0 7 8 9 1
6 1 7 8 9
5 0 2 7 8
'+ 6 1 3 7
5
0
2
7
8
9
'+ 3
3 '+
'+ 5 6
6 0 1
0
7
8
9
1
3
5
2
6
1
7
8
9
2
'+ 9 8 7 1 2
6 5 9 8 2 3
1 0 6 9 3 '+
3 2 1 0 '+ 5
7 '+ 3 2 5 6
8 7 5 '+ 6 0
9 8 7 6 0 1
3
2
9
8
3
'+
5
6
0
1
2
5 2 '+ 6
'+ 3 5 0
3 '+ 6 1
9 5 0 2
8
7
6
0
1
2
6 1
0 2
1 3
7 8
9 7
8 9
3
'+
COM BIN A TOR I A L
5
9
8
7
MATHEMATICS
YEA R
5 6 o 1 7 8 9
0 1 2 3 8 9 7
2 3 '+ 5 9 7 8
TERNARY RIN~ OF A CLAss OF LINEARLY
It
REPRESENTABLE SEMI -TRANSLATlai PlANES
by
R. C. Bose and K. J. C. Smith
University of North CaroUna at Chapel Bin
Institute of Statistics Mimeo Series No. 600.33
November, 1970
It
This research was sponsored by the NationaZ Science Foundation under GPant No.
GP-23520 and the U. S. Air Force Office of Scientific Research under Grant No.
AFOSR-68-1415C.
Ternary Rings of a Class of linearly
Jt
Representable semi-translation Planes
by
R. C. Bose and K. J. C. Smith
Univezasity of Nozath CaroZina at Chapel, HiU
PasTRACT
The theory of linear representation of projective planes was developed by
Bruck and one of the authors (Bose) in two earlier papers [J. Algebra, 1 (1964).
85-102 and 4 (1966). 117-172].
Bose and Bar10tti obtained some new linear repre-
sentations by generalizing the concept of incidence in the representation [Ann.
Mat. Pura. App Zi •,
(197 ).
].
In this paper. we show that the 6-p1anes
of Bose and Bar10tti are semi-translation planes. and obtain the ternary rings of
these planes. where the ternary function of
6
is expressed explicitly in terms
of the addition and multiplication in the Veblen-trledderburn system. coordinatizing
the translation plane T.
from which
6
can be obtained by dualization and
derivation.
This reseazaoh was sponsored by the National Science Foundation undeza Grant No.
GP-23520 and the U. S. Aiza Force Oflice of Scientifio Reseazaoh undeza Gzaant No.
AFOSR-68-1415C.
2
§ 1.
1.
A (right) Veblen-Wedderbum system (or quasif1eld) is a set of elements
R, with two binary operations + and
x satisfying the following axioms:
(i)
(R,+)
is an Abelian group with zero element
(ii)
x x 0
=
(iii)
If
R,
0 x
X
for any
0
•
(v)
(x+y)xz
If
in
R.
is the set of non-zero elements of
with identity element
(iv)
x
=
x, y, z
for all
are elements of
only one element x
R then
(R"x)
is a loop
1.
xxy + yxz
a, b, c
O.
of
R,
with
a
R such that
in
~
R.
b,
x x a
there exists one and
=
x
~. b
+ c.
We shall denote the Veblen-Wedderburn system satisfying (i)-(v) by
or in short
Let
R.
K be the subset of all elements
fx(x+y)
(1.1)
(R,+,x),
for all x, y
in
= fxx+ f)(~
R.
(fxx)-
~
f
=
Y
of
R for which
f x (xxy)
Then we know that [see for example (5»), that
field, which is defined
K is a skew
to be the left operator skew field (or kernel) of
(R,+, x) •
2.
plane
It is well known [8] that we can coordinatize any affine translation
aT
by using a suitable Veblen-Wedderburn system
(R,+,x).
The ternary
ring of T is
(1.2)
F(x,m,b)
=
x x m + b.
The points of T are ordered pairs
lines of T are of two types.
pairs
[m,b),
where m and
specified by a single element
(x,y)
where
x
and yare in
R.
The
Lines of the first type are specified by ordered
b are in
[c]
of
R.
R.
The lines of the second type are
m is called the slope and b
the
3
y-intercept of the line
of the line
[m,b]
[c]
[m,b]
of the first type.
of the second type.
The point
c
is called the x-intercept
(x,y)
is incident with the line
of the first type if and only if
(1.3)
y
The point
only if
=
"" F(x,m,b)
x "" c.
(x,y)
x x m
+ b.
is incident with the line
The affine plane
aT
[c]
of the second type if and
can be extended to a projective plane
T
in the usual manner.
We shall show that if
tained in
K and satisfies an additional condition, then we can use
ordinatize another affine plane
6,
is two dimensional over some skew field con-
(R,+, x)
0.6
which when extended to a projective plane
is a semi-translation plane in the sense of Ostrom [9].
3.
If
(R,+,x)
is two dimensional over
a1ity choose the basis of
ment of
F and
R,
and
Then any element
x
in
in
is
x
To each element
x
F,
and x i
2
with coordinates from
(1.4)
Rover
i
F
F,
to be
(1,0)
are in
F.
of
R,
1
x
= x 1+x2i
is the unit e1eR,
not in
where
Xl' x
there corresponds the binary vector
This is a
F.
2
are
(1,1)
correspondence.
(x ,x )
l 2
If
x+y
of
Rare
"" z,
and
=
(0,1) •
F.
where
x i.
""
and
i
is a suitably chosen element of
R can be expressed as
2
we can without loss of gener-
1,
The vectors corresponding to the elements
m*2
R to co-
and
We shall say that the
Am
=
0, 1
and
im "" m* 1+tn* 2i,
2x2
matrix
i
where
(0,0),
ml , m2 , m*l'
4
corresponds to the element
responding to
R.
The first row of
A
m
is the vector cor-
m,
and the second row of A is the vector corresponding to im.
m
C be the collection of all 2x2 matrices over F, which correspond in this
Let
way to the elements of
F.
m of
Now if
y
R.
Then
C is a subset of the set of
2x2
matrices over
= xxm,
=
(1.5)
=
=
=
Hence the vector co-responding to
therefore alternatively represent
addition is the
or~inary
xxm
(R,+,x)
(x m +x2m*1' xl m2+x2m*2)' We can
l l
by binary vectors over F where the
is
vector addition given by (1.4) and multiplication is de-
fined by
=
(1.6)
In particular
=
The vector corresponding to the product
of the vector corresponding to
write the elements of
4.
(1. 7)
R as
x
xxm
is the ordinary matrix product
and the matrix corresponding to
x +x i
l 2
or
(x ,x )
l 2
m.
We shall
according to convenience.
A 2x2 matrix
U
=
with elements from the skew field
F is defined to be singular, if its row vec-
tors are independent from the left, i.e. there exist elements
r , r
l
2
of
F,
5
not both zero, such that
U is defined to be non-singular if
U is not singular.
We state here, for
subsequent use, the following two obvious Lemmas.
LEf+1A G.l).
The column vectors of the matrix
pendent from the right, i.e., there exist elements
U,
sl
given by (1.7) are deand
s2
of
F,
not both
zero, such that
if and only if
U is singular.
In other words, the dependence
(ind~pendence)
of the row vectors of U from
the left is equivalent to the dependence (independence) of the column vectors of
U from the right.
CoROUARY.
If
U is non-singular, it cannot have a null row or a null
column.
LE~
G.2).
If the matrix U given by (1. 7) is non-singular, then the
equations
...
have a unique solution
x
1
,Xz
and the equations
...
have a unique solution
5.
The collection
...
Y1' Y2.
C of 2x2 matrices corresponding to the elements of R
has the properties given by the following Lemma:
6
~
(b)
m m
(d)
If
A
m
If
m
(Y l ,Y 2)
(e)
belong to
C.
appears as the first (second) row of a mique
If
are two distinct matrices of C then their difference
and A ,
m
In particular, if A
m
(x ,x ) ; (0,0)
l 2
mique matrix A
(1.8)
R,
I
C.
is non-singular.
A -A ,
Am
The null matrix Q and the mit matrix
(a)
Each binary vector of
matrix of
(c)
0.3).
of
(YpY )
2
then A
m
is non-singular.
are binary vectors of
R,
there is a
C such that
=
(x l ,X2)Am
(Y ,Y )
l 2
and
(x ,x )
l 2
=
I~
(Xl,x2)~*1
(m ,m ) ; (0,0)
l 2
is the mique matrix of
unique vector
and
0,
~
of
m~
m*~·
are given binary vectors of
C whose first row is
(m ,m ),
l 2
R for which (1.8) holds.
