affine plane from a projective plane. Similarly, the process of constricting
about a regular ovoid in a GQ(q + 1, q 1) to form a GQ(q, q) is analogous to
forming a projective plane by adding in a line to an affine plane.
34


3.4 Regularity in Here the regularity of lines and points of regular line is found; next regular points are found in terms of pivotal ovoids
from S.
Proposition 3.4.1 /, v is a regular line of Proof: For this proof, incidences from both GQ,S and S^, are
considered. Let L*, Lj, L& be distinct lines of (O,). (Oj). (O/,.) respectively. In S, Lt = I) Lj = Tj1 where I) and 7} are
affine lines in distinct parallel classes of 71(0^). By the dual of observation
1.3.3 it suffices to show that {L,h Lj} is a regular pair.
Let L E {Li,Lj}; in S, L intersects O^ in some point x. Hence
x is collinear with all of and with all of T^. Let Rx be the set of lines
through x thought of as lines of then Rx C {Li,Lj}. In fact as \RX\ =
q = { L,. L,} {Loo}|, { /'/ } = Rx U {Loo}. If Lk ~ L then by similar
argument L is concurrent everything in /?,: i.e. any line concurrent with the
pair {Loo,L} C {Lj^Lj}1- is concurrent with all of {Li,Lj}. From the dual
of observation 1.3.1, this means that {Li, Lj} is a regular pair and hence L^
is regular.
For the next proposition the following lemma is needed.
35


Lemma 3.4.2 In any GQ{s,t), let L and M be distinct lines through 2 such
that each point on L {z} forms a regular pair with each point on M {z}.
In this case 2 is itself a regular point.
Proof: For any xel {). y e M {} observe that {.r. //}
{^}| = t. There are s choices for x and s choices for y, and hence there are
s2 traces {x, y}1- containing 2. This accounts for s2t points not collinear with
2. On the other hand, the total number of points of the GQ outside of ^ is
(s + l)(st + 1) s(t + 1) 1 = s2t. Now every point not collinear with 2 has
been accounted for in these traces. As {x, y} is assumed to be regular, every
pair from {x, y}1- must be regular. Specifically 2 forms a regular pair with
every point it is not collinear with, i.e. z is regular. Thus any noncollinear pair
in is regular.
We will now consider the relationship between regular points in
and pivotal ovoids in S.
Proposition 3.4.3 (Oj) is a regular point of for the fan M.
Proof: 1) {{Oj) is regular implies Oj is pivotal for M)
In what follows assume that x and y are points of S^ which were also
points of S; i.e. they are not points of the form {Oj).
36


{x,y} is a set of non-collinear points of (0,i)L in if and only if
x,y E Oi and x and y are in distinct traces, Tf1, of the hyperbolic lines, say
'/'i and T2, of C> v such that Tf1, C C>( and Ti fi T2 = 0.
(C?i) is a regular point of implies {x, y} is a regular pair in S^, and
hence {.r. //. } = {x\ = x,X2 = //. x:>. x\.....rq. i}. Each point of {.r. //}
is collinear with each point of {x,y}. Each line through (Oi) of the form
Tf- contains a unique point of {:c, t/}-11. As there are only q such lines, the
remaining point must be some (Oj). This gives {x, y}1- = {t/i, t/2,..., yq, yq+\ =
(Oi)} and {./. y}^ = {.r,. .....rq. xq+i}.
Thus |/i, y-2, , yq are all collinear with (Oj) in are all contained in the ovoid Oj in S. This shows that if (Oi) is a regular
point of S0o, any two points contained in Oi (in S) have their trace in an ovoid
Oj of M and have their span completely contained in Oj. This means Ot is
pivotal for M.
2) (Oi is pivotal for M in S implies (Oi) is a regular point of Soo.)
Assume Ot is pivotal for M. Let x1;x2 be collinear with (Oi) such
that Xi 9^ x2 and x\,x2 not on L v. Then x\,x2 e Oi, and if Oi = T-j1 U -UT^
with Ooo = Ti U U Tq, without loss of generality assume x\ E I\.x2 E T2.
In S, since Oi is pivotal, {.r t} = {.r t.....rq} C Oi and {.r t} =
{yi,...,yq} C Oj for some Oj E M. Then for 1 < i,j < q, xt ~ yj
37


in S and hence also in S^. Back in {.ri. = {x\,....rq. (Oj)}. Therefore {aq,^} is a regular pair of points
in S0c. By lemma 3.4.2, {aq, (Ok)} is regular for all (Ok) on L^. Therefore
(Oi) is a regular point of S^.
In addition to the relationship between regular points of and piv-
otal ovoids in S, there is also an interesting relationship between planes asso-
ciated with Soo and planes associated with S. Here, too, regularity plays a key
role.
For the moment, consider the plane ^(Ooo). Let £0 = { /'i 7L>...., Tq}
and E\ = {i?i, R2,..., Rq} be two distinct parallel classes of lines in ^(O^).
Each point of ,k(000) is on a unique hyperbolic line Tj and a unique
hyperbolic line Rj, hence these points may be labeled as aqj = Tj fi Rj.
Tj- _ Xi^}1- for any h ^ j. and Rj- = {aq(. ,r/iV} for any k ^ i.
In what follows the notation is abused somewhat, with the _L notation used
for whichever GQ is convenient. The context should make it clear what the
notation means. To aid in this, when Tj- and Rj- are viewed as lines of S^,
we will write I., = Xj- and .1 /( = Rj-.
Traces and spans of lines in can be determined as follows. In the
GQ 38


the line L00. Every line of {,l/(} can be viewed as the trace in S of a
hyperbolic line of which contains Xij.
From the regularity of L00 in S^, there is an affine plane tt(L(*,)
which arises as in construction 1.3.10. The point set here can be viewed as
Mj\ : 1 < i,j < q}; £(Too) = {HL : H is a hyperbolic line of C^}.
In this setting, if Hx is an affine line of ,k(L00), the affine points incident with
H1- are the hyperbolic points in Theorem 3.4.4 The affine planes ir(O^ and ^(Loo) are isomorphic.
Proof: Deline ip from V(Ox.) to V(L0c) by 'tp(xij) = { !) Hj } .
To see that ip preserves col linearity, first suppose that l = Tt = {xiA...., x,^q}.
In which case 'tp(l) = {{1) /t^ } . {1) ./?._;} ..{l) Iiq } i.e.. ex-
actly the set of points in P>(T00) on the line Tp-. Now suppose that l ^ £o, say
l = {xhri,x2,r2, ,%rj- m is then {ip(xhri), -ip(x2,r2), , ^(%rg)} which
equals { { /', } {7b . } ...., {Tq /i>(: } } which in turn is equal to
{{ /. i^}^, {7b . ___{7 / } }. But this is exactly the set of points
in 'P(Too) on the line lL E £(£oo). Thus ip maps lines to lines. Thus ip is an
isomorphism between ^(0^) and
39


Now suppose that the ovoid Oq is also pivotal for the fan M. so that
the point {Oq) of L00 is regular in Soq.
From the regularity of the point {Oq) in there is an affine plane
7r( {Oq) ) which arises as in construction 1.3.8. This new affine plane has point
set V{ {Oq) ) = {x : x e Oq} and line set £( {Oq) ) = {{rr, t/}JJ : x,y e Oq}.
But these are exactly the point- and line sets of k{Oq), and the incidences are
the same. This proves the following theorem.
Theorem 3.4.5 The two affine planes 7r( {Oq) ) and k{Oq) are identical (not
merely isomorphic).
In the case under consideration here, i.e., that both (D^ and Oq are
pivotal for M. in S, Payne has shown in [Pay72a], [Pay77] and Chapter 12
of [PT84] that details). In light of theorems 3.4.4 and 3.4.5 the following is an immediate
consequence
Theorem 3.4.6 if and only if the two affine planes 7r((9oo) and k{Oq) are desarguesian.
This theorem will be utilized in a characterization given later.
40


3.5 Grid-like Fans
Next consider the associated planes in the case where at least three
of the ovoids of M. are pivotal. To answer the question, Are these planes
isomorphic, a new axiom is introduced and examined. Let S be a GQ(q +
1 ,q 1) with a fan M.
Grid-Like Axiom: Let Ot, Oj, Ok be any triple of distinct pivotal ovoids for
M with ./ i. E O,. i/i, |/2 E Oj, z\,z2 E Ok such that {xi,x2} and {yi,y2} <£ Ot U Ok. If xi, 2/1, Z\ lie on a line and if x2,y2, z2 lie on a line,
then every line meeting {.rj. ) and {2/i,2/2} also meets {zi,z2}-
Any fan with at least pivotal ovoids satisfying this axiom will be called
a grid-like fan. In section 2.4, P(T2(Q), < A,B >) was shown to have a fan
M. = {Oab, Oba}UVa,b with every ovoid pivotal provided fi was a translation
oval, i.e. < a >= Aut(F). Additionally, the following theorem holds.
Theorem 3.5.1 M. = {Oab, Ob a} U Va,b is a grid-like fan.
Proof: Let I\. I2 E Oab, and let Qi,Q2 E Oba Let xi,x2 E
na, 2/1,2/2 ^ na+m, Z\, z2 E na+n with m, n, 0 all distinct such that X\, 2/1, Z\, P\,
and Qi are all collinear in S by Ai, and x2,y2, z2, P2,Q2 are all collinear in S
by A-_>. P\ =< 1. A1 2>. (J\ =< B, Ai ^>,P2 =< 1. A_> >, and (J2 =< B. A_> >.
Let Xq =< Xi,x2 > nll.v 2/0 =< 2/1, y2 > rll v A =< Zi,z2 > miv. Let
41


