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Generalized quadrangles of order 5

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Title:
Generalized quadrangles of order 5
Creator:
Barnhart, Shawna
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English
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48 leaves : ; 28 cm

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Subjects / Keywords:
Finite generalized quadrangles ( lcsh )
Finite generalized quadrangles ( fast )
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bibliography ( marcgt )
theses ( marcgt )
non-fiction ( marcgt )

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Bibliography:
Includes bibliographical references (leaf 48).
General Note:
Department of Mathematical and Statistical Sciences
Statement of Responsibility:
by Shawna Barnhart.

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|University of Colorado Denver
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Auraria Library
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ocm47832934
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Full Text
GENERALIZED QUADRANGLES OF ORDER 5
by
Shawna Barnhart
B.S., University of Colorado at Denver, 1998
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Masters of Science
Discrete Mathematics
2001


This thesis for the Masters of Science
degree by
Shawna Barnhart
has been approved
by
^4 "2^ -c (
Date


Barnhart, Shawna (M.S., Discrete Mathematics)
Generalized Quadrangles of Order 5
Thesis directed by Professor Stanley Payne
ABSTRACT
The subject of this thesis is generalized quadrangles with equal pa-
rameters, GQ(s,s). All GQs of order 2 and 3, GQ(2,2) and GQ(3,3), are
known. It is conjectured all the GQs of order 5 are known as well. The known
GQ of order 5 is the symplectic GQ, W(5). W(5) along with its dual are
conjectured to be the only GQs of order 5.
Examples of GQs of order 2 and 3 are given. The construction of a
symplectic GQ is given and then used to construct the known GQ of order
5. Information about GQs was gathered, including some basic combinatorics,
affine GQs, and incidence matrices in an attempt to prove the above conjec-
ture.
This abstract accurately represents the content of the candidates thesis. I
recommend its publication.
Signed
Stanley Payne
m


CONTENTS
Figures ............................................................ i
Tables.............................................................. i
Chapter 1. Introduction................................................ 1
Chapter 2. Definitions, Combinatorics, Examples........................ 2
2.1 Basic Combinatorics of Generalized Quadrangles..................... 5
2.2 Affine Generalized Quadrangles.................................... 12
2.3 Linear Algebra Definitions........................................ 15
2.4 Incidence Matrix.................................................. 18
Chapter 3. Classical Generalized Quadrangle........................... 23
3.1 The Sympletic GQ................................................. 23
3.2 GQ(2,2)........................................................... 25
3.3 GQ(3,3)........................................................... 27
3.4 Sympletic GQ of order 5........................................... 31
Chapter 4. Generalized Quadrangle of Order 5.......................... 34
4.1 Counting Triads with i Transversals............................... 34
Chapter 5. Conclusion................................................. 47
Notation Index......................................................... 48
References............................................................. 49
IV


FIGURES
Figure
3.1 zq, zi, z2, and z% are points in trace of x and y and w is collinear
with 3 of them..................................................... 27
3.2 Starting with a triad of points {x,y,w} where all pairs are anti-
regular you can see the lines {L\, L3, L5} form triad with all pairs
of lines regular......................................................... 30
4.1 Represents the pair of lines, {x,y} and their trace................. 34
4.2 n6 = 1 and TI4 > 1 but this violates the GQ axioms.................. 37
4.3 tiq = 3 and n3 > 1 but this violates the GQ axioms.................. 38
4.4 n6 = 3 and 712 > 1 but this violates the GQ axioms.................. 39
4.5 n6 = 2 and n3 > 1 but this violates the GQ axioms.................. 40
v


TABLES
Table
2.1 The results found from the incidence matrix.......................... 22
3.1 Labeling the lines of a triad in GQ(3,3) 30
4.1 The different cases of Uq = 1........................................ 42
4.2 The different cases of Uq = 0........................................ 44
vi


1. Introduction
A Generalized Quadrangle is a geometry made up of points and lines
satisifying a set of axioms. One of the main properties of a GQ that comes
from the axioms is there are no triangles in a GQ. Important characteristics of
GQ can be found by looking at pairs of points and lines and how they interact
with the rest of the GQ. There are a number of lemmas and corollaries about
these interactions. There are a number of basic combinatoral results that hold
for generalized quadrangles.
The focus of this thesis is generalized quadrangles of order s. This
means the parameters of a GQ(s,t) are equal, especially s = t = 5. There are
some known examples given of GQs of order s. It is conjectured the known
example of GQ(5,5) and its dual are the only GQs of order 5. There is a section
that narrows down the cases of GQs of order 5 and makes some suggestions
about how to prove or disprove the given conjecture.
1


2. Definitions, Combinatorics, Examples
The first and most important definition is generalized quadrangle.
From now on generalized quadrangles will be denoted by GQ. Let V be a set
of points, B a set of lines, and let X be the incidence between the points and
lines in V and B. Then S = (V,B,X) is a GQ(s,t) where s > 1 and t > 1, if
the following properties hold:
(1) Every pair of points is incident with at most one line.
(2) There are (1 + s) points on a line and (1 + i) lines through a point.
(3) Let x %L. Then there is a unique line M that connects i to a
point, y, on L, i.e. xXMXyXL.
(4) PUB? {ID}.
If a pair of points, x and y, are collinear this is denoted by x ~ y.
The point-line dual of a GQ is obtained by interchanging the roles of points
and lines and interchanging s and t. An important thing to note about the
concept of GQ is that it is self dual. In other words if you interchange the roles
of s and t all the properties that were true about the GQ(s, t) have duals that
are true for GQ(t, s).
2


