UNIVERSITY OF NORTH CAROLINA
Department of Statistics
Chapel Hill, N C.
Q
Mathematical Sciences Directorate
Air Force Office of Scientific Research
Washington 25, D. C.
F. . ,
AFOSR Report No.
APPLICATION OF THE GEOMETRY OF QUADRICS
FOR CONSTRUCTING PBIB DESIGNS
by
Do K. Ray-Cbaudhuri
June, 1961
Contract No. AF 49(638)-213
In this paper a number of families of PBIB designs
with two associate classes is constructed using
the properties of quadrics in finite projective
geometry.
Qualified requestors may obtain copies of this report from the
ASTIA Document Service Center, Arlington Hall Station, Arlington
12, Virginia. Department of Defense contractors must be established for ASTIA services, or have their "need-to-know" certified
by the cognizant military agency of their project or contract.
Institute of Statistics
Mimeo Series No o 294
1.
Introduction.
Partially balanced incomplete block
(PBIB) designs with two associate classes were introduced by Bose and
L-27.
Nair
stress
Bose and Shimamato ~~7 have rephrased the definition so as to
the distinction between the association scheme and the design.
Bose
and Shimamato definition for the PBIB design with m associate classes is substantiallyas follows.
A PBlB design with
in b
m associate classes is an arrangement of
v treatments
blocks such that
(1) Each of the v treatments is replicated r
of size
k
times in
b
blocks each
and no treatment occurs more than once in every block.
(2) There exists a relationship of association between every pair of the
v
treatments satisfying the following conditions:
(a)
Any two treatments are either
second associates
(b)
e ••
first associates or
or m-th associates.
Each treatment has
n
l
first associates,
n 2 second
associates, ••• n m-th associates.
m
(c)
Given any two treatments which are i-th associates,
the number
first and
i
Pjk
of treatments which are
k-th associates of the second is independent of the pair
of treatments with which we start.
i,j,k
j-th associates of the
i
Furthermore
i
Pjk = Pkj' for
= 1,2, •••m.
(3) Any pair of treatments which are
together in exactly
~1
blocks for
i-th associates occur
i=1,2, •••m.
Bose and Clatworthy ~g7 have given a less demanding definition for the PBIB design with two associate classes which is as follows:
A PBIB
ments in
b
design with two associate classes is an arrangement of
v treat-
blocks such that
(1) Each of the
v
treatments is replicated
r
times in
b blocks each
2
ot size
(2)
k
and no treatment occurs more than once in any block e
There exists a relationship of association between the
v treatments
satisfying the following conditions:
(a)
Any two treatments are either first associates or
second associates.
(b)
Each treatment has
n
l
first associates and
n2
second
treatments which are
i-th asso-
associates.
(c)
For any pair of the
ciate the number
i
Pll
v
of treatments common to the first associates of
the first and first associates of the second is independent of the
pair of treatments with which we start, i
(3) Any pair of treatments which are
~.
1.
= 1,2.
i-th associates occurs in exactly
blocks, i=1,2.
Since it is easier to check the conditions of Bose-Clatworthy definition, we
shall use this definition.
The following relations hold between the parameters
of a PBIB design and are useful for computing some parameters when others are
known.
vr = bk ,
(1.1)
m
i
E
P'k
k=l
J
for
i ::: j
for
i ~ j
=
The theory of linear spaces in finite projective geometry have been used
very profitably by several authors in constructing BIB and PBIB designs.
Bose ~17 first used the properties of quadric surfaces in finite projective
geometry of two and three dimensions for constructing experimental designs.
3
In ~§7 we have derived several properties of quadrics in finite ~rojective
spaces.
In this
series of
~aper
we have used the geometry of quadrics to construct several
P;BIB designs with two associate classes.
Let 'B denote the class consisting of the sets
B., j=I,2, •••b, is a set of points in FG(n,s).
Let V
J
consisting of fue sets
points in
FG(n,s).
where
B , B2 , ••• B
b
l
V.,
1.
where
denote another class
i=I,2,o.~v
These two classes generate a design
,<
is a set of
D(' B , V
.;
) with
1,
the following incidence matrix
N
= «n i )
(vxb)
vxb
where
n
ij
= 1, if the
sets
0, if the sets
Vi and Bj
intersect each other,
Vi and Bj do not intersect.