Ws Cl II x. A2 0 Iloo
Construct the hyperbolix lines H\ = {.r i. } . //_. = {//1. //_,} //.> =
{-i- c} //. = {/C/A} and //- = . Let H = {//,...........//-,}.
To show that M. is grid-like, we need to show that any line intersecting two
hyperbolic lines of % must intersect every line of %. To do this, we must con-
sider the various ways xQ,yQ, and zQ could be related to each other and to A
and B. The following lemma will assist in this.
Lemma 3.5.2 If x0 ^ yo, then in fact xQ ^ zQ ^ y0.
Proof of lemma: Suppose for contradiction that t/0 = %o- Let II* be
the plane < t/o, Vi, > Observe the following implications. z\,y\ E II* implies
Xi,ws E 11,: Z\, Zq E 1I* implies z2 E 11,,,: z2, //_> C 11* implies x2, wt E 11,: xi,x2 E
II* implies xQ E II*. Finally xQ,yQ E II* implies A,B E II*. Thus Iloo fi II*
is line containing A, B,ws,wt, a contradiction. Hence r/o ^ zq, and similarly
xQ ^ zQ. This proves the lemma, and return now to the proof of the theorem.
Consider the following cases and subcases for x0,yo, and z0:
(1) xq = A with
(a) x0 = y0 = z0, or
(b) xo,yo, zq are distinct members of < A, B >.
42


(2) xq = B (But results here will be like those above with A and B inter-
changed.)
(3) xQ = y0 = zQ G< A, B > \{. 1. B} with
(a) ws = wt, or
(b) ws wt.
Case 1: xQ = A. Let x\ = (a. b. c. I )../_. = (a,b + e,c,l),iJi =
(a + m,b + ms, c + msa, 1), t/2 = (a + m, b + e + mt, c + mta, l),z\ = (a + n,b +
ns, c + nsa, 1), 1J2 = (a + n,b + e + nt,c + nta, 1).
First consider A = xQ = y0 = zQ (this is true if and only if s = t). If
x?j G {xi, X2},'i/3 G {yi,'tj2} with xq ~ y$, then there exists some d G F
such that Xq = (a,b + d, c, 1) and y3 = (a + m,b + d + ms, c + msa, 1). Let
Zq = (a + n,b + d + ns, c + nsa, 1). Observe that z3 G {i. z2} and z3 G<
xq,ws >=< Xq, i)Q >. This means that any S line meeting {.r t} and
{yuU2} must also meet {z1,z2}. Also, observe that Pi = P2 G {xi,x2},
and that < B,XQ,yQ >G {Qi,Q2}-
Now consider xq = A,y0,ZQ as three distinct members of < A,B >,
and let x\. yx. //_,.i. be as above. Without loss of generality, consider
l/o = B, Then y2 = (a + m,b + ms, c + msa + h, 1) for some h. Comparing
this to y2 as written above observe t = em1 + s and h = ml^aea. So in fact
1/2 = (a + m, b + ms, c + msa +ml^aea, 1), and also z2 = (a + n, b + e, n{emTl +
43


s), c + n(em-1 + s), 1).
Again choose xz E {%i, X2},V3 £ {yi,V2} with xz ~ yz- Thus
Xz = (a, b + d, c,l),yz = (a + m,b + ms, c + msa + /i, 1) such that < .r3. |/3 >
intersects Lt^ at a point (m, d + ms, /i + ms, 0) E fi+ Dividing through by m
gives (dmrl+s)a = sa+hmrl which forces h = dam1-a. In fact y3 = (a+m, b+
ms, c + msa + dam1-a, 1), and < Xz, yz > Cin^ = (1, s + dm-1, sa + dam~a, 0).
From this it follows that < .r3. y3 > filla+n = (a + n,b + d + n(s + dm-1), c +
n(sa + dam~a), 1). Call this point z3.
Now observe that z3 E {zi,z2} = {(a + n, b + ns, c + nsa, 1), (a +
n, b + ns + (1 + nm-1)e, c + nsa + nm~aea, 1)}J~ = {(a + n, b + ns + (1 +
nm-1)er,c + nsa + nm-aeara, l)|r E F} (to see this let r = de-1). Any
Saline meeting {.rt) and {yi,y2} must also meet {+, z2}- Observe
that < A,xz,yz >e {Pi,P2} and < B,x3,yz >e {Qi,Q2}-
As was noted above, case 2 is similar to case 1.
Case 3: xq = y3 = zq = (0, //. 1.0). // ^ 0. Let x\ = (a, b, c, I). x2 =
(a, e, f, 1) which forces y = (b + e)(c+/)_1. Let y\ = (a + m, b + ms, c + msa, 1).
Consider ws = wy, thus y2 = (a + m, e + ms, f + msa, 1), Z\ = (a + n, b + ns, c +
nsa, 1), and z2 = (a + n, e + ns, / + ns, 1). Again let x3 e {.ri. } . y3 e
{di, y2}J- with x3 ~ y3. Xz = (a,b+(b + e)r, c + (c + /)r, 1), r ^ 0,1; y3 =
(a + m,b + ms + (6 + e)t>, c + ms + (c + /)f, 1) for some v.
44


x3 ~ 1/3 implies (1, s+m~l{b+e){r+v), sa+mr1(c+f)(r+v)a, 0) e
which in turn implies that [s + m^l{b + e)(r +1>)] = sa + m~l{c + f)(r + v)a.
Note that if r ^ r then rri = (c + f)y'x-T in which case < X2,yi > rffloo =
1 cx
(1 /y1^ + s/y1^ + s,0) e But this means that x2 ~ y\ and therefore
yi e {xi,x2\l.
The only case needing to be considered is when r = v. But in this case
< yz > nrioo = ws, and hence < x:>. y2j > nn3 = (a + n, b + ns + (b + e)r, c +
nsa + (c + f)ra, 1) e {zi, z2}- Once again, any S line meeting {.rt}
and {|/i, y2} must also meet {z\, z2}- Also note that < A, x?>. y3 > = <
A,xz,ws >G {Pi,P2} and likewise < B,x$,y$ >e {Qi,Q2}-
Finally consider ws ^ wt. Since wt G< xQ,X\,yi > Hi 1 v. then
, 1 a *. 1
wt G< ws, x0 >. Thus wt = (I. k //1 . sn //1 . 0). i.e, t = s + y1- .
1 cx
From this we see thatt/2 = (a + m,e + ms + my, / + msa + my, 1), z\ =
. , . 1 cx ,
(a + n,b + ns, c + nsa, 1), z2 = (a + n,e + ns + ny1-**, j + nsa + ny1-a 1).
As before, let x$ = (a, b + (b + e)r, c + (c + f)ra, 1), r ^ 0,1. If yz is again
chosen to be the point of {t/i, t/2}JJ which is on an S line with x$ then
1 cx
y?> = (a + m,b + ms + (b + e + m,yT^)r, c + msa(c + / + m,yT^)ra, 1) and
1
< £3,2/3 > intersects IIoo at the point wu where u = s + yl-ar.
1 cx
Thus z3 = (a+n,b+ns+(b+e+nyI^)r,c+nsa(c+f+nyI^)ra,l) =<
x3,y3 > n{^2}. Yet again (as has been the refrain) any S line meeting
45


{.ri. and {yi,'tj2} must also meet {zi,z2}- To see that < x^,y^ >
intersects {Pi, P2}, first observe that the point t>3 = (a + (c + fyy^1, (c +
f)y'^-Ts,c+ (c + f)y'^-Tsa, 1) is a point of Pi fi P2, and thus Pi =< A, v3, ws >
,P2 =< A. r3. ir, >. From this we see that {Pi. P2} is the set of planes
through the line < A, t>3 >. Finally observe that t>3 E P3 =< 1. .r3. n\, > and
P3 E {PL,P2}. Similarly, < x3,|/3 > is contained in a plane of {Qi,Q2}-
Hence M is grid-like.
Now that the concept of grid-like is seen to be non-vacuous, the axiom
is used to prove a theorem about the associated affine planes. Return to the
notation of letting S be any GQ(q + 1,5 1) with a fan M. If any ovoid of
M is pivotal then in fact M must have been constructed as in section 3.1. To
see this first note the following lemma.
Lemma 3.5.3 If Cl v is pivotal for M and £0 = {Ti,.. .,Tq} is a parallel class
of the associated affine plane 'k{O0O).j then Tf1 U ... U Tq is an ovoid of M.
Proof: doo is pivotal implies Tf1 C d* for some d* M. Let
x E Tj ^ Ti. Choose y E xL fi d* and choose x' E yL fl doo\{a;}. Suppose
z E {./. PjT-1 fl T\. Then z ~ y implies y is in Tf1 and x is in I\. a contradic-
tion. Therefore {./../'} E t\,. i.e. {./../'} = 7). As Tj1 fl dj is not empty,
Tj- must be contained in dj.
46


With this proved, it follows that given any parallel class of ^(Ooo)
the traces of the lines in the parallel class partition the points of an ovoid of
M. Because the number of ovoids in equals the number of parallel
classes in ^(Ooo), each ovoid of Ad\{(Aoo} can be viewed as the union of traces
of hyperbolic lines in a given parallel class of 71(0^). This proves the following
theorem.
Theorem 3.5.4 Every regular ovoid of S is pivotal for exactly one fan, namely
the fan constructed in the manner of Section 3.1.
We are now in a position to use the Grid-like Axiom to identify iso-
morphic planes.
Theorem 3.5.5 If M. is a grid-like fan, then the affine planes associated with
the pivotal ovoids of M. are all isomorphic.
Proof: Assume that M. is grid-like. Let O^.Oq.Oi be distinct
pivotal ovoids in M. Let £q,£i be parallel classes of 7r((Aoo) such that each
hyperbolic line of £0 has its trace in O0 and each hyperbolic line of £\ has its
trace in 0\. Furthermore let JF0 = {T^jT e £0} and T\ = { R A E £{\. Label
the members of £q by £q = {I\. '/_...., Tq} and let R\, i?2 be distinct members
of £\.
Identify particular points of Clx.. For i = 1,2 let = I) n R,.
47