The trace (or perp) of x denoted by x1 is the collection of all the
points collinear (~) with the point x, so
xL = {y : x ~ y}.
Note x G xL. The number of points collinear with a single point, x, is (1+s+st)
since there are (t +1) lines through x with s points on those lines not equal to
x. By the duality principle we can interchange the roles of s and t to obtain
|xJ-|=(l +1 + st) in the dual of a GQ(s,t). This number also represents the
number of lines intersecting a line L in a GQ(s, i).
i-
The trace of a pair of points, (z, y}1 is the set of all the points
collinear with both x and y, which is the same thing as the intersection of x1
and yx. If x ~ y then the number of points in the trace is (s + 1). These points
are collinear with the points on the line containing x and y. There cant be
any more points in the trace since this would form a triangle and violate axiom
3 of a GQ. If x 'ft y then the number of points in the trace is (1 + f). There are
(i + 1) lines through x and y those lines will meet exactly once from axiom 3
of a GQ so we get (t + 1) points in the trace.
The span of two points x and y denoted by {rr, y}J"L is equal to
{u eV : u E zxVz G {x, y}1}.
If we have a GQ(s,t) and if X'fty, then |{a;, y}J"L| < t + 1. If x 3


|{z, y}xx| = t + 1 then the pair {x, y} is regular. A pair of points {x, y}
can also be anti-regular, which means there is no point other than x and y
that is collinear with more than 2 points of {x,y}L. A point x is regular if it
forms a regular pair with each point in the GQ not equal to x. A point x is
anti-regular if it forms an anti-regular pair with each non-collinear point in the
GQ. Note the span of two non-collinear points can also be called a hyperbolic
line.
The same definitions hold for lines, a pair of nonconcurrent lines,
{L, M}, is regular if and only if \{L, M}\ = s +1. A regular line is a line that
is regular with every line in the GQ. The following are some oberservations
about regularity.
Observation 2.0.1 The pair of points {x, y} is regular if and only if whenever
{w,z} C {x,y}L then {w,z}L = {x,?/}xx
Proof: If x ~ y then {x, y}x = {x, y}xx so the observation is true. Now
consider the case where x gC y. Suppose {x,y} is regular and set {x, y}xx =
{z0,..., ztj. Let {z, w} be in the trace of x and y. Since {x, y} is regular each
Zi is in {z, w}-1. We know the order of {w, z} is t + 1 so {z, w}1 = {z, y}11.
Now assume {w,z} C {x,^}-1- and {wyz}-1 = {x,y}xx. The order of
{w, z}L = t + 1 which tells us |{x, y}xx| = t + 1. So the pair {x, y} is regular.

4


Observation 2.0.2 A point is z is regular if and only if every pair of points
in x1- is regular.
Proof: Let x be regular. Let yy and y2 be in zx. Take a point y3 in {yy, y2}L\
{z}. Using Observation 2.0.1, {yy, y2}1- = {x, yz}1-1-. So {yy, y2}x = {x, y3}1.
Hence the order of {yy, y2111- is t + 1 and the pair regular.
Let every pair of points in zx be regular. Take {yy,y2} G x1. Let
Xy be another point distinct from x in {yy,y2}L. From Observation 2.0.1 if
{x,xy} A triad of points is an unordered triple of pairwise noncollinear points.
Given a triad T = {z, y, z} a center of T is a point in Tx. There are different
types of triads, if |T^l = 0 then T is accentric, if \TX\ > 0 then T is centric,
and if IT^I = 1 then T is unicentric. The analog of a center in the dual is a
transversal.
2.1 Basic Combinatorics of Generalized Quadrangles
Let S = (V, 13,1) be a GQ of order (s, t). Let \P\ = v and |H| = b.
Theorem 2.1.1 This theorem counts different objects of a GQ.
1. v=(l+s)(l+st); b=(l+t)(1+st)
2. For distinct points z, y G V, |{z, y}L \ = s + lort + lifz~y and if x^y
respectively.
5


3. For each point x, Jrc-1 j = 1 + s + st.
4. For each x G V, \P \ x\ ts2
5. If x ~ y and x ^ y, \V \ (x1- U yL) \ ts2 st.
6. If x y, \P \ {xL U yL) \ =ts2 st s + t
Proof:
1. This counts the number of points and lines in S. Let L be a line in B. There
are s + 1 points on L and through a point, x, on L there are t lines through
x distinct from L. On each line through x not equal to L there are s points.
Combining all these together we get
(ts)(l + s) + (1 + s) = (1 + s)(l + si),
which gives us the number of points in S. Using duality we can interchange
the roles of s and t to give us the number of lines in S.
2. Let x ~ y. The line containing x and y has s + 1 points on it. If x and
y are collinear with any point off this line a triangle would be formed, so the
number of points in the trace is (s + 1).
Let x y. There are (t+1) lines through x and y. The lines through
x and y will meet exactly once from axiom 3 of a GQ so there are (t + 1) points
in the trace.
6


3. There are (t +1) lines through x with s points on those lines not equal to x,
multiply the two together and this is the number of points collinear with the
point x.
4. We know v = (l + s)(l + st) = 1 + st + s+ts2 and |:r-L| = 1 + s + st. Subtract
the two and we get \P/xL\ = ts2.
5. |rrJ-| = 1 + s + st, It/1! = 1 + s + st, and \xLr\y\ = s + 1. So the number of
points in la;-1 U yx\ = 2st + 2s + 2 (s + 1) = 2st + 1. Subtract this number
from the total number of points and we get ts2 st.
6. The proof of part 6 is similar to part 5 except lx1 fl?/-L| = t +1. The number
of points in Ire1 U yL\ = 2st + 2s + 2 (f +1) = 2st + 2s t + 1. Subtract this
from v to obtain ts2 st s + t.
by:
Prom now on assume s > 1 and t > 1. Let xqPy have the trace given
{x,y}L = {z0,...,zt}.
Let W = {wi,... ,Wd) be the set of points where {x,y,w} is a triad. Using
part 6 from Theorem 2.1.1 we know d = ts2 st s +1. Let U, for 1 < i < d,
be the number of Zj collinear with Let rij, for 0 < i < t + 1 be the number
of triads {re, y, w} with exactly i centers.
7


Lemma 2.1.1 Counting up the total number of triads we get:
t+1
Tlj = ts2 St S + t.
i=0
Lemma 2.1.2 J^i=i U = Eii = (i2 l)s
Proof: If we count the number of ordered pairs {to, zj where to G W and z G
{a;, y}x and to ~ z it will give us the total number of z* collinear with to G W.
Let Lx and Ly represent a line containing the point x and y respectively. A
point Z{ £ {x, 2/}x has t 1 lines through it not including the lines Lx and Ly.
Each of those lines has s points on it. There are t+1 points in {x, y}x. Putting
all those together the number of ordered pairs {to, z} is, (tl)(t+l)s = (t2l)s.