R and
then there is a
The property (a) is obvious since the null matrix
corresponds to the zero element of
corresponds to the unity of
R.
R,
and the unit matrix
The first part of (b) follows from definition.
The second part follows by noting that if the vector
from
F is given, and
termines the element
m*
= m*1+m*2i,
m = m +m i
l 2
of
(m*1,m*2)
then the relation
R.
im
with elements
= m*
uniquely de-
Since (1.8) can be written as
(d) and (e) follow from axiom (iii), which states that the non-zero elements of
form a loop with respect to multiplication.
property (c).
Let
We thus only need to prove the
R
7
=
A
m
A
m
be two distinct matrices belonging to
m'*~
~
(m ,m )
l 2
is singular, then by definition there exists a vector
Am-Am'
with coordinates from
F,
C.
m'~
=
t
From (a),
(m'l,m'Z).
If
such that
(1.9)
which is equivalent to
=
(1.10)
Since
(r ,r 2)
l
~
(0,0),
which is a contradiction.
(ml'~)
it follows from (iii), that
= (m'1,m'2)
This proves (c) and completes the proof of the
Lemma.
6.
We shall now assume that the Veblen-Wedderburn system
dimensional over the skew field
(kernel)
(vi)
K of
R,
R which is two
F contained in the left operator skew field
satisfies the following additional axiom:
For any given elements
there exists a tmique element
m
x
Q
Xl + x2i ~ 0
Q
~
+ m2i
of
and
y
= Yl + Y2i
of
R,
R, such that
(1.12)
l~ere
im
= m*1+m*21,
and
xl,x2'Yl'YZ,ml,m2,m*1,m*2
are in
F.
and
(Yl'Y2)
Axiom (vi) may alternatively be written as:
(vi,a)
given the binary vectors
ordinates from
(1.l3)
Am[:~]
F,
(x ,x )
l 2
~
(0,0)
there exists a unique matrix A of
m
=
~lm* m~ [:~]
1
=
[~~] .
with co-
C such that
8
LEr1'V\ 0.4).
Given any binary vector with coordinates from
as the first (second) column of a unique matrix
A
m
of
C,
F,
it appears
if axiom (vi) is
satisfied.
Let the binary vector be
"1
= Yl ,
m*l· Y2°
(Yl'YZ).
In (1.13), choose
Hence there is one matrix of
xl
= 1,
z = 0.
Then
X
C with the first column
[~~]o
There cannot be two such matrices, since their difference would be singular from
the corollary to Lemma (1.1), contradicting Lemma (1.3), part (c).
choosing
Xl
= 0,
second column is
7.
x2
= 1,
Similarly by
it follows that there is a unique matrix of
[~~]o
In view of Lemma (1.3) part b, we can define the functions
elements
C whose
of
F
F , F2 of
l
by setting
(1.14)
where
is the uniquely defined matrix of
C,
whose first row is
(Xl ,x 2) •
Similarly, when axiom (vi) is satis fied, we can define functions
elements
=
AX is the uniquely defined matrix of
LEfIto4A U.S).
to
F
(1.16)
of
F by setting
of
(1.15)
where
Gl' G2
C whose first column is [Xl
x*l ] •
Ifaxiom (vi) is satisfied and m , m :f: 0, b l , b 2 belonging
l
Z
are given, and
=
then anyone of the pairs
(x1'Yl)' (x 2 ,YZ)
uniquely determines the other.•
9
The relation (1.16) is equivalent to
=
(1.17)
where
x*1+x*2i
= i(x1+x2i),
i.e. the matrix on the right in (1.17) belongs to
Hence
C.
(1.18)
and
and therefore from Lemma (1.4)
x*
2
are uniquely determined if
the second equation in (1.18).
x 'Y1
l
are given.
Then
In terms of the functions
Y2
is determined by
F1 , F2 , G1 , G2
given
by (1.14) and (1.15) we can write
(1.19)
(1.20)
In the same way. we can show that if
x , Y2
2
are given, xl' Y1
are uniquely
determined.
8.
We note that if
F is finite, i.e.,
F is a Galois field
GF(q),
then
axiom (vi) is automatically satisfied.
This follows by observing that the right
hand side of (1.13) can assume exactly
q
matrices in
C.
If for some
must exist distinct matrices
(Yl'Y2)
Amr
and
2
values and there are exactly
there is no corresponding
AmII belonging to
Hence
(Am,-AmII) [Xl]
x
2
= o.
A,
m
C such that
q
2
then there
10
,x ) ~ (0,0), it follows from Lemma (1.1), that A ,-A" is
l 2
m m
singular, which contradicts Lemma (1. 3), part (c). Hence when F is finite, and
Since
(Xl'~) ~
(X
(O,O)
and
(Yl'Y2)
are given, there is a unique
A
m
c
belonging to
for which (1.13) is satisfied.
In §3, we shall give a geometrical interpretation of axiom (vi).
1.
sidered in §l, to obtain an affine plane
with the ternary ring (1.2).
a6
mensional over the skew field
the basis elements
is not in
F.
1
and
i,
As
We shall assume that
1
a6
before, we take
F contained in the kernel
where
con-
different from the affine plane
It will be shown in §5 that
lation plane in the sense of Ostrom [9].
i
(R, +, x)
We shall now use the (right) Veblen-Wedderburn system
aT
is a semi-transR to be two di-
K of
is the unit element of
R,
and having
F and
R,
and
R satisfies the standard axioms (i)-(v)
and the additional axiom (vi).
2.
points,
Let us consider an incidence structure
L
is a set of lines and
1
[P, L, I] where P is a set of
is an incidence relation between points and
lines, defined as follows:
Each point of
P is coordinatized by an ordered pair of elements of
The coordinates of a point can be taken as
y2 are elements of
F.
l,fu.en
{xl+Yli, x2+Y2il,
F is the Galois field
GF{q) ,
where
R.
xl' Yl' x2 '
there are
q
4
points.
Lines of
L are of several types. Each line of type 1 is coordinatized by
an ordered pair of elements of
R,
the first of which does not belong to
The coordinates of a line of type 1 can be taken as
IDz
~
O.
[m +m i, b +b 2i],
1
l 2
F.
where
Lines of type 2a are coordinatized by an ordered pair of elements of
R,
11
the first of which belongs to
taken as
ment of
[II],' b l +b 2i].
q
2
The coordinates of a line of type 2a can be
Each line of type 2b is coordinatized by a single ele-
Thus a line of type 2b has a coordinate
R.
Galois field
and
F.
GF(q),
there are
q4_q3
[c +C i]. When F is the
1 2
lines of type 1, q3 lines of type 2a
lines of type 2b.
The line
[m +m i, b +b i]
l
with the point
The line
l
of type 1,
2
{x +Yli, xZ+Y21}
1
[m , bl+bZi]
0,
is
defined to be incident
if and only if
if and only if
[c +c i]
l 2
{XI+Yli, x2+Y21}
roz:ls
of type 2a is defined to be incident with the point
l
{xI+Yli, x 2+YZi}
The line
2
of type 2b is said to be incident with the point
if and only if
We shall show that the incidence structure
(P,L,J)
so defined is an affine
plane.
We have already noted that the elements of
sented by binary vectors over
The point
{(xl+Yli),(xZ+Y2i)}
[m +m i, b +b i] becomes
l 2
l 2
the line [cl,cZ].
(2.4)
through the mapping
then becomes
[(ml'm ), (bl'b 2 )],
2
The condition for the line
cident with point
F
R can altematively be repre-
{(xl 'Yl)'(x2 ,yZ)}.
and the line
[(m ,roz),(b l ,b 2)],
1
{(X 1 'Yl),(X2 'Y2)}
(X +x i)
l 2
then becomes
~
(X ,x ).
l 2
The line
[c l +c 2i]
becomes
m2 :1s 0 of type 1 to be in-
12
(2.5)
belongs to
C.
The condition for the line
the point
{(x l 'Yl),(x 2 'Y2)}
[(m ,0),(b ,b )]
l
l 2
becomes
The condition for the line
point
{(x l 'Yl)'(x2 ,yZ)}
(2.7)
Two lines
[cl'cZ.l
of type Zb to be incident with the
becomes
y
3.
of type 2a to be incident with
1
=
[(m l ,m2),(b ,b )]
l Z
and
[(m'1,m'2),(b'1,b'Z)]
both of type 1
or both of type 2a will be defined to belong to the same parallel class if and
(ml'~)
only if
c
(m'l,m'2).