{.ri. meets each member of £Q and each member of E\. For i = 3............q
let Xi = Ti fl {.ri. } and let Ri be the unique line of E\ containing x%. For
each i = 1,..., q let {A^i,... Aij(?} be the set of lines from S which are incident
with Xi and let = A*j fl Oq, bi;j = A*j fl 0\.
Dehne the map

iai,j) = h,j- The claim is that

morphism from 7t( It remains to be shown that

7,(0,).
Let yi e 7'j //_. E T^,z\ = (f>(yi),Z2 = ^(1/2) Let T be the parallel
class of 7r((90) containing the affine line {t/i,2/2}JJ- For each xi E {.ri.
each line through Xi must meet a unique affine line of T. Let nii be the line
of S through Xi and incident with a point of {yi,'tj2}- Obviously y\,Z\ are
on ni\, and y2, z2 are on m2. By the grid-like axiom, since each ra* contains Xi
and some point of {yi,y2}, it follows that nii must also contain a point of
{z1,z2}.
By the construction if yt = nii fl {yi,y2}, then 4>{y.j) = nii f1
{z1,z2}.

fore

Corollary 3.5.6 If M is a grid-like fan, then either all or none of the affine
48


planes associated with the ovoids of M. are desarguesian.
Revisiting P(T2(0), < A, B >), observe that the affine plane 7t(Oab)
has the following structure. The affine points are q2 planes of Q which contain
A but not B. The affine lines can be viewed as the q2 + q lines of Q which
contain A but not B with incidence inherited from Q.
For each i e FU {oo}, the pencil of lines through A in 11* is a parallel
class. 7t(Oab) completes to a projective plane 7t(Oab) by including the line
Loo =< A, B > in the line set and all of the planes of Q through L0Q in the
point set. Let p be the polarity of Q defined by p : (a, b, c, d) [c, d, a, b]T.
7't(Oab) is a plane of Qp; hence 7t(Oab) is desarguesian.
Corollary 3.5.7 All of the planes associated with the fan M = {Oab, Ob a} U
Va,b of -P(^2(n), < A, B >) are desarguesian.
49


4. Regularity in S
As has been shown, GQ(q + l,q l) with regular ovoids yield GQ(q, q)
with regular points. In this chapter we will consider regularity of points and
lines in any GQ(q + 1,5 1). For q > 3 no known such GQ has a regular point.
In [TvM97] Thas and van Maldeghem give an interesting theorem greatly re-
stricting the existence of regular points. In the first section this theorem is
presented. While regular points are absent, regular pairs of points abound. In
the subsequent sections, these regular pairs and the regular ovoids are exam-
ined.
4.1 A theorem of Thas and van Maldeghem
The only known example of a GQ(q + 1,5 1) with a regular point
is are regular. What follows is an elaboration of a proof first given in [TvM97]
showing that this is in fact the only GQ(q + 1, q 1) with all points regular.
Because the proof is rather lengthy, the main steps in the proof are
outlined first.
(1) Construct a hyperbolic line L and its trace, LL.
50


(2) Construct VL, the set of points not collinear with L or ZA, and consider
some of its subsets.
(3) Partition VL UiUf1 into two ovoids.
(4) Partition these ovoids into hyperbolic lines.
(5) Find a hyperbolic line and its perp in one ovoid.
(6) Observe the contradiction as such an ovoid would then contain collinear
points and conclude that no such GQ(q + 1, q 1), q > 3, exists with
all points regular.
Start with a GQ(q + 1 ,q 1), say S = with all points
regular. Let y,z E V with y and z not collinear. Then {y, z}A- = L has size
q as y is regular. Also ZA = {y, z}L has size q. Since every point of S is on
q lines, each line incident with a point of L is also incident with a point of
ZA. Put another way: each point collinear with a point of L is collinear with
a point of ZA.
Let Vl be the set of points of V not collinear with any point of L.
(Then observe VL = Vl ) Each of the q points of L is on q lines, and each
of these lines has q + l + 1^2 = q points not in LUi1. Other than points
of L U ZA, there are qqq = q3 points which are collinear with points of L. All
together there are \L\ + |ZA| + q3 = 2q + q3 points of V collinear with points of
L. This means |Vx,| = (q + 1 + 1) ((q + 1 )(q 1) + 1) (2q + q3) = 2q(q 1).
51


Now pick x E VL. For each u E L, {.r. a} is a hyperbolic line. As
u ranges over L this gives q hyperbolic lines. For u' E L\{m}, suppose p E
{./. 11} n {./. a'} . Then as x is regular, {./. 11} = {./.p} = {./. a'} =
{u, = L. This is a contradiction as x 0 L. This proves the first lemma:
Lemma 4.1.1 Every pair of hyperbolic lines through x and a point of L has
only x in common.
Let W be the union of the hyperbolic lines through x and points of L,
with the points of L removed: W = U((, /.{./. 11} \L. Then \W\ = q(q^2)+l =
(q l)2. Similarly let W = UueL{x,u}\L, and \W\ = (q l)2. W and
W will later be shown to be contained in ovoids of S.
Lemma 4.1.2 (W U W) C VL.
Proof: Now let v E W, v ^ x. For contradiction, assume there is
some u' E L where v is collinear with u'. Observe that v E W implies there is
some u E L such that v E {.r. w}2-1. Let w be the unique point of vu' collinear
with u. Then the regularity of v (or u) implies that w is collinear with every-
thing in {-iyu}J- = {./. 11} . Specifically tv and x are collinear. But since w
is collinear with two points of L, it is collinear with all points of L, and every
line through w has a point of L. This means x is collinear with a point of L.
This contradicts x E VL. The contradiction came from assuming a point of W
was collinear with a point of L. Therefore W C Vl- Similarly W C Vl-
52


Lemma 4.1.3 W fl W = {./}. and hence |W U W\ = 2(q l)2 1.
Proof: Suppose there is some v' E W fl W with v' ^ x. Then
{./. c'} has a point a E L and a point b E ZA. By definition of ZA,a and b
are collinear. But this is a contradiction as no two points of {.r. r1} can be
collinear. Therefore | W U W\ = (q l)2 + (q l)2 1 = 2(q l)2 1
By construction no point of W is collinear with x. Suppose there were
two collinear points, v and v' E W, distinct from one another (and from x).
Then {r. ./} ^ {r'. ./} . The line vvf has a point v" collinear with x. Let
L fl {r../ } = {m} and let L fl {r1../ } = {'}. Since v" is collinear with v
and x, it is collinear with everything in their span; particularly, v" is collinear
with u. Likewise, v" is collinear with vf. But any point collinear with two
points of L is collinear with all of L and hence is in IT; i.e. v" E LL. This is
a contradiction as x is collinear with v" and x E Vl = VL. Observe then that
no two points of W are collinear. Similarly, no two points of W are collinear;
i.e. W and W are (q l)2-caps of S.
By construction, WC\L is empty, which means |ITUL| = (c/ I )2 f/ =
q2 q + 1. Note that no two points of L can be collinear. Since W C VL, no
point of W is collinear with a point of L. As above no two points of W are
53


collinear, in fact W U L (and similarly W U L) is a (q2 q + l)-cap of S.
Now examine the points of Vl outside of W U W. Let I = {1.2______q\
and let M = {Mj\j El} be the set of lines which contain x. Now hx some
% E I. Assume for contradiction that M, has some point p collinear with two
points of L. Then p is collinear with all points of L and hence p E LL. This
contradicts x E VL = VL. So different points of L are collinear with different
points of Mi.
There are g + 1 + 1 = q + 2 points of Mt and only q points of L.
This leaves exactly 2 points of \l; which are not collinear with anything in
L, i.e. |Mi fl Vl\ = 2. One such point is x\ call the other such point t*. By
construction ti cannot be in W U W, as both W and W are caps containing
x. For each j E 1 construct tj from .1 /( as above. Let '/' = {/, j E 1} and let
X = Vl\(WUWUT) in which case \X\ = 2q(q^ 1) (2(q l)2 1) ^q = q 1.
Again think of i as fixed in I. Let x' be a point of Vl collinear with ti
such that x' ^ x. Then x' cannot be on any Mj E M, else a triangle is formed.
Thus for every tj E T, x' ^ tj. Suppose for contradiction that x' E W and let
{./. fl L = {m}. As ti is collinear with x and ti must be collinear with
u. This contradicts ti E Vl. Hence x' 0 W. Similarly x' 0 W. Therefore x'
must be in X.
It has been shown that on each line through a point of Vl there is
54


a unique other point also in VL; this is how tj was constructed. And observe
that x1- fi Vl = T U {a;}. Now apply this argument to a fixed tj. Through tj
there are q lines all together. One line, Mj, contains x. The remaining q 1
lines, Ml, Mf,..., M?_1, each have a unique point other than tj in VL, say
xj,xf,..., Xj^1, and each of these x\ are in X, k = 1,..., q 1.
This argument could be applied to any of the q different from tj E T to
get x\,..., a;?-1, x\,..., a;!-1,... xlq,... xq~l, all in X. But as was seen earlier,
|AT| = q1. This means that for all j E /, {a *...., a-j 1} = {x\,..., xj 1} = X.
The set X U {a;} is a set of q points, all collinear with everything in T. In other
words, TL = X U {./} and TLL = T. T and X U {./} are then hyperbolic
lines. Writing VL = W U W U T U X almost gives a partition for VL. In fact
Vl = (W\{a;}) U (IT\{a;}) U T U A" U {a;} is a partition.
Lemma 4.1.4 = W U L U X and 0LyX = W U L1 U I are ovoids of S.
Proof: \Ol,x\ = (q l)2 + q + (q 1) = q2 = ((q 1) + l)2. Recall
that X U {a;} is a hyperbolic line and that W U L is a (q2 q + l)-cap. Since
X C Vl, no point of X is collinear with a point of L. It needs only to be shown
that no point of X is collinear with a point of IT\{a;}. Let x' E X, v E IT\{a;},
and suppose x' is collinear with v. Let {u, a;}J~ fl L = {}. Every line through
x' contains a point of T, say tj E vx'\ then tj is collinear with all of {r. a } .
Particularly, tj is collinear with u. This contradicts tj E Vl. Hence Ol,x is an
55


