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Journal of Geometry. Vol. 5/2 1974. Birkh~user Verlag Basel
ON THE GRASSMANIAN OF LINES IN PG(4,q)
AND R(I,2) REGULI
J. W. Freeman*
An R(I,2) regulus is a collection of qel mutually skew planes in PG(5,q) with the property that a line meeting three of the planes must meet all the planes. An (Z,~)-configuration is the collection of lines in PG(4,q) meeting a line s and a plane ~ skew to i. A correspondence between (s in PG(4,q) and R(I,2) reguli in the associated Grassmanian space G(I,4) is examined. Bose has shown that R(I,2) reguli represent Baer subplanes of a Desarguesian projective plane in a linear representation of the plane. With the purpose of examining the relations between two Baer sub- planes of PG(2,q2), the author examines the possible intersections of a 3-flat with an R(I,2) regulus.
1. Motivation and Main Results.
Let PG(n,q) or S(n) denote the n-dimensional pro- k
jective space over the Galois field GF(q), q = p , p odd
prime. The lines of S(3) correspond in a i-i manner onto
the points of a nondegenerate hyperbolic quadric K = G(I,3)
in S(5) under the Pl~cker coordinate map ~, [17, p.327].
If p and q are two skew lines of S(3) with ~(p) = P,
~(q) = Q, then P and Q are not conjugate with respect to
(w.r.t.) K. The lines of S(3) meeting both p and q
correspond under ~ to the points of a nondegenerate
*This work was supported (in part) by National Science Foundation Grant No. GP-40583.
159
2 Freeman
hyperbolic quadric section of K, Q(3). In fact Q(3) is
the intersection of K = G(I,3) with the intersection of
the tangent spaces of K at P and Q. The quadric Q(3) is
ruled by two collections of lines, termed reguli. Each
such collection is an R(l,t-l) regulus (see this section
below) of S(2t-l) in the simplest case t = 2. Hence the
configuration of lines meeting two skew lines in S(3)
corresponds under ~ to the points of an R(I,I) regulus.
Let ~ be a line skew to a plane ~ in S(4). The
collection of lines of S(4) which meet ~ and ~ in each a
point is called an (~,~)-configuration. The elements of
an (~,~)-configuration correspond under ~ to the points
of an R(I,2) regulus (see below) lying on the associated
Grassmanian space, G(I,4) = L(9), (proposition 6). -i
Conversely, let Q be a point on L(9) with q = ~ (Q)
a line of S(4). Let T(Q) be the tangent space to L(9) at
Q. Then T(Q) is a 6-dimensional projective subspace
(6-flat) and there is a i-i correspondence between R(I,2)
reguli in T(Q)~ L(9) and (q,~)-configurations of S(4),
(proposition 7, corollary 3).
The L(9) locus may be viewed as the intersection of
five degenerate quadrics in S(9). The polar properties
for a locus given by more than one quadric together with
the group of projective transformations of S(9) induced
by the projective group of S(5) are the basic tools for
the analysis of L(9).
The polar space [w.r.t. L(9)] of a point Q is a 6-flat
[4-flat] when Q is [is not] on L(9). Unlike the polar
space at a point w.r.t, a single quadric not all lines of
S(9) meeting L(9) in a single point Q are in the polar
space of L(9) at Q, (section S, remark i). From this point
160
Freeman 3
of view we discus~ a partition of the points of S(9).
The partition follows ~rom a property of the points of
S(9)'~L(9) given by H. R. Brahana [6, p.543]. Connections
with some properties of L(~) to the work of J. A. Todd [16]
are given in section 4.
Let C be a collection of mutually disjoint projective
subspaces (flats) of a projective space S. A transversal
to the collection is a (projective) line which has exactly
one point in common with each member of C. An R(l,t-l)
regulus is a collection of mutually skew (t-l)-dimensional
projective subspaces of a space, S(2t-l), with the following
properties: i) R(l,t-l) has at least 3 distinct members,
ii) every transversal to 3 distinct members of R(l,t-l) is
a transversal to R(l,t-l), iii) if L is a transversal to
R(l,t-l), each point of L is on a member of R(l,t-l).
Some basic properties of R(l,t-l) reguli are discussed
in [8, p.162]. Dembowski [9, p.220] calls an R(l,t-l)
regulus a (t-l)-regulus. Let E be a simple extension of
a field F with [E:F] = t. The points of the projective
line over E may be represented by the pairs [i,~] and [0,i]
with ~ e E, see [5]. The points [l,m] and [0,i] with m ~ F
give rise to an R(l,t-l) regulus in the (2t-l)-dimensional
projective space over F. The connection between R(l,t-l)
reguli, regular spreads and the linear representation of
finite Desarguesian projective planes is discussed in [8]
(see corollary to Thm. 12.1, p.164).