Then we have the following.
b
ri
=
k.
=
E
j=l
n ij
= Number of sets of the class
B'which intersect
Vi·
n ij
= Number of sets of the class
V which intersect
B .•
v
J
E
i=l
b
=
E nijn i • .
J
j=l
= Number
of sets of the
c~ass
B which intersect
both V. and
1.
Let
C , C2 ,
I
0 ••
C be
m
m classes of sets in FG(njs) such that
Vi ~ Vir
The sets
Vi
and
Vi'
€
for some
are said to be
€
Let
Cj ,
C.,
J
j,j=I,2, •••m.
j-th associates if
j=1,2,~ ••m.
t
puV(Vi,V it ) denote the number of sets of the class
associates of
Vi
and v-th associates of
Vi"
V which are
i = it, i,i'=1,2, •••v,
u-th
J
4
t,u,v
= l,2, •••m.
The following result, stated as a lemma for the sake of
reference, follows easily.
Lemma
l~
The design
D(B,V) is a PBIB design with two associate classes if the
following are true.
(1)
Any two sets are either first associates or second associates ••• or
m-th associates.
(2)
Each set
V (i=1,2, •••v) has
i
n
l
first associates, n2 second associates,
• • • nm m-th associates •
and
= r,
r.
~
i=1,2, •••v and k. = k
J
For any pair of
,
j=1,2, •••b.
t-th associate sets (vi,V i ,) Aii , = At' t = 1,2, •••m.
As a consequence of Bose-Clatworthy definition of PBIB designs with two associate
classes we have the following result.
Lemma 2.
The design
D(B,V)
is a PBIB design with two associate classes if the
conditions (1), (2), (4) and (5) are satisfied for m=2 and the following condi-
(3)' is satisfied instead of (3).
tion
(3)9.
The number
t
Pll (Vi,V ,) is equal for every pair of
i
t-th associate sets
(Vi,V i ,), t = 1,2.
2.
Preliminaries on quadrics in finite geometry.
.... 0:
is a prime power.
a point of
each
Xi
be the
s elements of a Galois field &F(s) where
(n+l)- tuple x
Any
-
=
(x , Xl' ••• x)
n
0
FG(n,s), the finite projective geometry of
is an element of the field and at least one
of the field, i
x'
S
= (x~, xi,
= 0,1,2,
eo. x~)
••• n.
Two (n+l) - tuples,
s
can be considered as
n dimensions, where
Xi
~
is a non null element
= (xo '
Xl' ••• xn ) and
are assumed to represent the same point if for some
000-
5
null element
of the field
p
A quadric
Q in
xi
= P Xi'
i=1,2, •••n.
PG(n,s) is the set of all points
X = (xo'xl, •••xn )
which satisfies an equation of the form
o
,
•
•
where A is
GF(s)
•
•
x
0
x'
A
=
0
a (0+1 x n+I) triangular matrix with elements from the field
is the transpose of the row vector !. The expression ! A !'
and x'
is called the form of the quadric
Q.
quadric
A
Q in
PG(n,s) is said to be
degenerate if there exists a nonsingular transformation ! = B 1 WQich transforms the form of
Q to
r
!: c " Yi y.
j~i i J
J
•
i=o
A quadric which is not degenerate will be said to be nondegenerateo' Two paiats
a = (ao , al,oo. a)
and
n
with respect to (woroto)
a point
The polar of a
If a
point a.
e
~
Q if
~(A
0
~n)
+ At)~'
~
are said to be mutually conjugate
=
O.
The polar space
which are conjugate to
0
T
(0) of
wor.t.
Q.
is the linear space determined by the equation
•
e
~
•
(A + At) !t
is a point of
If a
then the line
and
o~
~
Q,
T (0)
=
o.
is called the tangent space of
are two points of
Q is called a generator
PG(2k,s)
Q
at the
Q which are mutually conjugate,
determined by the two points 0
A line contained in
quadric in
= (~o' ~l' ••
is the set of all points
0
•
~
of
and
Q.
~
is contained in
Q.
Every nondegenerate
contains linear spaces of dimensionality (k-l) and does
not contain any linear space of higher dimensionality.