ovoid of S. Likewise O
L>x = W U L1 U I is an ovoid of S.
Let UL = \ /. UlU ZA. Then \UL\ = 2 q(q 1) + q + q = 2 q2, i.e.,
Ul is the size of two disjoint ovoids. In fact if Ol,x = (W U^U T)\{x} =
((9Lj:r\A"U{a;})U(A"U{a;}), then 0LjX is an ovoid having no point in common
with Ol,x- Ul can be partitioned into two disjoint ovoids as Ol^x U Ol,x-
We now proceed to partition the ovoids Ol,x and Ol,x into hyperbolic
lines. Notice that the construction of VL was independent of x, so one could
construct analogues of W,W,T and X using any point of VL. For any point
m E Vl, {mL fl UL)\{m} is a hyperbolic line S. Here S is to m as T is
to x. Also for any point m E L, (mL fl UL)\{m} = LL. Finally for any
m E ZA, {mL fl UL)\{m} = L. Hence for any point m E Ul, {mL fl Ul)\{tti}
is a hyperbolic line. Call such a hyperbolic line the corresponding hyperbolic
line to m.
Lemma 4.1.5 Let m, m! be a pair of distinct points of Ul contained in either
Ol,x or Ol,x- The two hyperbolic lines corresponding to m and m! are either
disjoint or identical.
Proof: Without loss of generality, assume m,m' E Ol,x Let N, N'
be the hyperbolic lines corresponding respectively to m and m'; observe N U
N' C Ol,x- Let n E iV fl N'. If n E ZA, then m,m' E L, which means
56


N = N' = Ll.
Now consider n 0 L1- which means n E Vl- Perform analogous con-
structions for n as were done for x above, in which case VL consists of the
following:
(1) The union of hyperbolic lines containing n and one point of L U ZA
(the analogue of W U W),
(2) The hyperbolic line S = n1- fl Vl (the analogue of T), and
(3) The hyperbolic line SL = (NL fl Vl) (the analogue of X U {a;}).
Since N, N' are hyperbolic lines through n, (N U N') fl S is empty. If N U N'
contains a point of ZA, either m or m! is collinear with a point of LL. But
the only points of UL collinear with points of Lx are points of L\ this gives a
contradiction as N U N' C Ol,x-
This says that N and N' are hyperbolic lines corresponding to points
of UL and containing no point of LU LL. But there is only one such hyperbolic
line, namely SL = N = N'. Thus in any case if n E NON' exists, then
N = N'. This proves the claim.
It has been shown that for any pair of points in one of the partitioning
ovoids, the corresponding hyperbolic lines are contained in the other partioning
ovoid and are either identical or pairwise disjoint. Since each partitioning
57


ovoid has q2 points and each hyperbolic line has q points, each partitioning
ovoid contains ^ = q disjoint hyperbolic lines corresponding to points from
the other partitioning ovoid.
Let L.L\...............Lq i be the q disjoint hyperbolic lines of Ol,x corre-
sponding to the points of Ol,x as formed above. Likewise let L1-, L\.....
be the q disjoint hyperbolic lines of Ol,x- Since the trace of a hyperbolic line
is itself a hyperbolic line, for any j < q there is some k < q with Lf = Lk.
Indices can be chosen such that Lf = Lj for each j < q. Therefore forming
the set £ consisting of the hyperbolic lines {L, ....Lq L Lf,..., Lf_^
is independent of the choice of x E Vl- Choosing some v E Ol,x\{L U {a;}}
would yield the same £. There is an ovoid OljV similarly determined by l and
v.
Lemma 4.1.6 £>,..< = 0,,.r.
Proof: Consider the hyperbolic line {x, First suppose {x,
0 {Li,..., L9_i}. In this case {./. r\ has a point in common with each of
L, Li,... Lq-i. Let u' = {x, Cl L. Ol,x can be viewed as the union of the
q hyperbolic lines in £ which have exactly one point in common with {./. m}j-
where u is any point of L.
Ol,v can now be viewed as the union of hyperbolic lines of £ which
have exactly one point in common with {tyu}-1-1 where u is any point of
58


L. For the specific case u = u' observe {r. a} = {.r. a} So OL,0 =
L U L\ U ... U L(j i = Ol,x- Now suppose {.r. r\ E {/. i..... L(j i}. say
{./. = /.i. Let ti' 6 t2. Then as above Cl/...,. = Cland Cl/..r = OLjVi, i.e.,
Ol,v = Ol,x- In either case, we have for any v E Oi,^\{Lyj{x}}, OL,0 = Ol,x-
Choose indices such that x E L\. Note that as above, OljX is the
union of the q hyperbolic lines from £ which have exactly one point in common
with {./. h}-11- for any u E L. Hence OljX = LL U L\ U L2 U ... U This
can be generalized.
Lemma 4.1.7 If u E L,h Cl/.. . can be partitioned as /, L /., I__/./ L.\,-L
T-L r-L
ni+i,. ., |.
Lemma 4.1.8 q < 3.
Proof: Suppose q > 3 and choose a point w E 0LyX such that
w 0 ZA U L\ U L2j say w E . By lemma 4.1.6, OljX = Ol>w. Let u E
applyingLemma 4.1.7 shows that OljU = ZA U L\ U L2 U U ... and
a- E Ol u. Again by lemma 4.1.6 Ol u = Olw and transitivly, Cl;. .,. =
Olu. This means that L2 C Ol x.
0LyX contains a hyperbolic line, L2, as well as its perp, L2. This is a
contradiction as all points of L2 are collinear with all points of L2 but no two
points of an ovoid, Cl/..may be collinear.
59


The contradiction above arose from the existence of a w E L%. Of
course such a w is available only if q > 3. In [PT84] it is shown that the only
GQ(4, 2) is classical and has all points regular. Hence the main theorem of this
section has now been proved:
Theorem 4.1.9 S is a thick GQ(q + 1, only if q = 3.
4.2 Regular pairs
In this section, pairs of regular points are examined. The goal of the
remainder of this chapter is to show that if a GQ(q + 1, regular ovoids pivotal for the same fan, then every regular ovoid of the GQ is
pivotal for that fan.
Let I be the index set {0,1,..., 5},and 1 = 1 U {00}. Let S be a thick
GQ(q + 1, q 1) with a fan M = {Oi\% E 1} for which O^ is pivotal. Recall
from section 3.3 that constricting about O^ yields a GQ(q,q), S^, and that
Loo is a regular line of Theorem 4.2.1 Suppose that {xi,x2} is a regular pair of points in S with
x\ 9^ Put T = {.r 1. = {.ri...., xq}, and TL = {//1...., yq}. Then
exactly one of the following must occur:
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(i) There are distinct i,j El such that T C Oj and T1- C Oj.
(ii) Exactly one of T, TL has a unique point in common with O^. Without
loss of generality, assume TL fi O^ = {t/i} and T fi O^ = 0. In this case
there is a unique O, E M for which T1-\ {t/i} C Oj. |Tfi Oj\ = 1 for each
J e A-
Proof: Since O^ is pivotal, if either T or TL has two points in
common with (D^, then it is contained in Ooq and its perp is contained in
some Oj. If T fl O^ = 0 = TL fi O^, then by proposition 3.2.2 there must
be distinct i,j El for which T C Oj and TL C Oj. In the remaining case,
exactly one of T, TL has a unique point in common with O^. Without loss of
generality assume that TL fl O^ = {t/i}.
There is some i E I for which TL \ {t/i} C Oj, since otherwise by
proposition 3.2.2 T would have to meet Ooo, clearly an impossibility. Also
1j2,,yq must be in traces of distinct hyperbolic lines of O^, which means
they are on distinct lines (^ L0c) through (Oj) in S^. By proposition 3.2.2
\T fl Oj\ = 1 for each j E
Hence, using the notation above, in Soo, {2/2,..., yq} C {(Oj),x 1,..., Xq}1-
and {2/2, , 2/q}^ = {(Oj),x 1,..., xq}. Let T0 be the line of through (Oj)
61


different from LOQ, and missing {y2,..., yq}. The lines of S through yi to-
gether with Lqo form one ruling of a grid Q. For ease of notation, suppose that
Oi = O0, and Xj E Oj, 1 < j < q. Then the other ruling of Q consists of
y^- Pi Ok = Tk, 0 < k < q, and Xj E Oj fi y^.
Observation 4.2.2 If yi E Ocq and y2 E Oj, and if {yi,y2} is regular in S,
then {yi,y2}L\{yi} C Oj.
Corollary 4.2.3 With the notation adopted above, if for some i E I, Oj is
regular, then it is pivotal for M.
Proof: If Oj is regular for some i E I, and x, y are distinct points of Oj, then
{x,y} is regular and {x^}^ C Oj. This does not permit the second case of
the above theorem to hold, hence the first case must hold.
Now suppose that O is a regular ovoid of S possibly different from
Ooo. Recall that 7r(Oqo) and ir(O) are affine planes whose lines are the
hyperbolic lines contained in them. Clearly 7r(Coo) Cl ir(O) is a subspace of
7r(Coo) and of tt(O). This proves the following lemma.
Lemma 4.2.4 Any regular ovoid of S different from O^ meets O^ in exactly
0, 1 or q points.
The three possibilities for a regular ovoid meeting Oa0 are considered
separately in the next three lemmas.
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Lemma 4.2.5 If O is a regular ovoid with |O fi = q. then O = Ti U
U U T^, where O fi Ox. = l\. and Ox. = Ti U T2 U U Tq is a
parallel class partition of the points of 71(0^). Hence Ok = T-j1 U U for
some k E I.
Proof: Suppose that Cl v = {i.... ,zq, xq+\,..., x2q,....../y< }. where
T\ = {1 , zq}, /_ = {xq+i.,..., x2q}, , Tq = {^(q-pq+i,..., x^i} are the
disjoint hyperbolic lines of one parallel class of ,k{0oq), and O = {z\,..., zq, yq+1
,... ,yq2}, with O fl Ooo = T\. Let y E O fi Ok for some k E I. Then for each
Zi e O, C Ok by theorem 4.2.1, so Uf=1 ({zi,y}L1- \ {zij'j gives
I 2) = Since 2(q2 2q + 1) > q2 q for q > 3, 0\ O^ must be contained in
just one Ok, i.e., \OnOk\ = q2^q, and hence OC\Ok = {yq+i,.. .,yq2}. Clearly
no yj is collinear with any z%. From this, yf fl Ooq = {a;(g+i),... ,a;(+i)g} is
disjoint from Ti. thus the parallel class of containing Tx must contain q 1
hyperbolic lines T2,... ,Tq whose traces cover OC\Ok. So Of'Ok = 74 U- / :
O was constructed to contain T\. Finally Ok E M implies Ok consists of the
traces of the hyperbolic lines contained in some specific parallel class of 77(0^).
Hence:
63