Bose and Barlotti [4] have given a representation of
a class of projective planes (delta-planes) in a 4-
dimensional projective space in which the definition of
incidence is not simply the inclusion relation. Incidence
in the representation of the delta-planes is given by how
161
4 Freeman
the objects which represent the 'delta-points' and 'delta-
lines' intersect. The representation of the delta plane
was originally developed in a 5-dimensional projective
space. With this point in view the present paper concludes
with an examination of how a 3-flat and an R(I,2) regulus
may intersect (proposition i0).
The facts of proposition i0 are used in the examination 2
of the relations between two Baer subplanes of PG(2,q ).
The intersection relations are discussed in a forthcoming
paper by R. C. Bose and the author.
2. A Representation of the Lines of S(4) and S(9).
A line s of S(4) may be determined by two linearly
independent points on s having vector representations, ~,
~, which correspond to a 2 x 5 matrix:
i) = 0 <i<4. Yi -- --
A line represented as in i) may be associated with the
vector (P(ij)), 0 < i < j < 4 where
ii) P(ij) =
Xo X, l ]
Yi Yj
It is well known [13] that the association (denoted as ~)
is well-defined on the lines of S(4) and i-i onto a locus
L(9) of points in S(9) defined by five (dependent) quadratic
relations labeled Qi in iii) below.
iii) Qi: P(jk)P(Zm) - P(js + P(jm)P(ks = 0
with {i,j,k,s = {0,1,2,3,4}, j < k < s < m.
162
Freeman 5
These quadrics are related by:
iv)
P(04)Q 0 - P(14)Q I + P(24)Q 2
p(03)Q0 - P(13)Q 1
p(02)Q0 - P(12)Q 1
P(OI)Q 0 - P(12)Q2
+ P(23)Q 2
+ P(23)Q 3
- P(S4)Q3 = 0
- P(34)Q 4 = 0
- P(24)Q 4 = 0
+ P(13)Q 3 - P(14)Q4 = 0
The five basis vectors (ei) , 0 _< i _< 4,
Ze. in the u n d e r l y i n g v e c t o r space g ive 1
for S(4). The ten lines of S(4), < el,
to a basis {< e . . >} f o r t h e u n d e r l y i n g 13
P(Ol)Q I - P(02)Q 2 + P(03)Q 3 - P(04)Q4 : 0.
and the vector
rise to a frame
ej >, map under
vector space of
S(9). To avoid unnecessary double subscripts the point
(P(01),P(02),P(03),P(04),P(12),P(13),P(14),P(23),P(24),P(34))
of S(9) is also denoted by (P0,PI,P2,P3,P4,P5,P6,P7,P8,Pg).
All calculations and matrices are taken w.r.t, these bases.
Since any line of S(9) meets a quadric in either 0,1,2 or
q+l points, it follows that any line of S(9) meets L(9)
in either 0,1,2, or q+l points.
3. Conjugacy w.r.t. L(9) and Polar Properties.
The equations iii) may be written in the form:
v) Qi: xA'xt = 0,
where A. is an upper triangular matrix and x = (x0,...,x 9) 1
is a coordinate vector of X e S(9). Let R be a point of
S(9) with coordinates r. A point S with coordinates s is
said to be conjugate to R w.r.t, the locus L(9) if
vi) [(Ai+At)f = 0
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6 Freeman
for each i : 0,1,2,3,4. Thus, conjugacy of a point w.r.t.
L(9) is equivalent to conjugacy w.r.t, each of the five
quadrics, Qi" Consequently, the conjugacy relation w.r.t.
L(9) is independent of coordinate system, symmetric, and
if R is conjugate to {Yk } w.r.t. L(9), then R is conjugate
to every point dependent on {Yk }. The set P(R) of points
conjugate to any point R E S(9) w.r.t. L(9) is a subspace
called the polar space of L(9) at R. If R ~ L(9) the
polar space has some properties of a tangent space for
one quadric (propositions 1,2). So, we also denote P(R)
by T(R) when R e L(9). For the remainder of this paper
the term conjugacy will only be used to mean conjugacy
w.r.t. L(9) as defined above. If the polar space of R
passes through S, then the polar space of S passes through
R.
LEMMA i. If R e L(9), then the polar space, T(R), is a
6-flat.
Pf: The polar flat w.r.t, each quadric Qi is a hyperplane
so the five relations iii) determine a system five linear
equations. It is sufficient to show the system of equations
has rank 3 when R s L(9). At least one coordinate, Ri, of
R is nonzero and by inspection of the associated 5 x i0
matrix a 3 x 3 subdeterminant containing R i may be found
which is not zero. Hence, the rank is at least three.
By using the identities iv) it follows that any one of the
linear equations is a linear combination of three of the
remaining four equations. Hence, the rank of the matrix
is at most three and the lemma is established.