Lemma 2.1.3 ZLi U{U 1) = Ei2 *(* ~ l)m = ~ 1)t
Proof: If we count the number of ordered triples {w,z, z'} where to G W,
z ^ z', z and z1 G {a;, y}x, w ~ z, and to ~ z1 it will give us the sums above.
There are t + 1 points in {rr, y}x and after we choose a point, z, there are t
other points that we can choose from. Since z^z' {z, z'}x = t + 1. If we take
away the two lines Lx and Ly we get t-1. Multiplying these numbers together
we get (t + l)(t 1 )t = (t2 l)t.
Lemma 2.1.4 nt 0.
8


Proof: Let {z0, zi,..., zt} G {x^}-1. Assume there is a point w G W that is
not collinear with z0 but collinear with all the other points in {s, y}1, z1;zt.
Let wzi represent the line that connects w to Zi G {x^y}^. The point zq is
incident with a unique line L* that intersects the line wzi. There are t of these
types of lines through zQ as well as Lx and Ly. This contradicts property (2) of
a GQ that says there are t +1 lines through a point in a GQ. Therefore nt = 0.

Theorem 2.1.2 The inequality of D. G. Higman says s < t2 if t > 1. The
dual also holds so t < s2 if s > 1.
Proof: Suppose t > 1. Add the results from lemmas 2.1.2 and 2.1.3 to get,
- 1) = = (* + 1)(f *)(* + s)-
i=l i 1 i=l
Let m be the average value of the Vs, som = J U- We know
0 < X(m ^i)2
Z=1
since it is a sum of non-negative terms. Multiplying the equation out and
because m is a constant the equation becomes,
d
0 < dm2 2m(md) + X^i-
t=i
We know m2d < 52f=i Substitute the s and t values of m and d to get
9


0 < (t + l)(i l)(t + s) (t2 l)s(s2i st s + t).
Simplify to get
0 < t(s2 t)(s 1).
We assumed 0 < t and 0 < (s 1) so t < s2.
Observation 2.1.1 Using Higmans Theorem the following are equivalent,
1. t = s2
2. For some x -ft y, each triad {x, y, z} has a constant number of centers, (s+1).
3. Every triad of points has a constant number of centers, (s + 1).
Proof: Suppose t = s2. Let m be the average value for t{. Then
. 1^. (s4-l)s
171 djz' * (s2S2 SS2 S + S2)
Simplifing the above equation gives us m = (s + 1) so if s = t2 then every pair
of points has a constant number of centers, a, with a = (s + 1).
Let A, B, and C be the equalities from Lemmas 2.1.1, 2.1.2, and 2.1.3
respectively.
Corollary 2.1.1 (t+ l)n0 = (s t)t(s 1)(< + 1) + ~ 1)(^ + 1 i)n,.
Proof: Take (t + \)A + (t +1 )B C and compute it to get the above result.
First take the parameters of the GQ and simplify. So (t + l)(s2t st s +
10


t) + (t + l)(i2 1 )s t3 -1) becomes (t + l)t(s -t)(l s). Rewrite the sums of
the above equalities starting at zero and set them equal to the simplification,
t+i
J2 [( 1(* ( !))*] = {* + !)*( -*)(!- s)
i=0
Note, when i is equal to 1 or (t + 1) the above sum is zero. The sum then
becomes
-(t + 1 )n0 + £[(i !)(*(* 1 ))rk\ = (* + !)*( <)(1 )
i=2
Subtracting the sum from both sides and multiplying by 1 gives us the result
from Corollary 2.1.1.
Corollary 2.1.2 n\ = Y,\=2 ^
Proof: To calculate the above result take and ^ and subtract one from the
other. Since = 0 we know that
S t
^ i(i l)wj ^ Q
i=l S i=2 ^
When i = 1 the left sum becomes Both the sums have the same index so
by combining them and doing some simplification we get,

Subtract n\/s from both sides and multiple by st to give the result from
Corollary 2.1.2.
11


Corollary 2.1.3 n0 = 0 if and only if
t-i
(t s)t(s l)(t + 1) = 53(i l)(f + 1 i)rii
i-2
So no = 0 implies t>s, with equality if and only if n* = 0 for 2 < i < t.
Note, when n0 = 0 each triad {a;, y, w} is centric.
Corollary 2.1.4 Computing C B we see that n\ = 0 if and only if
t+1
{t s)(t2 1) = 53(i 2)rii
i=3
in which case t > s. If t = s, then nt- = 0 for i ^ 0,2.
2.2 Affine Generalized Quadrangles
Let S' be a GQ(s,t). Remove one point, oo, all the lines through oo
and all the points on those lines to get an Affine GQ, AGQ(s,t). The lines
through oo are oo-lines. Let V and B' denote the set of points and lines in
AGQ respectively.
There are some basic combinatorics to compute with an AGQ. We
know S has (1 +1) (1 + st) lines and each point has (t +1) lines through it. To
get the number of lines in an AGQ take the total number of lines in a GQ and
subtract the oo-lines.
\B'\ = (1 + f)(l + st) (f + 1) = st + st2 (2.1)
12