ooo= L. u rq.
Ok= U U Tf,
0= T\ U U U Tf.
This completes the proof of the theorem.
Corollary 4.2.6 Continuing with the notation of the above proof, let X\ E Ti,
x-2 E T2^, {.ri. = T. Then for j E /\{/>}. |T1- C\Oj\ = 1.
Proof: By hypothesis {.r C O. If two points of {.r
belong to the same Tj~, then of course {.r i. would have to equal Tj~.
Hence T = {./-i..has one point in each of I\. 7h........Tq Now let
{.rj. = {wi,..., wq} = TC Clearly T1- n C\. = T1- n Ok = 0, since
both Cqo and Ok are ovoids. If two points of TL belong to the same Oj, then
since T n C v ^ 0, it would have to be that T C O^. Hence |TL fl Cj| = 1 for
each j E I\{k}. u
Theorem 4.2.7 If O is a regular ovoid with |CnCoo| = 1, then \OnOi \ = q 1
for 0 < i < q. If CflCoo = {z} and x E OflOi, then {.x} \ {} = CflCV
Proof: Suppose O fl = {z}. From the observation following
64


theorem 4.2.1 we see that if x E 0 \ 0V. say x E Oj. then {z, \ {} C 0,-.
First suppose that {z, x}J- \ {z} COD Oj, {z,y}J- \ {z} COD Oj, and
2: 0 {./. //} . hence {.r. //} fl 0V = 0. Say:
{z}J"L = {z, x = x2, , xq}\ {z, y\ = {z, y = y2,..., yq}.
Since {x.yj^nO^ = 0, if {x,y}Ci000 = 0 also, then {x.,y}L C Oj
for some j E and {x, y}^ C Oj. On the other hand, if {x, y}Ci000 ^ 0,
then {x,y} C Oj implies {x.,y}L C Oa0 and {x^}^ C Oj. In any case the
following holds : For x,y E O fl Oj, {x/y}^ C (On Oj) U {z}. Thus
(0 n 0j) U {2:} is a subspace of 7r(0). Hence if |0 n Oj| ^ 5, then O
(0n0,)u{2}.
Suppose Oj = Tf1 U U T^, where 0X. = '/'1 L L 7r Say
Ti = {zi,..., zq}., and zi is the point of Oj for which (0^(21 })U{2:} = 0. Then
the lines of S covered by zi are the same lines as those covered by z, an obvious
impossibility as {z, zi} determines at most one line. Hence it must be that for
each % E I, |(0n0j)| = q 1, and if x E OllOj, then (z, a;}J-\{2:} = 0n0j.
Theorem 4.2.8 Let 0 be a regular ovoid with 0 fl 0^1 = 0. Then exactly
one of the following two possibilities must hold:
65


(i) O is some O, £ M. in which case
Oi is pivotal for the fan M:
(ii) There is a unique O^i £ 7, for which
\On Ooo\ = \On Oi\ = 0, and \OnOj\ = q
for all j £ 7\{i}. In this case O U {( is an ovoid of S^.
Proof: Suppose O fi = 0. Let x,y be distinct points of O, thus
T = {./. //} C O, and T n Cl v = 0. If T1- fi C> v = 0, then 7' C O, for some
% e I, and T1- C Ok for some k £ 7\{7}. If T^fl ^ 0, then either T1- C Oa0
and T C Ojt for some i, or |T1- fi O^] = 1 and T1- \ O^ C Ot for some % £ I
while |Tn Oj\ = 1 for each j £ 7\{i}. In all cases, if x,y are distinct points
of O fl Oi for some i, then {./. y}^ C O O Oi. Hence O fi Oit is a subspace of
7r(O), and must have size 0,1, q or q2. One possibility is that O = Ot £ M.
Suppose that O 0 M; \0 fl Oi\ = 0,1 or q, for each i £ I. Let an be
the number of Oj with |dn Oj\ = n £ {0,1, q}, for j £ I. Clearly each an > 0,
and
(i) (1q + Oi + (Iq = q + 1,
(ii) 0 aQ + 1 a,i + q aq = q2,
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from which ai = q(q aq). Put m = q aq, so oq = mq. It follows that
q m = aq = q + 1 clq a\, implying oq + o,q = m + 1 = o,q + mq, or
k(q l) = 1 a0, which must be nonnegative. This allows only two possibilities:
aQ = 1 or 0.
If ao = 1, then m = 0 = 00o and from one other member of M. say Oi, and meets each of the others
Oi,... ,Oq in q points, i.e. in an affine line of 7r(O). It is straightforward to
check that O U {(C^)} is an ovoid of S^. For the other case, suppose a0 = 0.
Here m = 1 and q = 2. But as only thick GQ are under consideration, q > 2.
So in fact this case not occur.
Suppose now that in addition to having O^ pivotal for the fan .Vf.
the ovoid Oq is regular and hence also pivotal for M.
Theorem 4.2.9 Suppose {x,y} is a regular pair of points with x ^ y. Put
T = {./. //} . T1- = {./. //} Then there must exist distinct i. j e I for which
T C Oi, T 6 Oj.
Proof: Suppose not. Then one of TL, T has a unique point in
common with say |TL n Ooo\ = 1. Then there is some i E I for which
|TL fi Oj] = q 1. Clearly Oi is not regular, so i ^ 0. Then \T fl Oj\ = 1 for
each j E 7\{}, including j = 0. But \T fl Oq\ = 1 says there must be some
67


k £ I\{0} with \T fl Ok\ = q 1, an impossibility for q > 2.
This leads to the main theorem of this section.
Theorem 4.2.10 Let O^ and O0 be distinct pivotal ovoids of M; if O is a
regular ovoid of S, then in fact O is a pivotal member of M.
Proof: Suppose O is a regular ovoid with O 0 M. Then \OC\Ooo \ =
0,1 or q, and \0 fl Oq\ = 0,1, or q.
Case (i) \0 fl Oac| = q. In this case there is a k £ I with \0 O Ok\ =
q2 q\ k ^ 0 and \0 fl O0\ = 0. But theorem 4.2.8 applied to O0 in place of
Ooo says there is some j £ 7\{0} with \0 fl Oj\ = 0 and |O fl Om\ = q for
m £ Hence this case cannot arise.
Case (ii) \0 fl \ = 1. Here \0 fl 0,-\ = q 1 for 0 < i < q. But as
\0 fl Oq\ ^ q 1 if q >2, this case cannot arise.
Case (iii) |O fl = 0. Then also it follows that \0 fl Oq\ = 0.
If O M then \0 Pi Oj\ = q for all j £ 7\{0}. But suppose x £ O fl Oj,
y £ O fl Ok, with j and k distinct members of 7\{0}. Then {x^}^ does not
belong to a single member of M. contradicting theorem 4.2.9. Hence this case
does not arise, completing the proof.
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5. Characterizations
In [ST84], De Soete and Thas gave characterization axioms for GQ(q
1,5 + 1) (with no restriction on q). This inspired Payne to formulate axioms
Ai A7 characterizing the known GQ(q + 1,5 1). Namely the GQ satisfying
Ai A7 are exactly the known GQ(q +1, q 1) arising from a 5-arc as described
in section 2.4. De Soete and Thas then revised their approach [ST86] and
consequently Paynes A7 was made redundant. In [PM98] a relationship was
discovered between Paynes axioms and conditions on admissible permutations
giving plane amalgamations. In this chapter, these axiom systems are reviewed.
5.1 (0,2)-sets
The motivation for this and the next section is the desire to embed
an abstract GQ into affine or projective space. One of the main hurdles in
doing this is determining how to represent sets of non-collinear points of the
GQ as affine/projective lines. If the points in question form regular pairs,
then the hyperbolic line they determine may suffice. However as seen in the
previous chapter, not all pairs will be regular; hence a new notion is needed: an
abstraction of a hyperbolic line. De Soete and Thas give one such notion with
69


their interesting concept of a (0, 2)-set. What follows here is a compilation of
some of their ideas from [ST84] and [ST86].
If S = (V, B.X) is a GQ(s, t), then a nonempty subset K of V is said
to be a (0, 2)-set of S provided the points of K are pairwise noncollinear and
lx1- fl K\ E {0, 2} for all x E V\K. Observe that in the GQ T2*(fi+), if L is a
line of Q which meets in a point outside of fi+ then the points of
form a (0, 2)-set.
Lemma 5.1.1 If K is a (0,2)-set of S then \K\ = s + 1.
Proof: Let x E K, and let L be a line through x. For x ^ k E K, k
is collinear with a unique point p of L. As K is a (0, 2)-set, p is collinear with
no point of K\{x,k}. As there is a bijection between points of K and points
of L, \K\ = \L\ = s + 1
Observe that if L is a line which completely misses K then each point
of K is collinear with some point p of L. Since K is a (0, 2)-set, p is collinear
with exactly one other point of K. Thus as the points of K arise in disjoint
pairs, s + 1 must be even. This gives the following lemma.
Lemma 5.1.2 If K is a (0, 2)-set of S with t > 1 then s is odd. (Having t > 1
allows the existence of such an L mentioned above).
The first axiom used ensures a collection of (0, 2)-sets partitioning the
70