PROPOSITION i. Let i.) R e L(9); 2.) r = e-l(R). Then
for each point Q e T(R)~ L(9), Q # R, Q corresponds under -i
to a line in S(4) meeting the line r. For each line m
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Freeman 7
of S(4) meeting the line r, m corresponds under ~ to a
point M e T(R)~L(9).
Pf. Let Q ~ T(R)~ L($) with q = -I(Q). The coordinates
of Q then satisfy all five linear relations vi) of the
polar of R and also relations iii). Points of S(4) having
row vectors ~i' ~' [el' ~] may be chosen which span the
line r[q]. The five 4 • 4 subdeterminants of the 5 x 4 t t t)
matrix (r~, r2, ql' q2 are precisely the relations vi)
which Q satisfies. Hence, the rank of the matrix is at
most three. The rank of the matrix is at least three
since the two lines are distinct. Hence, the two lines,
r,q, meet in a point.
Conversely, suppose two distinct lines r,q of S(4)
meet in a point. Thus, the associated 5 x 4 matrix obtained
from any four points spanning the lines (as in the above
paragraph) has rank 3. Hence, the five 4 x 4 subdeterminants
are zero which are precisely the relations vi) evaluated at
R and Q. Thus, Q e T(R)~L(9). We note that for R e L(9),
R e T(R)~L(9).
PROPOSITION 2. Let i.) C e L(9); 2.) D ~ L(9); 3.) C and
conjugate w.r.t. L(9). Then the line CD meets L(9) in
exactly C.
Pf. Let c, d be coordinate vectors representing C, D,
respectively. Any point W on the line CD other than D has
a coordinate vector,
W: w = c + 6d, 6 e GF(q).
For each i, i = 0,...,4, cA.c t = O, and for at least one l
k, k ~ {0,1,2,,3,4},
d_~d_ t # O.
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8 Freeman
If W is to be on L(9) then in particular
However C and D are conjugate, hence the only solution
is 6 = 0. Thus, C is the only point of the line CD
meeting L(9).
Remark i. The above development shows that some properties
of the conjugacy relation for a single quadric extend to
conjugacy w.r.t. L(9). However~ not all properties extend.
For if R is on L(9) every line in the polar space of R
passing through R and not contained in L(9) meets L(9) in
only the point R. However, there are lines meeting L(9)
in only the point R and not in the polar space of R. With
C = < e01 > = < (i 0 0 0 0 0 0 0 0 0) >,
D = < (0 1 0 0 1 0 1 1 0 i) > it is easily checked that
C e L(9), D s L(9), the line CD meets L(9) in only C, and
yet C and D are not conjugate.
4. Generating Flats of L(9).
The correspondence between lines of S(3) and the
points of nondegenerate hyperbolic quadric in S(5), often
termed the Klein quadric is well known [17~ p. 327]. The
Grassmanian space of the lines of S(4) has analogous
properties. The facts developed in this section are
indicated in [16, p.521,526]. They are included here
only for continuity of exposition and to emphasize proper-
ties of ~.
A nondegenerate hyperbolic quadric Q(5) of S(5) is
ruled by two collections of plsnes (generating planes).
For, if P is a point of S(3) the lines of S(3) passing
through P correspond under e to the points of a plane,
166
Freeman 9
z(P), on Q(5) and dually, the lines in a plane A of S(3)
correspond under e to the points of a plane z(A) on Q(5).
Two distinct planes of the same collection meet in exactly
a point. If P,Q are points of S(3) then ~(P)~z(Q)
corresponds under -i to the line PQ of S(3). If A,B are
planes of S(3), then ~(A)~z(B) corresponds under -i to
the line A/~B of S(3). Two planes, one from each class,
either meet in a line or are skew depending whether the
associated point and plane in S(3) are incident or not.
The analogous situation occurs in the L(9) case.
There are at least two types of planes of S(9) which are
generating planes of L(9). The lines of a 3-flat passing
through a given point correspond to the points of a
generating plane of L(9). Label such planes type I planes.
The lines of S(4) belonging to a plane of S(4) correspond to
a generating plane of L(9) also. Label such a plane type
If.
PROPOSITION 3. Any generating ' plane of L(9) is either a
plane of type I or II.
Pf. Let z be a generating plane of L(9) with 3 noncollinear
points of ~ labeled P,Q,R. Since zCL(9) these 3 points
are mutually conjugate and correspond to 3 lines p,q,r of
S(4) no two of which are skew. Let p,q meet in the point 0.
Either r passes through 0 or not. The plane ~ is of type I
or type II if r passes through 0 or not.
A similar argument shows:
PROPOSITION 4. The lines of S(4) passing through~ point
of S(4) correspond under ~ to the points of a 3-flat of
S(9) lying in L(9). Conversely , if G(3) is a generating
3-flat of L(9), then the points of G(3) must correspond t o_o
the lines of S(4) passing through ~ point. Two generating
3-flats of L(9) meet in exactly ~ point.