If a nondegenerate quadric in
FG(2k-l,s) contains (k-l)-flats (linear
spaces of dimensionality k-l), it 1s called a hyperbolic quadric 1n PG(2k-l,s).
If a nondegenerate quadric in PG(2k-l,s) contains (k-2)-flats and no flats of
higher dimension,
it is called an elliptic quadric in
PG(2k-l,s). N(p,n) will
6
denote the number of
~(p,k),
~l(p,k)
and
p-flats contained in a nondegenerate ~uadric in
~2(p,k)
respectively denote the number of
tained in a nondegenerate quadric in PG(2k,s),
PG(2k-l,s)
and a hyperbolic quadric in
are obtained in theorem 3 02 of L:~7e
that for every nondegenerate quadric in
PG(n,s)
p-flats con-
an elliptic quadric in
PG(2k-l,s)0 Formulas for these functions
It is proved in theorem 4.1 of f§.7
PG(2k,s) there is a point
S such
that every line through S intersects the quadric in a single point.
point S is called the nucleus of polarity of the
~uadric.
Now we prove a few lemmas which will be useful later.
simple equations for hyperbolic and elliptic
This
~uadrics
Lemma:; gives
in PG(2k-l,s).
These
simple equations are useful for writing down the blocks of the designs which
are obtained-from the incidence properties of these quadrics o
Lemma 3.
Let
Ct
Let
GF(s)
be a Galois field with characteristic not equal to
be a non-zero element of
-13 is nct
a non-zero element such that
The quadric
-a is a square and 13 be
such that
a B:lusre.' Let ~
0
~kwl in
2
2
••• + x2k _ + etx
2k =
l
is a hyperbolic nondegenerate
(2)
The ~uadr1c
be an element of
is irreducible in GF(2m) Then
m
PG(2k-l,s), s ~ 2 , with the equation
such that
(1)
GF(s)
~k-l in
o
~uadric.
m
PG(2k-l,s), s , 2 ,
with the equation
222
2
f3 x2 + x + CtX4 + ••• +
3
is an elliptic nondegenerate quadric.
m
(3) The quadric ~k-l in PG(2k-l j 2 ) with the equation
Xl +
x
1 2 + x 3x4 + ••• + x?k_l 2k
is a hyperbolic nondegenerate quadric.
X X
(4)
The quadric
=
2.
o
o
m
~k-l in PG(2k-l, 2 ) with the equation
2
2
;"(X + x ) + x x + x x4 + ....... x2k_lx2k ::
2
l
l 2
3
0
m
GF(2 )
7
is an elliptic nondegenerate quadric o
(1)
Proof"
that
It is obvious that
Q-
"2k·,l
is nondegenerate" We shall show
is hyperbolic by finite induction on
for
k=l.
~
Since
of GF(s)
is a square element of
k"
First we prove the result
GF(s), there exists an element
A
such that
•
The equation of
Q
l
is
2
2
xl + aX2 = O.
It can be easily seen that ~ contains the two points (A, 1) and (-A, 1)0
~
Since
contains linear space of dimensionality
is hyperbolic
0
Assume that
Q2k-3
is hyperbolic.
O(=k-l)
ioe"
points,
Q
l
Consider the nonsingular
transformation
Y2k-l
Y2k
= x2k _l
= x2k _l
+ Ax 2k
- Ax2k
•
It is easy to see that under this transformation
Q2k-l
transforms to
Q2k-l
with the equation
=
0
•
Since the incidence properties in a projective geometry remain invariant over
nonsingular transformations, it is sufficient to show that
Consider the point
P
of
Q2k-l
e
=
The equation of
(0 0
-rep)
Y2k = 0
Let
Then
~
.0<>
o1
0)
is obviously
"
be the (n-l)-flat with the equation
~k-l
n
'T
n~
has the equation
~k-l
is hyperbolic.
8
222
2
2
2
Yl + aY2 + Y3 + aY4 + ... + Y2k - 3 + aY2k _2 = 0
Byassumption
~t-2.
Q2k-l
The point
n
-r (l
is hyperbolic and hence contains a (k-2)-flat
1C
and the (k-2 ~-flat Ik-2
P
Q2k-l and are mutually conjugate
determined by P and
bolic
Ik-2
•
are both contained in the quadric
Hence by Lemma 2.3 of
0
is contained in
Q;?k-l.