noncollinear point pairs in S:
Basic Axiom: S contains a collection B\ of (0, 2)-sets such that each
pair of noncollinear points of S is in a unique member of B\.
Observation 5.1.3 Let V = V. B' = B U B\. S' = (V, B', e) is a 2 ((s +
1) (st + 1), s +1,1) design with s odd. In the case where s = q 1 and t = q +1
observe that such a design has the parameters of affine 3-space (and that q is
even). Members of B\ will be referred to as blocks.
The goal of the remaining part of this section is to see that S' really
is affine 3-space. By the Basic Axiom if x E V and Lei? with x 0 L then
there exists a subset of s (0, 2)-sets in i?i, say CXjl = I-\_L-- containing x
and a point of L. The next axiom relates lines skew to L with these (0, 2)-sets
(Observe that if V is a line skew to L and meeting each member of £, then
B'x.L £x,L')-
Axiom T: Let x 0 M E B with M Cl L = 0. If there are two distinct
(0, 2)-sets in CXjL which are disjoint from M then for every Lt E Cx,l, is
disjoint from M.
Let i,j be distinct indices in {1,... s} with = Lj n L, yj = Lj fi L.
As Lj is a (0, 2)-set, yj- contains yj and some other point (not on L) y'- of Lj.
Likewise as Lt is a (0,2)-set, y'jL contains and some other point y\ of Lj.
71


Let M be the line of S determined by yj and yj and observe that LP M = 0,
otherwise a triangle is formed. This proves the following:
Lemma 5.1.4 For x E V, L E B with x 0 L and for each pair of distinct
indices i,j e {1,..., s} there is a line M E B which misses L and which meets
both Li and Lj.
The next lemma can be used to weaken the hypothesis of Axiom T.
Unless stated otherwise, for 1 < i < s, yt is defined as the intersection of L
with Li.
Lemma 5.1.5 If M E B and if at least two different i and j give M fi Lt ^
i ^ M fl Lj then for each h E {1,..., s}, M P Lh ^ 0.
Proof: If M misses L, then the conclusion follows from Axiom T.
Consider the case where M and L have a common point, say yj E Lj. Suppose
M misses some (0, 2)-set Lr E C. Let yr = Lr P L. Since yj is collinear with
yr, Lr must have some other point y'r in yj-. Likewise y'r must be collinear with
some other point yj of Lj. Let L' be the line of S determined by yj and yj. By
Axiom T, V must meet every member of £. Observing that L' PM = 0 and
applying Axiom T to V forces M to meet every member of £. This contradicts
M fl Lr = 0. Thus M must in fact meet all of the members of £.
With the above lemma proved, it makes sense to replace Axiom T
72


with the above lemma taken as an axiom referred to as Axiom T*. Now
observe that this axiom forces regularity of lines.
Lemma 5.1.6 If M and L are as in Axiom then they form a regular pair.
Proof: If M meets L then by definition they form a regular pair.
Assume M misses L. For distinct k,r E {l,...,s} there is a unique point
Vkr = vt n Lr, Vkr ^ Vr- By Axiom T* for a hxed k, {ykr\r ^ k} U {yk} is
a set of s mutually collinear points. Call the line they determine Nk. In this
way there arise s lines, Ni,..., Ns each of which have a point in common with
each Lir The lines Ni,...Ns then completely cover the points of the various
Li, and the points of M are likewise covered by the lines Ni.
Let z = LCi xL\ for 1 < i < s + 1 let = Nt fl M where As+i is the
line of S determined by x and To see that As+i really does meet M, observe
that for 1 < i < s, zit cannot be collinear with z (as z% is already collinear with
the point yi on L). x and z must both be collinear with the unique point of
M which is not a z% for 1 < % < s. This unique point is exactly the point zs+1.
For Nk E {L, .1/} different from As+i observe that ykr = NkCiLr, k ^
r, and yk = Nk fl Lk fl L. As each Lk is a (0, 2)-set, there exists some y'kr E Lk
different from yk such that ykr is collinear with y'kr. By Axiom /'. the line
determined by ykr and y'kr has one point in common with each of the
For a hxed k as r varies over this gives s 1 disjoint lines Kr.
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Let K. denote this set of lines.
Note that as these lines together with L form a set of s disjoint lines
which cover the s2 points of the various Lt, M must be in 1C. It follows that
K. U {L} form one ruling of an (s x s) grid G with the other ruling given by
Ni,... ,NS. As L and M both meet Ns+i, each line of K must also meet Ns+1.
Let tv be the unique point of Ni collinear with x (i.e. W is the only
point of N\ not on some L*. Let K be the line of S determined by x and tv,
then in fact K contains the unique point on each Nk which is not on some
Ljr Observe now that iVs+i and K extend G to an (s + 1 x s + 1) grid
G(x, L) = {L, .1/} U {L, M}J- with {L, M}J- = JC U {L, K} (where M is in
fact one of the s 1 lines of KC). Hence the pair {L, M} is regular.
With this proved, now an affine plane can be constructed.
Lemma 5.1.7 In the notation of the previous lemma, let V* = {y E V\y is on
a line of {L, .1/} } and let B* = G(x, L)\JB\ where B\ consists of all (0, 2)-sets
of B\ which contain at least two points of V*. The incidence structure given
by (V*, B*, e) is a 2 ((s + l)2, (s + 1), 1) design, i.e. an affine plane of order
s + 1 with s odd.
Proof: It is straightforward to see that \P*\ = (s + l)2 and that
each block of B* contains s + 1 points. It remains to be seen that any two
74


points V* are contained in a unique block of B*.
Let u, v £ V*. If u is collinear with v in in S is precisely the block of B* that contains them. Suppose now that u and
v are not collinear in which contains u and v. This (0, 2)-set is then necessarily in B\. What follows
demonstrates that this (0, 2)-set has all of its points contained in V*.
Without loss of generality say u 0 A. If either u or v is x, the block
containing u and v is obvious. Assume v. In what follows, we see that
the grids G(x, L),G(u, A), and G(v,L) are all the same.
Case 1: u and x are not collinear in S. Let Lk be the (0,2)-set of
£>i containing both u and x; observe that the points of Lk are in both grids
G(x, A) and G(u, A). Let Nr be the line through u meeting L and M; then Nr
is a member of both grids G(x, L) and G(u, L). Let y'r be a point of Nr distinct
from u and yr. As y'r is collinear with the point u of Lk, it is also collinear with
some other point u' of Lk. This puts L, Nr, y'r, and u' in both grids. Hence the
two grids G(x,L) and G(u,L) are actually identical.
Case 2: u and x lie on the line K. Pick w E G(x, L)\{L} such that
tv is not collinear with x nor with u. As seen above G(x, L) = G(w, A); hence
u E G(w, A). i
75


Observation 5.1.8 The affine plane above is a substructure of the design S']
for any non-incident pair (x,L) Furthermore, the following corollary arises.
Corollary 5.1.9 If x E V, L E £>i, x 0 L with \xL fl L\ = 2 then the substruc-
ture of S' generated by x and L is an affine plane of order s + 1.
Proof: Let xL fl L = {2/1,2/2} and let L' be the line of S through
x and 1/2. The substructure of S' generated by y\ and L' is contained in the
substructure of S' generated by x and L\ and vice-versa. But the first of these
is an affine plane of order s + 1 as indicated above.
The term Type 1 will be used to describe the affine planes of order
s + 1 as constructed in the preceding lemma, observation, and corollary. Each
point in such a plane is incident with two affine lines coming from V and s
affine lines coming from B\. Moreover for a non-incident point-line pair (x, L')
in such a plane, if L' is a block, then x is collinear in S with two points of L'.
Lemma 5.1.10 Let a be a Type 1 affine plane in S' whose point set is denoted
Va with a the projective completion of a. If u E V'\Pa and uL fl Va =
{x\,... xs+i} and if x' and x" are the two projective points of a defined by the
two parallel classes of lines from B in a, then the set O = {^i,... xs+i,x', x}
is a hyper oval in a
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Proof: First note that as x' and x" are both off of a, no line of a
through x' and x" can also contain any of the x-i. Let R be a line of a. If R E B
then there is a unique Xi on R collinear with u; furthermore exactly one of x'
and x" is incident with R in a. On the other hand if R E i?i then as R is a
(0, 2)-set, |uL fl R\ E {0,2}, i.e. R meets 0\{x',x"} in 0 or 2 points; further-
more as R is in not a line of B, its projective completion contains neither x'
nor x". Thus every line of a meets O in 0 or 2 points.
Up to this point it has been assumed that S is a GQ(s, t) with the only
restriction being s is odd. For the remainder of this section, further assume
that t = s + 2. For L, M E B, L and M will be said to be Type 1 parallel,
written L || M provided there is some Type 1 affine plane a in which L and
M are parallel, written L ||a M.
Lemma 5.1.11 For any point x and any Lei? there is a unique M E V such
that x E M and L || M.
Proof: Recall that \P\ = (1 + s)(l + s(s + 2)) = (1 + s)3; this gives
(1 + s)(s(s + 2)) points of V not on L. Each Type 1 affine plane containing
L has s(s + 1) points not on L. Any two of these planes meet exactly at L.
This gives = s + 2 Type 1 affine planes through L. In each of these
planes, there are s lines other than L in a parallel class with L. Including L
77