We may, therefore, speak of parallel classes
is said to be type 1 if m + 0, and type
2
All lines of type Zb are defined to belong to the same parallel
(m ,m ).
l 2
The parallel class
2a if m
Z
= O.
class which may be denoted by
the same parallel class.
(m ,m )
l 2
(\10).
Two lines of different types do not belong to
The two succeeding Lemmas provide the justification for
this nomenclature.
LEr-t\t\ <2.1).
Two distinct lines of the same parallel class are not both in-
cident with the same point.
Case I.
Let
[(ml ,m2 ),(b ,b 2)]
l
and
[(ml ,m2),(b'l,b'2)]'
distinct lines of type 1 belonging to the parallel class
they are incident with the same point
(bl,b )
Z
= (b'l,b'Z).
{(xl 'Yl),(x Z'Y2)}'
(ml,mZ)
m2
+0
be two
of type 1.
then from (Z.4),
This is a contradiction since the two lines are distinct.
If
13
[(m ,0),(b l ,b'2)] be two distinct
l
l
lines of type 2a belonging to the parallel class (m ,0). If they are incident
l
with the same point {(x l 'YI),(x2 'Y2)}' then from (2.6), b l • b l l , b 2 a b'2'
which is a contradiction.
Case II.
Let
Case III.
[(m ,0),(b ,b )]
l 2
l
and
From (2.7), two distinct lines
cannot be incident with the same point
lBJMA <2.2).
[c~,c2]'
[c l ,c'2]
l
of type 2b
{(xI'Yl)'(~'Y2)}.
Two lines belonging to different parallel class are both in-
cident with a single point.
Case I.
and the line
It follows from (2.6) and (2.7) that the line
[(m ,0),(b ,b )]
l
l 2
[c ,c ]
l 2
of type 2b,
of type 2a are both incident with the unique point
{(cl,cZ),(clml+bl' cZml +b 2)}·
Case II.
If
[c ,c2 ]
l
is a line of type 2b, and
is a line of type 1,
then if
lines,
C
xl = c ' Yl =
l
z
{(xI'YI)'(xZ'YZ)}
and (Z.4) is satisfied.
[(m ,m ),(b ,b 2)],
l
l 2
m2
~
°
is a point common to the two
It follows from Lemma (1.5),
that there is a single point with which both lines are incident.
Case III.
Let
[(m1 ,mZ),(b 1 ,b Z)]'
[(m'l,m'Z),(b'l,b'Z)]'
m2 1 0,
be two given lines of type 1, not belonging to the same parallel class.
(ml,IDz) 1 (m'l,m'2).
If
{(x ,y ),(x 'Y )}
2 Z
1 I
m'z r/
0,
Then
is a point incident with both lines,
it follows from (Z.4) that
(Z.8)
(0,0)
If
x
2
=0
a
(m1-m'l'
~-m'Z)[:;l
x::J + (bl-b'l' bZ-b'Z)'
(bl,b ) = (b'1,b'2) then from (Z.8) and Lemma (1.3) part (d), xl = 0,
Z
is the only solution of (Z.8). Hence from (2.4), {(0,0),(b ,b 2)} is
l
the unique point incident with both lines.
(bl,b ) 1 (bll,b'z)' then from (Z.8), (xl,xZ) ~ (0,0). Hence from
Z
Lenna (1.3) part (c), the matrix (Z.5) is non-singular. Hence from Lemma (1.3)
If
part (d),
(xI,x )
Z
is uniquely determined.
Then
(Yl,YZ)
is determined by
(2.4~
14
Thus there is a unique point
Case IV.
Let
{(xl,Y1) , (x ,Y2)} incident with both lines.
2
[(m l ,0),(b l ,b 2)], [(m'l,O),(b'l,b'2)] be two lines of type
Then m ~ m'l' If {(xl,y l ),
l
is a point incident with both lines, it follows from (2.5), that
2a, not belonging to the same parallel class.
(x 2 'Y2)}
(2.9)
x2
Hence
Xl
=
= -(bl-b'l)(ml-m'l)-l,
are uniquely determined by (2.9).
Case V.
Let
[(m l ,m2),(b l ,b 2 )],
[(m'1,0),(b'1,b'2)]
m
2
~
0 be a line of type 1,
be a line of type 2a.
If {(x l ,y ,(Xz'Y 2)}
l
cident with both lines, it follows from (2.4) and (2.6) that
(2.10)
Yl
=
(2.11)
Y2
= ml x2 + rnzx*2 + b 2 ,
(2.12)
x2
=
xlm'l + b'l'
(2.13)
Y2
=
ylm\ + b'2'
Let).
and
is a point in-
mlxl + m2 x*1 + b l ,
= m2-l(blm\-mlb\-b2+b'2)'
find tmique matrix Ax
belonging to
From axiom (vi,a) formula (1.13), we can
Ct
such that
(2.14)
Then (2.12) is satisfied, and
and (2.11).
(2.10) by
Y 'Y2 can be uniquely determined from (2.10)
l
We shall shaw that (2.13) is automatically satisfied. Multiplying
-m'l
from the right and adding (2.11) we have
15
Using (2.14) and substituting for
A,
this reduces to (2.13).
Thus there is
exactly one point incident with both the given lines.
l.a+1A <2.3>,
4.
Given any point
{(x 'Yl)'(x2 ,yZ)}'
l
there is exactly one
line in any given parallel class incident with the point.
Case I.
In the parallel class
there is only one line, viz
Case II.
lel class
[x 'Yl]
1
[(m ,O),(b ,b 2)]
l
l
Let
(IDI,O).
(~)
consisting of all lines of type 2b,
which is incident with the given point.
be a line of type 2a belonging to the paral-
If it is incident with the point
(b ,b ) is uniquely determined by (2.6).
l 2
Case III. Let [(m ,m ),(b ,b )], m
2
l 2
l 2
the point
{(x I 'Yl),(x 2 'Y2)}'
lEMMA <2.4),
(xI,Y I )
~
(x'l,Y'l)'
(2.15)
(2.16)
Then
For given
~
0
{(xl,Yt'~2'Y2)}'
be a line of type I incident with
is uniquely determined by (Z.4).
(bl,b Z)
(xI'YI)' (x Z'Y2)' (x'l,Y'I)' (x'Z,y'2)'
where
the equations
=
=
in which the matrices appearing on the right belong to C,
uniquely determine
if and only if the matrix
(2.17)
is singular.
From (2.15) and (2.16) we have
(2.18)
then
=
~
,
Xl - x 1
x* 1 - x'* 1
16
(x l ,x2)
Now
~
(x'1,x'2)'
the hypothesis.
otherwise
contradicting
determined is zero,
From Lemma (1.2),
m, m
2
l
are uniquely deter-
is determined from either (2.15) or (Z.16).
(b ,b )
l 2
Then
= (x'l,Y'l)
From Lemma (1.3), part (c), the matrix appearing on the
Light in (2.18) is non-singular.
mined.
(xl,Y l )
If mZ so
then
(Z.19)
Then by definition, the matrix (2.17) is singular.
Conversely, if (Z.17) is singular, we can find
Since
(xl,Y l )
~
If we set
Hence (2.18) is satisfied with
satisfied.
determined by (2.18),
uniquely determine
5.
r Z ~ 0.
(x'l,y'l)'
{(x'l,Y'l)'(x'Z'Y'Z)}
Let
2
m1
~
(0,0)
= -r Z-1 r 1 ,
such that
(Z.19) is
m = 0.
Since (ml,m ) is uniquely
2
Z
Thus in this case (Z.15), (Z.16)
must be zero.
(ml,mZ)' (bl,b Z)
lEMMA (2.5).
Case I.
m
(r ,r )
1 Z
with
m2
Any two distinct points
= 0.
{(xl,y1),(xZ'YZ)}
and
are incident with exactly one line.
(x l ,Y 1 )
= (x'l,y'l)'
clearly incident with both points.
Then the line
If the line
[xl,Y I ]
of type Zb is
[(m ,m ),(b ,b »),
l 2
l 2
IDz
~
°
of
type 1 is incident with both points, then from (2.4) and Lemma (1.5), the two
points would coincide.
Similarly, if the line
[(m ,O),(b 1,b 2)]
l
of type 2a is
incident with both points, then from (2.6) the two points with coincide.
the line
[x ,Y ]
l 1
Case II.
of type 2b is the only line incident with both points.
Suppose
(xl,Y )
l
From Lemma (1.1) we can find
(2.20)
Hence
~
(x'l,y'l)'
(sl,s2)
~
(0,0)
and the matrix (2.17) is singular.
such that
17
Now
s2
hypothesis.
is non-zero,
-1
Let
ml = -sIs 2' b l
[(m ,0),(b ,b )]
l
l 2
Then from (2.6),
points.
othen~ise
(xl'Yl)
= x2-x l ml
= (x'l,Y'l)
= x'2-x 'lml ,
contradicting the
b2
= y 2-y l ml
= y'2-Y'lm l •
is the only line of type 2a incident with both
From (2.7) there is no line of type 2b incident with both points.