£§.7
Hence
the (k-l )-flat
Q;?k-l
is hyper-
0
(2) We shall prove the result by induction of kG First we prove the result
for k
= 1.
The quadric
Q
l
2
in
2
xl + I3 x 2
FG(l, s)
=
with the equation
0
will be elliptic if
contains a point
Q does not contain any point. If possible, suppose
l
(xi, x2 ) Then x I o. We can easily get
Q
l
2
0
,2
xl
:xr=
2
-13
-13 is a non-square element. Hence Q
which contradicts the assumption that
l
is elliptic.
Assume that the result is true for (k - 1)
erate elliptic quadric.
so that
Q2k-3
is a nondegen-
Applying the nonsingular transformation used in part 1,
Q;?k-l with the equation
we can transform ~k-l to
..
222
2
YI + /3Y2 + Y3 + aY4 +
.. 0
2
2
+ Y2k - 3 + aY2k_2 + Y2k-I Y2k = 0
As before it will be sufficient to show that
suppose
Q;?k-l
is hyperbolic.
Then
The equation of
-rep)
.0. o I
Let
1C
be the
0)
is
Y2k = 0
0
(n-l)-flat with the equation
Y2k-1
=0
is elliptic.
If possible,
Q2k-l contains (k-l)-flats. Consider
the point
P = (0 0
Q2k-1
•
•
9
Then the equation of
~k-l
nT
(\ 1f
is
2
2
+ Y2k-3 + O:Y2k_2=
• co.
which is ell;tptic by induction assumption.
of
p-flats passing through a point
By theorem
P of a
,
°
of f~7 the number
302
nondegenerate quadric and con-
tained in the quadric is equal for every point
P of the quadric.
Since
Q~
~k-l
contains (k-l)-flats, it follows that there exists a (k-l)-flat Lk-l contained
in Q2k-l and passing through
in
~k-l
Lk-2"
n T(P)
So
~k-l
Hence
co
n
the assumption that
T(P)
Pc>
Lk-l
n
~k-l
11
intersects
Q2k-l
n
T(P)
n
is elliptic
1f
.£8_7
T(P)
contains a (k-2) -flat
1f
Parts' (3) and (4)
of part (2)..
By theorem 2.2 of
0
Lk-2"
0
Ik-l is contained
1f
in a (k-2 )-flat
This contradicts
This completes the proof'
of the theorem can be proved by arguments
exactly similar to arguments used in parts (1) and (2) respectively.
30 PBIB designs from the configuration of generators for blocks and points for
treatments of the quadric.
Definition.
a generator of the
Generator.
A line which is contained in the Qil.dr1C is called
quadric~
P and P be two points of a nondegenerate quadric ~ in
l
2
FG(n,s) such that the line P P is not a generator.. The number of points P
1 2
such that both the lines PP and PP are generators of the quadric is
l
2
N(O,n-2) where N(p,n) denotes the number of p-flats contained in a non-degen-
4
Lemma
Let
0
erate quadric of the type of
Proof co
P and
l
tangent spaces at
lemma
and
PP
2
2 0 3 of
f§.7
l
P P is not a . genevator, by lemma 2.3 of f§.7
1 2
P are not mutually conjugate.. Let T and T denote the
2
l
2
Pl and P respectively. Let P be a point of ~ such that
2
are generators of ~. Since PPl is a generator, by
P must be conjugate to Pl· Hence P must be a point of
Since the line
the points
both PP
Qn (elliptic hyperbolic) in PG(n,s)o
Tl • Similarly, P must be a point of
T2
0
Hence
P is a point of
~nTl
()T2 •
10
Conversely if
P
l
and
P
2
in
Qn"
is a point of
P
Q ('\ T
n
l
n T2 "
is a conjugate to both
P
and hence both
PP
and PP
are generators of Q " It follows
n
2
l
from the above argument that the required number is equal to the number of points
n T2 "
Tl
Since
So
Tlo
and
P
l
is a tangent space at
T
l
through
hyperbolic~
elliptic or
Lemma 5"
Let
P
l
such that both
Proof"
Let
~
P •
2
Let
Q
n
PP
Let
e
P
2
P
P
PP
Qn
in
T and
l
T
2
PG(n"s)
other than
is a point of
PP1
n
Q
n
1'1
2 ..2
of
f§7"
n 1'(I1) n
1'(~)
n
1"1
~n-2 is
T
2
I1
n 1'2~
n
= T(~l)"
Qn-4'
is a cone with
and
and
P
l
PP2
P1
and
P2
~ is (s-l)+ s2N(0"n_4).