Each oo-line has s points on it not including oo. Since there are t+ 1
oo-lines we are removing s(t + 1) + 1 points from the GQ.
\P'\ (1 + s)(.l + st) s(t + 1) + 1 = st^ (2-2)
The trace of a point, x, in AGQ is
(t + l)(s 1) = st t + s. (2.3)
x has (t + 1) lines through it and those lines contain (s 1) points
not equal to x. Multiple the two together to provide the number of points in
the trace of re. A line L in AGQ has s points on it. Those points have t lines
through them not equal to L. So
L1- = ts. (2.4)
This means that the concept of an AGQ is not self-dual because
interchanging the roles of s and t from xx would have to give us the number
of lines in a trace of a line.
Do the axioms of a GQ still hold? The first axiom, every pair of
points is incident with at most one line, still holds. Axiom 2 talks about the
number of lines through points and points on lines of a GQ. The number of
lines through a point will remain the same, (t + 1). The number of points on
13


a line has changed, a line could have had a point on it that was removed. So
there are some lines with s points. There are (t + 1) oo-lines. Each line has
s points on them not equal to oo with t lines through them not equal to an
oo-line. So there are (t + 1 )ts lines with s points. Since this is equal to \B'\ all
the lines in a AGQ have s points on them.
Axiom 3 does not hold for all the lines and points of a AGQ. In GQ
a line, L, and a point not on that line, x, are connected by a line M to unique
point on L. Let y be the unique point on L. If y is on an oo-line then axiom
3 of a GQ no longer holds. If y is not on an oo line axiom three holds.
Now some new terminolgy is needed with respect to AGQs. A point,
x, is joined to a line L if there is a line, M such that x _L M _L y _L L where y
is a point on L. A pair of lines L, M is parallel is there is no point of L joined
to M. Every line is considered parallel to itself. If two lines, L and M, are
parrallel this will be denoted by M || L. A line L is said to be totally connected
to another line M if every point on L is joined to M, denoted by L = M.
From the obeservations above we can see the axioms of a GQ no
longer hold for a AGQ. The following describes a new set axioms specific to
an AGQ [7].
(1) Every line contains at least two points and every point is on at
least three lines.
14


(2) For every line, L, if x in not on that line then IA-1 fl L\ < 1.
(3) For any two lines L and M if L = M then M = L.
(4) For two nonconcurrent lines L, M if neither L || M or L = M
then L ~ M.
(5) Given a line L and a point x not on L or joined to L, then there
exists a unique line through x parallel to L.
(6) Given two lines L, M and a point x such that L _L x _L M there
exists exactly one line line N through x where L = N = M.
2.3 Linear Algebra Definitions
First here are some definitions we will need to work with the incidence
matrix. A basic definition that we will use is the trace of a matrix denoted by
tr(A). Let A be a square matrix then tr(A) is the sum of the entries on the
main diagonal of A.
Theorem 2.3.1 Let A and B be square matices of equal dimension then
tr(AB)=tr(BA).
15


Proof: Suppose
1,1 ^1,71
A =
Onti 7i
£>1,1 frl.n
B =
bn, 1 bn,n
The entries on the diagonal matrix AB are
ABjj ^ Q'jkbkj-
k=1
Since we know what the entries along the diagonal look like we can calculate
the trace of AB:
n n
X X ajkbkj-
j=1k=1
Which equals
n n
X) X bJkakj = trBA
fc=i j=i
Therefore when A and B are square matrices of the same size the trace of AB
equals the trace of BA
16


Corollary 2.3.1 Let A be a matrix where there is a matrix P such that
P~lAP = D and D is a diagonal matrix. Then tr(P~lAP) tr(A).
Proof:
tr[P~lAP) = tr{APP~l) = tr(A).
The minimal polynomial of a matrix A is the monic polynomial p 6
P(f) of the smallest degree such that p(A) = 0. To find the minimal polynomial
of A take the powers of A until the list is linearly dependent,
(I,A,A2,A3,...,Am).
You can then find the coefficients by finding the scalars of the following equation
o0/ + giA + 02 A. ^ + ... + om_iA.m ^ T -Am = 0
Theorem 2.3.2 Let A. be a matrix. The roots of the minimal polynomial of
A are the eigenvalues of A.
Proof: Let
p{x) a0 + aix + a2x2 + ... + am_ irr771-1 + xm
be the minimal polynomial of A. Let A be a root of p(x). We can rewrite the
minimal polynomial p(x) = (x A)q(x) where q is a monic polynomial. Since
p(A) = 0 we have,
0 = (A \I)(q(A)v)
17


where v is a vector where dimension makes sense.
The degree of q(x) is less than the degree of p(x) so there must exist
one vector,v, such that q(A)v ^ 0. That makes the root A an eigenvalue of A.

Here are a couple more definitions. A square matrix is symmetric if
AT = A. Two square matrices A and B are similar if there is an invertible
matrix where A = PBP~1.
2.4 Incidence Matrix
Let S = (V, B, X) where V be denoted by {aq,..., oq} and let the lines in B be denoted by {Iq,..., Lb}.
Let A be the line-point incidence matrix of S. This is a b x v matrix with
entries of 0s and ls
/
1
aij <
0
if XjTLi;
otherwise
What happens if we take ATA? If i j then we get row i multiplied
by column i. This gives us the number of lines through a point. Since we are
dealing with a GQ we know this number is t +1. If x^xj and Xi ~ Xj then we
get 1 because a pair of points is on at most 1 line. If i ^ j and Xi ^ Xj then
the entry (ATAij) is 0.
18


1 + t if i = j;
(ATAis) = \ j
if Xi ~xj,i ^ i;
(2.5)
0 if Xi Xj.
ATA is a, v xv matrix. tr(j4T.A) = (1 + t)v (1 + £)(1 + s)(l + st).
ATA is also symmetric because (ATA)T = AT(AT)T = ATA. Since
ATA is symmetric a matrix P can be found where
( \
Xi 0
P{ATA)P~1 =
^ ^ ^v )
What is the number of paths from a point z* to a line Lj? We can
find this by looking the the matrix ATAAT. If z* is not on Lj then we know
there is a unique line connecting Zj to Lj from property (3) of a GQ, so there
is one path from z* to Lj. If Z; is incident with Lj then z; has t lines through
it not equal to Lj. There are (1 + s) points on Lj including z* so the number
of paths is (1 + s + t).
(AtAj%) =
(2.6)
1 + s +1 if X{XLj
1 otherwise.
Let JmXn be the mxn matrix of all ones. (s + t)AT will be a matrix
with (s + t)Afj = (s + t) entries whenever x^ILj and zeros every place else. It
19


follows that
AtAAt (s + t)AT Jvxb.
(2,7)
Jb*vATA is b x v matrix that counts up the entries in each column of
ATA. Fix j
y, ATAij 1 + £ + s + st (1 + s)(l + t)Jbxv (2-8)
i
Similarly JVXVA represents the column sums of A. Fix j
]T>y = (i+l) (2.9)
i
therefore JVXVA = (t + 1 )JVXV.
(AtA)2 = AtAAtA (Jvxb + (s + t)AT)A = JvxbA + (s + t)ATA.
Using the information above we can find the following equations,
(A^ A)2 (s+t) A? A = JvxbA + (s -\-£)oF A (s -ht)A^' A = JvxbA = (t + l)Jvxv-
(2.10)
AtA(AtA (s + t)I)(ATA (1 + s)(l + t)I) = 0 (2.11)
Let p(x) = x(x s + t)(x (1 + s)(l +1)). From the above equation
we know p(ATA) = 0. From this it follows p(x) is also the minimal polynomial
of ATA.
20