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i0 Freeman
COROLLARY I. If a flat of dimension n of S(9) is a
generating flat of L(9), then n ~ 3; generating t-flats
occur for t = 0,1,2,3.
5. A Group of PFojective Transformations , a Partition of
s(9), and the polar Space of a Point n~ ~ L(9 )'
The group of projective transformations of S(4)
induces a group G' of projective transformations on S(9)
which leave L(9) invariant. In this section we show that
the group G' acts transitively on the points of L(9) and
on the points of S(9)~L(9). We also examine the polar
flat of a point of S(9)~L(9). See [i, p.92] for defini-
tions.
We first obtain a partition of the points of S(9).
Let F(3) be a 3-flat of S(4). Since PGL(5,q) is transitive
on the 3-flats of S(4)~ F(3) may be taken to be
< eo,el,e2,e3 >. The six lines < ei,e j > map under ~ to
six independent points < e.. >, 0 < i < j < 3 which span i] -- --
a 5-flat, G(5), of S(9). This 5-flat sections L(9) in
exactly a nondegenerate hyperbolic quadric Q(5) since
is i-i. Consider another 3-flat F'(3) of S(4). Hence
F(3)~F'(3) is a plane F(2). Similarly the lines of
F'(3) correspond to the points of a quadric Q'(5) in a
5-flat G'(5). The lines of F(2) correspond to the points
of a plane G(2) in L(9) so G'(5)~G(5) is at least a plane.
Since PGL(5,q) is transitive on pairs of 3-flats of S(4)
a routine coordinate calculation shows:
LEMMA 2. For any pai r f distinct 3-flats F(3), F'(3) of
S(4), the associated 5-flats G(5), G'(5) under ~ span an
8-flat of S(9).
168
Freeman ii
COROLLARY 2. If G(5), G'(5) is a pair of 5-flats of S(9)
obtained as in lemma 2, then G(5)/~G'(5)~L(9).
The (q5-1)/(q-l) 3-flats of S(4) determine a corre-
sponding set of 5-flats,
vii) {G(5)}
of S(9) each meeting L(9) in a quadric Q(5). The number
of points in one of the 5-flats not on the corresponding
Q(5) is q2(q3-1). Any two 5-flats in vii) have no point
in common in S(9)~L(9). The total number of points
either in one of the 5-flats and not in L(9), or on L(9)
is seen to be the number of points of S(9). This proves:
LEMMA 3. The points of S(9) may be partitioned into the
following sets of points: i.) th e points of L(9),
2.) the points not on L(9) and belonging to the 5-flats
associated with the 3-flats of S(4) given in vii). Each
point of S(9)~L(9) is in exactly one such 5-flat.
COROLLARY 3. Let G' < PGL(10,q) be the Kroup of ppojective
transformations T' of S(9) induced from the projective
transformations T of S(4). Then G' acts transitively on
the points of L(9) and acts tpansitively on the points of
S(9) not in L(9).
Pf. Since PGL(5,q) acts transitively on the lines of S(4),
the induced group G' leaves L(9) invariant and acts tran-
sitively on the points of L(9). It suffices to show that
G' is transitive on the points of S(9)~L(9).
Let R be a point of S(9)~L(9). There is a unique
5-flat G(5) of the collection in vii) passing through R
(lemma 3). This flat meets L(9) in a quadric Q(5). There
is a secant line of Q(5) passing through R. This is also
a secant line of L(9) since G(5) meets L(9) in Q(5). The
169
12 Freeman
two points of Q(5) on a secant line correspond to two skew
lines of S(4). Since PGL(5,q) is transitive on pairs of
skew lines in S(4), G' is transitive on secant lines of
L(9). By Witt's theorem [14, p.98] the group of projective
transformations of G(5) leaving invariant Q(5) and a
secant line is transitive on the points of the line not
in Q(5). G' is transitive on the collection vii) since
G is transitive as 3-flats of S(4). Hence G' is transitive
on the points of S(9)~L(9).
COROLLARY 4. The polar flat of any point R not on L(9) is
Pf.
be
a 4-flat.
By Corollary 3 the coordinates of R may be taken to
viii) R: p(l 0 0 0 0 0 0 1 0 0) = < e01 + e23 >.
By vi) the polar flat is given by five linearly independent
equations. Since the polar relation is preserved under
nonsingular linear transformations, the statement follows
from Corollary 3.
REMARK 2. Let P(R) be the polar 4-flat of R w.r.t. L(9).
Then P(R) is the same 4-flat as the polar space of R w.r.t.
Q(5) = GR(5)~L(9) if the associated 5-flat GR(5) in vii)
is considered as the entire space.
By proposition 2 every line RQ with Q e L(9) ~P(R)
is a tangent line to L(9). By Remark I, there is a line
passing through R meeting L(9) in a single point Q where
Q ~ P(R). A line of S(9) passing through R meeting L(9)
in exactly one point Q where Q ~ P(R) is said to be a
pseudo-tangent. Through points not on L(9) there pass
secants, tangents and pseudo-tangents to L(9).