P
l
be the tangent spaces at
~
be a point of
such that P1 P2
other than
P
are generators of
2
is equal to the number of points of
By lemma
~
Then the number of points
and
l
~ is
Q
be two points of
4" we can show that both
1\
is a nondegenerate
~n "1"1 r,1"2
P •
2
P
2
and
~
be an (n-2)-flat not intersecting the line
n-2
only if
Qn
f§7
of
not passing
is an (n-l)-flat
T
2
Hence the lemma follows
and
is a generator of
£§7"
and
P
l
does not pass through
Qn-2 in PG(n-2"s) which is elliptic or hyperbolic according
quadric
lemma
2,,1
By theorem
Pl "
-r2
are mutually not conjugate
P
2
respectively.
determined by P and
l
As in the proof
are generators of
Q
n
of
if and
Hence the required number of points
T
l
n
1'2
other than
the polar of
~l'
l
and
P "
2
By theorem 2001 of
a nondegenerate quadric in
as the vestex and
P
Q () ,.(~)
n
FG(n-4,s) and
n
~n-2
as the
base"
To count the
of
Qn
n 1'~ ~) n
Qn,n 1'(~1)"
2
s "
re~uired
~n-2
number of points" we notice that for every point
the plane determined by
P and
L).
is contained in
The number of points of such a plane which do not lie on
Hence it follows easily that the required number of points is
s (a-I) + s2 N(0,n-4).
P
S
is
11
Theorem 1. Let B
in
PO(n,s) and
a pointset.
be the class of generators of
V be the class of points of
The D(
B ,
V )
~,
~,
a nondegenerate quadric
each point being regarded as
is a PBIB design with two associate classes
with the following parameters.
v = N(O,n), b = N(l,n), k = s+l, r = N(0,n-2),
2
2
Al = 1, A2 = 0, n1 = s N(0,n-2), pil = (s-l)+s N(0,n-4) and P11 = N(?,n-2).
Proof.
It will be sufficient to check the conditions of lemma 2. It
follows easily that
b
= number
of generators of
points on a generator = s+l, v = number of points
= N(l,n), k
at
Q ;.= N(O~n) and r == ntirlber
n
.
.
of generamrs passing through a point = N(0,n-2) (theorem 3.3 of
association scheme is defined as follows.
The
Q are
n
is a generator and are second
P P
I 2
is not a generator.
P P
I 2
be at most one generator passing through two points of
A2 = O.
L:§7).
of
Two points
first associates of each other if the line
associates of each other if the line
= number
~
Since there can
we have
n'
The number of points
Let
Q
~=land
P1 be a particular point of ~.
P which
are first associates to P is equal to the number of points P such that PP
1
1
is a generator and hence equal to the number of points lying on the generators
passing through
and
Pl.
From the above argument, we have n
P2 be two first associate points.
= sN(O,n...2}.
l
P
1
The
P P is a generator of Qn.
l 2
number of points P which are first associates of both P and P2 is equal
1
to the number of points P other than P and P such that both PP and
l
2
l
PP2 are generators. By lemma 5 this number is equal to (s-1)+s2N(o,n-4).
From the above argument it follows that
Then
Let
2
pi1 = (s-l) + s N(O,n-4) ..
Let
Pl and
is not a generator.
P2 be two second associate points. Then the line P P
I 2
The number of points P which are first associates to both
P and P2 is
1
equal to the number of points P which are such that both PP and PP2 are
l
generators. By lemma 4 this number does not depend on the particular pair of
second associate points and is equal to
follows that
2
Pll
= N(O,n-2).
N(o,n-2).
From the above argument it
This completes the proof of the theorem.
12
Specializing the quadric
series of PBIB designss
=
s2t_ 1
s - 1
b
=
(s2~ 1) (62t-~ 1)
.