If the
°
{(m ,m ),(b ,b )}, m ~
of type 1 is incident with both points, then
2
l 2
l 2
from (2.4), equations (2.15) and (2.16) are satisfied. Hence from Lemma (2.4),
line
(m ,m ), (b ,b 2)
l 2
l
are uniquely determined with
m
2
=
° which is a contradiction.
Thus there is exactly one line which is incident with both the points.
It is of
type 2a.
Case III.
singular.
Suppose
(xl'Yl)
~
(x'l,y'l)
and the matrix (2.17) is non-
From Lemma (2.4), we can uniquely determine
satisfying (2.15) and (2.16).
~ ~
(m l ,m2), (bl,b Z)'
Hence from (2.4) there is a unique line
°
[(ml ,m2),
(b ,b )] of type 1 incident with both points. From (2.7) there is no line of
l 2
type 2b incident Witll both points. Again, if there is a line of type 2a incident
with both points then from (2.6), the equation (2.20) i9 satisfied with
9
1
= ml ,
= -1. Hence from Lemma (1.1), the matrix (2.17) is singular, which is a conZ
tradiction. Thus there is exactly one line incident with both points. It is of
9
type 1.
7.
THEOREM
<2.1).
The incidence structure
[P, L, I]
defined in §2,
paragraph 2, is an affine plane.
In view of the Lemmas (2.1), (2.2), (2.3), (2.5) it is sufficient to verify
the existence of three non-collinear points.
(1,0)}
point
The points
are from (2.6), incident with the line
{(l,O),(O.O)}
{(O,O),(O,O)}, {(O,O),
[(1,0),(0,0)]
a~-points,
and its lines will be called
to a projective plane
The
is not incident with this line.
The affine plane of Theorem (2.1) will be denoted by
called
of type 2a.
~
in the usual manner.
a~-line9.
ao.
Its points will be
It can be completed
Corresponding to each parallel
18
(m ,m )
l 2
class
of type 1 or 2a, we take a point at infinity denoted by
Similarly corresponding to the parallel class
which consists of all a8-lines
(=)
{=}.
of type 2b, we take a point at infinity denoted by
[=].
at infinity denoted by
{m ,m }.
l 2
Finally, we take a line
Each point at infinity is defined to be incident
with the line at infinity, and all lines of the parallel class corresponding to
it (and with no other line).
Thus 8-points consist of a8-points and the points
at infinity, and 8-lines consist of a8-lines and the line at infinity.
3.
§
1.
Let
F be a skewfield.
F,
space based on
operators.
84 be a four dimensional projective
Let
with
8
specified basis
V.
V.
Points of
8
4
and write down
correspond to one
Thus the coordinates of a point
P which
are elements of
where
are
not all zero.
8 ,
4
in the usual manner, by choosing a
4
of
dimensional vector subspaces of
F
The coordinates of a point are arbitrary up to a non-zero multiple
of an element of
F
from the left.
(rxl , rx2 , rYl' ryZ' rz)
=0
~lal+~2aZ+nlbl+n2bZ+~c
and only if
Thus if
r
~
0,
then the point with coordi-
is identical with
(where
all zero) represents a 3-space
8
3
aI'
8
P.
then the equation
A linear equation
2, b l , b2, c
are elements of
which contains the point
=
0.
F not
(xl,xZ'Yl'YZ'z)
if
The equation of a 3-space is
arbitrary up to a non-zero multiple of an element of
k 1: 0,
F as a ring of left
We can then assign coordinates to the points of
the equations of linear subspaces of
nates
V,
regarded as a vector space
F
from the right.
Thus if
19
also represents
8 •
3
Dependence and independence of vectors or equations can be
defined in the usual manner with the convention that multiples of vectors are to
be taken from the left and multiples of equations from the right.
Dependence or
independence of points is defined by the dependence or independence of the corresponding coordinate vectors.
corresponds to a
t+l
A
t
dimensional subspace
dimensional vector subspace
5
veSt)
of
t
of
54
V,
and may there-
fore be defined as the set of points dependent on a given set of
points corresponding to a basis of
by a set of
4-t
veSt).
independent equations.
Let
S.
line of
A plane in
E
S
can be represented
St
E
S
if and only if
is defined as a set of
is contained in exactly one
cannot contain more than one line of
F
S
(otherwise
S is called a dual spread if each
can be taken as the Galois field
must necessarily be a dual spread.
may not be a dual spread [7].
then every other line
sao;
independent
E contains at least (and therefore exactly) one line of S.
a finite space,
spread
A spread
The spread
t+l
i.e., the point is contained in
such that every point of
I:,
they would intersect in a point).
plane of
St'
E be a fixed 3-space of 54.
lines contained in
St
A point belongs to
its coordinates satisfy the equations of
2.
Alternatively
(0 S t S 4)
s
of
S
S.
s ao
S
of the plane, not in
In the finite case,
S
is
mayor
S,
belonging to
meets the plane in a unique point
S
S4
In this case, the
In the infinite case,
If a plane contains the line
and conversely through every point
passes a unique line of
GF(q).
If
S,
soo'
contains exactly
not in
there
q2+l
lines.
For further properties of spreads, see references [4], [5], [6] and [12].
3.
Let us identify
F,
Tl1ith skew field
F
of §l, which is contained in the
(R, x, +)
left operator skew field (kernel)
K of the Veblen-Hedderburn system
which is two dimensional over
and which satisf1.es the standard axioms
F,
20
(i)-(v).
then
We shall continue to use the notation of §1, i.e., if
im· m* 1+m* Zi,
m , m2 , m*l' m* Z are in F. We shall also continue
1
x1~i of R,
where xl and X
are in F, with the
(x ,x ).
1 Z
be distinct points of
a point
z
E , EZ ' E'l' E'l
1
Let
m1~i,
where
to identify the element
binary vector
m=
S4'
be four independent points of
and let
L where the line
not contained in
Uo which is not contained in any of the planes
ElE'lE'Z' EZE'lE'Z.
E,
We can now choose a basis of
EOU
EO
and
meets
E in
ElEZE'l' E1EZE'Z'
V so that the coordinates of
E , E , E'l' E'Z' EO and U are (1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0),
2
l
(0,0,0,1,0), (0,0,0,0,1) and (1,1,1,1,1) respectively. Then the equation of
is
~
= O.
by
TI
and
and
s 00
TI*
Let the planes
TI*
respectively.
by the equations
= E' 1E' Z'
in
soo'
= 0,
~l
EZE'lE'Z'
= O.
r,;
TI
and TI*
~l
soo.
= 0,
Similarly, if TI o*
(O,1'Y*1'Y*2'0).
(1,0,1,0,0)
E be denoted
~2
= 0,
~
=0
contain the common line
~2 •
0,
~
= O.
Let
TI o be
Then the E;l-coordinate of any point
is the set of points in
then the coordinates of any point y*
coordinates
E
Hence the coordinates of Y can be written in the cannon-
(1,0'Y1'YZ'0).
nonical form
contained in
is represented by the equations
TI which are not in
TI o is non-zero.
ica1 form
TI
and
represented by the equations
the set of points in
Y in
ElE'lE'Z
U
and
in
In particular, let
U*1
the point of
IT O*
TI*
but not
can be written in the can-
U be the point of
1
ITo
with
TI o* with coordinates
(0,1,0,1,0) •
(3.1)
be the unique matrix of
element, m of
(l,O,m ,IDz'O)
1
C,
corresponding to
m.
R then corresponds a unique point
and a unique point
M*
of
Then
im
= m*1+m*2i.
M of ITo
To each
with coordinates
TI o* with coordinates
(O,l,m*l,m*Z'O).
21
To any element
m of
we shall denote by
R,
then
M and
R,
m
become
to the zero element of
we choose
U*
l
In particular if we choose
s •
~*
therefore corresponds a unique line
R
1
is thus
E E
1 2
m to be the unit element of
respectively.
The line corresponding
which can be denoted by
R,
then
M and
sO.
1
and
R is thus
sl.
So be the set of all lines corresponding to different elements of R,
Let
S be obtained from So by adjoining the additional line
and let
Again if
become U
M*
The line corresponding to the unit element of
U1U*1 which may be denoted by
\ u __~l
:..)
m to be the zero element of
E respectively.
Z
and
E
1.~1
1:"'::-."
s~.
