* Lectures given at the University of North Carolina at Chapel Hill
supported by the U.S. Air Force Office of Scientific Research under Grant
No. AFOSR-68-1406.
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COM BIN A TOR 1 A L
MAT HEM A TIC S
YEA R
February 1969 - June 1970
CONSTRUCTION OF RULED SURFACES IN 5-DIMENSIONAL FINITE
PROJECTIVE GEOMETRIES *
by
R. H. BRUCK
University of Wisconsin
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 600.18
NOVEMBER 1969
Construction of Ruled Surfaces in 5-dimensional Finite
Projective Geometry *
by
R.H. Bruck
University of Wisconsin
1.
Introduction.
I want to describe, as briefly as possible, the
methods for constructing one of my ruled surfaces in
any prime-power.
q=2.
PG(5,q).
Here
However, the theory becomes trivial (and exceptional) for
q > 3
2.
(q
Preliminary.
Lemma 2.1.
Let
a prime-power).
Nothing is said in my paper on ruled surfaces (the
Luxembourg paper) about reguli.
S
I need the following as background:
~
be a surface in
= PG(5,q)
planes, belonging, say, to classes I, II and III.
in a ruling plane of (say) class I.
Let
Let
ruled by 3 systems of
L
R be the set of
be a line lying
q+l
ruling
planes of (say) class II each of which contains exactly one point of
Then: (i)
(ii)
R is a regulus of planes of
Each of the
planes of
(iii)
q+l
planes of
q
2
R in the
Each of the
Proof.
q
2
R in the
L.
~.
and
q+l
ruling planes of class I meets the
q+l
points of a line -- a transversal line to
and
q+l
q+l
points of a conic.
ruling planes of class III meets the
It is to be understood that
Luxembourg paper.
*
is
So assume
(1.1)
q+l
q
S
is constructed as in my
Set
Lectures given at the University of North Carolina at Chapel Hill
supported by the U.S. Air Force Office of Scientific Research under Grant
No. AFOSR-68-l406.
R.
2
GF(q),
F
and let
V
=K
(x,y), x,y
E
K,
f(x,y)
y+y,)} v x,y,x',y'
V
(fx, fy)
consists of all points
N(x)
F
consisting
with
(x+x' ,
(x,y) + (x' ,y')
S
3
GF(q ),
x K be the 6-dimensional vector space over
of ordered pairs
Then
K
«x,y»
f
E
E
K
F.
such that
N(y);
equivalently
2
x
The classes
2
q +q+l
Y
q +q+l
(I), (II), (III) are defined as the set of planes of the fol-
lowing sorts:
(I)
y
kx
(II)
y
kx
(III)
y
kx
For any fixed k such that
2
N(k) = k q +q+l
l.
q
q
2
For the proof, we may assume that
< (1,1) ,
L
for some fixed
q+l
is the 2-dimensional vector space
(a, a) >
a,
a
Then the
L
points of
L
E
K-F.
are the following:
< (1, 1) >
the point
3
and
q
points of form
<f(l,l) + (a,a»
<(f+a, f+a) > ,
The plane of class (II) through
«1,1»
The plane of class (II) through
«f+a, f+a»
f+a
We may check that, for each
kEK,
E
F.
is
--x
f+a
f
is
q
q
k ~ 0,
the line
transversal to
Furthermore,
L
k
lies in the following plane of class I,
y
k
q-l
x.
We also note that
L
so that the total number
k'
<=>
3
q-l
R
lines
L
k
q
2
L
k
is
+q+l.
is a regulus of (skew) planes of
as its transversal lines.
F
E
k'
of distinct lines
3.....:.!.
Thus:
k'
L::PG(5,q),
with the
2
q +q+l
4
Next consider a typical ruling plane of class III, say
a:
where
k
y
=
is some fixed element of
k.x
q
2
K with
2
k q +q+1
We can choose
b
(in
q-1
ways) so that
k
Thus the equation of
a
1.
=
b
q-1
•
becomes
a:
y
b
q-1 q
2
x
Then we may check that
2
< (b -q , b -1) >
2
< (b -q
2
(f+a) -q ,
b
-1
(f+a)
To complete the proof, we must show that the
lie on a conic (in
a).
If
q
-1
>,
q+1
V
f E
F.
points so obtained
is odd, it is enough to show that no
three of the points lie on a line.