Let 6i be an eigenvalue of ATA and dj be the multiplicity of 0*. Since
p(x) is the minimal polynomial of ATA, every eigenvalue of ATA is a root
of the minimal polynomial, p(x). There are three eigenvalues of ATA. Let
0o = (1 + s)(l + t), 6i = 0, and 02 = {s + t). We still need to find the
multiplicity of each 0*. We do know v = do + di + d2 and
tr(ATA) = d00o T d\d\ + d202 = (1 + t)(l + s)(l + sf). (2.12)
Looking at equation (6), (0j)2 (s+t)9i is an eigenvalue of (1+t) JVXv,
in other words it is either (1 + t)(l + s)(l + st) or 0. The multiplicity of the
eigenvalue 0 with respect to (1 + t)Jv) has multiplicity 1. 0O = (1 + s)(l +1) corresponds to (1 + s)(l +1)( 1 + st) and
do = 1.
Subsituting this information into equation (8) we get
(l +1) (1 + s) (l + st) = (1 + s) (1 +1) + d]0 + d2(s +1)
The above equation is gives us d2 in terms of s and t,
, (l + s)(l + t)st
do -----;----:-----.
(s + t)
We can subsitute d2 into
v = (1 + s)(l + st) = 1 + di + d2 = 1 + di + ^
21


Table 2.1. The results found from the incidence matrix
@0 (1 + s)(l + t) II O ^3
o II os J (l + st)s2 1 ~ (s+t)
&2 (s + t) J (l+*)(l+i)t d2 (s+t)
Solving the above equation for di in terms of s and t gives us,
, (1 + st)s2
(s + t) '
We have now established the following information
Theorem 2.4.1 must be an integer.
Proof: The mulitplicity of an eigenvalue has to be an integer therefore we
know from above
(l + s)(l+t)si
do -----:---------
{s + t)
has to be an integer.
22


the incidence inherited from PG(3,y). We know that the number of points on
a line is going to be q + 1 from PG(3,g). The number of points in PG(3,q)
is q3 + q2 + q + 1 so \P\ = q3 + q2 + q + 1. The lines through a point x are
just the lines lying in the plane a;-1 which gives us {q + 1). Let x, y, and w
be three distinct points with y,w £ x1. Then y ~ w if and only if x, y, and
w lie on the same line. To get this result we know if y ~ x ~ z ~ w then it
implies y £< x, y, w >x so yL =< x,y,w >= x then x = y. This shows that
the above construction is a GQ of order q.
Theorem 3.1.1 W(q) is a regular GQ(q, q), in other words every pair of points
in W (q) is regular.
Proof: Let x and y be distinct points of W(q). Two things can happen, either
x ~ y and = {x,y}XJ- is line in W(q) through x and y or x 76 y and
{x, y}1- is a line in PG(3, q). Then Lp = LL = {x, y}11 is the span of {x,y}.
Therefore each pair of points in W(q) is regular.
3.2 GQ(2,2)
The symplectic GQ tells us we can construct at least one GQ of order
s where all the pairs of points are regular and all pairs of lines are anti-regular.
We also know the dual of W(q) is a GQ as well. The next sections will construct
some examples of known GQs of order s, GQ(s).
25


Consider the case where GQ has order 2, GQ(2,2). We can use the
sympletic GQ but there is another construction using duads and synthemes.
We know the number of points in GQ(2,2) is 15,. the number of lines is 15,
each point has three lines through, and each line contains 3 points.
Let D = (V,B,X). Let S = {1, 2,3,4,5,6}. V is the set of points
equal to the set of duads from S i.e. the subsets of of order two. So V =
{{i,j} ' h3 £ S'}- A syntheme is a set of 3 disjoint duads. £ is a set of
lines equal to the set of syntheme formed from V. For example the set
{{1,2}, {3,4}, {5,6}} is a syntheme and a line in £. The incidence, X, of
the points and lines is containment, the point x = {1,2} is on the line formed
by the syntheme above.
The construction above yields a GQ of order 2. From the definition
we get that each line in £ contains three points. Given two points in a line
the remaining point is determined so those points distinguish a line in the GQ.
Each point is contained in exactly three lines. For example, {1,2} is disjoint
with 6 duads. The syntheme has to contain a 3 and there are three duads
disjoint from {1,2} with 3 in them. Once this duad is chosen the last duad is
forced, so {1,2} is contained in three lines. No two lines are incident with the
same pair of points. If there is a line and point not on that line is it easy to
see there is a line containing that point that intersects the orginal line.
26


3.3 GQ(3,3)
Let GQ(s, s) = G. All the GQs of order 3 are known [5]. The GQ(3)
comes from the sympletic GQ or the dual of that. From now on consider only
the case where s=3 to get the following results.
Lemma 3.3.1 Let x and y be points in G, then the pair is either regular or
antiregular.
Proof: Suppose the pair {x, y} in G is a neither regular or anti-regular. Let
{x, y}L = {z0, zi, z2, Z3}. Since {x,y} is not regular or anti-regular there is a
point w that meets the trace of x and y in exactly three points (see figure 3.1).
Figure 3.1. Zo, zi, Z2, and z3 are points in trace of x and y and w is collinear
with 3 of them.
27


Note any point in G has to have 4 lines through it. The point w is
not on the lines xz3 or yx2. From axiom 3 of a GQ we know w has to be on
a line that intersects both xz3 and yx3. So w cannot be collinear with exactly
three points in the trace of {x,y}. So the pair {x,y} regular or anti-regular.