170
Freeman 13
PROPOSITION 5. Let R { L(9) and consider the associated
5-flat GR(5) given in vii).
rt angent, if S e P(R)
A line RS with S ~ L(9) is a Jsecant, if S ~ GR(5)~P(R)
Lpseudo-tangent, if S s GR(5).
Pf. By Corollary 3, R may be taken to have coordinates
as in viii). Let S e L(9). Then S satisfies all 5
equations in ill). A point W # R on the line RS has
coordinates
W = S + 8R = (S0+8 S I S 2 S 3 S 4 S 5 S 6 $7+8 S 8 $9) ,
for some 8 e GF(q). Using relation iii) and the hypothesis
that S e L(9) it follows that W e L(9) if and only if
ix) 368=0, S36=0,$9i8=0 ,S88=0; 6(6+(S0+$7) )=0. By remark 2, P(R)~ GR(5) and the equations of G5(R) are
X 9 = X 8 = X 6 = X 3 = 0. There are three cases to consider
for S e L(9).
Case i: If S e P(R), then S 6 = S 3 = S 9 = S 8 = 0 and
S O + S 7 = 0. Hence, by ix), W e L(9) if and only if
6 = O. Hence, as expected, if S ~ P(R)~L(9), then RS
is a tangent line.
Case 2: If S ~ G5(R)'~P(R) , then S 9 = S 8 = S 6 = S 3 = 0
and S O + S 7 # 0. Thus 6 = 0 and 6 = -(S O + S 7) ~ 0 give
two distinct points on the line RS and on L(9). Hence, if
S e G5(R)~-P(R) , then RS is a secant line.
Case 3: If S { Gs(R) and on L(9), then at least one
of $9,$8,$3,S 6 is nonzero. Hence from ix), if W is to be
on L(9), then 6 = 0 is the only solution. This establishes
the proposition.
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14 Freeman
The properties given in lemma 3 and proposition 5 are
indicated in Brahana's work on metabelian groups [6].
However Brahana's statement on tangency [6, p.548] seems
unnecessarily vague and proposition 5 regarding lines
meeting L(9) and passing through a point R not on L(9) is
much more pPecise.
6. R(I,2) Reguli and (w,~)-configu~ations.
Let w be a line and z be a plane of S(4). Suppose
w and ~ are skew. The collection of lines of S(4) meeting
both w and ~ is called a (w,~)-configuration. Recall
(section i) that an R(I,2) regulus is a collection of
q+l skew planes in S(5) such that a line meeting three of
the planes meets all q+l planes. We call these planes the
axis-plangs of the R(I,2) regulus.
PROPOSITION 6. ~ (w~)Tconfiguration corresponds under
to the points of an R(I,2) regulus on L(9).
Pf__. Label the points of w, P0,...,Pq and let W = ~(w).
The lines of S(4) meeting w in P. correspond under e to i
the points of a generating 3-flat, St(3) , in the tangent
space, T(W), (propositions 1 and 4). Any two of the q+l
3-flats S.(3) meet in exactly W since w is the only line 1
of S(4) passing through all the Pi" Since < P''~l > is a
3-flat of S(4) the lines of S(4) passing through P. and l
meeting ~ correspond under ~ to the points of a plane,
~i' of St(3). Any two of these planes are skew since no
line of S(4) meets ~ in a point and s in two points.
Consider the collection of {~i }, of planes in
T(W)~L(9) obtained from an (w,~)-configuration. Let S
be a point on a plane ~k of the collection with s = ~-I(S).
The line s of S(4) meets w in the point Pk and ~ in a
172
Freeman 15
point, Q(s). The planar pencil < w,Q(S) > with vertex
Q(S) corresponds under ~ to a generating line of L(9).
This line meets each element of {~.} in a point. Hence, 1
the line is a transversal to the collection through S.
Thus, at each point of the collection of skew planes
there passes a transversal to the collection. Any two
planes of {~.} span a 5-flat in the tangent 6-flat, T(W). m
A dimension argument using the transversals to the collection
{~.} shows that the collection spans a 5-flat. Hence, the i
collection {~k } is an R(I,2) regulus.
REMARK 3. The R(I,2) regulus on L(9) determined by a
(w,~)-configuration spans a 5-flat of T(W) not passing
through W.
Pf. As in the proof of proposition 6, the axis planes do
span a 5-flat S(5) in T(W). Suppose S(5) passes through
W. Then the two generating 3-flats of L(9), < W,~. > and ]
< W,~ k > with j # k are in a 5-flat. Hence, they meet in
at a line which contradicts proposition 4.