itself, there are 1 + s(s + 2) = (1 + s)2 lines M in V such that M ||a L for some
a. As these lines are disjoint, they cover all (1 + s)3 points of V exactly once.
Lemma 5.1.12 The parallelism given by || is in fact an equivalence relation
on the lines of B.
Proof: That || is both reflexive and symmetric is straightforward.
To see that || is transitive requires effort. Let a, fi be two affine planes of Type
1 and let L, M, N e B such that L ||a M and M ||/j N. If L, M, N, a, fi are not
all distinct, transitivity follows immediately. Assume instead they are in fact
all distinct. What is now needed is the demonstration of a Type 1 affine plane
7 in which L ||7 M.
Let x N, and let 7 be the Type 1 affine plane generated by x and L.
Let N' be the unique line 7 through x which is parallel (in 7) to L as ensured
by the previous lemma. If N = N', transitivity is demonstrated.
Assume for contradiction that N ^ N', and let K be a line of a in
{L, M jW As was demonstrated above, xL fi a is an (s + l)-arc. Let R be a line
through x meeting a at a point y. In a, y is incident with just one line parallel
to K, say K'. If R || K, this would give two lines of B through y parallel
to K, a contradiction. Hence none of the s + 1 lines joining x to a point of
a are parallel to K. As K contains a point of L and as L is the unique line
78


through that point and parallel to Nr, N' cannot be parallel to K. Similarly
as K contains a point of M. N cannot be parallel to K. This gives s + 3 lines
through x, none of which are parallel to K, contradicting the previous lemma.
Hence N = N'; i.e. L || N.
As S is a GQ(s, s + 2), there are s + 3 parallel classes with (s + l)2
lines in each parallel class. Let M = {7Z,7Z\,... ,77s+2} be the collection of
parallel classes. Observe that M is a packing of spreads for S. To see that
in fact 1Z is pivotal for this packing note the following. If L, M E 1Z then by
the definition of parallel class, {L,M} is a regular pair and {/.. M\ C 1Z.
Furthermore {L, .1/} is a set of mutually parallel lines; i.e. {L, .1/} C 7Zt for
some 6 {l,...s + 2}. Now constrict S about 1Z to get a new GQ(s +1, s +1),
say S* with a regular point (oo) such that S = P( S*, (oo) ). Let x be a point
of S not in a given Type 1 affine plane a; then ja;2- fl a\ = s + 1. Note that x
is on some line L of 7Z\] likewise x is on some line M of 7Z. In S the lines L
and M generate a Type 1 affine plane (3.
With this parallelism of lines shown to be an equivalence relation, a
similar relation can be defined on Type 1 affine planes. Two Type 1 affine
planes a, fl are said to be parallel provided there exist distinct concurrent
affine lines L, M in a and distinct concurrent affine lines L', M' in fl with
79


L || V and M || M'.
The symbol || is again used to denote this parallelism as well as paral-
lelism of lines. The context should make clear which sense is intended. Let a, (3
be parallel Type 1 affine planes with L, L', M, M' as in the above definition.
a and (3 have no common points. If they shared some point x, then L
would be parallel to some .Y in n through x and L' would be parallel to some
N' through x. Hence N || N' with x E N fl N', a contradiction.
If J, K are distinct concurrent lines in a then there exist a pair of
distinct concurrent lines J',K' in (3 such that ,J || J', K || K'. To see this let
y E (3 and consider the Type 1 affine plane generated by y and J and that
generated by y and K\ let J', K' be the respective intersections of these planes
with (3.
The relation || is an equivalence relation on Type 1 affine planes in
S. Reflexivity and symmetry are immediate. Transitivity follows from the
previous paragraph.
The Type 1 affine planes in a parallel class partition the points of S.
Hence each parallel class consists of = s + 1 planes.
Let (\. i be distinct parallel planes with x a point of (3. Of the s + 3
lines (in S) through x, s+1 meet lines of a. This gives (l+s)2^(l+s) = s(s+l)
points of a which are coincident with x from blocks of £>i, i.e. (0,2)-sets, no
80


two of which have points other than x in common.
If N is a line of (3 which misses x, there are s points of N not collinear
in S with x. These s points determine s (0, 2)-sets with x. This accounts for
s(s + 1) + s = s2 + 2s (0, 2)-sets through x, i.e. this accounts for all (0, 2)-sets
through x. To see that each point is on s2 + 2s (0, 2)-sets, recall that S' is a
2 ((s + l)3, s +1,1) design, and hence each point is on s2 + 3s + 3 blocks. Of
these, s + 3 are lines of B, and the remaining s2 + 2s are (0, 2)-sets.
Now let a, (3 be non-parallel affine planes of Type 1 with T = {a =
Q!i, Q!2, , Q!s+i} the parallel class containing a. For e T, any points of
l3na,i must all lie in a common block of £>': For suppose x, y, z are three points
of (3Dai not contained in a block of £>'. The blocks [xy] and [xz] can be thought
of as affine lines in both affine planes having a common point, forcing || (3,
a contradiction. This means that (3 O has at most s + 1 points. On the
other hand as T partitions V and as (3 has (s + l)2 points. Each must have
exactly s + 1 points of (3. Thus 1 P n, f B'. and in fact a and (3 must meet in
a common block of £>'.
The contrapositive of this is stated as the following observation.
Observation 5.1.13 If two Type 1 affine planes in S do not have a common
block then they are parallel.
Let a' E T\{n} with L = (3 fl a and L* = (3 f1 o'. As L and L* are
81


disjoint lines in (3, they are parallel and hence are either both in B or both in
B\. Further let (3' be a Type 1 affine plane parallel to (3 with V = a' fl (3'. By
the previous argument L* and V are parallel lines from either B or B\. This
forces L and V to be parallel lines from either B or B\.
Lemma 5.1.14 Let 7 be a Type 1 affine plane distinct from a and (3 which
contains L. V is in some plane 7' parallel to 7.
Proof: Let x E L' and let 7' be the unique plane through x parallel
to 7. Put M = 7' fl a and N = 7' fl (3. This forces L \\@ N and L ||a M. If
7 = 7' then M = L = N. If 7 ^ 7' suppose M fl N = Because z E a and
z E f3, z must be a point of L, but L ||a M (and L \\p N), a contradiction.
M fl N is then empty, and hence M ||y N. Let M' = 7' fl a', N' = 7' fl (3',
in which case M ||y Mr,N ||y N'. Thus M' ||y Nr,x e M',x e N' forces
M' = N' C a' fl (3' = //. Hence // is in 7'.
At this point define a parallelism on B\ as follows. For L, M e B\, L
and M are said to be parallel (again written L || M) provided there exist Type
1 affine planes a, a(3, (3' with a || o'. (3 || (3' such that a fl (3 = L, a' fl (3' = M.
Lemma 5.1.15 The parallelism given by || is an equivalence relation on £>,
Proof: Reflexivity follows by letting a = a', (3 = (3'; symmetry is
immediate. For transitivity let L,M,N E B\ with L || M. M || N in which
82


case there are Type 1 affine planes a, a', (3, (3r, tt, tt!, v, v' with a || a', [3 || (3r, tt ||
-,t'. /' || u such that a fl (3 = L, a' fl (3' = M = ir fi z/, N = ir' fi u'. Letting 7r, /'. a1
play the respective roles of a, (3, 7 in the previous lemma forces the existence
of some Type 1 affine plane a" which is parallel to a' and which contains N.
Likewise, letting (3' play the role of 7 forces the existence of some Type 1 affine
plane 3" which is parallel to (3' and which contains N. In fact N = a" f1 (3" and
by the transitivity of || on Type 1 affine planes a || a", (3 || (3". Hence L || N.
Corollary 5.1.16 Each parallel class of £7 partitions the points of V.
Proof: Let L e B\,x E V, and let a, (3 be distinct Type 1 affine
planes which meet at L. Furthermore let ar, (3' be distinct Type 1 affine planes
containing x such that a || o', i || (3'. Put // = n' P i'\ so ./ 7 // and L || Lf.
Suppose there is some L" e B\ which contains x and which is parallel to L.
This forces the existence of planes 7 and 8 through L" with 7 || a, 8 || (3. But
as classes of parallel planes partition P', 7 = a', 8 = (3'. Hence L" = 13: i.e.
there is exactly one member of B\ parallel to L which contains x (namely L').
Observe that if x 7 /. then L = V.
Corollary 5.1.17 If two members of B\ are parallel as lines in a Type 1 affine
plane then they are parallel under || as defined above.
83


Proof: Let L, M E £7 with L ||a M for some Type 1 affine plane a.
Let /? be some other Type 1 affine plane through L, let y be a point of M, let
(Â¥ be the Type 1 affine plane through y and parallel to /3, and let M' be the
common block on a and f3'. Observe then that M' || L, and hence M' ||a L.
This forces M' = M. and therefore L || M. m
The following lemmas and corollaries relate parallel blocks and planes.
Lemma 5.1.18 Let a be a Type 1 affine plane which does not contain the
block L. a and L have no common point if and only if L belongs to one of the
parallel classes defined by the blocks of a.
Proof: First assume that L is parallel to some block M E a. If
L,M E B then they are parallel in an affine plane which meets a exactly at
M and hence L has no point of a. If L, M E B\ then there are type 1 affine
planes 7,7', <)<)' with 7 || 7', S || S', L = 7 fi S, M = 7' fi S'. Suppose L met a in
a point x. This would force 7 to meet a in a block N through ./ If .Y met M
then 7 could not be parallel to 7', but N ||a M implies N || M which implies
N || L, a contradiction. Hence L must miss a.
On the other hand, assume now that a and L have no common points.
Then L is in a parallel class containing a block of a as follows. If L E B choose
x to be any point of a. If L E B\, choose x to be a point of a collinear in S
84