2
(s-l) (s+l)
,
s
=
r
= 2,
t
= 2t,
n
2t-2
-- 1
1
,
~
=1
s
,
Putting
n
Taking
v
of theorem 1, we shall get a number of
Q
,
we get the following series.
k = s+l,
,
2
pI
11
2t-4
= (6-1) + s (6 --
1)
(s - 1)
s
and
2t-2
- 1
s - 1
e
we get the symmetric series with
v ::: b ::: s 3 + S 2 + S + 1, r = k = s + 1, A ::: 1, A = 0
2
1
2
2
n1 ::: s + s , pil = s - 1 and Pll::: s + 10 This series was obtained by
Clatworthy ~§7 by a different method o
~
Taking n::: 2t - 1, t
2t-l
v :::
s
k=
s+l,
~
3, and
s
t-l
elliptic, we get the series with
s 2t-3
_
,
-1
s - 1
s
6
pil
2
Pll
:::
1) + s
2
2t-5
(6
-
( s 2t-3
-
S
t-2
S -
s 2t-3
_
6
t-l +
s - 1
S t-2
.,..
s
+
t-l
-
6
1
1
t-2
s -
1,
+ s t-2
1 )
+
t-l
(s + 1)
6
(8 _
t-l
s - 1
b=
(s - 1)2
=
s
=
r
•
,
+
1
t-3
-
1)
and
,
13
= 2t
Taking n
v
=
s
k
=
8
- 1, t
>2
2t-l + s t
and
s
Q
n
t-l
hyperbolic, we get the series with
,
1
r
s - 1
b = (s
+ 1 ,
2t-l .
s
,..
t
=s
t-3+
s t-l_
S - 1
s t-2
1
.;;;.-.--~--:::-~--...;;;...
1) ( s2t-3 + s t-l
(s + 1)
t-l
s (8 _ 1)2
s t-2
s
,
t-2
1)
,
1)
s - 1
pil
=
2
Pll
=
Putting
t
S2(S2t-5 + s t-2
s - 1
(s - 1) +
s 2t-3 +
= 2,
s t-l
s - 1
s
t-2
1
s
t-3
-
1)
and
c
in the above series, we get the family of simple lattice designs
4. PBIB designs from the configuration of points of a quadric for blocks and
generators of a quadric for treatments.
In this section we consider a nondegenerate quadric
contains lines but does not contain planes.
generate quadric in
PG(4,s) or an elliptic quadric in
nondegenerate quadric in
Lemma 6
0
Let
£1
and
~
can be a nonde-
PG(5,s) or a hyperbolic
£2 be two intersecting generators of
£1
and
Qno
Then the
£2 which intersect both
£1
and
£2
N(O,n-2) - 2.
Proof
Po
n
FG(3,s).
number of generators other than
is
Therefore,
Q in FG(n,s) which
0
Suppose the two generators in
£1
and
£2
intersect at the point
Then there cannot be any generator which intersects both
£1
and
£2 at
points other than
P.
If possible, suppose there exists a generator which
intersects
Pl
and
£1 at
£2 at
P2 •
are mutually conjugate and are points of
Then the three points
~..
By lemma
P, PI and
2 ..3 of
P2
f§7, the
0
14
plane determined by them is contained in
that
~
does not contain any planes.
tors
are those which intersect both
£§.7
3.3 of
which
number
So it follows that the required genera£1 and
£2 are two generators o
Let
£2 at the point
£1 and
Let
which intersect
P is
N(0,n-2)
£2 be two nonintersecting generators of
£1 and
P be a point of the generator
£2 at
By theorem
P.
of
Hence the lemma follows.
of generators which intersect both
.!!.9.2!.
This contradicts the assumption
the number of generators passing through
£1 and
Lemma 7.
~.
P and also intersect
£2
~2Q
Qne
Then the
is (s+l) •
Consider the generators
£10 Any such generator will lie
in the plane
2:
contained in
~, and contains a generator £1 of Qn and a point P of Qn
not lying on
£1 () From these facts it follows easily that
2 determined by
£1 and the point
P 0 The plane
is not
L
2
in a pair of lines.
Q
n
Hence there exists one and only one line passing through
P and intersecting
£1.
points of
£2
Theorem 2.
is
This is true for every point
(s+l).