PROPOSITION 7. Let i.) W be a point of L(9), 2.) ~ = e-!(W),
3.) T(W) be the tangent 6-flat of L(9) at W, 4.) S(5) be
any 5-flat not passing through W. Then , S(5) ~L(9) is an
R(I~2) regulus and the points of R(l~2) correspond under -i
to a (w~)-configuration for a unique plane ~ skew to
W,
Pf. As before, label the points of w, P0,...,Pq.
L(9) AT(W) is a collection of q+l 3-flats S.(3) pairwise 1
meeting in W, ( p r o p o s i t i o n s 1 and 4). Since S(5) does
not pass through W, (q+l) skew planes ~. are determined in z S(5) where
x) ~. : S ( 5 ) ~ S . ( 3 ) . 1 1
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16 Freeman
-! The points of T 0 correspond under ~ to the lines of S(4)
meeting w in the point P0 and which are in a 3-flat~ F0(3) ,
of S(4). This 3-flat F0(3) does not contain w since W { T 0
on L(9). Hence, F0(3)~w = P0' That is, the lines of
F0(3) through P0 correspond under ~ to the points of the
plane T 0 .
For any plane ~' of F0(3) with P0 ~ ~'' the image
under e of the (w,~')-configuration is an R(I,2) regulus
on L(9) having T 0 as an axis-plane (proposition 6). We
now need to find a unique plane ~, of F0(3) so that the
(w,~)-configuration corresponds under ~ to the points of
the planes {~.} of (x). To determine this plane in S(4) i
consider another plane 71 # T 0 of the collection in (x).
The points of ~i correspond under -i to lines of S(4)
meeting w in P1 # P0 and lying in some 3-flat Fl(3) not
containing w. The 3-flats F0(3) and FI(3) are distinct
since T 0 and ~i are distinct generating 3-flats S0(3) ,
SI(3) of L(9). Hence, F0(3) and FI(3) meet in a plane,
7. Since P0 and P1 are not in F0(3)~FI(3)=~ is skew to
w,
We now show that for this choice of ~, the (w,~)-
configuration corresponds under ~ to the points of the
planes in (x), so the collection (x) is an R(I,2) regulus
(proposition 6). Let Q be any point of ~. The planar
pencil < w,Q > with vertex Q corresponds to a generating
line, g, of L(9). In particular, the two lines QP0' QPI
of the pencil correspond to two points Q0' QI' on 70, ~l'
respectively. Hence, the line QoQI = g is in S(5). The
line g must meet each plane 7. in a point, since S(5) 1
meets L(9) in exactly the skew planes of (x) and g, a
generating line, is in S(5). Hence, g is a transversal
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Freeman 17
to the collection {~.}. The planar pencil < w,Q > with 1
vertex Q corresponds under ~ to a transversal of the
collection. Thus, the collection {~.} is an R(I,2) i
regulus.
COROLLARY 5. Let i.) F(3) b_~e ~ 3-flat of S(4) meeting
line w in exactly ~ oip_qi~ , 2.) ~ be the generating
plane of type ~ o__nn L(9) cprresponding under ~ to the lines
of F(3) passing through PO , 3.) ~(w) = W e L(9). Then
there is a i-i correspondence between (w~)-confi~urations
in S(4) and R(I~2)reguli of T(W) on L(9) where ~ is any
plane of F(3) not passing through P~ and ~ is an axis
plane of each of the R(I~2) reguli.
REMARK 4. Given two R(I,2) reguli in S(5) two frames for
S(5) may be chosen showing that PGL(6,q) is transitive on
R(I,2) reguli of S(5). By the natural embedding of S(5)
in S(9) it follows that any R(I,2) regulus may be viewed
as the image of an (w,~)-configuration in S(4).
7. R(I,2) Reguli and 3-flats.
Bose has shown that R(I,2) reguli represent Baer
subplanes in a linear representation of PG(2,q2), [3,11].
With this fact in mind and with the purpose of examining
intersection relations of Baer subplanes of PG(2,q 2) we
now investigate how a 3-flat of S(5) may intersect an
R(I,2) regulus.
An elementary coordinate argument following [5] may
be used to show that any three skew planes of S(5) deter-
mine a unique R(I,2) regulus. Let R(I,2) be an R(I,2)
regulus with axis-planes labeled ~.. Since the axis-planes l
are skew each point P of R(I,2) is on a unique transversal
line t(P), in R(I,2). It follows that:
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18 Freeman
REMARK 5. A. Consider three transversals of R(I,2) that
are not in the same 3-flat. Then any two of the three
3-flats spanned by pairs of the 3 transversals meet in
exactly the common transversal.
B. Let P, Q e ~i' P # Q" Then the two 3-flats
< t(P), ~i >' < t(Q), Z'l > are distinct.
PROPOSITION 8. Let t(P), t(Q) be distinct transversals
of R(I,2) meeting the axis planes -m ~" . . . . in ~i' Q-i' respectively.