We shall
S is a spread.
show that
Clearly
s
m
cannot intersect
s~,
=
(3.2)
and is represented by the equations
=
...
0,
which may be written as
(3.3)
~
=
If
s,
m
l
m~
m* 1 m* 2
is another line of S,
=
z;
corresponding to the element
s ,
then the equations of the corresponding line
...
(3.4)
where
A ,
m
is the matrix of
(3.5)
(a ,a ,b ,b ,C)
1 2 l 2
=
m
of
S
c
corresponding to
z.;
m'.
m'
~
m of
R,
are
=
cannot have any point in connnon.
ordinates
= O.
= 0,
The lines
If there is a common point
must satisfy both (3.3) and (3.4).
P,
s m and
then its co-
Hence
c'" 0,
and
22
Hence
=
(3.6)
Since P is not in
(0,0).
s 00 ,
singular, which contradicts Lemma (1.3) part (c).
lines of S
s.
We have thus shown that any two
= (0,0),
r with coordinates
be any point of
Q
then
(Y1' Y2) ~ (0,0)
and
Q
It cannot be contained in any other line
s
nates would satisfy (3.3), making
m
of S,
= (0,0).
(Y1'Y2)
If
m
coordinates of
Q satisfy (3.3).
which contains
Q.
This completes the proof that S
(R, +, x)
E.
If
s 00
of
otherwise its coordi-;
(x ,x )
l 2
~
of
such that the
C,
Hence there is a unique line
We have now shown that starting from
sponding spread S in
(x l ,x2 'YI'Y2'0).
is contained in the line
from Lemma (1. 3) part (d), there is a unique matrix A
S,
is
are non-intersecting.
Again, let
(x ,x 2)
l
Am-Am,
Hence the matrix
s
(0,0),
m
then
belonging to
is a spread.
we can construct a corre-
This is a special case of a result proved in [5].
Note that the necessary and sufficient condition for the point
to be contained in the line
m
of S
is
=
(3.8)
where
A
m
4.
S
s
is the matrix of
C corresponding to m.
We shall now show that
R will satisfy axiom (vi) of
n, precisely when
is a dual spread.
Let
S be a dual spread and let
nm be the plane represented by the
equations
(3.9)
1;
=
O.
23
then
a unique line
sm of S,
contains the points
where
A
m
nm does not contain sw. Hence it must contain
corresponding to the element m of
M and M* with coordinates
given by (3.1) is the matrix of
(l,O,m ,m ,0), (O,1,m*1,m*2'O),
l 2
C corresponding to m. Hence
=
(3.10)
(x ,x )
l 2
+ (0,0)
there exists a unique matrix
Am of
Hence given the binary vectors
nates from
R. Therefore nm
F,
and
(Yl'Y2)'
C,
such that
with coordi-
(3.11)
We have thus proved the alternative form of axiom (vi).
(vi) or alternatively (vi,a) holds, and
sented by the equations (3.9), then if
SQO
of S.
nm is an arbitrary plane of t,
(x ,x )
l 2
= (0,0),
Otherwise (3.11) determines a unique matrix
tains the corresponding line
s
m
Conversely if axiom
nm contains the line
A
m
of
C,
of S.
Note that the necessary and sufficient condition that the plane
=
contains the line
(3.12)
s
m
of the spread S
z;
is
=0
repre-
and n
m
con-
24
We have shown in
1.
§)
how to construct a spread S
Veblen-Wedderbum system
(R, +, x)
contained in its kemel.
This process is reversible.
Let
E
S4 meeting E in the plane
sco.
Let
a spread in
Let
E.
Sco be a
S, and let So be the set of lines of S other than sco.
be a plane contained in
TI
and passing through
E,
E'l' E'Z
Let
TI.
E1 be any point of II not in
sco'
Let
through the points
E
l
and
U
1
respectively.
and
which intersects
sco'
and
Q
a)-space
be any two distinct points on
and
U
l
any point of the line
and
sl
be the lines of S
passing
l
passing
There is a unique line
Let
be the plane containing i
through
E'Z
and
sco'
and
s1
E and U*
be the points of intersection of l with So
l
1
respectively. Let EO by any point of Q not in E, and U any
o
II*
and let
point of
sl
the line
EOUO distinct from
distinct from U
l
such that the points
and
and
EO
Finally, let
U*l"
U be a point of
He can then choose a basis of
UO•
respectively.
sm
Then the equation of
SCO and the coordinates of
is any line of SO'
which
U
l
n is
and
~2
U*l
= 0,
TI
and
II*
and
are as in paragraph 3 of §3.
are
M and M*
(l,O,m ,m2 ,O)
1
respectively, then we can associate with the line
(4.1)
A
m
m1
= m*
~
1
m~ •
m*
2
V,
(1,1,1,1,1)
and the equations of
and if the coordinates of the points
sm intersects the planes
S4 on
and U have the coordinates
E , E ' E' l' E'Z' EO
Z
1
(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)
II*,
F
S4 be a four dimensional projective space based on the skew field F,
particular line of
in
which is two dimensional over a skew field
a particular 3-space contained in S4' and S
Let
corresponding to a
E,
TI,
If
in
and
s
m
the matrix
25
Let
C be the set of all possible matrices Am corresponding in this way
to all possible lines of SO'
C satisfy the properties
Then the matrices of
(a)-(e) of Lemma (1.3).
The property (a) is obvious since the null matrix 0
and the unit matrix
for any binary vector
(l,O,ml,mZ'O).
Let
corresponding to
row.
If A
m
A -A
m m'
sm be the line of So
and A,
s
m
m
C,
(m*l'm*2)
with
of
n with coordinates
then the matrix
(ml,mZ)
A
m
as its first
occurs as the second row of a
C, then equations of the
are given by (3.3) and (3.4) respectively.
and
we can find
(al,a )
Z
be determined by (3.5).
s
m
and
of S are non-intersecting.
A
M,
through
are two distinct matrices of
m
is singular,
(b ,b )
l 2
M
C.
the equations of both
then
there is unique point
sm is the unique matrix of
corresponding lines
Let
(m ,m )
l 2
Similarly, any binary vector
unique matrix of
For (b), we observe that
corresponds to the line
I
corresponds to the line
~
(0,0)
If
such that (3.6) is satisfied.
Then the point
(a ,a ,b ,b 2 ,O)
l 2 l
satisfies
This is a contradiction since any two lines
This establishes (c).
In particular, if
A = 0,
m
is non-singular.
The property (d) follows because given
is a unique line
sm of So
(x ,x )
l 2
~
(0,0)
which passes through the point
and
(Yl'Y2)
there
(xl,xZ'Yl'YZ'O).
The property (e) is equivalent to the assertion that the equations
have a unique solution
(x l ,x 2), given (m ,m 2)
l
from Lemma (1.2) and the non-singularity of A.
~
0
and
(Yl'YZ).
This follows
m
We can now take
sider the system
(4.Z)
R to be the set of binary vectors
(R, +, x)
(ml'm2)
over
with addition and multiplication defined by
=
F.
Con-
26
(4.3)
where
Am
is the unique matrix of
(m ,m ) • Then
l 2
is a right Veblen-Wedderburn system with zero element (0,0) and unity
(R, +, x)
C whose first row is
(l,0) •
The axioms (i) and (ii) follow directly from definitions.
lows from properties (d) and (e) of Lemma (1.3).
definition.
Axiom (iii) fol-
Axiom (iv) follows directly from
Axiom (v) is a consequence of the property (c) of Lemma (1.3).
also readily seen that the subset of elements
R,
in the kernel of
form a basis of
and isomorphic with
Rover
F.
(ml,O)
It is
form a skew field contained
The elements
(1,0)
and
(0,1)
F since
=
(4.4)
(4.5)
(1,0)
= 1,
(0,1)
= i,
=
im
=
(m*1,m*2)
=
Finally, as we have shown in §3, axiom (vi) or (vi,a)
(R, +, x)
if S is a dual spread.
m*l + m*2i.
is satisfied by
In what follows, we shall always take S to
be a dual spread.
2.
\.Je shall now obtain a linear representation of the affine plane
S4'
tained in §2, in the projective space
sponding Veb1en-lvedderburn system
Let
X and X* be points of
(O,1,x*1,x*2'0).
Sx
= XX*
(4.6)
Then
belongs to SO'
=
are as in paragraph l.
S4 with coordinates
TI,
and X*
(l,0,x ,x ,0)
l 2
is point of
and from (3.2) its equation is
=
ob-
where the spread S and the corre-
(R,+,x)
X is a point of
a6
=
0,
TI*.
and
The line
27
which may be written as
=
=
(4,7)
where A
x
is the ma trix belonging to
C,
z;
(0,0'-Y1'-YZ,1).
sented by the plane
(s ,Y).