Let B
L
intersects
2
£1 and the number of
Hence the lemma follows.
be the class of points of a nondegenerate quadric
which does not contain planes and V be the class of generators of
D(
B ,
Qn.
~
Then
V ) is a PBIB design with two associate classes with the following
parameters
V ::::
n l :::: (N(0,n-2) - 1) (s+l),
easily that
:I
It is sufficient to check the conditions of lemma 2
Proof"
Qn
b = N(O,n) , A. = 1, A. = 0,
l
2
2
pil:::: N(0,n-2) - 2 and Pn = (s+l).
N(l,n) , r = (s+l) ,k:::: N(0,n-2)
v
= number
of generators of
intersecting a generator = s+l, k
secting a point
=
N(O,n-2) and
b
Qn = N(l,n),
= number
= number
association scheme is defined as follows.
I'
= number of points of
of generators of
of points of
Two generators
It follows
0
Qn
Qn
inter-
= N(O,n).
£1 and
£2
The
of
Qn
are first associates if they intersect and are second associates if they do not
intersect.
Since two generators can intersect in at most one point, we have
15
~
= 1 and
= O.
~2
Let
£1 be a given generator.
which are first associates of
intersect
Thro~gh
£10
of which
£1
is one.
£2
is equal to the number of generators which
every point of
£1 and
tersect both
£1 and
£1 and
£2-
= N(0,n-2)
- 2.
nl
N(0,n-2) generators
= (N(0,n-2)
~
1) (s+l).
£2
Con-
By definition
The number of generators which are first
is equal to the number of generators which in-
By lemma
6, this number is independent of the par-
ticular pair of intersecting generators
Hence pi~
2
Pll = s+l.
there passes
£2 which are first associates.
intersect each other v
associates of both
£1
Hence it follows that
sider two generators
£1 and
£1
The number of generators
£1 and
£2
and is equal to N(0,n-2)-2.
Similarly using lemma 7, we can show that
The completes the proof of the theorem.
Specializing
in theorem 2, we shall get a number of series of
~
with two associate classes.
Taking n=5 and
Q
5
PBIB designs
an elliptic quadric, we get the
series with
= s 2 (s+l)
nl
Taking n
=3
, pil
and
Q
3
= s2
2
- 1 and
Pll
= s+l.
a hyperbolic nondegenerate quadric in
PG(3,s), we get
the series with
v
= 2(s+1)
, r
= (8+1)
2
Pll
,
k
= 2,
b
= (8+1) 2
, ~
= 1,
generate quadric with
4
=
pi
11
b
=
= ss--S -
1
1;
1 and
,
l
= s+l,
= 8+1.
This SEries is also given by Clatworthy ~17. Taking n
v
~2 = 0, n
r
2
Pll
=k = s
=
+ 1,
~l = 1, ~2
= 0,
= 4 and
nl
=
4 a nonde-
Q
s(s + 1),
s + 1.
5. PBIB designs from the configuxation of generators on generators.
Theorem 3.
Let B
be the class of generators of a nondegenerate quadric
~
16
~
which contains lines but does not contain planes c> ,·-If· a genera1:iox is C01l61-d rail as
non-.intersecting with itsell',
classes with the following
=k
v = b = N(l,n), r
D(B, V) . is a PBIB design with two associate
parameters~
~l =
1) (s+l),
= (N(O,n-2) -
= s+l,
2
n l = (N(0,n-2) - l)(s+l), pil = N(O,n-2) - 2 and
Proof",
N(O,n-2) - 2, ~2
Pll = 6+1.
Two generators are defined to be first associates if they intersect
each othero
If two generators do not intersect, they are defined to be second
associates. With the association scheme so defined, the proof of the theorem
70
easily from lemma 6 and
follows
PG(4~s),
Taking :Qn , a nondegenerate quadric in
+ 1,
v = b = s3 + s2 + s
= s(s
nl
=s
+ 1), pil
Taking n = 5 and
=k
r
- 1 and
= s(s
2
.f-
=s
Pll
we get the following series
~ = s -
1),
1,
~2
=s
+ 1,
+ 1 ..