The following lemma follows directly from the above lemma.
Lemma 3.3.2 Let T = {x,y,w} be a triad in G, then
(1) if T has 1 or 4 centers then (x, y), (x, w), and (y, w) are all regular pairs,
(2) if T has 0 or 2 centers then (x,y), (x,w), and (y,w) are all anti-regular
pairs and,
(3) the is no T with 3 centers.
Corollary 3.3.1 If T = {x,y,w} is triad then the pairs of T are either all
regular or all anti-regular.
Lemma 3.3.3 All pairs of non-collinear points in G are regular or they are
all anti-regular. The same is true for the dual.
Proof: Assume not every pair of points are anti-regular. So there exists a pair
of points {x, y} that is regular. Take a point w not equal to x or y then w
can be related to the pair {x, y} in three ways. Case one is when w ~ x and
w ~ y then the pairs {x, w} and {y, w} are regular. The point w can also be
non-collinear with both x and y. From Corollary 3.2.1 we know {w,y} and
28


{w,x} form regular pairs.
The last case is when w ~ x andw is regular. Since s > 2 we can find a point 2 not collinear with x, y, or w.
Since the pair, {u;,?/}is regular we know from Corollary 3.2.1 all the pairs in
the triad {y,w,z} are regular. So every point not collinear with y is regular
with y. It now easy to see every pair of points is regular if only one pair of
non-collinear points is regular.
Theorem 3.3.1 Any GQ of order 3 has all pairs of points regular and all lines
anti-regular or is the dual of such a GQ.
Proof: Suppose all pairs of non-collinear points are anti-regular. Let T =
(x, y, w) be a centric triad in G. Since T has at least one center and each pair
of points is anti-regular T has to have exactly two centers, wx and w2. Pick a
line Li through x that does not intersect wx or w2. Using L\ as basis label 6
more lines as described in table 3.1.
T has two centers so Lx does not equal L4. Lx, L4, and L7 cannot
intersect or w2- This tells us Lx = L7. So (LX,L3, L$) is a triad with centers
L2, L4, and Le (See figure 3.2). Since it has three centers that means it has to
have 4 centers and each pair of lines in the triad forms a regular pair. Therefore
all pair of lines form a regular pair.
29


Table 3.1. Labeling the lines of a triad in GQ(3,3)
Line Through Point Meeting Line
l2 y u
U w l2
U X Lz
U y U
U w U
l7 X U
Ll L3 L5
xi h L4
c w wi L6
% 9 y. L2
1 9
Figure 3.2. Starting with a triad of points {x,y,w} where all pairs are anti-
regular you can see the lines {Li, L3, L^} form triad with all pairs of lines
regular.
30


Since we know a GQ of order 3 has all pairs of points regular and all
pair of lines anti-regular or is the dual of a such a GQ, then a GQ(3,3) or its
dual is isomorphic to W(3).
3.4 Sympletic GQ of order 5
A known construction of a GQ of order 5 is the sympletic GQ. This
section will give examples of points and lines in W(5).
Let F be the Galois field of order 5, G.F(5). F* denotes the non-zero
elements of F. Let V be the four dimensional vector space of 4-tuples over F.
A vector in V will have this form, (x0, xi,%2, x3) where x* G {0,..., 4}. Using
the same method as in section 3.1 construct the associated projective sapce of
V, PG(3,5).
Let W(5) be the sympletic GQ constructed from PG(3,5). We know
the set of points is W(5) is the same as the set of point of PG(3,5). The number
of points in PG(3,5) is qz + q2 + q + 1 = 53 + 52 + 5 + 1, therefore the number
of points in W(5) is 156.
Let x be a point in PG(3,5) and in W(5). Set x = {Ax : A F*} =
{A, 0,0,0}. A line, L, in PG(3,5) is a line in W(5) when x,y are points on L
and x o y = 0. So the point x together with a point y forms a line in W(5)
31


when
xoy = x0yx Xiy0 + x2y3 x3y2 = lyi 0y0 + 0y3 0y2 = yi = Q.
It is easy to see rrx forms a plane in PG(3,5). The lines on this plane are the
lines in W(5).
Let y be the point in W(5) where y = {Ay : A G F*} = (0, A, 0,0}.
The points x and y are non-collinear in W(5) since x o y does not equal zero.
A point in W(5), w from w = {wq,Wi,w2,w3}, is in the trace of x and y if w0
and Wi are zero. There are six points in W(5) where this is true, {0,0,1,1},
{0,0,0,1}, {0,0,1,0}, {0,0,1,2}, {0,0,1,3}, and {0,0,1,4}. So {x,y}x =
s + 1 = 6.
A point in W(5) forms a triad with the points x and y if it is non-
collinear with both x and y so let z be the point that comes from the vector
z = {1,1,0,0}. Now look at the triad {x,y,z}. The point z is non-collinear
with {a;,y} but it is collinear with all the points in {x,y}.
We also know all the pairs of points in W (5) are going to be regular.
32


4. Generalized Quadrangle of Order 5
Let of a GQ we know there are 6 lines through each point and 6 points on a line. S
has \P\ = (1 + 5)(1 + 5 5) = 156 and \B\ = 156. The order of the perp x E P
is 31. The order of the trace of {rr, y} is s +1 = 6 or t +1 = 6 when x ~ y and
x f^y respectively. We also know the order of the span, |{rr, y}| < t +1 = 6.
If the span is equal to 6 then the pair is regular. Let x and y be two distinct
points, we can make 100 different triads using the pair {x,y} together with
one of points in the set {itq, w2,..., tw100} where wiy with 1 < i < 100, is not
collinear with x or y.
4.1 Counting Triads with i Transversals
Let x, y be a pair of lines that do not intersect. Let W = {uii, W2, , udoo}
be the set of lines that do not intersect x or y. From this set we can form the
triads {a;, y, Wi} where 1 < i < 100. We know {re, y}1 contains 6 lines.
A t6 will be a transversal through those 6 lines, a is will be a transver-
sal through 5 of those 6 lines, etc. until we get to t0 which will represent the
lines that do not intersect any of the lines in {x, y}-1.
33