54'
The equations of the plane
~
=
(4.8)
(1,1)
passing through
..
0,
(sx,Y)
1
Sx
in the point
{(x1 'Y1),(xZ'Y2)}
and
Y
be repre-
correspondence between
84 passing through the lines of SO'
a6-points, and planes of
in
Let the a6-point
This sets up a
x
0
corresponding to the element
x .. x 1+x2i of R. Denote by (sx,Y) the plane of 54
meeting the plane E'lE'2EO with equations n1 = 0, n2
with coordinates
=
and not contained
are
x~
x* 1 x* 2
Let the a6-1ine
[(m ,mZ),(b ,b )]
1
1 2
of type 1, be represented by the point
(m ,mZ-b ,-b ,l), m ~ 0, of 54. This sets up a (1,1) correspondence beZ
Z
1
1
tween a6-1ines of type 1, and points of 54 not contained in E or n.
Again let the a6-1ine
[(m ,O),(b ,b 2 )]
l
1
of type Za be represented by the
plane
.
(4,9)
of
54"
This sets up a
planes of
..
(1,1)
correspondence between a6-1ines of type Za, and
S4 which are contained in
n,
and which do not pass through the
E'2'
point
Finally, let the a6-line
(4.10)
of
0
8 ,
4
=
=
This sets up a
[cl,cZl
(1,1)
of type 2b be represented by the plane
0
correspondence between a6-lines of type Zb, and
28
planes of
8 ,
4
which are contained in
do not contain the line
0,
and pass through the point
E' 2 but
s 00 •
Thus the a6-lines of type 2a and 2b taken together are represented by planes
n but not containing the line s co •
contained in
lEMMA (4.1>.
{(xl 'Yl),(x2 'Y2)}
An a6-point
{(ml,~),(bl,b2)}' ~ ~
°
of type 1, if and only if the plane of
ing the a6-point, passes through the point of
The condition for the plane
senting the a6-point
-b 2 ,1)
where
of
84
(x ,x ).
l 2
(sx,Y)
{(x ,y l ),(x2 'Y2)}
l
representing the a6-line
(x*l'x*2)
is incident with the a6-line
of
8 ,
4
8
4
84
represent-
representing the a6-line.
with equations (4.8), repre-
to pass through the point
{(m ,m ),(b ,b )}
l 2
l 2
is the uniquely determined matrix of
C,
(m ,m2 ,-b ,
l
l
is
with the first row
This is precisely the same as the condition (2.4) for the a6-point
{(xl 'Yl),(x 2 'Y2)}
to be incident with the a6-line
[(m ,m2),(b l ,b 2)], m2 ~
l
°
of
type 1.
lEMMA (4.2).
[(m ,O),(b ,b )]
1
l 2
An
a6-point
{(x ,y ),(x 'Y2)}
l l
2
is incident with the a6-line
of type 2a if and only if the plane of
a6-point intersects the plane of
The a6-point
8
4
{(x ,y ),(x ,y )}
l l
2 2
8
4
representing the
representing the a6-line.
is represented by the plane
(sx,Y)
with
equations (4.8), and the a6-line
with equations (4.9).
elements
[(m ,O),(b ,b )] is represented by the plane
l
l 2
These planes intersect in a line if and only if there exist
r , r , sl' 52
l
2
of
F,
not all zero, such that the result of adding the
four equations after multiplying them successively from the right by
s2
is an identity.
(4.11)
This gives
r , r , sl'
2
1
29
(4.12)
=
(4.13)
0.
If the above conditions are satisfied,
zero.
sl
~
0,
otherwise
Hence without loss of generality, we can take
-m , r 2
1
= 1.
1, sl
=:
[c l ,c2 ]
r
1
=:
Then (4.11), (4.12), (4.13) are satisfied.
lEMMA (4.3).
=:
1.
are all
This gives
-m , r
l
2
r
=:
1
Conversely, if
Then the equations (4.12) reduce to (2.6).
the conditions (2.6) are satisfied, we can choose
x*lm1-x*2.
sl
r ,r ,sl,s2
1 2
= 1,
51
=:
1,
8
2
=
This proves the Lemma.
An a6-point
{(x 'Yl),(x2 'Y2)} is incident with the a6-1ine
1
of type 2b, if and only if the plane of 54 representing the a6-point,
intersects the plane representing the a6-line in a line.
Proof is similar to Lemma (4.2).
The affine plane
a6
can be completed to the projective plane
now obtain linear representations for the elements at infinity in
6.
6,
We shall
denoting
them as in §2, paragraph 7.
Let the point at infinity
54'
where
infinity in
m
2
6
~
0.
This sets up a
0,
corresponding to the parallel
(1,1)
~l
= ~ml'
The point at infinity
= ~m2
of
correspondence between the points at
s ClO
and are not contained in
{m ,0}
1
{ClO},
or
Q.
(O,O,l,m ,O)
l
of
which corresponds to the parallel class
sisting of all lines of type 2b, will be represented by the point
viz. the point E'2.
54.
which corresponds to the parallel class
of type 2a, will be represented by the point
the point at infinity
S4'
~2
which correspond to parallel classes of type 1, and planes of
which pass through the line
(~,O)
~
of type 1, be represented by the plane
(m ,m2)
l
class
{m ,m }, m
2
1 2
We thus have a
(1,1)
54'
(ClO)
Again
con-
(0,0,0,1,0)
of
correspondence between the points
30
~t
infinity in
~,
which correspond to parallel classes of type 2a or 2b, and the
points of the line
SCll
represented by the line
called a
~-point
8 " Finally
of
4
SCll
of
8 "
4
the line at infinity in
[CIl]
~,
A point at infinity in
at infinity, and the line at infinity in
~,
6 will be
may also be
may be called the
6-line at infinity.
The following Lemmas are now easy to verify"
lEMMA (4.4).
A 6-point at infinity is incident with an a6-1ine if and only
if the linear space of
54
representing the
or contains the linear space of
54
~-point
at infinity is contained in
representing the
case when both the 6-point at infinity and the
a~-line
a~
line, except in the
are represented by planes
in which case they are incident if and only if the planes representing them intersect in a line.
LEM-'A (4.5).
The line
s CIl
representing the
~-line
at infinity either con-
tains or is contained in the linear space of 54 representing any 6--pafrtt at infinity.
Thus we have arrived at the same linear representation of a projective plane
~,
which was obtained by Bose and Bar10tti in [3].
We thus have an alternative
proof of the following theorem due to them.
THEOREM
E and
(4.1).
Let
54
be a projective space based on a skew field
n be distinct 3-spaces of 54'
a spread of lines in
contained in
IT,
E,
and let
which is also a dual spread.
~-points
54
54
Let
s 00
IT.
Let
Let
S be
be the line of S
So be the set of lines of S not contained in IT.
Let us consider an incidence structure
by linear spaces of
intersecting in the plane
F.
~
whose points and lines are represented
as follows:
are of the following types:
passing through lines of SO'
(i) Type
0,
and not contained in
represented by planes of
E;
(ii)
Type 11'
31
represeut",d by planes of
or
0;
nreof the follcwingtypes:
not contained in
tained in
passing through
0,
Soo
represented by points of
(iii)
[,-·:.:IlC::':
54'
E or
OJ
(ii)
(i)
c• .
c•.
represented by points of
represented by planes of
s 00'.
but not containing the line
sented by the line
nut
s 00 •
Type 1,
Type 2,
a:1d
(iii)
S4
8 ,
4
con-
A single 6-line repre-
s 00 •
A 6-point and 6-line both of which are not represented by palnes are inci-
54
dent if and only if the linear space of
representing the 6-point either con-
tains or is contained in the linear space representing the 6-line.
A 6-point and
a 6-line bodl of which are represented by planes are incident, if the planes representing them intersect in a line.
0 are the a6-points.
Note that the 6-points of type
6-points of type
11
are the 6-points at infinity corresponding to parallel classes of type 1, and
6-points of type
1
2
are 6-points at infinity corresponding to parallel classes
of type 2a or 2b.
Bose and Barlott! [3] showed that the plane
6
is the derived plane in the
sense of Ostrom [10,11], and Albert [1] of the dual translation plane
can be linearly represented as follows:
Points of
oT-points of type 1 are represented by planes of
S,
and not contained in
represented by
E;
Lines of
of type 1 are represented by points of
54
of type 2 are represented by the lines of
containing contained relation.
(i)
aT
8
are of two types:
not belonging to
S.