Then, ! transversal of R(I,2) passinK through ! point of
the line PiQ i is in the 3-flat < t(P), t(Q) > and is a
transversal to the R(I~I) regulus determined b_flany three
distinct lines PjQj, PkQk , PmQ m.
Pf. The 3-flat < t(P), t(Q) > does not contain any plane
~'i (B, remark 5) and thus meets ~.l in the line PiQ i. The
3 lines PiQi, i = 0,1,2 determine a unique R(I,I) regulus,
R(I,I). Since a line of the opposite regulus meets each
of these three lines in a point, it meets each of the
planes T0, ~i' ~2 in a point. Hence, for each point R
on PoQ0 , the line of this opposite regulus through R must
be a transversal to R(I,2) and may be properly labeled t(R).
Now t(R) must meet each of the remaining axis planes.
Since t(R) is in < t(P), t(Q) > the remaining lines of
R(I,I) are precisely the lines PiQi , i = 3,...,q.
This shows that two transversal lines of R(I,2)
determine a 3-flat which meets R(I,2) in exactly a non-
degenerate hyperbolic quadric.
PROPOSITION 9. Let i.) two lines si, ~ lie in two
distinct axis planes, ~i' ~ of R(I~2), 2.) F(3) = < ~i' ~ >"
Then either F(3) meets R(I,2) in a nonde~enerat e hyperboli c
quadric, or F(3) meets R(I,2) exactly i__n_n ~i , ~ and a
transversal t.
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Freeman 19
Pf. The transversals of R(I,2) through s. meet 7. in a
line (proposition 8) which meets s. in at least one point.
Thus, F(3) contains at least one transversal line, t.
If F(3) meets R(I,2) in one further point, R of 7., l
then R and s. span z.. Hence, ~. is in F(3). This implies 1 l l
that s. meets 7. in a point which contradicts the fact that
the axis planes are skew. Hence, if F(3) meets R(I,2) in
a further point R, then R must be in a plane, ~k' with
i ~ k ~ j. Now t meets ~k in a point, Tk, whence the line
TkR is in F(3). The three lines, si, s. and TkR are skew,
share a common 3-flat, and thus determine an R(I,I) regulus.
Each line of the opposite regulus meets ~i' ~j' ~k and
hence is a transversal to R(I,2). Hence F(3) must be the
unique 3-flat meeting R(I,2) in the R(I,I) regulus passing
through s.. l
Our calculation of the possible intersections of a
3-flat with an R(I,2) regulus depends on the following
lemma.
LEMMA 4. Let i.) the axis planes of an R(I,2)regulus b__ee
labeled ~., i = 0~i~ ,q-l~ 2.) ~ be a nonruling plane --9- "'" '
of R(l~2) which meets R(I~2) in at least three noncollinear
points, ~, [i' P-~' so that no two are on the same axis
plane of R(l~2), 3.) t(P0) , t(Pl) , t(P ) be the associated
transversal lines passing through the respective points.
Then, i.) If the transversals lie in the same 3-flat, then
meets R(I~2) in a (degenerat e or nondegenerate) conic.
2.) If the transversals are not in the same 3-flat, then
meets R(I~2) i_nnexactly the three points.
Pf. Since the three points are not collinear at least two
distinct transversals are determined. The first statement
then follows from proposition 8.
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20 Freeman
To establish the second statement we may suppose
that the three points determine three distinct trans-
versals which do not lie in the same 3-flat. Using the
idea given in [5, p.88], coordinates may be introduced
so that,
P0 = < e0 > e T0: < e0, el, e 2 >
PI = < el+e4 > ~ ~i: < e0+e3' el+e4' e2+e5 >
P = < e 5 > e ~: < e 3, e 4, e 5 >
Thus t(P 0) = < e0, e 3 >, t(P I) = <el, e 4 >,
t(P ) = < e2, e 5 >, no transversal of these three is in
the 3-flat determined by the remaining two, and
= < e 0, el+e 4, e 5 >,
~k = < (i 0 0 k 0 0), (0 1 0 0 k 0), (0 0 1 0 0 k) >,
for ~k distinct from ~ . By linear algebra it follows
that there is no point distinct from P0' PI' P~ which on
and ~k for any k, k = 0,1,. .. ,q-l,~.
Since any 3-flat meets each axis plane of R(I,2) in
at least one point, the following proposition gives the
possible intersections of a 3-flat with an R(I,2) regulus.
PROPOSITION i0. A 3-flat 7 of S(5) is in exactly one of
the followin~ classes w.r.t, an R(!,2)" regulus.
I. ) 7. contains one axis plane ~. of R(I,2) and meets
the remainin ~ planes i__nn each a point. The points are on
a transversal to R(I,2).