X
Y
Figure 4.1. Represents the pair of lines, {x,y} and their trace
We can say the following about the triads with i centers using the
corollaries and lemmas from the introduction. Lemma 2.1.1 counts up the
total number of triads with i centers. In the case of S we have,
Tto + nl + n2 + ^3 + n4 + TI5 + 71q = 100. (4-1)
From Corollary 2.1.1 we know 6n0 4n2 + 6/13 + 6/14 which is the same as
n0 = |^2 + n3 + n4. ni is calculated in Corollary 1.0.2,
Tii = 3n3 + 8n4 + 24n6 (4.2)
Insert the values of n0 and into equation 4.1 to get |n2 + 5n3 + 10n4+25n6 =
34


100. Solving this equation in terms of n2 we get,
n<2 = 60 3 n3 67*4 15n6. (4.3)
Subsitute the values of n2 and n 1 into equation 4.1 and solve for n0 to get
n0 = 40 n3 3 n4 10n6. (4.4)
Observe that 0 < n6 < 4 and that n2 is a multiplie of 3. Let n2 = 3h2.
Let x and y be two distinct lines. The trace of x and y consists of six
lines. We can now focus on the values for n6 which is the number of triads,
x, y, Wi, where the line Wi intersects all six lines in{a:, y}x The largest value for
n& is 4 which happens if and only if the pair (x, y) is regular. Using the above
equations we can see a triad containing the pair x and y is either unicentric or
centric with 6 centers. This tells us n 1 is equal to 96 and n6 is 4.
Let w be a line that does not intersect x or y and where the triad
{rr, y, w} is unicentric. Let r be a line where the number of centers for the triad
{x, y, r} is 6. Label the lines in the trace of x and y by q where 1 < < 6.
Assume the line w meets the line cx where the line r intersects ci. Then the
line Ci is in the trace of x and w. Let yc\ be the point where y and Ci intersect.
The point yc\ needs to be connected to the five other lines in the trace of x
and w. The only way to connect yc\ is with a 6-line. This tells us there is at
least one triad containing the pair {x,w} that has 6 centers, i.e. n6 > 1. We
35


will soon see when uq > 1 then it implies n6 = 4. This implies the pair {a;, w}
is regular since
{x, w}J"L = t + 1 = 6.
It follows easily all pairs of lines in S are regular. Therefore when n& is equal
to 4 S = W{5).
Lemma 4.1.1 Let S be a GQ of order 5 where n6 > 0 then n4 = 0.
Proof: Let uq = 4 then from the above result we know every pair of lines in
S is regular. The triads containg the pair of line x and y have either 6 centers
or 1 center.
Consider the case where n6 = 1. There is exactly one line that inter-
sects all the lines in the trace {x, y} not equal to the lines x and y. Assume
w is a line in n4. Let W be the set of four points on w that are in the trace
of x and y. There are two lines in the trace of {x, y} that w do not intersect
w. Axiom 2 of a GQ tells us each of the 4 points in W have to be on exactly
one line that intersects those two lines. Connect three of the four points to the
two lines following the axioms of a GQ (See figure 4.2). Now there is no way
to connect the last point, A, in W without forming a triangle. Therefore when
ri6 > 0, n4 = 0.
36


Figure 4.2. n§ = 1 and n4 > 1 but this violates the GQ axioms
Consider the case where ne = 3. We know n4 = 0 from Lemma 4.2.1.
Calculating equations 4.2 through 4.4 using the known values gives us
n0 = 10 n3, (4.5)
n i = 3n3 + 72, (4.6)
and
n2 = 60 3n3 45 = 15 3n3. (4.7)
From that we know 0 < n3 < 5. There are 6 points not on the
lines x, y or on the 3 f6 lines, so the only way to connect those points without
37


violating axiom 3 of a GQ is with another t$ line. This argument is similar to
the one for n4 > 0 and is shown in figure 4.3. So n3 = 0.
Figure 4.3. n6 = 3 and n3 > 1 but this violates the GQ axioms
Now recompute equation 4.7 to get n2 = 15. Look at the points on
one i2 n2 Those two points need to be on a line that connects them to a
point on the remaining 4 lines in the trace of X and Y. The only way to do
this without violating the GQ axioms is to use a t6 line, so n2 = 0. Figure 4.4
shows the trace of x and y with at least one n2.
38


Figure 4.4. n& = 3 and n2 > 1 but this violates the GQ axioms
This is a contradiction since n2 is equal 15. So uq does not equal 3.
Since n6 / 3 we have narrowed down the size of n6, it is either 0, 1, 2,
or 4. Now let n6 = 2. Therefore n4 = 0. We can substitute the known values
into equations 4.2 through 4.4 again to give us
n0 = 40 n3,
rii = 3n3 + 48,
and
n2 = 30 3n3.
Again n3 = 0 because we have no 714 lines and we cant create a n3 line without
forcing a n4 line. This is shown in figure 4.5.
39