Q an elliptic nondegenerate quadric in
5
PG(5,s), we get
the series with
v
=b
= (s3 + 1) (s2 + 1)
r
2
2
n l = s (s + 1), pil = s - 1
Taking n
v
=3
=b =
and
= k = s 2 (s +
2
and
Pll
=s
1), ~l
= s2
- 1, ~2
=s
+ 1
+ 10
Q a hyperbolic nongenerate quadric, we get the series with
3
2(s + 1)
=s
r = k
~l
+ 1,
= 0,
~2
=s
+ 1,
nl
=s
+ 1 ,
2
Pll = s + 1
c>
Now we shall give an example to illustrate the actual method of constructing the
blocks of the design.
take
v
n
=4
= b = 15,
and
r
=k
Consider a design of the series given in theorem 1. We
Q4 a non-degenerate quadric and
= 3, ~ = 1, ~2
= 0,
is obtained by taking generators of
ments.
The equation of
Q4
nl
= 6,
pil
s
= 2~
=1
The parameters are
and
Q4 as blocks and points of
can be taken as
The design
Q4 as treat-
1T"
2
o + x1x2 + x 3x4
Q4 are
X
The
15 points of
=
00
P1
= ( 00001 ),
P9 =
(11101 ),
P2
=
=
(000l0),
PIO
(00110),
(10011),
Pll
=
=
P12 =
(11110),
=
=
=
(10111),
P
3
P4
P5
P6
= (00100) ,
= (01000) ,
= (11100),
P
13
P14
P = (00101),
7
P8 = (01001),
P
15
(01010),
(11011),
(Ol1l1)g
To write down the blocks systematically we can proceed as follows o
treatment
1. The blocks which contain treatment
tors containing the point
Pl.
Consider the
1 correspond to the genera-
To find out the generators passing through
Pl ,
we find out
Q4 () ,-(P ) where ,-(P ) is the tangent space at Pl. For any
l
l
point P of Q4 I~ ,-(Pl ), PP
is a generator. In this way we can exhaust all
l
the blocks containing treatment 1. Next we proceed to treatment 2 and by a
similar procedure can find out the blocks containing treatment
2 which are not
already included. We continue in this manner until all the blocks are obtained.
P1P4 , PI P , PIP~, P~P4' P2P , P2P6 ,
5
5
P P4 , P P , P P6 , P Pll , P P12 , PaPIO' P8 P12 , P PIO' and P P <> So the blocks
3
3 5
3
7
7
9 11
9
of the designs are
In our example the 15 generators are
(1,4,7),
(1,5,8),
(1,6,9),
(2,4,10),
(2,5,11 ),
(2,6,12),
(3,4,13) ,
(3,5,14) ,
(3,6,15) ,
(7,11,15),
(7,12,14),
(8,10,15),
(8,12,13) ,
(9,10,14), and«9,11,13).
.
.
18 '
Acknowledgment.
I wish to express my thanks to Professor Ho C. Bose for
suggesting the problem and for many stimulating discussions o
HEFEBENCES
R. C. Bose, "Mathematical theory of the symmetrical factorial designs,"
~an~hy&, vOla 8 (~947), pp. 107-166.
£g7H. C. Bose, and Wo Ht. Clatworthy, "Some classes of partially balanced
designs," Anno Math., Stat., vol. 26 (1955), pp. 212-232.
£"2.7 Ho C. Bose, and K'. Ho Nair, "Partially balanced incomplete block designs,"
Sankhyg, vol., 4 (1939), 337-372.
R. C. Bose, and To Shimamato, "Classification and analysis of partially
balanced incomplete block designs with two associate classes,"
Jour. Am., Stato Assn., vol. 47 (1952), pp. 151-184.
W. Ho C1atworthy, "Contributions on partially balanced incomplete block
designs with two associate Classes," National Bureau of Standards,
Applied Mathematical Series, 47, 1956
0
£f! w
o
Ho Clatworthy, "A geometrical configuration which is a partially
balanced incomplete block design," Proceedings of the American
Mathematical Society, vol .. 5 (195 4 ), 47-55.
117 W. H. Clatworthy, "Partially balanced incomplete block designs with two
associate classes and two treatments per'block," Journal of Research
of the National Bureau of Standards, vol. 54 (l955J.
Do Kg
Ray-Chaudhuri, "Some results on quadrics in finite projective
geometry," Accepted for publication in Canadian Mathematical Journal.