Incidence in
E;
aT
E
s 00
and contained in
(i)
aT-lines
(ii)
aT-lines
is given by the
The derivation set consists of all
represented by planes passing through
and the aT-point represented by
are of two types:
which
passing through the lines of
4
There is only a single aT point of type 2
(ii)
E.
oT
aT
aT
points
n but not in
32
§
In §2, we have obtained a projective plane
1.
Wedderbum system
of
6,
fixes
5.
(U, +, x)
eation of
6,
and So
based on a Veblen-
and in §4, we have obtained a linear representation
in a projective space
t, fl, Sco
6,
(0
54'
Any linear transformation
0
of
may interchange lines of SO) will give a collin-
in terms of the representation in
54'
We want to consider in particular a linear transformation of
a group of elations of
by the points on
6,
54 which
with axis represented by
soo'
Consider the group
~1'
~' 2
s co
54 which gives
and centres represented
G of linear transformations
0(k ,k )
l 2
given by
=
~' 1
(5.1)
to
~lal
Thus
0(k ,k )
l 2
+
=
+
~2a2
~2a2
+ n1bl + n 2b 2 + Z;;(kl b l +k2b 2+c)
6
The equations of
are
by planes
6
s 00
remains fixed.
= z;;m l ,
and So
=
o.
and transforms a linear space of
S4
into a linear space representing some element of .6.
~1
= 0,
~2
= 0,
Z;;
= O.
Hence
6-points at infinity of type
~2
(5.2)
=
is transformed by
0(k l ,k2)
2
and
the line at
[00],
11
are represented
are repre2
sented by points on sco' They remain fixed, i.e., the 6-points {m ,m } and {oo}
l 2
remain fixed. A 6-point of type 0, i.e. an a6-point represented by the plane,
~1
= z;;m ,
z;;.
+ n 1b 1 + n 2b 2 + Z;;c = 0
t 1 , fl, 5 00
fixes
z;; ,
=
n' 2
~lal
representing an element of
infinity in
n' 1 =
~2'
6 points at infinity of type
to the 6-point
r~presented
by the plane
1
33
(5.3)
=
Hence the 6-point
(x ,y +k )}.
2 2 2
{(x l 'Yl)'(x2 ,yZ)}
It is fixed only hy the identity of
{(m ,m ),(b ,b )}, m
l 2
l 2
2
The 6-line
1 ,m2 ,-b l ,-b 2 ,1)
of
(ID
S4
represented by the point
identity of
+0
G.
of type 1, represented by the point
is transformed to the 6-line
(ml,m2,-bl-kl,-b2-kZ,1).
[(m ,O),(b ,b )]
l 2
l
{(ml,m2),(bl+kl,b2+k2)}
It is fixed only by the
of type 2a represented by the plane
=
=
(5.4)
O(kl ,k 2)
is transformed by
by the plane
of
{(xl'Yl+k l ),
G.
The 6-line
Gut 1
is transformed to
[(ml,O),(bl,k2-klml+b2)]
= -~lbl+~(k2-~ml+b2)'
n1rnl-n2
k 2 = klm ,
l
G for which
Now the 6-line
to the 6-line
l~here
[(m ,O),(b ,b )]
l
l 2
0
m
l
~2
= O.
represented
It is fixed by the subgroup
is fixed, but
k
l
is arbitrary.
is incident with the 6-point at infinity
which is represented by the point
(O,O,l,ml,O)
of
s=,
{ml'O~
and conversely every
6-line incident with this 6-point at infinity is of the form
[(m .O),(b ,b )]
l
l 2
where
F.
m
l
is a fixed and b , b
are arbitrary elements of
l
2
a subgroup of elations for which the line at infinity in
6-point at infinity
to
F,
we get a subgroup
The 6-line
(5.5)
[cl,c ]
Z
is the center.
'
l
of type 2b
~
o(k l ,k 2)
n1
= ~lcl-~(kl+c2)'
k
= 0,
but
k
Z
~2
= O.
Gut
is
l
is the axis and the
Corresponding to each m
l
belonging
Gut
=
is transformed by
l
{ID1,O}
6
Thus
2
represented by the plane
=0
to the line
[cl,cZ+k ],
l
represented by the plane
It is fixed by the subgroup
is arbitrary.
~ow
the 6-line
[cl,c ]
Z
G=
of
G for which
is incident with the
34
~-point
every
{~},
~-line
represented by the point
incident with
belonging to
[~]
(0,0,0,1,0)
s~,
of
and conversely
[c ,c ] of type 2b for some c ' c
2
l
l 2
Hence the subgroup under consideration is a s tbgroup of elations
F.
is a line
~
for which the line at infinity in
~-point
is the axis and the
{~}
is the
center.
Any two subgroups have only the identity in common.
the line
represents a
s~
~-point
Each point of
soo'
When
2
is of order
q,
F is the field
and there are
q+l
GF{q),
~
~,
is a finite plane.
subgroups each of order
semitranslation plane with respect to the line at infinity in
Ostrom [9].
on
at infinity, which is the center for a sub-
group of elations, for which the axis is the line at infinity in
sented by
S4
q.
repreThe group
Hence
~
G
is a
in the sense of
~
Of course, the situation is completely analogous in the infinite
case.
Z.
~.
points and lines considered in this paragraph belong to
only of points and lines instead of
[(0,0),(0,0)]
y-axis.
~.
We shall now obtain the ternary ring of the projective plane
~-points
and
From (2.6) and (2.7), the point
[0,0]
{(O,O),(O,O)}
of type 2b as the
is incident both with the
x-axis and y-axis.
This will be chosen to be the origin.
the parallel class
(0,0)
of type 2a.
Hence we shall speak
Let us choose the line
~-lines.
of type 2a to be the x-axis and the line
All
The x-axis belongs to
Lines parallel to the x-axis have co-
[(0,0),(b ,b )]. The y-axis belongs to the parallel class [~] conl 2
sisting of all lines of type 2b. Hence lines parallel to the y-axis have coordinates
ordinates
[cl,c Z].
unit line.
(x 2 ,yZ)}
Let us choose the line
[(1,0).(0,0)]
From (2.6) it is incident with the origin.
of type 2a to be the
If the point
is an incident with the unit line, then from (2.6),
Xl
Hence a general point incident with the unit line has coordinates
(xl'YI) }.
1s a
(1,1)
We may associate to this point the element
correspondence between the elements of
(xl'YI)
of
{(xl'Yl)'
=xz,
YI
= yZ'
{(x1'YI) ,
R.
Thus there
R and the points on the unit
35
line, the element
(x,y)
of
R corresponding to the point with coordinates
{(x,y),(x,y)}
lEMMA (5.1),
3.
(x
Given two points
{(xl,yl),(xl,y l )}
and
{(x 2 'Y2)'
incident with the unit line corresponding to the elements
Z'Y2)}
of
(x ,yZ)
2
(xl,y ),
l
R the line through the first point parallel to the y-axis, meets the
line through the second point parallel to the x-axis in the point whose coordinates
are
{(xl ,y l ),(xZ'Y2)}
{(x ,y ),(x 'Y )}
2 Z
l l
The line parallel to the y-axis and incident with
(2.7),
[xl,y ].
l
{(x 2 ,y 2),(x2 ,y Z)}
Again the line parallel to the x-axis and incident with
is from (2.6),
[(O,0),(x 'Y2)].
2
[xl,y ]
l
see that the point incident with both
(x ,Y )}.
2 2
and
Then from (2.6) and (2.7), we
[(O,0),(x2 ,yZ)]
y-coordinate of the point
However,
m-coordinate, and
Since
(1,0)
(xl,Y )
l
{(x ,y ),(x ,y )}.
l l
2 2
(m ,m )
l 2
[(ml,~),(bl,b2)]
the line
(b ,b )
l 2
and
(bl,b )
Z
the x-coordinate and
{(xl,y l ),
are not the slope and the y-intercept of
We will call
the b-coordinate of the line
is the unity of
R,
The line
[1,0]
{«1,0),(1,0)}
(b ,b )].
l 2
Case I.
[(m ,m ),
l 2
Then the line parallel to this and inci-
[(m ,m ),(O,0)].
l 2
If it meets the slope line in the point
its slope is
then
Let the line
(u ,v ),
2 2
incident wi th the line
[(m ,m2),(b ,b )]
l
l 2
then the point
[(ml,rnz), (0,0)].
on
Let
(u2 ,v ) is defined to be the slope of the line
2
We have to consider two seperate cases.
{(1,0),(u2 ,v2)},
the
of type 2b, parallel to the
y-axis through the unit point will be called the slope line.
dent with the origin is
(m l ,m2)
[(m ,m ),(b ,b )1.