Figure 4.5. n6 = 2 and n3 > 1 but this violates the GQ axioms
This then determines all the centers of the triads so n0 = 20, ni = 48,
n2 = 30, n3 0, Hi = 0, and n6 = 2. Take the point P not on the lines x
or y or on one of the ne lines through the trace of {x, y} and let it have one
line through it that intersects the trace of {x,y} uniquely at P. This is one of
the ni lines. We know n3 = 0 so the only way to connect this point with all
the other lines in the trace is by using n2s. This would create 5 lines through
P that are n2s, one line through P from the ni, and one line through P that
intersects the pair of lines {z, y}. Then P would have 7 lines through it which
is a contradiction. Therefore n^2.
We know n6 = 0, 1, or 4. Let n6 = 1. Which gives us the following
40


equations:
no = 30 n3 (4.8)
n i = 3 n3 + 24 (4.9)
3 to II CO 0 1 CO 3 CO (4.10)
with n4 = 0 and n6 = 1. Let P be the points on the trace of {x,y} not on
an line in n6. Let P G P. We need to introduce some terminology: an i-line
through P is a line that intersects P in i points. For example a 1-line of P
intersects only the point P in P. We can classify the points in P by counting
the different i-lines intersecting them. There are three ways to connect P with
the other points in the trace of x and y.
One way to connect P to the other lines in the trace is with two 3-lines
and one 2-line. Now axiom 3 is satisfied with the lines in the trace and two
lines have been unaccounted for. Since there are 6 lines through P we know
the remaining two lines are 1-lines. If P has one 3-line through then there are
three lines in the trace it still needs to be connected to. P will be connected to
the three remaining lines by three 2-lines. If P has no 3-lines through it then
it has to have five 2-lines to connect it to the 5 other lines in the trace. That
gives us 6 lines so there are no 1-lines. It is easy to see there are no other ways
41


to draw the lines through P.
The following notation talks about the three possible cases. Let hi be
the number of points P with i 1-lines, i 3-lines, and 5 2i 2-lines. The three
cases are, h0, hi, and h3. The different cases are described in table 4.1.
Table 4.1. The different cases of n6 = 1
Number of i-lines Case One: ho Case Two: hi Case Three: /i3
1-line 1 2 0
2-line 3 1 5
3-line 1 2 0
P has 18 points in it. Since we know all the cases for a point in P we
get the following equation,
ho + hi + /12 18. (4-11)
Let (P, l3) be a 3-line on P. There are three points on each line so
3n3 gives us the number of points on 3-lines. Looking at table 4.1 we can also
count the number of points on 3-lines by hi + 2/12- Let (P,h) be a 1-line on
P. This tells us ni = hi + 2ft.2- So
ni = hi + 2h 2 = 3 n3.
42


This is a contradiction because we know from equation 4.9 ni = 3n3 + 24. So
n6^l.
From the above cases the next lemma arises.
Lemma 4.1.2 The order n6 for a pair of lines in GQ(5) is either 4 or 0.
The last case to look at is n6 = 0. Using the corollaries from the basic
combinatorics section we get the following equations
n0 = 40 ri3 3 n4 (4.12)
n i = 3 n3 + 8 n4 (4.13)
n2 = 60 3r&3 6714 (4.14)
Again let P be the points on the trace of {x, y} not on the lines x or
y. Let <7j be the number of points on 2-lines through the point P. |P| is equal
to 24.
Let P G P. There are five ways to connect P to the other lines in
the traces of x and y. We know a point can have at most one 4-line so assume
P has one 4-line. There are two lines in the trace of x and y that the point P
needs to be connected to by a line. This can be done in two ways, with one
3-line or with two 2-lines. Then P is an go or an gi point respectively.
Now let P have no 4-lines through it. If P has two 3-lines through it
43


then it has one 2-line as well. The remaining three lines are 1-lines. Then P
is an gi. P could also have only one 3-line. Then it has three 2-lines and P is
an g3.
Now let P have no 3-lines or 4-lines. The only way to connect P to
the five other lines in the trace is using 2-lines. So there are five 2-lines and P
is an g5. This exhausts all the possible cases which shows there are only five
differeny types of points in P. Table 4.2 shows all the cases above.
Table 4.2. The different cases of n6 = 0
No. i-lines 9o 5i 92 33 35
1-line 3 2 2 1 0
2-line 0 1 2 3 5
3-line 1 2 0 1 0
4-line 1 0 1 0 0
Since P = 24 we know g0 + gi + g2 + 9a + 3s = 24. Let (P, la): where
la is 3-line through a point P. Counting up all the points on the lines in n3 we
get
3n3 = go + 2 gx + Qg2 + ga + 0#s = <7o + 2 te


Do the same thing with the 1-lines, 2-lines, and 4-lines to get;
ni 3<7o + 2(?i + 2g2 + <73 + 0,
2n2 = gi + 2g2 + Sg^ + 5^5,
4 n4 = g0 + g2.
If all the pairs of points on the lines in the trace of {re, y} are g^s then
each pair of points will be anti-regular. So every pair of points will be anti-
regular and every pair of lines will be regular giving us the dual of W(5). This
is the only GQ(5,5) known (for now) when n6 = 0. It is still an open question
if there other GQ(5,5)s or if a GQ of order 5 is unique up to isomorphism to
W(5) or W(5)s dual.
45


5. Conclusion
Generalized quadrangles of order s have some properties that hold
for them specifically. Using the sympletic construction we can construct GQs
of order s where every pair of points is regular and every pair of lines is anti-
regular. All the GQs of order 2 and 3 are known to be the symplectic GQ and
its dual. We know there is a GQ of order 5 of this form as well. Using the
basic combinatorics of generalized quadrangles we can eliminate a number of
possibilities for generalized quadrangles of order 5. We now know given a pair
of lines {x,y} we have only 2 possible values for ne, 4, 0. When n6 = 4 then
{x,y} is regular and GQ(5,5) is equal to W{5). If ne = 0 then one of the cases
left is the dual of W (q).
For further study I would suggest trying to reduce the number of cases
possible when n6 = 0. I tried reducing the known cases of GQ(5,5) into an
affine GQ to see if the AGQs axioms still held. This method was not fully
explored and it is recommended that it be looked at again. Also using the
incidence matrix of a GQ could help describe a GQ of order 5 and maybe rule
out certain possiblities.
46


NOTATION INDEX
a;x: the perp of a point x. The perp is all the points collinear with x.
x ~ y: The points x and y are collinear
x y: x and y are not collinear
{x,y}L\ the trace of two points x and y, this denotes all the points
collinear with both a; and y.
{a;, t/}xx: The span of x and y which is the set of points collinear with
the point the trace of (x, y).
47


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[41 Stanely E. Payne. Nonisomorphic generalized quadrangles. Journal of
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[5] Stanely E. Payne. All generalized quadrangles of order 3 are known. Jour-
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[7] Harm Pralle. Affine generalized quadranlges an axiomatization. Geome-
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