Consider the special case of these
points
i
where
f , f , f
1
2
3
1, 2, 3,
are distinct elements of
a line, we must have
F.
If these points lie on
5
2
2
2
2
u(b- q (fl+a)-q , b-l(fl+a)-l) + v(b- q (f +a)-q , b- l (f +a)-1)
2
2
2
2
+ w(b- q (f +a)-q , b- l (f +a)-1)
3
3
o
for elements
u, v, w
of
we get two equations for
F,
not all zero.
Considering components,
u, v, w which (after cancelling some non-
zero factors) become
2 2 2
(fl+a)-q u + (f +a)-q v + (f +a)-q w
2
3
(1)
0
(2)
We note that (2) implies (1).
all zero) means that
in
F.
Since
a
E
a
However, (2) (with
u, v, w
in
F,
not
satisfies a quadratic equation with coefficients
K-F,
K is three dimensional over
and
F,
this is
false.
IToona, IT f na, IT f na,
Similarly, three points
cannot be collinear.
1
2
We leave the proof of Lemma 2.1 at this point.
3.
Construction of surfaces
S.
(Some algebraic details.)
We suppose given a regulus
ting of
q+l
skew planes
every
IT.
1
in a point.
disjoint from the
q+l
of planes of
Ii
= 1,2, ... , q+ l}
L = PG(5,q),
consis-
IT',
R
such that each of the
R
{ IL
2
q +q+l
1
transversal lines to
We also suppose given one more plane,
planes
IT .•
1
IT,
6
We consider the problem of constructing a triply-ruled surface
of
S
such that the set
L
R u {II}
of
q+2
skew planes forms part of one of the three systems of
ruling planes of
2
q +q+l
S.
In the light of Lemma 2.1, we may assume that the planes of
belong to Class II, and that every plane of Class I contains a
Ru{n}
R.
(unique) transversal line to
We may suppose that
vector space
V over
L
= PG(5,q)
F = GF(q)
is given by a six-dimensional
with a basis
t ,
l
t 2 , t 3 , t{, t;, t;
chosen so that
Then every plane skew to
matrix
X=(x. )
1
over
J(X)
In particular,
II =J(oo)
1
R consists of
J(X)
for a unique
3x 3
where
F=GF(q),
has basis
has form
xllt l + x 12 t 2 + x13t 3 +
t{,
x
t + x t + x t +
23 3
22 2
2l l
t;,
x
t + x t + x t + t~.
33 3
3l l
32 2
J(oo)
and the
J(fI) ,
fEF,
where
7
Also
II
for some irreducible
Every vector
3x3
a
in
J(U)
matrix
J(oo)
has form
for unique elements
for every
3x3
Let
a
of the plane
F.
in
matrix
J(X)
U.
X.
In this notation
+ .{..1'.{..2
0'
oX + .{..O ', .{..oX + .{..
0 '.
2
3
3
oX
.{..l
has basis
J(oo) ,
be a non-zero vector in
J(oo).
We define
The transversal line,
so that
L ,
a
to
<a>
R
is a point
through
<a>
the two-dimensional vector space
L
a
=
<a, a'>
where we define
A ruling plane, of our proposed surface
S,
which has Class (I) and
contains
II
J(U)
L
a
must (in particular) meet
in a point, say in
<b U + b '>,
where
b
is some non-zero vector in
J(oo).
Then this ruling plane is
(say)
< a,
a',
bU+b '> •
is
8
The ruling plane
namely in a point of
<bU+b'>
of
IT
a
R in a point,
should meet each plane of
This puts some restrictions on the point
L •
a
or, equivalently, the point
we note that each point of
has form
a
<b>
<v>
of
J(oo).
where
v
To see this,
is a non-zero
vector of form
fa + gal + h(bU+b'),
v
where
f, g, h
are elements of
<v>
not all zero.
will be in
J(oo)
if and only if
ga + hb
We wish this condition to imply
<b>
(1)
If
tEF,
:f
so that
O.
Thus we want
h
O.
R,
the point
<0>.
J(tI)
is in
<v>
if and only if
fa + hb
U
t(ga+hb)
or
U
h (b -tb) = (-f+tg) a
or
(-f+tg)a.
We want this equation to imply
Thus we want
h=O.