II.) i) ~ R(I,2) is a nonde~enerate hyperbolic quadric.
ii) Z__A R(I,2) is 3 lines, si, ~, t__, with s{,
belonging to two distinct axis planes ~i' ~j' respectively'
The line t is a transversal of R(I,2) and si, s.. --j
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Freeman 21
iii) Z meets R(I~2) in a plane tangent to an R(I~I)
regulus o__nnR(l~2).
iv) ~ meets one axis plane ~. in a line s with I
the remaining ~ points of intersection coplanar add lying
on a nondegenerate conic with s meeting the plane of the
conic in exactly the '(q+l)-st' point of the conic.
IIl.) i) Z meets R(I~2) in exactly !transversal line.
ii) Z meets R(I~2) in a set of q+l points, no 4
coplanar. 2
Pf. Since there are q +q+l 3-flats containing a given
plane of S(5), if Z contains an axis plane of R(I,2) then
Z must be in class I (B, remark 5).
If Z does not contain an axis-plane and meets R(I,2)
in at least 3 lines, then the proof of proposition 8 implies
that Z meets R(I,2) in a nondegenerate hyperbolic quadric.
Hence Z is of class ll(1). If Z meets two axis planes in
lines, then Z is either class ll(i) or l!(iii) (proposition
9).
Suppose Z meets exactly one axis plane, n0, in exactly
a line s. Hence Z meets each remaining axis plane in
exactly a point. Thus q points are determined. If 3 of
these q points are collinear (q odd), then the common line
must be a transversal. Since this transversal is in Z,
then the q points of intersection must be on this trans-
versal, t. If t and s are skew, then Z would contain T 0
which contradicts the hypothesis of this paragraph. Hence,
t and s span a plane which is tangent to the nondegenerate
hyperbolic quadric determined by the transversals to s as
in proposition 8. Hence, if Z meets exactly one axis plane
in a line, s, and if three of the q remaining points are
collinear, then ~ is of class ll(iii).
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22 Freeman
Assume now that Z meets one axis plane in exactly
a line, s, and that no three of the q remaining points
of intersection with the axis planes are collinear. Let
be the plane determined by any three, R0, RI, R 2. If
contained s, then the line R0R I would meet s in a point.
Hence R0R I meets R(I,2) in three points and would then be
a transversal. Hence, Z is of class II(iii) and the q
points are collinear. Since we are assuming that no three
of the q points are collinear, z must not contain the line
s. Hence ~ meets s in a point, distinct from R0, RI, R 2.
Hence ~ meets R(I,2) in a nondegenerate conic (lemma 4)
and the line s must meet ~ in a point of the conic. Thus,
is of class il(iv).
Assume now that E meets each axis plane in exactly
a point. Thus, q+l points of ~ are determined. If three
such points were eollinear, then the common line is a
transversal and the remaining points of intersection would
be collinear. Hence, if three points of intersection are
collinear then Z is of class III(i).
Suppose that Z~R(I,2) is a set of q+l points, no
three collinear. If four points of these points were
coplanar then the plane would meet R(I,2) in a nondegenerate
conic (lemma 4). We will show now that all 3-flats containing
a plane, ~, which meets R(I,2) in a nondegenerate conic C
must be of class If(i) or II(iv).
The q+l distinct transversals determined by the points
of the conic C are all in the same 3-flat(lemma 4) and
determine an R(I,I) regulus with R(I,I)(-~ : C. Hence,
is contained in a 3-flat of class II(i), Z(~).
Label the points of the conic Qi where Qi = ~i ~ C.
Through Qi there pass q lines, s , of ~i such that <~,s
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Freeman 23
is a 3-flat distinct from Z (7). If two such 3-flats were
the same, say Z ', then Z ' would contain ~i" Hence Z ' would
be of class I. Since ~ contains C then Z' could not
contain ~. This contradiction establishes the fact that
the q 3-flats <7 ,s are distinct and distinct from Z(~).
Let s m, of ~i, ~j pass through Qi' Qj' respectively,
with s m, # ~(~). Similarly, the two 3-flats <~,~>
<7, m > are distinct. Hence, for a plane ~ meeting R(I,2)
in a nondegenerate conic, the q2+q+l 3-flats passing
through~ are either of class II(i),l, or of class ll(iv),
q(q+l). Each such 3-flat meets R(I,2) in a line. Thus,
if a 3-flat meets R(I,2) in exactly q+l points, no three
collinear, then it follows that no four are coplanar.
Hence, they lie on a rational cubic if q ~ 5115, p.311].
The number of 3-flats in each class described in
Proposition i0 is given below.
I. (q+l)(q2+q+l)
2 II. i) q +q+l
2 2 it) (q+l) (~)(q2+q+l) 2
iii) (q+l)2(q-l)(q2+q+l)
iv) (q+l)2(q-l)(q2)(q2+q+l)
Ill. i) (q-l)2(q2)(q2+q+l)
4 ,, 4 22 it) (q -q)~q -q )
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J.W. FREEMAN Department of Mathematics Colorado State University Fort Collins, Colorado 80521, USA.
(Eingegangen am 31. Juii 1974)
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