with two points of L. Let (3 be the Type 1 affine plane generated by L and x;
let M = a fl (3. Because L misses a, L \\p M, and hence L || M. m
Corollary 5.1.19 Parallel affine planes of Type 1 define the same s+2 parallel
classes of blocks.
Proof: This is immediate from the lemma.
Corollary 5.1.20 If a and (3 are Type 1 affine planes such that two distinct
intersecting blocks, L,M, of a are parallel respectively to two distinct inter-
secting blocks, of (3 then a || (3.
Proof: Suppose a and (3 are not parallel and hence they have a
common block N. If \LCiN\ = 1, then \L'CiN\ = 1. As L || //. there is a Type
1 affine plane 7 containing L, L', and N. But this is a contradiction as L and
N determine the Type 1 affine plane a. Hence L ||a N. This forces M to meet
N in exactly one point. A similar contradiction arises. Thus a and (3 cannot
have a common block; i.e. a || (3.
For the remainder of this section assume s > 3.
Lemma 5.1.21 Let y\Z\U\ and be two triangles of a Type 1 affine
plane a which are perspective from the point tv with w, yi, z\, u\, y^, Z2, and
85


distinct points and y\Z\, y\Ui, Z\U\, t/22; z2u2, wyi, wz\, and wu\ distinct
lines. If yiZi ||a 1/2^2 £ B and z\Ui ||a Z2U2 £ B then u\y\ ||a u^yi-
Proof: Note that since members of B are lines of S no three of them
can form a triangle; i.e. it/i,2?/2 £ B\. Let L £ B be a line meeting a at the
point w and construct a Type 1 affine plane f}\ from uyyi which meets L in
some point w\ different from w.
Count the number of different planes which could be constructed in
this manner: For pi £ uyyi there is a unique point w\ of L which is collinear
in S with pi. As upyi is a (0,2)-set, there is a unique other point p2 £ upyi
which is also collinear with w\. liw is collinear in S with both u\ and yi, this
leaves pairs of points in upyi which are collinear with a point of L\tv, if
w is not collinear in S with both u\ and yi, this leaves ^ pairs of points in
u\yi which are collinear with a point of L\w. This means there are at least ^
different f}\ which can be constructed in this manner. Similarly there are at
least ways to construct a Type 1 affine plane /32 through u2y2 which meets
L\{w,wi} in a point W2.
If Pi || P2 then upyi || u^y^- Assume f}\ and /32 are not parallel.
Suppose for contradiction that W2Z2 || w\Z\, this would force %02^21/2 || w\Ziyi
and W2Z2U2 || tjj\Z\U\. It has been shown that when two parallel planes are
intersected with a third plane, the resulting lines of intersection are parallel.
86


Specifically, (Lyi fi Wizyyi) || (Lyi fi w2Z2y2), i.e. y\Wi || y2w2- Similarly,
U\W\ || U2W2 By the previous corollary, f}\ || (h, a contradiction. Hence w2z2
and W\Z\ cannot be parallel. As %v,zi,z2 are all collinear in the plane a, this
argument shows also that Wit/i [\ w2y2 and w\U\ [fw2u2 because either of these
parallelism would imply w\Z\ || w2z2.
Label these points of intersection: z3 = W\Z\ fl w2z2,U3 = W\U\ fl
W2U2,y3 = w\yi fl W2I/2. Recall that by hypothesis u\Z\ || U2Z2 and observe
that u?jZ% is the line of intersection of the planes u\Z\Wi and U2Z2W2 From this
u3z3f)uizi = 0 = u3z3nu2z2; i.e. u3z3 ||1Z1TO1 uxzi and u3z3 \\U2Z2W2 u2z2. Thus
U1Z1 || ^32:3 and similarly y\Z\ || y^z?,-
This forces u?Jz?J and y3z3 to be members of £>, and hence by the pre-
vious corollary a || u3z3y3. Finally observe that the plane UiyiWi hits the
parallel planes a and uzz?/y?j in the respective parallel lines y\U\ and y?;uz, and
likewise u2y2W2 hits the parallel planes a and u^z^y^ in the respective parallel
lines y2u2 and y$u$. By transitivity y\U\ || y2u2-
One final lemma is needed before the main theorem of this section
will emerge.
Lemma 5.1.22 Let a and /? be distinct parallel Type 1 affine planes. Let u
be a point not in a or /3, and let L1, L2 be distinct blocks through u which do
87


not belong to to the parallel classes of blocks defined by a (or (3). If yt is the
common point of a and Li, and if z% be the common point of a and Lt, i = 1,2,
then |/i 1/2 || zi z2.
Proof: Since a || /3, y\y2 fl did2 = 0- If yiV2 £ £> then the points
u,y\,y2,z\,z2 all belong to a Type 1 affine plane, in which case y\y2 || dis-
similarly Z\Z2 G B implies yyy2 || Z\Z2. Assume that neither yyy2 nor Z\Z2
are in B. For i = 1,2 let M, 6 B be a line through yi such that M\ meets
M2 at some point w. Let w' be the point of (3 on uw, and observe that
M\ a fl (3 = Z\w', M2u fl (3 = z2w'. As a || (3, Mi ||Ml z\w' and M2 ||Miu z2w'.
Since Mi, M2 G B it follows that z\w', z2w' E B.
Now choose a line L G B through u which meets both a and (3 but
misses both z\w' and z2w'. If L meets Z\Z2 then the Type 1 affine plane
containing L and z\z2 must meet a at exactly the line y\y2 (as a || /3); in this
case yiy2 || Z\Z2 and the proof is completed. So now assume that L misses Z\Z2
and let u' = L fl (3. As shown with (32 in the previous lemma, there are ^
planes different from a which contain y\y2 and a point of L\{h'}. Let 7 be
such a plane.
If u G 7 then as above u,y\,y2,z\,z2 are all coplanar (here in 7) and
hence yyy2 || Z\Z2, so further assume u is not in 7 and let u" = L fl 7. Let
vi = yiu" fl (3, v2 = y2u" fl (3, w" = wu" fl (3. To see that such points actually


exists use lemma 5.1.18 applied to (3 and observe that the parallel classes of (3
are the same as those of a. The plane u"wyi contains the lines Mi of a and
V\w" of /?; hence M\ || v\w" and likewise .1 /_. || v2w", forcing v\w", v2w" e B.
By transitivity observe that zyM || V\w" and z2w' || v2w".
iji G Z\U implies y\ e Lz\ which implies y\u" C Lz\ and hence v\ e
Lzi. Clearly u' and z\ are in Lz\. But as these three points (iq, z\, and u')
are also in (3, they must lie in a common block. Similarly v2,z2,u', are on a
common block, as are w",w',u'.
The two triangles formed by vi,v2,w" and zi,z2,u/ are perspective
from the point u'. Since z\w' || v\w" and z2w' || v2w", lemma 5.1.21 indicates
that Z\z2 || V\V2. As 7 meets a and (3 in parallel lines, yyy2 || V\V2. Therefore
ym || Z1Z2.
This leads to the main theorem of this section:
Theorem 5.1.23 Let S be a GQ(s,s + 2 ),s^ 1 which contains a set B\ of
(0, 2)-sets such that every pair of non-collinear points is contained in a unique
member of B\ and such that Axiom (T) holds. It follows that S is isomorphic
to T2*(C>) (the GQ obtained by expanding T2(0) about a regular point) for
some oval O.
Proof: As was shown early on, under these hypotheses, s must be
89


odd. If s = 3 then by [PT84] the result follows. Assume that s > 3. S' as
derived above is shown to be affine three-space as follows (let V = V, B,B\,B'
be as defined above):
From the construction, two points are either in a unique member of
B or a unique member of B\.
The two block parallelisms have been shown to be equivalence rela-
tions on B and B\ respectively; jointly they form an equivalence relation on
V. Let x E V,L E B'. From lemma 5.1.11 and corollary 5.1.16 there exists a
unique M E £>' through x which is parallel to L.
Let L and M be parallel blocks with x\ E L,x2 E M. Let w E
XiX2\{x\,X2},x'l E L\{a;i}. Let a be a Type 1 affine plane through xix'x but
not through w; let (3 be the Type 1 affine plane through x2 and parallel to a.
From lemma 5.1.18 M C (3. Let x'2 = wx' fi (3. Such a point is also ensured by
lemma 5.1.18. Since M is the unique line of (3 which is parallel to L and which
contains x2 it follows that M = x2x'2, i.e. it is shown that for any x' E L\{a;i}
and w E x\x2\{x\,x2} the block wx' meets M (the Axiom of Veblen).
By construction each block contains s + 1 (i.e. more than 2) points.
By [Len54] S' is the design of an affine space. As the point set has
size (s + l)3, in fact this is AG(3, s + 1); thus S is embedded in AG(3, s + 1)
with s odd. By [Tha78] it follows that S is in fact isomorphic to T2(0) as s +1
90


is even.
5.2 Axioms of Payne
Whereas De Soete and Thas concerned themselves with characterizing
T2(0), i.e. the quadrangles obtained by expanding T2(0) about a regular
point, Payne gave seven axioms characterizing the quadrangles obtained by
expanding T2(0) about a regular line. The last of these was latter shown to
be redundant. In this section results from [Pay 85a] concerning the remaining
six axioms are presented.
Axioms Ai, A2, and A4 imply the existence of a GQ(q + 1, S = (V, B,l), with q even and having a fan of ovoids M = {Ooo, Oq, Oi,..., Oq}
for which O^ is pivotal. Furthermore propositions 3.2.1 and 3.2.2 hold in this
context and S can be constricted about the pivotal ovoid O^ to form GQ(q,q) as in section 3.3 with L^ regular.
I .el C> x. = l, 12 l, l, and O0 = U U U T^. For
a,\, a2 e Oqo we write ai = a2 provided a,i fi Oq Pi a2 ^ 0; in fact this is true
if and only if af- fi O0 = a2 fi O0. It is straightforward to check that = is an
equivalence relation. Let [a] be the equivalence class of a. Thus for a E [a]
is the hyperbolic line whose perp is in Oq. If [b] = {b\,... bq} = a1- fi Oq then
91


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