<a>
or, equivalently,
(2)
<b>
:f
Equivalently,
U
(fa+hb ) + (ga+hb)'.
v
The point
F,
<a
(U-tI)
-1
>,
V
t
E
F.
will be in
J(tI)
9
It may be shown that the point
<a>
of
J(oo)
together with the
q
points
<a
of
J(oo)
of
IT,
(U-tI) -1
J(oo).
form a conic of
>,
t
E
F
Hence, also, the corresponding points
namely
U
<a +a'>
form a conic of
<a
and
(U-tI)-~
+ ( (U-tI)-l),
a
>,
Also the point
IT.
points of the latter conic.
<bU+b'>
<bU+b'>
q+2
F,
must avoid the
q+1
q
2
so that
=
meets each of the
E
This gives
(q2+q + 1 ) - (q+1)
choices of the point
t
<a,
planes in
Ru{IT}
in (exactly) a point.
It is easy to check that the conditions on
b
are equivalent to
the following:
(3)
a, b, b
However, since
U
form a basis of
U
is irreducible and
J(oo)
over
(3' )
f, g, h
<a>
in
F
bfa,
F = GF(q).
fb + gb
for some
J(oo)
form a basis of
with
h'O.
U
over
F
the vectors
GF(q).
U
u2
b, b , b
Hence (3) means that
+ hb
2
U
Equivalently (the case
h=l)
10
for some
q
chosen in
2
F.
in
f O' go
(Note that the ordered pair
can be
ways. )
Assuming (3'), we can write the plane
as the plane
a
a(b)
where
a(b)
(4)
Here
go
f O' go
b
in
f '
can be any non-zero vector in
J(oo).
If, now, for fixed
we consider all the planes
a(b),
we check easily that they
F,
O
form a set of
distinct, mutually skew, planes, each of which meets every plane of
Ru{n}
in exactly a point, and each of which contains a unique trans-
versal line to
R,
namely the line
Although we are not supplying a proof here, these
2
q +q+l
planes
a(b)
constitute the desired (complete) collection of planes of Class I.
We still have to supply the
q
2
+ q + 1 - (q+2)
2
missing planes of Class II, and the
4.
Geometric approach.
•
q
q +q+l
The material of
2
- 1
planes of Class III.
Section 3 may be summarized
as follows:
Suppose given a regulus,
and a plane
precisely
(4.1)
IT
q
2
L
R,
of
q+l
skew planes of
disjoint from each of the planes in
R.
L
= PG(5,q)
Then, in
distinct ways, we can set up a one-to-one correspondence
*-+
P
11
2
q +q+l
between the
P
of
II
in
2
R are the q +q+l
to
L
points
such that:
(i)
II
transversal lines
If
L, P
are a corresponding pair, the plane
and intersects each plane
P
2
The q +q+l
(ii)
planes
L+P
of
II.
1
L+P
R in the point
obtained by letting
L, P
intersects
LnIl ..
1
range over
all corresponding pairs are structually skew.
This is essentially all we need to complete the construction of the
surface
S.
We assume that some fixed correspondence (4.1) has been chosen.
Consider any corresponding pair
L, P.
We note first that the projective
space
(4.2)
L
H
P
+
IT
~
is four-dimensional and hence is a hyperplane of
PG(5,q).
Next we
construct a projective 3-space,
T ,
p
in the following manner:
R
and let
meeting
l
~
Ill' Il
L ,
2
be two distinct planes of
M be the unique transversal line through
in points
2
transversal lines to
L
IT , Il
2
l
Let
R
PI' P
through
else the point
P
respectively.
2
PI' P
of
II
2
P
Let
to
L , L2
l
respectively.
Ill' Il ,
2
be the
Note that
would lie on a transversal to
and hence on one of the planes of
R,
a contradiction.
Since
R
Ll~L2'
then the space
(4.3)
is a projective 3-space.
is on
L ,
l
P,
versal through
Since
is on
L ,
l
P
L , L
l
2
to
contains
M
then
in
T
p
T
p
contains
P, PI' P
2
P and
M
and since
PI
is the trans-
12
We need to note the following:
(a)
The 3-space
T
the choice of the planes
(b)
T
Ill' II
and contains precisely
q+l
R
not on
in a line, giving
R,
transversals to
versals to the first set of lines.