1 2
l 2
we may take the point
be any line of type I or 2a.
(x 2 ,Y 2) the
(See Albert [1].)
in the standard sense.
the unit line, to be the unit point.
(b ,b )]
l 2
is
This proves the Lemma.
Thus we may appropriately call
4.
is from
be of type 1, Le.,
[(ml,m )'
Z
m2 :/: 0.
If
{ (1, 0) , (u2 ' v2) } of the slope line is
From (2.4) the condition for this is
3:i
=
(5.6)
where the matrix appearing on the right hand side of (5.6) belongs to
(~1*'~2*)
= (0,1)x(1'~2)
C,
i.e.,
Fl , FZ' G , GZ defined
l
or in terms of the functions
by (1.14) and {I. 15)
*
(5.7)
~1
~2'
From Lemma (1.5),
v
are uniquely determined.
2
From (1.19) and (1.20),
(5.8)
Also
~
v2
O.
If
= 0,
v2
then from Lemma (1.3), part (d), the matrix on
the right in (5.6) would be the null matrix which is a contradiction.
(~2,v2)'
Conversely, given
determined, and since
press
ml ,
m
2
~
0,
from Lemma (1.3) part (e),
it follows that
v
2
~
O.
(ml,mZ)
is uniquely
We can explicitly ex-
IDz as
(5.9)
where
~1*' ~2*
Note that
are given by (5.7) and
d
~
0,
d
otherwise the column vectors of the matrix on the right in
(5.6), would be dependent from the right.
be singular.
null.
m
2
~
= ~1*~2-~2*'
From Lemma (1.1), the matrix would then
This would contradict Lemma (1.3) part (c), since the matrix is non-
Thus there is a
(1,1)
correspondence between m-coordinates
(ml,rnz),
characterizing parallel classes of type 1, and the corresponding slopes
0
(~2,v2)' v2 ~
Case II.
(~2tV2)'
o.
Consider the line
then the point
line
[(ml,O),{O,O)].
line
[(ml .O).(b 1 ,b 2 )]
[(ml,O),(bl,b Z)]
{(1,O)'(~2,v2)}
Hence from (2.6),
of type 2a is
of type 2a.
If its slope is
of the slope line is incident with the
~2
= ml ,
(ml,O).
v2
= O.
Hence the slope of the
37
The slope of any line
[c ,c ]
l 2
5. Given any two elements
multiplication
(~2,v2)
(xl,y ),
l
=.
R we can define a new
of
by setting
@
(5.10)
if
of type 2b may be defined to be
=
(x2 'Y2)
is the y-coordinate of the uniquely determined point with x-coordi-
nate
(xl,Y ) and incident with the line through the origin whose slope is
l
(~2,v2). Then this multiplication is a loop with unity
(1,0).
(xl'Yl)$(~2,v2)
We can express
(R, x, +)
as follows:
(~2,v2).
slope
vious paragraph.
If
[(ml'~)'(O,O)]
Let
Then if
v2
~
v2
(x 'Y2)} is incident with
2
(1.120) of Lemma (1.5), if
in terms of the operations
= 0,
(m ,m )
l 2
0, (m ,m )
l 2
v2
~
in
from the Case I I of the pre-
is given by (5.9).
Now the point
{(xl'Yl)'
Hence from (2.4), and formulae
(l.l~
0,
=
(5.11)
and +
be the line through the origin with
= (~2'0).
[(~,m2)'(0,0)].
x
=
where
(5.12)
and the functions
~2,v2
terms of
when
v2
If
(5.13)
~
G , G
2
l
are defined by (1.15).
Substituting for
from (5.9), we have an explicit expression for
m , m2
l
in
(xl'Yl)@(~2,v2)
O.
v2
= 0,
then
(ml ,m2 )
= (~2'0).
Then from 2.6),
=
Note that from Lemma (2.4) and (2.5) the line through the origin with slope
is of type 1 or 2a according as
v 2 '" 0
or
v
2
= o.
Hence if we wish to
38
determine
(~2,v2)
given
(Xl,Y ), (X 'Y2),
2
l
valid according as the matrix
the formula (5.11) or (5.13) is
is non-singular or singular.
6.
The equation of any line of
the slope
Let
(~2,v2)
6,
passing through the origin and having
is given by (5.8).
[(ml,~),(bl,b2»)
be any line of type 1.
sufficient condition for the point
{(x 'Yl),(x2 'Y2)}
l
From (2.4), the necessary and
to be incident with it is
=
(5.14)
(x*1,x*2) = (O,1)x(x l ,x2). It follows that the point {(xl,yl-b ),
l
(x2'Y2-b2)} is incident with the line [(m ,ID ),(O,O») through the origin.
l 2
(~2,v2) is the slope of the line, we have
where
•
(5.15)
This is the equation of a line of type 1, with slope
nate
If
(~2,v2)
and b-coordi-
(b ,b ).
l 2
Similarly, the equation of a line of type 2a with slope
ordinate
(5.16)
7.
(b ,b )
l 2
(~2'O)
and b-co-
is
=
Finally we note that the b-coordinate of a line is not the y-intercept of
a line in the standard sense.
He may define the y-intercept of a line of type 1
or type 2a, to be the y-coordinate of the point in which the line intersects the
y-axis.
39
consider two cases.
Case I.
point
The line
{(O,0),(6 z 'YZ)}
[(ml,~),(bl,b2)]
is of type 1.
Hence
m2 ~ O.
The
Hence from (2.4),
is incident with it.
•
(5.17)
where the functions
F1
From Lennna (1.5),
is given.
and
2
(8 ,y )
2
Conversely given
Case II.
F
The line
2
are given by
is uniquely determined by
(8 2 'Y2)'
(5.17) determines
[(ml ,O),(b ,b )]
1 2
is incident with it, we have from (2.6),
the y-intercept is
(1.14).
is of type 2a.
(8 2 ,y 2)
(b ,b ).
1 2
Since
= (b 1 ,b 2).
{(O,O),(S2'Y2)}
Thus in this case,
(b ,b ).
l 2
We are now in a position to write down the ternary ring of
(5.18)
~.
Let
=
if
=
where in the case
\}2
~
0,
we determine
Then (5.18) is the ternary ring of
(b ,b )
1 2
\}2
= 0,
from (5.15) and (5.7).
6.
The equations of lines of type 1 or
y
coordinates of a point incident with
type 2a are given by
=
(5.19)
where
a
(x1'Yl)' (x2 'Y2)
line ",hich has
slope
are the
(1l20vZ)
x
and
and y-intercept
(8 ,y ).
2 2
40
The lines of type 2b, parallel to the y-axis have equations
=
(5.20)
f?cFERENCES
1.
Albert, A.A.
"The finite planes of Ostrom."
Mathematiaa
2.
Bose, R.C.
Me~aana,
BoZetin de la Soaiedad
11 (1966), 1-13.
"Coordinatization of a class of projective planes which can be
linearly represented in a projective space of four dimensions."
Proa. 2nd Chapel Hi lZ Conferenae on Combinatorial Mathematias and Its
Appliaations, University of North Carolina, Chapel Hill, N.C., (1970),
37-56.
3.
Bose, R.C. and Barlotti, A.
"Linear representation of a class of projective
planes in a four dimensional projective space."
Ann. Mat. Pum. Appl.,
4.
Bruck, R.H.
"Construction problems of finite projective planes."
Combinatorial Mathematias and Its Appliaatio71s,
Chapter 27, 416-514.
5.
Bruck, R.H. and Bose, R.C.
UNC Press (1969),
"The construction of translation planes from
proj ective spaces." J. Algebra, 1 (1964), 85-102.
6.
Bruck, R.H. and Bose, R.C.
"Linear representations of projective planes in
projective spaces." J. Algebra, 4 (1966), 117-172.
7.
Bruen, A. and Fisher, J. C.
"Spreads which are not dual spreads."
Canad. Math. BuZZ., 12 (1969), 801-803.
8.
The Theory of Groups,
Chapter 20.
Hall, M.
The MacMillan Co., New York, (1959),
41
9.
Ostrom, T.G.
"Semi-translation planes."
Trans.
Am.
Math. Soc., III (1964),
1-18.
10.
Ostrom, T.G.
''Translation planes of order
p2."
Combinatorial Mathematics
and Its Applications, UNC Press (1969), Chapter 27, 416-425.
11.
Ostrom, T.G.
"Vector spaces and construction of finite projective planes."
Ann. Math., 19 (1968), 1-25.
12.
Segre, B.
"Teoria di Galois, fibrazioni proiettive e geometrie non
desarguesiane."
Ann. Mat. Pura. Appl., 64 (1964), 1-76.