P,
II
or on
R.
of
2
meets each plane of
p
R and
depends only on
P
q+l
such lines,
namely the trans-
These two sets of
q+l
lines form
the two sets of rulings of a doubly-ruled quadric
~
of
~
and
T
p'
( c)
P
depends only on
lies on exactly
q+l
P
R in a line.
~
meets
plane of
(d)
~
planes to
R
contains exactly one transversal line to
of
R.
and
Each of the remaining
q
in
T
and meets exactly one plane
2
planes of
in a line.
T
n II
p
T
p
through
P
R,
in a conic, contains no transversal line to
R
Each of these
p
and meets no
P.
Next we need the following:
(4.4)
T
(4.5)
III
p
n L
P
is the empty point-set.
T n H
p
P
is a plane (disjoint from
To prove (4.4), first suppose that
line.
that
pI
R.
~
is a tangent plane to
P+L
:f P.)
Since
T
p
is contained in
and hence meets some plane of
This is a contradiction.
L
L
L).
Hence
L
is not in
T
p
.
.
Then
R in a
Next suppose
T
in a point pl. (Since P is not on L, necessarily
p
Since pI is on L, then pI is on some plane, say IIi' of
meets
pI
is in
II.nT
1
transversal line through
p'
pI
the
to
pI
R
is on
~.
is a ruling of
In particular, the
~,
and is in
But this transversal, being the unique transversal line through
pI
T
p
to
13
R,
must be
Hence
L.
is in
L
T ,
a contradiction.
P
Therefore, (4.4)
must be true.
T +L
In view of (4.4), the projective space
3+1-(-1)
=
5.
p
has dimension
Therefore,
T + L
P
(4.6)
Since
H
p
L,
contains
we see from (4.6) that
T
p
I.
+ Hp
Hence
dim T + dim H - dim
p
p
dim(III )
p
Thus
III
p
is a plane.
(4.7)
III
P
p
n L
T
This proves (4.5).
+ L
3 + 4 - 5
2.
Moreover
III
is empty by (4.4).
I
p
n H
P
n L
From (4.3) we get
H .
P
Indeed, the left-hand side of (4.7) is contained in the right-hand side,
and both sides have projective dimension 4.
that
P
is in
III
p
we get
III
P
and hence
(4.8)
III
P
From (4.7) and the fact
n (P+L)
=
P.
+ (P+L)
H
P
14
Next we need
(4.9)
If
L
containing
Ii,
L
contains one and only one transversal line to
R.
4
is a projective 4-space of
4
To see this, we note that there are precisely
R and precisely q 2+q+l
versal lines to
of
containing
L
IT
tains
I
is a line, say
points of M,
IT
distinct trans-
If (4.9) is false, there must be a
be one of the planes of
I
2
q +q+1
M.
The
L
each meet
R.
Let
R.
Then
transversal lines to
q+l
L
L
be such a
4
L
which con-
4
4
intersects
R through the
in a single point, manely a point of
4
The same process, carried out for the
then
distinct projective 4-spaces
but contains no transversal line to
4-space, and let
IT
IT.
L:
q+l
MI'
R,
distinct planes of
shows
that there must be at least
(q+l)
R.
distinct transversal lines to
is true.
2
>
2
q +q+l
This is a contradiction.
Hence (4.9)
As a special case of (4.9),
L
(4.10)
is the only transversal line to
R contained in H =L+TI.
P
By (4.10) and (4.4), (4.5),
The plane
(4.11)
III
In view of (4.11),
p
(The conic lies on
of a conic.
The
III
2
q +q+l
planes
R.
contains no transversal line to
P
III ,
P
meets the
q+l
R in the points
planes of
~.)
one for each point
candidates for the ruling planes of Class III.
p
of
IT ,
are our
(Compare Lemma 2.1.)
omit the proof that every two of these are disjoint.
We
15
Of course, the planes of class I are the planes
I
where
L, P
P
P + L
is a corresponding pair in the sense of (4.1).
It has to
be shown that every plane of class I meets every plane of class III in
a point.
In addition, we have in class II only the set
R u {IT}
of
2
q -1
q+2
planes.
But it should be clear at this point that the missing
planes are uniquely determined by the planes of Classes I and III.
We have only to consider Lemma 2.1 with classes I, II, III replaced
(for example) by classes III, I, II respectively.
I will stop here.
Note that one must prove that the construction
can actually be completed consistently.
(This is, in fact, true.)